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"When a shoe costs $80.00, there are 300 sales. Every
ahiahysel 2022-03-04 Answered
"When a shoe costs $80.00, there are 300 sales. Every $5.00 increase in price will result in 10 fewer sales. Find the price that will maximize income."
zahrkao8vm
x is not the price increase, $5 is the price increase.
x is just the number of price increases (like, say, number of five-dollar bills customer would have to pay extra) and also the corresponding number of sales decreases (each decrease is 10 sales).
One price increase brings one sales decrease, so:
y=\left(80+5\cdot 1\right)\left(300-10\cdot 1\right).
Two price increases bring two sales decreases:
y=\left(80+5\cdot 2\right)\left(300-10\cdot 2\right).
An egg distributer determines that the probability that any individual egg has a crack is 0.15.
a) Write the binomial probability formula to determine the probability that exactly x eggs of n eggs are cracked.
b) Write the binomial probability formula to determine the probability that exactly 2 eggs in a one-dozen egg carton are cracked. Do not evaluate.
\forall x \in \mathrm{ℝ}, x<2 ⇒{x}^{2}<4
Solve the following linear and quadratic systems of equations:
y + 3x = -2 or 3x +2 = y
{x}^{2}
Find all values of a that satisfy the inequality for all (x,y).
|a|+\sqrt{{a}^{2}+|b|}\le \left(1+\sqrt{2}\right)|{r}_{1}|
for the larger root
{r}_{1}\text{or}\text{ }{x}^{2}-2ax+b
\frac{1}{2}<\mathrm{cos}2A<1
6\mathrm{tan}A-6{\mathrm{tan}}^{3}A={\mathrm{tan}}^{4}A+2{\mathrm{tan}}^{2}A+1
\mathrm{tan}2A
How do you find the y intercept, the equation of the axis of symmetry and the x-coodinate of the vertex
f\left(x\right)={x}^{2}-10x+5? |
Alternating sign matrix - Wikipedia
Not to be confused with Alternant matrix.
{\displaystyle {\begin{matrix}{\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}\qquad {\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}}\\{\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}}\qquad {\begin{bmatrix}0&1&0\\1&-1&1\\0&1&0\end{bmatrix}}\qquad {\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}}\\{\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}}\qquad {\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}}\end{matrix}}}
The seven alternating sign matrices of size 3
In mathematics, an alternating sign matrix is a square matrix of 0s, 1s, and −1s such that the sum of each row and column is 1 and the nonzero entries in each row and column alternate in sign. These matrices generalize permutation matrices and arise naturally when using Dodgson condensation to compute a determinant.[citation needed] They are also closely related to the six-vertex model with domain wall boundary conditions from statistical mechanics. They were first defined by William Mills, David Robbins, and Howard Rumsey in the former context.
2 Alternating sign matrix theorem
3 Razumov–Stroganov problem
A permutation matrix is an alternating sign matrix, and an alternating sign matrix is a permutation matrix if and only if no entry equals −1.
An example of an alternating sign matrix that is not a permutation matrix is
{\displaystyle {\begin{bmatrix}0&0&1&0\\1&0&0&0\\0&1&-1&1\\0&0&1&0\end{bmatrix}}.}
Alternating sign matrix theorem[edit]
The alternating sign matrix theorem states that the number of
{\displaystyle n\times n}
alternating sign matrices is
{\displaystyle \prod _{k=0}^{n-1}{\frac {(3k+1)!}{(n+k)!}}={\frac {1!\,4!\,7!\cdots (3n-2)!}{n!\,(n+1)!\cdots (2n-1)!}}.}
The first few terms in this sequence for n = 0, 1, 2, 3, … are
1, 1, 2, 7, 42, 429, 7436, 218348, … (sequence A005130 in the OEIS).
This theorem was first proved by Doron Zeilberger in 1992.[1] In 1995, Greg Kuperberg gave a short proof[2] based on the Yang–Baxter equation for the six-vertex model with domain-wall boundary conditions, that uses a determinant calculation due to Anatoli Izergin.[3] In 2005, a third proof was given by Ilse Fischer using what is called the operator method.[4]
Razumov–Stroganov problem[edit]
In 2001, A. Razumov and Y. Stroganov conjectured a connection between O(1) loop model, fully packed loop model (FPL) and ASMs.[5] This conjecture was proved in 2010 by Cantini and Sportiello.[6]
^ Zeilberger, Doron, "Proof of the alternating sign matrix conjecture", Electronic Journal of Combinatorics 3 (1996), R13.
^ Kuperberg, Greg, "Another proof of the alternating sign matrix conjecture", International Mathematics Research Notes (1996), 139-150.
^ "Determinant formula for the six-vertex model", A. G. Izergin et al. 1992 J. Phys. A: Math. Gen. 25 4315.
^ Fischer, Ilse (2005). "A new proof of the refined alternating sign matrix theorem". Journal of Combinatorial Theory, Series A. 114 (2): 253–264. arXiv:math/0507270. Bibcode:2005math......7270F. doi:10.1016/j.jcta.2006.04.004.
^ Razumov, A.V., Stroganov Yu.G., Spin chains and combinatorics, Journal of Physics A, 34 (2001), 3185-3190.
^ L. Cantini and A. Sportiello, Proof of the Razumov-Stroganov conjectureJournal of Combinatorial Theory, Series A, 118 (5), (2011) 1549–1574,
Bressoud, David M., Proofs and Confirmations, MAA Spectrum, Mathematical Associations of America, Washington, D.C., 1999.
Bressoud, David M. and Propp, James, How the alternating sign matrix conjecture was solved, Notices of the American Mathematical Society, 46 (1999), 637–646.
Mills, William H., Robbins, David P., and Rumsey, Howard Jr., Proof of the Macdonald conjecture, Inventiones Mathematicae, 66 (1982), 73–87.
Mills, William H., Robbins, David P., and Rumsey, Howard Jr., Alternating sign matrices and descending plane partitions, Journal of Combinatorial Theory, Series A, 34 (1983), 340–359.
Propp, James, The many faces of alternating-sign matrices, Discrete Mathematics and Theoretical Computer Science, Special issue on Discrete Models: Combinatorics, Computation, and Geometry (July 2001).
Razumov, A. V., Stroganov Yu. G., Combinatorial nature of ground state vector of O(1) loop model, Theor. Math. Phys., 138 (2004), 333–337.
Razumov, A. V., Stroganov Yu. G., O(1) loop model with different boundary conditions and symmetry classes of alternating-sign matrices], Theor. Math. Phys., 142 (2005), 237–243, arXiv:cond-mat/0108103
Robbins, David P., The story of
{\displaystyle 1,2,7,42,429,7436,\dots }
, The Mathematical Intelligencer, 13 (2), 12–19 (1991), doi:10.1007/BF03024081.
Zeilberger, Doron, Proof of the refined alternating sign matrix conjecture, New York Journal of Mathematics 2 (1996), 59–68.
Alternating sign matrix entry in MathWorld
Alternating sign matrices entry in the FindStat database
Retrieved from "https://en.wikipedia.org/w/index.php?title=Alternating_sign_matrix&oldid=1082515817" |
Hooke's Law | Brilliant Math & Science Wiki
Matt DeCross, Andrew Ellinor, Christopher Williams, and
Restorative force of a spring opposes the force of gravity pulling a mass downward [2].
Hooke's law is an empirical physical law describing the linear relationship between the restorative force exerted by a spring and the distance by which the spring is displaced from its equilibrium length. A spring which obeys Hooke's law is said to be Hookean. In addition to springs, Hooke's law is often a good model for arbitrary physical systems that exhibit a tendency to return to a state of equilibrium quickly after perturbation.
Spring Force and Energy
Potential Minima and Small Perturbations
If a Hookean spring is compressed or extended by some displacement
x
from equilibrium, the spring will exert a force proportional to this displacement in the opposite direction:
F = -kx.
The proportionality constant
k
, called the spring constant, is dependent on the stiffness of the spring, which in turn depends on its shape and the material from which the spring is made.
How can Hooke's law be used to determine the mass of an object?
Given a Hookean spring of spring constant
k
, fix one end of the spring to the ceiling and the other end to the object. In equilibrium, the spring force will balance the downward force of gravity on the object, which allows computation of the mass
m
from the displacement
x
of the spring.
Gravity exerts a force
F_g = mg
downward proportional to the mass
m
of the object, which is perfectly matched by the spring force
F_s = -kx
in equilibrium, where the negative sign indicates that the spring force acts in the opposite direction. The mass is obtained by setting these forces equal:
m = \frac{kx}{g}.\ _\square
The linear relationship of Hooke's law empirically holds only for small displacements
x
. For large deformations, a spring or other Hookean material can be permanently distorted and exhibit a nonlinear restorative force.
In one dimension, the potential energy of a spring can be obtained from Hooke's law by integration:
U = -\int F \,dx = \int kx\,dx = \frac12 kx^2.
x^2\sqrt{k/m}
xk/m
kx^2/m
x\sqrt{k/m}
m
attached to a Hookean spring of spring constant
k
is pulled so that the spring is displaced a distance
x
from equilibrium. The mass is then released and allowed to oscillate at the end of the spring. What is the maximum velocity of the mass?
If a spring is compressed or extended and then released, it will oscillate (indefinitely if friction is neglected; otherwise, the spring will eventually return to equilibrium). The oscillation of a mass
m
on a spring can be derived directly from Newton's second law of motion,
F = ma
. Since acceleration is the second derivative of velocity, setting the force equal to the spring force
F = -kx
m\ddot{x} = ma = F = -kx.
So the equation of motion of the mass on the spring is
\ddot{x} + \frac{k}{m} x = 0,
where dots indicate time derivatives. The coefficient
\frac{k}{m}
is often denoted as
\omega_0^2
, the square of the natural frequency of oscillation of the spring. This is because the general solution to this equation of motion is
x(t) = A \cos (\omega_0 t) + B \sin (\omega_0 t)
A
B
depending on initial conditions.
The equation of motion above is often called the equation of motion of the simple harmonic oscillator, and systems obeying a similar equation of motion to the above are therefore said to exhibit simple harmonic motion.
M
suspended by two springs of spring constant
k
M,
suspended by two springs each of spring constant
k
as in the diagram, is compressed upwards a displacement
L
from the equilibrium length of the springs and allowed to fall under the influence of gravity. Find the subsequent displacement as a function of time.
The mass obeys the equation of motion of the simple harmonic oscillator where the coefficient
k
2k
since there are two spring forces acting directly on the mass. The displacement
x
therefore obeys the equation
x(t) = A \cos \left(\sqrt{\frac{2k}{m}} t \right) + B\sin \left(\sqrt{\frac{2k}{m}} t \right)
A
B
to be fixed by initial conditions. These initial conditions are
x(0) = L, \quad \dot{x} (0) = 0.
Plugging into the general solution above, these initial conditions yield
x(0) = A = L, \quad \dot{x} (0) = B\sqrt{\frac{2k}{m}} = 0 \implies B = 0.
The solution is thus uniquely fixed to be
x(t) = L\cos \left(\sqrt{\frac{2k}{m}}t \right).\ _\square
Show that a circuit with an inductor and capacitor, called an LC circuit, obeys the simple harmonic oscillator equation.
According to Kirchoff's laws, the voltage around a closed loop must sum to zero. The voltage across the capacitor is
V_c = Q/C,
C
Q
is the charge stored on the capacitor. The voltage across the inductor is
V = L\frac{dI}{dt},
I
is the current in the circuit and
L
is the inductance. Since the current
I = \frac{dQ}{dt}
is the time rate of change of the charge on the capacitor, Kirchhoff's loop rule reduces to
0 = -L \frac{dI}{dt} - \frac{Q}{C} \implies L\ddot{Q} + \frac{1}{C} Q = 0.
This is the simple harmonic oscillator equation where the inductance
L
plays the role of the mass and the inverse of the capacitance
\frac{1}{C}
plays the role of the spring constant. The linear voltage response with increasing charge stored on the capacitor is analogous to Hooke's law, with the rate of change of current through the circuit analogous to the acceleration.
_\square
The solution to the equations of motion above, in terms of cosines and sines, will continue to oscillate forever. However, real oscillators eventually return to a stable equilibrium. This is because real oscillators feel damping forces which remove energy from the system.
Almost any physical system with a stable equilibrium state is described well by Hooke's law when it is displaced slightly from equilibrium. To see why, it is useful to consider potential energy diagrams, which graphically display the potential energy of a system in different states.
In blue, a potential energy diagram for some physical system. In red, the harmonic oscillator potential approximation about the stable equilibrium point.
The above diagram plots in blue the potential energy
U (x)
of some system as a function of location
x
. A state of sufficiently low total energy in this system will not have enough kinetic energy to escape the potential well around the minimum of
U(x)
. The minimum
x_0
U(x)
is thus an equilibrium state. Futhermore, it is stable: if the system starts at
x_0
and is perturbed slightly, it will tend to exert a restorative force towards
x_0
As a result, Taylor expanding the potential energy about the minimum yields
U(x) \approx U(x_0) + \left. \frac12 (x-x_0)^2 \frac{d^2 U}{dx^2} \right|_{x=x_0}.
U(x_0)
, is just a constant energy shift. The quadratic term is the lowest-order term that describes how
U
x
. This quadratic potential exactly matches the form of the spring potential
U_s = \frac12 kx^2
. Therefore, near the equilibrium point
x_0
, the behavior of physical systems can be well approximated by simple harmonic motion, i.e. a force obeying Hooke's law. Above, the quadratic approximation to a potential at its minimum is plotted in red.
For a system with multiple interacting masses, it is useful to define the reduced mass
\mu
\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2} + \cdots.
The frequency of oscillation
\omega_0
about the minimum of a potential is then, in analogy to the Hookean spring,
\omega_0^2 = \sqrt{\frac{1}{\mu} \left. \frac{d^2 U}{dx^2} \right|_{x=x_0}}.
The Lennard-Jones 6,12 potential is used to model the interactions between two neutral atoms. The potential contains an attractive
\frac{1}{r^6}
term for the van der Waals interaction and a repulsive
\frac{1}{r^{12}}
term that models the exchange force due to the Pauli exclusion principle:
U = \epsilon \left[ \left(\frac{r_m}{r} \right)^{12} - 2\left( \frac{r_m}{r} \right)^6 \right],
\epsilon
controls the depth of the potential well and
r_m
is the minimum of the potential well. Find the frequency of oscillation about equilibrium for two atoms of mass
m
whose interactions are modeled by this potential.
U (r)
r_m
to second order is
U(r) \approx -\epsilon + 36 \frac{\epsilon}{r_m^2} (r-r_m)^2.
The reduced mass of the system is
\mu = \frac{m}{2}
. The frequency of oscillation about equilibrium is therefore
\omega_0^2 = \sqrt{\frac{2}{m} 72 \frac{\epsilon}{r_m^2}} = \frac{12}{r_m} \sqrt{\frac{\epsilon}{m}}.\ _\square
\sqrt{\frac{1}{8m}}
\sqrt{\frac{1}{m}}
\sqrt{\frac{1}{4m}}
\sqrt{\frac{1}{2m}}
moves in a potential given by
U(x) = \frac{1}{x^2} - \frac{1}{x}
. Find the (angular) frequency of small oscillations about equilibrium.
[1] D. Klepper and R. Kolenkow, An Introduction to Mechanics. McGraw-Hill, 1973.
[2] Image from https://upload.wikimedia.org/wikipedia/commons/5/5a/Mass-spring-system.png under Creative Commons licensing for reuse and modification.
Cite as: Hooke's Law. Brilliant.org. Retrieved from https://brilliant.org/wiki/hookes-law/ |
The particle travels along the path defined by the parabola
The particle travels along the path defined by the parabola y=0.5x^2. If t
y=0.5{x}^{2}
. If the component of velocity along the x axis is vx = (5t) ft/s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = 1 s. When t = 0, x = 0, y = 0.
{v}_{x}
we can get an equation for x:
x=2.5{t}^{2}
We can insert that to find the equation for y as a function of time:
y=0.5\cdot {2.5}^{2}\cdot {t}^{4}=3.125{t}^{4}
Distance at the t=1 is:
{d}_{x}=2.5
{d}_{y}=3.125ft
d=\sqrt{{d}_{x}+{d}_{y}}=\sqrt{{2.5}^{2}+{3.125}^{2}}
d=4
Acceleration in t=1 is:
{a}_{y}=37.5t=37.5\text{ }f\frac{t}{{s}^{2}}
{a}_{x}=5\text{ }f\frac{t}{{s}^{2}}
a=\sqrt{{a}_{x}+{a}_{y}}=\sqrt{{5}^{2}+{37.5}^{2}}
a=37.83\text{ }f\frac{t}{{s}^{2}}
d=4\text{ }ft
a=37.83\text{ }f\frac{t}{{s}^{2}}
A 0.50 kg ball that is tied to the end of a 1.1 m light cord is revolved in a horizontal planewith the cord making a
{30}^{\circ }
angle, with the vertical.
a) Determine the ball speed.
b) If, instead, the ball is revolved so that its speed is 4.0 m/s,what angle does the cord make with the vertical?
c) If the cord can withstand a maximum tension of 9.1 N, what is the highest speed at which the ballcan move?
Three Children are trying to balance on a seesaw, which consists of a fulcrum rock, acting as a pivot at the center, and avery light board 3.6 m long. Two playmates are already on either end. Boy A has a mass of 50kg, and girl B a mass of 35 kg. Where should girl C, whose mass is 25 kg, placed herself so as to balance the seesaw.
3.00\frac{m}{s}
Use the two-way table of data from another student survey to answer the following question.
\begin{array}{|cccc|}\hline \text{ }& \text{ Like Aerobic }& \text{ Exercise }& \\ \text{ Like Weight Lifting }& \text{ Yes }& \text{ No }& \text{ Total }\\ \text{ Yes }& 7& 14& 21\\ \text{ No }& 12& 7& 19\\ \text{ Total }& 19& 21& 40\\ \hline\end{array}
What is the conditional relative frequency that a student likes to lift weights, given that the student does not like aerobics?
A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws stones vertically downward 1.0 s apart and observes that the cause a single splash. The initial speed of the first stone was 2.0 m/s.
a. How long after the release of the first stone does the second stone hit the water?
b. What was the initial speed of the second stone?
c. What is the speed of each stone as it hits the water?
An ice cube tray of negligible mass contains 0.315 kg of water at
{17.7}^{\circ }
. How much heat must be removed to cool the water to
{0.0}^{\circ }
and freeze it? |
Give the first six terms of this sequence c_1=4, \ c_2=5,
Give the first six terms of this sequence
{c}_{1}=4,\text{ }{c}_{2}=5,\text{ }\text{and}\text{ }{c}_{n}={c}_{n-1}.\text{ }{c}_{n-2}\text{ }\text{for}\text{ }n\ge 3
{c}_{1}=4,\text{ }{c}_{2}=5,\text{ }{c}_{n}={c}_{n-1}{c}_{n-1},\text{ }\text{for}\text{ }n\ge 3
{c}_{3}={c}_{2}{c}_{1}=5\cdot 4=20
{c}_{4}={c}_{3}{c}_{2}=20\cdot 5=100
{c}_{5}={c}_{4}{c}_{3}=100\cdot 20=2000
{c}_{6}={c}_{5}{c}_{4}=2000\cdot 100=200000
Describe a sequence of translations that can be used to show that Figures A and A
Explain why there are no simple graphs with these degree sequences.
(а) 6,6,5,3,2,2,2
Construct multigraphs with these degree sequences. Are there degree sequences that are realizable as simple graphs but not as multigraphs? Why or why not?
Substitute n=1, 2, 3, 4, 5 to find the first five sequences in the given sequence
\left\{\frac{\left(-1{\right)}^{n-1}}{2\cdot 4\cdot 6\cdots 2n}\right\}
the sequence that lists the odd positive integers in increasing order, listing each odd integer twice.
The sum of three numbers is 74. The second number is 10 more than the first. The third number is 2 times the first. What are the numbers?
\underset{n\to \mathrm{\infty }}{lim}{e}^{\frac{n+1}{n}}=e
\left(\frac{n+1}{n}\right)
is the exponential of e
Show that the following sequences converge
{a}_{n}=5+\frac{1}{n} |
\textcolor[rgb]{0.470588235294118,0,0.0549019607843137}{\mathbf{ω}}
\mathrm{π}:E→M
be a fiber bundle, with base dimension
m
{\mathrm{π}}^{\mathrm{∞}}:{J}^{\mathrm{∞}}\left(E\right) → M
E
({x}^{i}, {u}^{\mathrm{α}}, {u}_{{i}_{}}^{\mathrm{α}}, {u}_{{i}_{}j}^{\mathrm{α}}
{u}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{α}}, ....)
{J}^{\infty }\left(E\right)
{\mathrm{dx}}^{i}
M
{\mathrm{Θ}}^{\mathrm{α}} = {\mathrm{du}}^{\mathrm{α}}-{u}_{\mathrm{ℓ}}^{\mathrm{α}}{\mathrm{dx}}^{\mathrm{ℓ}},
{\mathrm{Θ}}_{i}^{\mathrm{α}} = {\mathrm{du}}_{i}^{\mathrm{α}}-{u}_{i\mathrm{ℓ}}^{\mathrm{α}}{\mathrm{dx}}^{\mathrm{ℓ}\mathit{ }}, .... ,
{\mathrm{Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{α}} = {\mathrm{du}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{α}}-{u}_{\mathrm{ij}\cdot \cdot \cdot \mathrm{kℓ}}^{\mathrm{α}} {\mathrm{dx}}^{\mathrm{ℓ}} , ....
{\mathrm{dΘ}}^{\mathrm{α}} = {\mathrm{dx}}^{\mathrm{ℓ}} ∧ {\mathrm{Θ}}_{\mathrm{ℓ}\mathit{ }}^{\mathrm{\alpha }},
{\mathrm{Θ}}_{i}^{\mathrm{α}} = {\mathrm{dx}}^{\mathrm{ℓ}} ∧ {\mathrm{\Theta }}_{\mathrm{iℓ}\mathit{ }}^{\mathrm{α}}, .... ,
{\mathrm{dΘ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{α}} = {\mathrm{dx}}^{\mathrm{ℓ}} ∧ {\mathrm{\Theta }}_{\mathrm{ij}\cdot \cdot \cdot \mathrm{kℓ}\mathit{ }}^{\mathrm{α}}.
p-
\mathrm{ω} ∈ {\mathrm{Ω}}^{p}\left({J}^{\mathrm{∞}}\right)
\left(r,s\right)
r 1
M
s
\mathrm{ω} = {A}_{{i}_{1}{i}_{2}\cdot \cdot \cdot {i}_{r} {a}_{1} \cdot \cdot \cdot {a}_{s}}^{ }{\mathrm{dx}}^{{i}_{1}}∧{\mathrm{dx}}^{{i}_{2} }∧ \cdot \cdot \cdot ∧{\mathrm{dx}}^{{i}_{r}} ∧ {C}^{{a}_{1}}∧{C}^{{a}_{2}} \cdot \cdot \cdot ∧{C}^{{a}_{s}},
{C}^{{a}_{k}}
p
{\mathrm{Ω}}^{p}\left({J}^{\mathrm{∞}} \right) = \underset{r+s =p}{\overset{}{⨁}} {\mathrm{Ω}}^{\left(r,s\right)}\left({J}^{\mathrm{∞}}\left(E\right)\right)
d:{\mathrm{\Omega }}^{\left(r,s\right)}\left({J}^{\infty }\left(E\right)\right)→ {\mathrm{\Omega }}^{\left(r+1,s\right)}\left({J}^{\infty }\left(E\right)\right) ⊕{\mathrm{\Omega }}^{\left(r,s+1\right)}\left({J}^{\infty }\left(E\right)\right)
d = {d}_{H } + {d}_{V},
{d}_{H }:{\mathrm{\Omega }}^{\left(r,s\right)}\left({J}^{\infty }\left(E\right)\right)→ {\mathrm{\Omega }}^{\left(r+1,s\right)}\left({J}^{\infty }\left(E\right)\right)
{d}_{V }:{\mathrm{\Omega }}^{\left(r,s\right)}\left({J}^{\infty }\left(E\right)\right)→ {\mathrm{\Omega }}^{\left(r,s+1\right)}\left({J}^{\infty }\left(E\right)\right).
{d}_{H }
{d}_{V}
{d}_{H}∘{d}_{H} =0, {d}_{H}∘{d}_{V} + {d}_{V}∘{d}_{H} =0,
{d}_{V}∘{d}_{V} =0
{d}_{H}\left({x}^{i}\right) = {\mathrm{dx}}^{i }, {d}_{H}\left({u}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) = {u}_{\mathrm{ij} \cdot \cdot \cdot k\mathrm{ℓ}}^{\mathrm{α}} {\mathrm{dx}}^{\mathrm{ℓ}}, {d}_{H}\left({\mathrm{dx}}^{i}\right) = 0, {d}_{H}\left({\mathrm{Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) = {\mathrm{dx}}^{\mathrm{ℓ}} ∧ {\mathrm{Θ}}_{\mathrm{ij}\cdot \cdot \cdot k\mathrm{ℓ}}^{\mathrm{\alpha }}
{d}_{V}\left({x}^{i}\right) =0, {d}_{V}\left({u}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) = {\mathrm{Θ}}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{α}} , {d}_{V}\left({\mathrm{dx}}^{i}\right) = 0, {d}_{V}\left({\mathrm{Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) = 0.
\textcolor[rgb]{0.470588235294118,0,0.0549019607843137}{\mathrm{ω}}
{d}_{H}\left(\mathrm{ω}\right)
\mathrm{ω}
M
\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):
\mathrm{with}\left(\mathrm{JetCalculus}\right):
{J}^{2}\left(E\right)
E
\left(x, y, u, v\right) → \left(x, y\right)
\mathrm{DGsetup}\left([x,y],[u,v],E,2\right):
F≔f\left(x,y,u[],u[1],u[2]\right):
\mathrm{PDEtools}[\mathrm{declare}]\left(F,\mathrm{quiet}\right):
\mathrm{HorizontalExteriorDerivative}\left(F\right)
\left({\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{[]}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{\textcolor[rgb]{0,0,1}{x}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{+}\left({\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{[]}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{\textcolor[rgb]{0,0,1}{y}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}
\mathrm{ω1}≔A\left(x,y,u[],u[1],u[2]\right)\mathrm{Dx}+B\left(x,y,u[],u[1],u[2]\right)\mathrm{Dy}
\textcolor[rgb]{0,0,1}{\mathrm{ω1}}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{[}\textcolor[rgb]{0,0,1}{]}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{[}\textcolor[rgb]{0,0,1}{]}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}
\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω1}\right)
\textcolor[rgb]{0,0,1}{-}\left({\textcolor[rgb]{0,0,1}{A}}_{{\textcolor[rgb]{0,0,1}{u}}_{[]}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{A}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{A}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{B}}_{{\textcolor[rgb]{0,0,1}{u}}_{[]}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{B}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{B}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{A}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{B}}_{\textcolor[rgb]{0,0,1}{x}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}
\mathrm{ω2}≔\mathrm{Cu}[2]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&wedge\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{Cv}[2]
\textcolor[rgb]{0,0,1}{\mathrm{ω2}}\textcolor[rgb]{0,0,1}{:=}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{2}}
\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω2}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}} |
\frac{a}{b}=\left(\frac{a}{b}×100\right)%
\frac{1}{4}=\left(\frac{1}{4}×100\right)%=25%
\left[\frac{R}{\left(100+R\right)}×100\right]%
\left[\frac{R}{\left(100-R\right)}×100\right]%
P{\left(1+\frac{R}{100}\right)}^{n}
\frac{P}{{\left(1+\frac{R}{100}\right)}^{n}}
P{\left(1-\frac{R}{100}\right)}^{n}
\frac{P}{{\left(1-\frac{R}{100}\right)}^{n}}
\left[\frac{R}{\left(100+R\right)}×100\right]%
\left[\frac{R}{\left(100-R\right)}×100\right]%
If 15% of x = 20% of y, then x:y is ___?
Given 15% of x = 20% of y
=> x/y = 20/15
The ratio 5 : 4 expressed as a percent equals :
C) 80 % D) 125 %
D) 125 %
Answer & Explanation Answer: D) 125 %
5 : 4 = 5/4 = ( (5/4) x 100 )% = 125%.
A) (91 + 1/3)% B) (93 + 1/3 )%
C) (97 + 1/3 )% D) (98 + 1/3) %
A) (91 + 1/3)%
B) (93 + 1/3 )%
C) (97 + 1/3 )%
D) (98 + 1/3) %
Answer & Explanation Answer: B) (93 + 1/3 )%
Pass percentage =[(252/270)*100] % =280/3 % = 9 %
1100 boys and 700 girls are examined in a test; 42% of the boys and 30% of the girls pass. The percentage of the total who failed is :
A) 58 B) 62 *2/3
B) 62 *2/3
Answer & Explanation Answer: B) 62 *2/3
Total number of students = 1100 + 700 = 1800.
Number of students passed = (42% of 1100 + 30% of 700) - (462 + 210) = 672.
Number of failues = 1800-672 = 1128.
Percentage failure = (1128/1800 * 100 )% = 62 * 2/3 %.
What will come in place of the question mark(?) in the following question ?
56% of 870 + 82% of 180 = 32% of 90 + ?
56% of 870 = 56x870/100 = 487.20
32% of 90 = 32x90/100
487.20 + 147.6 - 28.8 = ?
? = 634.8 - 28.8
? = 606.
In a History examination, the average for the entire class was 80 marks. If 10% of the students scored 35 marks and 20% scored 90 marks, what was the average marks of the remaining students of the class ?
Let the number of students in the class be 100 and let this required average be x.
Then, (10 * 95) + (20 * 90) + (70 * x) = (100 * 80)
=> 70x = 8000 - (950 + 1800) = 5250
What will come in place of the question mark (?) in the following question ?
52.5% of 800 + 30.5% of 2800 = ? + 87.30
A) 1096.30 B) 1226.70
C) 1124.20 D) 1186.70
Answer & Explanation Answer: D) 1186.70
420 + 854 - 87.30 = ?
1274 - 87.30 = ?
? = 1186.70
If the numerator of a fraction is increased by 150% and the denominator of the fraction is increased by 350%, the resultant fraction is 25/51. What is the original fraction ?
Answer & Explanation Answer: B) 15/17
The original fraction is = 15/17. |
Engel expansion - Wikipedia
The Engel expansion of a positive real number x is the unique non-decreasing sequence of positive integers
{\displaystyle \{a_{1},a_{2},a_{3},\dots \}}
{\displaystyle x={\frac {1}{a_{1}}}+{\frac {1}{a_{1}a_{2}}}+{\frac {1}{a_{1}a_{2}a_{3}}}+\cdots .}
For instance, Euler's constant e has the Engel expansion[1]
corresponding to the infinite series
{\displaystyle e={\frac {1}{1}}+{\frac {1}{1}}+{\frac {1}{1\cdot 2}}+{\frac {1}{1\cdot 2\cdot 3}}+{\frac {1}{1\cdot 2\cdot 3\cdot 4}}+\cdots }
Rational numbers have a finite Engel expansion, while irrational numbers have an infinite Engel expansion. If x is rational, its Engel expansion provides a representation of x as an Egyptian fraction. Engel expansions are named after Friedrich Engel, who studied them in 1913.
An expansion analogous to an Engel expansion, in which alternating terms are negative, is called a Pierce expansion.
1 Engel expansions, continued fractions, and Fibonacci
2 Algorithm for computing Engel expansions
3 Iterated functions for computing Engel expansions
3.1 The transfer operator of the Engel map
4 Relation to the Riemann '"`UNIQ--postMath-00000018-QINU`"' function
6 Engel expansions of rational numbers
7 Engel expansions for some well-known constants
8 Growth rate of the expansion terms
Engel expansions, continued fractions, and Fibonacci[edit]
Kraaikamp & Wu (2004) observe that an Engel expansion can also be written as an ascending variant of a continued fraction:
{\displaystyle x={\cfrac {1+{\cfrac {1+{\cfrac {1+\cdots }{a_{3}}}}{a_{2}}}}{a_{1}}}.}
They claim that ascending continued fractions such as this have been studied as early as Fibonacci's Liber Abaci (1202). This claim appears to refer to Fibonacci's compound fraction notation in which a sequence of numerators and denominators sharing the same fraction bar represents an ascending continued fraction:
{\displaystyle {\frac {a\ b\ c\ d}{e\ f\ g\ h}}={\dfrac {d+{\cfrac {c+{\cfrac {b+{\cfrac {a}{e}}}{f}}}{g}}}{h}}.}
If such a notation has all numerators 0 or 1, as occurs in several instances in Liber Abaci, the result is an Engel expansion. However, Engel expansion as a general technique does not seem to be described by Fibonacci.
Algorithm for computing Engel expansions[edit]
To find the Engel expansion of x, let
{\displaystyle u_{1}=x,}
{\displaystyle a_{k}=\left\lceil {\frac {1}{u_{k}}}\right\rceil ,}
{\displaystyle u_{k+1}=u_{k}a_{k}-1}
{\displaystyle \left\lceil r\right\rceil }
is the ceiling function (the smallest integer not less than r).
{\displaystyle u_{i}=0}
for any i, halt the algorithm.
Iterated functions for computing Engel expansions[edit]
Another equivalent method is to consider the map [2]
{\displaystyle g(x)=x\left(1+\left\lfloor {x}^{-1}\right\rfloor \right)-1}
{\displaystyle u_{k}=1+\left\lfloor {\frac {1}{g^{(n-1)}(x)}}\right\rfloor }
{\displaystyle g^{(n)}(x)=g(g^{(n-1)}(x))}
{\displaystyle g^{(0)}(x)=x}
Yet another equivalent method, called the modified Engel expansion calculated by
{\displaystyle h(x)=\left\lfloor {\frac {1}{x}}\right\rfloor g(x)=\left\lfloor {\frac {1}{x}}\right\rfloor \left(x\left\lfloor {\frac {1}{x}}\right\rfloor +x-1\right)}
{\displaystyle u_{k}={\begin{cases}1+\left\lfloor {\frac {1}{x}}\right\rfloor &n=1\\\left\lfloor {\frac {1}{h^{(k-2)}(x)}}\right\rfloor \left(1+\left\lfloor {\frac {1}{h^{(k-1)}(x)}}\right\rfloor \right)&n\geqslant 2\end{cases}}}
The transfer operator of the Engel map[edit]
The Frobenius–Perron transfer operator of the Engel map
{\displaystyle g(x)}
acts on functions
{\displaystyle f(x)}
{\displaystyle [{\mathcal {L}}_{g}f](x)=\sum _{y:g(y)=x}{\frac {f(y)}{\left|{\frac {d}{dz}}g(z)\right|_{z=y}}}=\sum _{n=1}^{\infty }{\frac {f\left({\frac {x+1}{n+1}}\right)}{n+1}}}
{\displaystyle {\frac {d}{dx}}(x(n+1)-1)=n+1}
and the inverse of the n-th component is
{\displaystyle {\frac {x+1}{n+1}}}
which is found by solving
{\displaystyle x(n+1)-1=y}
{\displaystyle x}
Relation to the Riemann
{\displaystyle \zeta }
The Mellin transform of the map
{\displaystyle g(x)}
is related to the Riemann zeta function by the formula
{\displaystyle {\begin{aligned}\int _{0}^{1}g(x)x^{s-1}\,dx&=\sum _{n=1}^{\infty }\int _{\frac {1}{n+1}}^{\frac {1}{n}}(x(n+1)-1)x^{s-1}\,dx\\[5pt]&=\sum _{n=1}^{\infty }{\frac {n^{-s}(s-1)+(n+1)^{-s-1}(n^{2}+2n+1)+n^{-s-1}s-n^{1-s}}{(s+1)s(n+1)}}\\[5pt]&={\frac {\zeta (s+1)}{s+1}}-{\frac {1}{s(s+1)}}\end{aligned}}}
To find the Engel expansion of 1.175, we perform the following steps.
{\displaystyle u_{1}=1.175,a_{1}=\left\lceil {\frac {1}{1.175}}\right\rceil =1;}
{\displaystyle u_{2}=u_{1}a_{1}-1=1.175\cdot 1-1=0.175,a_{2}=\left\lceil {\frac {1}{0.175}}\right\rceil =6}
{\displaystyle u_{3}=u_{2}a_{2}-1=0.175\cdot 6-1=0.05,a_{3}=\left\lceil {\frac {1}{0.05}}\right\rceil =20}
{\displaystyle u_{4}=u_{3}a_{3}-1=0.05\cdot 20-1=0}
The series ends here. Thus,
{\displaystyle 1.175={\frac {1}{1}}+{\frac {1}{1\cdot 6}}+{\frac {1}{1\cdot 6\cdot 20}}}
and the Engel expansion of 1.175 is {1, 6, 20}.
Engel expansions of rational numbers[edit]
Every positive rational number has a unique finite Engel expansion. In the algorithm for Engel expansion, if ui is a rational number x/y, then ui+1 = (−y mod x)/y. Therefore, at each step, the numerator in the remaining fraction ui decreases and the process of constructing the Engel expansion must terminate in a finite number of steps. Every rational number also has a unique infinite Engel expansion: using the identity
{\displaystyle {\frac {1}{n}}=\sum _{r=1}^{\infty }{\frac {1}{(n+1)^{r}}}.}
the final digit n in a finite Engel expansion can be replaced by an infinite sequence of (n + 1)s without changing its value. For example,
{\displaystyle 1.175=\{1,6,20\}=\{1,6,21,21,21,\dots \}.}
This is analogous to the fact that any rational number with a finite decimal representation also has an infinite decimal representation (see 0.999...). An infinite Engel expansion in which all terms are equal is a geometric series.
Erdős, Rényi, and Szüsz asked for nontrivial bounds on the length of the finite Engel expansion of a rational number x/y; this question was answered by Erdős and Shallit, who proved that the number of terms in the expansion is O(y1/3 + ε) for any ε > 0.[3]
Engel expansions for some well-known constants[edit]
{\displaystyle \pi }
= {1, 1, 1, 8, 8, 17, 19, 300, 1991, 2492,...} (sequence A006784 in the OEIS)
{\displaystyle {\sqrt {2}}}
= {1, 3, 5, 5, 16, 18, 78, 102, 120, 144,...} (sequence A028254 in the OEIS)
{\displaystyle e}
= {1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,...} (sequence A028310 in the OEIS)
{\displaystyle e^{1/r}-1=\{1r,2r,3r,4r,5r,6r,\dots \}}
More Engel expansions for constants can be found here.
Growth rate of the expansion terms[edit]
The coefficients ai of the Engel expansion typically exhibit exponential growth; more precisely, for almost all numbers in the interval (0,1], the limit
{\displaystyle \lim _{n\rightarrow \infty }a_{n}^{1/n}}
exists and is equal to e. However, the subset of the interval for which this is not the case is still large enough that its Hausdorff dimension is one.[4]
The same typical growth rate applies to the terms in expansion generated by the greedy algorithm for Egyptian fractions. However, the set of real numbers in the interval (0,1] whose Engel expansions coincide with their greedy expansions has measure zero, and Hausdorff dimension 1/2.[5]
^ Erdős, Rényi & Szüsz (1958); Erdős & Shallit (1991).
^ Wu (2000). Wu credits the result that the limit is almost always e to Janos Galambos.
^ Wu (2003).
Engel, F. (1913), "Entwicklung der Zahlen nach Stammbruechen", Verhandlungen der 52. Versammlung deutscher Philologen und Schulmaenner in Marburg, pp. 190–191 .
Pierce, T. A. (1929), "On an algorithm and its use in approximating roots of algebraic equations", American Mathematical Monthly, 36 (10): 523–525, doi:10.2307/2299963, JSTOR 2299963
Erdős, Paul; Rényi, Alfréd; Szüsz, Peter (1958), "On Engel's and Sylvester's series" (PDF), Ann. Univ. Sci. Budapest. Eötvös Sect. Math., 1: 7–32 .
Erdős, Paul; Shallit, Jeffrey (1991), "New bounds on the length of finite Pierce and Engel series", Journal de théorie des nombres de Bordeaux, 3 (1): 43–53, doi:10.5802/jtnb.41, MR 1116100 .
Paradis, J.; Viader, P.; Bibiloni, L. (1998), "Approximation to quadratic irrationals and their Pierce expansions", Fibonacci Quarterly, 36 (2): 146–153
Kraaikamp, Cor; Wu, Jun (2004), "On a new continued fraction expansion with non-decreasing partial quotients", Monatshefte für Mathematik, 143 (4): 285–298, doi:10.1007/s00605-004-0246-3 .
Wu, Jun (2000), "A problem of Galambos on Engel expansions", Acta Arithmetica, 92 (4): 383–386, doi:10.4064/aa-92-4-383-386, MR 1760244 .
Wu, Jun (2003), "How many points have the same Engel and Sylvester expansions?", Journal of Number Theory, 103 (1): 16–26, doi:10.1016/S0022-314X(03)00017-9, MR 2008063 .
Weisstein, Eric W. "Engel Expansion". MathWorld–A Wolfram Web Resource.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Engel_expansion&oldid=1071659355" |
The classification of Lagrangians nearby the Whitney immersion
The Whitney immersion is a Lagrangian sphere inside the four-dimensional symplectic vector space which has a single transverse double point of Whitney self-intersection number
+1
. This Lagrangian also arises as the Weinstein skeleton of the complement of a binodal cubic curve inside the projective plane, and the latter Weinstein manifold is thus the “standard” neighbourhood of Lagrangian immersions of this type. We classify the Lagrangians inside such a neighbourhood which are homologically essential, and which are either embedded or immersed with a single double point; they are shown to be Hamiltonian isotopic to either product tori, Chekanov tori, or rescalings of the Whitney immersion.
nearby Lagrangian conjecture, Lagrangian fibration, Clifford torus, Chekanov torus, Whitney immersion, Whitney sphere
Seconded: Ciprian Manolescu, Leonid Polterovich |
Using the Variables Palette in the Simulation Results Tab - MapleSim Help
Home : Support : Online Help : MapleSim : Using MapleSim : Simulating a Model : Managing Simulation Results : Using the Variables Palette in the Simulation Results Tab
Using Variables Palette in the Simulation Results Tab
In the Simulation Results tab, the Variables palette lists probes and variables for the current model in a tree. This palette is used to create new simulation graphs and to add a new variable to a simulation graph. You can also change what variable is graphed on the x-axis or add a second vertical axis.
Use the search field in the Variables palette to find a variable quickly. Use the Display probes/variables button ( ) to toggle between showing probed/plotted variables and showing all variables. Enter a name or partial name. The variables tree list shows all matches.
To clear a search, click the Clear Search button ( ).
Note: If you do not see all the expected probes or variables, it may be because the filter from your last search is still active. Clear the current search to see all the variables.
When multiple models are open, you can change the current model by clicking on the name of a model in the Stored Results palette.
The following tools are displayed in the Variables palette.
Create new plot. Create a new plot for the selected variable.
By default, this creates the plot in a new plot window. To instead add the new plot to the existing plot window, hold the Shift key and click the Create new plot button.
Add variable to existing plot. Add the selected variable to the selected plot.
x
-axis variable. Place the selected variable on the
x
-axis of the selected plot.
Display probes/variables. Toggle the variables displayed in this palette between all model variables or only variables that have been probed variables or added to a custom plot.
Adding a Variable to a Simulation Graph
Adding a Second Vertical Axis to a Simulation Graph
Generating a New Plot Window Configuration
Setting the x-Axis Variable in a Simulation Graph
Viewing Extended Simulation Results After Simulating a Model |
Free solutions to electromagnetism problems
Electromagnetism problems with answers
Recent questions in Electromagnetism
Magnetic force on open circuit?
Let's say we have a straight horizontal wire and we let it drop inside a magnetic field which will be parallel to the ground(coming out of the screen). Charges inside the wire feel a force due to their movement inside the magnetic field .
Let's say the field is coming out of the screen. Positive charges gather on the left and negative on the right. If we had a loop I would have no trouble with this but during the charges' movement do we consider that we have a current? Therefore leading to a magnetic force opposing the bar's drop or not?
Ashley Fritz 2022-05-15 Answered
Does magnetic force act along the line joining the centres like gravitational and electric forces do?
Are the directions of magnetic field and magnetic lines of force the same? I have read that the direction of the field is tangential to the direction of the line of force.
What are the specific electronic properties that make an atom ferromagnetic versus simply paramagnetic?
The area of a rectangular loop is , and the plane of the loop makes an angle of
{41}^{\circ }
with a 0.28-T magnetic field. What is the magnetic flux penetrating the loop?
Is magnetic force pseudo?
Magnetic force exist only if charge is moving, so it must be pseudo. Imagine, a positively charged man who has the same speed as electron (charge). So, he doesn't feel any magnetic force as charge is at rest with respect to him. Therefore, he only experience electric force.
However a man who is at rest or has different speed than electron feels a magnetic force
Therefore magnetic force must be pseudo. Pls answer me
Jayden Mckay 2022-05-14 Answered
Given the quantum Heisenberg model with Hamiltonian
\stackrel{^}{H}=-\frac{1}{2}\sum _{i,j}{J}_{ij}{\stackrel{^}{\mathbf{\text{S}}}}_{i}\cdot {\stackrel{^}{\mathbf{\text{S}}}}_{j}
the uniform mean-field approximation
{\stackrel{^}{\mathbf{\text{S}}}}_{i}=⟨\stackrel{^}{\mathbf{\text{S}}}⟩+\left({\stackrel{^}{\mathbf{\text{S}}}}_{i}-⟨\stackrel{^}{\mathbf{\text{S}}}⟩\right)
allows to rewrite it as
{\stackrel{^}{H}}_{MF}=\sum _{i}{\mathbf{\text{B}}}_{eff}\cdot {\stackrel{^}{\mathbf{\text{S}}}}_{i}+\text{const.}
in order to perform a diagonalization by means of a Fourier transform. To do so, I am told to choose the z-axis so to align with the effective field
{\mathbf{\text{B}}}_{eff}
, which I find to be
{\mathbf{\text{B}}}_{eff}=-⟨\stackrel{^}{\mathbf{\text{S}}}⟩\sum _{j}\left({J}_{ij}+{J}_{ji}\right)
. That's where I remain stuck. First of all, how should the Fourier transform which allows me to diagonalize
{\stackrel{^}{H}}_{MF}
look like? And how is this related to the alignment of the effective field?
Gauss's law for magnetism is stated as followed with the beautiful closed surface double integral :
\underset{S}{\text{∯}\phantom{\rule{thinmathspace}{0ex}}}\mathbf{B}\cdot \text{d}\mathbf{A}=0
As I understand, the idea is to say that if we sum (continuous sum since integral) all the scalar products between the vector field B (i.e., magnetic field) and surface elements dA defined by their surface normals, we get 0?
othereyeshmt4l 2022-05-14 Answered
Can magnetic force do work?
I have been told numerous times that magnetic force do no work at all but I have some trouble digesting this fact. Now suppose we have two straight wire with some current, they certainly can feel force which may be repulsive or attractive depending upon current direction, can magnetic force do work? We also have magnetic potential energy defined to
U=-\stackrel{\to }{\mu }\cdot \stackrel{\to }{B}
which suggests magnetic field can store energy and hence do some kind of work.
Gauss's Law of Magnetism shows us that the divergence of Magnetic field is 0,
▽\cdot \stackrel{\to }{B}=0
Then how do you derive that statement by showing the divergence of a magnetic field upon an axis of a current carrying coil where radius is much smaller that distance so that we can use,
{B}_{z}=\frac{{\mu }_{o}I}{2{z}^{3}}\stackrel{^}{z}
\therefore
▽\cdot B\equiv \frac{\mathrm{\partial }}{\mathrm{\partial }z}\cdot \frac{{u}_{o}I}{2{z}^{3}}\stackrel{^}{z}\ne 0
This doesn't equal zero? What am I missing?
In explaining/introducing second-order phase transition using Ising system as an example, it is shown via mean-field theory that there are two magnetized phases below the critical temperature. This derivation is done for zero external magnetic field B=0 and termed spontaneous symmetry breaking The magnetic field is then called the symmetry breaking field. But, if the symmetry breaking occurs "spontaneously" at zero external field why do we need to call the external magnetic field the symmetry breaking field? I am confused by the terminology.
Stoyanovahvsbh 2022-05-13 Answered
Direction of magnetic force on a permanent magnet
How would one calculate the direction of the magnetic force on a permanent magnet? I have read about the poles of ferromagnets aligning in a magnetic field, so would the magnetic force just create a torque on the ferromagnet until it aligns with the field?
Why are the electric force and magnetic force classified as electromagnetism?
I confuse the four kinds of fundamental interactions, so I think the electric force and magnetic force should not be classified as a big class called electromagnetism.
1.The Gauss law of electric force is related to the surface integration but the Ampere's law corresponds with path integration.
2.The electric field can be caused by a single static charge while the magnetic force is caused by a moving charge or two moving infinitesimal current.
3.The electric field line is never closed, but the magnetic field line (except those to infinity) is a closed curve.
Blaine Stein 2022-05-13 Answered
If Div B = 0, where B = magnetic field intensity, then B must be a Curl of a some vector function. What is that vector function?
Solve the equations for
{v}_{x}
{v}_{y}
m\frac{d\left({v}_{x}\right)}{dt}=q{v}_{y}B\phantom{\rule{2em}{0ex}}m\frac{d\left({v}_{y}\right)}{dt}=-q{v}_{x}B
by differentiating them with respect to time to obtain two equations of the form:
\frac{{d}^{2}u}{d{t}^{2}}+{\alpha }^{2}u=0
u={v}_{x}
{v}_{y}
{\alpha }^{2}=qB/m
. Then show that
u=C\mathrm{cos}\alpha t
u=D\mathrm{sin}\alpha t
, where C and D are constants, satisfy this equation
Whenever I differentiate the first equation with respect to time, I get a resulting equation with the form:
\frac{{d}^{2}u}{d{t}^{2}}+{\alpha }^{2}\frac{du}{dt}=0
The current in a conductor is found to be moving from (-1, 2, -3) m to (-4, 5, -6) m. Given that the force the conductor experiences is and the magnetic field present in the region is , determine the current through the conductor. a.) -1 A;
b.) -2 A;
c.) This problem has no solution.
d.) 2 A;
Magnetic force between moving charges
Given two infinite parallel charged rods with equal charge density
\lambda
. They are moving with same constant velocity
\stackrel{\to }{v}
parallel to the rods. Find the speed
v
for which the magnetic attraction is equal to the electrostatic repulsion.
Well, I know how to solve this problem: we first find out the magnetic field created by one rod on the other using Biot's and Savart's law, then we use the definition of
\stackrel{\to }{B}
d\stackrel{\to }{F}=\stackrel{\to }{v}dq×\stackrel{\to }{B}
) to find the magnetic force, then equate magnetic and electrostatic forces to find
v
, which will be greater than or equal to
c
, thus conclude it is impossible for the forces to be equal.
However, one can argue as the following:
We all know that "same laws of physics apply in all inertial frames". With a constant velocity
\stackrel{\to }{v}
,the rest frame of the rods is an inertial frame. Therefore, if Biot-Savart law applies in our frame, it has to apply in the rest frame. If so, none of the rods will feel a magnetic field from the other one because their relative speed is zero, and there will be no magnetic force between the rods.
I've seen this question several times before in references, exams, exercise sheets,and in many different forms (parallel planes, beam of electrons ...),but no one ever used this argument.What is the problem in it ? Is it something related to Maxwell's equations or special relativity ? Or what else ?
I know a similar question was asked before, but the answers weren't satisfying. Please provide your answers with necessary mathematics.
In Maxwell's equations, I understand intuitively how:
\oint B\cdot da=0
(because there are no monopoles and so equal number of field lines going in and coming out of the surface).
And then using the divergence theorem:
{\int }_{V}\left(\mathrm{\nabla }\cdot B\right)d\tau ={\oint }_{S}B\cdot da
{\int }_{V}\left(\mathrm{\nabla }\cdot B\right)d\tau
must be = 0.
But then I'm not sure why I can say:
\mathrm{\nabla }\cdot B=0
and forget about the integral.
Does it just mean that
\mathrm{\nabla }\cdot B
must be zero everywhere?
Electric Force is to Magnetic Force as Gravitational Force is to ...?
One can no nothing about the magnetic force and yet arrive at it by taking the relativistic effects of a current and a moving charge system into account. I ask whether there exists such an inherent force in case of gravity.\
Yasmine Larson 2022-05-10 Answered
Magnetic force and work
If the magnetic force does no work on a particle with electric charge, then: How can you influence the motion of the particle? Is there perhaps another example of the work force but do not have a significant effect on the motion of the particle?
M=\sqrt{\frac{4U}{3}}\varphi
where M is the ferromagnetic order parameter and
\varphi
is the auxiliary field from the Hubbard-Stratonovich transformation. The book argues that because the above equation is correct, the mean field theory which is derived from the Hartree-Fock approach is equivalent to the the saddle point approximation formalism for H-S transformation auxiliary field Lagrangian. But I can not understand the equation. |
{\frac{1}{8}}^{\mathrm{th}}
{\frac{1}{16}}^{\mathrm{th}}
{\frac{1}{16}}^{\mathrm{th}}
\frac{1}{8}\mathrm{th}
Prem and Suresh were partners in a firm sharing profits in the ratio of 7 : 8. On 1.4.2015 their firm was dissolved. After transferring assets (other than cash) and outsider's liabilities to realisation account, you given the following information :
(a) Raman, a creditor of Rs 4,00,000 accepted land valued at Rs 7,00,000 and paid Rs 3,00,000 to the firm.
Nardeep, Hardeep and Gagandeep were partners in a firm sharing profits in 2:1:3 ratio. Their Balance sheet as on 31.3.2015 was as follows:
Balance Sheet of Nardeep, Hardeep and Gagandeep
From 1-4-2015 Nardeep, Hardeep and Gagandeep decided to share the future profits equally.
For this purpose it was decided that:
Prepare, Revaluation Account, Partners Capital Accounts and the Balance Sheet of the reconstituted firm. VIEW SOLUTION
On 1.4.2013 JMR Ltd. had 20,000, 9% debentures of Rs 100 each outstanding.
(i) On 1.4.2014 the company purchased in the open market 6,000 of its own debentures for Rs 98 each and cancelled the same immediately.
(iii) On 1.3.2016 the remaining debentures were purchased for immediate cancellation for Rs 3,99,000.
Ignoring interest on debentures and debenture redemption reserve, pass necessary journal entries for the above transactions in the books of JMR Ltd. VIEW SOLUTION
State any two objectives of preparing 'Cash Flow Statement'. VIEW SOLUTION
(a) 'One of the objectives of analysis of financial statements is to ascertain the relative importance of the different components of the financial position of the firm'. State two other objectives of this analysis.
(b) List any four items of 'reserves' that are shown under the heading 'Rserves and Surplus' in the Balance Sheet of a company as per schedule III of the Companies Act 2013. VIEW SOLUTION
(a) What is meant by 'Profitability Ratios'?
(b) From the following information calculate inventory turnover ratio; Revenue from operations Rs 16,00,000; Average Inventory Rs 2,20,000; Gross Loss Ratio 5%. |
Simulation of control system for resonant vibrating machines with two unbalanced exciters | JVE Journals
Grigory Panovko1 , Alexander Shokhin2 , Sergey Eremeykin3
Problems related to design of high-performance vibrating machines are described in this paper. For this purpose, numerical simulation of flat solid body oscillations excited by two unbalanced exciters was carried out. The developed model takes into account AC motor characteristics. A control system for automatic resonant tuning under varying system parameters is developed.
Keywords: control system, resonant tuning, vibrating machine, limited power.
Modern vibrating machines typically operate on above-resonant or below-resonant modes. But it was shown that energy is used in the most efficient way on resonant mode [1]. Practical application of resonant vibrating machines is associated with certain engineering problems. One of this problems is that resonant mode is unstable in nonlinear systems such as a typical vibrating machine. Moreover, some parameters of a real vibrating machine (such as operation load) could change during operation. Hence it is necessary to use a control system to tune such a machine to resonant mode continuously.
The issue of stabilizing resonant vibrating machines with unbalanced exciters driven by AC motors is studied insufficiently so far.
This paper is devoted to the development of a control system which keeps a machine with two unbalanced exciters and 3 DOF of working body in resonant mode. Design scheme that corresponds to the experimental setup described in [2] is considered.
2. Design scheme of the machine
Design scheme of a typical vibrating machine with working body that has 3 DOF in plane
XY
is shown in Fig. 1. Working body (referred as platform further) is modeled as a rigid body on viscoelastic support that has linear characteristics. Two unbalanced rotors (exciters) are arranged symmetrically about the axis passing through the center of mass of the machine. Rotors’ axes are parallel to each other and perpendicular to the plane
XY
(Fig. 1).
Fig. 1. Design scheme
yOx
is used to describe motion of the machine. Origin of the coordinate system is aligned with static equilibrium position of the platform’s center of mass. Axis
Oy
is directed upwards.
Unbalanced exciters are driven by identical AC motors. The motors are connected to three-phase AC mains via a single frequency converter so that its rotors rotate in opposite directions.
Rotors are driven by torques
{M}_{1}
{M}_{2}
. Thus, the system oscillates in the plane
XY
Motion of the system is described by five generalized coordinates: the linear displacements of the platform’s center of mass in the
Ox
Oy
directions, rotation angle
\phi
of the platform and rotors’ rotation angles
{\phi }_{1}
{\phi }_{2}
. All angular coordinates mentioned here are measured from the
Ox
axis counterclockwise. Differential equations of motion for the system have been derived using Lagrange equations of the second kind [3]:
m\stackrel{¨}{x}+{k}_{x}\stackrel{˙}{x}+{c}_{x}x={m}_{r1}{r}_{1}\left({\stackrel{˙}{\phi }}_{1}^{2}\mathrm{c}\mathrm{o}\mathrm{s}{\phi }_{1}+{\stackrel{¨}{\phi }}_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\phi }_{1}\right)+{m}_{r2}{r}_{2}\left({\stackrel{˙}{\phi }}_{2}^{2}\mathrm{c}\mathrm{o}\mathrm{s}{\phi }_{2}+{\stackrel{¨}{\phi }}_{2}\mathrm{s}\mathrm{i}\mathrm{n}{\phi }_{2}\right)
\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }+\left({m}_{r1}{\rho }_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\delta }_{1}+{m}_{r2}{\rho }_{2}\mathrm{s}\mathrm{i}\mathrm{n}{\delta }_{2}\right)\stackrel{¨}{\phi },
m\stackrel{¨}{y}+{k}_{y}\stackrel{˙}{y}+{c}_{y}y={m}_{r1}{r}_{1}\left(-{\stackrel{˙}{\phi }}_{1}^{2}\mathrm{c}\mathrm{o}\mathrm{s}{\phi }_{1}+{\stackrel{¨}{\phi }}_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\phi }_{1}\right)+{m}_{r2}{r}_{2}\left(-{\stackrel{˙}{\phi }}_{2}^{2}\mathrm{c}\mathrm{o}\mathrm{s}{\phi }_{2}+{\stackrel{¨}{\phi }}_{2}\mathrm{s}\mathrm{i}\mathrm{n}{\phi }_{2}\right),
J\stackrel{¨}{\phi }+{k}_{\phi }\stackrel{˙}{\phi }+{c}_{\phi }\phi =\left({m}_{r1}{\rho }_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\delta }_{1}+{m}_{r2}{\rho }_{2}\mathrm{s}\mathrm{i}\mathrm{n}{\delta }_{2}\right)\stackrel{¨}{x}
\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }+{m}_{r1}{\rho }_{1}{r}_{1}\left[{\stackrel{˙}{\phi }}_{1}^{2}\mathrm{s}\mathrm{i}\mathrm{n}\left({\phi }_{1}-{\delta }_{1}\right)-{\stackrel{¨}{\phi }}_{1}\mathrm{c}\mathrm{o}\mathrm{s}\left({\phi }_{1}-{\delta }_{1}\right)\right]
\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }+{m}_{r2}{\rho }_{2}{r}_{2}\left[{\stackrel{˙}{\phi }}_{2}^{2}\mathrm{s}\mathrm{i}\mathrm{n}\left({\phi }_{2}-{\delta }_{2}\right)-{\stackrel{¨}{\phi }}_{2}\mathrm{c}\mathrm{o}\mathrm{s}\left({\phi }_{2}-{\delta }_{2}\right)\right],
{J}_{1}{\stackrel{¨}{\phi }}_{1}-{m}_{r1}{r}_{1}\left[\stackrel{¨}{x}\mathrm{s}\mathrm{i}\mathrm{n}{\phi }_{1}-\left(\stackrel{¨}{y}+g\right)\mathrm{c}\mathrm{o}\mathrm{s}{\phi }_{1}-\stackrel{¨}{\phi }\mathrm{ }{\rho }_{1}\mathrm{c}\mathrm{o}\mathrm{s}\left({\phi }_{1}-{\delta }_{1}\right)\right]={\sigma }_{1}\left({M}_{1}-{M}_{C}\right),
{J}_{2}{\stackrel{¨}{\phi }}_{2}-{m}_{r2}{r}_{2}\left[\stackrel{¨}{x}\mathrm{s}\mathrm{i}\mathrm{n}{\phi }_{2}-\left(\stackrel{¨}{y}+g\right)\mathrm{c}\mathrm{o}\mathrm{s}{\phi }_{2}-\stackrel{¨}{\phi }\mathrm{ }{\rho }_{2}\mathrm{c}\mathrm{o}\mathrm{s}\left({\phi }_{2}-{\delta }_{2}\right)\right]={\sigma }_{2}\left({M}_{2}-{M}_{C}\right),
{m}_{r1}
{m}_{r2}
– unbalanced masses of rotors,
{r}_{1}
{r}_{2}
– eccentricities of unbalanced masses,
{J}_{r1}
{J}_{r2}
– moments of inertia for unbalanced rotors;
m={m}_{0}+{m}_{r1}+{m}_{r2}
– full mass of the system;
{m}_{0}
– mass of the platform;
{k}_{x}
{k}_{y}
{k}_{\phi }
– damping coefficients of supports in a horizontal, vertical and angular directions respectively;
{c}_{x}
{c}_{y}
{c}_{\phi }
– stiffness coefficients of the supports in horizontal, vertical, and angular directions respectively;
{\rho }_{1}
{\rho }_{2}
– distance from the platform’s center of mass to the axes of rotors respectively;
{\delta }_{1}
{\delta }_{2}=\pi -{\delta }_{1}
– angles between the
x
axis and the axis, which pass through platform’s center of mass and axis of rotors in plane
XY
(counted counterclockwise), and
{\delta }_{1}=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}\left(h/a\right)
{\delta }_{2}=\pi -\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}\left(h/a\right)
h
is distance between the axis of rotor and axis Ox;
2a=2l
is distance between springs,
b=0
(see Fig. 1);
J={J}_{0}+{m}_{r1}{\rho }_{1}^{2}+{m}_{r2}{\rho }_{2}^{2}
– moment of inertia of the system;
{J}_{0}
– moment of inertia of the platform;
g
– gravitational acceleration;
{\sigma }_{1}=+1\text{,}
{\sigma }_{2}=-1
– constants that define the direction of rotors’ rotation;
{M}_{C}
– resistance moment for the rotors.
{M}_{1}
{M}_{2}
in right parts of Eq. (1) could be described by static characteristic of motors. These characteristics are obtained using simplified Kloss formula:
{M}_{1}={M}_{1}\left({s}_{1}\right)=\frac{2\mathrm{ }{M}_{cr1\mathrm{ }}}{{s}_{1}/{s}_{cr1}+{s}_{cr1}/{s}_{1}},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{M}_{2}={M}_{2}\left({s}_{2}\right)=\frac{2\mathrm{ }{M}_{cr2\mathrm{ }}}{{s}_{2}/{s}_{cr2}+{s}_{cr2}/{s}_{2}},
{M}_{cr1}
{M}_{cr2\mathrm{ }}
– critical (maximum) torques for each motor,
{s}_{cr1}
{s}_{cr2\mathrm{ }}
– slip at critical torque,
{s}_{1}=1-P\left|{\stackrel{˙}{\phi }}_{1}/f\right|
{s}_{2}=1-P\left|{\stackrel{˙}{\phi }}_{2}/f\right|
– current slip determined by frequency
f
and angular velocity of rotors
{\stackrel{˙}{\phi }}_{1}
{\stackrel{˙}{\phi }}_{2}
P=\text{2}
is number of poles pairs.
4. Mathematical model verification
It is necessary to specify numerical values for the parameters in Eqs. (1-2). Numerical values that correspond to the experimental setup described in [2] are used for calculations: platform mass
{m}_{0}=\text{13.25}
kg; unbalanced mass of rotors
{m}_{r1}={m}_{r2}=\text{0.0807}
kg; eccentricities of unbalanced masses
{r}_{1}={r}_{2}=\text{0.005}
m; geometrical parameters of rotors' axes
{\rho }_{1}={\rho }_{2}=\text{0.257}
{\delta }_{1}=\mathrm{ }
13.71°,
{\delta }_{2}=180-{\delta }_{1}=\mathrm{ }
166.29°; platform length parameter
l=0.175
m; inertia moment of the platform
{J}_{0}=\text{0.229301}
kg∙m2; inertia moment of the rotors
{J}_{r1}={J}_{r2}=\mathrm{ }
2.0175∙10-6 kg∙m2.
Here motors with slightly different parameters are considered to figure out all possible normal modes of the oscillating system. Maximum torques
{M}_{cr\mathrm{ }1}=\text{2,64}
N∙m;
{M}_{cr\mathrm{ }2}=\text{1,01}{M}_{cr\mathrm{ }1}\text{;}
resistance moments for the rotors
{M}_{1c}={M}_{2c}=\text{0.02}
N∙m; maximum slips
{s}_{cr\mathrm{ }1}={s}_{cr\mathrm{ }2}={s}_{n}\left(\lambda +\sqrt{{\lambda }^{2}-1}\right)\approx \mathrm{ }
41.6 %, here
\lambda =\text{2.2}
is motors’ overload capacity,
{s}_{n}=\mathrm{ }
10 % – nominal slip.
Also it is necessary to specify stiffness parameters
{c}_{x}
{c}_{y}
{c}_{\phi }
and damping parameters
{k}_{x}
{k}_{y}
{k}_{\phi }
. All these parameters could be determined by comparing calculated values of resonant amplitudes and resonant frequencies with experimental data obtained in [2].
As a result of the verification, the following values of unknown parameters were set:
{c}_{y}=\mathrm{ }
420 kN/m,
{c}_{x}=\text{844}
kN/m,
{c}_{\phi }=\text{4375}
{k}_{x}=\text{7.9}
N∙s/m,
{k}_{y}=\text{0.68}
{k}_{\phi }=\text{1}
N∙s2/m.
5. Control system synthesis
The control system is designed to tune a vibrating machine to resonant mode when some system parameters change slowly.
Usually it is difficult to measure some system parameters like operation load. So the control system based on operation principle described in [4] was developed. This principle is based on controlling phase shift
\mathrm{\Delta }\epsilon
between platform oscillation law
y\left(t\right)
and driving force
F\left(t\right)
acting on the platform. This phase shift
\mathrm{\Delta }\epsilon
\pi /2
in resonant mode. In this paper a 3 DOF system is described in contrast to the paper [4], where a 1 DOF system is described. The system shown in Fig. 1 has three resonant states and three normal modes correspondingly. In this paper only tuning to the vertical resonant oscillation mode is described.
It is necessary to increase supply frequency slowly to reach maximum possible magnitudes in described case. So below-resonant mode must be initial state of the system. If the system is in above-resonant mode it should be tuned to bellow-resonant mode first. The system is considered to be close to resonant mode if the following condition is satisfied:
\left(\frac{\pi }{2}-{\delta }_{\epsilon }\right)<\mathrm{\Delta }\epsilon <\frac{\pi }{2},
{\delta }_{\epsilon }
is pre-specified accuracy of resonant tuning.
A control algorithm based on the solution of linearized equations of motion assuming unlimited power of electric motors (
{\stackrel{˙}{\phi }}_{1}=-{\stackrel{˙}{\phi }}_{2}=\omega =\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}
) could be proposed [5].
One can easily derive expression for phase shift in linearized system [6]:
\mathrm{\Delta }\epsilon =\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}\left(\frac{{k}_{y}\omega }{{p}^{2}-{\omega }^{2}}\right).
p
is natural frequency of the linearized system. Thus, approximation for
p
could be derived from Eq. (4):
p=\sqrt{{\omega }^{2}+{k}_{y}\omega ctg\left(\mathrm{\Delta }\epsilon \right)}.
It is assumed here that
\mathrm{\Delta }\epsilon
\omega
could be obtained from experimental data. So it is not necessary to know mass
m
to estimate eigenfrequency of the system.
To tune the system to resonant mode supply frequency
{f}_{s}={p}_{s}/\left(2\pi P\right)
should be set via frequency converter. And then supply frequency
{f}_{s+i}={p}_{s+i}/\left(2\pi P\right)
should be set cyclically on the (
s+i
)-th step of regulation until Eq. (3) is satisfied.
If the system is tuned to above-resonant mode during regulation on, suppose, (
s+i+1
)-th step of regulation (
\mathrm{\Delta }{\epsilon }_{s+i+1}>\pi /2
) than it should be returned to the previous step of regulation with supply frequency
{f}_{s+i}
and then supply frequency should be set to the new value
{f}_{s+i+2}=\left({f}_{s+i}+{f}_{s+i+1}\right)/K
K>1
. The cycle repeats again until Eq. (3) is satisfied.
It should be noted that only calculated in steady state values of the phase shift are correct. Steady state identification is a separate problem. We propose to track changes of current phase shift
\mathrm{\Delta }\epsilon
to identify steady state.
Phase shift could be calculated using some experimental data: platform displacement and rotors’ angular positions. For any moment
{t}_{i}
, when the platform is in the static equilibrium position, phase shift is determined by the formula:
\mathrm{\Delta }{\epsilon }_{i}={\phi }^{\mathrm{*}}-2\mathrm{ }\pi \mathrm{ }n,
{\phi }^{\mathrm{*}}=\left({\phi }_{1}+{\phi }_{2}\right)/2
is angle of total driving force direction, n is number of full revolutions in
{\phi }^{\mathrm{*}}
Numerical values for simulation correspond to the values described in Mathematical model verification part of the paper. Initial condition for numerical integration is state of static equilibrium.
The simulations were carried out for above-resonant mode as initial state (
f=\text{77}
Hz). This case is more general because the control system must tune the machine to the below-resonant mode first.
Simulation results are presented in the form of time diagrams of power supply frequency
f
(Fig. 2), generalized coordinate of the platform
y
(Fig. 3), phase shift between the estimated value of
y\left(t\right)
and the driving force (Fig. 4).
Fig. 2. Supply frequency law of change
Fig. 3. Platform oscillation law
One can find out that tuning process takes quite large number of iterations in this case, with jumps to above-resonant modes.
Tuning time expected to decrease if to increase the value of the damping parameter
{k}_{y}
Fig. 4 shows that the developed control system tunes the machine to near-resonant mode successfully (here 90° corresponds to resonant mode). Thus, simulation shows that the developed algorithm for control system allows to tune the machine to near-resonant mode.
Fig. 4. Phase shift
\mathrm{\Delta }\epsilon
The mathematical model for the controlled vibrating machine was developed. The model could be used to research some dynamic phenomenon in vibrating machines and to simulate controlled machine operating.
The developed algorithm for control system is based on linearized equations. Simulation shows that the developed algorithm tunes the machine to near-resonant mode and could keep the machine in this mode even if system parameters change over time.
The study was performed account for the grant of Russian Science Foundation (Project No. 15-19-30026).
Astashev V., Babitsky V., Kolovsky M. Dynamics and Control of Machines. Springer, 2000. [Search CrossRef]
Panovko G., Shohin A., Eremeykin S. Experimental analysis of the oscillations of a mechanical system with self-synchronized inertial vibration exciters. Journal of Machinery Manufacture and Reliability, Vol. 44, Issue 6, 2015, p. 492-496. [Search CrossRef]
Blekhman I. Synchronization of Dynamic Systems. Nauka, Moscow, 1971, (in Russian). [Search CrossRef]
Panovko G., Shohin A., Eremeykin S. The control of the resonant mode of a vibrating machine that is driven by an asynchronous electro motor. Journal of Machinery Manufacture and Reliability, Vol. 44, Issue 2, 2015, p. 109-113. [Search CrossRef]
Panovko G., Shohin A., Eremeykin S., Gorbunov A. Comparative analysis of two control algorithms of resonant oscillations of the vibration machine driven by an asynchronous AC motor. Journal of Vibroengineering, Vol. 17, Issue 4, 2015, p. 1903-1911. [Search CrossRef]
Biderman V. The Theory of Mechanical Oscillations. High School, Moscow, 1980, (in Russian). [Search CrossRef] |
An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white?
An urn contains 5 white and 10 black balls. Fair die is rolled and that number of balls is randomly chosen
An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?
insonsipthinye
1.Events:
A- ll of the choose balls are white
{E}_{i}
- result of the die rill is i,
i=1,2,3,4,5,6
Since the die is fair:
P\left({E}_{i}\right)=\frac{1}{6}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}i=\left\{1,2,3,4,5,6\right\}
If the die rolls i we choose a combination of i balls, among black and five white balls, therefore
P\left(A\mid {E}_{1}\right)=\frac{\left(\begin{array}{c}5\\ 1\end{array}\right)}{\left(\begin{array}{c}15\\ 1\end{array}\right)}=\frac{5}{15}=\frac{1}{3}
P\left(A\mid {E}_{2}\right)=\frac{\left(\begin{array}{c}5\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 2\end{array}\right)}=\frac{10}{105}=\frac{2}{21}
P\left(A\mid {E}_{3}\right)=\frac{\left(\begin{array}{c}5\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}=\frac{10}{455}=\frac{2}{91}
P\left(A\mid {E}_{4}\right)=\frac{\left(\begin{array}{c}5\\ 4\end{array}\right)}{\left(\begin{array}{c}15\\ 4\end{array}\right)}=\frac{1}{273}
P\left(A\mid {E}_{5}\right)=\frac{\left(\begin{array}{c}5\\ 5\end{array}\right)}{\left(\begin{array}{c}15\\ 5\end{array}\right)}=\frac{1}{3003}
P\left(A\mid {E}_{6}\right)=\frac{\left(\begin{array}{c}5\\ 6\end{array}\right)}{\left(\begin{array}{c}15\\ 6\end{array}\right)}=0
P\left(A\right),P\left({E}_{3}\mid A\right)
{E}_{1},{E}_{2},{E}_{3},{E}_{4},{E}_{5},{E}_{6}
are competing hypothesis, that is, mutually exclusive events, union of which is the whole outcome space, so conditioning on the roll of the die:
P\left(A\right)=\sum _{i=1}^{6}P\left(A\mid {E}_{i}\right)P\left({E}_{i}\right)
P\left({E}_{i}\right),P\left(A\mid {E}_{i}\right)
P\left(A\right)=\frac{1}{6}\left(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}\right)=\frac{5}{66}
P\left({E}_{3}\mid A\right)
can be calculated from the definition if we note the identity
P\left({}_{}
A-event that all balls drawn are white
{D}_{i}
- outcome of roll of the die is
i,i=1,2,\dots ,6
P\left(A\right)=\sum _{i=1}^{6}P\left(A|{D}_{i}\right)P\left({D}_{i}\right)
=\frac{1}{6}\left(P\left(A|{D}_{i}\right)+\dots +P\left(A|{D}_{6}\right)\right)
=\frac{1}{6}\left(\frac{5}{15}+\frac{{5}_{{C}_{2}}}{{15}_{{C}_{2}}}+\frac{{5}_{{C}_{3}}}{{15}_{{C}_{3}}}+\frac{{5}_{{C}_{4}}}{{15}_{{C}_{4}}}+\frac{{5}_{{C}_{5}}}{{15}_{{C}_{5}}}+0\right)=\frac{5}{66}
P\left({D}_{3}|A\right)=\frac{P\left(A|{D}_{3}\right)P\left({D}_{3}\right)}{P\left(A\right)}
=\frac{\frac{{5}_{{C}_{3}}}{{15}_{{C}_{3}}}}{\sum _{i=1}^{5}\frac{{5}_{{C}_{i}}}{{15}_{{C}_{i}}}}=\frac{22}{455}
The spring of a spring gun has force constant k =400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring.The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. Calculate the speed with which the ballleaves the barrel if you can ignore friction. Calculate the speed of the ball as it leavesthe barrel if a constant resisting force of 6.00 Nacts on the ball as it moves along the barrel. For the situation in part (b), at what position along the barrel does the ball have the greatest speed?
A 5.00kg sack of flour is lifted vertically at a constant speed of 3.50m/s through a height of 15.0m. a) How great a force is required? b) How much work is done on the sack by the lifting force? What becomes of this work?
A uniform plank of length 2.00 m and mass 30.0 kg is supported by three ropes. Find the tension in each rope when a 700-N person is 0.500 m from the left end.
A 1000 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 50 cm. What is the spring constant of the spring?
A glass bottle of soda is sealed with a screw cap. Theabsolute pressure of the carbon dioxide inside the bottle is
1.90×{10}^{5}
Pa. Assuming that the top and bottom surfaces of the cap eachhave an area of
3.60×{10}^{-4}{m}^{2}
,obtain the magnitude of the force that the screw thread exerts onthe cap in order to keep it on the bottle. The air pressure outsidethe bottle is one atmosphere.
A 755 N diver drops fromw a board 10.0 m aabove watersserface. Find the divers speed 5.00m above the waters serfface.Then find the divers speed just before hitting the water.
If a 4 by 4 matrix has
deta=\frac{1}{2},\text{ }find\text{ }det\left(2A\right),det\left(-A\right),det\left({A}^{2}\right),\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }det\left({A}^{-1}\right) |
Gap metric and Vinnicombe (nu-gap) metric for distance between two systems - MATLAB gapmetric - MathWorks í•œêµ
gapmetric
Compute Gap Metrics for Stable and Unstable Plant Models
Compute Gap Metric and Stability Margin
Gap Metric
Vinnicombe Gap Metric
Gap Metrics and Stability Margins
Gap Metrics in Robust Design
Gap metric and Vinnicombe (nu-gap) metric for distance between two systems
[gap,nugap] = gapmetric(P1,P2)
[gap,nugap] = gapmetric(P1,P2,tol)
[gap,nugap] = gapmetric(P1,P2) computes the gap and Vinnicombe (ν-gap) metrics for the distance between dynamic systems P1 and P2. The gap metric values satisfy 0 ≤ nugap ≤ gap ≤ 1. Values close to zero imply that any controller that stabilizes P1 also stabilizes P2 with similar closed-loop gains.
[gap,nugap] = gapmetric(P1,P2,tol) specifies a relative accuracy for calculating the gaps.
Create two plant models. One plant, P1, is an unstable first-order system with transfer function 1/(s–0.001). The other plant, P2, is stable, with transfer function 1/(s +0.001).
P1 = tf(1,[1 -0.001]);
P2 = tf(1,[1 0.001]);
Despite the fact that one plant is unstable and the other is stable, these plants are close as measured by the gap and nugap metrics.
gap = 0.0021
nugap = 0.0020
The gap is very small compared to 1. Thus a controller that yields a stable closed-loop system with P2 also tends to stabilize P1. For instance, the feedback controller C = 1 stabilizes both plants and renders nearly identical closed-loop gains. To see this, examine the sensitivity functions of the two closed-loop systems.
H1 = loopsens(P1,C);
subplot(2,2,1); bode(H1.Si,'-',H2.Si,'r--');
subplot(2,2,2); bode(H1.Ti,'-',H2.Ti,'r--');
subplot(2,2,3); bode(H1.PSi,'-',H2.PSi,'r--');
subplot(2,2,4); bode(H1.CSo,'-',H2.CSo,'r--');
Next, consider two stable plant models that differ by a first-order system. One plant, P3, is the transfer function 50/(s+50), and the other plant, P4, is the transfer function [50/(s+50)]*8/(s+8).
P3 = tf(50,[1 50]);
P4 = tf(8,[1 8])*P3;
bode(P3,P4)
Although the two systems have similar high-frequency dynamics and the same unity gain at low frequency, by the gap and nugap metrics, the plants are fairly far apart.
Consider a plant and a stabilizing controller.
P1 = tf([1 2],[1 5 10]);
C = tf(4.4,[1 0]);
Compute the stability margin for this plant and controller.
b1 = ncfmargin(P1,C)
Next, compute the gap between P1 and the perturbed plant, P2.
Because the stability margin b1 = b(P1,C) is greater than the gap between the two plants, C also stabilizes P2. As discussed in Gap Metrics and Stability Margins, the stability margin b2 = b(P2,C) satisfies the inequality asin(b(P2,C)) ≥ asin(b1)-asin(gap). Confirm this result.
b2 = ncfmargin(P2,C);
[asin(b2) asin(b1)-asin(gap)]
P1,P2 — Input systems
Input systems, specified as dynamic system models. P1 and P2 must have the same input and output dimensions. If P1 or P2 is a generalized state-space model (genss or uss) then gapmetric uses the current or nominal value of all control design blocks.
Relative accuracy for computing the gap metrics, specified as a positive scalar. If gapactual is the true value of the gap (or the Vinnicombe gap), the returned value gap (or nugap) is guaranteed to satisfy
|1 – gap/gapactual| < tol.
gap — Gap between P1 and P2
scalar in [0,1]
Gap between P1 and P2, returned as a scalar in the range [0,1]. A value close to zero implies that any controller that stabilizes P1 also stabilizes P2 with similar closed-loop gains. A value close to 1 means that P1 and P2 are far apart. A value of 0 means that the two systems are identical.
nugap — Vinnicombe gap (ν-gap) between P1 and P2
Vinnicombe gap (ν-gap) between P1 and P2, returned as a scalar value in the range [0,1]. As with gap, a value close to zero implies that any controller that stabilizes P1 also stabilizes P2 with similar closed-loop gains. A value close to 1 means that P1 and P2 are far apart. A value of 0 means that the two systems are identical. Because 0 ≤ nugap ≤ gap ≤ 1, the ν-gap can provide a more stringent test for robustness as described in Gap Metrics and Stability Margins.
For plants P1 and P2, let
{P}_{1}={N}_{1}{M}_{1}^{â1}
{P}_{2}={N}_{2}{M}_{2}^{â1}
be right normalized coprime factorizations (see rncf). Then the gap metric δg is given by:
{\mathrm{δ}}_{g}\left({P}_{1},{P}_{2}\right)=\mathrm{max}\left\{{\stackrel{â}{\mathrm{δ}}}_{g}\left({P}_{1},{P}_{2}\right),{\stackrel{â}{\mathrm{δ}}}_{g}\left({P}_{2},{P}_{1}\right)\right\}.
{\stackrel{â}{\mathrm{δ}}}_{g}\left({P}_{1},{P}_{2}\right)
is the directed gap, given by
{\stackrel{â}{\mathrm{δ}}}_{g}\left({P}_{1},{P}_{2}\right)=\underset{\text{stable }Q\left(s\right)}{\mathrm{min}}{â\left[\begin{array}{c}{M}_{1}\\ {N}_{1}\end{array}\right]â\left[\begin{array}{c}{M}_{2}\\ N2\end{array}\right]Qâ}_{\infty }.
For more information, see [1] and Chapter 17 of [2].
For P1 and P2, the Vinnicombe gap metric is given by
{\mathrm{δ}}_{\mathrm{ν}}\left({P}_{1},{P}_{2}\right)=\underset{\mathrm{Ï}}{\mathrm{max}}{â{\left(I+{P}_{2}{P}_{2}^{*}\right)}^{â1/2}\left({P}_{1}â{P}_{2}\right){\left(I+{P}_{1}{P}_{1}^{*}\right)}^{â1/2}â}_{\infty },
\mathrm{det}\left(I+{P}_{2}^{*}{P}_{1}\right)
has the right winding number. Here, * denotes the conjugate (see ctranspose). This expression is a weighted difference between the two frequency responses P1(jω) and P2(jω). For more information, see Chapter 17 of [2].
The gap and ν-gap metrics give a numerical value δ(P1,P2) for the distance between two LTI systems. For both metrics, the following robust performance result holds:
arcsin b(P2,C2) ≥ arcsin b(P1,C1) – arcsin δ(P1,P2) – arcsin δ(C1,C2),
where the stability margin b (see ncfmargin), assuming negative-feedback architecture, is given by
b\left(P,C\right)={â\left[\begin{array}{c}I\\ C\end{array}\right]{\left(I+PC\right)}^{â1}\left[\begin{array}{cc}I& P\end{array}\right]â}_{\infty }^{â1}={â\left[\begin{array}{c}I\\ P\end{array}\right]{\left(I+CP\right)}^{â1}\left[\begin{array}{cc}I& C\end{array}\right]â}_{\infty }^{â1}.
To interpret this result, suppose that a nominal plant P1 is stabilized by controller C1 with stability margin b(P1,C1). Then, if P1 is perturbed to P2 and C1 is perturbed to C2, the stability margin is degraded by no more than the above formula. For an example, see Compute Gap Metric and Stability Margin.
The ν-gap is always less than or equal to the gap, so its predictions using the above robustness result are tighter.
The quantity b(P,C)–1 is the signal gain from disturbances on the plant input and output to the input and output of the controller.
To make use of the gap metrics in robust design, you must introduce weighting functions. In the robust performance formula, replace P by W2PW1, and replace C by
{W}_{1}^{â1}C{W}_{2}^{â1}
. You can make similar substitutions for P1, P2, C1 and C2. This form makes the weighting functions compatible with the weighting structure in the H∞ loop shaping control design procedure used by functions such as loopsyn and ncfsyn.
[1] Georgiou, Tryphon T. “On the Computation of the Gap Metric.†Systems & Control Letters 11, no. 4 (October 1988): 253–57. https://doi.org/10.1016/0167-6911(88)90067-9.
[2] Zhou, K., Doyle, J.C., Essentials of Robust Control. London, UK: Pearson, 1997.
ncfmargin | loopsyn | ncfsyn | robstab | wcdiskmargin | wcgain |
Find an answer to any descriptive statistics problems
Get help with descriptive statistics problems
Recent questions in Descriptive Statistics
X\sim N\left(15,4\right)
Find P( X > 18.7 | X > 11.7 ).
Attempt: Rewrite as P ( x > 18.7) - P( x > 11.7)
Using the phi function
\varphi \left(1.85\right)-\varphi \left(-1.65\right)=\left(0.9678\right)-\left(0.0495\right)=0.9183
I'm new here and was hoping you guys could help me with a statistics problem that I don't quite understand. I'm not sure if it's proper etiquette to ask for help on a specific homework problem here, so I apologize if this question is out of line.
Suppose a variable of a population has mean 5 and standard deviation 11. For samples of size 121, find
c
P\left(X>2c\right)=0.3300
britesoulusjhq 2022-05-13 Answered
Mentioned that the mode of negative binomial distribution can be found by: t = 1 + ((r-1)/p) where t is some number, r is the number of successes and p is the probability of success If t is an integer, there will be 2 modes at t and t-1 If t is not an integer, the integer part of t is the mode.
This is what I have worked on so far: ((r-1)/p) is the expected number of attempts to achieve all required successes excluding the final success The expected number of attempts + the definite final success = the total number of expected attempts.The total number of expected attempts is the mode.
Why could there be two modes of t-1 and t?
P\left(X=x\right)=\left(\left(x-1\right)choose\left(r-1\right)\right)\left({p}^{r}\right)\left({q}^{\left(x-r\right)}\right)
\left(\left(t-2\right)choose\left(r-1\right)\right)\left({p}^{r}\right)\left({q}^{\left(t-r-1\right)}\right)=\left(\left(t-1\right)choose\left(r-1\right)\right)\left({p}^{r}\right)\left({q}^{\left(t-r\right)}\right)
\left(\left(t-2\right)choose\left(r-1\right)\right)\left({q}^{-1}\right)=\left(\left(t-1\right)choose\left(r-1\right)\right)
\left(\left(t-2\right)\left(t-3\right)...\left(t-r\right)\right)/\left(r-1\right)!x\left({q}^{-1}\right)=\left(\left(t-1\right)\left(t-2\right)...\left(t-r+1\right)\right)/\left(r-1\right)!
\left(t-r\right)/q=t-1
[I am stuck here]
I have a set of 1000 data points. I would like to estimate their mean using a confidence interval. I read somewhere that if the sample size,
n
, is bigger than 30 you should use a t-score, and else use a z-score.
The probability density function of the random variable
X
f\left(x\right)=\left\{\begin{array}{ll}4\left(x-{x}^{3}\right),& 0\le x\le 1\\ 0,& \text{elsewhere}\end{array}
What is the probability that three independent observations from the distribution of
X
are all less than the mode of
X
This is a question I got incorrect. I have very little experience with mode and I understand that it's the value of the random variable with the highest probability. The solution for finding the mode is as follows:
The max point occurs when
{f}^{\prime }\left(x\right)=0
{f}^{\prime }\left(x\right)=4-12{x}^{2}=0
x=\sqrt{\frac{1}{3}}
This is the max point or mode because
{f}^{″}\left(x\right)
is a negative number.
How do you define a variable and write an algebraic expression for the phrase: 3 increased by a number?
affizupdaftf3opg 2022-05-08 Answered
How can I calculate the mode in a grouped frequency distribution when the largest frequency occurs in two or more classes?
Deshawn Cabrera 2022-05-08 Answered
My main goal is to calculate the sampling variance. But I will start with the standard deviation. I am doing a meta-analysis and need to calculate the variance for EACH effect size (prevalence in this case) in each study.
For example, I have 10 positive cases out of 1000 people that were tested. This gives a prevalence of 0.01 (or 1%). How do I find the standard deviation from this information?
Reese Estes 2022-05-07 Answered
What is the mean score for the 20 rolls?
A fair die is rolled twenty times. The results are shown in the bar graph. What is the mean score for the 20 rolls?
The scores obtained on a Test can be normally distributed with mean
\mu
100
\sigma
=15.What percentage of scores lie: Below 85?
Progress: used formula
\frac{𝑥-𝜇}{𝜎}
Got -1 which corresponds to 0.15866 on z score table
Since it asks for scores below 85 do i minus 1 (the sum of all values) from the value i got ?
A class average for a test is 75 with a standard deviation of 6. How can I use this to calculate the percent of ata that are below the z score of
z=-1.50
. The possible solutions are &6.81%, 6.68%, 7.35%, 8.08%&
Carrie is performing an experiment on acids and bases, so she forms the following hypothesis: If an acidic solution is added to a basic solution, the pH of the resultant solution will decrease. What is the dependent variable in Carrie's experiment?
A student took a math exam and scored 77. If the class exam scores were mound-shaped with a mean score of 70 and standard deviation of 16, use z-score to determine how the student placed comparing with the class average i.e. is the student's mark an outlier or not?
So my attempt: I used this equation
n=\mu +\sigma z
n
is the data point (77)
\mu
is the mean (70)
\sigma
is the standard deviation (16)
4. and
z
z
-score (unknown)
To solve for the unknown:
\begin{array}{rl}77& =70+16z\\ 7& =16z\\ 0.4375& =z\end{array}
This value is equivalent to 0.6700 on the z table. I just don't know what to do with this information and what steps I would take next to figure out the question.
x
is normal with mean
\mu
\sigma
. Then I see how to derive mode of
f\left(y\right)
(distribution of
y
), as we need to find the value y that makes
{f}^{\prime }\left(y\right)==0
However, why is mode not simply
{e}^{\mu }
y
is a monotonic function of
x
, and so when
x
reaches its mode, then y should also reach its mode. The mode of
x
is its mean (
\mu
) hence
y
's mode should be
{e}^{\mu }
what mistake have I made?
Caitlyn Cole 2022-05-03 Answered
Find the critical value that corresponds to a 90% confidence level.
I know that the answer is 1.645. But i tried to look it up on the Z score table, but i couldn't figure out how to read it. I did the following: 0.9+1 = 1.90.
I then look at the left side for 1.9 and on the top i search for 0.0, the number that is stated there is 0.9713. Which isnt true.
Can someone explain me please how to use the Z-table
Maeve Bowers 2022-05-03 Answered
X
has the Binomial distribution with parameters
n,p
. How can I show that if
\left(n+1\right)p
is integer then
X
has two mode that is
\left(n+1\right)p
\left(n+1\right)p-1?
Taliyah Spencer 2022-05-03 Answered
Looking for some real world examples for mode in Statistics involving topics which students like say Football or Social networks. Also they need to clearly identify differences in the usefulness of mode and mean. For example which player to pick for a football match depending on scores against a particular team while playing against that team. Mean doesnt make sense here. Any thoughts ?
Bailee Ortiz 2022-05-03 Answered
Trying to find the probability
b/w
two scores. I know mapping to
z
-score is:
{z}_{11}=\left(11-10\right)/1.5=.67
{z}_{14}=\left(14-10\right)/1.5=2.67
\overline{x}=10,\text{ }\sigma =1.5.
Would I simple subtract the two from each other? and so on the real number line it would range from 0 to 100 and that is how the distribution would be set up. So, 2.67−2=2?
On the z-table 2 is given a value of 0.9772.
My teacher likes to give things as either a proportion or a percentage so how might one be able to address the two. Like if this question was asking for the proportion of scores that fall between 11 and 14 or if the question said what's the percentage of scores that fall between 11 and 14. How might be able to address both scenarios?
Just confused on the ideas of proportion, probability, and percentage and how they are relate.
airesex2 2022-05-01 Answered
For z score, you are taking the sample value subtracting population mean and dividing it by std deviation. Is that correct so far?
Now, the "sample value" is defined by an equation. In my scenario, I have a bunch of stores that are reviewed by their customers. The customer amount varies. One store has 2 customers who give review while other store has 19 people who give reviews. For the 1st store in my example, both customers give 5/5. while the 2nd store customers, avg score given by all 19 customers is 4.
Sample value is calculated as score/customer count. So for the 1st store, it will be 5/2 while the other store will be 4/19. If I use that in z score eq. it will mean that first store is doing much better than the last store. I would like to know how to make the score "proportional""normalized""fair" etc...
Oh my many gods, I can't believe I ended like that. Thanks for pointing that out.
Magdalena Norton 2022-05-01 Answered
Without assuming that the diameters of apple pies are distributed according to the normal distributions, estimated the probability that the mean diameter is larger than 32 cm. The sample standard deviation is estimated to be 2. The sample mean is 28 and the sample size is 100.
When I used CLT (because the sample size is >30). I am getting a z score of 20? Is this correct? |
This is the second part of a work aimed at establishing that for solutions to Cauchy–Dirichlet problems involving general non-linear systems of parabolic type, almost every parabolic boundary point is a Hölder continuity point for the spatial gradient of solutions. Here we establish higher fractional differentiability of solutions up to the boundary. Based on the necessary and sufficient condition for regular boundary points from the first part of Bögelein et al. (in this issue) [7] we achieve dimension estimates for the boundary singular set and eventually the almost everywhere regularity of solutions at the boundary.
title = {The boundary regularity of non-linear parabolic systems {II}},
TI - The boundary regularity of non-linear parabolic systems II
Bögelein, Verena; Duzaar, Frank; Mingione, Giuseppe. The boundary regularity of non-linear parabolic systems II. Annales de l'I.H.P. Analyse non linéaire, Tome 27 (2010) no. 1, pp. 145-200. doi : 10.1016/j.anihpc.2009.09.002. http://archive.numdam.org/articles/10.1016/j.anihpc.2009.09.002/
[2] E. Acerbi, G. Mingione, G.A. Seregin, Regularity results for parabolic systems related to a class of non-newtonian fluids, Ann. Inst. H. Poincaré Anal. Non Linéaire 21 no. 1 (2004), 25-60 | EuDML 78611 | Numdam | Zbl 1052.76004
[5] C. Bennett, R. Sharpley, Interpolation of Operators, Academic Press, Boston (1988) | MR 928802 | Zbl 0647.46057
[6] V. Bögelein, Partial regularity and singular sets of solutions of higher order parabolic systems, Ann. Mat. Pura Appl. 188 (2009), 61-122 | Zbl 1183.35158
[7] V. Bögelein, F. Duzaar, G. Mingione, The boundary regularity of non-linear parabolic systems I, Ann. Inst. H. Poincaré Anal. Non Linéaire 27 no. 1 (2010), 201-255 | Numdam | Zbl 1194.35086
[8] V. Bögelein, M. Parviainen, Self-improving property of nonlinear higher order parabolic systems near the boundary, NoDEA Nonlinear Differential Equations Appl., doi:10.1007/s00030-009-0038-5 | MR 2596493
[9] B. Bojarski, T. Iwaniec, Analytical foundations of the theory of quasiconformal mappings in
{ℝ}^{n}
, Ann. Acad. Sci. Fenn. Ser. A I 8 (1983), 257-324 | MR 731786 | Zbl 0548.30016
[10] E. Dibenedetto, Degenerate Parabolic Equations, Universitext, Springer-Verlag, New York (1993) | Zbl 0794.35090
[11] Y.Z. Chen, E. Dibenedetto, Boundary estimates for solutions of nonlinear degenerate parabolic systems, J. Reine Angew. Math. 395 (1989), 102-131 | EuDML 153113 | Zbl 0661.35052
[12] A. Domokos, Differentiability of solutions for the non-degenerate p-Laplacian in the Heisenberg group, J. Differential Equations 204 (2004), 439-470 | MR 2085543 | Zbl 1065.35103
[13] F. Duzaar, J.F. Grotowski, Optimal interior partial regularity for nonlinear elliptic systems: The method of a-harmonic approximation, Manuscripta Math. 103 (2000), 267-298 | Zbl 0971.35025
[14] F. Duzaar, J.F. Grotowski, M. Kronz, Partial and full boundary regularity for minimizers of functionals with nonquadratic growth, J. Convex Anal. 11 (2004), 437-476 | Zbl 1066.49022
[15] F. Duzaar, J. Kristensen, G. Mingione, The existence of regular boundary points for non-linear elliptic systems, J. Reine Angew. Math. (Crelles J.) 602 (2007), 17-58 | Zbl 1214.35021
[16] F. Duzaar, G. Mingione, Second order parabolic systems, optimal regularity, and singular sets of solutions, Ann. Inst. H. Poincaré Anal. Non Linéaire 22 (2005), 705-751 | EuDML 78676 | Numdam | Zbl 1099.35042
[17] F. Duzaar, G. Mingione, Harmonic type approximation lemmas, J. Math. Anal. Appl. 352 (2009), 301-335 | Zbl 1172.35002
[18] F. Duzaar, G. Mingione, K. Steffen, Parabolic systems with polynomial growth and regularity, Mem. Amer. Math. Soc., in press
[19] F.G. Duzaar, K. Steffen, Optimal interior and boundary regularity for almost minimizers to elliptic variational integrals, J. Reine Angew. Math. 546 (2002) | Zbl 0999.49024
[20] C. Fefferman, E.M. Stein,
{\mathrm{H}}^{p}
[21] M. Giaquinta, A counter-example to the boundary regularity of solutions to quasilinear systems, Manuscripta Math. 24 (1978), 217-220 | EuDML 154543 | Zbl 0373.35027
[22] M. Giaquinta, Multiple Integrals in the Calculus of Variations and Nonlinear Elliptic Systems, Princeton Univ. Press, Princeton, NJ (1983) | Zbl 0516.49003
[23] E. Giusti, Direct Methods in the Calculus of Variations, World Scientific Publishing Company, Singapore (2003) | MR 1962933 | Zbl 1028.49001
{L}^{p}
-integrability in PDE's and quasiregular mappings for large exponents, Ann. Acad. Sci. Fenn. Ser. A I 7 no. 2 (1982), 301-322 | MR 686647 | Zbl 0505.30011
[25] J. Kristensen, G. Mingione, The singular set of minima of integral functionals, Arch. Ration. Mech. Anal. 180 (2006), 331-398 | Zbl 1116.49010
[26] J. Kristensen, G. Mingione, Boundary regularity in variational problems, in press
[27] J. Kristensen, G. Mingione, Boundary regularity of minima, Rend. Lincei Mat. Appl. 19 (2008), 265-277 | Zbl 1194.49048
[28] G. Mingione, The singular set of solutions to non-differentiable elliptic systems, Arch. Ration. Mech. Anal. 166 (2003), 287-301 | Zbl 1142.35391
[29] G. Mingione, Bounds for the singular set of solutions to non linear elliptic systems, Calc. Var. Partial Differential Equations 18 (2003), 373-400 | Zbl 1045.35024
[30] G. Mingione, Regularity of minima: An invitation to the dark side of the calculus of variations, Appl. Math. 51 (2006), 355-425 | EuDML 33259 | Zbl 1164.49324
[31] M. Parviainen, Global gradient estimates for degenerate parabolic equations in nonsmooth domains, Ann. Mat. Pura Appl. 188 no. 2 (2009), 333-358 | MR 2491806 | Zbl 1179.35080
[32] J. Stará, O. John, J. Malý, Counterexamples to the regularity of weak solutions of the quasilinear parabolic system, Comment. Math. Univ. Carolin. 27 (1986), 123-136 | EuDML 17443 | Zbl 0625.35047
[33] E.W. Stredulinsky, Higher integrability from reverse Hölder inequalities, Indiana Univ. Math. J. 29 (1980), 407-413 | MR 570689 | Zbl 0442.35064
[34] A. Zygmund, Trigonometric Series I, Cambridge Univ. Press, Cambridge (1977) | Zbl 0367.42001 |
(Redirected from Fibonacci–Sylvester expansion)
Simple method for finding Egyptian fractions.
In mathematics, the greedy algorithm for Egyptian fractions is a greedy algorithm, first described by Fibonacci, for transforming rational numbers into Egyptian fractions. An Egyptian fraction is a representation of an irreducible fraction as a sum of distinct unit fractions, such as 5/6 = 1/2 + 1/3. As the name indicates, these representations have been used as long ago as ancient Egypt, but the first published systematic method for constructing such expansions is described in the Liber Abaci (1202) of Leonardo of Pisa (Fibonacci). It is called a greedy algorithm because at each step the algorithm chooses greedily the largest possible unit fraction that can be used in any representation of the remaining fraction.
Fibonacci actually lists several different methods for constructing Egyptian fraction representations (Sigler 2002, chapter II.7). He includes the greedy method as a last resort for situations when several simpler methods fail; see Egyptian fraction for a more detailed listing of these methods. As Salzer (1948) details, the greedy method, and extensions of it for the approximation of irrational numbers, have been rediscovered several times by modern mathematicians, earliest and most notably by J. J. Sylvester (1880); see for instance Cahen (1891) and Spiess (1907). A closely related expansion method that produces closer approximations at each step by allowing some unit fractions in the sum to be negative dates back to Lambert (1770).
The expansion produced by this method for a number x is called the greedy Egyptian expansion, Sylvester expansion, or Fibonacci–Sylvester expansion of x. However, the term Fibonacci expansion usually refers, not to this method, but to representation of integers as sums of Fibonacci numbers.
1 Algorithm and examples
2 Sylvester's sequence and closest approximation
3 Maximum-length expansions and congruence conditions
4 Approximation of polynomial roots
5 Other integer sequences
6 Related expansions
Algorithm and examples[edit]
Fibonacci's algorithm expands the fraction x/y to be represented, by repeatedly performing the replacement
{\displaystyle {\frac {x}{y}}={\frac {1}{\left\lceil {\frac {y}{x}}\right\rceil }}+{\frac {(-y){\bmod {x}}}{y\left\lceil {\frac {y}{x}}\right\rceil }}}
(simplifying the second term in this replacement as necessary). For instance:
{\displaystyle {\frac {7}{15}}={\frac {1}{3}}+{\frac {2}{15}}={\frac {1}{3}}+{\frac {1}{8}}+{\frac {1}{120}}.}
in this expansion, the denominator 3 of the first unit fraction is the result of rounding 15/7 up to the next larger integer, and the remaining fraction 2/15 is the result of simplifying −15 mod 7/15 × 3 = 6/45. The denominator of the second unit fraction, 8, is the result of rounding 15/2 up to the next larger integer, and the remaining fraction 1/120 is what is left from 7/15 after subtracting both 1/3 and 1/8.
As each expansion step reduces the numerator of the remaining fraction to be expanded, this method always terminates with a finite expansion; however, compared to ancient Egyptian expansions or to more modern methods, this method may produce expansions that are quite long, with large denominators. For instance, this method expands
{\displaystyle {\frac {5}{121}}={\frac {1}{25}}+{\frac {1}{757}}+{\frac {1}{763\,309}}+{\frac {1}{873\,960\,180\,913}}+{\frac {1}{1\,527\,612\,795\,642\,093\,418\,846\,225}},}
while other methods lead to the much better expansion
{\displaystyle {\frac {5}{121}}={\frac {1}{33}}+{\frac {1}{121}}+{\frac {1}{363}}.}
Wagon (1991) suggests an even more badly-behaved example, 31/311. The greedy method leads to an expansion with ten terms, the last of which has over 500 digits in its denominator; however, 31/311 has a much shorter non-greedy representation, 1/12 + 1/63 + 1/2799 + 1/8708.
Sylvester's sequence and closest approximation[edit]
Sylvester's sequence 2, 3, 7, 43, 1807, ... (OEIS: A000058) can be viewed as generated by an infinite greedy expansion of this type for the number 1, where at each step we choose the denominator ⌊ y/x ⌋ + 1 instead of ⌈ y/x ⌉. Truncating this sequence to k terms and forming the corresponding Egyptian fraction, e.g. (for k = 4)
{\displaystyle {\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{7}}+{\frac {1}{43}}={\frac {1805}{1806}}}
results in the closest possible underestimate of 1 by any k-term Egyptian fraction (Curtiss 1922; Soundararajan 2005). That is, for example, any Egyptian fraction for a number in the open interval (1805/1806, 1) requires at least five terms. Curtiss (1922) describes an application of these closest-approximation results in lower-bounding the number of divisors of a perfect number, while Stong (1983) describes applications in group theory.
Maximum-length expansions and congruence conditions[edit]
Any fraction x/y requires at most x terms in its greedy expansion. Mays (1987) and Freitag & Phillips (1999) examine the conditions under which the greedy method produces an expansion of x/y with exactly x terms; these can be described in terms of congruence conditions on y.
Every fraction 1/y requires one term in its greedy expansion; the simplest such fraction is 1/1.
Every fraction 2/y requires two terms in its greedy expansion if and only if y ≡ 1 (mod 2); the simplest such fraction is 2/3.
A fraction 3/y requires three terms in its greedy expansion if and only if y ≡ 1 (mod 6), for then −y mod x = 2 and y(y + 2)/3 is odd, so the fraction remaining after a single step of the greedy expansion,
{\displaystyle {\frac {(-y){\bmod {x}}}{y\left\lceil {\frac {y}{x}}\right\rceil }}={\frac {2}{\,{\frac {y(y+2)}{3}}\,}}}
is in simplest terms. The simplest fraction 3/y with a three-term expansion is 3/7.
A fraction 4/y requires four terms in its greedy expansion if and only if y ≡ 1 or 17 (mod 24), for then the numerator −y mod x of the remaining fraction is 3 and the denominator is 1 (mod 6). The simplest fraction 4/y with a four-term expansion is 4/17. The Erdős–Straus conjecture states that all fractions 4/y have an expansion with three or fewer terms, but when y ≡ 1 or 17 (mod 24) such expansions must be found by methods other than the greedy algorithm, with the 17 (mod 24) case being covered by the congruence relationship 2 (mod 3).
More generally the sequence of fractions x/y that have x-term greedy expansions and that have the smallest possible denominator y for each x is
{\displaystyle 1,{\frac {2}{3}},{\frac {3}{7}},{\frac {4}{17}},{\frac {5}{31}},{\frac {6}{109}},{\frac {7}{253}},{\frac {8}{97}},{\frac {9}{271}},\dots }
Approximation of polynomial roots[edit]
Stratemeyer (1930) and Salzer (1947) describe a method of finding an accurate approximation for the roots of a polynomial based on the greedy method. Their algorithm computes the greedy expansion of a root; at each step in this expansion it maintains an auxiliary polynomial that has as its root the remaining fraction to be expanded. Consider as an example applying this method to find the greedy expansion of the golden ratio, one of the two solutions of the polynomial equation P0(x) = x2 − x − 1 = 0. The algorithm of Stratemeyer and Salzer performs the following sequence of steps:
Since P0(x) < 0 for x = 1, and P0(x) > 0 for all x ≥ 2, there must be a root of P0(x) between 1 and 2. That is, the first term of the greedy expansion of the golden ratio is 1/1. If x1 is the remaining fraction after the first step of the greedy expansion, it satisfies the equation P0(x1 + 1) = 0, which can be expanded as P1(x1) = x2
1 + x1 − 1 = 0.
Since P1(x) < 0 for x = 1/2, and P1(x) > 0 for all x > 1, the root of P1 lies between 1/2 and 1, and the first term in its greedy expansion (the second term in the greedy expansion for the golden ratio) is 1/2. If x2 is the remaining fraction after this step of the greedy expansion, it satisfies the equation P1(x2 + 1/2) = 0, which can be expanded as P2(x2) = 4x2
2 + 8x2 − 1 = 0.
Since P2(x) < 0 for x = 1/9, and P2(x) > 0 for all x > 1/8, the next term in the greedy expansion is 1/9. If x3 is the remaining fraction after this step of the greedy expansion, it satisfies the equation P2(x3 + 1/9) = 0, which can again be expanded as a polynomial equation with integer coefficients, P3(x3) = 324x2
3 + 720x3 − 5 = 0.
Continuing this approximation process eventually produces the greedy expansion for the golden ratio,
{\displaystyle \varphi ={\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{9}}+{\frac {1}{145}}+{\frac {1}{37986}}+\cdots }
Other integer sequences[edit]
The length, minimum denominator, and maximum denominator of the greedy expansion for all fractions with small numerators and denominators can be found in the On-Line Encyclopedia of Integer Sequences as sequences OEIS: A050205, OEIS: A050206, and OEIS: A050210, respectively. In addition, the greedy expansion of any irrational number leads to an infinite increasing sequence of integers, and the OEIS contains expansions of several well known constants. Some additional entries in the OEIS, though not labeled as being produced by the greedy algorithm, appear to be of the same type.
Related expansions[edit]
In general, if one wants an Egyptian fraction expansion in which the denominators are constrained in some way, it is possible to define a greedy algorithm in which at each step one chooses the expansion
{\displaystyle {\frac {x}{y}}={\frac {1}{d}}+{\frac {xd-y}{yd}},}
where d is chosen, among all possible values satisfying the constraints, as small as possible such that xd > y and such that d is distinct from all previously chosen denominators. For instance, the Engel expansion can be viewed as an algorithm of this type in which each successive denominator must be a multiple of the previous one. However, it may be difficult to determine whether an algorithm of this type can always succeed in finding a finite expansion. In particular, the odd greedy expansion of a fraction x/y is formed by a greedy algorithm of this type in which all denominators are constrained to be odd numbers; it is known that, whenever y is odd, there is a finite Egyptian fraction expansion in which all denominators are odd, but it is not known whether the odd greedy expansion is always finite.
Cahen, E. (1891), "Note sur un développement des quantités numériques, qui presente quelque analogie avec celui en fractions continues", Nouvelles Annales des Mathématiques, Ser. 3, 10: 508–514 .
Curtiss, D. R. (1922), "On Kellogg's diophantine problem", American Mathematical Monthly, 29 (10): 380–387, doi:10.2307/2299023, JSTOR 2299023 .
Freitag, H. T.; Phillips, G. M. (1999), "Sylvester's algorithm and Fibonacci numbers", Applications of Fibonacci numbers, Vol. 8 (Rochester, NY, 1998), Dordrecht: Kluwer Acad. Publ., pp. 155–163, MR 1737669 .
Lambert, J. H. (1770), Beyträge zum Gebrauche der Mathematik und deren Anwendung, Berlin: Zweyter Theil, pp. 99–104 .
Mays, Michael (1987), "A worst case of the Fibonacci–Sylvester expansion", Journal of Combinatorial Mathematics and Combinatorial Computing, 1: 141–148, MR 0888838 .
Salzer, H. E. (1947), "The approximation of numbers as sums of reciprocals", American Mathematical Monthly, 54 (3): 135–142, doi:10.2307/2305906, JSTOR 2305906, MR 0020339 .
Salzer, H. E. (1948), "Further remarks on the approximation of numbers as sums of reciprocals", American Mathematical Monthly, 55 (6): 350–356, doi:10.2307/2304960, JSTOR 2304960, MR 0025512 .
Soundararajan, K. (2005), Approximating 1 from below using n Egyptian fractions, arXiv:math.CA/0502247 .
Spiess, O. (1907), "Über eine Klasse unendlicher Reihen", Archiv der Mathematik und Physik, Third Series, 12: 124–134 .
Stong, R. E. (1983), "Pseudofree actions and the greedy algorithm", Mathematische Annalen, 265 (4): 501–512, doi:10.1007/BF01455950, MR 0721884, S2CID 120347233 .
Stratemeyer, G. (1930), "Stammbruchentwickelungen für die Quadratwurzel aus einer rationalen Zahl", Mathematische Zeitschrift, 31: 767–768, doi:10.1007/BF01246446, S2CID 120956180 .
Sylvester, J. J. (1880), "On a point in the theory of vulgar fractions", American Journal of Mathematics, 3 (4): 332–335, doi:10.2307/2369261, JSTOR 2369261 .
Wagon, S. (1991), Mathematica in Action, W. H. Freeman, pp. 271–277 .
Retrieved from "https://en.wikipedia.org/w/index.php?title=Greedy_algorithm_for_Egyptian_fractions&oldid=1059591299" |
How to find expressions for f_{x x} and f_{y y}
How to find expressions for
{f}_{xx}
{f}_{yy}
for the multivariable function
f\left(x,y\right)=\mathrm{ln}\left({x}^{2}y\right)+{y}^{3}{x}^{2}
f\left(x,y\right)=\mathrm{ln}\left({x}^{2}y\right)+{y}^{3}{x}^{2}
For finding
{f}_{x}
, we have to differentiate
f\left(x,y\right)
with respect to x as follows:
{f}_{x}=\frac{2xy}{{x}^{2}y}+{y}^{3}
Now, for finding
{f}_{xx}
, differentiate
{f}_{x}
x
Try in the same way for
{f}_{yy}
raefx88y
Thanks, it's helpful because it shows the way. Anyway, there is
{a}^{2}
after x at last.
Use polar coordinates to find the limit. [Hint: Let
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
, and note that (x, y) (0, 0) implies r 0.]
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
Lynbrook West , an apartment complex , has 100 two-bedroom units. The montly profit (in dollars) realized from renting out x apartments is given by the following function.
P\left(x\right)=-12{x}^{2}+2136x-41000
To maximize the monthly rental profit , how many units should be rented out?
What is the maximum monthly profit realizable?
FILL IN THE BLANKS(EXPLAIN IN ONE OR TWO LINES)
imagine that the true number of variables should be included in a logistic regression model is 7 out of ten variables available .
then , in order to find the optimal model with 7 variables , the number of varaibels included in the training should be higher than _______
Relative extrema of multivariables:
f\left(x,y\right)=\frac{xy}{7}
find critical points and relative extrema, given an open region.
The analysis of tooth shrinkage by
Loring Brace and colleagues at the University of Michigan’s Museum of Anthropology indicates that human tooth size is continuing to decrease and that the evolutionary process has not yet come to a halt. In northern Europeans, for example, tooth size reduction now has a rate of 1% per 1000 years. In about how many years will human teeth be 90% of their present size?
{X}_{1},{X}_{2},\stackrel{˙}{s},{X}_{n}
be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables. Find n such that Pr
\left(X>2000\right)\ge 0.95
Solve the system of equation by the method of your choice.
{x}^{2}+{\left(y-9\right)}^{2}=49
{x}^{2}-7y=-14
Green's, Stokes', and the divergence theorem |
Plot 3-D implicit equation or function - MATLAB fimplicit3
Plot 3-D Implicit Symbolic Equation
Plot 3-D Implicit Symbolic Function
Line Style and Width for Implicit Surface Plot
Control Resolution of Implicit Surface Plot
Apply Rotation and Translation to Implicit Surface Plot
Plot 3-D implicit equation or function
fimplicit3(f,[min max])
fimplicit3(f,[xmin xmax ymin ymax zmin zmax])
fi = fimplicit3(___)
fimplicit3(f) plots the 3-D implicit equation or function f(x,y,z) over the default interval [-5 5] for x, y, and z.
fimplicit3(f,[min max]) plots f(x,y,z) over the interval [min max] for x, y, and z.
fimplicit3(f,[xmin xmax ymin ymax zmin zmax]) plots f(x,y,z) over the interval [xmin xmax] for x, [ymin ymax] for y, and [zmin zmax] for z. The fimplicit3 function uses symvar to order the variables and assign intervals.
fimplicit3(___,LineSpec) uses LineSpec to set the line style, marker symbol, and face color.
fimplicit3(___,Name,Value) specifies line properties using one or more Name,Value pair arguments. Use this option with any of the input argument combinations in the previous syntaxes.
fimplicit3(ax,___) plots into the axes with the object ax instead of the current axes object gca.
fi = fimplicit3(___) returns an implicit function surface object. Use the object to query and modify properties of a specific surface. For details, see ImplicitFunctionSurface Properties.
{x}^{2}+{y}^{2}-{z}^{2}=0
by using fimplicit3. The fimplicit3 function plots over the default interval of
\left[-5,5\right]
x
y
z
fimplicit3(x^2 + y^2 - z^2)
Plot the hyperboloid specified by the function
f\left(x,y,z\right)={x}^{2}+{y}^{2}-{z}^{2}
. The fimplicit3 function plots over the default interval of
\left[-5,5\right]
x
y
z
f(x,y,z) = x^2 + y^2 - z^2;
Specify the plotting interval by specifying the second argument to fimplicit3. Plot the upper half of the hyperboloid
{x}^{2}+{y}^{2}-{z}^{2}=0
by specifying the interval
0<z<5
x
y
, use the default interval
\left[-5,5\right]
f = x^2 + y^2 - z^2;
fimplicit3(f, interval)
Plot the implicit equation
x\mathrm{sin}\left(y\right)+z\mathrm{cos}\left(x\right)=0
\left(-2\pi ,2\pi \right)
for all axes.
eqn = x*sin(y) + z*cos(x);
fimplicit3(eqn,[-2*pi 2*pi])
title('xsin(y) + zcos(x) for -2\pi < x < 2\pi and -2\pi < y < 2\pi')
ax.YTickLabel = arrayfun(@texlabel, S, 'UniformOutput', false);
{x}^{2}+{y}^{2}-{z}^{2}=0
z
-5<z<-2
-2<z<2
2<z<5
fimplicit3(f,[-5 5 -5 5 -5 -2],'--.','MarkerEdgeColor','g')
fimplicit3(f,[-5 5 -5 5 -2 2],'LineWidth',1,'FaceColor','g')
fimplicit3(f,[-5 5 -5 5 2 5],'EdgeColor','none')
1/{x}^{2}-1/{y}^{2}+1/{z}^{2}=0
. Specify an output to make fimplicit3 return the plot object.
fi = fimplicit3(f)
Function: 1/x^2 - 1/y^2 + 1/z^2
Show only the positive x-axis by setting the XRange property of fi to [0 5]. Remove the lines by setting the EdgeColor property to 'none'. Visualize the hidden surfaces by making the plot transparent by setting the FaceAlpha property to 0.8.
fi.XRange = [0 5];
fi.EdgeColor = 'none';
fi.FaceAlpha = 0.8;
Control the resolution of an implicit surface plot by using the 'MeshDensity' option. Increasing 'MeshDensity' can make smoother, more accurate plots while decreasing 'MeshDensity' can increase plotting speed.
Divide a figure into two by using subplot. In the first subplot, plot the implicit surface
\mathrm{sin}\left(1/\left(xyz\right)\right)
. The surface has large gaps. Fix this issue by increasing the 'MeshDensity' to 40 in the second subplot. fimplicit3 fills the gaps showing that by increasing 'MeshDensity' you increased the resolution of the plot.
f = sin(1/(x*y*z));
fimplicit3(f,'MeshDensity',40)
Apply rotation and translation to the implicit surface plot of a torus.
A torus can be defined by an implicit equation in Cartesian coordinates as
\mathit{f}\left(\mathit{x},\mathit{y},\mathit{z}\right)={\left({\mathit{x}}^{2}+{\mathit{y}}^{2}+{\mathit{z}}^{2}+{\mathit{R}}^{2}-{\mathit{a}}^{2}\right)}^{2}-4{\mathit{R}}^{2}\left({\mathit{x}}^{2}+{\mathit{y}}^{2}\right)
a
R
and
R
as 1 and 5, respectively. Plot the torus using fimplicit3.
f(x,y,z) = (x^2+y^2+z^2+R^2-a^2)^2 - 4*R^2*(x^2+y^2);
x
\pi /2
radians. Shift the center of the torus by 5 along the
x
0 cos(alpha) sin(alpha);
0 -sin(alpha) cos(alpha)];
g = subs(f,[x,y,z],[r_90(1)-5,r_90(2),r_90(3)]);
Add a second plot of the rotated and translated torus to the existing graph.
fimplicit3(g)
f — 3-D implicit equation or function to plot
symbolic equation | symbolic expression | symbolic function
3-D implicit equation or function to plot, specified as a symbolic equation, expression, or function. If an expression or function is specified, then fimplicit3 assumes the right-hand size to be 0.
[min max] — Plotting interval for x-, y- and z- axes
Plotting interval for x-, y- and z- axes, specified as a vector of two numbers. The default is [-5 5].
[xmin xmax ymin ymax zmin zmax] — Plotting interval for x-, y- and z- axes
[–5 5 –5 5 –5 5] (default) | vector of six numbers
Plotting interval for x-, y- and z- axes, specified as a vector of six numbers. The default is [-5 5 -5 5 -5 5].
Axes object. If you do not specify an axes object, then fimplicit3 uses the current axes.
The properties listed here are only a subset. For a complete list, see ImplicitFunctionSurface Properties.
Number of evaluation points per direction, specified as a number. The default is 35.
fi — One or more objects
One or more objects, returned as a scalar or a vector. The object is an implicit function surface object. You can use these objects to query and modify properties of a specific line. For details, see ImplicitFunctionSurface Properties.
fimplicit3 assigns the symbolic variables in f to the x axis, the y axis, then the z axis, and symvar determines the order of the variables to be assigned. Therefore, variable and axis names might not correspond. To force fimplicit3 to assign x, y, or z to its corresponding axis, create the symbolic function to plot, then pass the symbolic function to fimplicit3.
For example, the following code plots the roots of the implicit function f(x,y,z) = x + z in two ways. The first way forces fimplicit3 to assign x and z to their corresponding axes. In the second way, fimplicit3 defers to symvar to determine variable order and axis assignment: fimplicit3 assigns x and z to the x and y axes, respectively.
syms x y z;
f(x,y,z) = x + z;
fimplicit3(f);
fimplicit3(f(x,y,z)); % Or fimplicit3(x + z);
fcontour | fimplicit | fmesh | fplot | fplot3 | fsurf |
The coefficient matrix for a system of linear differential equations of the form
The coefficient matrix for a system of linear differential equations of the form y^{1} = A_{y} has the given eigenvalues
{y}^{1}={A}_{y}
has the given eigenvalues and eigenspace bases. Find the general solution for the system.
\left[{\lambda }_{1}=-1⇒\left\{\left[\begin{array}{c}103\end{array}\right]\right\},{\lambda }_{2}=3i⇒\left\{\left[\begin{array}{c}2-i1+i7i\end{array}\right]\right\},{\lambda }_{3}=-3i⇒\left\{\left[\begin{array}{c}2+i1-i-7i\end{array}\right]\right\}\right]
y={c}_{1}{y}_{1}+{c}_{2}{y}_{2}+{c}_{3}{y}_{3}
{y}_{1}={e}^{{\lambda }_{1}t}u
{y}_{2}={e}^{at}\left(\mathrm{sin}\left(bt\right)Re\left(u\right)+\mathrm{cos}\left(bt\right)Im\left(u\right)\right)
{y}_{3}={e}^{at}\left(\mathrm{cos}\left(bt\right)Re\left(u\right)-\mathrm{sin}\left(bt\right)Im\left(u\right)\right)
\left[{\lambda }_{1}=-1⇒\left\{\left[\begin{array}{c}103\end{array}\right]\right\},{\lambda }_{2}=3i⇒\left\{\left[\begin{array}{c}2-i1+i7i\end{array}\right]\right\},{\lambda }_{3}=-3i⇒\left\{\left[\begin{array}{c}2+i1-i-7i\end{array}\right]\right\}\right]
{y}_{1}={e}^{-t}\left(\left[\begin{array}{c}103\end{array}\right]\right)
{y}_{2}={e}^{0t}\left(\left(\mathrm{sin}\left(3t\right)\left[\begin{array}{c}212\end{array}\right]+\mathrm{cos}\left(3t\right)\left[\begin{array}{c}-111\end{array}\right]\right)
=\left(\left(\mathrm{sin}\left(3t\right)\left[\begin{array}{c}212\end{array}\right]+\left(\mathrm{cos}\left(3t\right)\left[\begin{array}{c}-111\end{array}\right]\right)
{y}_{3}={e}^{0t}\left(\mathrm{cos}\left(-3t\right)\left[\begin{array}{c}210\end{array}\right]-\mathrm{sin}\left(-3t\right)\left[\begin{array}{c}1-17\end{array}\right]\right)
=\left(\mathrm{cos}\left(-3t\right)\left[\begin{array}{c}210\end{array}\right]-\mathrm{sin}\left(-3t\right)\left[\begin{array}{c}1-17\end{array}\right]\right)
y={c}_{1}{y}_{1}+{c}_{2}{y}_{2}+{c}_{3}{y}_{3}
={c}_{1}{e}^{-t}\left(\left[\begin{array}{c}103\end{array}\right]\right)+{c}_{2}\left(\left(\mathrm{sin}\left(3t\right)\left[\begin{array}{c}212\end{array}\right]+\left(\mathrm{cos}\left(3t\right)\left[\begin{array}{c}-111\end{array}\right]\right)+{c}_{3}\left(\mathrm{cos}\left(-3t\right)\left[\begin{array}{c}210\end{array}\right]-\mathrm{sin}\left(-3t\right)\left[\begin{array}{c}1-17\end{array}\right]\right)
{y}_{1}={c}_{1}{e}^{t}\left({c}_{2}\left(2\mathrm{sin}\left(3t\right)-\mathrm{cos}\left(3t\right)\right)+{c}_{3}\left(2\mathrm{cos}\left(-3t\right)-\mathrm{sin}\left(-3t\right)\right)\right)
The integrating factor method, which was an effective method for solving first-order differential equations, is not a viable approach for solving second-order equstions. To see what happens, even for the simplest equation, consider the differential equation
y{}^{″}+3{y}^{\prime }+2y=f\left(t\right)
. Lagrange sought a function
\mu \left(t\right)\mu \left(t\right)
such that if one multiplied the left-hand side of
y{}^{″}+3{y}^{\prime }+2y=f\left(t\right)
\mu \left(t\right)\mu \left(t\right)
, one would get
\mu \left(t\right)\left[y{}^{″}+{y}^{\prime }+y\right]=d\frac{d}{dt}\left[\mu \left(t\right)y+g\left(t\right)y\right]
where g(t)g(t) is to be determined. In this way, the given differential equation would be converted to
\frac{d}{dt}\left[\mu \left(t\right){y}^{\prime }+g\left(t\right)y\right]=\mu \left(t\right)f\left(t\right)
, which could be integrated, giving the first-order equation
\mu \left(t\right){y}^{\prime }+g\left(t\right)y=\int \mu \left(t\right)f\left(t\right)dt+c
which could be solved by first-order methods. (a) Differentate the right-hand side of
\mu \left(t\right)\left[y{}^{″}+{y}^{\prime }+y\right]=d\frac{d}{dt}\left[\mu \left(t\right)y+g\left(t\right)y\right]
and set the coefficients of y,y' and y'' equal to each other to find g(t). (b) Show that the integrating factor
\mu \left(t\right)\mu \left(t\right)
satisfies the second-order homogeneous equation
\mu {}^{″}-{\mu }^{\prime }+\mu =0
called the adjoint equation of
y{}^{″}+3{y}^{\prime }+2y=f\left(t\right)
. In other words, althought it is possible to find an "integrating factor" for second-order differential equations, to find it one must solve a new second-order equation for the integrating factor μ, which might be every bit as hard as the original equation. (c) Show that the adjoint equation of the general second-order linear equation
y{}^{″}+p\left(t\right){y}^{\prime }+q\left(t\right)y=f\left(t\right)
is the homogeneous equation
\mu {}^{″}-p\left(t\right){\mu }^{\prime }+\left[q\left(t\right)-{p}^{\prime }\left(t\right)\right]\mu =0
Find the general solution to the equation
x\left(\frac{dy}{dx}\right)+3\left(y+{x}^{2}\right)=\frac{\mathrm{sin}x}{x}
d\frac{{d}^{2}y}{d{t}^{2}}\text{ }-\text{ }8d\frac{dy}{dt}\text{ }+\text{ }15y=9t{e}^{3t}\text{ }with\text{ }y\left(0\right)=5,\text{ }{y}^{\prime }\left(0\right)=10
I got the sum of A is 0? There is no solution to this? Can someone please help. Tha
y{}^{″}-4{y}^{\prime }+4y=-6{e}^{2t}
y=3{e}^{3x}
is a solution of a second order linear homogeneous differential equation with constant coefficients. The equation is:
{y}^{″}-\left(3+a\right){y}^{\prime }+3ay=0
, a any real number.
{y}^{″}+{y}^{\prime }-6y=0
{y}^{″}+3{y}^{\prime }=0
{y}^{″}+\left(3-a\right){y}^{\prime }+3ay=0
y{}^{″}+4{y}^{\prime }=\mathrm{tan}\left(t\right)
I have used the method of variation of parameters. Currently I am at a point in the equation where I have this:
{u}_{1}=\int \frac{\mathrm{tan}t\mathrm{cos}2t}{2}
Deterrmine the first derivative
\left(\frac{dy}{dx}\right)
y=2{e}^{2}x+In{x}^{3}-2{e}^{x} |
The analysis of the ventricle assist device controlled rotor dynamics | JVE Journals
Elena Ovsyannikova1 , Alexander M. Gouskov2
1, 2Moscow Bauman State Technical University, Moscow, Russia
The analysis of dynamics of rotor-driven artificial ventricle (VAD) was conducted in the work. A comparison of two types of control is given: a linear-quadratic (LC) optimization and PID-regulator. It was shown that LC – control allows the pump rotor positioning with an accuracy of 0.2 mm at speeds ranging from 5,000 to 12,000 rev/min.
Keywords: mechanical circulatory support, ventricle assist device, artificial heart, active magnetic bearings, dynamics of the control system, LC-control, PID controller, rotor stabilization.
The problems of heart failure are particularly acute in recent years. One of the variants of solution became circulatory support devices –ventricle assist devices (VADs). In accordance with articles [1-3], the axial VADs are preferred nowadays. The work is devoted to describing the VAD rotor dynamics: the theoretical bases were shown; the equation of motion was obtained; active magnetic bearings were considered. The question of the rotor movement control was considered: two types of control were selected for comparison, PID control and LC-management. The numerical simulation of the rotor stabilization was realized. The rotor positioning error had not to exceed 0.2 mm. The behavior of the rotor and the control response is examined in the speed range from 5000 rev/min to 12000 rev/min.
The symmetric homogeneous rigid rotor rotating along the longitudinal axis at a constant angular velocity in two radial active magnetic bearings AMP1 (A) and AMP2 (B) of axial pump VAD is considered. Specifications are given in the tables (Table 1).
Table 1. Rotor technical characteristics
D
l
m
\epsilon
Rotation frequency, rev./min.
\mathrm{\Omega }
Length from rotor center mass to bearings, m
a
b
Length from rotor center mass to censors, m
c
d
The assumptions for the equation of motion are the following [4]:
1) The rotor is symmetric and rigid.
2) Deviations from the reference position are small in comparison with the rotor dimensions.
3) The angular velocity
\mathrm{\Omega }
of the rotor about its longitudinal axis
z
is assumed tobe constant.
The inclinations and the angular motion around the rotor spin axis are described by the three so-called Cardan angles
\alpha
\beta
\gamma
. Linearization leads to characterizing the angles
\alpha
\beta
as inclinations about the
X
Y\text{.}
The equations of motion are given for the variables
\mathbf{q}={\left\{\beta ,{x}_{S},-\alpha ,{y}_{S}\right\}}^{T}
The equations of motion follow from Lagrange’s equations:
\frac{d}{dt}\left(\frac{\partial T}{\partial {\stackrel{˙}{q}}_{i}}\right)-\frac{\partial T}{\partial {q}_{i}}={Q}_{i},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }i=1...4,
with the kinetic energy
T
and generalized forces
{Q}_{i}
{q}_{i}
– the generalized coordinates. The kinetic energy
T
T=\frac{1}{2}m\left({{\stackrel{˙}{x}}_{S}}^{2}+{{\stackrel{˙}{y}}_{S}}^{2}+{{\stackrel{˙}{z}}_{S}}^{2}\right)+\frac{1}{2}{\mathrm{\Omega }}^{T}{Ι}_{S}\mathrm{\Omega },
{\stackrel{˙}{x}}_{S}
{\stackrel{˙}{y}}_{S}
{\stackrel{˙}{z}}_{S}
– components of center mass velocity,
{Ι}_{S}
– rotor inertia tensor
\mathrm{\Omega }
– angular velocity vector.
Equation of motion of the rotor with active magnetic bearings [4]:
\mathbf{M}\stackrel{¨}{\mathbf{q}}\left(t\right)+\mathbf{G}\stackrel{˙}{\mathbf{q}}\left(t\right)+\mathbf{K}\mathbf{q}\left(t\right)={\mathbf{B}}_{q}{\mathbf{K}}_{i}\mathbf{i}\left(t\right)+{\mathbf{F}}_{ext},
\mathbf{M}
[4×4] – symmetric positive definite mass matrix,
\mathbf{G}\mathbf{ }
[4×4] - skew-symmetric gyroscopic matrix,
\mathbf{K}
[4×4] – stiffness matrix,
{\mathbf{B}}_{q}
[4×4] – transformation matrix which relates the generalized coordinates of the rotor center mass to the rotor displacements in magnetic bearings,
{\mathbf{K}}_{i}
[4×4] – matrix of current stiffness’s,
\mathbf{i}\left(t\right)
[4×1] – vector of currents in magnets,
{\mathbf{F}}_{ext}
[4×1] – vector of generalized external forces.
The rotor receives the load
{\mathbf{F}}_{ext}
as the force of gravity and the moments from the hydrodynamic forces in the fluid flow:
{\mathbf{F}}_{ext}\left(t\right)=\left\{\begin{array}{l}C\mu \pi {R}^{2}l\left(-\stackrel{˙}{\beta }\left(t\right)+\mathrm{\Omega }\alpha \left(t\right)\right),\\ {A}_{1}\mathrm{c}\mathrm{o}\mathrm{s}\left(\omega t\right),\\ C\mu \pi {R}^{2}l\left(-\stackrel{˙}{\alpha }\left(t\right)-\mathrm{\Omega }\beta \left(t\right)\right),\\ -mg+{A}_{2}\mathrm{c}\mathrm{o}\mathrm{s}\left(\omega t\right),\end{array}\right\
\mu
– the blood viscosity,
\mu =
(3-4)×10-3 Pa∙s at 37 °С,
C
– the drag coefficient,
C=
0,91-0,85,
R
– radius,
l
– length,
\mathrm{\Omega }
– rotation frequency. The impact of external influences on a person is taken into account in the form of vibration, expressed by harmonic functions, acting at the
x
y
{A}_{1}
{A}_{2}
– the amplitudes of the oscillation of transport.
4.1. The decentralized PID-control
The local control shown in Fig. 1 feeds each local sensor signal back to the corresponding bearing control current using the feedback gains [5]. The four output signals
\mathbf{i}\left(t\right)
combine in the output vector
\mathbf{q}
\mathbf{i}\left(t\right)=-\left(\mathbf{P}\mathbf{q}\left(t\right)+\mathbf{D}\stackrel{˙}{\mathbf{q}}\left(t\right)+\mathbf{I}\underset{{t}_{0}}{\overset{{t}_{1}}{\int }}\mathbf{q}\left(t\right)d\tau \right),
\mathbf{P}
\mathbf{D}
\mathbf{I}
– the diagonal matrixes of control coefficients. The equation of motion then takes the form:
\mathbf{M}\stackrel{¨}{\mathbf{q}}\left(t\right)+\mathbf{G}\stackrel{˙}{\mathbf{q}}\left(t\right)+\mathbf{K}\mathbf{q}\left(t\right)+{\mathbf{K}}_{C}\mathbf{q}\left(t\right)+{\mathbf{D}}_{C}\stackrel{˙}{\mathbf{q}}\left(t\right)+{\mathbf{I}}_{C}\underset{{t}_{0}}{\overset{{t}_{1}}{\int }}\mathbf{q}\left(t\right)d\tau ={\mathbf{F}}_{ext}\left(t\right),
{\mathbf{K}}_{C}={\mathbf{B}}_{q}{\mathbf{K}}_{i}\mathbf{P}\mathbf{C}
{\mathbf{D}}_{C}={\mathbf{B}}_{q}{\mathbit{K}}_{i}\mathbf{D}\mathbf{C}
are stiffness and damping matrixes respectively and
{\mathrm{Ι}}_{C}={\mathbf{B}}_{q}{\mathbf{K}}_{i}\mathbf{I}\mathbf{C}
– matrix of integral components,
\mathbf{C}
– transformation matrix.
4.2. The linear quadratic method
Linear-quadratic regulator (LQR) – optimal control algorithm based on the idea of minimizing a certain functional [6]. The novelty of this method is based on varying parameters which can be chosen for different types of conditions. Besides, for this type of control the equation of motion will change, because it should be implemented by bearing coordinates:
{\mathbf{M}}_{b}{\stackrel{¨}{\mathbf{q}}}_{b}+{\mathbf{G}}_{b}{\stackrel{˙}{\mathbf{q}}}_{b}={\mathbf{B}}_{q}\left(-{\mathbf{K}}_{S}{\mathbf{q}}_{b}+{\mathbf{K}}_{i}\mathbf{i}\right)+{\mathbf{F}}_{ext}, {\mathbf{M}}_{b}=\mathbf{M}{\left({\mathbf{B}}_{q}^{T}\right)}^{-1},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathbf{G}}_{b}=\mathbf{G}{\left({\mathbf{B}}_{q}^{T}\right)}^{-1}.
In the standard form:
\stackrel{˙}{\mathbf{z}}\left(t\right)=\mathbf{A}\mathbf{z}\left(t\right)+\mathbf{B}\mathbf{i}\left(t\right)+{\mathbf{C}}^{\mathrm{*}}{\mathbf{F}}_{ext},
\mathbf{A}=\left[\begin{array}{cc}0& \mathbf{E}\\ -{\mathbf{M}}_{b}^{-1}{\mathbf{B}}_{q}{\mathbf{K}}_{S}& -{\mathbf{M}}_{b}^{-1}{\mathbf{G}}_{b}\end{array}\right], \mathbf{B}=\left[\begin{array}{c}0\\ -{\mathbf{M}}_{b}^{-1}{\mathbf{B}}_{q}{\mathbf{K}}_{i}\end{array}\right], {\mathbf{C}}^{\mathrm{*}}=\left[\begin{array}{c}0\\ -{\mathbf{M}}_{b}^{-1}\end{array}\right].
Formally control problem can be represented as follows [6]. It is required to find a control law
\mathbf{i}\left(t\right)
that minimizes the objective function:
J\left(\mathbf{y}\left(t\right),\mathbf{i}\left(t\right)\right)=\underset{0}{\overset{\mathrm{\infty }}{\int }}\left[{\mathbf{z}}^{T}\left(t\right)\mathbf{Q}\mathbf{z}\left(t\right)+\rho {\mathbf{i}}^{T}\left(t\right)\mathbf{R}\mathbf{i}\left(t\right)\right]dt,
\mathbf{z}\left(t\right)={\left\{{\mathbf{q}}_{b}^{T}\left(t\right){\stackrel{˙}{\mathbf{q}}}_{b}^{T}\left(t\right)\right\}}^{T}
– is the solution of the system Eq. (9).
\mathbf{Q}\in {R}^{8×8}
is positive semi-definite and
\mathbf{R}\in {R}^{8×8}
– positive definite matrixes, respectively, obtained in accordance with Bryson rule [10]. The control law will be found as:
\mathbf{i}\left(t\right)=-\mathbf{K}\mathbf{z}\left(t\right),\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathbf{K}={\mathbf{R}}^{-1}{\mathbf{B}}^{T}\mathbf{X}.
\stackrel{˙}{\mathbf{z}}\left(t\right)=\mathbf{A}\mathbf{z}\left(t\right)-\mathbf{B}\mathbf{K}\mathbf{z}\left(t\right)+\mathbf{C}{\mathbf{F}}_{ext},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathbf{q}}_{b}={\mathbf{C}}^{\mathrm{*}}\mathbf{z}\left(t\right).
5. Modeling based on PID control
The study of rotor oscillations at fixed positional stiffness of bearings for different rotor speeds was conducted. Yellow line shows oscillations at a rotor speed of 5000 rev/min, the purple – at a speed of 10000 rev/min, green – 12000 rev/min.
For a fixed positional stiffness
{k}_{S}=
104N∙m-1 the results are illustrated with Fig. 1.
6. Modeling based on LQR method
The comparison of the rotor displacements was held at different speeds and constant stiffness. The value of the positional stiffness
{k}_{S}
is 105N∙m-1, of the current stiffness
{k}_{i}
– 10 N∙А-1. The black line shows changes of displacements at 5000 rev/min, blue line – at 10000 rev/min, red line at – 12000 rev/min (Fig. 2).
Table 2. Comparison of the displacement amplitude values for different rotation speeds and positional bearing stiffnesses
Rotation speed, rev/min
Max values of displacements
{k}_{S}=
104 N×m-1, m
{k}_{S}=
{k}_{S}=
Fig. 1. Displacements of the section centers A and B
It can be concluded that with the increase of rotor speed, values of the oscillation amplitudes and of the control currents increase too. However, the obtained values are permissible. The values of currents in the magnets arranged along the
y
-axis are greater than in magnets along the
x
-axis. It is due to the fact that in addition to the influence of the hydrodynamic moments of the blood flow the force of gravity acts along the
y
The results obtained for the three types of control have been tabulated (Table 2) for the velocity of 10000 rev/min and 12000 rev/min respectively.
Table 3. Comparison of rotor center displacements at different control types at 10000 rev/min. The value of the positional stiffness
{k}_{S}
is 105N∙m-1
LQR-method
{x}_{bAmax}
{y}_{bAmax}
{i}_{bxmax}
, mА
{i}_{bymax}
The simulation results show that LQR method meets the requirements of the rotor control the best way. It provides the position of the centers of rotor sections within the permissible error – 0.2 mm and allows to optimize the control on several criteria, in this case, criteria were: the position of the bearing section centers and control currents. The results of these studies can be used for real axial pump design.
This work was supported by the Russian Foundation for Basic Research (Grant No. 15-29-01085 ofi_m).
Birks E. J. Left ventricular assist devices. Heart, Vol. 96, 2010, p. 63-71. [Search CrossRef]
Thunberg Christopher A., Gaitan Brantley Dollar, Arabia Francisco A., Cole Daniel J., Grigore Alina M. Ventricular assist devices today and tomorrow. Journal of Cardiothoracic and Vascular Anesthesia, Vol. 24, Issue 4, 2010, p. 656-680. [Search CrossRef]
Griffith B. P., Kormos R. L., Borovetz H. S., Litwak K., Antaki J. F., Poirier V. L., et al. HeartMate II left ventricular assist system: from concept to first clinical use. The Annals of Thoracic Surgery, Vol. 71, 2001, p. 16-20. [Search CrossRef]
Schweizer G., Maslen E. H. Chapter 7: Dynamics of the Rigid Rotorsa; Chapter 10: Dynamics of Flexible Rotors. Magnetic Bearings. Theory, Design and Application to Rotating Machinery, Springer, Berlin Heidelberg, 2009, p. 167-189, p. 251-297. [Search CrossRef]
Franklin G. F., Powell J. D., Emami-Naeini A. Feedback Control of Dynamic Systems. 4th Edition. Prentice Hall, Upper Saddle River, NJ, 2002. [Search CrossRef]
Barbaraci G., Virzì Mariotti G. Sub-optimal control law for active magnetic bearings suspension. Journal of Control Engineering and Technology, Vol. 2, Issue 1, 2012, p. 1-10. [Search CrossRef] |
Hong Kong and Macao Economic Research Institute, College of Economics, Jinan University, Guangzhou, China.
Fan, D. (2019) The Measurement of Competitiveness of Hong Kong International Shipping Center and Its Promotion Strategies. Modern Economy, 10, 853-871. doi: 10.4236/me.2019.103057.
X={\left({x}_{ij}\right)}_{m\times n}
{X}^{\prime }={\left({{x}^{\prime }}_{ij}\right)}_{m\times n}
{{x}^{\prime }}_{ij}=\frac{{x}_{ij}-\mathrm{min}{x}_{ij}}{\mathrm{max}{x}_{ij}-\mathrm{min}{x}_{ij}}
P={\left({p}_{ij}\right)}_{m\times n}
{p}_{ij}=\frac{{x}_{ij}}{\underset{i=1}{\overset{m}{\sum }}{x}_{ij}},j=1,2,\cdots ,n
{e}_{j}=-k\underset{i=1}{\overset{m}{\sum }}{p}_{ij}\mathrm{ln}\left({p}_{ij}\right),k>0,k=1/\mathrm{ln}\left(m\right),{e}_{j}\ge 0
{g}_{j}=1-{e}_{j}
w=\frac{{g}_{j}}{\underset{j=1}{\overset{n}{\sum }}{g}_{j}},\left(1\le j\le n\right)
R={\left({r}_{ij}\right)}_{m\times n}=\left[\begin{array}{cccc}{w}_{1}{p}_{11}& {w}_{2}{p}_{12}& \cdots & {w}_{n}{p}_{1n}\\ {w}_{1}{p}_{21}& {w}_{2}{p}_{22}& \cdots & {w}_{n}{p}_{2n}\\ \vdots & \vdots & & \vdots \\ {w}_{1}{p}_{m1}& {w}_{2}{p}_{m2}& \cdots & {w}_{n}{p}_{mn}\end{array}\right]
{D}_{i}^{+}=\sqrt{\underset{j=1}{\overset{n}{\sum }}{\left({r}_{ij}-{r}_{j}^{+}\right)}^{2}}
{D}_{i}^{-}=\sqrt{\underset{j=1}{\overset{n}{\sum }}{\left({r}_{ij}-{r}_{j}^{-}\right)}^{2}}
{r}_{j}^{+}=\left\{\underset{j}{\mathrm{max}}{r}_{j}|j=1,2,\cdots ,n\right\}
{r}_{j}^{-}=\left\{\underset{j}{\mathrm{min}}{r}_{j}|j=1,2,\cdots ,n\right\}
{D}_{i}=\frac{{d}_{i}^{-}}{{d}_{i}^{+}+{d}_{i}^{-}},i=1,2,\cdots ,m
[1] Wang, L. and Ning, Y. (2010) The Trend of International High-End Shipping Service Industry and Ningbo’s Strategy. Economic Geography, Changsha, 2.
[2] Wang, Y. (2014) The Development Status of Hong Kong’s Shipping Industry and the Way out for the Construction of International Shipping Centers. World Shipping, Dalian, 10.
[3] Qu, J. (2007) The Significance of Hong Kong and Shenzhen Building an International Shipping Center. Open Herald, Shenzhen, 2.
[4] Zeng, J. (2013) Analysis of the Transformation and Upgrading of Hong Kong International Shipping Center and Its Influencing Factors. Master’s Thesis of Jinan University, Guangzhou.
[5] Xu, X. (2003) Analysis of Competitive Advantages of Shanghai International Shipping Center. PhD Thesis of Hohai University, Nanjing.
[6] Wu, X. (2004) The Soft Environment Construction of Shanghai International Shipping Center and the Role of Government. Master’s Thesis of Shanghai Maritime University, Shanghai.
[7] Yang, J. (2005) Theory and Practice of Modern Port Development. PhD Thesis of Shanghai Maritime University, Shanghai.
[8] Ma, S. (2011) What Kind of International Shipping Center Should Shanghai Be. Pearl River Water Transport, Guangzhou, Z1.
[9] Li, J. (2003) The Construction of Shanghai International Shipping Center Has Joys and Worries. China Port, Shanghai, 5.
[10] Ma, S. (2007) Soft Power Is the Key to Building Shanghai International Shipping Center. Water Transportation Management, Shanghai, 5.
[11] Zhang, L. (2008) The Experience and Enlightenment of London’s Development of International Shipping Center. Port Economy, Tianjin, 9.
[12] Wang, X. (2006) Discussion on the Factors of Competitiveness of International Shipping Centers. Port Economy, Tianjin, 2.
[13] Li, W. and Yang, G. (2014) Comprehensive Comparison and Time Evolution Analysis of Urban Competitiveness Based on TOPSIS Method of Entropy Right—Taking “Central Four Corners” Urban Agglomeration as an Example. Philosophy and Social Sciences, Guangzhou, 10.
[14] Gao, W. (2012) Development Trends and Empirical Research of International Shipping Centers—Taking the Construction of Tianjin Northern International Shipping Center as an Example. Exploration of Modern Economy, Nanjing, 7. |
Frequency synthesizer with accumulator based fractional N PLL architecture - Simulink - MathWorks Benelux
Fractional N PLL with Accumulator
Buffer size for loop filter
Buffer size for PFD, charge pump, VCO, prescaler
Output amplitude gain
Fractional clock divider value
Min clock divider value
Filter component values
Loop bandwidth (Hz)
Phase margin (degrees)
Export Loop Filter Component Values
PFD up and PFD down (pfd_up and pfd_down)
Charge pump output (cp_out)
Loop filter output (lf_out)
Prescaler output (ps_out)
Plot Loop Dynamics
Frequency synthesizer with accumulator based fractional N PLL architecture
Mixed-Signal Blockset / PLL / Architectures
The Fractional N PLL with Accumulator reference architecture uses a Fractional Clock Divider with Accumulator block as the frequency divider in a PLL system. The frequency divider divides the frequency of the VCO output signal by a fractional value to make it comparable to a PFD reference signal frequency.
clk in — Input clock signal
Input clock signal, specified as a scalar. The signal at the clk in port is used as the reference signal for the PFD block in a PLL system.
clk out — Output clock signal
Output clock signal, specified as a scalar. The signal at the clk out port is the output of the VCO block in a PLL system.
Select to enable increased buffer size during the simulation. This increases the buffer size of all the building blocks in the PLL model that belong to the Mixed-Signal Blockset™/PLL/Building Blocks Simulink® library. The building blocks are PFD, Charge Pump, Loop Filter, VCO, and Fractional Clock Divider with Accumulator. By default, this option is deselected.
Buffer size for loop filter — Buffer size for loop filter
Buffer size for the loop filter, specified as a positive integer scalar. This sets the number of extra buffer samples available during the simulation to the Convert Sample Time subsystem inside the loop filter.
Selecting different simulation solver or sampling strategies can change the number of input samples needed to produce an accurate output sample. Set the Buffer size for loop filter to a large enough value so that the input buffer contains all the input samples required.
Use get_param(gcb,'NBufferFilter') to view the current value of Buffer size for loop filter.
Use set_param(gcb,'NBufferFilter',value) to set Buffer size for loop filter to a specific value.
Buffer size for PFD, charge pump, VCO, prescaler — Buffer size for PFD, charge pump, VCO, and prescaler
Buffer size for the PFD, charge pump, VCO, and prescaler, specified as a positive integer scalar. This sets the buffer size of the PFD, Charge Pump, VCO, and Fractional Clock Divider with Accumulator blocks inside the PLL model.
Selecting different simulation solver or sampling strategies can change the number of input samples needed to produce an accurate output sample. Set the Buffer size for PFD, charge pump, VCO, prescaler to a large enough value so that the input buffer contains all the input samples required.
Use get_param(gcb,'NBuffer') to view the current value of Buffer size for PFD, charge pump, VCO, prescaler.
Use set_param(gcb,'NBuffer',value) to set Buffer size for PFD, charge pump, VCO, prescaler to a specific value.
Use get_param(gcb,'DeadbandCompensation') to view the current value of Deadband compensation (s).
Use set_param(gcb,'DeadbandCompensation',value) to set Deadband compensation (s) to a specific value.
Select to add circuit impairments such as rise/fall time and propagation delay to simulation. By default, this option is deselected.
\Delta \text{T}=\frac{{\left(\text{Rise/fall time}\right)}^{2}}{6\text{ }·\text{ 0}\text{.22}}
\Delta \text{T}=\frac{\text{Rise/fall time}}{6\text{ }·\text{ Maximum frequency of interest}}
To enable this parameter, select Enable Impairments in the PFD tab.
To enable this parameter, select Enable Impairments in the PFD tab and choose Advanced for Output step size calculation.
Use set_param(gcb,'RiseFallTime',value) to set Rise/fall time (s) to a specific value.
Full scale magnitude of design output current, specified as a positive real scalar in amperes. This parameter is also reported as Charge pump current in the Loop Filter tab and is used to automatically calculate the filter component values of the loop filter.
Select to add current impairments such as current imbalance and leakage current to simulation. By default, this option is deselected.
To enable this parameter, select Enable current impairments in the Charge pump tab.
Select to add timing impairments such as rise/fall time and propagation delay to simulation. By default, this option is deselected.
\Delta \text{T}=\frac{{\left(\text{Rise/fall time}\right)}^{2}}{6\text{ }·\text{ 0}\text{.22}}
\Delta \text{T}=\frac{\text{Rise/fall time}}{6\text{ }·\text{ Maximum frequency of interest}}
To enable this parameter, select Enable timing impairments in the Charge Pump tab.
To enable this parameter, select Enable timing impairments in the Charge Pump tab and choose Advanced for Output step size calculation.
Use get_param(gcb,'MaxFreqInterestCp') to view the current value of Maximum frequency of interest (Hz).
Use set_param(gcb,'MaxFreqInterestCp',value) to set Maximum frequency of interest (Hz) to a specific value.
Rise/fall time (s) — 20% – 80% rise/fall time for up input port
20% – 80% rise/fall time for the up input port of the charge pump, specified as a positive real scalar in seconds.
Use get_param(gcb,'RiseFallUp') to view the current value of Up Rise/fall time (s).
Use set_param(gcb,'RiseFallUp',value) to set Up Rise/fall time (s) to a specific value.
Propagation delay (s) — Total propagation delay from up input port to output port of charge pump
Total propagation delay from the up input port to output port of the charge pump, specified as a positive real scalar in seconds.
Use get_param(gcb,'PropDelayUp') to view the current value of Up Propagation delay (s).
Use set_param(gcb,'PropDelayUp',value) to set Up Propagation delay (s) to a specific value.
Rise/fall time — 20% – 80% rise/fall time for down input port
20% – 80% rise/fall time for down input port of charge pump.
Use get_param(gcb,'RiseFallDown') to view the current value of Down Rise/fall time (s).
Use set_param(gcb,'RiseFallDown',value) to set Down Rise/fall time (s) to a specific value.
Use get_param(gcb,'PropDelayUp') to view the current value of Down Propagation delay (s).
Use set_param(gcb,'PropDelayUp',value) to set Down Propagation delay (s) to a specific value.
Define how VCO output frequency is specified:
Select Voltage sensitivity to specify output frequency from Voltage sensitivity (Hz/V) and Free running frequency (Hz).
Select Output frequency vs. control voltage to interpolate output frequency from Control voltage (V) vector versus Output frequency (Hz) vector.
Use set_param(gcb,'SpecifyUsing','Voltage sensitivity') to set Specify using to Voltage sensitivity.
Use set_param(gcb,'SpecifyUsing', 'Output frequency vs. control voltage') to set Specify using to Output frequency vs. control voltage.
To enable this parameter, select Voltage sensitivity in Specify using in the VCO tab.
Use get_param(gcb,'Kvco') to view the current Voltage sensitivity (Hz/V) value.
Use set_param(gcb,'Kvco',value) to set Voltage sensitivity (Hz/V) to a specific value.
Frequency of the VCO without any control voltage input (0 V), or the quiescent frequency, specified as a positive real scalar in Hz.
Use get_param(gcb,'Fo') to view current Free running frequency (Hz) value.
Use set_param(gcb,'Fo',value) to set Free running frequency (Hz) to a specific value.
To enable this parameter, select Output frequency vs. control voltage in Specify using in the VCO tab.
Use get_param(gcb,'ControlVoltage') to view current Control voltage (V) value.
Use set_param(gcb,'ControlVoltage',value) to set Control voltage (V) to a specific value.
[2e9 2.5e9 3e9] (default) | real valued vector
Output frequency of the values of the VCO, corresponding to the Control voltage (V) vector, specified in Hz.
Use get_param(gcb,'OutputFrequency') to view current Output frequency (Hz) value.
Use set_param(gcb,'OutputFrequency',value) to set Output frequency (Hz) to a specific value.
Output amplitude gain — Ratio of VCO output voltage to input voltage
Ratio of VCO output voltage to input voltage, specified as a positive real scalar. The input voltage has a nontunable value of 1 V.
Use get_param(gcb,'Amplitude') to view current Output amplitude gain value.
Use set_param(gcb,'Amplitude',value) to set Output amplitude gain to a specific value.
Add Phase-noise — Add phase noise as function of frequency
Select to introduce phase noise as a function of frequency to the VCO. By default, this option is deselected.
Phase noise frequency offset (Hz) — Frequency offsets of phase noise from carrier frequency
[30e3 100e3 1e6 3e6 10e6] (default) | real valued vector
Frequency offsets of the phase noise from the carrier frequency, specified as a real valued vector in Hz.
To enable this parameter, select Add phase noise in the VCO tab.
Use get_param(gcb,'Foffset') to view the current Phase noise frequency offset (Hz) metric.
Use set_param(gcb,'Foffset',value) to set Phase noise frequency offset (Hz) to a specific metric.
Phase noise level (dBc/Hz) — Phase noise power at specified frequency offsets relative to the carrier
[-56 -106 -132 -143 -152] (default) | real valued vector
Real valued vector specifying the phase noise power in a 1 Hz bandwidth centered at the specified frequency offsets relative to the carrier. The value is specified in dBc/Hz.
Use get_param(gcb,'PhaseNoise') to view the current Phase noise level (dBc/Hz) metric.
Use set_param(gcb,'PhaseNoise',value) to set Phase noise level (dBc/Hz) to a specific metric.
Fractional clock divider value — Value by which the clock divider divides the input frequency
70.20 (default) | positive real scalar
Value by which the clock divider divides the input frequency, specified as a positive real scalar.
Use get_param(gcb,'N') to view the current value of Fractional clock divider value.
Use set_param(gcb,'N',value) to set Fractional clock divider value to a specific value.
Min clock divider value — Minimum value by which clock divider can divide input frequency
Minimum value by which the clock divider can divide input frequency, specified as a positive real scalar. This parameter is also reported in the Loop Filter tab and is used to automatically calculate the filter component values of the loop filter.
Use get_param(gcb,'Nmin') to view the current value of Min clock divider value.
Use set_param(gcb,'Nmin',value) to set Min clock divider value to a specific value.
Filter component values — Determines how filter components are computed
Automatic (default) | Manual
Select how filter components for the loop filter are computed:
Select Automatic to automatically compute filter components from system specifications. Resistance and capacitance edit boxes in the Loop Filter tab are not editable if this option is selected. Rather, the filter component values are calculated from Loop bandwidth (Hz), Phase margin (degrees), VCO voltage sensitivity, Charge pump current, and Min clock divider value. By default, this option is selected.
Select Manual to manually enter the resistance and capacitance values to design a customized loop filter.
Loop bandwidth (Hz) — Frequency at which magnitude of open loop transfer function becomes 1
Frequency at which the magnitude of the open loop transfer function becomes 1, specified as a positive real scalar in Hz. Lower values of Loop bandwidth (Hz) result in reduced phase noise and reference spurs at the expense of longer lock time and less phase margin.
This parameter is only available when Automatic is selected for the Filter Component values parameter in the Loop Filter tab.
Use get_param(gcb,'Fc') to view the current value of Loop bandwidth (Hz).
Use set_param(gcb,'Fc',value) to set Loop bandwidth (Hz) to a specific value.
Phase margin (degrees) — Phase of open loop transfer function at loop bandwidth subtracted from 180°
Phase of the open loop transfer function at the loop bandwidth subtracted from 180°, specified as a positive real scalar in degrees. For optimum lock time, select a phase margin between 40° and 55°.
Use get_param(gcb,'Phi') to view the current value of Phase margin (degrees).
Use set_param(gcb,'Phi',value) to set Phase margin (degrees) to a specific value.
Order of the loop filter. Applies a second-, third-, or fourth-order passive RC loop filter in the PLL system.
Capacitor value C1, specified as a positive real scalar in farad.
This parameter is only editable when Manual is selected for the Filter Component values parameter in the Loop Filter tab.
2.33e+06 (default) | positive real scalar
Select to add circuit impairments such as operating temperature to determine thermal noise to simulation. By default, this option is deselected.
Temperature of the resistor, specified as a real scalar in ℃. Operating temperature determines the level of thermal (Johnson) noise.
To enable this parameter, select Enable impairments in the Loop Filter tab.
Use get_param(gcb,'Temperature') to view the current value of Operating temperature.
Use set_param(gcb,'Temperature',value) to set Operating temperature to a specific value.
Export Loop Filter Component Values — Export loop filter component values
Click to export loop filter component values to a spreadsheet (XLS file) or as comma-separated values (CSV file).
PFD up and PFD down (pfd_up and pfd_down) — Select to probe PFD outputs
Select to probe the PFD output wires (pfd_up and pfd_down) to view the response of the PFD.
Charge pump output (cp_out) — Select to probe charge pump output
Select to probe the charge pump output wire (cp_out) to view the response of the Charge Pump.
Loop filter output (lf_out) — Select to probe loop filter output
Select to probe loop filter output wire (lf_out) to view the response of the Loop Filter. The loop filter output provides the control voltage to the VCO.
Prescaler output (ps_out) — Select to probe prescaler output
Select to probe the prescaler output wire (ps_out) to view the response of the Fractional Clock Divider with Accumulator.
Open Loop Analysis — Plot the presimulation open loop analysis
Select to plot the gain margin and phase margin of the PLL system before simulation. By default, this option is selected.
Closed Loop Analysis — Plot the presimulation closed loop analysis
Select to plot the pole-zero map, loop bandwidth, step response, and impulse response of the PLL system before simulation. You must have a license to Control System Toolbox™ to plot the step response and impulse response of the PLL system. By default, this option is deselected.
Plot Loop Dynamics — Plot loop dynamics of PLL system
Click to plot the presimulation loop dynamics of the PLL system.
PFD | Charge Pump | Loop Filter | Fractional Clock Divider with Accumulator | VCO |
Difference between revisions of "Problem 2.c observer state plot" - Murray Wiki
Difference between revisions of "Problem 2.c observer state plot"
[[Category:CDS110b FAQ - Homework 1]]
Q: In Problem 2.c it is asked to plot the observer state with a certain initial condition, but
{\displaystyle u}
is not known. What is one supposed to do?
A: It is possible to either plot the observer state with
{\displaystyle u=0}
and the given initial conditions, or plot the state estimation error (here you do not need to know the input), with the same initial conditions
Retrieved from "https://murray.cds.caltech.edu/index.php?title=Problem_2.c_observer_state_plot&oldid=5451" |
(Redirected from Angular unit)
{\displaystyle {\widehat {\rm {BAC}}}}
{\displaystyle {\begin{aligned}&\sin ^{2}A+\sin ^{2}B=1&&\cos ^{2}A+\cos ^{2}B=1\\[3pt]&\tan A=\cot B&&\sec A=\csc B\end{aligned}}}
{\displaystyle \theta ={\frac {k}{2\pi }}\cdot {\frac {s}{r}}.}
{\displaystyle m\angle \mathrm {AOC} =m\angle \mathrm {AOB} +m\angle \mathrm {BOC} }
{\displaystyle \mathbf {u} \cdot \mathbf {v} =\cos(\theta )\left\|\mathbf {u} \right\|\left\|\mathbf {v} \right\|.}
{\displaystyle \langle \cdot ,\cdot \rangle }
{\displaystyle \langle \mathbf {u} ,\mathbf {v} \rangle =\cos(\theta )\ \left\|\mathbf {u} \right\|\left\|\mathbf {v} \right\|.}
{\displaystyle \operatorname {Re} \left(\langle \mathbf {u} ,\mathbf {v} \rangle \right)=\cos(\theta )\left\|\mathbf {u} \right\|\left\|\mathbf {v} \right\|.}
{\displaystyle \left|\langle \mathbf {u} ,\mathbf {v} \rangle \right|=\left|\cos(\theta )\right|\left\|\mathbf {u} \right\|\left\|\mathbf {v} \right\|.}
{\displaystyle \operatorname {span} (\mathbf {u} )}
{\displaystyle \operatorname {span} (\mathbf {v} )}
{\displaystyle \mathbf {u} }
{\displaystyle \mathbf {v} }
{\displaystyle \operatorname {span} (\mathbf {u} )}
{\displaystyle \operatorname {span} (\mathbf {v} )}
{\displaystyle \left|\langle \mathbf {u} ,\mathbf {v} \rangle \right|=\left|\cos(\theta )\right|\left\|\mathbf {u} \right\|\left\|\mathbf {v} \right\|}
{\displaystyle {\mathcal {U}}}
{\displaystyle {\mathcal {W}}}
{\displaystyle \dim({\mathcal {U}}):=k\leq \dim({\mathcal {W}}):=l}
{\displaystyle k}
{\displaystyle \cos \theta ={\frac {g_{ij}U^{i}V^{j}}{\sqrt {\left|g_{ij}U^{i}U^{j}\right|\left|g_{ij}V^{i}V^{j}\right|}}}.}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Angle&oldid=1088898411#Units" |
Relative change and difference - Wikipedia
In any quantitative science, the terms relative change and relative difference are used to compare two quantities while taking into account the "sizes" of the things being compared. The comparison is expressed as a ratio and is a unitless number. By multiplying these ratios by 100 they can be expressed as percentages so the terms percentage change, percent(age) difference, or relative percentage difference are also commonly used. The distinction between "change" and "difference" depends on whether or not one of the quantities being compared is considered a standard or reference or starting value. When this occurs, the term relative change (with respect to the reference value) is used and otherwise the term relative difference is preferred. Relative difference is often used as a quantitative indicator of quality assurance and quality control for repeated measurements where the outcomes are expected to be the same. A special case of percent change (relative change expressed as a percentage) called percent error occurs in measuring situations where the reference value is the accepted or actual value (perhaps theoretically determined) and the value being compared to it is experimentally determined (by measurement).
3 Percent error
4.1 Example of percentages of percentages
6.2 Uniqueness and extensions
Given two numerical quantities, x and y, their difference, Δ = x − y, can be called their actual difference. When y is a reference value (a theoretical/actual/correct/accepted/optimal/starting, etc. value; the value that x is being compared to) then Δ is called their actual change. When there is no reference value, the sign of Δ has little meaning in the comparison of the two values since it doesn't matter which of the two values is written first, so one often works with |Δ| = |x − y|, the absolute difference instead of Δ, in these situations. Even when there is a reference value, if it doesn't matter whether the compared value is larger or smaller than the reference value, the absolute difference can be considered in place of the actual change.
The absolute difference between two values is not always a good way to compare the numbers. For instance, the absolute difference of 1 between 6 and 5 is more significant than the same absolute difference between 100,000,001 and 100,000,000. We can adjust the comparison to take into account the "size" of the quantities involved, by defining, for positive values of xreference:
{\displaystyle {\text{Relative change}}(x,x_{\text{reference}})={\frac {\text{Actual change}}{x_{\text{reference}}}}={\frac {\Delta }{x_{\text{reference}}}}={\frac {x-x_{\text{reference}}}{x_{\text{reference}}}}.}
The relative change is not defined if the reference value (xreference) is zero.
For values greater than the reference value, the relative change should be a positive number and for values that are smaller, the relative change should be negative. The formula given above behaves in this way only if xreference is positive, and reverses this behavior if xreference is negative. For example, if we are calibrating a thermometer which reads −6 °C when it should read −10 °C, this formula for relative change (which would be called relative error in this application) gives ((−6) − (−10)) / (−10) = 4 / −10 = −0.4, yet the reading is too high. To fix this problem we alter the definition of relative change so that it works correctly for all nonzero values of xreference:
{\displaystyle {\text{Relative change}}(x,x_{\text{reference}})={\frac {\text{Actual change}}{|x_{\text{reference}}|}}={\frac {\Delta }{|x_{\text{reference}}|}}={\frac {x-x_{\text{reference}}}{|x_{\text{reference}}|}}.}
If the relationship of the value with respect to the reference value (that is, larger or smaller) does not matter in a particular application, the absolute difference may be used in place of the actual change in the above formula to produce a value for the relative change which is always non-negative.
Defining relative difference is not as easy as defining relative change since there is no "correct" value to scale the absolute difference with. As a result, there are many options for how to define relative difference and which one is used depends on what the comparison is being used for. In general we can say that the absolute difference |Δ| is being scaled by some function of the values x and y, say f(x, y).[1]
{\displaystyle {\text{Relative difference}}(x,y)={\frac {\text{Absolute difference}}{|f(x,y)|}}={\frac {|\Delta |}{|f(x,y)|}}=\left|{\frac {x-y}{f(x,y)}}\right|.}
As with relative change, the relative difference is undefined if f(x, y) is zero.
Several common choices for the function f(x, y) would be:
max(|x|, |y|),
max(x, y),
min(|x|, |y|),
min (x, y),
(x + y)/2, and
(|x| + |y|)/2.
Measures of relative difference are unitless numbers expressed as a fraction. Corresponding values of percent difference would be obtained by multiplying these values by 100 (and appending the % sign to indicate that the value is a percentage).
One way to define the relative difference of two numbers is to take their absolute difference divided by the maximum absolute value of the two numbers.
{\displaystyle d_{r}={\frac {|x-y|}{\max(|x|,|y|)}}}
if at least one of the values does not equal zero. This approach is especially useful when comparing floating point values in programming languages for equality with a certain tolerance.[2] Another application is in the computation of approximation errors when the relative error of a measurement is required.
Another way to define the relative difference of two numbers is to take their absolute difference divided by some functional value of the two numbers, for example, the absolute value of their arithmetic mean:
{\displaystyle d_{r}={\frac {|x-y|}{\left({\frac {|x+y|}{2}}\right)}}\,.}
This approach is often used when the two numbers reflect a change in some single underlying entity.[citation needed] A problem with the above approach arises when the functional value is zero. In this example, if x and y have the same magnitude but opposite sign, then
{\displaystyle {\frac {|x+y|}{2}}=0,}
which causes division by 0. So it may be better to replace the denominator with the average of the absolute values of x and y:[citation needed]
{\displaystyle d_{r}={\frac {|x-y|}{\left({\frac {|x|+|y|}{2}}\right)}}\,.}
Percent errorEdit
The percent error is a special case of the percentage form of relative change calculated from the absolute change between the experimental (measured) and theoretical (accepted) values, and dividing by the theoretical (accepted) value.
{\displaystyle \%{\text{ Error}}={\frac {|{\text{Experimental}}-{\text{Theoretical}}|}{|{\text{Theoretical}}|}}\times 100.}
The terms "Experimental" and "Theoretical" used in the equation above are commonly replaced with similar terms. Other terms used for experimental could be "measured," "calculated," or "actual" and another term used for theoretical could be "accepted." Experimental value is what has been derived by use of calculation and/or measurement and is having its accuracy tested against the theoretical value, a value that is accepted by the scientific community or a value that could be seen as a goal for a successful result.
Although it is common practice to use the absolute value version of relative change when discussing percent error, in some situations, it can be beneficial to remove the absolute values to provide more information about the result. Thus, if an experimental value is less than the theoretical value, the percent error will be negative. This negative result provides additional information about the experimental result. For example, experimentally calculating the speed of light and coming up with a negative percent error says that the experimental value is a velocity that is less than the speed of light. This is a big difference from getting a positive percent error, which means the experimental value is a velocity that is greater than the speed of light (violating the theory of relativity) and is a newsworthy result.
The percent error equation, when rewritten by removing the absolute values, becomes:
{\displaystyle \%{\text{ Error}}={\frac {{\text{Experimental}}-{\text{Theoretical}}}{|{\text{Theoretical}}|}}\times 100.}
It is important to note that the two values in the numerator do not commute. Therefore, it is vital to preserve the order as above: subtract the theoretical value from the experimental value and not vice versa.
Percentage changeEdit
A percentage change is a way to express a change in a variable. It represents the relative change between the old value and the new one.[3]
For example, if a house is worth $100,000 today and the year after its value goes up to $110,000, the percentage change of its value can be expressed as
{\displaystyle {\frac {110000-100000}{100000}}=0.1=10\%.}
It can then be said that the worth of the house went up by 10%.
More generally, if V1 represents the old value and V2 the new one,
{\displaystyle {\text{Percentage change}}={\frac {\Delta V}{V_{1}}}={\frac {V_{2}-V_{1}}{V_{1}}}\times 100\%.}
Some calculators directly support this via a %CH or Δ% function.
When the variable in question is a percentage itself, it is better to talk about its change by using percentage points, to avoid confusion between relative difference and absolute difference.
Example of percentages of percentagesEdit
If a bank were to raise the interest rate on a savings account from 3% to 4%, the statement that "the interest rate was increased by 1%" would be ambiguous. The absolute change in this situation is 1 percentage point (4% − 3%), but the relative change in the interest rate is:
{\displaystyle {\frac {4\%-3\%}{3\%}}=0.333\ldots =33{\frac {1}{3}}\%.}
In general, the term "percentage point(s)" indicates an absolute change or difference of percentages, while the percent sign or the word "percentage" refers to the relative change or difference.[4]
Car M costs $50,000 and car L costs $40,000. We wish to compare these costs.[5] With respect to car L, the absolute difference is $10,000 = $50,000 − $40,000. That is, car M costs $10,000 more than car L. The relative difference is,
{\displaystyle {\frac {\$10,000}{\$40,000}}=0.25=25\%,}
and we say that car M costs 25% more than car L. It is also common to express the comparison as a ratio, which in this example is,
{\displaystyle {\frac {\$50,000}{\$40,000}}=1.25=125\%,}
and we say that car M costs 125% of the cost of car L.
In this example the cost of car L was considered the reference value, but we could have made the choice the other way and considered the cost of car M as the reference value. The absolute difference is now −$10,000 = $40,000 − $50,000 since car L costs $10,000 less than car M. The relative difference,
{\displaystyle {\frac {-\$10,000}{\$50,000}}=-0.20=-20\%}
is also negative since car L costs 20% less than car M. The ratio form of the comparison,
{\displaystyle {\frac {\$40,000}{\$50,000}}=0.8=80\%}
says that car L costs 80% of what car M costs.
It is the use of the words "of" and "less/more than" that distinguish between ratios and relative differences.[6]
Logarithmic scaleEdit
Change in a quantity can also be expressed as the natural logarithm (ln) of the ratio of the two numbers, called log change.[1] Indeed, when
{\displaystyle \left|{\frac {V_{1}-V_{0}}{V_{0}}}\right|\ll 1}
, the following approximation holds:
{\displaystyle \ln {\frac {V_{1}}{V_{0}}}=\int _{V_{0}}^{V_{1}}{\frac {{\mathrm {d} }V}{V}}\approx \int _{V_{0}}^{V_{1}}{\frac {{\mathrm {d} }V}{V_{0}}}={\frac {V_{1}-V_{0}}{V_{0}}}={\text{relative change}}}
In the same way that relative change is scaled by 100 to get percentages,
{\displaystyle \ln {\frac {V_{1}}{V_{0}}}}
can be scaled by 100 to get what is commonly called log points.[7] Log points are equivalent to the unit centinepers (cNp) when measured for root-power quantities.[8][9] This quantity has also been referred to as a log percentage and denoted L%.[1] Since the derivative of the natural log at 1 is 1, log points are approximately equal to percentage difference for small differences – for example an increase of 1% equals an increase of 0.995 cNp, and a 5% increase gives a 4.88 cNp increase. This approximation property does not hold for other choices of logarithm base, which introduce a scaling factor due to the derivative not being 1. Log points can thus be used as a replacement for percentage differences.[10][8]
Using log change has the advantages of additivity compared to relative change.[1][8]
AdditivityEdit
When using log change, the total change after a series of changes equals the sum of the changes. With percent, summing the changes is only an approximation, with larger error for larger changes.[8] For example:
Log change 0 (cNp)
Total log change (cNp)
Relative change 0 (%)
Total relative change (%)
10 −5 5 10 −5 4.5
10 −10 0 10 −10 −1
50 −50 0 50 −50 −25
Note that in the above table, since relative change 0 (respectively relative change 1) has the same numerical value as log change 0 (respectively log change 1), it does not correspond to the same variation. The conversion between relative and log changes may be computed as
{\displaystyle {\text{log change}}=\ln(1+{\text{relative change}})}
By additivity,
{\displaystyle \ln {\frac {V_{1}}{V_{0}}}+\ln {\frac {V_{0}}{V_{1}}}=0}
, and therefore additivity implies a sort of symmetry property, namely
{\displaystyle \ln {\frac {V_{1}}{V_{0}}}=-\ln {\frac {V_{0}}{V_{1}}}}
and thus the magnitude of a change expressed in log change is the same whether V0 or V1 is chosen as the reference.[8] In contrast, for relative change,
{\displaystyle {\frac {V_{1}-V_{0}}{V_{0}}}\neq -{\frac {V_{0}-V_{1}}{V_{1}}}}
, with the difference
{\displaystyle {\frac {(V_{1}-V_{0})^{2}}{V_{0}V_{1}}}}
becoming larger as V1 or V0 approaches 0 while the other remains fixed. For example:
Log change (cNp)
10 9 −10.5 −10.0
9 10 +10.5 +11.1
10 1 −230 −90
1 10 +230 +900
10 0+ −∞ −100
0+ 10 +∞ +∞
Here 0+ means taking the limit from above towards 0.
Uniqueness and extensionsEdit
The log change is the unique two-variable function that is additive, and whose linearization matches relative change. There is a family of additive difference functions
{\displaystyle F_{\lambda }(x,y)}
{\displaystyle \lambda \in \mathbb {R} }
, such that absolute change is
{\displaystyle F_{0}}
and log change is
{\displaystyle F_{1}}
^ a b c d Törnqvist, Vartia & Vartia 1985.
^ What's a good way to check for close enough floating-point equality
^ Kazmi, Kumail (March 26, 2021). "Percentage Increase Calculator". Smadent - Best Educational Website of Pakistan. Smadent Publishing. Retrieved March 26, 2021.
^ Bennett & Briggs 2005, p. 141
^ Bennett & Briggs 2005, pp. 137–139
^ Bennett & Briggs 2005, p.140
^ Békés, Gábor; Kézdi, Gábor (6 May 2021). Data Analysis for Business, Economics, and Policy. Cambridge University Press. p. 203. ISBN 978-1-108-48301-8.
^ a b c d e Karjus, Andres; Blythe, Richard A.; Kirby, Simon; Smith, Kenny (10 February 2020). "Quantifying the dynamics of topical fluctuations in language". Language Dynamics and Change. 10 (1). Section A.3.1. doi:10.1163/22105832-01001200.
^ Roe, John; deForest, Russ; Jamshidi, Sara (26 April 2018). Mathematics for Sustainability. Springer. p. 190. doi:10.1007/978-3-319-76660-7_4. ISBN 978-3-319-76660-7.
^ Doyle, Patrick (2016-08-24). "The Case for a Logarithmic Performance Metric". Vena Solutions.
^ Brauen, Silvan; Erpf, Philipp; Wasem, Micha (2020). "On Absolute and Relative Change" (PDF). SSRN Electronic Journal. doi:10.2139/ssrn.3739890.
Bennett, Jeffrey; Briggs, William (2005), Using and Understanding Mathematics: A Quantitative Reasoning Approach (3rd ed.), Boston: Pearson, ISBN 0-321-22773-5
"Understanding Measurement and Graphing" (PDF). North Carolina State University. 2008-08-20. Archived from the original (PDF) on 2010-06-15. Retrieved 2010-05-05.
"Percent Difference – Percent Error" (PDF). Illinois State University, Dept of Physics. 2004-07-20. Retrieved 2010-05-05.
Törnqvist, Leo; Vartia, Pentti; Vartia, Yrjö (1985), "How Should Relative Changes Be Measured?", The American Statistician, 39 (1): 43–46, doi:10.2307/2683905
Retrieved from "https://en.wikipedia.org/w/index.php?title=Relative_change_and_difference&oldid=1085142584" |
Let us have 3 vectors in \mathbb R^3: (a_1,b_1,c_1),(a_2,b_2,c_2) and (a_3,b_3,
Let us have 3 vectors in
{\mathbb{R}}^{3}
\left({a}_{1},{b}_{1},{c}_{1}\right),\left({a}_{2},{b}_{2},{c}_{2}\right)
\left({a}_{3},{b}_{3},{c}_{3}\right)
These vectors are linearly dependent if and only if the following scalar equation holds:
{a}_{1}\left({b}_{2}{c}_{3}-{b}_{3}{c}_{2}\right)-{b}_{1}\left({a}_{2}{c}_{3}-{a}_{3}{c}_{2}\right)+{c}_{1}\left({a}_{2}{b}_{3}-{a}_{3}{b}_{2}\right)=0
1. This equation is a necessary and sufficient condition of lineary dependence.
2. This criterion is a scalar equation (not a pair of equations; not a vector form).
3. This equation is a linear equation as to
\left({a}_{i},{b}_{i},{c}_{i}\right)
i\in \left\{1,2,3\right\}
Now let us have 2 strings in
{\mathbb{R}}^{3}
\left({a}_{1},{b}_{1},{c}_{1}\right)
\left({a}_{2},{b}_{2},{c}_{2}\right)
Is there a similar way to determine whether this set of vectors in
{\mathbb{R}}^{3}
is linearly dependent?
I.e. the way such as the following scalar equation:
f\left({a}_{1},{b}_{1},{c}_{1},{a}_{2},{b}_{2},{c}_{2}\right)=0
1.This equation is a necessary and sufficient condition of linear dependence.
2.This criterion is a scalar equation (not a pair of equations; not a vector form).
\left({a}_{i},{b}_{i},{c}_{i}\right)
i\in \left\{1,2\right\}
avalescogzw
No, this is not possible. Let's assume we have such an f which satisfies all three of your properties. Fix some non-zero vector
\left({a}_{2},{b}_{2},{c}_{2}\right)
and consider the function
\left(a,b,c\right)↦f\left(a,b,c,{a}_{2},{b}_{2},{c}_{2}\right)
. Let's call this function
gcolon{\mathbb{R}}^{3}\to \mathbb{R}
. You want g to be a linear function and that
g\left(a,b,c\right)=0
if (a,b,c) linearly depends on
\left({a}_{2},{b}_{2},{c}_{2}\right)
. However, the set of all vectors which linearly depend on
\left({a}_{2},{b}_{2},{c}_{2}\right)
is a line in
{\mathbb{R}}^{3}
(all scalar multiples of
\left({a}_{2},{b}_{2},{c}_{2}\right)
) while the zero set of a linear function g is either a plane (if the function is not trivial) or
{\mathbb{R}}^{3}
(if the function is identically zero).
However, if you are willing to relax your conditions, you can get what you are looking for. The vectors
\left({a}_{1},{b}_{1},{c}_{1}\right)
\left({a}_{2},{b}_{2},{c}_{2}\right)
are linearly dependent if and only if the matrix
A=\left(\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right)
\le 1
. This can be checked using determinants. The matrix A has rank
\le 1
iff all
2×2
minors have determinant zero. This gives you three linear equations (not a scalar equation) which give you a sufficient and necessary condition for linear dependance:
{a}_{1}{b}_{2}-{b}_{1}{a}_{2}=0,\text{ }\text{ }{a}_{1}{c}_{2}-{c}_{1}{a}_{2}=0,\text{ }\text{ }{b}_{1}{c}_{2}-{c}_{1}{b}_{2}=0
If you want, you can combine them into a single equation
f\left({a}_{1},{b}_{1},{c}_{1},{a}_{2},{b}_{2},{c}_{2}\right)={\left({a}_{1}{b}_{2}-{b}_{1}{a}_{2}\right)}^{2}+{\left({a}_{1}{c}_{2}-{c}_{1}{a}_{2}\right)}^{2}+{\left({b}_{1}{c}_{2}-{c}_{1}{b}_{2}\right)}^{2}=0
However, this f is not linear in
\left({a}_{1},{b}_{1},{c}_{1}\right)
(or in
\left({a}_{2},{b}_{2},{c}_{2}\right)
It might be easier to see the argument in the case of one vector in
{\mathbb{R}}^{2}
(instead of two vectors in
{\mathbb{R}}^{3}
). A single vector (a,b) in
{\mathbb{R}}^{2}
is "linearly dependent" iff a=b=0 which gives you two scalar linear equations. You can combine them into a single equation
{a}_{2}+{b}^{2}=0
but this is not linear in (a,b).
Find the linear approximation of the function
f\left(x\right)=\sqrt{4-x}
a=0
Use L(x) to approximate the numbers
\sqrt{3.9}
\sqrt{3.99}
Round to four decimal places
Express the equations of the lines
r · (i −j) + 7 = 0,r · (i + 3j) − 5 = 0 in parametric forms and hence find the position
vector of their point of intersection.
Consider a system of three linear equations in three variables. Give examples of two reduced forms that are not row-equivalent if the system is: a) onsistent and dependent. b) Inconsistent
f\left(x\right)=\sqrt{4-x}
at a=0
Use L(x) to approximate the numbers sqrt (3.9) and sqrt (3.99) Round to four decimal places
I am stucked on the following challenge: "If the line determined by two distinct points
\left({x}_{1},{y}_{1}\right)
\left({x}_{2},{y}_{2}\right)
is not vertical, and therefore has slope
\frac{\left({y}_{2}-{y}_{1}\right)}{\left({x}_{2}-{x}_{1}\right)}
, show that the point-slope form of its equation is the same regardless of which point is used as the given point." Okay, we can separate
\left({x}_{0},{y}_{0}\right)
from the form to get:
y\left({x}_{2}-{x}_{1}\right)-x\left({y}_{2}-{y}_{1}\right)={y}_{0}\left({x}_{2}-{x}_{1}\right)-{x}_{0}\left({y}_{2}-{y}_{1}\right)
But how exclude this point
\left({x}_{0},{y}_{0}\right)
and leave only
x,y,{x}_{1},{y}_{1},{x}_{2},{y}_{2}
in the equation? UPDATE: There is a solution for this challenge:
\left({y}_{1}-{y}_{2}\right)x+\left({x}_{2}-{x}_{1}\right)y={x}_{2}{y}_{1}-{x}_{1}{y}_{2}
From the answer I found that
{y}_{2}\left(x-{x}_{1}\right)-{y}_{1}\left(x-{x}_{2}\right)=y\left({x}_{2}-{x}_{1}\right)
but why this is true?
Write the equation in standard form of linear equation and solve it for y(e) = 0.
x{y}_{1}+2y=\frac{1}{{x}^{2}\mathrm{ln}x}
x-y-z=0
x+2y-z=6
2x-z=5 |
The ScalarPotential(v) command computes the scalar potential of the vector field v. This is a function f such that
\mathrm{Gradient}\left(f\right)=v
\mathrm{with}\left(\mathrm{VectorCalculus}\right):
\mathrm{SetCoordinates}\left('\mathrm{cartesian}'[x,y,z]\right)
{\textcolor[rgb]{0,0,1}{\mathrm{cartesian}}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}}
v≔\mathrm{VectorField}\left(〈x,y,z〉\right)
\textcolor[rgb]{0,0,1}{v}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{x}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{y}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{z}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{z}}
\mathrm{ScalarPotential}\left(v\right)
\frac{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{2}}
v≔\mathrm{VectorField}\left(〈y,-x,0〉\right)
\textcolor[rgb]{0,0,1}{v}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{y}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{0}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{z}}
\mathrm{ScalarPotential}\left(v\right)
\mathrm{ScalarPotential}\left(\left(x,y,z\right)↦\frac{〈x,y,z〉}{{x}^{2}+{y}^{2}+{z}^{2}}\right)
\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{↦}\frac{\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\right)}{\textcolor[rgb]{0,0,1}{2}}
\mathrm{SetCoordinates}\left('\mathrm{spherical}'[r,\mathrm{\phi },\mathrm{\theta }]\right)
{\textcolor[rgb]{0,0,1}{\mathrm{spherical}}}_{\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\phi }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\theta }}}
v≔\mathrm{VectorField}\left(〈r,0,0〉\right)
\textcolor[rgb]{0,0,1}{v}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{r}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{r}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{0}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{\mathrm{φ}}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{0}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{\mathrm{θ}}}
\mathrm{ScalarPotential}\left(v\right)
\frac{{\textcolor[rgb]{0,0,1}{r}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{2}}
\mathrm{Gradient}\left(\right)
\left(\textcolor[rgb]{0,0,1}{r}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{r}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{0}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{\mathrm{φ}}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{0}\right){\stackrel{\textcolor[rgb]{0,0,1}{_}}{\textcolor[rgb]{0,0,1}{e}}}_{\textcolor[rgb]{0,0,1}{\mathrm{θ}}} |
Waves: Level 3-5 Challenges Practice Problems Online | Brilliant
Recordings that get played speakers can be very detailed, consisting of tens or hundred of distinct sound sources, each of which contribute a continuum of frequency components. Yet there is only one speaker. How is it possible for a single speaker to faithfully reproduce complex sounds?
Speakers play the most intense sound at any given time, our brains piece it back together. The speaker plays all the frequencies, this is why speakers usually have rings. Each frequency is played for a very small amount of time, in sequence with the others. The speaker has a time-dependent position, which is set by the sum of the component soundwaves.
An ambulance car is going with a speed of
60~\mbox{km/h}
, while a car is trying to go around it in a circle with a speed
50~\mbox{km/h}
. If the sound that ambulance car emits has a frequency of
1~\mbox{kHz}
, which frequency does the driver of the car hear in Hz when the ambulance car is in the center of the circle it makes, and the car makes an angle
\theta = 30^\circ
with the direction of the car?
Speed of sound in the air is
c = 1235~\mbox{km/h}
I'm in a spaceship very far away from Earth but traveling straight towards earth with a speed v. It's boring out here, so I decide to try and tune in to some of my favorite earthly radio stations. I remember that my favorite station has a frequency of 100.3 MHz and so tune my radio to exactly this frequency. Amazingly, I hear the radio station just like I do on Earth! How fast is my spaceship going in m/s? (Hint: it's not that fast... I think I should check whether my engines are on).
Photons of electromagnetic radiation have an intrinsic kinetic energy related to their frequency by
E=hf
h
The gravitational interaction between photons and Earth can be treated via usual Newtonian gravity and
E=mc^2
to convert between energy and mass.
The total energy of the photons is conserved.
The Earth can be treated as a sphere of radius 6370 km and mass
6 \times 10^{24}~kg
. You can ignore rotation of the earth.
3 \times 10^8~m/s
Assume the earth is at rest.
Airplanes are noisy things. For example, if you stand really close to a runway, a military jet can produce a sound intensity of about
100\text{ W/m}^2
at a distance of 30 m from the plane (this is beyond painful noise). You live near a military airport and want to work on physics with your friend. Since you don't want to shout, you decide to find a spot far enough away from the airport that the airplane noise is below the threshold of conversation
(10^{-6}\text{ W/m}^2)
. How far do you need to go from the airport in kilometers to accomplish this? |
Orbital period Wikipedia
Not to be confused with Rotation period.
For the music album, see Orbital Period (album).
Time an astronomical object takes to complete one orbit around another object
The orbital period (also revolution period) is the amount of time a given astronomical object takes to complete one orbit around another object. In astronomy, it usually applies to planets or asteroids orbiting the Sun, moons orbiting planets, exoplanets orbiting other stars, or binary stars.
For celestial objects in general, the sidereal period (sidereal year) is referred to by the orbital period, determined by a 360° revolution of one body around its primary, e.g. Earth around the Sun, relative to the fixed stars projected in the sky. Orbital periods can be defined in several ways. The tropical period is more particularly about the position of the parent star. It is the basis for the solar year, and respectively the calendar year.
The synodic period incorporates not only the orbital relation to the parent star, but also to other celestial objects, making it not a mere different approach to the orbit of an object around its parent, but a period of orbital relations with other objects, normally Earth and their orbits around the Sun. It applies to the elapsed time where planets return to the same kind of phenomena or location, such as when any planet returns between its consecutive observed conjunctions with or oppositions to the Sun. For example, Jupiter has a synodic period of 398.8 days from Earth; thus, Jupiter's opposition occurs once roughly every 13 months.
Periods in astronomy are conveniently expressed in various units of time, often in hours, days, or years. They can be also defined under different specific astronomical definitions that are mostly caused by the small complex external gravitational influences of other celestial objects. Such variations also include the true placement of the centre of gravity between two astronomical bodies (barycenter), perturbations by other planets or bodies, orbital resonance, general relativity, etc. Most are investigated by detailed complex astronomical theories using celestial mechanics using precise positional observations of celestial objects via astrometry.
1 Related periods
2 Small body orbiting a central body
3 Effect of central body's density
4 Two bodies orbiting each other
5 Synodic period
6 Examples of sidereal and synodic periods
6.1 Synodic periods relative to other planets
Related periods[]
See also: Lunar month § Types
There are many periods related to the orbits of objects, each of which are often used in the various fields of astronomy and astrophysics, particularly they must not be confused with other revolving periods like rotational periods. Examples of some of the common orbital ones include the following:
The sidereal period is the amount of time that it takes an object to make a full orbit, relative to the stars, the sidereal year. This is the orbital period in an inertial (non-rotating) frame of reference.
The synodic period is the amount of time that it takes for an object to reappear at the same point in relation to two or more other objects. In common usage, these two objects are typically the Earth and the Sun. The time between two successive oppositions or two successive conjunctions is also equal to the synodic period. For celestial bodies in the solar system, the synodic period (with respect to Earth and the Sun) differs from the tropical period owing to the Earth's motion around the Sun. For example, the synodic period of the Moon's orbit as seen from the Earth, relative to the Sun, is 29.5 mean solar days, since the Moon's phase and position relative to the Sun and Earth repeats after this period. This is longer than the sidereal period of its orbit around the Earth, which is 27.3 mean solar days, owing to the motion of the Earth around the Sun.
The draconitic period (also draconic period or nodal period), is the time that elapses between two passages of the object through its ascending node, the point of its orbit where it crosses the ecliptic from the southern to the northern hemisphere. This period differs from the sidereal period because both the orbital plane of the object and the plane of the ecliptic precess with respect to the fixed stars, so their intersection, the line of nodes, also precesses with respect to the fixed stars. Although the plane of the ecliptic is often held fixed at the position it occupied at a specific epoch, the orbital plane of the object still precesses, causing the draconitic period to differ from the sidereal period.[1]
The anomalistic period is the time that elapses between two passages of an object at its periapsis (in the case of the planets in the Solar System, called the perihelion), the point of its closest approach to the attracting body. It differs from the sidereal period because the object's semi-major axis typically advances slowly.
Also, the tropical period of Earth (a tropical year) is the interval between two alignments of its rotational axis with the Sun, also viewed as two passages of the object at a right ascension of 0 hr. One Earth year is slightly shorter than the period for the Sun to complete one circuit along the ecliptic (a sidereal year) because the inclined axis and equatorial plane slowly precess (rotate with respect to reference stars), realigning with the Sun before the orbit completes. This cycle of axial precession for Earth, known as precession of the equinoxes, recurs roughly every 25,770 years.[citation needed]
Small body orbiting a central body[]
The semi-major axis (a) and semi-minor axis (b) of an ellipse
According to Kepler's Third Law, the orbital period T of two point masses orbiting each other in a circular or elliptic orbit is:[2]
{\displaystyle T=2\pi {\sqrt {\frac {a^{3}}{\mu }}}}
a is the orbit's semi-major axis
μ = GM is the standard gravitational parameter
M is the mass of the more massive body.
For all ellipses with a given semi-major axis the orbital period is the same, regardless of eccentricity.
Inversely, for calculating the distance where a body has to orbit in order to have a given orbital period:
{\displaystyle a={\sqrt[{3}]{\frac {GMT^{2}}{4\pi ^{2}}}}}
a is the orbit's semi-major axis,
M is the mass of the more massive body,
T is the orbital period.
For instance, for completing an orbit every 24 hours around a mass of 100 kg, a small body has to orbit at a distance of 1.08 meters from the central body's center of mass.
In the special case of perfectly circular orbits, the orbital velocity is constant and equal (in m/s) to
{\displaystyle v_{\text{o}}={\sqrt {\frac {GM}{r}}}}
r is the circular orbit's radius in meters,
M is the mass of the central body.
This corresponds to 1⁄√2 times (≈ 0.707 times) the escape velocity.
Effect of central body's density[]
For a perfect sphere of uniform density, it is possible to rewrite the first equation without measuring the mass as:
{\displaystyle T={\sqrt {{\frac {a^{3}}{r^{3}}}{\frac {3\pi }{G\rho }}}}}
r is the sphere's radius
a is the orbit's semi-major axis in metres,
ρ is the density of the sphere in kilograms per cubic metre.
For instance, a small body in circular orbit 10.5 cm above the surface of a sphere of tungsten half a metre in radius would travel at slightly more than 1 mm/s, completing an orbit every hour. If the same sphere were made of lead the small body would need to orbit just 6.7 mm above the surface for sustaining the same orbital period.
When a very small body is in a circular orbit barely above the surface of a sphere of any radius and mean density ρ (in kg/m3), the above equation simplifies to (since M = Vρ = 4/3πa3ρ)
{\displaystyle T={\sqrt {\frac {3\pi }{G\rho }}}}
Thus the orbital period in low orbit depends only on the density of the central body, regardless of its size.
So, for the Earth as the central body (or any other spherically symmetric body with the same mean density, about 5,515 kg/m3,[3] e.g. Mercury with 5,427 kg/m3 and Venus with 5,243 kg/m3) we get:
and for a body made of water (ρ ≈ 1,000 kg/m3),[4] or bodies with a similar density, e.g. Saturn's moons Iapetus with 1,088 kg/m3 and Tethys with 984 kg/m3 we get:
Thus, as an alternative for using a very small number like G, the strength of universal gravity can be described using some reference material, such as water: the orbital period for an orbit just above the surface of a spherical body of water is 3 hours and 18 minutes. Conversely, this can be used as a kind of "universal" unit of time if we have a unit of mass, a unit of length, and a unit of density.
Two bodies orbiting each other[]
In celestial mechanics, when both orbiting bodies' masses have to be taken into account, the orbital period T can be calculated as follows:[5]
{\displaystyle T=2\pi {\sqrt {\frac {a^{3}}{G\left(M_{1}+M_{2}\right)}}}}
a is the sum of the semi-major axes of the ellipses in which the centers of the bodies move, or equivalently, the semi-major axis of the ellipse in which one body moves, in the frame of reference with the other body at the origin (which is equal to their constant separation for circular orbits),
M1 + M2 is the sum of the masses of the two bodies,
Note that the orbital period is independent of size: for a scale model it would be the same, when densities are the same, as M scales linearly with a3 (see also Orbit § Scaling in gravity).
In a parabolic or hyperbolic trajectory, the motion is not periodic, and the duration of the full trajectory is infinite.
Synodic period[]
One of the observable characteristics of two bodies which orbit a third body in different orbits, and thus have different orbital periods, is their synodic period, which is the time between conjunctions.
An example of this related period description is the repeated cycles for celestial bodies as observed from the Earth's surface, the synodic period, applying to the elapsed time where planets return to the same kind of phenomenon or location. For example, when any planet returns between its consecutive observed conjunctions with or oppositions to the Sun. For example, Jupiter has a synodic period of 398.8 days from Earth; thus, Jupiter's opposition occurs once roughly every 13 months.
If the orbital periods of the two bodies around the third are called T1 and T2, so that T1 < T2, their synodic period is given by:[6]
{\displaystyle {\frac {1}{T_{\mathrm {syn} }}}={\frac {1}{T_{1}}}-{\frac {1}{T_{2}}}}
Examples of sidereal and synodic periods[]
Table of synodic periods in the Solar System, relative to Earth:[citation needed]
Mercury 0.240846 87.9691 days 0.317 115.88
Venus 0.615 224.7 days[8] 1.599 583.9
Earth 1 365.25636 solar days —
Mars 1.881 687.0[9] 2.135 779.9
Jupiter 11.86 4331[10] 1.092 398.9
Saturn 29.46 10,747[11] 1.035 378.1
Uranus 84.01 30,589[12] 1.012 369.7
Neptune 164.8 59,800[13] 1.006 367.5
134340 Pluto 248.1 90,560[14] 1.004 366.7
Moon 0.0748 27.32 days 0.0809 29.5306
99942 Apophis (near-Earth asteroid) 0.886 7.769 2,837.6
90377 Sedna 12050 1.0001 365.3[citation needed]
In the case of a planet's moon, the synodic period usually means the Sun-synodic period, namely, the time it takes the moon to complete its illumination phases, completing the solar phases for an astronomer on the planet's surface. The Earth's motion does not determine this value for other planets because an Earth observer is not orbited by the moons in question. For example, Deimos's synodic period is 1.2648 days, 0.18% longer than Deimos's sidereal period of 1.2624 d.[citation needed]
Synodic periods relative to other planets[]
The concept of synodic period applies not just to the Earth, but also to other planets as well, and the formula for computation is the same as the one given above. Here is a table which lists the synodic periods of some planets relative to each other:
1.881 11.86 29.46 50.42 84.01 164.8 248.1 287.5 557.0
Binary stars[]
AM Canum Venaticorum 17.146 minutes
Beta Lyrae AB 12.9075 days
Alpha Centauri AB 79.91 years
Proxima Centauri – Alpha Centauri AB 500,000 years or more
Rotation period – time that it takes to complete one revolution around its axis of rotation
Satellite revisit period
^ Oliver Montenbruck, Eberhard Gill (2000). Satellite Orbits: Models, Methods, and Applications. Springer Science & Business Media. p. 50. ISBN 978-3-540-67280-7.
^ Bate, Mueller & White (1971), p. 33.
^ Density of the Earth, wolframalpha.com
^ Density of water, wolframalpha.com
^ Bradley W. Carroll, Dale A. Ostlie. An introduction to modern astrophysics. 2nd ion. Pearson 2007.
^ Hannu Karttunen; et al. (2016). Fundamental Astronomy (6th ed.). Springer. p. 145. ISBN 9783662530450. Retrieved December 7, 2018.
^ "Questions and Answers - Sten's Space Blog". www.astronomycafe.net.
^ "Planetary Fact Sheet". nssdc.gsfc.nasa.gov.
Look up synodic in Wiktionary, the free dictionary. |
Sum and Difference Formulas | Brilliant Math & Science Wiki
Hemang Agarwal, Siddhartha Srivastava, L N, and
The sum and difference formulas state that
\begin{aligned} \sin(a+b) &= \sin a \cos b + \cos a \sin b \\ \sin(a-b) &= \sin a \cos b - \cos a \sin b \end{aligned}
\begin{aligned} \cos(a+b) &= \cos a \cos b - \sin a \sin b \\ \cos(a-b) &= \cos a \cos b + \sin a \sin b. \end{aligned}
Derive the sum formulas
\begin{aligned} \sin(a+b) &= \sin a \cos b + \cos a \sin b \\ \cos(a+b) &= \cos a \cos b - \sin a \sin b.\end{aligned}
By Euler's formula we know that
e^{i\theta} = \cos \theta + i\sin \theta.
\theta = a + b
e^{i(a+b)} = \cos(a+b) + i\sin(a+b). \qquad (1)
We also know from the algebraic properties of exponentials that
e^{i(a+b)} = e^{ia}e^{ib}.
\begin{aligned} e^{ia}e^{ib} &= (\cos a + i\sin a)(\cos b + i\sin b)\\ &=\cos a \cos b + i\sin a \cos b + i\sin b\cos a + i^2\sin a \sin b \\ &= \cos a \cos b - \sin a \sin b + i(\sin a \cos b + \sin b \cos a). \end{aligned}
Now, from
(1)
\begin{aligned} e^{i(a+b)} &= \cos(a+b) + i\sin(a+b)\\ &=\cos a \cos b - \sin a \sin b + i(\sin a \cos b + \sin b \cos a). \end{aligned}
\cos
\sin
are real-valued functions, it must be true that
\begin{aligned} \cos(a+b) &= \cos a \cos b - \sin a \sin b \\ i\sin(a+b) &= i(\sin a \cos b + \sin b \cos a), \end{aligned}
\begin{aligned} \cos(a+b) &= \cos a\cos b - \sin a \sin b \\ \sin(a+b) &= \sin a \cos b + \sin b \cos a. \ _\square \end{aligned}
In the diagram, let point
A
revolve to points
B
C,
and let the angles
\alpha
\beta
\angle AOB = \alpha, \quad \angle BOC = \beta.
Also, let both
\overline{CD}
\overline{FG}
be perpendicular to
\overline{OA},
E
\overline{CD}
\lvert \overline{ED}\rvert=\lvert\overline{FG}\rvert.
Then the formula for cosine-sum
\cos (\alpha+\beta) ,
\frac{\lvert \overline{OD}\rvert}{\lvert \overline{OC}\rvert} ,
\begin{aligned} \cos (\alpha + \beta) &= \frac{\lvert \overline{OD}\rvert}{\lvert \overline{OC}\rvert} \\ &= \frac{\lvert \overline{OG}\rvert}{\lvert \overline{OC}\rvert} - \frac{\lvert \overline{EF}\rvert}{\lvert \overline{OC}\rvert} \\ &= \frac{\lvert \overline{OG}\rvert}{\lvert \overline{OF}\rvert} \cdot \frac{\lvert \overline{OF}\rvert}{\lvert \overline{OC}\rvert} - \frac{\lvert \overline{EF}\rvert}{\lvert \overline{CF}\rvert} \cdot \frac{\lvert \overline{CF}\rvert}{\lvert \overline{OC}\rvert} \qquad \left(\text{since } \lvert \overline{OD}\rvert = \lvert \overline{OG}\rvert-\lvert \overline{EF}\rvert\right)\\ &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta. \end{aligned}
The cosine-difference formula can be obtained from the cosine-sum formula by replacing
\beta
- \beta,
\cos( -\beta) = \cos \beta
\sin(-\beta) = -\sin \beta:
\begin{aligned} \cos(\alpha + \beta) &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \\ \Rightarrow \cos(\alpha - \beta) &= \cos \alpha \cdot \cos (-\beta) - \sin \alpha \cdot \sin (-\beta) \\ &= \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta. \end{aligned}
In summary, we have the following two formulas of cosine-sum and cosine-difference:
Cosine-sum formula:
\cos(\alpha + \beta)= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta ,
Cosine-difference formula:
\cos(\alpha - \beta) = \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta .
\cos 75^\circ ?
From cosine-sum formula, we have
\begin{aligned} \cos 75^\circ &= \cos (45^\circ + 30^\circ) \\ &= \cos 45^\circ \cdot \cos 30^\circ - \sin 45^\circ \cdot \sin 30^\circ \\ &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\ &= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{6}-\sqrt{2}}{4}.\ _\square \end{aligned}
\cos 15^\circ ?
From cosine-difference formula, we have
\begin{aligned} \cos 15^\circ &= \cos (45^\circ - 30^\circ) \\ &= \cos 45^\circ \cdot \cos 30^\circ + \sin 45^\circ \cdot \sin 30^\circ \\ &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\ &= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{6}+\sqrt{2}}{4}.\ _\square \end{aligned}
\cos 105^\circ ?
From cosine-sum formula,
\cos 105^\circ
\begin{aligned} \cos 105^\circ &= \cos (60^\circ + 45^\circ) \\ &= \cos 60^\circ \cdot \cos 45^\circ - \sin 60^\circ \cdot \sin 45^\circ \\ &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \\ &= \frac{\sqrt{2}-\sqrt{6}}{4}.\ _\square \end{aligned}
\cos 140^\circ \cdot \cos 50^\circ + \sin 140^\circ \cdot \sin 50^\circ .
\begin{aligned} \cos 140^\circ \cdot \cos 50^\circ + \sin 140^\circ \cdot \sin 50^\circ &= \cos(140^\circ-50^\circ) \\ &= \cos 90^\circ \\ &= 0. \ _\square \end{aligned}
\sin \alpha = \frac{13}{14}
\sin \beta = \frac{11}{14}
0 < \alpha < \frac{\pi}{2}
0< \beta < \frac{\pi}{2} ,
\alpha + \beta ?
\sin^{2} x + \cos^{2}x =1,
\begin{aligned} \cos \alpha &= \sqrt{1-\sin^{2}\alpha} = \sqrt{1-\frac{13^2}{14^2}} = \frac{3\sqrt{3}}{14}, \\ \cos \beta &= \sqrt{1-\sin^{2}\beta} = \sqrt{1-\frac{11^2}{14^2}} = \frac{5\sqrt{3}}{14}. \end{aligned}
Thus, from cosine-sum formula, we have
\begin{aligned} \cos (\alpha + \beta) &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \\ &= \frac{3\sqrt{3}}{14} \times \frac{5\sqrt{3}}{14} - \frac{13}{14} \times \frac{11}{14} \\ &= -\frac{1}{2}. \end{aligned}
Hence, since
0< \alpha + \beta < \pi ,
\alpha + \beta
\begin{aligned} \cos (\alpha + \beta) &= -\frac{1}{2} \\ \Rightarrow \alpha + \beta &= \frac{2}{3} \pi. \ _\square \end{aligned}
The tangent sum and difference formulas are
\begin{aligned} \tan(A+B) &= \dfrac{\tan A + \tan B}{1 - \tan A \tan B} \\\\ \tan(A-B) &= \dfrac{\tan A - \tan B}{1 + \tan A \tan B}. \end{aligned}
Derive the tangent sum formula.
\begin{aligned} \sin(a+b) &= \sin a \cos b + \cos a \sin b &\qquad (1) \\ \cos(a+b) &= \cos a \cos b - \sin a \sin b. &\qquad (2) \end{aligned}
(1)
(2)
\dfrac{\sin(a+b)}{\cos(a+b)} = \dfrac{ \sin a \cos b + \cos a \sin b}{\cos a \cos b - \sin a \sin b}.
Dividing the right side of this by
\cos a \cos b
\tan(a+b) = \dfrac{\dfrac{ \sin a \cos b}{\cos a \cos b} + \dfrac{\cos a \sin b}{\cos a \cos b}}{\dfrac{\cos a \cos b}{\cos a \cos b} - \dfrac{\sin a \sin b}{\cos a \cos b}} = \dfrac{\tan a + \tan b}{1 - \tan a \tan b},
which is the sum formula.
_\square
Derive the tangent difference formula.
\tan(-a) = -\tan a .
b = -b
in the tangent sum formula, we get
\begin{aligned} \tan\big(a+(-b)\big) &= \dfrac{\tan(a) + \tan(-b)}{1 - \tan a \tan(-b)} \\ \tan(a-b) &= \dfrac{\tan a - \tan b}{1 + \tan a \tan b}. \ _\square \end{aligned}
These identities are useful for finding the values of tangents of angles which are not known.
\tan 75^{\circ}
We want to break
75^{\circ}
into two angles whose tangents we know. One obvious pair is
(30^{\circ}, 45^{\circ})
Using the tangent sum formula, we have
\begin{aligned} \tan 75^{\circ} = \tan( 30^{\circ} + 45^{\circ}) &= \dfrac{\tan 30^{\circ} + \tan 45^{\circ}}{1 - \tan 30^{\circ} \tan 45^{\circ}} \\\\ &= \frac{ \frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}\cdot 1} \\\\ &= \frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{\ \ \ \frac{\sqrt{3} - 1}{\sqrt{3}}\ \ \ } \\\\ &= \frac{\sqrt{3}+1}{\sqrt3 - 1}. \ _\square \end{aligned}
\sin(18^\circ) = \frac14\big(\sqrt5-1\big).
\dfrac{\tan(x + 120^{\circ})}{\tan(x - 30^{\circ})} = \dfrac{11}{2}
x
is a solution to the above equation and
\cos(4x) = \dfrac{a}{b},
and
b
are coprime positive integers, then find
a + b.
\tan (a) \tan (2a) + \tan(2a) \tan(3a) +\cdots+ \tan(8a) \tan(9a),
a=\frac{\pi}{5}.
\large \tan(63^\circ) = \sqrt{\sqrt a-\sqrt b} + \sqrt{\sqrt c-\sqrt b}
The equation above is true for positive integers
a,b,
c.
a+b+c?
Cite as: Sum and Difference Formulas. Brilliant.org. Retrieved from https://brilliant.org/wiki/sum-and-difference-formulas/ |
Spark-ignition engine controller that uses the driver torque request - Simulink - MathWorks France
{\phi }_{ICP}
{\phi }_{ICPCMD}
{\phi }_{ECP}
{\phi }_{ECPCMD}
P{w}_{inj}
{L}_{cmd}={f}_{Lcmd}\left({T}_{cmd},N\right)
TA{P}_{cmd}={f}_{TAPcmd}\left({L}_{cmd},N\right)
TP{P}_{cmd}={f}_{TPPcmd}\left(TA{P}_{cmd}\right)
WA{P}_{cmd}={f}_{WAPcmd}\left({L}_{cmd},N\right)
{\phi }_{ICPCMD}={f}_{ICPCMD}\left({L}_{est},N\right)
{\phi }_{ECPCMD}={f}_{ECPCMD}\left({L}_{est},N\right)
{L}_{est}=\frac{Cps{R}_{air}{T}_{std}{\stackrel{˙}{m}}_{air,est}}{{P}_{std}{V}_{d}N}
Cps
{P}_{std}
{T}_{std}
{R}_{air}
{V}_{d}
{\stackrel{˙}{m}}_{air,est}
{f}_{TAPcmd}
TA{P}_{cmd}={f}_{TAPcmd}\left({L}_{cmd},N\right)
{f}_{TPPcmd}
TP{P}_{cmd}={f}_{TPPcmd}\left(TA{P}_{cmd}\right)
{f}_{WAPcmd}
WA{P}_{cmd}={f}_{WAPcmd}\left({L}_{cmd},N\right)
{f}_{Lcmd}
{L}_{cmd}={f}_{Lcmd}\left({T}_{cmd},N\right)
{f}_{ICPCMD}
{\phi }_{ICPCMD}={f}_{ICPCMD}\left({L}_{est},N\right)
{\phi }_{ICPCMD}
{f}_{ECPCMD}
{\phi }_{ECPCMD}={f}_{ECPCMD}\left({L}_{est},N\right)
{\phi }_{ECPCMD}
EG{R}_{pct}=100\frac{{\stackrel{˙}{m}}_{EGR}}{{\stackrel{˙}{m}}_{EGR}+{\stackrel{˙}{m}}_{air}}
\begin{array}{l}{\stackrel{˙}{m}}_{EGRstd,cmd}={\stackrel{˙}{m}}_{EGR,cmd}\frac{{P}_{std}}{{P}_{in,EGR}}\sqrt{\frac{{T}_{in,EGR}}{{T}_{std}}}\\ \\ {\stackrel{˙}{m}}_{EGRstd,max}={f}_{EGRstd,max}\left(\frac{{P}_{out,EGR}}{{P}_{in,EGR}}\right)\\ \\ {\stackrel{˙}{m}}_{EGR,cmd}=EG{R}_{pct,cmd}{\stackrel{˙}{m}}_{intk,est}\end{array}
EGRa{p}_{cmd}={f}_{EGRap,cmd}\left(\frac{{\stackrel{˙}{m}}_{EGRstd,cmd}}{{\stackrel{˙}{m}}_{EGRstd,max}},\frac{{P}_{out,EGR}}{{P}_{in,EGR}}\right)
\frac{{\stackrel{˙}{m}}_{EGRstd,cmd}}{{\stackrel{˙}{m}}_{EGRstd,max}}
\frac{{P}_{out,EGR}}{{P}_{in,EGR}}
{\stackrel{˙}{m}}_{EGRstd,cmd}
{\stackrel{˙}{m}}_{EGRstd,max}
{\stackrel{˙}{m}}_{EGR,cmd}
{\stackrel{˙}{m}}_{intk,est}
{\lambda }_{cmd}
{\lambda }_{cmd}=\frac{AF{R}_{cmd}}{AF{R}_{stoich}}
AF{R}_{cmd}=\frac{{\stackrel{˙}{m}}_{air,est}}{{\stackrel{˙}{m}}_{fuel,cmd}}
{\lambda }_{cmd}
{\lambda }_{cmd}={f}_{\lambda cmd}\left({L}_{est},N\right)
{\lambda }_{cmd}
{\lambda }_{cmd}
{\stackrel{˙}{m}}_{fuel,cmd}=\frac{{\stackrel{˙}{m}}_{air,est}}{AF{R}_{cmd}}=\frac{{\stackrel{˙}{m}}_{air,est}}{{\lambda }_{cmd}AF{R}_{stoich}}
P{w}_{inj}=\left\{\begin{array}{cc}\frac{{\stackrel{˙}{m}}_{fuel,cmd}Cps\left(\frac{60s}{min}\right)\left(\frac{1000mg}{g}\right)\left(\frac{1000g}{kg}\right)}{N{S}_{inj}{N}_{cyl}}& \text{when }Tr{q}_{cmd}>0\\ 0& \text{when }Tr{q}_{cmd}\le 0\end{array}
SA={f}_{SA}\left({L}_{est},N\right)
{f}_{SA}
{C}_{idle}\left(z\right)={K}_{p,idle}+{K}_{i,idle}\frac{{t}_{s}}{z-1}
{\stackrel{˙}{m}}_{air,est}
{\stackrel{˙}{m}}_{EGR,est}
{y}_{intk,EGR,est}=\frac{{\stackrel{˙}{m}}_{EGR,est}}{{\stackrel{˙}{m}}_{intk,est}}\frac{{t}_{s}z}{{\tau }_{EGR}z+{t}_{s}-{\tau }_{EGR}}
{y}_{intk,air,est}=1-{y}_{intk,EGR,est}
{\stackrel{˙}{m}}_{EGR,est}
{\stackrel{˙}{m}}_{intk,est}
\begin{array}{l}{\stackrel{˙}{m}}_{air,std}={\stackrel{˙}{m}}_{air,est}\frac{{P}_{std}}{{P}_{amb}}\sqrt{\frac{IAT}{{T}_{std}}}\\ \\ {P}_{in,EGR}={P}_{out,EGR}+\Delta {P}_{EGR}\\ \\ {\stackrel{˙}{m}}_{EGR,est}={\stackrel{˙}{m}}_{EGR,std}\frac{{P}_{in,EGR}}{{P}_{std}}\sqrt{\frac{{T}_{std}}{{T}_{in,EGR}}}\end{array}
{\stackrel{˙}{m}}_{EGR,std}={f}_{EGR,std}\left(EGRap,\frac{{P}_{out,EGR}}{{P}_{in,EGR}}\right)
{\stackrel{˙}{m}}_{EGR,std}
\frac{{P}_{out,EGR}}{{P}_{in,EGR}}
\frac{{P}_{out,EGR}}{{P}_{amb}}={f}_{intksys,pr}\left({\stackrel{˙}{m}}_{air,std}\right)
{\stackrel{˙}{m}}_{air,std}
\frac{{P}_{out,EGR}}{{P}_{amb}}
{\stackrel{˙}{m}}_{air,std}
{\stackrel{˙}{m}}_{EGR,std}
{\stackrel{˙}{m}}_{air,est}
{\stackrel{˙}{m}}_{EGR,est}
{f}_{Texh}
{T}_{exh}={f}_{Texh}\left(L,N\right)
{P}_{Amb}
MAP
MAT
{\phi }_{ICP}
{\phi }_{ECP}
P{w}_{inj}
SA
{\phi }_{ICPCMD}
{\phi }_{ECPCMD}
{\stackrel{˙}{m}}_{fuel,cmd}
{\stackrel{˙}{m}}_{intk,est}
{\stackrel{˙}{m}}_{air,est}
{\stackrel{˙}{m}}_{EGR,est}
P{w}_{inj}
SA
{\phi }_{ICPCMD}
{\phi }_{ECPCMD}
{f}_{Lcmd}
{L}_{cmd}={f}_{Lcmd}\left({T}_{cmd},N\right)
{f}_{TAPcmd}
TA{P}_{cmd}={f}_{TAPcmd}\left({L}_{cmd},N\right)
{f}_{TPPcmd}
TP{P}_{cmd}={f}_{TPPcmd}\left(TA{P}_{cmd}\right)
{f}_{WAPcmd}
WA{P}_{cmd}={f}_{WAPcmd}\left({L}_{cmd},N\right)
{f}_{ICPCMD}
{\phi }_{ICPCMD}={f}_{ICPCMD}\left({L}_{est},N\right)
{\phi }_{ICPCMD}
{f}_{ECPCMD}
{\phi }_{ECPCMD}={f}_{ECPCMD}\left({L}_{est},N\right)
{\phi }_{ECPCMD}
EG{R}_{pct,cmd}={f}_{EGRpct,cmd}\left({L}_{est},N\right)
EGRa{p}_{cmd}={f}_{EGRap,cmd}\left(\frac{{\stackrel{˙}{m}}_{EGRstd,cmd}}{{\stackrel{˙}{m}}_{EGRstd,max}},\frac{{P}_{out,EGR}}{{P}_{in,EGR}}\right)
\frac{{\stackrel{˙}{m}}_{EGRstd,cmd}}{{\stackrel{˙}{m}}_{EGRstd,max}}
\frac{{P}_{out,EGR}}{{P}_{in,EGR}}
{\stackrel{˙}{m}}_{EGRstd,max}
\frac{{\stackrel{˙}{m}}_{EGRstd,cmd}}{{\stackrel{˙}{m}}_{EGRstd,max}}
\frac{{P}_{out,EGR}}{{P}_{in,EGR}}
{S}_{inj}
{\lambda }_{cmd}
{\lambda }_{cmd}={f}_{\lambda cmd}\left({L}_{est},N\right)
{\lambda }_{cmd}
SA={f}_{SA}\left({L}_{est},N\right)
{N}_{cyl}
Cps
{V}_{d}
{R}_{air}
{P}_{std}
{T}_{std}
{f}_{{\eta }_{v}}
{\eta }_{v}={f}_{{\eta }_{v}}\left(MAP,N\right)
{\eta }_{v}
{f}_{Vivc}
{V}_{IVC}={f}_{Vivc}\left({\phi }_{ICP}\right)
{V}_{IVC}
{\phi }_{ICP}
{f}_{TMcorr}
T{M}_{corr}={f}_{TMcorr}\left({\rho }_{norm}, N\right)
T{M}_{corr}
{\rho }_{norm}
{\stackrel{˙}{m}}_{intkideal}={f}_{intkideal}\left({\phi }_{ECP},T{M}_{flow}\right)
{\stackrel{˙}{m}}_{intkideal}
{\phi }_{ECP}
T{M}_{flow}
{f}_{aircorr}
{\stackrel{˙}{m}}_{air}={\stackrel{˙}{m}}_{intkideal}{f}_{aircorr}\left({L}_{ideal},N\right)
{L}_{ideal}
{\stackrel{˙}{m}}_{air}
{\stackrel{˙}{m}}_{intkideal}
\frac{{P}_{out,EGR}}{{P}_{amb}}={f}_{intksys,pr}\left({\stackrel{˙}{m}}_{air,std}\right)
{\stackrel{˙}{m}}_{air,std}
\frac{{P}_{out,EGR}}{{P}_{amb}}
{\stackrel{˙}{m}}_{air,std}
{\stackrel{˙}{m}}_{EGR,std}={f}_{EGR,std}\left(EGRap,\frac{{P}_{out,EGR}}{{P}_{in,EGR}}\right)
{\stackrel{˙}{m}}_{EGR,std}
\frac{{P}_{out,EGR}}{{P}_{in,EGR}}
\frac{{P}_{out,EGR}}{{P}_{in,EGR}}
{T}_{brake}={f}_{TnL}\left(L,N\right)
{T}_{brake}
{f}_{Tqinr}
T{q}_{inr}={f}_{Tqinr}\left(L,N\right)
T{q}_{inr}
{f}_{Tfric}
{T}_{fric}={f}_{Tfric}\left(L,N\right)
{T}_{fric}
{f}_{SAopt}
S{A}_{opt}={f}_{SAopt}\left(L,N\right)
{f}_{Msa}
\begin{array}{l}{M}_{sa}={f}_{Msa}\left(\Delta SA\right)\\ \Delta SA=S{A}_{opt}-SA\end{array}
{M}_{sa}
\Delta SA
{f}_{M\lambda }
{M}_{\lambda }={f}_{M\lambda }\left(\lambda \right)
{M}_{\lambda }
\lambda
{f}_{Texh}
{T}_{exh}={f}_{Texh}\left(L,N\right) |
a) Find a weak formulation for the partial differential equation {\partial u\over\partial t
a) Find a weak formulation for the partial differential equation{\partial u\over\partial t
rhita3yp 2022-02-18 Answered
a) Find a weak formulation for the partial differential equation
\frac{\partial u}{\partial t\text{ }}+c\frac{\partial u}{\partial x\text{ }}=0
b) Show that
u=f\left(x-ct\right)
is a generalized solution of
\frac{\partial u}{\partial t\text{ }}+c\frac{\partial u}{\partial x\text{ }}=0
for any distribution
f
What i already haveI know that in order to find a weak form of a pde, we need to multiply it by a test function, then integrate it. Also, to find a generalized solution, we need to find a weak solution and just multiply it by the Heaviside function.Let's take any test function
\varphi
, then we have (integrating by parts second part of the integral)
{\int }_{\mathrm{\Omega }}\left(\frac{\partial u}{\partial t\text{ }}+c\frac{\partial u}{\partial x\text{ }}\right)\ast \varphi \left(x\right)dx=
={\int }_{\mathrm{\Omega }}\frac{\partial u}{\partial t}\varphi \left(x\right)dx-c{\int }_{\mathrm{\Omega }}u\left(x,t\right){\varphi }^{\prime }\left(x\right)dx
\varphi
vanishes at boundaries. So, is it the final form or can we proceed further? And how am I supposed to find a generalized solution?
Tate Puckett
a.) The idea of integral solutions is a little more complicated than just integration with a test function. Its
koffiejkl
b.) Use the weak formulation to integrate
u=f\left(x-ct\right)
. Youre
f\left(x\right)=\sqrt{4-x}
a=0
\sqrt{3.9}
\sqrt{3.99}
Differentiate the Gaussian elimination and LU- Factorization in solving system of linear equations.
Do the equations 5y−2x=18
6x=−4y−10
form a system of linear equations? Explain.
Find values of a and b such that the system of linear equations has (a) no solution, (b) exactly one solution, and (c) infinitely many solutions.
ax + by = −9
Do the equations 4x−3y=54x−3y=5 and 7y+2x=−87y+2x=−8 form a system of linear equations? Explain
n\in \mathbb{N}
W\le {\mathbb{F}}^{n}
, show that there exists a homogeneous system of linear equations whose solution space is W.
W\le {\mathbb{F}}^{n},k=\mathrm{dim}\left(W\right)\le \mathrm{dim}\left({\mathbb{F}}^{n}\right)
. Let's say that
\left\{{w}_{1},{w}_{2},\dots ,{w}_{k}\right\}
is a basis of W. Now, construct a matrix A (of size
k×n
) such that its rows are elements from the basis of W, stacked together. The row space of A is W, so the row space of its row-echelon form is W too. At this point, I'm stuck! I'm trying to come up with a homogeneous system with the help of A, though there may exist other easier ways of approaching this problem.
Could someone show me the light?
W\le {\mathbb{F}}^{n}
stands for W is a subspace of
{\mathbb{F}}^{n}
P.P.S. Isn't this equivalent to saying that W is the null-space of some matrix? Can we go ahead along these lines, and construct a matrix P such that
Pw=0
w\in W
Write the correct equation for this line.
y–{y}_{1}=m\left(x–{x}_{1}\right)
, when the slope is 4 and a line passes through the point (-3. 3). |
How do you know when to use a specific distribution?
How do you know when to use a specific distribution? For example, when would we use the Normal, binomial, grometric, hypergeometric, or negative binomial distribution? When doing so, when do we know if is is a discrete or continuous distribution?
Szeteib
To use a specific distribution one should know the properties of that distribution. For example if the data distribution is symmetric and have bell shape curve then we use a normal distribution. If the data distribution have two outcomes one success and one failure out of n trials then we use binomial distribution etc.
The probability distributions are classified as discrete and continuous probability distributions based on the random variable.
If the variable can take any value between two particular values then we say it as continuous distribution. If the variable cannot take any value between particular values then we say it is discrete probability distribution.
For example in binomial probability distribution where n is total trials and p is probability of success and x is number of success. Here x can take only values 0,1,2,3,4.....n. x cannot take values between 1 and 2. So we can say that the binomial distribution is discrete probability distribution.
For example if we find a distribution with data of weights of students in a school between 60lbs and 120 lbs then the weights can take any value between them. In this case we call the distribution to be continuous probability distribution.
Normal DIstribution, Exponential Distribution, - Continuous Distribution
Binomial Distribution, geometric Distribution, Hypergeometric Distribution, Negative binomial Distribution Poisson Distribution - Discrete Distribution
At a certain college, 6% of all students come from outside the United States. Incoming students there are assigned at random to freshman dorms, where students live in residential clusters of 40 freshmen sharing a common lounge area. How many international students would you expect to find in a typical cluster? With what standard deviation?
Convert the binomial probability to a normal distribution probability using continuity correction. P (x < 25).
What is the probability that if you threw 20 darts, that 8 of them would hit the same area?
a) To find the probability of winning at least one prize if you purchase 10 tickets.
b) To find the probability of winning at least one prize if you purchase 15 tickets.
The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1034 games played (as of a certain date). An upcoming monthly schedule contains 12 games.
What is the probability that the San Jose Sharks win 5 games in that upcoming month? Let
X=
number of games won in that upcoming month. (Round your answer to four decimal places.)
A robot has a probability to solve a puzzle of 47% if the robot tries to solve 10 puzzles. 1. What is the probability that the robot to solve less than 2 puzzles ( not including 2)? Use 4 dig 2. On average, how many puzzles will be solved by the robot:
According to a study by the Bureau of Justice Statistics, approximately 5% of the nations |
Numerical simulation and measurement of car strut under shock vibration | JVE Journals
Ondřej Novák1 , Michal Petrů2 , Aleš Lufinka3
1, 2, 3Technical University of Liberec, Liberec, Czech Republic
This article focuses on the analysis of the properties of the car strut used to capture the engine torque reaction generated during fast acceleration. There is described the methodology of measuring the parameters needed to assess the force effect and also a simulation of the properties of a redesigned strut using the finite element method. The simulation confirmed the correctness of the design and also showed that very small asymmetry of cross section cause significant reducing of the load capacity of the strut and its easy damage.
Keywords: strut, momentum reaction, impact, shock, Ansys, FEM.
A mounting of the drive unit in the car is an important element not only for the comfort but also for safety. A properly designed strut ensures smooth transmission of the torque of the engine to the car body without shocks and also allows to the drive unit to move during the crash in the desired direction, it means outside the car's cab. The part that realizes the mounting is a strut. Its proper dimensioning and quality of production are crucial for the resulting vehicle behavior in different situations. The torque strut consists of a metal cast in which the silent blocks are located in the mounting eyes. During strut prototype testing were problems with permanent bending deformation or strut breaking. The aim of analyzes were to determine the real forces acting in the strut, to design a suitable strut geometry and to assess its properties by numerical simulation. Recent analysis has shown that the problem of bending is not only due to the magnitude of the force but also to the asymmetry of the cross section that leads to bending deformation when the tensile stress is applied. Therefore, a variant simulating the asymmetric cross-section was also analyzed.
Fig. 1. Sample of strut with tensiometers
When the impulse force acts on the body, it causes its deflection from the equilibrium position and then it freely vibrates at one or more own frequencies. A simple example is tapping on a glass (the undamaged glass generates another sound in comparison with broken glass). When applied to strut impact excitation is used. Methods based on the deconvolution of the measured response in the time domain are often used to obtain the force course. These methods can generally be used for problems that correspond with linear time invariant – LTI. For such a problem can be applied the principle of the superposition, which can be described by the equation in the form of Eq. (1):
u\left(t\right)={\int }_{-\infty }^{\infty }f\left(\tau \right)\bullet g\left(t-\tau \right)d\tau ,
f\left(\tau \right)
is the driving force or generally, input of the system,
u\left(t\right)
is the response of the structure or generally, output of the system and
g\left(t\right)
is the impulse response of the system to the Dirac unit pulse. Problems with impulse force are poorly conditioned, however, a number of theories dealing with improving of the conditionality have been compiled [1]. Jacquelin et al. dealt with the identification of the impact force on the aluminum plate, when they evaluated several methods for improving of the task conditionality (so-called regularization methods) and showed their parameter setting [2]. Jang et al. dealt with the general solvability of the inverse task for impact loads and performed numerical simulation for selected tasks using the Tikhonov and Landwerber-Fridmen regularization [3]. Furthermore, Kim and Lee have considered using the singular value decomposition to improve conditionality of the task and the method was verified experimentally on cantilever beam [4]. Wang and Xie proposed [5] a new method of regularization and performed its verification on the numerical model of the laminate cylinder with a slight improvement in results compared to the Tichon’s regularization [6].
The force that acts in the strut during the launch was measured directly in the car. On the strut tensiometers on opposite side of a flange were glued. Then the tensiometers were verified and calibrated under known loading force. There tensile and bending loading was applied (Fig. 2). Then the strut was mounted to the car and the loading force was created by the fast releasing of a clutch under engine speeds up to a rev limiter. Totally five launches in first gear and reverse gear was carried out. The record of corresponding electrical voltage and forces are shown in Fig. 3. The maximum in first gear was 12,3, in reverse gear 18,8 kN. This curve was scaled for the obtaining of resulting force 45 kN – the force was multiplied by the constant 2,39, because higher loading capacity of new strut was expected. This adjusted curve was applied for ANSYS input in the form of data tab.
Fig. 2. Calibration of tensiometers
a) Bending loading
b) Tensile loading
4. FE model
A new geometry of the strut was designed (Fig. 4). The design has respected a requirement of higher structural strength. Therefore, walls of the strut have a higher thickness; applied material is same as in previous type of the strut – aluminum alloy. Its mechanical parameters are shown in Table 1. Border conditions are seen also in Fig. 4. One side of the strut is fixed against a displacement and rotation in all directions by the help of cylindrical constrain. On opposite side a tensile force on cylindrical sector surface is applied. The direction of the force coincides with the symmetry axis of the strut. Initial conditions were following. The applied force that was measured in the car during extreme launch was assigned and applied in the form of input data tab in ANSYS. As a maximal stress criterion the tensile ultimate strength 310 MPa was taken. Under this stress the strut will be broken. Results of the simulation show a principal stress and deformation for symmetrical (Fig. 5) and asymmetrical strut (Fig. 6). As seen, in the case of symmetrical deformation tensile force leads to a symmetrical elongation of the strut beam. The force that causes achieving of material ultimate strength corresponds to 43,200 kN. Total elongation is 0,23 mm. Asymmetrical strut indicates ultimate strength at force about 18,9 kN. It means that relatively low change in symmetry caused by a different flange area about 10 %, reduces highest achieved force more than two times. The bending deformation on the end of the beam was 0,35 mm. It can be said that the accuracy of a mold used for the strut casting is very important and low differences in geometry of the strut causes extreme decreasing of applicable loading force.
Fig. 3. Experimental data
a) Time response of deformation [μm/m]
b) Time response of force [kN]
Fig. 4. A new strut design; Border conditions. The cross-section of this design was enlarged, the main dimension of strut geometry – a pitch and total length were maintained
Fig. 5. FE model of symmetrical strut
a) Maximum stress corresponding to ultimate strength
b) Total deformation of symmetrical strut
Fig. 6. FE model of asymmetrical strut
b) Bending deformation on the of the strut beam
Table 1. Properties of applied material
Poisson’s ratio [GPa]
Bulk modulus [GPa]
Shear modulus [GPa]
Tensile yield strength [MPa]
Tensile ultimate strength [MPa]
Aluminum alloy-
The new geometry of the strut was designed. The design has been subjected to investigation using the finite element method. The strut with asymmetric cross section was also proposed, where the one flange of the I-beam had about 10 % smaller area compared to the second one. The problem with asymmetric cross section was discovered during the production. Loading force was obtained by measuring directly in the car. The simulation showed that the redesigned strut would transfer significantly higher force, more than twice compared with the force measured in the car. However, the strut with the asymmetric cross-section will break near to value measured in the car. Therefore, it is necessary to check the casting thoroughly to ensure that the problem does not appear again during production. It can be concluded that proposed strut is suitably dimensioned with a sufficient degree of safety.
Petrů M., Kovář R., Martinec T., Srb P., Lufinka Kulhavý A. P. Analysis and study of vibrations of a clamping device used for winding carbon fibers into the core of a frame. 53rd Conference on Experimental Stress Analysis, 2015, p. 287-292. [Search CrossRef]
Jacquelin E., Bennani A., Hameli P. Force reconstruction: analysis and regularization of a deconvolution problem. Journal of Sound and Vibration, Vol. 265, Issue 1, 2003, p. 81-107. [Search CrossRef]
Jang T. S., Baek H., Han S. L., Kinoshita T. Indirect measurement of the impulsive load to a nonlinear system from dynamic responses: inverse problem formulation. Mechanical Systems and Signal Processing, Vol. 24, Issue 6, 2010, p. 1665-1681. [Search CrossRef]
Kim S. J., Lee S. K. Experimental identification for inverse problem of a mechanical system with a non-minimum phase based on singular value decomposition. Journal of Mechanical Science and Technology, Vol. 22, Issue 8, 2008, p. 1504-1509. [Search CrossRef]
Choi I. H., Lim C. H. Low-velocity impact analysis of composite laminates using linearized contact law. Composite Structures, Vol. 66, Issues 1-4, 2004, p. 125-132. [Search CrossRef]
Wang L., Xie Y. A novel regularization method and application to load identification of composite laminated cylindrical shell. Journal of Applied Analysis and Computation, Vol. 5, Issue 4, 2015, p. 570-580. [Search CrossRef] |
Virtual knot - Wikipedia
Virtual knot
Generalization of knots in 3-dimensional Euclidean space
[Extension of Jones polynomial to general 3-manifolds.] Can the original Jones polynomial, which is defined for 1-links in the 3-sphere (the 3-ball, the 3-space R3), be extended for 1-links in any 3-manifold?
In knot theory, a virtual knot is a generalization of knots in 3-dimensional Euclidean space, R3, to knots in thickened surfaces
{\displaystyle \Sigma \times [0,1]}
modulo an equivalence relation called stabilization/destabilization. Here
{\displaystyle \Sigma }
is required to be closed and oriented. Virtual knots were first introduced by Kauffman (1999).
In the theory of classical knots, knots can be considered equivalence classes of knot diagrams under the Reidemeister moves. Likewise a virtual knot can be considered an equivalence of virtual knot diagrams that are equivalent under generalized Reidemeister moves. Virtual knots allow for the existence of, for example, knots whose Gauss codes which could not exist in 3-dimensional Euclidean space. A virtual knot diagram is a 4-valent planar graph, but each vertex is now allowed to be a classical crossing or a new type called virtual. The generalized moves show how to manipulate such diagrams to obtain an equivalent diagram; one move called the semi-virtual move involves both classical and virtual crossings, but all the other moves involve only one variety of crossing.
Virtual knots are important, and there is a strong relation between Quantum Field Theory and virtual knots.
Virtual knots themselves are fascinating objects, and having many connections to other areas of mathematics. Virtual knots have many exciting connections with other fields of knots theory. The unsolved problem shown is an important motivation to the study of virtual knots.
See section 1.1 of this paper [KOS] [1] for the background and the history of this problem. Kauffman submitted a solution in the case of the product manifold of closed oriented surface and the closed interval, by introducing virtual 1-knots .[2] It is open in the other cases. Witten’s path integral for Jones polynomial is written for links in any compact 3-manifold formally, but the calculus is not done even in physics level in any case other than the 3-sphere (the 3-ball, the 3-space R3). This problem is also open in physics level. In the case of Alexander polynomial, this problem is solved.
A classical knot can also be considered an equivalence class of Gauss diagrams under certain moves coming from the Reidemeister moves. Not all Gauss diagrams are realizable as knot diagrams, but by considering all equivalence classes of Gauss diagrams we obtain virtual knots.
A classical knot can be considered an ambient isotopy class of embeddings of the circle into a thickened 2-sphere. This can be generalized by considering such classes of embeddings into thickened higher-genus surfaces. This is not quite what we want since adding a handle to a (thick) surface will create a higher-genus embedding of the original knot. The adding of a handle is called stabilization and the reverse process destabilization. Thus a virtual knot can be considered an ambient isotopy class of embeddings of the circle into thickened surfaces with the equivalence given by (de)stabilization.
Some basic theorems relating classical and virtual knots:
If two classical knots are equivalent as virtual knots, they are equivalent as classical knots.
There is an algorithm to determine if a virtual knot is classical.
There is an algorithm to determine if two virtual knots are equivalent.
It is important that there is a relation among the following. See the paper [KOS] cited above and below.
Virtual equivalence of virtual 1-knot diagrams, which is a set of virtual 1-knots.
Welded equivalence of virtual 1-knot diagrams
Rotational welded equivalence of virtual 1-knot diagrams
Fiberwise equivalence of virtual 1-knot diagrams
Virtual 2-knots are also defined. See the paper cited above.
Knots and graphs
^ Kauffman, L.H; Ogasa, E; Schneider, J (2018), A spinning construction for virtual 1-knots and 2-knots, and the fiberwise and welded equivalence of virtual 1-knots, arXiv:1808.03023
^ Kauffman, L.E. (1998), Talks at MSRI Meeting in January 1997, AMS Meeting at University of Maryland, College Park in March 1997, Isaac Newton Institute Lecture in November 1997, Knots in Hellas Meeting in Delphi, Greece in July 1998, APCTP-NANKAI Symposium on Yang-Baxter Systems, Non-Linear Models and Applications at Seoul, Korea in October 1998, and Kauffman's paper1999 cited below., arXiv:math/9811028
Boden, Hans; Nagel, Matthias (2017). "Concordance group of virtual knots". Proceedings of the American Mathematical Society. 145 (12): 5451–5461. doi:10.1090/proc/13667. S2CID 119139769.
Carter, J. Scott; Kamada, Seiichi; Saito, Masahico (2002). "Stable equivalence of knots on surfaces and virtual knot cobordisms. Knots 2000 Korea, Vol. 1 (Yongpyong)". J. Knot Theory Ramifications. 11 (3): 311–322.
Carter, J. Scott; Silver, Daniel; Williams, Susan (2014). "Invariants of links in thickened surfaces". Algebraic & Geometric Topology. 14 (3): 1377–1394. doi:10.2140/agt.2014.14.1377. S2CID 53137201.
Dye, Heather A (2016). An Invitation to Knot Theory : Virtual and Classical (First ed.). Chapman and Hall/CRC. ISBN 9781315370750.
Goussarov, Mikhail; Polyak, Michael; Viro, Oleg (2000). "Finite-type invariants of classical and virtual knots". Topology. 39 (5): 1045–1068. arXiv:math/9810073. doi:10.1016/S0040-9383(99)00054-3. S2CID 8871411.
Kamada, Naoko; Kamda, Seiichi (2000). "Abstract link diagrams and virtual knots". Journal of Knot Theory and Its Ramifications. 9 (1): 93–106. doi:10.1142/S0218216500000049.
Kauffman, Louis H. (1999). "Virtual knot theory" (PDF). European Journal of Combinatorics. 20 (7): 663–690. doi:10.1006/eujc.1999.0314. ISSN 0195-6698. MR 1721925. S2CID 5993431.
Kauffman, Louis H.; Manturov, Vassily Olegovich (2005). "Virtual Knots and Links". arXiv:math.GT/0502014.
Kuperberg, Greg (2003). "What is a virtual link?". Algebraic & Geometric Topology. 3: 587–591. doi:10.2140/agt.2003.3.587. S2CID 16803280.
Manturov, Vassily (2004). Knot Theory. CRC Press. ISBN 978-0-415-31001-7.
Manturov, Vassily Olegovich (2004). "Virtual knots and infinite dimensional Lie algebras". Acta Applicandae Mathematicae. 83 (3): 221–233. doi:10.1023/B:ACAP.0000038944.29820.5e. S2CID 124019548.
Turaev, Vladimir (2008). "Cobordism of knots on surfaces". Journal of Topology. 1 (2): 285–305. arXiv:math/0703055. doi:10.1112/jtopol/jtn002. S2CID 17888102. </ref>
A Table of Virtual Knots
Elementary explanation with diagrams
Retrieved from "https://en.wikipedia.org/w/index.php?title=Virtual_knot&oldid=1073415834" |
\frac{1}{2},\frac{3}{10}
\frac{1}{5}.
On 1st April, 2012 Micro-tech Ltd. was formed with an authorised capital of Rs 50,00,000 divided into 5,00,000 equity shares of Rs 10 each. The company issued prospectus inviting applications for 4,50,000 equity shares. The company received applications for 4,20,000 equity shares.
During the first year, Rs 8 per share were called. Trilok holding 1,000 shares and Rajesh holding 2,000 shares did not pay the first call of Rs 2 per share. Rajesh's shares were forfeited after the first call and later on 1,500 of the forfeited shares were re-issued at Rs 6 per share, Rs 8 called up.
(a) Share Capital in the Balance Sheet of the company as per revised Schedule VI Part I of the Companies Act. 1956.
Rajeev, Sanjeev and Jatin were partners in a firm manufacturing blankets. They were sharing profits in the ratio of 5 : 3 : 2. Their capitals on 1st April, 2012 were Rs 1,00,000, Rs 2,00,000 and Rs 4,00,000 respectively. After the flood in Uttarakhand, all partners decided to help the flood victims personally.
For this Rajeev withdrew Rs 10,000 from the firm on 1st October, 2012. Sanjeev instead of withdrawing cash from the firm took blankets amounting to Rs 14,000 from the firm and distributed those to the flood victims. On the other hand, Jatin withdrew Rs 1,50,000 from his capital on 31st December, 2012 and set up a centre to provide medical facilities in the flood affected area.
NY Ltd. invited applications for issuing 90,000 equity shares of Rs 10 each at a premium of Rs 5 per share. The amount was payable as follows:
On applications and allotment − Rs 10 per share (including premium)
Applications for 2,70,000 shares were received. Applications for 90,000 shares were rejected and money refunded. Shares were allotted on pro-rata basis to the remaining applicants. The first and final call was made. The amount was duly received except on 1,800 shares applied by Govind. His shares were forfeited. The forfeited shares were re-issued at Rs 8 per share fully paid-up.
GY Ltd. invited applications for issuing 85,000 equity shares of Rs 10 each at a discount of 10%. The amount was payable as follows:
Application for 2,00,000 shares were received. Applications for 30,000 shares were rejected and money refunded. Shares were allotted on pro-rata basis to the remaining applicants. The first and final call was made. All money was received except on 1,700 shares applied by Hari. His shares were forfeited. The forfeited shares were re-issued at the maximum discount permissible under the law.
Why is 'Cash Flow Statement' prepared? State.
(vi) Loose tools VIEW SOLUTION
From the following Statement of Profit and Loss of Ajanta Ltd. for the year ended 31st March, 2013, prepare a Comparative Statement of Profit and Loss : |
Euler's Formula | Brilliant Math & Science Wiki
Hamza A, Sandeep Bhardwaj, A Former Brilliant Member, and
neelesh vij
In complex analysis, Euler's formula provides a fundamental bridge between the exponential function and the trigonometric functions. For complex numbers
x
, Euler's formula says that
e^{ix} = \cos{x} + i \sin{x}.
In addition to its role as a fundamental mathematical result, Euler's formula has numerous applications in physics and engineering.
A straightforward proof of Euler's formula can be had simply by equating the power series representations of the terms in the formula:
\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots
\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots,
\begin{aligned} \cos{x} + i \sin{x} &= 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots \\ &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \cdots \\ &= e^{ix}. \end{aligned}
x
is complex. Then
e^{ix} = \cos{x} + i \sin{x}.
e^{i \pi}
e^{i \pi} = \cos{\pi} + i \sin{\pi} = -1,
which leads to the very famous Euler's identity:
e^{i \pi} + 1 = 0.\ _\square
i^i
\forall k \in \mathbb{N}
\begin{aligned} e^{i (\pi / 2+2\pi k)} &= i \\ \Rightarrow \left(e^{i (\pi / 2+2\pi k)}\right)^i &= i^i. \end{aligned}
i^i = e^{i^2 (\pi / 2+2\pi k)} = e^{- \pi / 2-2\pi k}.\ _\square
Note: This means that
i^i
is not a well defined (unique) quantity. To remedy this, one needs to specify a branch cut. For example, we can define the argument of
e^{i \theta}
to be defined for
\theta \in [0, 2\pi)
, in which case we have that
i^i = e^{-\pi / 2}
. That is, this forces
k = 0
. Of course, different branch cut can be chosen yielding different values for
k
Euler's formula allows for any complex number
x
to be represented as
e^{ix}
, which sits on a unit circle with real and imaginary components
\cos{x}
\sin{x}
, respectively. Various operations (such as finding the roots of unity) can then be viewed as rotations along the unit circle.
One immediate application of Euler's formula is to extend the definition of the trigonometric functions to allow for arguments that extend the range of the functions beyond what is allowed under the real numbers.
A couple useful results to have at hand are the facts that
e^{-ix} = \cos{x} - i \sin{x},
e^{ix} + e^{-ix} = 2 \cos{x}.
\cos{x} = \frac{e^{ix} + e^{-ix}}{2},
\sin{x} = \frac{e^{ix} - e^{-ix}}{2i}
\tan{x} = \frac{e^{ix} - e^{-ix}}{i(e^{ix} + e^{-ix})}.
\cos{x} = 2
in the complex numbers.
We first note that if
x = x_0
is a solution, then so is
x = 2\pi k \pm x_0
k
\cos x
is an even function with a fundamental period of
2\pi
\cos{x} = \dfrac{e^{ix} + e^{-ix}}{2}
\begin{aligned} e^{ix}+e^{-ix} &= 4 \\ \left(e^{ix}\right)^2-4e^{ix}+1 &= 0 \\ e^{ix} &= 2\pm \sqrt{3} \\ \Rightarrow x &= \dfrac1i \ln \left (2 \pm \sqrt 3\right) \\ & =- i \ln \left (2 \pm \sqrt 3\right). \end{aligned}
x = 2 \pi k \pm i \ln \left ( 2\pm \sqrt{3} \right), 2 \pi k \mp i \ln \left ( 2\pm \sqrt{3} \right)
k
\ _\square
\frac\pi2 \pm i \ln\big(2 + \sqrt{3}\big)
- \frac\pi4 \pm i \ln\big(2 + \sqrt{3}\big)
\frac\pi4 \pm i \ln\big(2 + \sqrt{3}\big)
\frac\pi2 \pm i \ln{3}
\frac\pi4 \pm i \ln{3}
\sin{x} = 2
in the complex numbers?
Euler’s formula also allows for the derivation of several trigonometric identities quite easily. Starting with
e^{i(x \pm y)} = \cos(x \pm y) + i \sin(x \pm y),
\begin{aligned} e^{i(x \pm y)} &= e^{ix} e^{\pm iy} \\ &= (\cos{x} + i\sin{x})(\cos{y} \pm i \sin{y}) \\ &= \cos{x} \cos{y} \mp \sin{x} \sin{y} + i(\sin{x} \cos{y} \pm \cos{x} \sin{y}). \end{aligned}
Equating the real and imaginary parts, respectively, yields the familiar sum and difference formulas
\cos(x \pm y) = \cos{x} \cos{y} \mp \sin{x} \sin{y}
\sin(x \pm y) = \sin{x} \cos{y} \pm \cos{x} \sin{y}.
An important corollary of Euler's theorem is de Moivre's theorem.
(\cos{x} + i \sin x)^\phi = \cos{\phi x} + i \sin{\phi x}
\begin{aligned} (\cos{a\phi}+i\sin{a\phi}) \times (\cos{b\phi}+i\sin{b\phi}) &=\ \cos{a\phi}\cos{b\phi}+i\cos{a\phi}\sin{b\phi}+i\sin{a\phi}\cos{b\phi}-\sin{a\phi}\sin{b\phi} \\ (\cos{a\phi}+i\sin{a\phi})(\cos{b\phi}+i\sin{b\phi}) &= \cos{a\phi}\cos{b\phi}-\sin{a\phi}\sin{b\phi}+i(\cos{a\phi}\sin{b\phi}+\sin{a\phi}\cos{b\phi}) \\ \Rightarrow \cos{(a\phi+b\phi)}+i\sin{(a\phi+b\phi)} &= \cos{\big((a+b)\phi\big)}+i\sin{\big((a+b)\phi\big)}. \end{aligned}
a = b
(\cos{a\phi}+i\sin{a\phi})(\cos{a\phi}+i\sin{a\phi}) = \cos{(a\phi+a\phi)}+i\sin{(a\phi+a\phi)} = \cos{(2a\phi)}+i\sin{(2a\phi)}.
\begin{aligned} (\cos{a\phi}+i\sin{a\phi})^n &=\ \underbrace{(\cos{a\phi}+i\sin{a\phi}) \times \cdots \times (\cos{a\phi}+i\sin{a\phi})}_{n\text{ times}} \\ &= \cos{\big(\underbrace{(a+\cdots +a)}_{n\ a\text{'s}}\phi\big)}+i\sin{\big(\underbrace{(a+\cdots +a)}_{n\ a\text{'s}}\phi\big)} \\ \Rightarrow (\cos{a\phi}+i\sin{a\phi})^n &= \cos{(na\phi)}+i\sin{(na\phi)}. \end{aligned}
(\cos{x} + i \sin x)^\phi = e^{ix\phi} = e^{i(\phi x)} = \cos{\phi x} + i \sin{\phi x}.\ _\square
De Moivre's theorem has many applications. As an example, one may wish to compute the roots of unity, or the complex solution set to the equation
x^n = 1
n
e^{2\pi ki}
1
k
an integer, so the
n^\text{th}
roots of unity must be
e^{2\pi ki / n} = \cos\left(\frac{2\pi k}n\right) + i \sin\left(\frac{2 \pi k}n\right).
This process is akin to dividing the unit circle up into
n
equally spaced wedges.
Find the cube roots of unity.
The cube roots of unity are
\cos\left(\frac{2 \pi }{3}\right) + i \sin\left(\frac{2 \pi }{3}\right) = -\frac{1}{2} + i \frac{\sqrt{3}}{2}
\cos\left(\frac{4 \pi }{3}\right) + i \sin\left(\frac{4 \pi }{3}\right) = -\frac{1}{2} - i \frac{\sqrt{3}}{2}
\cos\left(\frac{6 \pi }{3}\right) + i \sin\left(\frac{6 \pi }{3}\right) = 1.\ _\square
z = \left[ \sin\left(\frac{\pi}{2014}\right) + i \cos\left(\frac{\pi}{2014}\right)\right]^{2014},
\dfrac1{z^{2014}}
14^\text{th}
Cite as: Euler's Formula. Brilliant.org. Retrieved from https://brilliant.org/wiki/eulers-formula/ |
Erratum to: DNA methylation and transcriptional noise | Epigenetics & Chromatin | Full Text
Erratum to: DNA methylation and transcriptional noise
After the publication of this work [1] it was brought to the authors’ attention that there was an error in the equation (1). The last ‘y’ term in the equation should have one bar on top instead of two bars. Please see below for the correctly displayed equation.
{\sum _{\mathit{i}=1}^{2}\sum _{\mathit{j}=1}^{2}\left({\mathit{y}}_{\mathit{ij}}-\overline{\overline{\mathit{y}}}\right)}^{2}=2×{\sum _{\mathit{i}=1}^{2}\left({\overline{\mathit{y}}}_{\mathit{i}}-\overline{\overline{\mathit{y}}}\right)}^{2}+{\sum _{\mathit{i}=1}^{2}\sum _{\mathit{j}=1}^{2}\left({\mathit{y}}_{\mathit{ij}}-{\overline{\mathit{y}}}_{\mathit{i}}\right)}^{2}
Huh I, Zeng J, Park T, Yi SV: DNA methylation and transcriptional noise. Epigenetics & Chromatin. 2014, 6: 9.
Department of Statistics, Bioinformatics and Biostatistics Laboratory, Interdisciplinary Program in Bioinformatics, Seoul National University, Seoul, 151-742, South Korea
Huh, I., Zeng, J., Park, T. et al. Erratum to: DNA methylation and transcriptional noise. Epigenetics & Chromatin 7, 13 (2014). https://doi.org/10.1186/1756-8935-7-13 |
stats(deprecated)/importdata - Maple Help
Home : Support : Online Help : stats(deprecated)/importdata
Read Statistical Data from a File
stats[importdata](filename, n)
importdata(filename, n)
name of the file to be read
(optional, default=1) number of streams into which to split the file data
The function importdata of the stats package reads statistical data from a file.
The data in the file are just a sequence of numbers, separated by spaces or the end of line. They will be processed by a sscanf(..., `%f`).
The character # introduces a comment that ends at the end of that line.
Missing data are represented in the data file by the * character. It will be converted to the keyword missing.
If the number of streams, which is indicated by the parameter n, is one, the data in the file given by filename are returned as an expression sequence. If the number of streams is greater than one then the data in the file are returned as an expression sequence of n lists. The first item in the file goes into the first list, the second item goes into the second list, and so on, until there is an item in each list. The next item in the file then goes at the end of the first list, and so on until the whole file is imported.
More sophisticated data files can be read using readline and sscanf. The function {readdata} is very similar to stats[importdata].
The command with(stats,importdata) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{stats}\right):
T≔\mathrm{importdata}\left(\mathrm{datafile},2\right)
where the datafile could contain the following ------------------------
| 6 * # one missing data
| # another comment
At the end T will be
\mathrm{T_received}≔[2.0,4.0,6.0],[3.0,5.0,'\mathrm{missing}']
transform(deprecated)[split] |
Interpolate 2-D or 3-D scattered data - MATLAB griddata - MathWorks Italia
Interpolate Scattered Data Over Uniform Grid
Interpolate 4-D Data Set Over Grid
Comparison of Scattered Data Interpolation Methods
vq = griddata(x,y,v,xq,yq) fits a surface of the form v = f(x,y) to the scattered data in the vectors (x,y,v). The griddata function interpolates the surface at the query points specified by (xq,yq) and returns the interpolated values, vq. The surface always passes through the data points defined by x and y.
vq = griddata(x,y,z,v,xq,yq,zq) fits a hypersurface of the form v = f(x,y,z).
vq = griddata(___,method) specifies the interpolation method used to compute vq using any of the input arguments in the previous syntaxes. method can be 'linear', 'nearest', 'natural', 'cubic', or 'v4'. The default method is 'linear'.
[Xq,Yq,vq] = griddata(x,y,v,xq,yq) and [Xq,Yq,vq] = griddata(x,y,v,xq,yq,method) additionally return Xq and Yq, which contain the grid coordinates for the query points.
Interpolate randomly scattered data on a uniform grid of query points.
Sample a function at 200 random points between -2.5 and 2.5.
x, y, and v are vectors containing scattered (nonuniform) sample points and data.
Define a regular grid and interpolate the scattered data over the grid.
Plot the gridded data as a mesh and the scattered data as dots.
Interpolate a 3-D slice of a 4-D function that is sampled at randomly scattered points.
Sample a 4-D function
\mathit{v}\left(\mathit{x},\mathit{y},\mathit{z}\right)
at 2500 random points between -1 and 1. The vectors x, y, and z contain the nonuniform sample points.
Define a regular grid with xy points in the range [-1, 1], and set
\mathit{z}=0
. Interpolating on this grid of 2-D query points (xq,yq,0) produces a 3-D interpolated slice (xq,yq,0,vq) of the 4-D data set (x,y,z,v).
Interpolate the scattered data on the grid. Plot the results.
Compare the results of several different interpolation algorithms offered by griddata.
Create a grid of query points.
Interpolate the sample data using the 'nearest', 'linear', 'natural', and 'cubic' methods. Plot the results for comparison.
x, y, z — Sample point coordinates
Sample point coordinates, specified as vectors. Corresponding elements in x, y, and z specify the xyz coordinates of points where the sample values v are known. The sample points must be unique.
Sample values, specified as a vector. The sample values in v correspond to the sample points in x, y, and z.
If v contains complex numbers, then griddata interpolates the real and imaginary parts separately.
xq, yq, zq — Query points
Query points, specified as vectors or arrays. Corresponding elements in the vectors or arrays specify the xyz coordinates of the query points. The query points are the locations where griddata performs interpolation.
Specify arrays if you want to pass a grid of query points. Use ndgrid or meshgrid to construct the arrays.
Specify vectors if you want to pass a collection of scattered points.
The specified query points must lie inside the convex hull of the sample data points. griddata returns NaN for query points outside of the convex hull.
'linear' (default) | 'nearest' | 'natural' | 'cubic' | 'v4'
Interpolation method, specified as one of the methods in this table.
'linear' Triangulation-based linear interpolation (default) supporting 2-D and 3-D interpolation. C0
'nearest' Triangulation-based nearest neighbor interpolation supporting 2-D and 3-D interpolation. Discontinuous
'natural' Triangulation-based natural neighbor interpolation supporting 2-D and 3-D interpolation. This method is an efficient tradeoff between linear and cubic. C1 except at sample points
'cubic' Triangulation-based cubic interpolation supporting 2-D interpolation only. C2
Biharmonic spline interpolation (MATLAB® 4 griddata method) supporting 2-D interpolation only. Unlike the other methods, this interpolation is not based on a triangulation.
Interpolated values, returned as a vector or array. The size of vq depends on the size of the query point inputs xq, yq, and zq:
For 2-D interpolation, where xq and yq specify an m-by-n grid of query points, vq is an m-by-n array.
For 3-D interpolation, where xq, yq, and zq specify an m-by-n-by-p grid of query points, vq is an m-by-n-by-p array.
If xq, yq, (and zq for 3-D interpolation) are vectors that specify scattered points, then vq is a vector of the same length.
For all interpolation methods other than 'v4', the output vq contains NaN values for query points outside the convex hull of the sample data. The 'v4' method performs the same calculation for all points regardless of location.
Xq, Yq — Grid coordinates for query points
Grid coordinates for query points, returned as vectors or matrices. The shape of Xq and Yq depends on how you specify xq and yq:
If you specify xq as a row vector and yq as a column vector, then griddata uses those grid vectors to form a full grid with [Xq,Yq] = meshgrid(xq,yq). In this case, the Xq and Yq outputs are returned as matrices that contain the full grid coordinates for the query points.
If xq and yq are both row vectors or both column vectors, then Xq = xq and Yq = yq.
Scattered data interpolation with griddata uses a Delaunay triangulation of the data, so can be sensitive to scaling issues in x, y, and z. When this occurs, you can use normalize to rescale the data and improve the results. See Normalize Data with Differing Magnitudes for more information. |
Analyses usually entail the application of various tools, algorithms and scripts.
handles boilerplate:
Snakemake infers dependencies and execution order.
Python + domain specific syntax
Decompose workflow into rules.
Rules define how to obtain output files from input files.
"path/to/dataset.txt"
"dataset.sorted.txt"
"sort {input} > {output}"
refer to input and output from shell command
"{dataset}.sorted.txt"
generalize rules with
rule sort_and_annotate:
"path/to/{dataset}.txt",
"path/to/annotation.txt"
"paste <(sort {input[0]}) {input[1]} > {output}"
multiple input or output files
refer by index
a="path/to/{dataset}.txt",
b="path/to/annotation.txt"
"paste <(sort {input.a}) {input.b} > {output}"
name input and output files
refer by name
a="path/to/{dataset}.txt"
b="{dataset}.sorted.txt"
with open(output.b, "w") as out:
for l in sorted(open(input.a)):
print(l, file=out)
use Python in rules
refer to Python or R scripts
(in version 3.5)
Dependencies are determined top-down
For a given target, a rule that can be applied to create it is determined (a job).
For the input files of the rule, go on recursively.
If no target is specified, Snakemake tries to apply the first rule in the workflow.
DATASETS = ["D1", "D2", "D3"]
expand("{dataset}.sorted.txt", dataset=DATASETS)
Job 1: apply rule all
(a target rule that just collects results)
Job i: apply rule sort to create i-th input of job 1
use arbitrary Python code in your workflow
Directed acyclic graph (DAG) of jobs
A job is executed if and only if
output file is target and does not exist
output file needed by another executed job and does not exist
input file newer than output file
input file will be updated by other job
execution is enforced
determined via breadth-first-search on DAG of jobs
Assumption: workflow defined in a Snakefile in the same directory.
# execute the workflow with target D1.sorted.txt
snakemake D1.sorted.txt
# execute the workflow without target: first rule defines target
# dry-run, print shell commands
snakemake -n -p
# dry-run, print execution reason for each job
snakemake -n -r
# visualize the DAG of jobs using the Graphviz dot command
snakemake --dag | dot -Tsvg > dag.svg
Disjoint paths in the DAG of jobs can be executed in parallel.
# execute the workflow with 8 cores
execute 8 jobs in parallel?
resources: mem_mb=100
"sort --parallel {threads} {input} > {output}"
refer to defined thread number
define arbitrary additional resources
define used threads
# prioritize the creation of a certain file
snakemake --prioritize D1.sorted.txt --cores 8
# execute the workflow with 8 cores and 100MB memory
snakemake --cores 8 --resources mem_mb=100
can execute 2 sort jobs in parallel
can execute only 1 sort job in parallel
Available jobs are scheduled to
maximize parallelization
prefer high priority jobs
while satisfying resource constraints.
\max_{E \subseteq J} \sum_{j \in E}\, (p_j, d_j, i_j)^T
\max_{E \subseteq J} \sum_{j \in E}\, (p_j, d_j, i_j)^T
\sum_{j \in E} r_{ij} \leq R_i \text{ for } i=1,2,...,n
\sum_{j \in E} r_{ij} \leq R_i \text{ for } i=1,2,...,n
free resource (e.g. CPU cores)
expand("{dataset}.sorted.txt", dataset=config["datasets"])
define config file
refer to config values
Workflows are executed in three phases
initialization phase (parsing)
DAG phase (DAG is built)
scheduling phase (execution of DAG)
Input functions defer determination of input files to the DAG phase
(when wildcard values are known).
lambda wildcards: config["datasets"][wildcards.dataset]
input functions take the determined wildcard values as only argument
"logs/sort/{dataset}.log"
"sort --parallel {threads} {input} > {output} 2> {log}"
define log file
refer to log file from shell command
# execute the workflow on cluster with qsub submission command
# (and up to 100 parallel jobs)
snakemake --cluster qsub --jobs 100
# tell the cluster system about the used threads
snakemake --cluster "qsub -pe threaded {threads}" --jobs 100
# execute the workflow with synchronized qsub
snakemake --cluster-sync "qsub -sync yes" --jobs 100
# execute the workflow with DRMAA
snakemake --drmaa --jobs 100
handling of temporary and protected files
HTML5 reports
tracking of tool versions and code changes
per file data provenance information
a Python API for embedding Snakemake in other tools
Distribution of Snakemake workflows
Solution 1: Git repository with
│ ├── script1.py
│ └── script2.R
# clone workflow into working directory
git clone https://bitbucket.org/user/myworkflow.git path/to/workdir
cd path/to/workdir
# edit config and workflow as needed
# install dependencies into isolated environment
conda create -n myworkflow --file requirements.txt
source activate myworkflow
Solution 2: Python/Conda package
# install workflow with all dependencies into isolated environment
conda create -n myworkflow myworkflow
# copy Snakefile and config file into working directory
myworkflow init path/to/workdir
Solution 3: Hide workflow in package
# copy only config file into working directory
# edit config as needed
# execute workflow with a wrapper that uses a Snakefile
# hidden in the package and delegates execution to Snakemake
myworkflow run -n
Sven Rahmann, Universität Duisburg-Essen
Christopher Schröder, Universität Duisburg-Essen
Marcel Martin, SciLifeLab Stockholm
Tobias Marschall, Max Planck Institute for Informatics
Sean Davis, NIH
David Koppstein, MIT
Ryan Dale, NIH
Chris Tomkins-Tinch, Broad Institute
Hyeshik Chang, Seoul National University
Karel Brinda, Université Paris-Est Marne-la-Vallée
Anthony Underwood, Public Health England
Elias Kuthe, TU Dortmund
Paul Moore, Atos SE
Mattias Frånberg, Karolinska Institute
Simon Ye, MIT
Willem Ligtenberg, Open Analytics
Per Unneberg, SciLifeLab Stockholm
Matthew Shirley, Johns Hopkins School of Medicine
Jeremy Leipzig, Childrens Hospital of Philadelphia
all users and supporters
https://bitbucket.org/johanneskoester/snakemake
Köster, Johannes and Rahmann, Sven. "Snakemake - A scalable bioinformatics workflow engine". Bioinformatics 2012.
Köster, Johannes. "Parallelization, Scalability, and Reproducibility in Next-Generation Sequencing Analysis", PhD thesis, TU Dortmund 2014.
Tutorial slides for GCB 2015
Snakemake Poster |
Map integer symbols from one coding scheme to another - Simulink - MathWorks Switzerland
Symbol set size (M)
Mapping vector
Map integer symbols from one coding scheme to another
The Data Mapper block accepts integer inputs and maps them to integer outputs. The mapping types include: binary to Gray coded, Gray coded to binary, and user defined. Additionally, a pass through option is available.
Gray coding is an ordering of binary numbers such that all adjacent numbers differ by only one bit.
Input signal, specified as a scalar, vector, or matrix of integers. Elements of the input signal must be nonnegative values. The block truncates noninteger values to integer values. When the input is a matrix, the columns are treated as independent channels.
Output signal, returned as a scalar, column vector, or matrix. The dimensions of the output signal match those of the input signal.
Mapping mode — Mapping mode
Binary to Gray (default) | Gray to Binary | User Defined | Straight through
Mapping mode, specified as one of the four options. The mapping for the Binary to Gray and the Gray to Binary modes are shown in the following table when the inputs range from 0 to 7.
Binary to Gray Mode
Gray to Binary Mode
0 0 (000) 0 (000) 0
When you select the User Defined mode, you can use any arbitrary mapping by providing a vector to specify the output ordering. When you select the Straight Through mode, the output equals the input.
Symbol set size (M) — Symbol set size
Symbol set size, specified as a positive integer. This parameter restricts the inputs and outputs to integers in the range of 0 to M-1.
Mapping vector — Maps input elements to the output elements
[0 1 3 2 7 6 4 5] (default) | vector
Mapping vector, specified as vector of nonnegative integers whose length equals . This parameter defines the relationship between the input and output integers. For example, the vector [1 5 0 4 2 3] defines the following mapping:
\begin{array}{l}0\to 1\hfill \\ 1\to 5\hfill \\ 2\to 0\hfill \\ 3\to 4\hfill \\ 4\to 2\hfill \\ 5\to 3\hfill \end{array}
Binary to Gray Conversion in Simulink
Use the Open model button to open the Binary-to-Gray model. The model converts a binary sequence to a Gray-coded sequence and vice versa by using Data Mapper blocks.
bin2gray | gray2bin |
Compound annual growth rate (CAGR) is a business and investing specific term for the geometric progression ratio that provides a constant rate of return over the time period.[1][2] CAGR is not an accounting term, but it is often used to describe some element of the business, for example revenue, units delivered, registered users, etc. CAGR dampens the effect of volatility of periodic returns that can render arithmetic means irrelevant. It is particularly useful to compare growth rates from various data sets of common domain such as revenue growth of companies in the same industry or sector.[3]
CAGR is equivalent to the more generic exponential growth rate when the exponential growth interval is one year.
CAGR is defined as:
{\displaystyle \mathrm {CAGR} (t_{0},t_{n})=\left({\frac {V(t_{n})}{V(t_{0})}}\right)^{\frac {1}{t_{n}-t_{0}}}-1}
{\displaystyle V(t_{0})}
is the initial value,
{\displaystyle V(t_{n})}
is the end value, and
{\displaystyle t_{n}-t_{0}}
is the number of years.
Actual or normalized values may be used for calculation as long as they retain the same mathematical proportion.
In this example, we will compute the CAGR over a three-year period. Assume that the year-end revenues of a business over a three-year period,
{\displaystyle V(t)}
, have been:
Year-End Revenue 9,000 13,000
Therefore, to calculate the CAGR of the revenues over the three-year period spanning the "end" of 2004 to the "end" of 2007 is:
{\displaystyle {\rm {CAGR}}(0,3)=\left({\frac {13000}{9000}}\right)^{\frac {1}{3}}-1=0.13=13\%}
Note that this is a smoothed growth rate per year. This rate of growth would take you to the ending value, from the starting value, in the number of years given, if growth had been at the same rate every year.
Multiply the initial value (2004 year-end revenue) by (1 + CAGR) three times (because we calculated for 3 years). The product will equal the year-end revenue for 2007. This shows the compound growth rate:
{\displaystyle V(t_{n})=V(t_{0})\times (1+{\rm {CAGR}})^{n}}
{\displaystyle =V(t_{0})\times (1+{\rm {CAGR}})\times (1+{\rm {CAGR}})\times (1+{\rm {CAGR}})}
{\displaystyle =9000\times 1.1304\times 1.1304\times 1.1304=13000}
the Arithmetic Mean Return (AMR) would be the sum of annual revenue changes (compared with the previous year) divided by number of years, or:
{\displaystyle {\text{AMR}}={\bar {x}}={\frac {1}{n}}\sum _{i=1}^{n}x_{i}={\frac {1}{n}}(x_{1}+\cdots +x_{n})={\frac {11.11\%+10\%+8.33\%}{3}}=9.81\%.}
In contrast to CAGR, you cannot obtain
{\displaystyle V(t_{n})}
by multiplying the initial value,
{\displaystyle V(t_{0})}
, three times by (1 + AMR) (unless all annual growth rates are the same).
the arithmetic return (AR) or simple return would be the ending value minus beginning value divided by the beginning value:
{\displaystyle {\text{AR}}={\frac {V(t_{n})-V(t_{0})}{V(t_{0})}}={\frac {13000-9000}{9000}}=44.44\%.}
These are some of the common CAGR applications:
Calculating and communicating the average returns of investment funds[4]
Demonstrating and comparing the performance of investment advisors[4]
Comparing the historical returns of stocks with bonds or with a savings account[4]
Forecasting future values based on the CAGR of a data series (you find future values by multiplying the last datum of the series by (1 + CAGR) as many times as years required). As with every forecasting method, this method has a calculation error associated.
Analyzing and communicating the behavior, over a series of years, of different business measures such as sales, market share, costs, customer satisfaction, and performance.
^ Mark J. P. Anson; Frank J. Fabozzi; Frank J. Jones (3 December 2010). The Handbook of Traditional and Alternative Investment Vehicles: Investment Characteristics and Strategies. John Wiley & Sons. pp. 489–. ISBN 978-1-118-00869-0.
^ root. "Compound Annual Growth Rate (CAGR) Definition | Investopedia". Investopedia. Retrieved 2016-03-04.
^ Emily Chan (27 November 2012). Harvard Business School Confidential: Secrets of Success. John Wiley & Sons. pp. 185–. ISBN 978-1-118-58344-9.
^ a b c "Compound Annual Growth Rate CAGR: Summary and Forum". www.12manage.com. Retrieved 2019-05-02.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Compound_annual_growth_rate&oldid=1080213512" |
Is the acceleration of an object at rest zero? | Brilliant Math & Science Wiki
Rohit Gupta, Tim O'Brien, Josh Silverman, and
Our basic question is: if an object is at rest, is its acceleration necessarily zero?
For example, if a car sits at rest its velocity is, by definition, equal to zero. But what about its acceleration?
To answer this question, we will need to look at what velocity and acceleration really mean in terms of the motion of an object. We will use both conceptual and mathematical analyses to determine the correct answer: the object's acceleration is not necessarily zero just because its velocity is zero. This may seem strange at first, but if we unpack it a bit, it should start to make sense.
Why do some people say it's zero?
Why acceleration is not necessarily zero (logical)
Why acceleration is not necessarily zero (quantitative)
If we think about the problem quickly, it might seem the acceleration must be zero. At one moment, we're not moving, and a small time later we're still not moving, so there has not been a change in speed. Therefore, the acceleration has to be zero.
There is a mistake here that we can see without doing any calculations.
It's clear that on a regular basis, objects that start out at rest end up in motion. For example, a person standing up from a chair or a plane taking off from a runway. In these cases, there is a clear change from zero velocity to non-zero velocity even though the object starts out at rest. This implies that there must be a moment where the object's acceleration is non-zero although the object remains in the same position.
That was a logical argument for why acceleration in a state of rest must be possible. We can do a better job with a rigorous quantitative argument.
Let's start by looking at the object's initial velocity, and confirm that it must be zero. When an object starts from rest (at
x(0)=0
) and starts to accelerate at the rate
\gamma
, its position a time
\Delta T
later is
x(\Delta T) = \frac12\gamma\left(\Delta T\right)^2.
With this in hand, we can do a straightforward calculation of the velocity at time zero,
v(0) = \lim_{\Delta T \rightarrow 0} \Delta x/\Delta T
\begin{aligned} v(0) &= \lim_{\Delta T\rightarrow 0} \frac{\frac12\gamma\left(\Delta T\right)^2 - 0}{\Delta T} \\ &= \lim_{\Delta T\rightarrow 0} \frac12\frac{\gamma\left(\Delta T\right)^2}{\Delta T} \\ &= \lim_{\Delta T\rightarrow 0} \frac12 \gamma \Delta T \\ &= 0 \end{aligned}
So that the initial velocity is zero, like we supposed.
Finding the acceleration
Now, let's look at our object's acceleration over the time from the beginning of its acceleration, and a time
\Delta T
later. At time
t = \Delta T,
its velocity has increased to
\gamma \Delta T
\begin{aligned} a(0) &= \lim_{\Delta T\rightarrow 0} \frac{v(\Delta T)- v(0)}{\Delta T} \\ &= \lim_{\Delta T\rightarrow 0} \frac{\gamma \Delta T - 0}{\Delta T} \\ &= \gamma, \end{aligned}
which is exactly what we expected to find.
Thus, even though the velocity of an object at rest must be zero, acceleration can clearly be non-zero for objects at rest.
Much more than the acceleration due to gravity
g
. Almost equal to the acceleration due to gravity
g
. Much less than the acceleration due to gravity
g
. Zero
A particle that is thrown vertically upwards stops momentarily at the highest point of motion. What is the acceleration of the particle at the highest point of motion?
Assume the size of the particle to be negligible.
Cite as: Is the acceleration of an object at rest zero?. Brilliant.org. Retrieved from https://brilliant.org/wiki/is-the-acceleration-of-an-object-at-rest-zero/
Learn more in our Physics of the Everyday course, built by experts for you. |
Backers - Goldfinch Docs | Goldfinch
Backers evaluate Borrowers and supply first-loss capital on their Borrower Pools. Backers can achieve higher returns when the Senior Pool leverages them with additional senior tranche capital. Currently, Backer rewards and Backer staking are not yet live, but it is expected that the community will introduce and vote on a proposal for these in the coming months.
Backers look at Borrower Pools as investment opportunities. They evaluate the information Borrowers provide and decide if they want to supply capital to the junior tranche of a Borrower Pool.
The Senior Pool provides additional senior tranche capital to the Borrower Pool according to the Leverage Model. To account for the lower risk of the senior tranche, 20% of the senior tranche’s nominal interest is reallocated to the junior tranche. In addition, the protocol retains 10% of all interest payments as reserves, which are managed by the decentralized Governance.
As a result, the Senior Pool earns an effective interest rate equal to 70% of the nominal interest rate. Or, in terms of the nominal interest rate,
i_{n}
, protocol reserve allocation,
p
, and junior reallocation percent,
j
i_{senior}=i_n*(1-p-j)
Accordingly, based on these same inputs and the leverage ratio,
r
, Backers receive an effective interest rate of:
i_{junior}=i_n*(1-p+r*j)
For example: Consider a Borrower Pool with a 15% interest rate and 4.0X leverage ratio. If the Backers supply $200K, the Senior Pool will allocate another $800K. Assuming the Borrower borrows the full $1M for one year, they will pay $1M * 15% = $150K in interest. Of that, the Senior Pool receives 0.15*(1 - 0.1 - 0.2) = 10.5% interest, or $800K * 0.105 = $84K. The Backers receive 0.15*(1 - 0.1 + 4*0.2) = 25.5% interest, or $200K * 0.255 = $51K. The remaining $15K is the 10% protocol reserve allocation.
It is easier to feel confident supplying to a Borrower Pool when a lot of other Backers are already supplying to it and the Senior Pool is already adding leverage. It is riskier to be the first one in a Borrower Pool. To incentivize Backers to supply early on, the protocol provides an additional GFI reward to all Backers who contribute early on, with the reward amount decreasing for later Backers as the Borrower Pool reaches its limit.
The protocol assigns the reward when a Backer supplies, but the reward is not immediately claimable. The percent of the reward that is claimable is proportional to the percentage of the full expected repayment of principal plus interest that the Borrower successfully repays. This ensures the Backer only receives the early Backer reward after the Borrower Pool proves valuable to the protocol.
In addition to evaluating individual Borrower Pools, Backers may also evaluate other Backers in order to give them leverage. Backers can do this by staking GFI directly on another Backer.
Based on the amount of GFI staked on a given Backer, the Senior Pool uses the Leverage Model to calculate a leverage ratio and allocate capital whenever that Backer supplies to Borrower Pools. For example, if a Backer has a leverage ratio of 4.0X based on who has staked GFI on them, then anytime they supply to a Borrower Pool, the Senior Pool will allocate 4.0X of that amount.
When GFI is staked on a Backer, that GFI serves as collateral against potential defaults for that Backer’s positions in Borrower Pools. When a Borrower defaults, the GFI staked on all the Backers in that pool are reallocated to the senior tranche until the senior tranche is made whole on their expected payments. This incentives Backers to stake on other Backers who supply to safe Borrower Pools.
Backers have an incentive to provide first-loss capital to Borrower Pools because they can receive both early Backer rewards and higher effective yields based on the Senior Pool leverage. They also have an incentive to stake GFI on other Backers because they can earn additional rewards when that Backer supplies to Borrower Pools |
Hockey Stick Identity | Brilliant Math & Science Wiki
Andy Hayes, Rohit Udaiwal, Mateo Matijasevick, and
The hockey stick identity is an identity regarding sums of binomial coefficients.
For whole numbers
n
r\ (n \ge r),
\sum_{k=r}^{n}\binom{k}{r} = \binom{n+1}{r+1}. \ _\square
The hockey stick identity gets its name by how it is represented in Pascal's triangle.
In Pascal's triangle, the sum of the elements in a diagonal line starting with
1
is equal to the next element down diagonally in the opposite direction. Circling these elements creates a "hockey stick" shape:
1+3+6+10=20.
The hockey stick identity is a special case of Vandermonde's identity. It is useful when a problem requires you to count the number of ways to select the same number of objects from different-sized groups. It is also useful in some problems involving sums of powers of natural numbers.
Sums of Powers of Natural Numbers
The hockey stick identity is often used in counting problems in which the same amount of objects is selected from different-sized groups.
Treating each of the balls in the figure below as distinct, how many ways are there to select 3 balls from the same horizontal row?
The smallest row has 3 balls and the largest row has 9 balls. The number of ways to select 3 balls from the same row can be expressed as a sum of binomial coefficients. This can then be computed with the hockey stick identity:
\sum\limits_{k=3}^{9}\binom{k}{3} = \binom{10}{4} = 210.
\color{#D61F06}{210}
ways to select 3 balls from the same row.
_\square
Mabel is playing a carnival game in which she throws balls into a triangular array of cups.
She will throw three balls, and she will win the game if the three balls are in a straight line (not necessarily adjacent) and in different cups.
Disregarding the order the balls go into the cups, how many ways can she win the game?
You might have noticed that Pascal's triangle contains all of the positive integers in a diagonal line.
Each of these elements corresponds to the binomial coefficient
\binom{n}{1},
n
is the row of Pascal's triangle. The sum of all positive integers up to
n
n^\text{th}
triangular number. It can be represented as
\sum\limits_{k=1}^{n}{k}=\sum\limits_{k=1}^{n}\binom{k}{1}.
These two expressions are equivalent because
k=\binom{k}{1}.
As this sum can be expressed as the sum of binomial coefficients, it can be computed with the hockey stick identity:
n
positive integers is
\sum\limits_{k=1}^{n}{k}=\sum\limits_{k=1}^{n}\binom{k}{1}=\binom{n+1}{2}.\ _\square
This leads to the more well-known formula for triangular numbers.
n
\sum\limits_{k=1}^{n}{k}=\binom{n+1}{2}=\frac{(n+1)!}{(n-1)!(2)!}=\frac{n(n+1)}{2}.\ _\square
Since each triangular number can be represented with a binomial coefficient, the hockey stick identity can be used to calculate the sum of triangular numbers.
n
triangular numbers is
\sum\limits_{k=1}^{n}\sum\limits_{j=1}^{k}{j}=\sum\limits_{k=1}^{n}\binom{k+1}{2}=\binom{n+2}{3}.\ _\square
This can also be represented algebraically.
n
\sum\limits_{k=1}^{n}\sum\limits_{j=1}^{k}{j}=\binom{n+2}{3}=\frac{(n+2)!}{(n-1)!(3)!}=\frac{n(n+1)(n+2)}{6}.\ _\square
100^{\text{th}}
The hockey stick identity can be used to develop the identities for sums of powers of natural numbers.
The sum of the squares of the first
n
\sum\limits_{k=1}^{n}{k^2}=\frac{n(n+1)(2n+1)}{6}=2\binom{n+2}{3}-\binom{n+1}{2}.\ _\square
Recall from the previous example that the identities for the sum of the first
n
\sum\limits_{k=1}^{n}k=\binom{n+1}{2}=\frac{n(n+1)}{2}.
n
triangular numbers can be expressed as
\sum\limits_{k=1}^{n}\frac{k(k+1)}{2}=\frac{1}{2}\sum_{k=1}^{n}{k^2}+\frac{1}{2}\sum\limits_{k=1}^{n}{k}.
n
triangular numbers was found previously using the hockey stick identity:
\sum\limits_{k=1}^{n}\frac{k(k+1)}{2}=\frac{n(n+1)(n+2)}{6}.
Substituting these identities, the identity for the sum of squares of the first
n
positive integers can be developed:
\begin{aligned} \frac{n(n+1)(n+2)}{6}&=\frac{1}{2}\sum_{k=1}^{n}{k^2}+\frac{1}{2}\left(\frac{n(n+1)}{2}\right)\\\\ \frac{1}{2}\sum_{k=1}^{n}{k^2}&=\frac{n(n+1)(n+2)}{6}-\frac{n(n+1)}{4}\\\\ \sum_{k=1}^{n}{k^2}&=\frac{n(n+1)(2n+1)}{6}. \end{aligned}
This can also be written in terms of the binomial coefficient:
\sum\limits_{k=1}^{n}{k^2}=2\binom{n+2}{3}-\binom{n+1}{2}.\ _\square
This method can be continued indefinitely to develop an identity for the sum of any power of natural numbers.
The sum of the cubes of the first
natural numbers is
\sum\limits_{k=1}^{n}{k^3}=\frac{n^2(n+1)^2}{4}=6\binom{n+3}{4}-6\binom{n+2}{3}+\binom{n+1}{2}.\ _\square
Consider the previous identity for the sum of squares of positive integers:
\sum\limits_{k=1}^{n}{k^2}=\frac{n(n+1)(2n+1)}{6}=2\binom{n+2}{3}-\binom{n+1}{2}.
Now consider the sum of the sum of squares of positive integers:
\begin{aligned} \sum\limits_{k=1}^{n}\sum\limits_{j=1}^{k}{k^2} &= \sum\limits_{k=1}^{n}\frac{k(k+1)(2k+1)}{6} \\ \\ &= \frac{1}{3}\sum\limits_{k=1}^{n}{k^3}+\frac{1}{2}\sum\limits_{k=1}^{n}{k^2}+\frac{1}{6}\sum\limits_{k=1}^{n}{k}. \end{aligned}
This sum can be alternatively computed using binomial coefficients and the hockey stick identity:
\begin{aligned} \sum\limits_{k=1}^{n}\sum\limits_{j=1}^{k}{k^2} &=\sum\limits_{k=1}^{n}\left[2\binom{k+2}{3}-\binom{k+1}{2}\right] \\ \\ &= 2\binom{n+3}{4}-\binom{n+2}{3} \\ \\ &= \frac{n(n+1)(n+2)(n+3)}{12}-\frac{n(n+1)(n+2)}{6} \\ \\ &= \frac{n(n+1)^2(n+2)}{12}. \end{aligned}
\begin{aligned} \frac{n(n+1)^2(n+2)}{12} &=\frac{1}{3}\sum\limits_{k=1}^{n}{k^3}+\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{12}\\\\ &=\frac{1}{3}\sum\limits_{k=1}^{n}{k^3}+\frac{2n(n+1)^2}{12}\\\\ \sum\limits_{k=1}^{n}{k^3}&=\frac{n^2(n+1)^2}{4}. \end{aligned}
This can also be expressed with binomial coefficients:
\sum\limits_{k=1}^{n}{k^3}=6\binom{n+3}{4}-6\binom{n+2}{3}+\binom{n+1}{2}.\ _\square
For the past decade, the king of Mathlandia has forced his subjects to build a pyramid in his honor. The king decreed the pyramid to be constructed with cubic stone slabs. The king also decreed the pyramid to be built in 100 square levels, with each subsequent level 2 units less on a side than the previous level. The top level was to be constructed with a single cube. (See the picture for an example of a pyramid 3 levels high constructed in the same way).
Moments after the final cube was placed, the king changed his mind. He ordered the pyramid to be taken down, and in its place, a cubic monolith was to be built. He ordered the monolith to be built as large as possible with the same stone slabs the pyramid was made of.
The king would tolerate no waste, so he ordered one of his subjects to be sacrificed for each leftover slab of stone.
How many of the king's subjects will be sacrificed?
Inductive Proof of Hockey Stick Identity:
r=n
\begin{aligned} \sum_{k=n}^{n}\binom{k}{n} = \binom{n}{n}&=1\\\\ \binom{n+1}{n+1}&=1. \end{aligned}
Suppose that for whole numbers
n
r \ (n \ge r),
\sum_{k=r}^{n}\binom{k}{r} = \binom{n+1}{r+1}.
\begin{aligned} \sum_{k=r}^{n+1}\binom{k}{r} &= \binom{n+1}{r+1}+\binom{n+1}{r} \\ \\ &= \frac{(n+1)!}{(n-r)!(r+1)!}+\frac{(n+1)!}{(n-r+1)!r!} \\ \\ &= \frac{(n-r+1)(n+1)!}{(n-r+1)!(r+1)!}+\frac{(r+1)(n+1)!}{(n-r+1)!(r+1)!} \\ \\ &= \frac{(n+2)!}{(n-r+1)!(r+1)!} \\ \\ &= \binom{n+2}{r+1}.\ _\square \end{aligned}
Combinatorial Proof using Identical Objects into Distinct Bins
Imagine that there are
m
identical objects to be distributed into
q
distinct bins such that some bins can be left empty. Using the stars and bars approach outlined on the linked wiki page above, this can be done in
\displaystyle\binom{m+q-1}{q-1}
Now consider a slightly different approach to compute this same result. Distribute
j
objects among the first
q-1
bins, and then distribute the remaining
m-j
objects into the last bin. This can be done in
\displaystyle\binom{j+q-2}{q-2}
ways for each value of
j.
Count all of the distributions among all possible values of
j
m
\sum\limits_{j=0}^{m}\binom{j+q-2}{q-2}.
These two methods for counting the distributions of
m
q
bins are equivalent, so the expressions which give the results are equal:
\sum\limits_{j=0}^{m}\binom{j+q-2}{q-2}=\binom{m+q-1}{q-1}.
k=j+q-2,
r=q-2,
n=m+q-2.
Substituting these variables in the identity above gives the hockey stick identity:
\sum\limits_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}.\ _\square
Cite as: Hockey Stick Identity. Brilliant.org. Retrieved from https://brilliant.org/wiki/hockey-stick-identity/ |
A principal made the histogram below to analyze how many years teachers had been teaching at her school.
How many teachers work at her school?
Add up the frequencies represented by the bars. How many total teachers are there?
If the principal randomly chose one teacher to represent the school at a conference, what is the probability that the teacher would have been teaching at the school for
10
or more years? Write the probability in two different ways.
Based on the histogram, how many teachers have been teaching for
10
years or more?
How can this probability be represented as a fraction and a percent?
\frac{7}{24}\text{ or}\approx29 \%
What is the probability that a teacher on the staff has been there for fewer than
5
Based on the histogram, how many teachers have been teaching for less than
5 |
GlideReflection(Q, P, p, AB)
A glide-reflection is the product of a reflection in a plane p and a translation of directed segment AB, where AB lies in the plane.
The command with(geom3d,GlideReflection) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{geom3d}\right):
\mathrm{point}\left(F,0,0,0\right),\mathrm{point}\left(A,1,0,0\right),\mathrm{point}\left(B,0,0,1\right):
Define the plane oxz
\mathrm{line}\left(\mathrm{l1},[F,A]\right),\mathrm{line}\left(\mathrm{l2},[F,B]\right),\mathrm{plane}\left(p,[\mathrm{l1},\mathrm{l2}]\right):
\mathrm{point}\left(C,1,1,0\right),\mathrm{triangle}\left(\mathrm{T1},[A,B,C]\right):
Apply glide-reflection on triangle ABC
\mathrm{dsegment}\left(\mathrm{AB},[A,B]\right):
\mathbf{for}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}i\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{from}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}2\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{to}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}5\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}\mathrm{GlideReflection}\left(T‖i,T‖\left(i-1\right),p,\mathrm{AB}\right)\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathbf{end}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{do}
\textcolor[rgb]{0,0,1}{\mathrm{T5}}
\mathrm{draw}\left({\mathrm{seq}\left(T‖i,i=1..5\right)},\mathrm{scaling}=\mathrm{constrained},\mathrm{style}=\mathrm{patch},\mathrm{title}=\mathrm{`Glide Reflection`}\right)
geom3d[translation] |
Let V and W be vector spaces, let T : V \rightarrow W be linear, and let \{w_1 , w_2 , ...
Joyce Smith 2022-01-04 Answered
Let V and W be vector spaces, let
T:V\to W
be linear, and let
\left\{{w}_{1},{w}_{2},\dots ,{w}_{k}\right\}
be a linearly independent set of k vectors from R(T). Prove that if
S=\left\{{v}_{1},{v}_{2},...,{v}_{k}\right\}
T\left({v}_{i}\right)={W}_{i}
i=1,2,\dots ,k,
then S is linearly independent.
1) The set of vectors
\left\{{v}_{1},{v}_{2},...,{v}_{k}\right\}
from a vector space V is said to be linearly independent if the linear combination of vectors
{\alpha }_{1}{v}_{1}+{\alpha }_{2}{v}_{2}+\dots +{\alpha }_{k}{v}_{k}=0
⇒{\alpha }_{1}={\alpha }_{2}=\dots ={\alpha }_{k}=0
2) Let V and W be vector spaces over the same field F.
T:V\to W
is said to be linear transformation from V to W if
T\left({0}_{v}\right)={0}_{w}
T\left(\alpha u+\beta v\right)=\alpha T\left(u\right)+\beta T\left(v\right)
\alpha ,\beta \in F
u,v\in V
Let us consider an arbitrary representation of zero vector of V as a linear combination of vectors from S.
{0}_{v}={\alpha }_{1}{v}_{1}+{\alpha }_{2}{v}_{2}+\dots +{\alpha }_{k}{v}_{k}=\sum _{i=1}^{k}{\alpha }_{i}{v}_{i}
...(1) where
{\alpha }_{i}\in FS
i=1,2,\dots ,k
Since T is linear,
{0}_{v}\in V
will always gets mapped to
{0}_{w}\in W
T\left({0}_{v}\right)={0}_{w}
{0}_{w}=T\left({0}_{v}\right)=T\left(\sum _{i=1}^{k}{\alpha }_{i}{v}_{i}\right)
By the linearity of T we have
{0}_{w}=T\left(\sum _{i=1}^{k}{\alpha }_{i}{v}_{i}\right)=\sum _{i=1}^{k}{\alpha }_{i}T\left({v}_{i}\right)
\sum _{i=1}^{k}{\alpha }_{i}{w}_{i}
T\left({v}_{i}\right)={w}_{i}
={\alpha }_{1}{w}_{1}+{\alpha }_{2}{w}_{2}+\dots +{\alpha }_{k}{w}_{k}
since the set
\left\{{w}_{1},{w}_{2},\dots ,{w}_{k}\right\}
is linearly independent
⇒{\alpha }_{1}={\alpha }_{2}=\dots ={\alpha }_{k}=0
{\alpha }_{1}={\alpha }_{2}=...={\alpha }_{k}=0
in (1) we have, the set
\left\{{v}_{1},{v}_{2},...,{v}_{k}\right\}
Thus (1) will be the trivial representation of zero vector of V.
S=\left\{{v}_{1},{v}_{2},...,{v}_{k}\right\}
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
Suppose f is a differentiable function of x and y, and
g(r, s) = f(2r − s, s2 − 7r).
Use the table of values below to calculate
gr(4, 3)
gs(4, 3).
f g fx fy
gr(4, 3) =
gs(4, 3) =
Find the dimension of the subspace spanned by the given vectors.
\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right],\left[\begin{array}{c}3\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}9\\ 4\\ -2\end{array}\right],\left[\begin{array}{c}-7\\ -3\\ 1\end{array}\right]
if x, y belong to
{R}^{p}
, than is it true that the relation norm
\left(x+y\right)=norm\left(x\right)+norm\left(y\right)
x=cy\text{ }or\text{ }y=cx\text{ }with\text{ }c>0
Is the vector space C∞[a,b] of infinitely differentiable functions on the interval [a,b], consider the derivate transformation D and the definite integral transformation I defined by D(f)(x)=f′(x)D(f)(x)=f′(x) and I(f)(x)=∫xaf(t)dt f(t)dt. (a) Compute (DI)(f)=D(I(f))(DI)(f)=D(I(f)). (b) Compute (ID)(f)=I(D(f))(ID)(f)=I(D(f)). (c) Do this transformations commute? That is to say, is it true that (DI)(f)=(ID)(f)(DI)(f)=(ID)(f) for all vectors f in the space? |
The Life of Tolstoy/List of Tolstoy's Works - Wikisource, the free online library
The Life of Tolstoy/List of Tolstoy's Works
1481378The Life of Tolstoy — List of Tolstoy's Works1911Paul Birukoff
Those works which are generally accepted as the most important are printed in blacker type. The dates show when the works were first published.
Anna Karenin 1873-76
Hadji Murat Not yet published
Father Sergius Not yet published
The Power of Darkness (drama) 1886
The Fruits of Enlightenment (comedy) 1889
The Corpse (unfinished drama) Not yet published
A Project for a General Plan for Elementary Schools
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}}
Esarlieddon 1903
Three Questions 1903
The Restoration of Hell 1903
Work, Death and Sickness 1903
Komey Vasilyeff 1906
The Divine and the Human 1906
A Letter on Science to a Peasant 1909
Published by Posrednik after Tolstoy's death:
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \\\ \\\ \\\ \\\ \\\ \\\ \\\ \\\ \\\ \\\ \\\ \\\ \ \end{matrix}}\right\}\,}}
Sins, Temptation and Superstitions
The Similarity of Men's Souls
Mementoes for Officers 1901
On the Working-Class Problem 1902
Letters to the Tsar 1902
To the Working People 1902
To Men of Politics 1903
To Social Reformers 1903
Letter to Pietro Mazzini 1903
Bethink Yourselves 1904
In the Russian Revolution 1904
How to Emancipate the Working Classes 1905
A Great Injustice (on the land problem) 1905
On the Social Movement in Russia 1905
The End of the Age 1905
An Appeal to the People 1906
On Military Service 1906
On the Meaning of the Russian Revolution 1906
What Must be Done? 1906
An Appeal to the Government, the Revolutionists and the People 1907
The Only Solution of the Land Question 1907
I Cannot be Silent (a protest against the wholesale executions) 1908
Concerning Molochnikoff's Arrest 1908
The Annexation of Bosnia and Herzegovina 1908
The Inevitable Revolution 1909
An Address to the Stockholm Peace Conference 1909
An Efficient Remedy (last article, published three days after his death by the St. Petersburg daily paper Rietch) 1910
Retrieved from "https://en.wikisource.org/w/index.php?title=The_Life_of_Tolstoy/List_of_Tolstoy%27s_Works&oldid=9989793" |
Effect of change in position of particle dampers on wind turbine blade for vibration suppression | JVE Journals
Santosh R. Sandanshiv1 , Umesh S. Chavan2
1Genba Sopanrao Moze College of Engineering, Pune, India
2Vishwakarma Institute of Technology, Pune, India
Received 25 July 2019; accepted 2 August 2019; published 28 November 2019
Copyright © 2019 Santosh R. Sandanshiv, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Wind turbine energy minimizes due to vibration of blade. In this research we focus on vibration suppression by using particle damping technique. Containers are used for fill the particles and mounted on the blade. As vibration of blade increases, it increases the movement of containers along with particles this brings particle to particle and particle to container wall collision takes place which results to energy loss. In this study we use four different positions for mounting containers, firstly on all four different positions we mount containers simultaneously and take three readings for three different ball sizes respectively, keeping 50 % fill constant in all readings. Then we reduce one container among four and take the readings. Repeat this procedure up to single container. Compare with damping results with without damping results and finding out optimum locations for mounting of dampers.
7 mm particle size gives optimum conditions for damper location as compare to 5 mm
3 position mounting with 7 mm particles with 9 mm particles gives optimum conditions
2 position mount with 9 mm particles gives optimum conditions
Keywords: vibration, particle damping, wind turbine blade, energy loss.
In wind energy power generation, if vibrations of the blade are high then it adversely effect on electricity generation [1]. Edgewise vibration and flapwise vibrations were the two main modes of vibration in blade [2]. Edgewise vibration is the main concern in this work. Structural vibration had been control by three methods namely active, semi active and passive control according to Dapeng [3]. Krenk [4] introduced active struts mounted near the blade root. Fitzgerald introduced edgewise vibration mitigation by using active tuned mass damper [5]. According to Box [6] active tendons was inserted inside blade for vibration control. Box [7] was introducing roller damper in blade. The new concept as tuned liquid column damper (TLCD) for blade vibration reduction, passive damping was used by Colwell [8] and Murtagh [9] in wind turbine tower for control of vibrations.
In this paper particle damping technique is used to check the effect of different parameters on 1 kW wind turbine blade for vibration suppression. Three parameters are used such as particle size variation, change in position and change in percentage fill. At this moment it is not possible to make pocket inside the blade for inserting balls. But for checking the effect we attach external container on blade.
Electrodynamic shaker (EEV 060) having force rating of 600 Kgf is use for generating frequency in the range of 10 Hz to 2000 Hz and Acceleration is consider for the first two modes. We consider 1 Kw wind turbine blade for testing which is mounted at the root location and hermetically sealed type piezo electric accelerometer is mounted at the position of 600 mm randomly considering the maximum displacement location. Output of accelerometer is connected to single channel, digital vibration controller (EESC – 04), generating low electric signals. These electric signals amplify with power amplifier from the vibration controller to a proportionally high voltage and high current output. Acceleration range is kept as 8 g and frequency range up to 2000 Hz. Materials and different parameters are explained as follows.
2.1. Parameters for testing
2.1.1. Ball size
Three different particle sizes are used as 3 mm, 5 mm and 9 mm for effect of vibration suppression.
2.1.2. Container position on blade
Total four positions taken for the testing as 300 mm, 600 mm, 900 mm and 1200 mm.
(a) Particle: Testing of particle is done by Wet Method: ALS:SOP:05-TM 503, REST by IS 228:2010 for particle damping having chemical composition of 0.010 % Mo, 0.050 % Ni, 0.98 % C, 0.33 % Mn, 0.25 % Si, 0.010 % S, 0.012 % P, 1.40 % Cr.
(b) Container: Container is made with Poly-propylene (PP) material, tested by method of ASTM D-792, D-297 and IS 13360.
(c) Blade: Glass fibre blade having length of 1525 mm.
Fig. 1(a) and (b) shows spring mass diagram for ball-wall and ball-ball collisions, where,
{K}_{N}
{K}_{S}
{C}_{N}
{C}_{s}
are stiffness’s and damping constants.
Fig. 1. a) Ball-wall spring mass diagram, b) ball-ball spring mass diagram
Fig. 2. Block diagram of experimental set up
Fig. 3. Actual diagram of experimental set up
Fig. 2 shows block diagram and Fig. 3 shows Actual experimental set up. Fig. 4 shows four different mounting conditions of containers as single, 2 positions, 3 positions and 4 positions mounting.
Fig. 4. Containers at different positions (single, two, three, four containers respectively)
All results are plotted as acceleration in gravity (g) versus frequency (Hz) for without damping and with damping conditions.
Fig. 5. Without damping
Fig. 6. Four position damper, 5 mm ball, 50 % fill
Fig. 5 show with damping results as 12.895 g and 2.753 g for 1st and 2nd modes respectively. All with damping results are compared with without damping and finding out the optical conditions for mounting dampers. All containers are fill with 50 % particles, and taking readings for three different ball sizes as 5 mm, 7 mm and 9 mm respectively.
Fig. 9. Three position damper, 5 mm ball, 50 % fill
Fig. 10. Three position damper, 7 mm ball, 50 % fill
Fig. 12. Two position damper, 5 mm ball, 50 % fill
Table 1 gives results of containers mounting at 4 position, 3 position, 2 position, 1 position and without damping conditions. Fig. 6-8 shows graphs for 4 position mount, Figs. 9-11 gives graphs for 3 position mount , 5, 7 and 9 mm ball sizes graphs. Figs. 12-14 shows graphs for 2 position mount and Fig. 15-17 shows graphs for single position mount.
Table 1 shows that in 4 position mount 7 mm ball gives optimum results as 8.565 g and 1.830 g acceleration for 1st and 2nd modes. All results are compared with without damping results as in 1st mode 12.895 g is very large acceleration as compared to 4 position damping and in second mode 2.753 g is also high acceleration value. In 3 position and single position also 7 mm ball size gives optimum results. But, in case of 2 position mount 9 mm ball size gives optimum result as shown in Table 1. Same results are shown in cumulative result graph.
Table 1. Testing results for different positions
Position of damper
Particle (Ball) size
4 Position damping
1st Mode: 12.895, 2nd Mode: 2.753
Fig. 15. Single position damper, 5 mm ball, 50 % fill
Fig. 18. Cumulative results for all positions and all particle sizes
At 7 mm particle size we get optimum conditions for damper location as compare to 5 mm and 9 mm ball size. In overall results 3 position mounting with 7 mm particles and 2 position mount with 9 mm particles gives optimum conditions.
Mina Ghassempour, Giuseppe Failla Vibration mitigation in offshore wind turbines via tuned mass damper. Engineering Structures, Vol. 183, 2019, p. 610-636. [Publisher]
Thomsen K., Petersen J. T., Nim E., Øye S., Petersen B. A method for determination of damping for edgewise blade vibrations. Wind Energy, Vol. 3, Issue 4, 2000, p. 233-246. [Publisher]
Qiu Dapeng, Chen Jianyun Dynamic responses and damage forms analysis of underground large scale frame structures under oblique SV seismic waves. Soil Dynamics and Earthquake Engineering, Vol. 117, 2019, p. 216-220. [Publisher]
Krenk S., Svendsen M. N., Høgsberg J. Resonant vibration control of three-bladed wind turbine rotors. AIAA Journal, Vol. 50, Issue 1, 2012, p. 148-161. [Publisher]
Fitzgerald B., Basu B., Nielsen S. R. K. Active tuned mass dampers for control of in-plane vibrations of wind turbine blades. Structural Control and Health Monitoring, Vol. 20, Issue 12, 2013, p. 1377-1396. [Publisher]
George Box E. P., Norman Draper R. Empirical Model Building and Response Surfaces. John Wiley and Sons, New York, 1987. [Search CrossRef]
George Box E. P., Norman Draper R. Response Surfaces, Mixtures, and Ridge Analysis. Second Edition, John Wiley and Sons, Hoboken, New Jersey, USA. [Search CrossRef]
Colwell S., Basu B. Tuned liquid column dampers in offshore wind turbines for structural control. Engineering Structures, Vol. 31, Issue 2, 2009, p. 358-368. [Publisher]
Murtagh P. J., Ghosh A., Basu B., Broderick B. M. Passive control of wind turbine vibrations including blade/tower interaction and rotationally sampled turbulence. Wind Energy, Vol. 11, Issue 4, 2007, https://doi.org/10.1002/we.249. [Publisher] |
We consider the nonlinear Klein–Gordon equations coupled with the Born–Infeld theory under the electrostatic solitary wave ansatz. The existence of the least-action solitary waves is proved in both bounded smooth domain case and
{ℝ}^{3}
case. In particular, for bounded smooth domain case, we study the asymptotic behaviors and profiles of the positive least-action solitary waves with respect to the frequency parameter ω. We show that when κ and ω are suitably large, the least-action solitary waves admit only one local maximum point. When
\omega \to \infty
, the point-condensation phenomenon occurs if we consider the normalized least-action solitary waves.
author = {Yu, Yong},
title = {Solitary waves for nonlinear {Klein{\textendash}Gordon} equations coupled with {Born{\textendash}Infeld} theory},
AU - Yu, Yong
TI - Solitary waves for nonlinear Klein–Gordon equations coupled with Born–Infeld theory
Yu, Yong. Solitary waves for nonlinear Klein–Gordon equations coupled with Born–Infeld theory. Annales de l'I.H.P. Analyse non linéaire, Tome 27 (2010) no. 1, pp. 351-376. doi : 10.1016/j.anihpc.2009.11.001. http://archive.numdam.org/articles/10.1016/j.anihpc.2009.11.001/
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Measurement of speed irregularities | JVE Journals
1Faculty of Mechanical Engineering, Technical University of Liberec, Liberec, Czech Republic
This article describes possible ways of measuring rotation irregularities. It focuses on hardware and software equipment necessary for measuring speed irregularities. The article contains examples of measurements and describes the reasons why it is necessary to know about possibilities of evaluating speed irregularities.
Keywords: vibration, torsional vibration, speed rotation, speed irregularities.
1. Speed irregularities
During measurements associated with acoustic or vibration signal diagnostics, determination of exact speed of the device being measured is sometimes essential. For example, vibration measurements on combustion engines or on electrically-powered drives with uneven load (i.e. compressors, presses, variable load hydraulic pumps, etc.) depend on the speed. However, these types of machines do not guarantee even revolutions. Changing the angle of rotation of the shaft during one single revolution is not linear. Diversion of angle is affected by actions inside the drive or its load. In a four-stroke combustion engine, the cycle depends on a single cycle of the combustion engine, which is equivalent to 720°.
Actions inside the drive or load lead to overloading the entire system with additional torsional vibrations, which may lead to shorter life of couplings, cogs or clutch shaft parts. This rotation unevenness has a major impact on the results of frequency analyses performed on drives or loads. Most analyses assume that the speed may vary during the measurement, but each individual revolution is linear.
The following part describes the drawbacks of using classic FFT analysis when measuring devices with irregular speed. The right choice of a current measurement system for the measured shaft is the key factor of correct evaluation of speed irregularity. Complete evaluation of rotational irregularity is a matter of our software solution for the PULSE system. After eliminating synchronization errors in speed measurement, it is possible to perform synchronous averaging of time data followed by FFT analysis.
2. FFT frequency analysis
This commonly used basic frequency analysis does not require specific introduction. This is the frequency analysis that is mostly used in vibrational signal processing. FFT analysis uses a modified Fourier transformation of the time signal. This is a transformation of the time signal into individual harmonic frequencies represented by
\mathrm{s}\mathrm{i}\mathrm{n}\left(x\right)
function. The result is shown in the form of frequency spectrum.
In technical diagnostics, it is advantageous for the frequency to have linear scale with very small step. This is advantageous for exact determination of the first rotation frequency. To set it, it is important to specify the required range of frequency, number of lines, time weighting function, averaging type and overlap.
The disadvantage of the FFT analysis is the need to have a long time signal of the measured signal to be able to perform it. The length of this signal depends on the frequency range and number of lines. The length is usually from 0.25 to 8 s. The signal must be invariable during this time, otherwise a deformation of measured spectrum will occur.
Fig. 1 shows two signals. One signal (thicker blue) was stable during the measurement while the other (thinner red) had irregular speed due to variable load. Application of this result to further analysis is rather limited.
Fig. 1. FFT spectrums of a first harmonic frequency
3. Speed measurement options
The speed can be easily detected directly from the measured vibration signal. In majority of cases, there is a disbalance at the first rotation frequency. This is theoretically possible, but in practice, this method of measuring actual speeds is not much used because of very good balance of the measured machine, or poor resolution in the frequency in the measured FFT spectrum. This poor resolution would lead to a major error in speed measurement.
Current speed measurement options:
– from measured FFT spectrum of sound or vibration;
– direct tacho probe measurement (rotational probe);
– non-contact measurement with a Hall sensor [1];
– speed sensor for rotating shaft [2].
From the above-mentioned methods of speed measurement, we will only focus on the use of probes which have a measurable output voltage corresponding to the revolution or its fractions. A suitable option to measure the actual speed is to use a tacho probe. The tacho probe usually works on the optical principle. The reflection of the transmitted signal is reflected from the reflective mark on the shaft and causes a step change in the output voltage. These voltage peaks can be used to display actual speed in the measurement system.
This speed measurement system gives us additional information. The tacho probe responds to the reflection from a particular point on the shaft that has constant angle to the measured system. Therefore, it is possible to state that the tacho probe can be used to determine the moment when the measured shaft is positioned in a particular angle. This is used for example by order analysis, which synchronizes and resamples the measured signal so that the result is independent from the rotation speed.
However, there is a major problem. The tacho probe is often positioned at a greater distance from the measured object, which is often placed on a flexible base. The tacho probe is often attached to a tripod and is usually standing on a common floor. This type of positioning allows relative movement between the shaft with the reflective mark on it and the tacho probe.
Despite this disadvantage, the use of tacho probe has other advantages, such as short installation time, ability to measure speed over longer distances (for example, through the glass in the testing room) and the possibility of focusing the ray of light on the inside parts of the machine (e.g. through a small hole).
In order to minimize relative movement between the measured object and the shaft, the use of measuring systems attached directly on the shaft is therefore more appropriate.
4. Software solutions for PULSE system
Brüel Kjær PULSE measuring system can be adapted to measure the irregularity of the speed. Our software module performs its own measurements. In the next step, it makes appropriate adjustments to the time signal and the individual revolutions. Subsequently, an angle is calculated in relation to the time corresponding to the regular revolution. This assumption based on probability is compared to real-time measured data. The detected differences show as deviations in degrees for given shaft position.
The following illustration shows an example of angle irregularity of the crankshaft during one work cycle. For piston combustion engines, typical crankshaft speed irregularity is 720° length of working cycle. Depending on the load and rotational speed of the internal combustion engine, the irregularity of crankshaft rotation speed varies considerably.
Fig. 2. Deviations in degrees depending on the shaft position. One signal (thicker blue) was for a small load and second signal (thinner red) had irregular speed due to variable heavy load
The designed algorithm evaluates the measured differences of shaft rotation from theoretical position which would correspond to a regular rotation of the crankshaft. The evaluation of speed rotational irregularity is also important in cases where torsional vibrations can occur, and it is necessary to know the frequencies and amplitudes of these oscillations. Torsional vibrations may cause additional unnecessary strain on couplings and cogs.
5. Synchronous averaging
Elimination of rotational speed problems enables us, for example, to perform synchronous averaging. The result of FFT analysis is only one spectrum. Classic averaging for FFT analysis is based on the calculation of a certain number of individual spectra, which are then averaged for individual frequencies.
This classic averaging is not always satisfactory. In some cases, it is better to avert the measured time signal itself. It is then possible to perform FFT analysis on this average time signal. By averaging the measured signal appropriately, other sources of vibration on the measured object may be eliminated [3]. For this averaging, we need to define a repeatable trigger (mostly performed by an optical probe responding to a pre-defined shaft rotation position) and a suitable object to be measured. If the measured part of the machine reaches a speed rotation that is no longer present on other parts of the machine, the following effect will occur at each subsequent measured data. Other unsynchronized sources will show in each measured data with a different phase. After some time, their influence is completely eliminated.
This method is particularly suitable for measuring vibrations in gearboxes and various types of drives. All other sources that do not rotate at the same speed rotation as the measured shaft will be eliminated by averaging after a while and their effect on the measured signal will be minimized.
Fig. 3 shows the time course of vibrations of the gearbox in a very poor technical condition. Both the measured time and the resulting spectrum were poorly readable. After applying synchronous averaging of the time measured data, the time in the middle of the image was calculated. It is even possible to identify strokes from individual cogs (bottom zoomed time).
Fig. 3. Application of the synchronous averaging on measured data
From the findings mentioned above, it is important to measure speed rotation for evaluation in technical diagnostics. This measurement should identify the possible speed irregularities of the shaft rotation. In the case of the speed irregularities, it is inappropriate to use the measurement for FFT analysis with a long-time data record. Torsional vibrations can be identified from measuring of the speed irregularities too.
This article was written at the Technical University of Liberec, Faculty of Mechanical Engineering with the support of the Institutional Endowment for the Long Term Conceptual Development of Research Institutes, as provided by the Ministry of Education, Youth and Sports of the Czech Republic in the year 2017.
Sheu G.-J., Sie Soedel M.-J.-W. Design and application of a novel contactless sensor. IEEE International Conference on Mechatronics and Automation, ICMA, 2012, p. 702-707. [Search CrossRef]
Kroening M., Hite J. Velocity sensor for rotating shafts. Sensors, Vol. 15, Issue 9, 1998, p. 89-91. [Search CrossRef]
Mark W. D. Time-synchronous-averaging of gear-meshing-vibration transducer responses for elimination of harmonic contributions from the mating gear and the gear pair. Mechanical Systems and Signal Processing, Vol. 62, 2015, p. 21-29. [Search CrossRef] |
If a single marble is chosen at random from jar
If a single marble is chosen at random from jar and it's color is recorded, give the sample space.What is the probability of each possible outcome?
A jar contains 3 red, 3 green, 7 blue, and 5 yellow marbles. If a single marble is chosen at random from the jar and it's color is recorded, give the sample space. What is the probability of each possible outcome?
pattererX
What is the mean of the probability distribution? A. 2.10 Cc. 2.01 B. 1.20 D. 1.02
P\left(A\mid B\right)<P\left(A\right)
Suppose a die is weighted such that the probability of rolling a three is the same as rolling a six, the probability of rolling a one, two, or four is 3 times that of six, and the probability of rolling a five is 5 times that of rolling a three. Find the probability of 1. rolling a one 2. rolling a two 3. rolling a three 4. rolling a four 5. rolling a five 6. rolling a six
If there are 7 bands and 3 floats, in how many different ways can they be arranged? |
Derivative-free Optimization (DFO) | nag
Optimizing complex numerical models is one of the most common problems found in the industry (finance, multi-physics simulations, engineering, etc.). To solve these optimization problems with a standard optimization algorithm such as Gauss–Newton (for problems with a nonlinear least squares structure) or CG (for unstructured nonlinear objective) requires good estimates of the model's derivatives. They can be computed by:
explicitly written derivatives
algorithmic differentiation (see NAG AD tools)
finite differences (bumping), \(\frac{\partial \phi}{\partial x_i} \approx \frac{\phi(x+he_i) - \phi(x)}{h}\)
If exact derivatives are easy to compute then using derivative-based methods is preferable. However, explicitly writing the derivatives or applying AD methods might be impossible if the model is a black box. The alternative, estimating derivatives via finite differences, can quickly become impractical or too computationally expensive as it presents several issues:
Expensive, one gradient evaluation requires at least \(n+1\) model evaluations;
Inaccurate, the size of the model perturbations \(h\) influences greatly the quality of the derivative estimations and is not easy to choose;
Sensitive to noise, if the model is subject to some randomness (e.g. Monte Carlo simulations) or is computed to low accuracy to save computing time, then finite differences estimations will be highly inaccurate;
Poor utilization of model evaluations, each evaluation is only used for one element of one gradient and the information is discarded as soon as that gradient is no longer useful to the solver.
These issues can greatly slow down the convergence of the optimization solver or even completely prevent it. Conversely, DFO solvers are designed to get good improvements of the objective in these situations. They are able to reach convergence with a lot fewer function evaluations and are naturally quite robust to noise in the model evaluations.
NAG introduces, at Mark 27, a complete update of its model-based DFO solvers for nonlinear least squares problems, nag_opt_handle_solve_dfls (e04ff) and nag_opt_handle_solve_dfls_rcomm (e04fg), and unstructured nonlinear problems, nag_opt_handle_solve_dfno (e04jd) and nag_opt_handle_solve_dfno_rcomm (e04je). They present a number of attractive features:
Integrated to the NAG Optimization Modeling Suite with simple interfaces for the solvers and related routines;
Optional reverse communication interface (e04fg, e04je);
Able to start making progress with as few as two objective evaluations;
Improved resilience to noise.
DFO documentation
Problems structure
One frequent problem in practice is tuning a model's parameters to fit real world observations as well as possible. Let us consider a process that is observed at times \(t_i\) and measured with results \(y_i\), for \(i=1,2,\dots,m\). Furthermore, the process is assumed to behave according to a numerical model \(\phi(t,x)\) where \(x\) are parameters of the model. Given the fact that the measurements might be inaccurate and the process might not exactly follow the model, it is beneficial to find model parameters \(x\) so that the error of the fit of the model to the measurements is minimized. This can be formulated as an optimization problem in which \(x\) is the decision variables and the objective function is the sum of squared errors of the fit at each individual measurement, thus:
\begin{array}{ll}\underset{x\in {ℝ}^{n}}{\mathrm{minimize}}\phantom{\rule{0.25em}{0ex}}& \sum _{i=1}^{m}{\left(\varphi \left({t}_{i},x\right)-{y}_{i}\right)}^{2}\\ \text{subject to}& l\le x\le u\end{array}
When the optimization problem cannot be written as nonlinear least squares, a more general formulation has to be used:
\begin{array}{ll}\underset{x\in {ℝ}^{n}}{\mathrm{minimize}}\phantom{\rule{0.25em}{0ex}}& f\left(x\right)\\ \text{subject to}& l\le x\le u\end{array}
The NAG solvers accommodate both of these formulations.
Noise robustness: illustration
Consider the following unbounded problem where \(\epsilon\) is some random uniform noise in the interval \(\left[-\nu,\nu\right]\) and \(r_i\) are the residuals of the Rosenbrock test function.
\underset{x\in {ℝ}^{n}}{\mathrm{minimize}}f\left(x\right)=\sum _{i=1}^{m}{\left({r}_{i}\left(x\right)+\epsilon \right)}^{2}
Let us solve this problem with a Gauss–Newton method combined with finite differences, nag_opt_lsq_uncon_mod_func_comp (e04fc) and the corresponding derivative-free solver (e04ff). For various noise level \(\nu\), we present in the following table the number of model evaluations needed to find a solution with an 'achievable' precision with respect to the noise level: \[ f(x) < \max (10^{-8}, 10 \times \nu^2) \]
The figures in Table 1 show the average (on 50 runs) of number objective evaluations to solve the problem. The number in brackets represent the number of failed runs out of 50.
Table 1: Number of objective evaluations required to reach the desired accuracy. Numbers in the brackets represent amount of failed runs.
noise level \(\nu\) \(0\) \(10^{-10}\) \(10^{-8}\) \(10^{-6}\) \(10^{-4}\) \(10^{-2}\) \(10^{-1}\)
e04fc 90 (0) 93 (0) 240 (23) \(\infty\) (50) \(\infty\) (50) \(\infty\) (50) \(\infty\) (50)
e04ff 22 (0) 22 (0) 22 (0) 22 (0) 22 (0) 17 (0) 15 (0)
On this example, the new derivative-free solver is both cheaper in terms of model evaluations and far more robust with respect to noise.
DFO Poster
[1] C. Cartis, J. Fiala, B. Marteau, and L. Roberts Improving the Flexibility and robustness of model-based derivative-free optimization solvers ACM Transactions On Numerical Software. 2019.
[2] C. Cartis and L. Roberts A derivative-free Gauss–Newton method Mathematical Programming Computation. 2017.
[3] Powell M. J. D. The BOBYQA algorithm for bound constrained optimization without derivatives Report DAMTP 2009/NA06 University of Cambridge. 2009.
Webinar On-demand: Calibrate models faster and improve performance 20%+ with DFO |
Solvability of a Class of Operator-Differential Equations of Third Order with Complicated Characteristic on the Whole Real Axis
Solvability of a Class of Operator-Differential Equations of Third Order with Complicated Characteristic on the Whole Real Axis ()
Egyptian Russian University, Badr City, Egypt.
On the whole real axis, we demonstrate sufficient conditions of regular solvability of third order operator-differential equations with complicated characteristics. These conditions were formulated only by the operator coefficients of the equation. In addition, by the principal part of the equation, the norms of the operators of intermediate derivative were estimated.
Operator-Differential Equation, Hilbert Space, Self-Adjoint Operator, Intermediate Derivative Operator
Ahmed, A. and Labeeb, M. (2018) Solvability of a Class of Operator-Differential Equations of Third Order with Complicated Characteristic on the Whole Real Axis. Open Access Library Journal, 5, 1-5. doi: 10.4236/oalib.1104631.
In a separable Hilbert space H, we have the following equation:
Pu\left(x\right)\equiv {p}_{0}u\left(x\right)+{p}_{1}u\left(x\right)=f\left(x\right),x\in R,
\begin{array}{l}{p}_{0}u\left(x\right)=\left(\frac{\text{d}}{\text{d}t}-A\right){\left(\frac{\text{d}}{\text{d}t}+A\right)}^{2}u\left(x\right).\\ {p}_{1}u\left(x\right)=\underset{s=1}{\overset{2}{\sum }}\text{ }{A}_{s}\frac{{\text{d}}^{3-s}u\left(x\right)}{\text{d}{x}^{3-s}},\end{array}
A is a self-adjoint positive-definite operator, and
{A}_{s},s=1,2
are generally linear unbounded operators. All derivatives are understood in the sense of distributions theory.
f\left(x\right)\in {L}_{2}\left(R;H\right)
{L}_{2}\left(R;H\right)=\left\{f\left(x\right):{‖f\left(x\right)‖}_{{L}_{2}\left(R;H\right)}^{2}{\left({\int }_{-\infty }^{+\infty }{‖f\left(x\right)‖}_{H}^{2}\text{d}t\right)}^{\frac{1}{2}}<+\infty \right\}
(see [1] [2] ), and
u\left(x\right)\in {W}_{2}^{3}\left(R;H\right)
, which are determined as follows:
{W}_{2}^{3}\left(R;H\right)=\left\{u\left(x\right):\frac{{\text{d}}^{3}u\left(x\right)}{\text{d}{x}^{3}}\in {L}_{2}\left({R}_{+};H\right),{A}^{3}u\left(x\right)\in {L}_{2}\left(R;H\right)\right\}.
{‖u‖}_{{W}_{2}^{3}\left({R}_{+};H\right)}={\left({‖\frac{{\text{d}}^{3}u}{\text{d}{x}^{3}}‖}_{{L}_{2}\left(R;H\right)}^{2}+{‖{A}^{3}u‖}_{{L}_{2}\left(R;H\right)}^{2}\right)}^{\frac{1}{2}},
See [2] .
Notice that the principal part of the investigated equation possesses complicated characteristic, not multiple characteristics as in [3] .
Definition 1. If for any
f\left(x\right)\in {L}_{2}\left(R;H\right)
u\left(x\right)\in {W}_{2}^{2}\left(R;H\right)
that satisfies Equation (1) almost everywhere in R, then it is called a regular solution of Equation (1)
f\left(x\right)\in {L}_{2}\left(R;H\right)
there exists a regular solution of Equation (1), and satisfies the inequality
{‖u‖}_{{W}_{2}^{3}\left(R;H\right)}\le const{‖f‖}_{{L}_{2}\left(R;H\right)},
then Equation (1) is called regularly solvable.
{p}_{1}:{W}_{2}^{3}\left(R;H\right)\to {L}_{2}\left(R;H\right)
{A}^{3-s}\frac{{\text{d}}^{s}u\left(x\right)}{\text{d}{x}^{s}}\in {L}_{2}\left(R;H\right)
s=1,2
And the following inequalities are valid (see [2] ).
{‖{A}^{3-s}\frac{{\text{d}}^{s}u\left(x\right)}{\text{d}{x}^{s}}‖}_{{L}_{2}\left(R;H\right)}\le {c}_{s}{‖u‖}_{{W}_{2}^{2}\left(R;H\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}s=1,2.
Definition 3. Parseval’s equality
{\int }_{-\infty }^{+\infty }{|f\left(x\right)|}^{2}\text{d}x=\frac{1}{\text{2π}}{\int }_{-\infty }^{+\infty }{|\stackrel{˜}{f}\left(\zeta \right)|}^{2}\text{d}\zeta
\stackrel{˜}{f}\left(\zeta \right)={\int }_{-\infty }^{+\infty }f\left(x\right){\text{e}}^{-ix\zeta }\text{d}x.
Theorem 1. The operator
{P}_{0}
is an isomorphism from the space
{W}_{2}^{3}\left(R;H\right)
{L}_{2}\left(R;H\right)
Proof. From (2), it is easy to prove that the operator
{P}_{0}
acts from
{W}_{2}^{3}\left(R;H\right)
{L}_{2}\left(R;H\right)
be bounded. Using Fourier transforms for the equation
{P}_{0}u\left(x\right)=f\left(x\right)
\left(-i\xi E-A\right){\left(-i\xi E+A\right)}^{2}\stackrel{˜}{u}\left(\xi \right)=\stackrel{˜}{f}\left(\xi \right).
(E is the unit operator), where
\stackrel{˜}{u}\left(\xi \right),\stackrel{˜}{f}\left(\xi \right)
are Fourier transform for the functions
u\left(x\right),f\left(x\right)
, respectively. The operator pencil
\left(-i\xi E-A\right){\left(-i\xi E+A\right)}^{2}
is invertible and moreover
\stackrel{˜}{u}\left(\xi \right)={\left(-i\xi E-A\right)}^{-1}{\left(-i\xi E+A\right)}^{-2}\stackrel{˜}{f}\left(\xi \right),
u\left(x\right)=\frac{1}{\text{2π}}{\int }_{-\infty }^{+\infty }{\left(-i\xi E-A\right)}^{-1}{\left(-i\xi E+A\right)}^{-2}\stackrel{˜}{f}\left(\xi \right){\text{e}}^{i\zeta x}\text{d}\zeta .
u\left(x\right)\in {W}_{2}^{3}\left(R;H\right)
. By using the Parseval equality and (3), we obtain:
\begin{array}{c}{‖u‖}_{{W}_{2}^{3}\left(R;H\right)}^{2}={‖\frac{{\text{d}}^{3}u}{\text{d}{t}^{3}}‖}_{{L}_{2}\left(R;H\right)}^{2}+{‖{A}^{3}u‖}_{{L}_{2}\left(R;H\right)}^{2}={‖-i{\zeta }^{3}\stackrel{˜}{u}\left(\xi \right)‖}_{{L}_{2}\left(R;H\right)}^{2}+{‖{A}^{3}\stackrel{˜}{u}\left(\xi \right)‖}_{{L}_{2}\left(R;H\right)}^{2}\\ ={‖-i{\zeta }^{3}{\left(-i\xi E-A\right)}^{-1}{\left(-i\xi E+A\right)}^{-2}\stackrel{˜}{f}\left(\xi \right)‖}_{{L}_{2}\left(R;H\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{‖{A}^{3}{\left(-i\xi E-A\right)}^{-1}{\left(-i\xi E+A\right)}^{-2}\stackrel{˜}{f}\left(\xi \right)‖}_{{L}_{2}\left(R;H\right)}^{2}\\ \le \underset{\zeta \in R}{\mathrm{sup}}{‖-i{\zeta }^{3}{\left(-i\zeta E-A\right)}^{-1}{\left(-i\zeta E+A\right)}^{-2}‖}_{H\to H}^{2}{‖\stackrel{˜}{f}\left(\zeta \right)‖}_{{L}_{2}\left(R;H\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{\zeta \in R}{\mathrm{sup}}{‖{A}^{3}{\left(-i\zeta E-A\right)}^{-1}{\left(-i\zeta E+A\right)}^{-2}‖}_{H\to H}^{2}{‖\stackrel{˜}{f}\left(\zeta \right)‖}_{{L}_{2}\left(R;H\right)}^{2}\end{array}
\sigma \left(A\right)
is a spectrum of the operator A, then we consider
\begin{array}{l}\underset{\xi \in R}{\mathrm{sup}}{‖-i{\zeta }^{3}{\left(-i\zeta E-A\right)}^{-1}{\left(-i\zeta E+A\right)}^{-2}‖}_{H\to H}\\ \le \underset{\xi \in R}{\mathrm{sup}}\underset{\sigma \in \sigma \left(A\right)}{\mathrm{sup}}|-i{\zeta }^{3}{\left(-i\zeta -\sigma \right)}^{-1}{\left(-i\zeta +\sigma \right)}^{-2}|\\ =\underset{\xi \in R}{\mathrm{sup}}\frac{{|\zeta |}^{3}}{{}^{{\left({\zeta }^{2}+{\sigma }^{2}\right)}^{\frac{3}{2}}}}\le 1,\end{array}
\begin{array}{l}\underset{\xi \in R}{\mathrm{sup}}{‖{A}^{3}{\left(-i\zeta E-A\right)}^{-1}{\left(i\zeta E+A\right)}^{-2}‖}_{H\to H}\\ \le \underset{\xi \in R}{\mathrm{sup}}\underset{\sigma \in \sigma \left(A\right)}{\mathrm{sup}}|{\sigma }^{3}{\left(-i\zeta -\sigma \right)}^{-1}{\left(-i\zeta +\sigma \right)}^{-2}|\\ =\underset{\sigma \in \sigma \left(A\right)}{\mathrm{sup}}\frac{{\sigma }^{3}}{{}^{{\left({\zeta }^{2}+{\sigma }^{2}\right)}^{\frac{3}{2}}}}\le 1\end{array}
Taking into account (5) and (6) into (4) we obtain:
u\left(x\right)\in {W}_{2}^{3}\left(R;H\right)
Applying Banach theorem on the inverse operator, we get that the operator
{P}_{0}
{W}_{2}^{3}\left(R;H\right)
{L}_{2}\left(R;H\right)
Now, we estimate the norms of intermediate derivative operators participating in the main part of the Equation (1) for finding exact conditions on regular solvability of the given equation, expressed only by its operator coefficients.
From theorem 1, we have that the norms
{‖{p}_{0}u‖}_{{L}_{2}\left(R;H\right)}
{‖u‖}_{{W}_{2}^{3}\left(R;H\right)}
are equivalent in the space
{W}_{2}^{3}\left(R;H\right)
. Therefore by the norm
{‖{p}_{0}u‖}_{{L}_{2}\left(R;H\right)}
the theorem on intermediate derivatives is valid as well.
u\left(x\right)\in {W}_{2}^{3}\left(R;H\right)
. Then there hold the following inequalities:
{‖{A}^{3-s}\frac{{\text{d}}^{s}u\left(x\right)}{\text{d}{x}^{s}}‖}_{{L}_{2}\left(R;H\right)}\le {a}_{s}{‖{p}_{0}u‖}_{{L}_{2}\left(R;H\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}s=1,2.
{a}_{1}={a}_{2}=\frac{2}{3\sqrt{3}}
Proof. To establish the validity of inequality (11) we make change
{p}_{0}u\left(x\right)=f\left(x\right)
and apply the Fourier transformation. We get
\zeta \in R
we estimate the following norms:
\begin{array}{l}{‖{A}^{3-s}{\left(-i\zeta \right)}^{s}{\left(-i\zeta E-A\right)}^{-1}{\left(-i\zeta E+A\right)}^{-2}‖}_{H\to H}\\ \le \underset{\sigma \in \sigma \left(A\right)}{\mathrm{sup}}|{\sigma }^{3-s}{\left(-i\zeta \right)}^{s}{\left(-i\zeta -\sigma \right)}^{-1}{\left(i\zeta +\sigma \right)}^{-2}|\\ =\underset{\sigma \in \sigma \left(A\right)}{\mathrm{sup}}|{\sigma }^{-s}{\left(-i\zeta \right)}^{s}{\left(-i\frac{\zeta }{\sigma }-1\right)}^{-1}{\left(-i\frac{\zeta }{\sigma }+1\right)}^{-2}|\\ \le \underset{\mu =\frac{{\xi }^{2}}{{\sigma }^{2}}\ge 0}{\mathrm{sup}}\frac{{\mu }^{s/2}}{{\left(\mu +1\right)}^{3/2}}=\frac{1}{3\sqrt{3}}{s}^{s/2}{\left(3-s\right)}^{\left(3-s\right)/2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}s=1,2.\end{array}
Finally, from (12), we have
{‖{A}^{3-s}{\left(-i\zeta \right)}^{s}{\left(-i\zeta E-A\right)}^{-1}{\left(-i\zeta E+A\right)}^{-2}\stackrel{˜}{f}\left(\zeta \right)‖}_{{L}_{2}\left(R;H\right)}\le {a}_{s}{‖\stackrel{˜}{f}\left(\xi \right)‖}_{{L}_{2}\left(R;H\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}s=1,2.
Lemma. The operator
{P}_{1}
continuously acts from
{W}_{2}^{3}\left(R;H\right)
{L}_{2}\left(R;H\right)
provided that the operators
{A}_{s}{A}^{-s},s=1,2
are bounded in H.
Taking into account the results found up [4] to now we get possibility to establish regular solvability conditions of Equation (1).
Theorem 3. Let the operators
{A}_{s}{A}^{-s},s=1,2
be bounded in H and it holds the inequality
\underset{s=1}{\overset{2}{\sum }}\text{ }{a}_{s}{‖{A}_{3-s}{A}^{-\left(3-s\right)}‖}_{H\to H}\prec 1
, where the numbers
{a}_{s},s=1,2
are determined in theorem 2. Then the Equation (1) is regularly solvable.
Proof. By theorem 1, provided that the operator
{P}_{0}
has a bounded inverse operator
{P}_{0}^{-1}
acting from
{L}_{2}\left(R;H\right)
{W}_{2}^{3}\left(R;H\right)
, then after replacing
{p}_{0}u\left(x\right)=v\left(x\right)
in Equation (1) can be written as
\left(E+{p}_{1}{p}_{0}^{-1}\right)v\left(x\right)=f\left(x\right)
Now we prove under the theorem conditions (see [5] ), that the norm
{‖{p}_{1}{p}_{0}^{-1}‖}_{{L}_{2}\left(R;H\right)\to {L}_{2}\left(R;H\right)}<1.
By theorem (2), we have:
\begin{array}{c}{‖{p}_{1}{p}_{0}^{-1}v‖}_{{L}_{2}\left(R;H\right)}={‖{p}_{1}u‖}_{{L}_{{}_{2}}\left(R;H\right)}\le \underset{s=1}{\overset{2}{\sum }}{‖{A}^{s}\frac{{\text{d}}^{3-s}u}{\text{d}{x}^{3-s}}‖}_{{L}_{2}\left(R;H\right)}\\ \le \underset{s=1}{\overset{2}{\sum }}{‖{A}_{s}{A}^{-s}‖}_{H\to H}{‖{A}^{s}\frac{{\text{d}}^{3-s}u}{\text{d}{x}^{3-s}}‖}_{{L}_{2}\left(R;H\right)}\\ \le \underset{s=1}{\overset{2}{\sum }}\text{ }{a}_{3-s}{‖{A}_{s}{A}^{-s}‖}_{H\to H}{‖{p}_{0}u‖}_{{L}_{2}\left(R;H\right)}\\ =\underset{s=1}{\overset{2}{\sum }}\text{ }{a}_{3-s}{‖{A}_{s}{A}^{-s}‖}_{H\to H}{‖v‖}_{{L}_{2}\left(R;H\right)}\end{array}
{‖{p}_{1}{p}_{0}^{-1}‖}_{{L}_{2}\left(R;H\right)\to {L}_{2}\left(R;H\right)}\le \underset{s=1}{\overset{2}{\sum }}\text{ }{a}_{3-s}{‖{A}_{s}{A}^{-s}‖}_{H\to H}\prec 1.
Thus, the operator
E+{p}_{1}{p}_{0}^{-1}
{L}_{2}\left(R;H\right)
u\left(x\right)
u\left(x\right)={p}_{0}^{-1}{\left(E+{p}_{1}{p}_{0}^{-1}\right)}^{-1}f\left(x\right)
, moreover
\begin{array}{c}{‖u‖}_{{W}_{2}^{3}\left(R;H\right)}\le {‖{p}_{0}^{-1}‖}_{{L}_{2}\left(R;H\right)\to {W}_{2}^{3}\left(R;H\right)}{‖{\left(\left(E+{p}_{1}{p}_{0}^{-1}\right)\right)}^{-1}‖}_{{L}_{2}\left(R;H\right)\to {L}_{2}\left(R;H\right)}{‖f‖}_{{L}_{2}\left(R;H\right)}\\ \le const{‖f‖}_{{L}_{2}\left(R;H\right)}.\end{array}
We formulated exact conditions on regular solvability of Equation (1), expressed only by its operator coefficients. We estimated the norms of intermediate derivative operators participating in the principle part of the given equation. In the case when in the perturbed part of Equation (1), the participant variable operator coefficients, i.e.
{A}_{s}\left(x\right),s=1,2
are linear operators, which determined for all
x\in R
, are investigated in a similar way.
[1] Hille, E. and Phillips, R. (1962) Functional Analysis and Semi-Groups. IL, Moscow, 829 p. (In Russian)
[2] Lions, J.L. and Majenes, E. (1971) Inhomogeneous Boundary Value Problems and Their Applications. Mir, Moscow, 371 p. (In Russian)
[3] Aliev, A.R. and Elbably, A.L. (2012) On the Solvability in a Weight Space of a Third-Order Operator-Differential Equation with Multiple Characteristic. Doklady Mathematics, 85, 233-235.
[4] Gasymov, M.G. and Mirzoev, S.S. (1992) On Solvability Boundary Value Problems for Elliptic Type Operator Differential Equations of Second Order. Different Uravnenie, 28, 651-666. (In Russian)
[5] Mirzoyev, S.S. (2003) On the Norms of Operators of Intermediate Derivatives, Transactions of NAS of Azerbaijan. Series Physical-Technical and Mathematical Sciences, 23, 157-164. |
SolveTools/Identity - Maple Help
Home : Support : Online Help : SolveTools/Identity
solve expressions with identities
Identity(eqns, ineqs, vars)
system of equations with identities to solve
variables to solve for
The Identity command solves a system containing one or more identities. The difference between calling Identity and calling solve on a system of identities, is that Identity does only enough processing to remove the identities before attempting to solve the system. The solve command does much more processing and so may return more complete answers for a system with a mix of identities and other equations or inequations.
\mathrm{SolveTools}:-\mathrm{Identity}\left({\mathrm{identity}\left({t}^{z}-{x}^{t}+{5}^{t}=\frac{1}{t},t\right)},\varnothing ,{x,z}\right)
[{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-1}}]
\mathrm{SolveTools}:-\mathrm{Identity}\left({\mathrm{identity}\left(\frac{y}{x}=a{\left(x+c\right)}^{b},x\right)},\varnothing ,{a,b,c}\right)
[{\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}}] |
Subbase - formulasearchengine
In topology, a subbase (or subbasis) for a topological space X with topology T is a subcollection B of T that generates T, in the sense that T is the smallest topology containing B. A slightly different definition is used by some authors, and there are other useful equivalent formulations of the definition; these are discussed below.
3 Results using subbases
3.1 Alexander subbase theorem
Let X be a topological space with topology T. A subbase of T is usually defined as a subcollection B of T satisfying one of the two following equivalent conditions:
The subcollection B generates the topology T. This means that T is the smallest topology containing B: any topology U on X containing B must also contain T.
The collection of open sets consisting of all finite intersections of elements of B, together with the set X and the empty set, forms a basis for T. This means that every non-empty proper open set in T can be written as a union of finite intersections of elements of B. Explicitly, given a point x in a proper open set U, there are finitely many sets S1, …, Sn of B, such that the intersection of these sets contains x and is contained in U.
(Note that if we use the nullary intersection convention, then there is no need to include X in the second definition.)
For any subcollection S of the power set P(X), there is a unique topology having S as a subbase. In particular, the intersection of all topologies on X containing S satisfies this condition. In general, however, there is no unique subbasis for a given topology.
Thus, we can start with a fixed topology and find subbases for that topology, and we can also start with an arbitrary subcollection of the power set P(X) and form the topology generated by that subcollection. We can freely use either equivalent definition above; indeed, in many cases, one of the two conditions is more useful than the other.
Sometimes, a slightly different definition of subbase is given which requires that the subbase B cover X.[1] In this case, X is an open set in the topology generated, because it is the union of all the {Bi} as Bi ranges over B. This means that there can be no confusion regarding the use of nullary intersections in the definition.
However, with this definition, the two definitions above are not always equivalent. In other words, there exist spaces X with topology T, such that there exists a subcollection B of T such that T is the smallest topology containing B, yet B does not cover X. In practice, this is a rare occurrence; e.g. a subbase of a space satisfying the T1 separation axiom must be a cover of that space.
The usual topology on the real numbers R has a subbase consisting of all semi-infinite open intervals either of the form (−∞,a) or (b,∞), where a and b are real numbers. Together, these generate the usual topology, since the intersections
{\displaystyle (a,b)=(-\infty ,b)\cap (a,\infty )}
for a < b generate the usual topology. A second subbase is formed by taking the subfamily where a and b are rational. The second subbase generates the usual topology as well, since the open intervals (a,b) with a, b rational, are a basis for the usual Euclidean topology.
The subbase consisting of all semi-infinite open intervals of the form (−∞,a) alone, where a is a real number, does not generate the usual topology. The resulting topology does not satisfy the T1 separation axiom, since all open sets have a non-empty intersection.
The initial topology on X defined by a family of functions fi : X → Yi, where each Yi has a topology, is the coarsest topology on X such that each fi is continuous. Because continuity can be defined in terms of the inverse images of open sets, this means that the initial topology on X is given by taking all fi−1(U), where U ranges over all open subsets of Yi, as a subbasis.
Two important special cases of the initial topology are the product topology, where the family of functions is the set of projections from the product to each factor, and the subspace topology, where the family consists of just one function, the inclusion map.
The compact-open topology on the space of continuous functions from X to Y has for a subbase the set of functions
{\displaystyle V(K,U)=\{f\colon X\to Y\mid f[K]\subset U\}}
where K is compact and U is open Y.
Results using subbases
One nice fact about subbases is that continuity of a function need only be checked on a subbase of the range. That is, if Template:Mvar is a subbase for Template:Mvar, a function f : X → Y is continuous iff f −1(U) is open in Template:Mvar for each Template:Mvar in Template:Mvar.
There is one significant result concerning subbases, due to James Waddell Alexander II.
Alexander Subbase Theorem. Let Template:Mvar be a topological space with a subbasis Template:Mvar. If every cover by elements from Template:Mvar has a finite subcover, then the space is compact.
Note that the corresponding result for basic covers is trivial.
Proof Outline: Assume by way of contradiction that the space Template:Mvar is not compact, yet every subbasic cover from Template:Mvar has a finite subcover. Use Zorn's Lemma to find an open cover Template:Mvar without finite subcover that is maximal amongst such covers. That means that if Template:Mvar is not in Template:Mvar, then C ∪ {V} has a finite subcover, necessarily of the form C0 ∪ {V}.
Consider C ∩ B, that is, the subbasic subfamily of Template:Mvar. If it covered Template:Mvar, then by hypothesis, it would have a finite subcover. But Template:Mvar does not have such, so C ∩ B does not cover Template:Mvar. Let Template:Mvar in Template:Mvar be uncovered. Template:Mvar covers Template:Mvar, so x ∈ U for some U ∈ C. Template:Mvar is a subbasis, so for some S1, ..., Sn ∈ B, we have: x ∈ S1 ∩ ... ∩ Sn ⊆ U.
Since Template:Mvar is uncovered, Si ∉ C. As noted above, this means that for each Template:Mvar, Si along with a finite subfamily Ci of Template:Mvar, covers Template:Mvar. But then Template:Mvar and all the Ci cover Template:Mvar, so Template:Mvar has a finite subcover after all. Q.E.D.
Although this proof makes use of Zorn's Lemma, the proof does not need the full strength of choice. Instead, it relies on the intermediate Ultrafilter principle.
Using this theorem with the subbase for R above, one can give a very easy proof that bounded closed intervals in R are compact.
Tychonoff's theorem, that the product of compact spaces is compact, also has a short proof. The product topology on ∏i Xi has, by definition, a subbase consisting of cylinder sets that are the inverse projections of an open set in one factor. Given a subbasic family Template:Mvar of the product that does not have a finite subcover, we can partition C = ∪i Ci into subfamilies that consist of exactly those cylinder sets corresponding to a given factor space. By assumption, no Ci has a finite subcover. Being cylinder sets, this means their projections onto Xi have no finite subcover, and since each Xi is compact, we can find a point xi ∈ Xi that is not covered by the projections of Ci onto Xi. But then xi is not covered by Template:Mvar.
Note, that in the last step we implicitly used the axiom of choice (which is actually equivalent to Zorn's lemma) to ensure the existence of xi.
de:Basis (Topologie)
Retrieved from "https://en.formulasearchengine.com/index.php?title=Subbase&oldid=229679" |
Two-sample t-test - MATLAB ttest2 - MathWorks Benelux
Two-Sample t-Test for Equal Means
t-Test for Equal Means Without Assuming Equal Variances
h = ttest2(x,y)
h = ttest2(x,y,Name,Value)
[h,p] = ttest2(___)
[h,p,ci,stats] = ttest2(___)
h = ttest2(x,y) returns a test decision for the null hypothesis that the data in vectors x and y comes from independent random samples from normal distributions with equal means and equal but unknown variances, using the two-sample t-test. The alternative hypothesis is that the data in x and y comes from populations with unequal means. The result h is 1 if the test rejects the null hypothesis at the 5% significance level, and 0 otherwise.
h = ttest2(x,y,Name,Value) returns a test decision for the two-sample t-test with additional options specified by one or more name-value pair arguments. For example, you can change the significance level or conduct the test without assuming equal variances.
[h,p] = ttest2(___) also returns the p-value, p, of the test, using any of the input arguments in the previous syntaxes.
[h,p,ci,stats] = ttest2(___) also returns the confidence interval on the difference of the population means, ci, and the structure stats containing information about the test statistic.
Load the data set. Create vectors containing the first and second columns of the data matrix to represent students’ grades on two exams.
Test the null hypothesis that the two data samples are from populations with equal means.
tstat: 0.0167
The returned value of h = 0 indicates that ttest2 does not reject the null hypothesis at the default 5% significance level.
Test the null hypothesis that the two data vectors are from populations with equal means, without assuming that the populations also have equal variances.
[h,p] = ttest2(x,y,'Vartype','unequal')
The returned value of h = 0 indicates that ttest2 does not reject the null hypothesis at the default 5% significance level even if equal variances are not assumed.
Sample data, specified as a vector, matrix, or multidimensional array. ttest2 treats NaN values as missing data and ignores them.
If x and y are specified as vectors, they do not need to be the same length.
If x and y are specified as matrices, they must have the same number of columns. ttest2 performs a separate t-test along each column and returns a vector of results.
If x and y are specified as multidimensional arrays, they must have the same size along all but the first nonsingleton dimension.
If x and y are specified as multidimensional arrays, they must have the same size along all but the first nonsingleton dimension. ttest2 works along the first nonsingleton dimension.
Example: 'Tail','right','Alpha',0.01,'Vartype','unequal' specifies a right-tailed test at the 1% significance level, and does not assume that x and y have equal population variances.
'both' — Test against the alternative hypothesis that the population means are not equal.
'right' — Test against the alternative hypothesis that the population mean of x is greater than the population mean of y.
'left' — Test against the alternative hypothesis that the population mean of x is less than the population mean of y.
ttest2 tests the null hypothesis that the population means are equal against the specified alternative hypothesis.
Vartype — Variance type
'equal' (default) | 'unequal'
Variance type, specified as the comma-separated pair consisting of 'Vartype' and one of the following.
'equal' Conduct test using the assumption that x and y are from normal distributions with unknown but equal variances.
'unequal' Conduct test using the assumption that x and y are from normal distributions with unknown and unequal variances. This is called the Behrens-Fisher problem. ttest2 uses Satterthwaite’s approximation for the effective degrees of freedom.
Vartype must be a single variance type, even when x is a matrix or a multidimensional array.
Example: 'Vartype','unequal'
Confidence interval for the difference in population means of x and y, returned as a two-element vector containing the lower and upper boundaries of the 100 × (1 – Alpha)% confidence interval.
Test statistics for the two-sample t-test, returned as a structure containing the following:
sd — Pooled estimate of the population standard deviation (for the equal variance case) or a vector containing the unpooled estimates of the population standard deviations (for the unequal variance case).
The two-sample t-test is a parametric test that compares the location parameter of two independent data samples.
t=\frac{\overline{x}-\overline{y}}{\sqrt{\frac{{s}_{x}^{2}}{n}+\frac{{s}_{y}^{2}}{m}}},
\overline{x}
\overline{y}
are the sample means, sx and sy are the sample standard deviations, and n and m are the sample sizes.
In the case where it is assumed that the two data samples are from populations with equal variances, the test statistic under the null hypothesis has Student's t distribution with n + m – 2 degrees of freedom, and the sample standard deviations are replaced by the pooled standard deviation
s=\sqrt{\frac{\left(n-1\right){s}_{x}^{2}+\left(m-1\right){s}_{y}^{2}}{n+m-2}.}
In the case where it is not assumed that the two data samples are from populations with equal variances, the test statistic under the null hypothesis has an approximate Student's t distribution with a number of degrees of freedom given by Satterthwaite's approximation. This test is sometimes called Welch’s t-test.
ttest | ztest | sampsizepwr |
Visual Basic - Maple Help
Home : Support : Online Help : Programming : Code Generation Package : Visual Basic : Visual Basic
translate Maple code to Visual Basic code
VisualBasic(x, cgopts)
The VisualBasic command translates Maple code to Visual Basic code.
- If the parameter x is an algebraic expression, then a Visual Basic statement assigning the expression to a variable is generated.
- If x is a list, Maple Array or rtable of algebraic expressions, then a sequence of Visual Basic statements assigning the elements to a Visual Basic array is produced. Only the initialized elements of the rtable or Maple Array are translated.
\mathrm{nm}=\mathrm{expr}
\mathrm{nm}
\mathrm{expr}
is an algebraic expression, this is understood to mean a sequence of assignment statements. In this case, the equivalent sequence of Visual Basic assignment statements is generated.
- If x is a procedure, then a Visual Basic module is generated containing a function equivalent to the procedure, along with any necessary import statements.
- If x is a module, then a Visual Basic module is generated, as described on the VisualBasicDetails help page.
For more information about how the CodeGeneration package translates Maple code to other languages, see Translation Details. For more information about translation to VisualBasic in particular, see VisualBasicDetails.
\mathrm{with}\left(\mathrm{CodeGeneration}\right):
\mathrm{VisualBasic}\left(x+yz-2xz,\mathrm{resultname}="w"\right)
\mathrm{VisualBasic}\left([[x,2y],[5,z]],\mathrm{resultname}="w"\right)
\mathrm{cs}≔[s=1.0+x,t=\mathrm{ln}\left(s\right)\mathrm{exp}\left(-x\right),r=\mathrm{exp}\left(-x\right)+xt]:
\mathrm{VisualBasic}\left(\mathrm{cs},\mathrm{optimize}\right)
s≔\mathrm{VisualBasic}\left(x+y+1,\mathrm{declare}=[x::\mathrm{float},y::'\mathrm{integer}'],\mathrm{output}=\mathrm{string}\right)
\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{"cg = x + CDbl\left(y\right) + 0.1E1"}
\mathrm{VisualBasic}\left(f,\mathrm{defaulttype}=\mathrm{integer}\right)
Public Module CodeGenerationModule
Public Function f( _
ByVal z As Integer) As Integer
Return y * x - y * z + x * z
\mathrm{VisualBasic}\left(f\right)
Public Function f(ByVal n As Integer) As Double
Dim cgret As Double
x = x + CDbl(i)
Return cgret
Translate a procedure accepting an Array as a parameter. Note that the indices are renumbered so that the Visual Basic array starts at index 0.
\mathrm{VisualBasic}\left(f\right)
Public Function f(ByVal x() As Double) As Double
Return x(0) + x(1) + x(2)
\mathrm{VisualBasic}\left(m,\mathrm{resultname}=\mathrm{t0}\right)
Public Module m
Public Function p(ByVal x As Double, ByVal y As Integer) As Integer
If (0 < y) Then
Return Fix(x)
Return Ceiling(x)
Private Function q(ByVal x As Double) As Double
Return Pow(Sin(x), 0.2E1)
\mathrm{VisualBasic}\left(2\mathrm{cosh}\left(x\right)-7\mathrm{tanh}\left(x\right)\right)
cg0 = 0.2E1 * Cosh(x) - 0.7E1 * Tanh(x)
\mathrm{VisualBasic}\left(f\right)
Public Sub f(ByVal a As Integer, ByVal p As Integer)
System.Console.WriteLine("The integer remainder of " & a & " divided by " & p & " is: " & a Mod p)
VisualBasicDetails |
Frequency synthesizer with single modulus prescaler based integer N PLL architecture - Simulink - MathWorks Benelux
Integer N PLL with Single Modulus Prescaler
Frequency synthesizer with single modulus prescaler based integer N PLL architecture
The Integer N PLL with Single Modulus Prescaler reference architecture uses a Single Modulus Prescaler block as the frequency divider in a PLL system. The frequency divider divides the frequency of the VCO output signal by an integer value to make it comparable to a PFD reference signal frequency.
Select to enable increased buffer size during the simulation. This increases the buffer size of all the building blocks in the PLL model that belong to the Mixed-Signal Blockset™/PLL/Building Blocks Simulink® library. The building blocks are PFD, Charge Pump, Loop Filter, VCO, and Single Modulus Prescaler. By default, this option is deselected.
Buffer size for the PFD, charge pump, VCO, and prescaler, specified as a positive integer scalar. This sets the buffer size of the PFD, Charge Pump, VCO, and Single Modulus Prescaler blocks inside the PLL model.
\Delta \text{T}=\frac{{\left(\text{Rise/fall time}\right)}^{2}}{6\text{ }·\text{ 0}\text{.22}}
\Delta \text{T}=\frac{\text{Rise/fall time}}{6\text{ }·\text{ Maximum frequency of interest}}
20% – 80% rise/fall time for the up output port of the PFD, specified as a real positive scalar in seconds.
\Delta \text{T}=\frac{{\left(\text{Rise/fall time}\right)}^{2}}{6\text{ }·\text{ 0}\text{.22}}
\Delta \text{T}=\frac{\text{Rise/fall time}}{6\text{ }·\text{ Maximum frequency of interest}}
To enable this parameter, select Add Phase-noise in the VCO tab.
Clock divider value — Value by which the clock divider divides the input frequency
100 (default) | real positive scalar
Value by which the clock divider divides the input frequency, specified as a real positive scalar.
Use get_param(gcb,'N') to view the current value of Clock divider value.
Use set_param(gcb,'N',value) to set Clock divider value to a specific value.
Minimum value by which the clock divider can divide input frequency, specified as a real positive scalar. This parameter is also reported in the Loop Filter tab and is used to automatically calculate the filter component values of the loop filter.
331.5752 (default) | positive real scalar
1.7e+08 (default) | positive real scalar
28.1695 (default) | positive real scalar
PFD | Charge Pump | Loop Filter | Single Modulus Prescaler | VCO |
Bose-Einstein statistics - Simple English Wikipedia, the free encyclopedia
statistical description for the behaviour of bosons
In statistical mechanics, Bose-Einstein statistics means the statistics of a system where you can not tell the difference between any of the particles, and the particles are bosons. Bosons are fundamental particles like the photon.[1]
The Bose-Einstein distribution tells you how many particles have a certain energy. The formula is
{\displaystyle n(\varepsilon )={\frac {1}{e^{(\varepsilon -\mu )/kT}-1}}}
{\displaystyle \varepsilon >\mu }
and where:
n(ε) is the number of particles which have energy ε
ε is the energy
{\displaystyle \varepsilon -\mu \gg kT}
, then the Maxwell–Boltzmann statistics is a good approximation.
Griffiths, David J. (2005). Introduction to quantum mechanics (2nd ed.). Upper Saddle River, NJ: Pearson, Prentice Hall. ISBN 0131911759.
↑ Bosons have integer (whole number) spin and the Pauli exclusion principle is not true for them.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Bose-Einstein_statistics&oldid=7295488" |
dynamic viscosity - Maple Help
Home : Support : Online Help : Science and Engineering : Units : Known Units : dynamic viscosity
Units of Dynamic Viscosity
Dynamic viscosity has the dimension mass per length time. The SI composite unit of dynamic viscosity is the kilogram per meter second.
Maple knows the units of dynamic viscosity listed in the following table.
An asterisk ( * ) indicates the default context, an at sign (@) indicates an abbreviation, and under the prefixes column, SI indicates that the unit takes all SI prefixes, IEC indicates that the unit takes IEC prefixes, and SI+ and SI- indicate that the unit takes only positive and negative SI prefixes, respectively. Refer to a unit in the Units package by indexing the name or symbol with the context, for example, poise[standard]; or, if the context is indicated as the default, by using only the unit name or symbol, for example, poise.
The units of dynamic viscosity are defined as follows.
A poise is defined as a dyne second per square meter.
A reyn is defined as a poundforce second per square inch.
A rhe is defined as an inverse poise.
\mathrm{convert}\left('\mathrm{poise}','\mathrm{dimensions}','\mathrm{base}'=\mathrm{true}\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{mass}}}{\textcolor[rgb]{0,0,1}{\mathrm{length}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{time}}}
\mathrm{convert}\left(1,'\mathrm{units}','\mathrm{poise}',\frac{'\mathrm{kg}'}{'m''s'}\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{10}}
\mathrm{convert}\left(1,'\mathrm{units}','\mathrm{poise}','\mathrm{reyn}'\right)
\frac{\textcolor[rgb]{0,0,1}{129032000}}{\textcolor[rgb]{0,0,1}{8896443230521}}
\mathrm{convert}\left(1,'\mathrm{units}',\frac{1}{'\mathrm{poise}'},'\mathrm{rhe}'\right)
\textcolor[rgb]{0,0,1}{1} |
EUDML | Monochromatic forests of finite subsets of . EuDML | Monochromatic forests of finite subsets of .
Monochromatic forests of finite subsets of
ℕ
Brown, Tom C.. "Monochromatic forests of finite subsets of .." Integers 0 (2000): Paper A04, 7 p., electronic only-Paper A04, 7 p., electronic only. <http://eudml.org/doc/120809>.
author = {Brown, Tom C.},
keywords = {Ramsey theory; piecewise syndetic; algorithmic progression},
title = {Monochromatic forests of finite subsets of .},
TI - Monochromatic forests of finite subsets of .
KW - Ramsey theory; piecewise syndetic; algorithmic progression
Diana Piguetová, A canonical Ramsey-type theorem for finite subsets of
ℕ
Ramsey theory, piecewise syndetic, algorithmic progression |
Mr. Takaya can eat three slices of pizza in five minutes. If he continues to eat at the same rate, how long will it take him to eat the whole pizza, which has twelve slices? How many slices could he eat in half of an hour?
If Mr. Takaya’s rate is:
\frac{3\ \text{slices}}{5\ \text{minutes}}
What can you multiply that by to get
12
slices?
\frac{3\text{ slices}}{5 \text{ minutes}}\cdot \frac{4}{4}=\frac{12 \text{ slices}}{? \text{ minutes}}
5\ \text{minutes}\ (4)=20\ \text{minutes}
Mr. Takaya can eat the whole pizza in
20
Try using this same method to find the number of slices Mr. Takaya can eat in half an hour (
30
minutes). |
Heron's Formula | Brilliant Math & Science Wiki
Aditya Virani, Victor Loh, Niranjan Khanderia, and
Heron's formula is a formula that can be used to find the area of a triangle, when given its three side lengths. It can be applied to any shape of triangle, as long as we know its three side lengths. The formula is as follows:
The area of a triangle whose side lengths are
a, b,
c
A=\sqrt{s(s-a)(s-b)(s-c)},
s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}
, semi-perimeter of the triangle.
Other useful forms are
\begin{aligned} A&=\frac 1 4\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\\ \\ A&=\frac 1 4\sqrt{ \big[(a+b+c)(a+b-c) \big] \times \Big[\big(+(a-b)+c\big)\big(-(a-b)+c\big) \Big]}\\ A&=\frac 1 4\sqrt{\Big[(a+b)^2-c^2\Big] \times \ \Big[c^2-(a-b)^2\Big] }\\ \\ A&=\frac{1}{4}\sqrt{4a^2b^2-\big(a^2+b^2-c^2\big)^2}\\ A&=\frac{1}{4}\sqrt{2\left(a^2 b^2+a^2c^2+b^2c^2\right)-\left(a^4+b^4+c^4\right)} \\ A&=\frac{1}{4}\sqrt{\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)}. \end{aligned}
Although this seems to be a bit tricky (in fact, it is), it might come in handy when we have to find the area of a triangle, and we have no other information other than its three side lengths.
This formula follows from the area formula
A=\frac{1}{2}ab\sin C
\cos C=\frac{a^2+b^2-c^2}{2ab}
Substituting into the Pythagorean identity
\sin C=\sqrt{1-\cos^2 C}
yields Heron's formula (after a series of algebraic manipulations).
_\square
Find the area of the triangle below.
Since the three side lengths are all equal to 6, the semiperimeter is
s=\frac{6+6+6}{2}=9
. Therefore the area of the triangle is
A=\sqrt{9\times(9-6)\times(9-6)\times(9-6)}=9\sqrt{3}.\ _\square
Since the three side lengths are 4, 5, and 7, the semiperimeter is
s=\frac{4+5+7}{2}=8
A=\sqrt{8\times(8-4)\times(8-5)\times(8-7)}=4\sqrt{6}.\ _\square
Since the three side lengths are 13, 14, and 15, the semiperimeter is
s=\frac{13+14+15}{2}=21
A=\sqrt{21\times(21-13)\times(21-14)\times(21-15)}=84.\ _\square
Since the three side lengths are 6, 8, and 10, the semiperimeter is
s=\frac{6+8+10}{2}=12
A=\sqrt{12\times(12-6)\times(12-8)\times(12-10)}=24.\ _\square
Find the area of a triangle with side lengths
4,13
15
a=4, b=13, c=15
s=\frac{4+13+15}{2}=16
A = \sqrt{16(16-4)(16-13)(16-15)} = 24. \ _\square
We can use the Pythagorean theorem to find that the side lengths are
5, \sqrt{ 29}, 2 \sqrt{10}
If we used the direct form of
A = \sqrt{ s (s-a)(s-b)(s-c) }
, we will quickly get into a huge mess because these lengths are not integers.
Instead, we will use an alternate form of Heron's formula:
\begin{aligned} A & = \frac{1}{4}\sqrt{2\big(a^2 b^2+a^2c^2+b^2c^2\big)-\big(a^4+b^4+c^4\big)} \\ & = \frac{1}{4} \sqrt{ 2 ( 25 \times 29 + 25 \times 40 + 29 \times 40) - 25^2 - 29^2 - 40^2 } \\ & = \frac{1}{4} \sqrt{ 2704 } \\ & = 13. \ _\square \end{aligned}
Note: This triangle appears in Composite Figures, which is an easier approach.
s=\frac{6+6+6}{2}=9
A=\sqrt{9\times(9-6)\times(9-6)\times(9-6)}=9\sqrt{3}.\ _\square
s=\frac{4+5+7}{2}=8
A=\sqrt{8\times(8-4)\times(8-5)\times(8-7)}=4\sqrt{6}.\ _\square
s=\frac{13+14+15}{2}=21
A=\sqrt{21\times(21-13)\times(21-14)\times(21-15)}=84.\ _\square
s=\frac{6+8+10}{2}=12
A=\sqrt{12\times(12-6)\times(12-8)\times(12-10)}=24.\ _\square
What is the area of a triangle with sides of length 13, 14, and 15?
\triangle \text{JAY}
has side lengths 10, 8, and 4, then the area of the triangle can be expressed as
\sqrt{\, \overline{abc}\, }
\overline{abc}
3
-digit number. Find
a+b+c
In the figure to the right, the areas of the squares
A, B,
C
are 388, 153, and 61, respectively.
Find the area of the blue triangle.
Cite as: Heron's Formula. Brilliant.org. Retrieved from https://brilliant.org/wiki/herons-formula/ |
Organic Chemistry - Vocabulary - Course Hero
General Chemistry/Organic Chemistry/Vocabulary
hydrocarbon containing a hydroxyl functional group
organic compound that contains a carbonyl group (
{\rm{C}{=}{O}}
) bound to one alkyl (
{-\rm{R}}
{\rm{RC({=}O)H}}
{\rm{R}{-}{CHO}}
alicyclic molecule
cyclic hydrocarbon that contains nonaromatic rings
hydrocarbon that contains only straight or branched carbon-carbon chains
hydrocarbon containing only
{\rm{C}{-}{C}}
single bonds and hydrogen atoms with CnH2n+2 stoichiometry
hydrocarbon containing at least one carbon-carbon double bond (
{\rm{C}{=}{C}}
) with CnH2n stoichiometry
hydrocarbon that contains at least one carbon-carbon triple bond (
{\rm{C}{\equiv}{C}}
) with CnH2n-2 stoichiometry
organic compound with a general
{\rm{RC({=}O)NRR^\prime}}
{\rm{C}{=}{O}}
{\rm{C}{-}{N}}
organic compound that is a derivative of ammonia (NH3), in which one or more hydrogen atoms are replaced by alkyl or aryl units (R), forming
{\rm{N}{-}{R}}
single bonds
planar hydrocarbon of CnHn stoichiometry that consists of alternating
{\rm{C}{-}{C}}
{\rm{C}{=}{C}}
bonds. Benzene (C6H6) is the smallest neutral carbon-only aromatic compound.
functional group containing a
{\rm{C}{=}{O}}
{\rm{C}{=}{O}}
{{-}\rm{OH}}
{\rm{R{-}C({=}O)OH}}
\rm{R{-}COOH}
compound that contains an alkyl or aryl group (R) attached to a carboxyl group (
{{-}\rm{COOH}}
molecule that does not have a plane of symmetry and its isomers cannot be rotated or reflected to match
stereoisomer that has a mirror image that is not superimposable on itself
organic compound that contains a carboxyl unit in which a hydroxyl group is replaced by an alkyl or aryl group, giving it
{\rm{R{-}C({=}O)OR^\prime}}
{\rm{R{-}COOR^\prime}}
organic molecule containing an oxygen atom bound by two alkyl or aryl groups through
{\rm{C}{-}{O}}
group of atoms with specific physical, chemical, and reactivity properties
one of two or more molecules that have different spatial arrangements of functional groups around a double bond, ring, or other rigid structures
organic compound that contains only carbon-carbon and carbon-hydrogen bonds
addition of a hydroxyl group (
{{-}\rm{OH}}
) to a molecule, which can be accomplished by a substitution reaction with an alkyl halide
functional group with the formula
{{-}\rm{OH}}
{\rm{C}{-}{OH}}
fragments and characterizes molecules called alcohols and phenols
one of two or more molecules that have the same chemical formula but different molecular structures
benzene ring in which a
{\rm{C}{-}{H}}
{\rm{C}{-}{OH}}
organic compound that contains only carbon-carbon single bonds and
{\rm{C}{-}{H}}
carbon atom with four unique substituents
atom or group of atoms (functional group) that replaces a
{\rm{C}{-}{H}}
bond in an organic compound
organic compound that is derived from H2S. It contains an alkyl or aryl group covalently linked to a sulfhydryl group,
{{-}\rm{SH}}
, through
{\rm{C}{-}{S}}
bonds, with
{\rm{R}{-}{SH}}
organic compound that contains
{\rm{C}{-}{H}}
{\rm{C}{=}{C}}
{\rm{C}{\equiv}{C}}
<Overview>What is Organic Chemistry? |
Largest Nth Power - Maple Help
Home : Support : Online Help : Mathematics : Number Theory : Largest Nth Power
LargestNthPower(m, n)
The LargestNthPower(m, n) command computes the greatest positive integer
b
{b}^{n}
divides m.
Every positive integer is a divisor of 0, so there is no greatest positive integer
b
{b}^{n}
divides 0. For this reason, LargestNthPower(0, n) returns an error.
\mathrm{with}\left(\mathrm{NumberTheory}\right):
\mathrm{LargestNthPower}\left({m}^{2},1\right)
{\textcolor[rgb]{0,0,1}{m}}^{\textcolor[rgb]{0,0,1}{2}}
\mathrm{LargestNthPower}\left(-1,\mathrm{exp}\left(k\right)\right)
\textcolor[rgb]{0,0,1}{1}
\mathrm{LargestNthPower}\left(0,k\right)
Error, (in NumberTheory:-LargestNthPower) there is no largest integer which, raised to the power k, divides 0
The greatest integer power divisor can be seen from the prime factorization.
\mathrm{LargestNthPower}\left({2}^{2}{3}^{4}{5}^{3},2\right)
\textcolor[rgb]{0,0,1}{90}
\mathrm{ifactor}\left(90\right)
\left(\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{}{\left(\textcolor[rgb]{0,0,1}{3}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{5}\right)
The NumberTheory[LargestNthPower] command was introduced in Maple 2016. |
Translate each statement into a linear equality or inequality then solve algebraically. 1. Seven more than three ti
Translate each statement into a linear equality or inequalit
Translate each statement into a linear equality or inequality then solve algebraically.
1. Seven more than three times a number equals eight less than four times the number.
2. Twelve minus nine times a number is the same as two minus eight times the number.
3. Five minus two times x is less than or equal to seven minus three times x.
The objective is to solve the given statements algebraically.
1) It is given that, seven more than three times a number equals eight less than four times the number.
Consider the unknown number as x. Then the equation can be represented as,
7+3x=4x-8
On solving algebraically,
7+3x=4x-8
4x-3x=8+7
x=15
Hence, the value of the unknown number is 15.
2) It is given that twelve minus nine times a number is the same as two minus eight times a number.
12-9x=2-8x
12-9x=2-8
9x-8x=12-2
x=10
3) It is given that five minus two times x is less than or equal to seven minus three times x.
The given equation can be represented as,
5-2x\le 7-3x
3x-2x\le 7-5
x\le 2
Hence, the value of the unknown number is less than or equal to 2.
x=\frac{1}{7}
x=10
3x+7=8-4x
3x+7=8±4x
3x+7=-4x+8
3x+7+4x=-4x+8+4x
7x+7=8
7x+7-7=8-7
7x=1
7\frac{x}{7}=\frac{1}{7}
x=\frac{1}{7}
9x-12=8x-2
9x-12-8x=8x-2-8x
x-12=-2
x-12+12=-2+12
x=10
12-9x=2-8
5-2x\le 7-3x
3x-2x\le 7-5
x\le 2
Write the trigonometric expression as an algebraic expression in u.
\mathrm{sec}\left({\mathrm{cos}}^{-1}u\right)
A=P{e}^{rt}
A=3P
Find t if r is:
Find the value of x so that the slope of the line through the points
\left(x,-3\right)\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }\left(4,5\right)\text{ }is\text{ }\frac{2}{3}
Janine worked 15 hours last week. One job as a clerk in a sporting goods store paid her $8.46 per hour, while her job giving piano lessons paid $18 per hour. If she earned $184.14 between the two jobs, how many hours did she work at each job?
Write an algebraic expression for the verbal description. The total revenue obtained by selling x units at $10.74 per unit $?
Write an algebraic expression to represent each verbal expression. Fifteen less than the cube of a number. |
"Stronger" Induction | Brilliant Math & Science Wiki
"Stronger" Induction
Suparjo Tamin, Calvin Lin, and Jimin Khim contributed
Prove that for all positive integer
n
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right) \cdots \left(1+\frac{1}{2^n}\right) < \frac{5}{2}.
Notice that the LHS of the above inequality increases as
n
increases but the RHS remains constant. So induction doesn't immediately work here. To prove this inequality, we make a minor change on the expression as follows:
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right) \cdots \left(1+\frac{1}{2^n}\right) < \frac{5}{2}- \frac{m}{2^n}.
The goal is to determine a positive constant
m
that will allow us to prove the inequality by induction. If this inequality is true, we can conclude that the original inequality is also true!
First, let's determine what
m
should be. The induction step is:
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right) \cdots \left(1+\frac{1}{2^{k+1} }\right) < \frac{ 5}{2} - \frac{ m}{ 2^{k+1} }.
Applying the induction hypothesis, we obtain
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right) \cdots \left(1+\frac{1}{2^{k+1} }\right) < \left ( \frac{ 5}{2} - \frac{ m}{ 2^{k} } \right) \times \left( 1 + \frac{1}{ 2^{k+1 } } \right) < \frac{ 5}{2} - \frac{ m}{ 2^{k+1} }.
Expanding and simplifying, we need that
\frac{ 5 }{ 2^{k+2 } } < \frac{ 2m } { 2^{k+2} } + \frac{m} {2^{2k+1 } }.
Now, this is clearly satisfied for
m = 3
It now remains to find a base case to make the strengthened statement true. For
n = 3
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right) = \frac{135}{64} < \frac{136}{64} = \frac{5}{2} - \frac{3} { 2^3 }.
Thus, we will begin our induction on the strengthened statement and a base case of
n = 3
. We will deal with
n = 1, 2
separately, since they do not satisfy the strengthened statement.
We didn't need to focus on a constant
m
. We could make it a function
m(n)
and attempted to find a better shape. Of course, this additional specification leads to more complication, and it could no longer be so easy to find a suitable function.
In such problems where we "over-compensate," the strengthened induction hypothesis might not be true for some initial cases. We simply treat them as part of our base cases.
Cite as: "Stronger" Induction. Brilliant.org. Retrieved from https://brilliant.org/wiki/stronger-induction/ |
Create inverted-F antenna over rectangular ground plane - MATLAB - MathWorks France
Create and View Inverted-F Antenna
Plot Radiation Pattern of Inverted-F
Create inverted-F antenna over rectangular ground plane
The invertedF object is an inverted-F antenna mounted over a rectangular ground plane.
The width of the metal strip is related to the diameter of an equivalent cylinder by the equation
w=2d=4r
d is the diameter of equivalent cylinder
r is the radius of equivalent cylinder
For a given cylinder radius, use the utility function cylinder2strip to calculate the equivalent width. The default inverted-F antenna is center-fed. The feed point coincides with the origin. The origin is located on the xy- plane.
f = invertedF
f = invertedF(Name,Value)
f = invertedF creates an inverted-F antenna mounted over a rectangular ground plane. By default, the dimensions are chosen for an operating frequency of 1.7 GHz.
f = invertedF(Name,Value) creates an inverted-F antenna, with additional properties specified by one, or more name-value pair arguments. Name is the property name and Value is the corresponding value. You can specify several name-value pair arguments in any order as Name1, Value1, ..., NameN, ValueN. Properties not specified retain their default values.
Height — Vertical element height along z-axis
Vertical element height along z-axis, specified a scalar in meters.
Strip width should be less than 'Height'/4 and greater than 'Height'/1001. [2]
LengthToOpenEnd — Stub length from feed to open end
Stub length from feed to open end, specified as a scalar in meters.
Example: 'LengthToOpenEnd',0.05
LengthToShortEnd — Stub length from feed to shorting end
Stub length from feed to shorting end, specified as a scalar in meters.
Example: 'LengthToShortEnd',0.0050
Example: f.Load = lumpedElement('Impedance',75)
Create and view an inverted-F antenna with 14 mm height over a ground plane of dimensions 200 mm-by-200 mm.
f = invertedF('Height',14e-3, 'GroundPlaneLength',200e-3, ...
'GroundPlaneWidth',200e-3);
This example shows you how to plot the radiation pattern of an inverted-F antenna for a frequency of 1.3 GHz.
f = invertedF('Height',14e-3, 'GroundPlaneLength', 200e-3, ...
pattern(f,1.39e9)
invertedL | pifa | patchMicrostrip | cylinder2strip |
Comparative gallium-68 labeling of TRAP-, NOTA-, and DOTA-peptides: practical consequences for the future of gallium-68-PET | EJNMMI Research | Full Text
Comparative gallium-68 labeling of TRAP-, NOTA-, and DOTA-peptides: practical consequences for the future of gallium-68-PET
Karolin Pohle1,2 &
Currently, 68Ga-labeled 1,4,7,10-tetraazacyclododecane-tetraacetic acid (DOTA)-peptides are the most widely used class of 68Ga radiotracers for PET, although DOTA is not optimal for 68Ga complexation. More recently, 1,4,7-triazacyclononane-triacetic acid (NOTA) and particularly triazacyclononane-phosphinate (TRAP) chelators have been shown to possess superior 68Ga binding ability. Here, we report on the efficiency, reproducibility, and achievable specific activity for fully automated 68Ga labeling of DOTA-, NOTA-, and TRAP-peptide conjugates.
Compared to NOTA- and DOTA-peptides, achievable specific activity (A S) for TRAP-peptide is approximately 10 and 20 times higher, respectively. A S values in the range of 5,000 GBq/μmol were routinely obtained using 1 GBq of 68Ga, equivalent to 0.11 μg of cold mass for a 185-MBq patient dose of a 3-kDa conjugate. The TRAP-peptide could be 68Ga-labeled with excellent reproducibility and > 95% radiochemical yield for precursor amounts as low as 1 nmol.
High 68Ga labeling efficiency of TRAP-peptides could facilitate realization of kit labeling procedures. The good reproducibility of the automated synthesis is of relevance for GMP production, and the possibility to provide very high specific activities offers a high degree of safety in first clinical trials, due to reduction of cold mass content in tracer formulations.
With the commercial availability of 68Ge/68Ga generators, cyclotron-independent on-site production of tracers for positron-emission tomography (PET) has become widely feasible [1, 2]. Thus, in the near future, a ubiquitous implementation of PET and PET/CT even in regions with less well-developed infrastructure can be expected, similar to the global story of success of 99mTc-based scintigraphy which started half a century ago with the introduction of 99Mo/99mTc generators [1, 3]. In the long run, a partial substitution of single photon emission computed tomography (SPECT) by PET (and SPECT/CT by PET/CT, respectively) appears to be a realistic scenario in view of the advantages of PET, such as superior spatial resolution and sensitivity. Besides, in contrast to reactor-produced 99Mo, 68Ge is cyclotron-produced. This can be considered advantageous with regard to the recent insufficiency of global reactor capacity for reliable 99Mo supply [4], and independence of 68Ga-PET from nuclear reactors might positively influence the bias of its public perception.
Generally, 68Ga labeling is done by complexation of the 68Ga3+ ion. For this purpose, dedicated chelators usually have to be introduced into precursor molecules by bioconjugation, wherein they readily determine the labeling chemistry. To facilitate global implementation of 68Ga-PET, production of 68Ga radiopharmaceuticals must be simple, robust, and reliable; this demands highly efficient labeling chemistry and, therefore, highly efficient chelators. Recently, we have shown that the bifunctional triazacyclononane-phosphinate (TRAP) ligand [5–8] possesses markedly improved 68Ga labeling properties [6]. This applies also to TRAP-based peptide conjugates, the practical consequences of which we further elucidate in this study.
TRAP(RGD)3 was prepared as described before [6]. NODAGA-cyclo(RGDyK) (‘NODAGA-RGD’) was purchased from ABX GmbH (Radeberg, Germany). DOTATOC was obtained from Bachem (Bubendorf, Switzerland). Fully automated 68Ga labeling was performed using unpurified eluate fractions of a 68Ge/68Ga generator with SnO2 matrix (iThemba LABS, Somerset West, South Africa), as described previously [6, 9] (5 min reaction at 95°C, pH adjusted with HEPES, pH 3.2 for DOTATOC and NODAGA-RGD, pH 2 for TRAP(RGD)3, purification using C8 SPE cartridge). Radiochemical yield was calculated from decay-corrected product activity in relation to the sum of significant decay-corrected residual activities contained elsewhere, that is, in reaction vial, SPE cartridge, and non-product cartridge purging liquids.
Calculation of specific activities
Product activities (A P) were measured after the end of preparation (approximately 15 min after the start of syntheses) and decay corrected to a typical injection time, 30 min after the start of synthesis (A P,30). In order to be able to calculate corresponding specific activity values that are representative for the respective precursor amounts and independent from small deviations in the starting activity A 0 (in our experiments, ranging from 800 to 1,050 MBq, depending on the regeneration state of the 68Ga generator), product activities were normalized to a representative starting activity A N = 1 GBq, according to
{A}_{\text{P},30,\text{N}}={A}_{\text{P},30}\left(\frac{{A}_{\text{N}}}{{A}_{0}}\right)
. Specific activity (A S) values were calculated by the division of A P,30,N by the precursor amount used. It is assumed that all precursor peptide is actually retained on and subsequently eluted from the cartridge, and thus transferred into the formulation. This means that both retention and elution efficiency are considered 100%, both of which can be somewhat lower in practice. As a result, all given A S represent the lower bounds and will never overestimate actual values.
Although previous comparative studies focusing on the basic chelator structures TRAP, 1,4,7-triazacyclononane-triacetic acid (NOTA), and 1,4,7,10-tetraazacyclododecane-tetraacetic acid (DOTA) already proved superior Ga3+ complexation/68Ga labeling properties of TRAP [6, 7], these data are not sufficient to quantify the behavior of respective peptide conjugates, for two reasons: Firstly, functionalization of chelators with peptides, resulting in conjugates with a multiple of the molecular weight of the neat chelators, is definitely prone to change overall complexation properties with potentially unpredictable outcome. Secondly, the chelating moiety in compounds commonly dubbed ‘DOTA-peptides’ is actually not DOTA, but DOTA-monoamide (see Figure 1), which exhibits different Ga3+ complexation behavior [10]. In contrast, for TRAP and the bifunctional NOTA-derivative NODAGA, the structure of the chelating site is not affected by conjugation. To assess the impact of these effects, representative peptide conjugates (TRAP(RGD)3 [6] and the commercially available ‘NOTA’- and ‘DOTA’-peptides NODAGA-RGD and DOTATOC, respectively, see Figure 1) were labeled under similar conditions using our standard automated procedure [6, 9].
Structures of peptide conjugates used in this study. The complexation sites of TRAP, NOTA, and DOTA-monoamide, featured in TRAP(RGD)3, NODAGA-RGD, and DOTATOC, respectively, are highlighted in blue color.
Figure 2 shows that TRAP(RGD)3 allows to use much lower precursor concentrations for labeling than required for NODAGA-RGD and particularly DOTATOC, which is why 68Ga-TRAP(RGD)3 can be prepared with much higher A S (see also Table 1). Using 0.1 nmol of TRAP(RGD)3, almost 5,000 GBq/μmol was reached with a satisfying decay-corrected yield of 66 ± 6%. The use of even lower amounts of TRAP(RGD)3 (17 pmol) frequently resulted in preparations with extremely high A S of >10,000 GBq/μmol, although not reliably reproducible. The highest A S value observed during these experiments was 14,900 GBq/μmol (actual value, not normalized to starting activity), which is approximately 1/7 of the theoretically possible maximum value, that of carrier-free 68Ga. Although such high specific activities are not usually needed for clinical applications, we nevertheless, deem this feature of high practical value for the following reasons:
A hypothetical patient dose of 185 MBq (5 mCi) of a 5,000-GBq/μmol preparation contains only 37 pmol of peptide; for a compound like TRAP(RGD)3 with a molecular weight of ≈ 3 kDa, this calculates to a total of 0.11 μg of cold mass, or less than 2 ng/kg body weight for an average patient. Such tiny amounts are extremely unlikely to cause any pharmacological effects. Therefore, TRAP could facilitate the use of such biomolecules for imaging that possess very high pharmacological potential, and 68Ga-labeled TRAP conjugates could generally offer high safety when tested in first clinical trials.
As a 15-MBq dose of said preparation is equivalent to 3 pmol or 9 ng of our exemplary 3-kDa peptide, it can always directly be used for evaluation studies in rodents without having to separate off unlabeled precursor or, unfavorably, reduce the administered dose. High receptor occupancy or even saturation effects, which otherwise are frequently encountered in small animal imaging due to the necessity of applying much higher activity doses per kilogram body weight than in humans, can be practically ruled out.
Several studies have outlined that the amount of co-injected cold mass can have a significant influence on biodistribution and imaging results [11–14]. In clinical routine, it is therefore highly recommended to utilize radiopharmaceutical formulations with constant, optimized specific activity (i.e., well-defined cold mass content). Such productions could be done most conveniently and reliably by adding the desired amount of active compound to a pre-conditioned vial containing a fixed amount of cold standard. This approach, however, requires radiolabeled tracers with very high specific activity in order not to change the overall contained amount of cold mass significantly. 68Ga-labeled TRAP conjugates appear ideally suited for this purpose.
Radiochemical yields and corresponding calculated minimal A S of radiopharmaceutical formulations. Radiochemical yields (solid lines, %, mean ± SD, n ≥ 4) and corresponding calculated minimal A S (dashed lines, GBq/μmol, mean ± SD) of the formulations at typical time of injection (30 min after the start of synthesis) as functions of precursor amount for automated 68Ga labeling of TRAP(RGD)3 (T), NODAGA-RGD (N), and DOTATOC (D). A S for TRAP(RGD)3 concentrations > 1 nmol are not shown for clarity of presentation.
Table 1 Radiochemical yields of automated 68 Ga labeling, and corresponding calculated minimal A S of radiopharmaceutical formulations
Furthermore, regarding Figure 2, one notices that variation of radiochemical yields, reflected by the size of error bars, is the larger the lower precursor amounts are. This is because the generator eluate usually contains traces of ionic contaminants, such as Zn2+, Sn4+, Al3+, and Fe3+, the concentration of which in the individual eluates is varying. These can compete with 68Ga3+ at the chelating site of the precursor, which is naturally the more impacting on labeling yield the lower the stoichiometric excess of precursor over 68Ga3+ ion is. The error bars in Figure 2 show that except for precursor amounts exceeding 20 nmol, use of a TRAP peptide will result in a more reliable radiosynthesis, being less prone to be perturbed by variation of other parameters (reaction pH, eluate volume, trace metal contaminations, etc.). Differences in radiochemical yield and reproducibility are very pronounced for peptide amounts in the range of 1 to 10 nmol (e.g., equivalent to 1.4 to 14 μg of DOTATOC) which are frequently used in routine 68Ga labeling procedures; near-quantitative yields and excellent reproducibility can be expected here for a TRAP peptide. This is of high relevance for routine GMP tracer production, where reproducibility and robustness of procedures is crucial. In addition, we assume that due to higher labeling efficiency, realization of kit labeling procedures will be much simpler using TRAP conjugates, which we deem of importance for the aforementioned possibility of global implementation of 68Ga-PET. Finally, the recent introduction of NOPO, a TRAP variety designed specifically for monoconjugation, expands the portfolio of P-functionalized triazacyclononane-triphosphinate chelators, thus offering even more synthetic possibilities for development of 68Ga tracers [15].
1,4,7,10-tetraazacyclododecane-tetraacetic acid
Triazacyclononane-phosphinate.
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Fani M, André JP, Maecke HR: 68 Ga-PET: a powerful generator-based alternative to cyclotron-based PET radiopharmaceuticals. Contrast Media Mol Imaging 2008, 3: 67–77.
Bartholomä MD, Louie AS, Valliant JF, Zubieta J: Technetium and gallium derived radiopharmaceuticals: comparing and contrasting the chemistry of two important radiometals for the molecular imaging era. Chem Rev 2010, 110: 2903–2920. 10.1021/cr1000755
Ballinger JR: 99 Mo shortage in nuclear medicine: crisis or challenge? J Label Compd Radiopharm 2010, 53: 167–168.
Notni J, Hermann P, Havlíčková J, Kotek J, Kubíček V, Plutnar J, Loktionova N, Riss PJ, Rösch F, Lukeš I: A triazacyclononane-based bifunctional phosphinate ligand for the preparation of multimeric 68 Ga tracers for positron emission tomography. Chem Eur J 2010, 16: 7174–7185.
Notni J, Šimeček J, Hermann P, Wester HJ: TRAP, a powerful and versatile framework for gallium-68 radiopharmaceuticals. Chem Eur J 2011, 17: 14718–14722. 10.1002/chem.201103503
Notni J, Plutnar J, Wester HJ: Bone seeking TRAP conjugates: surprising observations and implications on development of gallium-68-labeled bisphosphonates. EJNMMI Res 2012, 2: 13. 10.1186/2191-219X-2-13
Pohle K, Notni J, Bussemer J, Kessler H, Schwaiger M, Beer AJ: 68 Ga-NODAGA-RGD is a suitable substitute for 18 F-Galacto-RGD and can be produced with high specific activity in a cGMP/GRP compliant automated process . Nucl Med Biol 2012. 10.1016/j.nucmedbio.2012.02.006
Kubíček V, Havlíčková J, Kotek J, Tircsó G, Hermann P, Tóth E, Lukeš I: Gallium(III) complexes of DOTA and DOTA-monoamide: kinetic and thermodynamic studies. Inorg Chem 2010, 49: 10960–10969. 10.1021/ic101378s
Froidevaux S, Calame-Christe M, Schuhmacher J, Tanner H, Saffrich R, Henze M, Eberle AN: A gallium-labeled DOTA-alpha-melanocyte-stimulating hormone analog for PET imaging of melanoma metastases. J Nucl Med 2004, 45: 116–123.
Breeman WAP, Kwekkeboom DK, Kooij PPM, Bakker WH, Hofland LJ, Visser TJ, Ensing GJ, Lamberts SWJ, Krenning EP: Effect of dose and specific activity on tissue distribution of indium-111-pentetreotide in rats. J Nucl Med 1995, 36: 623–627.
de Jong M, Breeman WAP, Bernard BF, van Gameren A, de Bruin E, Bakker WH, van der Pluijm ME, Visser TJ, Mäcke HR, Krenning EP: Tumour uptake of the radiolabelled somatostatin analogue [DOTA0 , TYR3 ]octreotide is dependent on the peptide amount. Eur J Nucl Med 1999, 26: 693–698. 10.1007/s002590050439
Velikyan I, Sundin A, Eriksson B, Lundqvist H, Sörensen J, Bergström M, Långström B: In vivo binding of [68 Ga]-DOTATOC to somatostatin receptors in neuroendocrine tumours—impact of peptide mass. Nucl Med Biol 2010, 37: 265–275. 10.1016/j.nucmedbio.2009.11.008
Simecek J, Zemek O, Hermann P, Wester HJ, Notni J: A monoreactive bifunctional triazacyclononane-phosphinate chelator with high selectivity for gallium-68. Chem Med Chem 2012. 10.1002/cmdc.201200261
Johannes Notni, Karolin Pohle & Hans-Jürgen Wester
Department of Nuclear Medicine, Klinikum rechts der Isar, Technische Universität München, Ismaninger Str. 22, Munich, 81675, Germany
JN developed the study concept, performed the radiolabeling of TRAP(RGD)3 and DOTATOC, and wrote the manuscript. KP performed the radiolabeling of NODAGA-RGD and critically reviewed the manuscript. HJW gave advice in the interpretation of the data and critically reviewed the manuscript. All authors approved the final manuscript.
Notni, J., Pohle, K. & Wester, HJ. Comparative gallium-68 labeling of TRAP-, NOTA-, and DOTA-peptides: practical consequences for the future of gallium-68-PET. EJNMMI Res 2, 28 (2012). https://doi.org/10.1186/2191-219X-2-28 |
Adding velocity vectors | Brilliant Math & Science Wiki
We've seen that we can repeat a displacement
n
times by multiplying its length by the scalar
n
\vec{m}_1 \rightarrow \vec{m}_n = n \vec{m}_1
Another way to think of this repetition is as the addition of
n
\vec{m}_1
\vec{m}_n = \overbrace{\vec{m}_1 + \ldots + \vec{m}_1}^{n\text{ times}}
For example, consider the case of
\vec{m}_2 = 2 \vec{m}_1
What if we have two distinct displacement vectors
\vec{m}_1
\vec{m}_2
that we wish to compose? How can we combine the two so that we undertake the displacement one
\vec{m}_1
, followed by displacement two
\vec{m}_2
Let's draw them out:
We can see that the way to add displacement vectors is to line them up tip to tail. The resultant displacement is the vector that spans from the tail of the first to the tip of the second. Notice that the resultant vector is independent of the order. We can do either first, and we always end up with the same sum.
As always, the result of a vector operation is independent of the representation, and so is our sum. However, in the Cartesian coordinate system, the sum of two vectors can be calculate in a particularly convenient way. Suppose we have two displacement vectors
\vec{m}_1 = \langle 1, 2, 3\rangle
\vec{m}_2 = \langle 2, -3, 7\rangle
. If we follow our arrow drawing carefully, we see that the components of the sum is given simply by adding the components of
\vec{m}_1
\vec{m}_2
in pairwise fashion:
\begin{aligned} \vec{m}_1 + \vec{m}_1 & = \langle x_1, y_1, z_1\rangle + \langle x_2, y_2, z_2\rangle \\ &= \langle x_1 + x_2, y_1 + y_2, z_1 + z_2 \rangle \end{aligned}
By composing displacement vectors, we can move up, down, to the side, diagonally, or backwards in a series of steps. It is even possible to go nowhere through a series of intermediate steps, which is what happens when we walk in a circle:
Breaking complex trajectories into a composition of vector displacements can give elegant arguments and explanations for some of the most complex rules of the universe, as Feynman showed in his book on quantum electrodynamics, QED.
Cite as: Adding velocity vectors. Brilliant.org. Retrieved from https://brilliant.org/wiki/adding_velocity_vectors/ |
Do heavier objects fall faster than lighter objects? | Brilliant Math & Science Wiki
Rohit Gupta, Mehul Arora, Calvin Lin, and
Why some people say it's true: If a feather and an egg are dropped, then the egg will reach the ground first.
Why some people say it's false: Acceleration due to gravity is independent of the mass of the object.
\color{#20A900}{\text{Reveal the Correct Answer:}}
\color{#D61F06}{\textbf{false}}
The above statement is not true in all situations. It is true that if from the top of a building, a feather and an egg are dropped then the egg will win the race. To understand why feather lost, let's think about the forces. During the fall, there are two forces acting on the body. One of the forces is the gravitational force, which depends on the mass of the object. The other force is air resistance which depends on the surface area of the objects. The net acceleration of an object is thus given by,
a = g - \frac{{{f_{air\,drag}}}}{M}.
g
is acceleration due to gravity,
{{f_{air\,drag}}}
is the air resistance,
M
is the mass of the object, and
a
is the acceleration of the object. It can be seen from the above equation that the larger the mass the greater the acceleration and the air drag will have lesser effects. Therefore, feather loses due to the air drag.
What if the air resistance is not present? Which one will win the race this time? In the above equation if
is put equal to zero, then the acceleration of the object equals acceleration due to gravity
g.
This acceleration is independent of the mass of the object, and thus both the feather and the egg will fall with the same acceleration and reach the ground at the same time.
\color{#3D99F6}{\text{See Further Discussions:}}
Query: What will happen if the air drag acting on the objects is different?
Reply: Air drag depends on the surface area and the speed of the object. It can be different, but the mass of the egg is much higher than a feather, which overpowers the effects of the difference in air drag. At the start when the speed is negligible, the air drag on the feather is greater as it has a larger surface area.
Query: What will be the effects of buoyancy of air that acts on the objects?
Reply: Yes, buoyancy of air acts on objects. But air has a very less density, and therefore the buoyancy effects of air can be neglected.
Iron Both will reach at same time Cotton
Suppose Galileo dropped a one-kilogram ball of cotton and one-kilogram ball of iron from the top of the Leaning Tower of PIsa, then which one will reach the ground first?
Assume that the cotton ball is tightly wadded up and that initially the bottoms of the cotton ball and iron ball are at the same horizontal level.
Cite as: Do heavier objects fall faster than lighter objects?. Brilliant.org. Retrieved from https://brilliant.org/wiki/do-heavier-objects-fall-faster-than-lighter/ |
Follow instructions to create a construction
Use precise mathematical language to describe a construction
Required materials: compass, straightedge
Explain why each statement is true.
Length of
\overline{EA}
is equal to the length of
\overline{EB}
\Delta ABF
is equilateral
AB=\frac{1}{3}CD
CB=DA
Swap instructions with a neighbor, but do not let them see your finished pattern!
What was difficult about attempting someone else's pattern?
What would you change to make it easier?
Compass and straightedge moves can be used to create interesting patterns. However, for the patterns to be recreated, precise instructions must be made. Labelling points and segments are helpful in creating clear directions. |
Begriffsschrift Knowpia
The title page of the original 1879 edition
Lubrecht & Cramer
Begriffsschrift is usually translated as concept writing or concept notation; the full title of the book identifies it as "a formula language, modeled on that of arithmetic, for pure thought." Frege's motivation for developing his formal approach to logic resembled Leibniz's motivation for his calculus ratiocinator (despite that, in the foreword Frege clearly denies that he achieved this aim, and also that his main aim would be constructing an ideal language like Leibniz's, which Frege declares to be a quite hard and idealistic—though not impossible—task). Frege went on to employ his logical calculus in his research on the foundations of mathematics, carried out over the next quarter century. This is the first work in Analytical Philosophy, a field that future British and Anglo philosophers such as Bertrand Russell further developed.
Notation and the systemEdit
The calculus contains the first appearance of quantified variables, and is essentially classical bivalent second-order logic with identity. It is bivalent in that sentences or formulas denote either True or False; second order because it includes relation variables in addition to object variables and allows quantification over both. The modifier "with identity" specifies that the language includes the identity relation, =. Frege stated that his book was his version of a characteristica universalis, a Leibnizian concept that would be applied in mathematics.[1]
Frege presents his calculus using idiosyncratic two-dimensional notation: connectives and quantifiers are written using lines connecting formulas, rather than the symbols ¬, ∧, and ∀ in use today. For example, that judgement B materially implies judgement A, i.e.
{\displaystyle B\rightarrow A}
In the first chapter, Frege defines basic ideas and notation, like proposition ("judgement"), the universal quantifier ("the generality"), the conditional, negation and the "sign for identity of content"
{\displaystyle \equiv }
(which he used to indicate both material equivalence and identity proper); in the second chapter he declares nine formalized propositions as axioms.
Modern notations
{\displaystyle \vdash A,\Vdash A}
{\displaystyle p(A)=1,}
{\displaystyle p(A)=i}
{\displaystyle \vdash A,\Vdash A}
{\displaystyle \neg A}
{\displaystyle {\sim }A}
Conditional (implication)
{\displaystyle B\rightarrow A}
{\displaystyle B\supset A}
{\displaystyle \forall x\,F(x)}
{\displaystyle \exists x\,F(x)}
Content identity (equivalence/identity)
{\displaystyle A\equiv B}
{\displaystyle A\leftrightarrow B}
{\displaystyle A\equiv B}
{\displaystyle A=B}
In chapter 1, §5, Frege defines the conditional as follows:
"Let A and B refer to judgeable contents, then the four possibilities are:
A is asserted, B is asserted;
A is asserted, B is negated;
A is negated, B is asserted;
A is negated, B is negated.
signify that the third of those possibilities does not obtain, but one of the three others does. So if we negate
, that means the third possibility is valid, i.e. we negate A and assert B."
The calculus in Frege's workEdit
Frege declared nine of his propositions to be axioms, and justified them by arguing informally that, given their intended meanings, they express self-evident truths. Re-expressed in contemporary notation, these axioms are:
{\displaystyle \vdash \ \ A\rightarrow \left(B\rightarrow A\right)}
{\displaystyle \vdash \ \ \left[\ A\rightarrow \left(B\rightarrow C\right)\ \right]\ \rightarrow \ \left[\ \left(A\rightarrow B\right)\rightarrow \left(A\rightarrow C\right)\ \right]}
{\displaystyle \vdash \ \ \left[\ D\rightarrow \left(B\rightarrow A\right)\ \right]\ \rightarrow \ \left[\ B\rightarrow \left(D\rightarrow A\right)\ \right]}
{\displaystyle \vdash \ \ \left(B\rightarrow A\right)\ \rightarrow \ \left(\lnot A\rightarrow \lnot B\right)}
{\displaystyle \vdash \ \ \lnot \lnot A\rightarrow A}
{\displaystyle \vdash \ \ A\rightarrow \lnot \lnot A}
{\displaystyle \vdash \ \ \left(c=d\right)\rightarrow \left(f\left(c\right)=f\left(d\right)\right)}
{\displaystyle \vdash \ \ c=c}
{\displaystyle \vdash \ \ \forall a\ f(a)\rightarrow \ f(c)}
These are propositions 1, 2, 8, 28, 31, 41, 52, 54, and 58 in the Begriffschrifft. (1)–(3) govern material implication, (4)–(6) negation, (7) and (8) identity, and (9) the universal quantifier. (7) expresses Leibniz's indiscernibility of identicals, and (8) asserts that identity is a reflexive relation.
All other propositions are deduced from (1)–(9) by invoking any of the following inference rules:
Modus ponens allows us to infer
{\displaystyle \vdash B}
{\displaystyle \vdash A\to B}
{\displaystyle \vdash A}
The rule of generalization allows us to infer
{\displaystyle \vdash P\to \forall xA(x)}
{\displaystyle \vdash P\to A(x)}
if x does not occur in P;
The rule of substitution, which Frege does not state explicitly. This rule is much harder to articulate precisely than the two preceding rules, and Frege invokes it in ways that are not obviously legitimate.
The main results of the third chapter, titled "Parts from a general series theory," concern what is now called the ancestral of a relation R. "a is an R-ancestor of b" is written "aR*b".
Frege applied the results from the Begriffsschrifft, including those on the ancestral of a relation, in his later work The Foundations of Arithmetic. Thus, if we take xRy to be the relation y = x + 1, then 0R*y is the predicate "y is a natural number." (133) says that if x, y, and z are natural numbers, then one of the following must hold: x < y, x = y, or y < x. This is the so-called "law of trichotomy".
"If the task of philosophy is to break the domination of words over the human mind [...], then my concept notation, being developed for these purposes, can be a useful instrument for philosophers [...] I believe the cause of logic has been advanced already by the invention of this concept notation."
— Preface to the Begriffsschrift
Influence on other worksEdit
For a careful recent study of how the Begriffsschrift was reviewed in the German mathematical literature, see Vilko (1998). Some reviewers, especially Ernst Schröder, were on the whole favorable. All work in formal logic subsequent to the Begriffsschrift is indebted to it, because its second-order logic was the first formal logic capable of representing a fair bit of mathematics and natural language.
Some vestige of Frege's notation survives in the "turnstile" symbol
{\displaystyle \vdash }
derived from his "Urteilsstrich" (judging/inferring stroke) │ and "Inhaltsstrich" (i.e. content stroke) ──. Frege used these symbols in the Begriffsschrift in the unified form ├─ for declaring that a proposition is true. In his later "Grundgesetze" he revises slightly his interpretation of the ├─ symbol.
In "Begriffsschrift" the "Definitionsdoppelstrich" (i.e. definition double stroke) │├─ indicates that a proposition is a definition. Furthermore, the negation sign
{\displaystyle \neg }
can be read as a combination of the horizontal Inhaltsstrich with a vertical negation stroke. This negation symbol was reintroduced by Arend Heyting[2] in 1930 to distinguish intuitionistic from classical negation. It also appears in Gerhard Gentzen's doctoral dissertation.
In the Tractatus Logico Philosophicus, Ludwig Wittgenstein pays homage to Frege by employing the term Begriffsschrift as a synonym for logical formalism.
Frege's 1892 essay, "On Sense and Reference," recants some of the conclusions of the Begriffsschrifft about identity (denoted in mathematics by the "=" sign). In particular, he rejects the "Begriffsschrift" view that the identity predicate expresses a relationship between names, in favor of the conclusion that it expresses a relationship between the objects that are denoted by those names.
Gottlob Frege. Begriffsschrift: eine der arithmetischen nachgebildete Formelsprache des reinen Denkens. Halle a/S: Verlag von Louis Nebert, 1879.
Bynum, Terrell Ward, trans. and ed., 1972. Conceptual notation and related articles, with a biography and introduction. Oxford Uni. Press.
Bauer-Mengelberg, Stefan, 1967, "Concept Script" in Jean van Heijenoort, ed., From Frege to Gödel: A Source Book in Mathematical Logic, 1879-1931. Harvard Uni. Press.
Beaney, Michael, 1997, "Begriffsschrift: Selections(Preface and Part I)" in The Frege Reader. Oxford: Blackwell.
Calculus of equivalent statements
^ Korte, Tapio (2008-10-22). "Frege's Begriffsschrift as a lingua characteristica". Synthese. 174 (2): 283–294. doi:10.1007/s11229-008-9422-7.
^ Arend Heyting: "Die formalen Regeln der intuitionistischen Logik," in: Sitzungsberichte der preußischen Akademie der Wissenschaften, phys.-math. Klasse, 1930, pp. 42–65.
George Boolos, 1985. "Reading the Begriffsschrift", Mind 94: 331–44.
Ivor Grattan-Guinness, 2000. In Search of Mathematical Roots. Princeton University Press.
Risto Vilkko, 1998, "The reception of Frege's Begriffsschrift," Historia Mathematica 25(4): 412–22.
Wikimedia Commons has media related to Begriffsschrift.
Zalta, Edward N. "Frege's Logic, Theorem, and Foundations for Arithmetic". In Zalta, Edward N. (ed.). Stanford Encyclopedia of Philosophy.
Begriffsschrift as facsimile for download (2.5 MB) |
Overshoot metrics of bilevel waveform transitions - MATLAB overshoot
Overshoot Percentage in Posttransition Aberration Region
Overshoot Percentage, Levels, and Time Instant in Posttransition Aberration Region
Overshoot Percentage, Levels, and Time Instant in Pretransition Aberration Region
oslev
osinst
Overshoot metrics of bilevel waveform transitions
os = overshoot(x)
os = overshoot(x,fs)
os = overshoot(x,t)
[os,oslev,osinst] = overshoot(___)
[___] = overshoot(___,Name,Value)
overshoot(___)
os = overshoot(x) returns overshoots expressed as a percentage of the difference between the low- and high-state levels in the input bilevel waveform. The values in os correspond to the greatest absolute deviations that are greater than the final state levels of each transition.
os = overshoot(x,fs) specifies the sample rate fs in hertz.
os = overshoot(x,t) specifies the sample instants t.
[os,oslev,osinst] = overshoot(___) returns the levels oslev and sample instants osinst of the overshoots for each transition. You can specify an input combination from any of the previous syntaxes.
[___] = overshoot(___,Name,Value) specifies additional options using one or more Name,Value arguments. You can use any of the output combinations from previous syntaxes.
overshoot(___) plots the bilevel waveform and marks the location of the overshoot of each transition. The function also plots the lower and upper reference-level instants and associated reference levels and the state levels and associated lower- and upper-state boundaries.
Determine the maximum percent overshoot relative to the high-state level in a 2.3 V clock waveform.
Load the 2.3 V clock data. Determine the maximum percent overshoot of the transition. Determine also the level and sample instant of the overshoot. In this example, the maximum overshoot in the posttransition region occurs near index 22.
[oo,lv,nst] = overshoot(x)
oo = 6.1798
overshoot(x);
Determine the maximum percent overshoot relative to the high-state level, the level of the overshoot, and the sample instant in a 2.3 V clock waveform.
Determine the maximum percent overshoot, the level of the overshoot in volts, and the time instant where the maximum overshoot occurs. Plot the result.
[os,oslev,osinst] = overshoot(x,t)
os = 6.1798
oslev = 2.4276
osinst = 5.2500e-06
overshoot(x,t);
Determine the maximum percent overshoot relative to the low-state level, the level of the overshoot, and the sample instant in a 2.3 V clock waveform. Specify the 'Region' as 'Preshoot' to output pretransition metrics.
Determine the maximum percent overshoot, the level of the overshoot in volts, and the sampling instant where the maximum overshoot occurs. Plot the result.
[os,oslev,osinst] = overshoot(x,t,'Region','Preshoot')
overshoot(x,t,'Region','Preshoot');
Bilevel waveform, specified as a real-valued vector. The sample instants in x correspond to the vector indices. The first sample instant in x corresponds to t = 0.
Example: 'Region','Preshoot' specifies the pretransition aberration region.
Reference levels as a percentage of the waveform amplitude, specified as a 1-by-2 real-valued vector. The function defines the lower-state level to be 0 percent and the upper-state level to be 100 percent. The first element corresponds to the lower percent reference level, and the second element corresponds to the upper percent reference level.
Aberration region over which to compute the overshoot, specified as 'Preshoot' or 'Postshoot'. If you specify 'Preshoot', the function defines the end of the pretransition aberration region as the last instant when the signal exits the first state. If you specify 'Postshoot', the function defines the start of the posttransition aberration region as the instant when the signal enters the second state. By default, the function computes overshoots for posttransition aberration regions.
Aberration region duration, specified as a real-valued scalar. The function computes the overshoot over the specified duration for each transition as a multiple of the corresponding transition duration. If the edge of the waveform is reached or a complete intervening transition is detected before the aberration region duration elapses, the duration is truncated to the edge of the waveform or the start of the intervening transition.
Tolerance level, specified as a real-valued scalar. The function expresses tolerance as a percentage of the difference between the upper and lower state levels. The initial and final levels of each transition must be within the respective state levels.
os — Overshoots
Overshoots expressed as a percentage of the state levels, returned as a vector. The length of OS corresponds to the number of transitions detected in the input signal. For more information, see Overshoot.
oslev — Overshoot level
Overshoot level, returned as a column vector.
osinst — Sample instants
Sample instants of pretransition or posttransition overshoots, returned as a column vector. If you specify fs or t, the overshoot instants are in seconds. If you do not specifyfs or t, the overshoot instants are the indices of the input vector.
To determine the transitions, the overshoot function estimates the state levels of the input bilevel waveform x by using a histogram method with these steps.
The function identifies all intervals which cross the upper-state boundary of the low state and the lower-state boundary of the high state. The low-state and high-state boundaries are expressed as the state level plus or minus a multiple of the difference between the state levels.
The function computes the overshoot percentages based on the greatest deviation from the final state level in each transition.
For a positive-going (positive-polarity) pulse, the overshoot is given by
100\frac{\left(O-{S}_{2}\right)}{\left({S}_{2}-{S}_{1}\right)}
where O is the maximum deviation greater than the high-state level, S2 is the high state, and S1 is the low state.
For a negative-going (negative-polarity) pulse, the overshoot is given by
100\frac{\left(O-{S}_{1}\right)}{\left({S}_{2}-{S}_{1}\right)}
This figure shows the calculation of overshoot for a positive-going transition.
The red dashed lines indicate the estimated state levels. The double-sided black arrow depicts the difference between the high- and low-state levels. The solid black line indicates the difference between the overshoot value and the high-state level.
settlingtime | statelevels |
Meromorphic function - Wikipedia
Class of mathematical function
In the mathematical field of complex analysis, a meromorphic function on an open subset D of the complex plane is a function that is holomorphic on all of D except for a set of isolated points, which are poles of the function.[1] The term comes from the Ancient Greek meros (μέρος), meaning "part".[a]
1 Heuristic description
2 Prior, alternate use
5 On Riemann surfaces
Heuristic description[edit]
Intuitively, a meromorphic function is a ratio of two well-behaved (holomorphic) functions. Such a function will still be well-behaved, except possibly at the points where the denominator of the fraction is zero. If the denominator has a zero at z and the numerator does not, then the value of the function will approach infinity; if both parts have a zero at z, then one must compare the multiplicity of these zeros.
From an algebraic point of view, if the function's domain is connected, then the set of meromorphic functions is the field of fractions of the integral domain of the set of holomorphic functions. This is analogous to the relationship between the rational numbers and the integers.
Prior, alternate use[edit]
Both the field of study wherein the term is used and the precise meaning of the term changed in the 20th century. In the 1930s, in group theory, a meromorphic function (or meromorph) was a function from a group G into itself that preserved the product on the group. The image of this function was called an automorphism of G.[2] Similarly, a homomorphic function (or homomorph) was a function between groups that preserved the product, while a homomorphism was the image of a homomorph. This form of the term is now obsolete, and the related term meromorph is no longer used in group theory. The term endomorphism is now used for the function itself, with no special name given to the image of the function.
A meromorphic function is not necessarily an endomorphism, since the complex points at its poles are not in its domain, but may be in its range.
Since the poles of a meromorphic function are isolated, there are at most countably many.[3] The set of poles can be infinite, as exemplified by the function
{\displaystyle f(z)=\csc z={\frac {1}{\sin z}}.}
By using analytic continuation to eliminate removable singularities, meromorphic functions can be added, subtracted, multiplied, and the quotient
{\displaystyle f/g}
can be formed unless
{\displaystyle g(z)=0}
on a connected component of D. Thus, if D is connected, the meromorphic functions form a field, in fact a field extension of the complex numbers.
In several complex variables, a meromorphic function is defined to be locally a quotient of two holomorphic functions. For example,
{\displaystyle f(z_{1},z_{2})=z_{1}/z_{2}}
is a meromorphic function on the two-dimensional complex affine space. Here it is no longer true that every meromorphic function can be regarded as a holomorphic function with values in the Riemann sphere: There is a set of "indeterminacy" of codimension two (in the given example this set consists of the origin
{\displaystyle (0,0)}
Unlike in dimension one, in higher dimensions there do exist compact complex manifolds on which there are no non-constant meromorphic functions, for example, most complex tori.
All rational functions,[3] for example
{\displaystyle f(z)={\frac {z^{3}-2z+10}{z^{5}+3z-1}},}
{\displaystyle f(z)={\frac {e^{z}}{z}}\quad {\text{and}}\quad f(z)={\frac {\sin {z}}{(z-1)^{2}}}}
as well as the gamma function and the Riemann zeta function are meromorphic on the whole complex plane.[3]
{\displaystyle f(z)=e^{\frac {1}{z}}}
is defined in the whole complex plane except for the origin, 0. However, 0 is not a pole of this function, rather an essential singularity. Thus, this function is not meromorphic in the whole complex plane. However, it is meromorphic (even holomorphic) on
{\displaystyle \mathbb {C} \setminus \{0\}}
{\displaystyle f(z)=\ln(z)}
is not meromorphic on the whole complex plane, as it cannot be defined on the whole complex plane while only excluding a set of isolated points.[3]
{\displaystyle f(z)=\csc {\frac {1}{z}}={\frac {1}{\sin \left({\frac {1}{z}}\right)}}}
is not meromorphic in the whole plane, since the point
{\displaystyle z=0}
is an accumulation point of poles and is thus not an isolated singularity.[3]
{\displaystyle f(z)=\sin {\frac {1}{z}}}
On Riemann surfaces[edit]
On a Riemann surface, every point admits an open neighborhood which is biholomorphic to an open subset of the complex plane. Thereby the notion of a meromorphic function can be defined for every Riemann surface.
For every Riemann surface, a meromorphic function is the same as a holomorphic function that maps to the Riemann sphere and which is not the constant function equal to ∞. The poles correspond to those complex numbers which are mapped to ∞.
On a non-compact Riemann surface, every meromorphic function can be realized as a quotient of two (globally defined) holomorphic functions. In contrast, on a compact Riemann surface, every holomorphic function is constant, while there always exist non-constant meromorphic functions.
^ Greek meros (μέρος) means "part", in contrast with the more commonly used holos (ὅλος), meaning "whole".
^ Hazewinkel, Michiel, ed. (2001) [1994]. "Meromorphic function". Encyclopedia of Mathematics. Springer Science+Business Media B.V. ; Kluwer Academic Publishers. ISBN 978-1-55608-010-4.
^ Zassenhaus, Hans (1937). Lehrbuch der Gruppentheorie (1st ed.). Leipzig; Berlin: B. G. Teubner Verlag. pp. 29, 41.
^ a b c d e Lang, Serge (1999). Complex analysis (4th ed.). Berlin; New York: Springer-Verlag. ISBN 978-0-387-98592-3.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Meromorphic_function&oldid=1072149735" |
Proximal Policy Optimization Agents - MATLAB & Simulink - MathWorks France
Proximal policy optimization (PPO) is a model-free, online, on-policy, policy gradient reinforcement learning method. This algorithm is a type of policy gradient training that alternates between sampling data through environmental interaction and optimizing a clipped surrogate objective function using stochastic gradient descent. The clipped surrogate objective function improves training stability by limiting the size of the policy change at each step [1].
PPO is a simplified version of TRPO. TRPO is more computationally expensive than PPO, but TRPO tends to be more robust than PPO if the environment dynamics are deterministic and the observation is low dimensional. For more information on TRPO agents, see Trust Region Policy Optimization Agents.
PPO agents can be trained in environments with the following observation and action spaces.
PPO agents use the following actor and critics.
During training, a PPO agent:
If the UseDeterministicExploitation option in rlPPOAgentOptions is set to true the action with maximum likelihood is always used in sim and generatePolicyFunction. As a result, the simulated agent and the generated policy behave deterministically.
To estimate the policy and value function, a PPO agent maintains two function approximators.
You can create and train PPO agents at the MATLAB® command line or using the Reinforcement Learning Designer app. For more information on creating agents using Reinforcement Learning Designer, see Create Agents Using Reinforcement Learning Designer.
At the command line, you can create a PPO agent with default actor and critic based on the observation and action specifications from the environment. To do so, perform the following steps.
Specify agent options using an rlPPOAgentOptions object.
Create the agent using an rlPPOAgent object.
Create an actor using an rlDiscreteCategoricalActor object (for discrete action spaces) or an rlContinuousGaussianActor object (for continuous action spaces).
If needed, specify agent options using an rlPPOAgentOptions object.
Create the agent using the rlPPOAgent function.
PPO agents support actors and critics that use recurrent deep neural networks as function approximators.
PPO agents use the following training algorithm. To configure the training algorithm, specify options using an rlPPOAgentOptions object.
{S}_{ts},{A}_{ts},{R}_{ts+1},{S}_{ts+1},\dots ,{S}_{ts+N-1},{A}_{ts+N-1},{R}_{ts+N},{S}_{ts+N}
{G}_{t}=\sum _{k=t}^{ts+N}\left({\gamma }^{k-t}{R}_{k}\right)+b{\gamma }^{N-t+1}V\left({S}_{ts+N};\varphi \right)
{D}_{t}={G}_{t}-V\left({S}_{t};\varphi \right)
\begin{array}{c}{D}_{t}=\sum _{k=t}^{ts+N-1}{\left(\gamma \lambda \right)}^{k-t}{\delta }_{k}\\ {\delta }_{k}={R}_{t}+b\gamma V\left({S}_{t};\varphi \right)\end{array}
{G}_{t}={D}_{t}+V\left({S}_{t};\varphi \right)
{L}_{critic}\left(\varphi \right)=\frac{1}{M}\sum _{i=1}^{M}{\left({G}_{i}-V\left({S}_{i};\varphi \right)\right)}^{2}
{\stackrel{^}{D}}_{i}←{D}_{i}
{\stackrel{^}{D}}_{i}←\frac{{D}_{i}-mean\left({D}_{1},{D}_{2},\dots ,{D}_{M}\right)}{std\left({D}_{1},{D}_{2},\dots ,{D}_{M}\right)}
{\stackrel{^}{D}}_{i}←\frac{{D}_{i}-mean\left({D}_{1},{D}_{2},\dots ,{D}_{N}\right)}{std\left({D}_{1},{D}_{2},\dots ,{D}_{N}\right)}
Update the actor parameters by minimizing the actor loss function Lactor across all sampled mini-batch data.
\begin{array}{c}{L}_{actor}\left(\theta \right)=\frac{1}{M}\sum _{i=1}^{M}\left(-\mathrm{min}\left({r}_{i}\left(\theta \right)\cdot {D}_{i},{c}_{i}\left(\theta \right)\cdot {D}_{i}\right)+w{ℋ}_{i}\left(\theta ,{S}_{i}\right)\right)\\ {r}_{i}\left(\theta \right)=\frac{\pi \left({A}_{i}|{S}_{i};\theta \right)}{\pi \left({A}_{i}|{S}_{i};{\theta }_{old}\right)}\\ {c}_{i}\left(\theta \right)=\mathrm{max}\left(\mathrm{min}\left({r}_{i}\left(\theta \right),1+\epsilon \right),1-\epsilon \right)\end{array}
Di and Gi are the advantage function and return value for the ith element of the mini-batch, respectively.
π(Ai|Si;θ) is the probability of taking action Ai when in state Si, given the updated policy parameters θ.
π(Ai|Si;θold) is the probability of taking action Ai when in state Si, given the previous policy parameters θold from before the current learning epoch.
ε is the clip factor specified using the ClipFactor option.
ℋi(θ) is the entropy loss and w is the entropy loss weight factor, specified using the EntropyLossWeight option. For more information on entropy loss, see Entropy Loss.
{ℋ}_{i}\left(\theta ,{S}_{i}\right)=-\sum _{k=1}^{P}\pi \left({A}_{k}|{S}_{i};\theta \right)\mathrm{ln}\pi \left({A}_{k}|{S}_{i};\theta \right)
{ℋ}_{i}\left(\theta ,{S}_{i}\right)=\frac{1}{2}\sum _{k=1}^{C}\mathrm{ln}\left(2\pi \cdot e\cdot {\sigma }_{k,i}^{2}\right)
[1] Schulman, John, Filip Wolski, Prafulla Dhariwal, Alec Radford, and Oleg Klimov. “Proximal Policy Optimization Algorithms.” ArXiv:1707.06347 [Cs], July 19, 2017. https://arxiv.org/abs/1707.06347.
rlPPOAgent | rlPPOAgentOptions |
oliviayychengwh 2021-12-12 Answered
\frac{3}{25}
3÷25=0.12
\frac{3}{25}
to a decimal by applying long division:
\frac{3}{25}=3÷25=0.12
In the given equation as follows, use partial fractions to find the indefinite integral :-
see the equation as attached here
\int \frac{{x}^{2}}{{x}^{2}-2x+1}dx
Fractions with radicals in the denominator
I'm working my way through the videos on the Khan Academy, and have a hit a road block. I can't understand why the following is true:
\frac{6}{\phantom{\rule{1em}{0ex}}\frac{6\sqrt{85}}{85}\phantom{\rule{1em}{0ex}}}=\sqrt{85}
y=?
\text{Add: }\text{ }\frac{7}{6{x}^{2}}+\frac{2}{9x} |
EUDML | Vector Topologies and Linear Maps on Products of Topological Vector Spaces. EuDML | Vector Topologies and Linear Maps on Products of Topological Vector Spaces.
Vector Topologies and Linear Maps on Products of Topological Vector Spaces.
Wilde, Marc de. "Vector Topologies and Linear Maps on Products of Topological Vector Spaces.." Mathematische Annalen 196 (1972): 117-128. <http://eudml.org/doc/162248>.
@article{Wilde1972,
author = {Wilde, Marc de},
title = {Vector Topologies and Linear Maps on Products of Topological Vector Spaces.},
AU - Wilde, Marc de
TI - Vector Topologies and Linear Maps on Products of Topological Vector Spaces.
Pedro Pérez Carreras, Some aspects of the theory of barreled spaces
B
{B}_{r}
Articles by Marc de Wilde |
EUDML | M-Groups and the Supersolvable Residual. EuDML | M-Groups and the Supersolvable Residual.
M-Groups and the Supersolvable Residual.
Gary M. Seitz
Seitz, Gary M.. "M-Groups and the Supersolvable Residual.." Mathematische Zeitschrift 110 (1969): 101-122. <http://eudml.org/doc/171158>.
@article{Seitz1969,
author = {Seitz, Gary M.},
title = {M-Groups and the Supersolvable Residual.},
AU - Seitz, Gary M.
TI - M-Groups and the Supersolvable Residual.
\pi
Formations of groups, Fitting classes
Articles by Gary M. Seitz |
EUDML | Structure of the set of all minimal total dominating functions of some classes of graphs EuDML | Structure of the set of all minimal total dominating functions of some classes of graphs
Structure of the set of all minimal total dominating functions of some classes of graphs
K. Reji Kumar; Gary MacGillivray
Discussiones Mathematicae Graph Theory (2010)
In this paper we study some of the structural properties of the set of all minimal total dominating functions (
{}_{T}
) of cycles and paths and introduce the idea of function reducible graphs and function separable graphs. It is proved that a function reducible graph is a function separable graph. We shall also see how the idea of function reducibility is used to study the structure of
{}_{T}\left(G\right)
for some classes of graphs.
K. Reji Kumar, and Gary MacGillivray. "Structure of the set of all minimal total dominating functions of some classes of graphs." Discussiones Mathematicae Graph Theory 30.3 (2010): 407-423. <http://eudml.org/doc/270796>.
abstract = {In this paper we study some of the structural properties of the set of all minimal total dominating functions ($_T$) of cycles and paths and introduce the idea of function reducible graphs and function separable graphs. It is proved that a function reducible graph is a function separable graph. We shall also see how the idea of function reducibility is used to study the structure of $_T(G)$ for some classes of graphs.},
author = {K. Reji Kumar, Gary MacGillivray},
journal = {Discussiones Mathematicae Graph Theory},
keywords = {minimal total dominating functions (MTDFs); convex combination of MTDFs; basic minimal total dominating functions (BMTDFs); simplex; polytope; simplicial complex; function separable graphs; function reducible graphs},
title = {Structure of the set of all minimal total dominating functions of some classes of graphs},
AU - K. Reji Kumar
AU - Gary MacGillivray
TI - Structure of the set of all minimal total dominating functions of some classes of graphs
AB - In this paper we study some of the structural properties of the set of all minimal total dominating functions ($_T$) of cycles and paths and introduce the idea of function reducible graphs and function separable graphs. It is proved that a function reducible graph is a function separable graph. We shall also see how the idea of function reducibility is used to study the structure of $_T(G)$ for some classes of graphs.
KW - minimal total dominating functions (MTDFs); convex combination of MTDFs; basic minimal total dominating functions (BMTDFs); simplex; polytope; simplicial complex; function separable graphs; function reducible graphs
[1] B. Grünbaum, Convex Polytopes (Interscience Publishers, 1967).
[2] E.J. Cockayne and C.M. Mynhardt, A characterization of universal minimal total dominating functions in trees, Discrete Math. 141 (1995) 75-84, doi: 10.1016/0012-365X(93)E0192-7. Zbl0824.05034
[3] E.J. Cockayne, C.M. Mynhardt and B. Yu, Universal minimal total dominating functions in graphs, Networks 24 (1994) 83-90, doi: 10.1002/net.3230240205. Zbl0804.90122
[4] E.J. Cockayne, C.M. Mynhardt and B. Yu, Total dominating functions in trees: Minimality and convexity, J. Graph Theory 19 (1995) 83-92, doi: 10.1002/jgt.3190190109. Zbl0819.05035
[5] T.W. Haynes, S.T. Hedetniemi and P.J. Slater, Fundamentals of Domination in Graphs (Marcel Dekker, Inc., New York, 1998). Zbl0890.05002
[6] T.W. Haynes, S.T. Hedetniemi and P.J. Slater, Domination in Graphs - Advanced Topics (Marcel Dekker, Inc., New York, 1998). Zbl0883.00011
[7] K. Reji Kumar, Studies in Graph Theory - Dominating functions, Ph.D Thesis (Manonmaniam Sundaranar University, Tirunelveli, India, 2004).
[8] K. Reji Kumar, G. MacGillivray and R.B. Bapat, Topological properties of the set of all minimal total dominating functions of a graph, manuscript. Zbl1244.05167
[9] D.B. West, Graph Theory : An introductory course (Prentice Hall, New York, 2002).
minimal total dominating functions (MTDFs), convex combination of MTDFs, basic minimal total dominating functions (BMTDFs), simplex, polytope, simplicial complex, function separable graphs, function reducible graphs
Articles by K. Reji Kumar
Articles by Gary MacGillivray |
Limits of Functions: Level 4 Challenges Practice Problems Online | Brilliant
\large \lim_{x \to a} \bigg( 2 - \frac{a}{x} \bigg)^{\tan \frac{\pi x}{2a}} = e^{n}
Find approximate value of
n
\large \displaystyle \lim_{n \to \infty} \left(\dfrac{a+\sqrt[n]{b}-1}{a}\right)^{n}
Calculate the limit above in terms of
and
b
and
b
b\geq0
\sqrt[a]{b}
e^{\pi}
\sqrt[b]{a}
e
\pi
f:(1,\infty) \rightarrow (0,\infty)
be a continuous decreasing function with
\large \lim_{x\to\infty} \dfrac{f(4x)}{f(8x)} = \, 1
\large \lim_{x\to\infty} \dfrac{f(6x)}{f(8x)} = \, ?
\{x_n\}
x_1=1,\ x_nx_{n+1}=2n
n\ge 1
\displaystyle \lim_{n\to\infty} \dfrac{|x_{n+1}-x_n|}{\sqrt{n}}.
by Khang Nguyen Thanh
\large \lim_{ n\to{\infty}}{\dfrac{ 2016(1^{2015}+2^{2015}+ \cdots +n^{2015}) - n^{2016}}{2016(1^{2014}+2^{2014}+\cdots+n^{2014})}}. |
EUDML | -cross products and a generalized quantum mechanical -body problem. EuDML | -cross products and a generalized quantum mechanical -body problem.
{C}^{*}
-cross products and a generalized quantum mechanical
N
-body problem.
Damak, Mondher; Georgescu, Vladimir
Damak, Mondher, and Georgescu, Vladimir. "-cross products and a generalized quantum mechanical -body problem.." Electronic Journal of Differential Equations (EJDE) [electronic only] 2000 (2000): 51-69. <http://eudml.org/doc/121208>.
@article{Damak2000,
author = {Damak, Mondher, Georgescu, Vladimir},
keywords = {-cross products; quantum mechanical -body problem; algebra of compact operators; essential spectrum; Mourre estimate; -cross products; quantum mechanical -body problem},
title = {-cross products and a generalized quantum mechanical -body problem.},
AU - Damak, Mondher
AU - Georgescu, Vladimir
TI - -cross products and a generalized quantum mechanical -body problem.
KW - -cross products; quantum mechanical -body problem; algebra of compact operators; essential spectrum; Mourre estimate; -cross products; quantum mechanical -body problem
{C}^{*}
-cross products, quantum mechanical
N
-body problem, algebra of compact operators, essential spectrum, Mourre estimate,
{C}^{*}
N
-body problem
Selfadjoint operator theory in quantum theory, including spectral analysis
{C}^{*}
{W}^{*}
Operator algebra methods
Many-body theory; quantum Hall effect
Crossed product algebras (analytic crossed products)
Applications of operator algebras to physics
Articles by Damak
Articles by Georgescu |
Atomic physics problems, solved and explained
Atomic physics questions and answers
Recent questions in Atomic Physics
What will be the wavelength of second line belonging to balmer series in hydrogen emission spectrum?
Do someone knows the units of the spectra provided here ? It seems obvious enough that it's said nowhere, but even Wikipedia and other sites are quite blurry on this point.
So, is it power (W), radiance(
\phantom{\rule{mediummathspace}{0ex}}\mathrm{W}/{\mathrm{m}}^{2}\mathrm{s}\mathrm{r}
), or something else ?
What is the third line in the Balmer series (m=2)?
Calculate the energy of the first 4 photons emitted in the balmer series of the hydrogen spectrum. The Balmer series corresponds to nf=2 and the first 4 photons are the lowest energy transitions from higher energy levels. These 4 photons all hvae a wavelength in the visible spectrum.
Iyana Macdonald 2022-05-15 Answered
From the figure find the shortest wavelength present in the Balmer series of hydrogen, If
R=1.097×{10}^{7}\text{ }{m}^{-1}
Lines in one spectral series can overlap lines in another. Use the Rydberg equation to see if the range of wavelengths in the n₁=1 series overlaps the range in the n₁=2 series.
Energy vs wavelength relation is given by,
E\left(eV\right)=\frac{1241}{nm}
. Energy of H-atom in the ground state is -13.6eV as form the relation
E\left(eV\right)=-13.6\frac{{Z}_{eff}^{2}}{{n}^{2}}\left(eV\right)
. Find that in which transition on hydrogen atom is the wavelength 486.1 nm produced? To which seriesdoes it belong ?
Why can one observe an electronic transition of the 3s state to the 3p state in the emission spectrum of the nitrogen atom, but not in its absorption spectrum?
I know that the selection rules technically don't forbid either one since
\mathrm{\Delta }\ell =±1
One of the visible lines in the hydrogen emission spectrum corresponds to the n = 6 to n = 2 electronic transition. What color light is this transition ?
How much does the gravitational redshift change a neutron star emission spectra disturbing so the measurement of its surface temperature?
I need to be able to convert an arbitrary emission spectrum in the visible spectrum range (i.e. for every wavelength between 380 and 780, I have a number between 0 and 1 that represents the "intensity" or dominance of that wavelength), and I need to be able to map any given spectrum into a particular color space (for now I need RGB or CIE-XYZ). Is it possible?
For the spectrum say I have the emission spectrum of a white light, then every wavelength in the spectrum will have an intensity of 1, whereas for a green-bluish light I'd have most of the wavelengths between 500 and 550 with an intensity close to 1, with other wavelengths gradually dropping in intensity. So the first spectrum should be converted to pure white whereas the other one would be converted to a green-bluish color in any color space.
A line of the Lyman series(nf=1) of the spectrum of hydrogen has a wavelength of 95nm. What was the "upper" quantum state (ni) involved in the associated electron transition?
For the hydrogen atom, the Balmer Series is the series of line spectra produced when an election falls from any higher energy level and has n=2 as its final energy state. The longest wavelength is the Balmer Series is 656.7 nm. what was the initial principal energy level (n) associated with this photon.
Will the energy value in the trasnistion be the geatest or smalles in the entie Balmer Series?
Leon Robinson 2022-05-13 Answered
Calculate the longest wavelength of the hydrogen Balmer series using the Bohr model of the atom.
anniferathonniz8km 2022-05-10 Answered
When sodium is bombarded with electrons accelerated through a potential difference
\mathrm{\Delta }V
, its x-ray spectrum contains emission peaks at 1.04 keV and 1.07 keV. Find the minimum value of
\mathrm{\Delta }V
required to produce both of these peaks.
The first member of Balmer series of hydrogen atom has a wavelength of 6561 Å. Find wavelength of the second member of the Balmer series (in nm)
I learnt that in astrophysical spectroscopy, the emission spectrum of distant stars is used to determine what they're made of. So why is it that our own Sun is emitting the whole spectrum ?
garcialdaria2zky1 2022-05-09 Answered
Calculate the two longest wavelength of the Balmer series of triply ionized berillium (Z=4)
When light hits an atom (I will use a carbon atom for simplicity), if it is not in the absorption and/or emission spectrum of carbon, it will simply pass through without interacting with the atom. Whereas if it is in the absorption or emission spectrum it will be absorbed and either re-emitted or it will decay to become thermal energy.
However, if there are a lot of (carbon)atoms in close proximity (like in a block of coal), the light will (obviously) not pass through it no matter where it is on the visible spectrum. Why does this happen? |
\begin{array}{cccccc}Ten-Crores& Crores& Ten-Lacs& Lacs& Ten-Thousands& Thousands\\ 6& 8& 9& 7& 4& 5\end{array}\begin{array}{ccc}Hundreds& Tens& Units\\ 1& 3& 2\end{array}
6×{10}^{8}
{5}^{n}
{2}^{n}
975436×{5}^{4}
\frac{9754360000}{{2}^{4}}
\left({x}^{n}-{a}^{n}\right)
\left({x}^{n}-{a}^{n}\right)
\left({x}^{n}+{a}^{n}\right)
\frac{n}{2}\left[2a+\left(n-1\right)d\right]
\frac{n}{2}\left[a+l\right]
\left(1+2+3+...+n\right)=\frac{1}{2}n\left(n+1\right)
\left({1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}\right)=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)
\left({1}^{3}+{2}^{3}+{3}^{3}+...+{n}^{3}\right)=\frac{1}{4}{n}^{2}\left(n+1\right){ }^{2}
a,ar,a{r}^{2},a{r}^{3},...
a{r}^{n-1}
\left\{\begin{array}{ll}\frac{a\left(1-{r}^{n}\right)}{\left(1-r\right)}& where r<1\\ \frac{a\left({r}^{n}-1\right)}{\left(r-1\right)}& where r>1\end{array}\right\
The product of three consecutive even numbers when divided by 8 is 720. The product of their square roots is :
A) 12 sqrt(10) B) 24 sqrt(10)
C) 120 D) none of these
A) 12 sqrt(10)
B) 24 sqrt(10)
Answer & Explanation Answer: B) 24 sqrt(10)
Let the numbers be x, x + 2 and X + 4.
Then, x(x + 2)(x +4) / 8 = 720 => x(x + 2)(x +4) = 5760.
√x * √(x + 2) * √(x +4) = √x(x +2)(x + 4 ) = √5760 = 24√10.
(385/1001) = 5/13
=> Fourth number is 13
Let the number be x. Then, 15x - x = 196 <=> 14x = 196 => x = 14..
C) 17 D) none of these answers can be determined
D) none of these answers can be determined
Answer & Explanation Answer: D) none of these answers can be determined
=> 9 (x - y) = 63
=> x - y = 7.
Thus, none of the numbers can be deermined..
Let x and y be the two parts of 96.
x/7 = y/9 => x:y = 7:9
=> The smallest part is = 7/16 x 96 = 42.
If 50 is subtracted from two-third of a number, the result is equal to sum of 40 and one-fourth of that number. What is the number ?
Then, 2/3x-50 = 1/4x +50
=> 5/12x = 90
x = (90 x 12) / 5 = 216
If the sum and diference of two numbers are 20 and 8 respectively, then the difference of their squares is :
Let the numbers be x and y. Then, x + y = 20 and x - y = 8.
x^2 - y^2 = (x + y) (x - y) = 20 x 8 = 160.
If a number 72k23l is divisible by 88. Then find value of k and l ?
A) k=8 & l=2 B) k=7 & l=2
C) k=8 & l=3 D) k=7 & l=1
A) k=8 & l=2
B) k=7 & l=2
C) k=8 & l=3
D) k=7 & l=1
Answer & Explanation Answer: B) k=7 & l=2
If a number to be divisile by 88, it should be divisible by both "8" and "11"
Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]
Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]
So x = 7 , y= 2. |
Demodulate broadcast FM-modulated signal - Simulink - MathWorks Switzerland
Demodulate broadcast FM-modulated signal
The FM Broadcast Demodulator Baseband block demodulates a complex baseband FM signal by using the conjugate delay method, and filters the signal by using a de-emphasis filter. To demodulate stereo audio using 38 kHz, enable stereo demodulation. To demodulate RBDS signals from the 57 kHz band, enable RBDS demodulation.
Specify the input signal sample rate as a positive real scalar.
De-emphasis filter time constant (s)
Specify the de-emphasis lowpass filter time constant in seconds as a positive real scalar. FM broadcast standards specify a value of 75 μs in the United States and 50 μs in Europe.
Output audio sample rate (Hz)
Specify the output audio sample rate as a positive real scalar.
Play audio device
Select this check box to play sound from a default audio device.
Specify the buffer size the block uses to communicate with an audio device as a positive integer scalar. This parameter is available only when the Play audio device check box is selected.
Select this check box to enable demodulation of a stereo audio signal. If not selected, the audio signal is assumed to be monophonic.
RBDS demodulation
Select this check box to demodulate the RBDS signal from the input complex baseband FM signal. By default, this check box is not selected.
Number of samples per RBDS symbol
Specify the number of samples of the RBDS output as a positive integer. The RBDS sample rate is given by Number of samples per RBDS symbol × 1187.5 Hz. According to the RBDS standard, the sample rate of each bit is 1187.5 Hz.
This parameter appears when you select the RBDS demodulation check box.
RBDS Costas loop
Specify whether a Costas loop is used to recover the phase of the RBDS signal. Select this check box for radio stations that do not lock the 57 kHz RBDS signal in phase with the third harmonic of the 19 kHz pilot tone.
By default, this check box is not selected.
The input length must be an integer multiple of the audio decimation factor. If the RBDS demodulation check box is selected, the input length in addition must be an integer multiple of the RBDS decimation factor.
The FM Broadcast demodulator includes the functionality of the baseband FM demodulator, de-emphasis filtering, and the ability to receive stereophonic signals. For more information about the algorithms used for basic FM modulation and demodulation, see the comm.FMDemodulator System object™.
{H}_{\text{p}}\left(f\right)=1+j2\pi f{\tau }_{\text{s}}\text{\hspace{0.17em}},
{H}_{\text{d}}\left(f\right)=\frac{1}{1+j2\pi f{\tau }_{\text{s}}}\text{\hspace{0.17em}}.
m\left(t\right)={C}_{0}\left[L\left(t\right)+R\left(t\right)\right]+{C}_{1}\mathrm{cos}\left(2\pi ×19kHz×t\right)+{C}_{0}\left[L\left(t\right)-R\left(t\right)\right]\mathrm{cos}\left(2\pi ×38kHz×t\right)+{C}_{2}RBDS\left(t\right)\mathrm{cos}\left(2\pi ×57kHz×t\right)\text{\hspace{0.17em}},
comm.RBDSWaveformGenerator | comm.FMBroadcastDemodulator | comm.FMDemodulator |
Poison - OSRS Wiki
(Redirected from Poisonous)
The RuneScape Wiki also has an article on: rsw:PoisonThe RuneScape Classic Wiki also has an article on: classicrsw:Poison (effect)
This article is about the detriment. For the item needed for several quests, see Poison (item).
Poison is a detriment that players and monsters suffer when they are attacked by a poisonous weapon. Many NPCs are able to inflict poison. Players can use poisonous weapons or the Smoke spells from the Ancient Magicks to cause their target(s) to become poisoned. The maximum damage from this can range from 1 to over 20 depending on the type of poison.
When a player or monster is poisoned, a number appears with a green hitsplat ( ) instead of a red or blue one, indicating the amount of damage the poison has dealt. The victim is then damaged by this poison once every 18 seconds unless it is cured or it wears off. Poison wears off over time, decreasing by a value of one after every 5 hits. Poison doesn't do damage (or wear off) when a player has an interface open. It will hit once immediately after the interface is closed, and continue damaging the player every 18 seconds after that (30 game ticks).
Poison works on an internal timer, this means that NPCs with their own timers for actions like speaking can't be poisoned, even if monster examine shows they're not immune.
Weak thieving poison by opening some thieving chests: 1-2
Converted venom to poison by drinking any antipoison potion once: 6-20 (the same amount of damage the venom did)
3 Weapon poisoning
List of monsters[edit | edit source]
Exhaustive lists of monsters susceptible and immune to poison.
Curing poison[edit | edit source]
Poison can be cured through antidotes in the form of anti-poison potions. Anti-poison potions come in various levels, providing progressively longer immunity to becoming poisoned in addition to curing it. Players who are poisoned can use one of the following remedies or simply left-click their Hitpoints orb on the minimap and their character will automatically use a remedy that they have. Just like how poison does not damage the player with an interface opened, the antipoison timer will not go down until the interface is closed.
For those who don't have any form of cure can simply exit your POH (Player Owned House)
Antipoison 1.5 2,170 0
Superantipoison 6.0 708 1
Antidote+ 9.0 3,177 1
Antidote++ 12.0 3,226 1
Anti-venom 12.0 5,740 1
First 18 seconds include immunity to venom
Anti-venom+ 15.0 13,224 1
First 3 minutes include immunity to venom
Sanfew serum 6.0 92,985 1
15 minutes immunity to disease
Prayer book N/A N/A 1
Requires prayer points
Strange fruit N/A N/A 1
Guthix rest N/A N/A 1
Does not cure, but reduces poison damage per hit by 1. It will downgrade venom to poison if the player is envenomed
Cure Me N/A N/A 1
Requires completion of Lunar Diplomacy
Serpentine helm ∞ N/A 1
Charged with Zulrah's scales. Provides full immunity to becoming poisoned or envenomed while worn, but does not cure preexisting poison or venom when equipped.
Weapon poisoning[edit | edit source]
Members can inflict poison by using weapons treated with weapon poisons, made primarily through the Herblore skill. There are several types of weapon poisons, which deal progressively more damage. Only spears, hastae, daggers, arrows, bolts, darts, throwing knives, and javelins can be poisoned. The abyssal tentacle and emerald bolts (e) are also capable of poisoning players. A weapon poison potion can only be used on five thrown objects before running out (e.g. 5 arrows or 5 darts). Poison can be removed from weapons by making use of a cleaning cloth, obtained by using Karamja rum with silk, by purchasing it directly from Tamayu after completing Tai Bwo Wannai Trio, or by purchasing it from the Grand Exchange for 347 coins.
A player must deal non-zero damage with the poisoned weapon to have a chance at poisoning their enemy. The chances of inflicting poison are 1/4 for melee and 1/8 for ranged and smoke spells.[1]
Weapon poison 4 2 50 15
Weapon poison+ 5 3 75 18
Weapon poison++ 6 4 105 34
Karambwan paste (poison)[2] 6 N/A 105 N/A
Abyssal tentacle 4 N/A 50 N/A
Emerald bolts (e) N/A 5 N/A 75
^ Jagex. Mod Ash's Twitter account. 5 July 2017. (Archived from the original on 29 May 2020.) Mod Ash: "1/4 for melee, 1/8 for ranged. [Poison chance is] not material-dependent. Damage must be non-zero."
^ Can only be applied to spears and hastae.
The total poison damage (Melee: DM and Ranged: DR) can be calculated by the sum of all numbers (n) up to the max hit (M), times the amount of hits per number (H=5).
{\displaystyle {\begin{aligned}D_{M}(M)&=H\cdot \sum _{n=1}^{M}n=H\cdot {\frac {1}{2}}M(M+1)=2.5\cdot M(M+1)\\D_{R}(M)&=H\cdot \sum _{n=1}^{M-1}n+M=H\cdot {\frac {1}{2}}M(M-1)+M=2.5\cdot M(M-1)+M\end{aligned}}}
The poison's duration can be reset while the enemy is poisoned with the same chance as on an unpoisoned enemy.
Many monsters, such as wall beasts, turoths, waterfiends, and the God Wars Dungeon bosses, are immune to poison.
Poison interrupts some actions, including most skilling actions (but not Woodcutting or Fishing) and the home and minigame teleports. Poison does not interrupt combat in any way.
If a poisoned monster goes out of combat (no longer trying to attack) the effect of poison is immediately removed from the monster.
The swamp lizard (salamander weapon) has a chance of inflicting poison equivalent to weapon poison++ when used in melee mode.
Monsters that appear for a limited amount of time, such as Zamorak wizards or quest monsters like the bandit champion are often immune to poison and venom. This is likely because their programming uses the poison/venom timer to calculate how long they should appear for.
Poison and venom are mutually exclusive, but it is possible to be poisoned or venomed and have disease at the same time, splitting your health orb in two to signify both applied effects.
Retrieved from ‘https://oldschool.runescape.wiki/w/Poison?oldid=14278721’ |
Decide if the equation defines an ellipse, a hyperbola, a parabola, or no conic
Decide if the equation defines an ellipse, a hyperbola, a parabola, or no conic section at all.displaystyle{left({a}right)}{4}{x}{2}-{9}{y}{2}={12}{left({b}right)}-{4}{x}+{9}{y}{2}={0}displaystyle{left({c}right)}{4}{y}{2}+{9}{x}{2}={12}{left({d}right)}{4}{x}{3}+{9}{y}{3}={12}
Decide if the equation defines an ellipse, a hyperbola, a parabola, or no conic section at all.
\left(a\right)4{x}^{2}-9{y}^{2}=12\left(b\right)-4x+9{y}^{2}=0
\left(c\right)4{y}^{2}+9{x}^{2}=12\left(d\right)4{x}^{3}+9{y}^{3}=12
Standard equation of ellipse:
\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1
Standard equation of a parabola:
{y}^{2}=4ax
Standard equation of a Hyperbola:
\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1
4{x}^{2}-9{y}^{2}=12
Divide by coefficient of square terms : 4
{x}^{2}-\frac{9}{4}{y}^{2}=3
\frac{1}{9}{x}^{2}-\frac{1}{4}{y}^{2}=\frac{1}{3}
\frac{1}{3}
\frac{{x}^{2}}{3}-\frac{{y}^{2}}{\frac{4}{3}}=1
So, this is the form of hyperbola
\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1
4{x}^{2}-9{y}^{2}=12
defines a hyperbola.
-4x+9{y}^{2}=0
9{y}^{2}=4x
{y}^{2}=4\frac{x}{9}
So, this is the form of parabola
{y}^{2}=4ax
-4x+9{y}^{2}=0
defines a parabola.
4{y}^{2}+9{x}^{2}=12
{x}^{2}+\frac{4}{9}{y}^{2}=\frac{4}{3}
\frac{1}{4}{x}^{2}+\frac{1}{9}{y}^{2}=\frac{1}{3}
\frac{1}{3}
\frac{{x}^{2}}{\frac{4}{3}}+\frac{{y}^{2}}{3}=1
This is the form of ellipse
\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1
4{y}^{2}+9{x}^{2}=12
defines an ellipse.
4{x}^{3}+9{y}^{3}=12
The above equation is not an ellipse, parabola and a hyperbola.
4{x}^{3}+9{y}^{3}=12
is not a conic section.
The radical expression
\sqrt{20}
How do you find the eccentricity, directrix, focus and classify the conic section
r=\frac{14.4}{2-4.8\mathrm{cos}\theta }
How do you name the curve given by the conic
r=6
Identify the conic section given by the polar equation
r=\frac{4}{1-\mathrm{cos}\theta }
and also determine its directrix.
\left(x,y\right)=\left(m\mathrm{cos}\left(\omega t\right),n\mathrm{cos}\left(\omega t-\varphi \right)\right)
describes an ellipse for all m, n,
\omega ,\varphi
\left(2x+\sqrt{4{x}^{2}+1}\right)\left(\sqrt{{y}^{2}+4}-2\right)\ge y>0
minimum vale of
x+y
How to solve system of equations
{A}_{1}{x}^{2}+{B}_{1}xy+{C}_{1}{y}^{2}+{D}_{1}x+{E}_{1}y+{F}_{1}=0
{A}_{2}{x}^{2}+{B}_{2}xy+{C}_{2}{y}^{2}+{D}_{2}x+{E}_{2}y+{F}_{2}=0
How to express x through y? |
Modular Arithmetic - Parity, Odd/Even Practice Problems Online | Brilliant
The sum of two odd numbers is odd.
The sum of the squares of any three odd numbers is odd.
The sum of the squares of three consecutive integers is always odd.
Determine the parity of
\large 101010_3,
that is, 101010 in base 3.
Note: In the context of number theory, the parity of a number is whether or not it is even or odd. This is different from the context of computer science, in which a parity bit counts the number of 1s in a binary code.
The sum of an odd number and an even number is odd. |
Effect of gait on the energy consumption of walking robots | JVE Journals
Eugene S. Briskin1 , Yaroslav V. Kalinin2
1, 2Volgograd State Technical University, Volgograd, Russia
The energy consumption of walking robots has been considered on the base of solution of variational problem. It is fixed that the gait is affected on the energy efficiency and must vary with speed.
Keywords: walking robot, walking mechanism, energy efficiency, energy consumption, heat losses, variational problem, gait of walking robot.
One of the problems of walking robots is their low energy efficiency. It is known [1] that energy consumptions grow in proportion to the cube of speed when walking robot moves with constant velocity of its body. This is due with the periodic acceleration and deceleration of unbalanced walking mechanisms in the transfer phase from one position to another. The use of energy recovery allowing to accumulate energy in the braking phase of the walking mechanism in the transfer and then to give it in the phase of accelerating, allows to increase the energy efficiency [2].
It is known methods to improve energy efficiency at the expense of motion comfort as bipedal [3] and multilegged robots [4]. For multilegged robots with dual walking mechanisms resulting equation gives you the opportunity to choose such law of motion, which ensures low energy consumption [5]. This equation for walking movers operating from a common motor has a form:
\stackrel{˙}{T}+QV={Q}_{a}{V}_{a},
T
is kinetic energy of robots body and its movers;
Q
V
{Q}_{a}
{V}_{a}
- according resistance force and the velocity of the body translational motion in the current time and their average values in the one cycle.
In the general case the Eq. (1) is nonlinear differential equation of the second order with respect to the coordinates.
As we know there is a fairly complete classification of possible gaits of multilegged walking robots during their translational motion with rectilinear and uniform motion of body center of mass. The ratio mode
\gamma
is the important element of classification:
\gamma =\frac{\tau -{t}_{i}}{\tau },
{t}_{i}
is the time spent of walking mechanism in the phase of interacting with supported surface,
\tau
is total cycle time.
The another important feature is the time
{\tau }_{j}
of the start transfer of
j
mover to a new position. The countdown is since the beginning of shifting phase of one of the walking mechanism.
This mechanism is considered as baseline
{\tau }_{j}
Given that of the beginning and end of the transfer,
j
mover it absolute speed:
{\stackrel{˙}{x}}_{j}=0,
the motion of the body is performed without sliding of its pads on the ground.
The possible dependence
{\stackrel{˙}{x}}_{j}={\stackrel{˙}{x}}_{j}\left(t\right)
are shown in the Fig. 1.
Fig. 1. Velocities of movers for example gait (
\tau
= 0.5 second): 1 – The baseline walking mechanism
{\tau }_{j}
= 0, 2 –
j
walking mechanism in case
{\tau }_{j}
> 0, 3 –
j
{\tau }_{j}
The simplest analytical dependence of the absolute speed
{\stackrel{˙}{x}}_{j}
with time and satisfying the set conditions in Eq. (3) has the form:
{\stackrel{˙}{x}}_{j}=\left\{\begin{array}{ll}-\frac{6S}{{t}_{i}^{3}}{\left(t-{\tau }_{j}\right)}^{2}+\frac{6S}{{t}_{i}^{2}}\left(t-{\tau }_{j}\right),& {\tau }_{j}<t<{\tau }_{j}+{t}_{i},\\ 0,& t<{\tau }_{j}, t>{\tau }_{j}+{t}_{i},\end{array}\right\
{\stackrel{˙}{x}}_{j}=\left\{\begin{array}{ll}-\frac{6S}{{t}_{i}^{3}}{\left(t-{\tau }_{j}\right)}^{2}+\frac{6S}{{t}_{i}^{2}}\left(t-{\tau }_{j}\right),& \tau +{\tau }_{j}<t, t<{\tau }_{j}+{t}_{i},\\ 0,& {\tau }_{j}+{t}_{i}<t<{\tau }_{j}+\tau ,\end{array}\right\
S
is the distance at which the walking mechanism pad is transferred over time
\tau -{t}_{i}
The main goal of submitted paper is to define such a low of rectilinear motion of the robot center of mass during its forward motion which provides the minimum of total heat losses all engines that control motion of all movers at various gaits and speeds.
3. Design scheme and mathematical model of walking robot
The translational motion of walking robot body of mass
M
N
its movers of mass
m
each are studied. These motions are described by absolute coordinates
{x}_{c}
{x}_{j}
(Fig. 2) at that
{a}_{j}
– coordinate of
j
mover installation,
{z}_{j}
– coordinate
j
mover with respect to the point of its installation.
Fig. 2. Design scheme of walking robot
The resistance force
Q
and forces of interactions between the body of walking robot and their movers
{F}_{j}
are generated by drives of the course motion. The interaction of
j
mover with supported surface is taken into account by tangent forces
{R}_{j}
. The supported surface is modeled by homogeneous elastic medium and the body of the robot is modeled by absolutely solid. For this reason, all reactions
{R}_{j}
for movers interacted with a ground are equal each other:
{R}_{j}=R.
Then the differential equations, described the motion of considered mechanical system have a form:
M{\stackrel{¨}{x}}_{c}=\sum {F}_{j}-Q,\mathrm{ }\mathrm{ }j=1,2,\dots ,N,
0=R-{F}_{j}, j=1,2,\dots ,k,
m{\stackrel{¨}{x}}_{j}=-{F}_{j},\mathrm{ }\mathrm{ }j=k+1,k+2,\dots ,N,
N
is the total number of movers,
k
is the number of movers interacted with the support surface in a given moment of time and dependent on the gait. It follows from Eq. (7) that the force, generated by actuator is determined from second equation for movers interactive with the support surface and for movers which are in phase of transfer the force is determined from third. The constant force
R
is determined from the first Eq. (7):
R=\frac{M{\stackrel{¨}{x}}_{c}+Q+\sum _{j=k+1}^{N}m{\stackrel{¨}{x}}_{j}}{k}.
The feature of the Eq. (8) is the fact that in the process of motion a number of movers
k
interact with the supporting surface may change. Hence considered mechanical system is system of variable structure.
In accordance with the task the method of solution is based on the requirement of the minimum of integral.
{I}_{\mathrm{*}}=\underset{0}{\overset{\tau }{\int }}\sum _{j=1}^{N}{F}_{j}^{2}dt.
It is taken into account the isoperimetric condition:
\underset{0}{\overset{\tau }{\int }}{\stackrel{˙}{x}}_{c}dt=S,
S
is the distance, traversed by the body of robot during the time of interacting of walking mechanism with a supported surface.
Then we can obtain a new functional
I
by combining Eq. (9) and Eq. (10):
I=\underset{0}{\overset{\tau }{\int }}\left(\sum _{j=1}^{N}{F}_{j}^{2}+\mu {\stackrel{˙}{x}}_{c}\right)dt=\underset{0}{\overset{\tau }{\int }}\mathrm{\Phi }dt.
\mathrm{\Phi }=\sum _{j=1}^{k}{\left(\frac{M{\stackrel{¨}{x}}_{c}+Q+\sum _{j=k+1}^{N}m{\stackrel{¨}{x}}_{j}}{k}\right)}^{2}+\sum _{j=k+1}^{N}{\left(m{\stackrel{¨}{x}}_{j}\right)}^{2}+\mu {\stackrel{˙}{x}}_{c},
\mu
is undetermined Lagrange multiplier.
{\stackrel{¨}{x}}_{j}
are defined as time function Eq. (4) and Eq. (5) and parameters
S
{t}_{i}
{\tau }_{j}
, the Euler-Poisson equations [6], providing the minimum of functional has a form:
\frac{d}{dt}\left(\frac{\partial \mathrm{\Phi }}{\partial {\stackrel{¨}{x}}_{c}}\right)-\frac{\partial \mathrm{\Phi }}{\partial {\stackrel{˙}{x}}_{c}}=C,
or in expended form:
2\frac{{M}^{2}}{k}{\stackrel{⃛}{x}}_{c}+2\frac{M}{k}\sum _{j=k+1}^{N}m{\stackrel{⃛}{x}}_{j}=\mu +C,
C
The Eq. (14) must satisfy the initial and final conditions:
{x}_{c}\left(0\right)=0, {x}_{c}\left(\mathrm{\tau }\right)=S, {\stackrel{˙}{x}}_{c}\left(0\right)={\stackrel{˙}{x}}_{c}\left(\mathrm{\tau }\right)={V}_{0},
which allow to determine three arbitrary constants of differential equation and the amount
\mu +C
{V}_{0}
– initial and final velocity of robot’s body in each cycle.
One more peculiarity of Eq. (14) is dependence of the amount
k
of movers interacting with a support surface from the gait.
5. Examples of the influence of gait on energy efficiency
The real walking robot “Ortonog” [7] (Fig. 3) have been considered as example of walking robot.
Fig. 3. Walking robot “Ortonog”
For this robot four mechanisms of walking always are interacted with the support surface. It is considered a such gait when the interacting with a support surface is carried out in pairs of two mechanisms of walking with ratio mode
\gamma
> 0.5. It is presented on chart in Fig. 4 the dimensionless parameter
\epsilon
\epsilon =\frac{I}{{m}^{2}{g}^{2}\tau }\mathrm{ },
\epsilon
is parameter proportional energy losses, depending on ratio mode and average speed of robot,
g
is acceleration of gravity.
The results show us that in order to minimize the energy consumption it is necessary to change the gait. This is confirmed by theory of synchronization [8], behavior of animals and investigation of multilegged walking machines and robots motion [9].
Fig. 4. Dependencies of the rate of energy loss of walking robot’s speed and gait 1 –
{V}_{0}
is 0.5 m/s, 2 –
{V}_{0}
is 0.33 m/s, 3 –
{V}_{0}
is 0.25 m/s
The gait of walking robot significant effects on the energy consumption. For each speed you should choose energy optimal gait, but the most effective use of such actuator together with control system which could independently change the gait.
The work was supported by the RFBR (Grants Nos. 14-01-00655-A, 14-08-01002-A) and by Grant of Ministry of Education and Science of Russian Federation No. 862.
Ohotsimsky D. Е., Platonov A. K., Kiril’chenko А. А., Lapshin V. V. Schagayuschie Maschiny (Walking Machines). Preprint IPM of M.V. Keldysch AN USSR, Moscow, 1989, (in Russian). [Search CrossRef]
Lapshin V. V. Modelnye Ocenki Energozatrat Schagayuschego Apparata (Model Estimates of Energy Walking Machine). Izvestia RAS, МТТ (News of RAS. Mechanics of Solid Body), Vol. 1, 1993, p. 65-74, (in Russian). [Search CrossRef]
Beletskiy V. V. Dvunogaya Hod’ba: Model’nye Zadachi Dinamiki i Upravleniya (Biped Walking: Modeling Problems of Dynamics and Control). Science, Moscow, 1984, (in Russian). [Search CrossRef]
Briskin Е. S., Kalinin Ya V. On energetically efficient algorithms of walking machines with cyclic drives. Journal of Computer and Systems Sciences International, Vol. 50, Issue 2, 2011, p. 348-354. [Search CrossRef]
Briskin Е. S., Chernyschev V. V., Kalinin Ya. V., Маloletov А. V. On the energy efficiency of cyclic mechanisms. Mechanic of Solids, Vol. 49, Issue 1, 2014, p. 11-17. [Search CrossRef]
Halfman R. L. Dynamics. Addison-Wesley Publishing Company, Inc. Reading, Massachusetts, Palo Alto, London, 1962. [Search CrossRef]
Briskin Е. S., Maloletov А. V., Sharonov N. G., Fomenko S. S., Kalinin Ya. V., Leonard А. V. Development of rotary type movers discretely interacting with supporting surface and problems of control their movement. Proceedings of the 21st CISM-IFToMM Symposium on Robot Design, Dynamics and Control, Udine, Italy, 2016, p. 351-359. [Search CrossRef]
Blekhman I. I. Synchronization in Science and Technology. ASME Press, New York, 1981. [Search CrossRef]
Briskin Е. S. Ob Obschei Dinamike i Povorote Shagayuschih Mashin (On General Dynamics and Turns of Walking Machines). Problemy Mashinostroeniya i Dinamiki Mashin (Problems of Engineering and Reliability of Machines), Vol. 6, 1997, p. 33-39, (in Russian). [Search CrossRef] |
EUDML | Six-dimensional considerations of Einstein's connection for the first two-classes. I: The recurrence relations in 6--UFT. EuDML | Six-dimensional considerations of Einstein's connection for the first two-classes. I: The recurrence relations in 6--UFT.
Six-dimensional considerations of Einstein's connection for the first two-classes. I: The recurrence relations in 6-
g
-UFT.
Chung, Kyung Tae; Yang, Gye Tak; Hwang, In Ho
Chung, Kyung Tae, Yang, Gye Tak, and Hwang, In Ho. "Six-dimensional considerations of Einstein's connection for the first two-classes. I: The recurrence relations in 6--UFT.." International Journal of Mathematics and Mathematical Sciences 22.3 (1999): 469-482. <http://eudml.org/doc/49097>.
@article{Chung1999,
author = {Chung, Kyung Tae, Yang, Gye Tak, Hwang, In Ho},
keywords = {generalized Riemannian manifold; recurrence relations in 6--UFT; 6-dimensional Einstein's unified field theory; recurrence relations in 6--UFT; 6-dimensional Einstein’s unified field theory},
title = {Six-dimensional considerations of Einstein's connection for the first two-classes. I: The recurrence relations in 6--UFT.},
AU - Chung, Kyung Tae
AU - Yang, Gye Tak
AU - Hwang, In Ho
TI - Six-dimensional considerations of Einstein's connection for the first two-classes. I: The recurrence relations in 6--UFT.
KW - generalized Riemannian manifold; recurrence relations in 6--UFT; 6-dimensional Einstein's unified field theory; recurrence relations in 6--UFT; 6-dimensional Einstein’s unified field theory
generalized Riemannian manifold, recurrence relations in 6-
g
-UFT, 6-dimensional Einstein's unified field theory, recurrence relations in 6-
g
-UFT, 6-dimensional Einstein’s unified field theory
Unified, higher-dimensional and super field theories
Differentiable manifolds, foundations
Articles by Chung |
(e) Sketch the graph of h by hand. (d) Use
(e) Sketch the graph of h by hand. (d) Use function notation to write h in terms of the parent function f. h(x)=(x−2)^3+5
h\left(x\right)={\left(x-2\right)}^{3}+5
cyhuddwyr9
a)Parent function:
f\left(x\right)={x}^{3}
b)-Horizontal shift 2 units to the right
-Vertical shift 5 units upward
c)The graph will be:
d)In function notation, h(x)=f(x-2)+5 <- Answer
Describe the transformations that must be applied to
y={x}^{2}
to create the graph of each of the following functions.
y=\frac{1}{4}{\left(x-3\right)}^{2}+9
y={\left(\left(\frac{1}{2}\right)x\right)}^{2}-7
Graph f and g in the same rectangular coordinate system. Use transformations of the graph of f to obtain the graph of g. Graph and give equations of all asymptotes. Use the graphs to determine each functions
Graph f and g in the same rectangular coordinate system. Use transformations of the graph off to obtain the graph of g. Graph and give equations of all asymptotes. Use the graphs to determine each function’s domain and range.
f\left(x\right)=\mathrm{ln}x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g⟨x\right)=-\mathrm{ln}\left(2x\right)
y=\text{ }-{\mathrm{log}}_{2}x
a) Use transformations of the graphs of
y={\mathrm{log}}_{2}x
y={\mathrm{log}}_{3}x
o graph the given functions.
b) Write the domain and range in interval notation.
c) Write an equation of the asymptote.
Write an equation for the graphed function by using transformations of the graphs of one of the toolkit functions.
An observational study is retrospective if it considers only existing data. It is prospective if the study design calls for data to be collected as time goes on. Tell which of the following observational studies are retrospective and which are prospective. To see whether crime mes are related to moon phase, a researcher looks at ten years of archived police blotter reports and compares them with moon charts from the same period.
Begin by graphing
f\left(x\right)=\text{ }{\mathrm{log}}_{2}x
Then use transformations of this graph to graph the given function. What is the graphs |
Fundamental_pair_of_periods Knowpia
In mathematics, a fundamental pair of periods is an ordered pair of complex numbers that define a lattice in the complex plane. This type of lattice is the underlying object with which elliptic functions and modular forms are defined.
Fundamental parallelogram defined by a pair of vectors in the complex plane.
A fundamental pair of periods is a pair of complex numbers
{\displaystyle \omega _{1},\omega _{2}\in \mathbb {C} }
such that their ratio ω2/ω1 is not real. If considered as vectors in
{\displaystyle \mathbb {R} ^{2}}
, the two are not collinear. The lattice generated by ω1 and ω2 is
{\displaystyle \Lambda =\left\{m\omega _{1}+n\omega _{2}\mid m,n\in \mathbb {Z} \right\}}
This lattice is also sometimes denoted as Λ(ω1, ω2) to make clear that it depends on ω1 and ω2. It is also sometimes denoted by Ω or Ω(ω1, ω2), or simply by ⟨ω1, ω2⟩. The two generators ω1 and ω2 are called the lattice basis. The parallelogram defined by the vertices 0,
{\displaystyle \omega _{1}}
{\displaystyle \omega _{2}}
is called the fundamental parallelogram.
While a fundamental pair generates a lattice, a lattice does not have any unique fundamental pair; in fact, an infinite number of fundamental pairs correspond to the same lattice.
A number of properties, listed below, can be seen.
EquivalenceEdit
A lattice spanned by periods ω1 and ω2, showing an equivalent pair of periods α1 and α2.
Two pairs of complex numbers (ω1,ω2) and (α1,α2) are called equivalent if they generate the same lattice: that is, if ⟨ω1,ω2⟩ = ⟨α1,α2⟩.
No interior pointsEdit
The fundamental parallelogram contains no further lattice points in its interior or boundary. Conversely, any pair of lattice points with this property constitute a fundamental pair, and furthermore, they generate the same lattice.
Modular symmetryEdit
{\displaystyle (\omega _{1},\omega _{2})}
{\displaystyle (\alpha _{1},\alpha _{2})}
are equivalent if and only if there exists a 2 × 2 matrix
{\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}}
with integer entries a, b, c and d and determinant ad − bc = ±1 such that
{\displaystyle {\begin{pmatrix}\alpha _{1}\\\alpha _{2}\end{pmatrix}}={\begin{pmatrix}a&b\\c&d\end{pmatrix}}{\begin{pmatrix}\omega _{1}\\\omega _{2}\end{pmatrix}},}
that is, so that
{\displaystyle \alpha _{1}=a\omega _{1}+b\omega _{2}}
{\displaystyle \alpha _{2}=c\omega _{1}+d\omega _{2}.}
This matrix belongs to the matrix group
{\displaystyle \mathrm {SL} (2,\mathbb {Z} )}
, which is known as the modular group. This equivalence of lattices can be thought of as underlying many of the properties of elliptic functions (especially the Weierstrass elliptic function) and modular forms.
Topological propertiesEdit
The abelian group
{\displaystyle \mathbb {Z} ^{2}}
maps the complex plane into the fundamental parallelogram. That is, every point
{\displaystyle z\in \mathbb {C} }
{\displaystyle z=p+m\omega _{1}+n\omega _{2}}
for integers m,n, with a point p in the fundamental parallelogram.
Since this mapping identifies opposite sides of the parallelogram as being the same, the fundamental parallelogram has the topology of a torus. Equivalently, one says that the quotient manifold
{\displaystyle \mathbb {C} /\Lambda }
Fundamental regionEdit
The grey depicts the canonical fundamental domain.
Define τ = ω2/ω1 to be the half-period ratio. Then the lattice basis can always be chosen so that τ lies in a special region, called the fundamental domain. Alternately, there always exists an element of PSL(2,Z) that maps a lattice basis to another basis so that τ lies in the fundamental domain.
The fundamental domain is given by the set D, which is composed of a set U plus a part of the boundary of U:
{\displaystyle U=\left\{z\in H:\left|z\right|>1,\,\left|\operatorname {Re} (z)\right|<{\tfrac {1}{2}}\right\}.}
where H is the upper half-plane.
The fundamental domain D is then built by adding the boundary on the left plus half the arc on the bottom:
{\displaystyle D=U\cup \left\{z\in H:\left|z\right|\geq 1,\,\operatorname {Re} (z)=-{\tfrac {1}{2}}\right\}\cup \left\{z\in H:\left|z\right|=1,\,\operatorname {Re} (z)\leq 0\right\}.}
Three cases pertain:
{\displaystyle \tau \neq i}
{\displaystyle \tau \neq e^{{\frac {1}{3}}i\pi }}
, then there are exactly two lattice bases with the same τ in the fundamental region:
{\displaystyle (\omega _{1},\omega _{2})}
{\displaystyle (-\omega _{1},-\omega _{2}).}
{\displaystyle \tau =i}
, then four lattice bases have the same τ: the above two
{\displaystyle (\omega _{1},\omega _{2})}
{\displaystyle (-\omega _{1},-\omega _{2})}
{\displaystyle (i\omega _{1},i\omega _{2})}
{\displaystyle (-i\omega _{1},-i\omega _{2}).}
{\displaystyle \tau =e^{{\frac {1}{3}}i\pi }}
, then there are six lattice bases with the same τ:
{\displaystyle (\omega _{1},\omega _{2})}
{\displaystyle (\tau \omega _{1},\tau \omega _{2})}
{\displaystyle (\tau ^{2}\omega _{1},\tau ^{2}\omega _{2})}
and their negatives.
In the closure of the fundamental domain:
{\displaystyle \tau =i}
{\displaystyle \tau =e^{{\frac {1}{3}}i\pi }.}
A number of alternative notations for the lattice and for the fundamental pair exist, and are often used in its place. See, for example, the articles on the nome, elliptic modulus, quarter period and half-period ratio.
Tom M. Apostol, Modular functions and Dirichlet Series in Number Theory (1990), Springer-Verlag, New York. ISBN 0-387-97127-0 (See chapters 1 and 2.)
Jurgen Jost, Compact Riemann Surfaces (2002), Springer-Verlag, New York. ISBN 3-540-43299-X (See chapter 2.) |
Prove that If W is a subspace of a vector space
killjoy1990xb9 2022-01-04 Answered
If W is a subspace of a vector space V and
{w}_{1},{w}_{2},\dots ,{w}_{n}
are in W, then
{a}_{1}{w}_{1}+{a}_{2}{w}_{2}+\dots +{a}_{n}{w}_{n}\in W
for any scalars
{a}_{1},{a}_{2},\dots ,{a}_{n}
Linda Birchfield
Here W is a subspace of a vector space V and
{w}_{1},{w}_{2},\dots ,{w}_{n}
are in W.
{w}_{1}\in W,{w}_{2}\in W,\dots .,{w}_{n}\in W
Since, W is closed under scalar multiplication
{a}_{1}{w}_{1}\in W,{a}_{1}{w}_{2}\in W,\dots .,{a}_{1}{w}_{n}\in W
{a}_{1},{a}_{2},\dots ,{a}_{n}
W is closed under addition.
{a}_{1}{w}_{1}+{a}_{2}{w}_{2}+\dots +{a}_{n}{w}_{n}\in W
{a}_{1},{a}_{2},\dots ,{a}_{n}
Suhadolahbb
I understand how you proved!
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
Is there a relationship between Facebook use and age among college students? The following two-way table displays data for the 219 students who responded to the survey.
\begin{array}{cc}& \text{Age}\\ \text{Facebook user?}& \begin{array}{lccc}& \begin{array}{c}\text{ Younger }\\ \left(18-22\right)\end{array}& \begin{array}{c}\text{ Middle }\\ \left(23-27\right)\end{array}& \begin{array}{c}\text{ Older }\\ \left(28\text{ and up)}\end{array}\\ \text{ Yes }& 78& 49& 21\\ \text{ No }& 4& 21& 46\end{array}\end{array}
What percent of the students who responded were older Facebook users?
Let S be the parallelogram determined by the vectors
{b}_{1}=\left[\begin{array}{c}-2\\ 3\end{array}\right]
{b}_{2}=\left[\begin{array}{c}-2\\ 5\end{array}\right]
A=\left[\begin{array}{cc}6& -3\\ -3& 2\end{array}\right]
Compute the area of the image of S under the mapping
x↦Ax
What is the angle between <1,3,−8> and <4,1,5>
Use determinants to decide if the set of vectors is linearly independent.
\left[\begin{array}{c}4\\ 6\\ 2\end{array}\right],\left[\begin{array}{c}-7\\ 0\\ 7\end{array}\right],\left[\begin{array}{c}-3\\ -5\\ -2\end{array}\right] |
Comparing Direct Observation of Strain, Rotation, and Displacement with Array Estimates at Piñon Flat Observatory, California | Seismological Research Letters | GeoScienceWorld
Department of Earth and Environmental Sciences, Ludwig Maximilian University Munich, Theresienstraße 41, 80333 Munich, Germany, donner@geophysik.uni-muenchen.delin@geophysik.uni-muenchen.dehadzii@geophysik.uni-muenchen.degebauer@geophysik.uni-muenchen.deigel@geophysik.uni-muenchen.dejowa@geophysik.uni-muenchen.de
Chin‐Jen Lin;
Frank Vernon;
Institute of Geophysics and Planetary Physics, Scripps Institution of Oceanography, University of California San Diego, 9500 Gilman Drive Number 0225, La Jolla, California 92093‐0225 U.S.A., flvernon@ucsd.edudagnew@ucsd.edu
Duncan Carr Agnew;
Duncan Carr Agnew
Heiner Igel;
Forschungseinrichtung Satellitengeodaesie, Technical University of Munich, Fundamentalstation Wettzell, Sackenrieder Straße 25, 93444 Bad Koetzting, Germany, ulrich.schreiber@bv.tum.de
Stefanie Donner, Chin‐Jen Lin, Céline Hadziioannou, André Gebauer, Frank Vernon, Duncan Carr Agnew, Heiner Igel, Ulrich Schreiber, Joachim Wassermann; Comparing Direct Observation of Strain, Rotation, and Displacement with Array Estimates at Piñon Flat Observatory, California. Seismological Research Letters 2017;; 88 (4): 1107–1116. doi: https://doi.org/10.1785/0220160216
The unique instrument setting at the Piñon Flat Observatory in California is used to simultaneously measure 10 out of the 12 components, completely describing the seismic‐wave field. We compare the direct measurements of rotation and strain for the 13 September 2015
Mw
6.7 Gulf of California earthquake with array‐derived observations using this configuration for the first time. In general, we find a very good fit between the observations of the two measurements with cross‐correlation coefficients up to 0.99. These promising results indicate that the direct and array‐derived measurements of rotation and strain are consistent. For the array‐based measurement, we derived a relation to estimate the frequency range within which the array‐derived observations provide reliable results. This relation depends on the phase velocity of the study area and the calibration error, as well as on the size of the array.
Pinon Flat Observatory
Gulf of California earthquake 2015
Computation of Vector Hazard Using Salient Features of Seismic Hazard Deaggregation |
Biped dynamic walker with alternate unpowered and partially powered steps in a gait cycle | JVE Journals
Krishna Prakash Yadav1 , Prasanth Kumar R2
1, 2Department of Mechanical and Aerospace Engineering, Indian Institute of Technology Hyderabad, Kandi, Telangana, 502285, India
Copyright © 2020 Krishna Prakash Yadav, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The biped dynamic walker considered in this paper has three actuators - two at the ankle joints and one at the hip joint. We consider the case of one of the two ankle actuators at fault. Despite having only two actuators operational, we show that successful gait is possible for a typical case of virtual passive dynamic walking. We analyze such gaits for local and global stability for a virtual slope and for the cases of completely unpowered or partially powered alternate steps. It is shown that completely unpowered alternate steps are preferred over partially powered alternate steps in the case of virtual passive dynamic walking for global stability, and the other way for local stability.
Keywords: virtual passive dynamic walker, limit cycle, hybrid dynamics, bifurcation, basin of attraction, Poincare map.
Biped walking imitates human locomotion mechanism through alternate stance and swing phases of the robot’s legs. This kind of walking generally involves sophisticated and complex nonlinear hybrid dynamics with multiple variables. However, a class of bipeds called passive dynamic walkers can walk down a gentle slope powered by gravity alone. In order to simplify analysis and control input generation for biped robots, dynamic walking involving passive dynamics is considered. Research in passive dynamic waking was initiated by McGeer et al. [1], who gave the first mathematical model to analyze dynamic walking. Later Goswami et al. [2] introduced the widely used compass gait model for passive dynamic walkers and gave the concept of limit cycle walking in terms of symmetry and chaos. He also proposed the control law based on energy, which is used in the application of active walking on level grounds. Asano et al. [3] proposed the concept of active level walking of a compass gait model based on virtual gravity concept. He explained the active level walking of compass gait with and without knee. Spong et al. [4], has also studied and proposed the concept of passivity control and energy shaping [4]. Asano analyzed under actuated virtual passive dynamic walking with semi-circular feet. Apart from that, he also covered the stability, dynamics, and model analysis using unified properties of virtual passive dynamics models [5-8]. To the best our knowledge, dynamic walking with actuator fault has not be studied by researchers in this field. In this paper, we propose an active dynamic walker model based on the virtual gravity concept for the case of one ankle actuator fault. The fault is assumed to make the joint free with zero torque or resistance applied by the faulty actuator. This situation is modeled as one step of the gait cycle fully powered and another step, where the leg with the faulty ankle actuator is in stance phase, partially powered by the hip actuator alone.
2. Biped dynamic walker modeling
We have considered the Virtual Passive Dynamic Walker (VPDW) model and chosen the model parameters from Asano et al. [9]. The schematic diagram of the virtual passive dynamic walker is shown in Fig. 1. The system dynamics consists of two-phases – swing phase and stance phase with an impulsive transition in between. It has to be satisfied with the geometrical conditions given by Asano et al. [9] for the impulsive transition. The equation of motion is given by:
M\left(\theta ,\stackrel{˙}{\theta }\right)\stackrel{¨}{\theta }+C\left(\theta ,\stackrel{˙}{\theta }\right)\stackrel{˙}{\theta }+G\left(\theta \right)=Bu,
where 𝑀 – mass matrix, 𝐶 – Coriolis or centripetal matrix, 𝐺 – gravitation matrix.
The impulsive transition during impact for virtual passive dynamic walker can be referred to as [5], i.e.:
{\theta }^{+}=R{\theta }^{-}, {\stackrel{˙}{\theta }}^{+}=S{\stackrel{˙}{\theta }}^{-},
S=Q{p}^{-1}Qm
. The mapping of position and velocity before and after the collision has been done by using these two matrices
R\in {\mathbb{R}}^{2}
S\in {\mathbb{R}}^{2}
Fig. 1. Schematic of the biped dynamic walker robot
Fig. 2. a) Limit cycle plot of VPDW, b) step length convergence plot of VPDW
From suitably chosen initial condition and slope value, the dynamic walker walks on flat surface robustly. We observe this from a numerical simulation which shows convergence for the given initial condition. As shown in Fig. 2(a), the system starts with some initial condition and approaches a fixed step length after 10th iteration, Fig. 2(b) shows limit cycle converging to the stable solution from its initial condition. The above two figures show the mathematical assurance of the robustness of the VPDW [9].
In active biped dynamic walking, we realize the case where one of the actuators undergoes a fault condition. Generally, in typical flat surface dynamic walking, all motors will be in active phase. Some faulty situations may arise on the hip joint or on one of the ankle joints in some adverse conditions. In this paper, we consider one of the ankle actuators to be non-functional due to fault. As discussed in the earlier section, dynamic walking is a nonlinear dynamical phenomenon where each link's dynamics is coupled with another one. We have chosen the virtual passive dynamic walker as a reference model for control input generation. Each gait cycle consists of two steps. The fault condition is analyzed in two cases: (a) alternate steps consist of powered (ankle and hip) and unpowered (none), and (b) alternate steps consist of powered (ankle and hip) and partially powered steps (only hip). We did numerical analysis of the above-mentioned cases in terms of global and local stability based on Poincare map and basin of attraction plot.
3.1. Alternate powered and unpowered steps dynamic walker
An Alternate Powered and Unpowered Step Dynamic Walker (APUSDW) is a special case of VPDW in which the first step is considered as powered, and the next one is unpowered. So whatever energy is pumped into the system during the powered cycle is not going to be lost completely during the impulsive transition. The dynamic walker is able to walk without any energy input during the unpowered cycle by using residual energy. The governing equations of motion is a combination of the passive and active dynamic models on the flat surface. These are given by Eq. (1) along with the transition condition Eq. (2) for powered step. The equation of motion for the unpowered step is similar to the powered step, but on the right-hand side of Eq. (1) would be
Bu=
0. One practical significance of the alternate powered walker is that while walking both motors stop supplying the power. However, the walker can still manage robust walking. Fig. 4 indicates that the system gets converged at two limit cycles for the powered and unpowered case.
Fig. 3. Schematic of alternate powered and unpowered steps dynamic walker
3.2. Alternate powered and partially powered steps dynamic walker
In this section, we consider the Alternate Powered and Partially Powered Steps Dynamic Walker (APPPSDW). The second step always behaves like an underactuated system with hip actuation alone active. The system dynamics is similar to APUSDW – equation of motion given by Eq. (1) along with the transition condition by Eq. (2). The equation of motion for the partially powered step is similar to the powered step but on the right-hand side of Eq. (1) will have
Bu ={\left[{u}_{2} -{u}_{2}\right]}^{T}
. The result is similar to that shown in Fig. 4. There is convergence of limit cycle and step length. But the robustness of walking for this model is found to be less as compared to the other model.
4. Model analysis and numerical validation
Our primary focus is to develop a robust walking model for the case of ankle actuator fault based on the virtual passive dynamic walking concept. We modelled APUSDW and APPPSDW to find out the better of the two. We have also performed numerical and graphical analysis for these proposed models which show similar behaviour to VPDW. To ensure stability, we need to perform Poincare analysis and analyze how it behaves over the variation in slope parameter 𝜙.
Fig. 4. Limit cycle plot of APUSDW
Many researchers use Lyapunov concept to find the stability of the nonlinear systems. However, the conventional definition of stability in the Lyapunov sense (about the fixed point) is not suitable for these nonlinear system models due to their hybrid nature, since the periodic solutions cannot be asymptotically stable in the Lyapunov sense. Therefore, we need to define the stability of the system in a periodic or orbital sense. A periodic orbit’s stability for an autonomous system can be calculated by considering Poincare map, which replaces the n-dimensional continuous vector field’s flow with an n-1 dimensional map [10]. The impact point of a dynamic walker is a natural choice for Poincare section, and its successive transition can be related by
{x}_{k+1}=P\left({x}_{k}\right)
x={\left[{\theta }_{1} {\theta }_{2} {\stackrel{˙}{\theta }}_{1} {\stackrel{˙}{\theta }}_{2}\right]}^{T}
represents the state vector of the robot. If the
{x}^{*}
is a fixed point for mapping, we can also write
{x}^{*}=P\left({x}^{*}\right)
. For small perturbation
\delta {x}^{*}
, using Taylor series expansion, we can elaborate
\nabla P ={\gamma }^{-1}\mathrm{\Phi }
. By referring to Guckenheimer et al. [10], we could conclude the stability by calculating the max singular value of ∇𝑃 and by plotting the eigenvalues of ∇𝑃. As shown in Fig. 5(a) and Fig. 5(b), the eigenvalues of APUSDW and APPPSDW are inside the unit circle [1, 11]. We can conclude that our model shows limit cycle stability. The maximum eigenvalue of alternate powered and unpowered is greater than the alternate powered and partially step model. So we can say that the alternate powered and unpowered step model is less stable locally in the neighbourhood of its fixed point.
Fig. 5. Eigenvalues plot for APUSDW and APPPSDW
a) APUSDW
b) APPPSDW
4.2. Basin of attraction
A set of points in state space for which the system can converge to a particular fixed point is known as the basin of attraction. For nonlinear dynamical systems, multiple solutions can exist which could be termed as stable if they fall in the region of basin of attraction. A stable attractor is a point, region, limit cycle, or orbit in state space where all trajectories tend. To better understand the system's global stability, we analyze a set of points around the fixed point. If the system is converging to the particular fixed points within that set, then we could say that the initial condition is stable. For our models, we have analyzed all possible combinations of
{\stackrel{˙}{\theta }}_{1}
{\stackrel{˙}{\theta }}_{2}
between 0.04-2° over the range of
\alpha
(included angle between legs at transition) from 0.01-0.5° for a given value of angular slope
\varphi =
1°. Among these initial conditions, some of them lie inside the basin of attraction, and some of them are not. Initial conditions outside of basin of attraction do not converge to any fixed points.
The basins of attraction for APUSDW and APPPSDW are respectively given in Fig. 6(a) and Fig. 6(b). It can be observed that Fig. 6(a) has a larger basin of attraction than that of Fig. 6(b).
Fig. 6. Basin of attraction APUSDW and APPPSDW
Therefore, the global stability of APUSDW is better as it covers larger region of initial conditions leading to convergence. As far as local stability is concerned, APPPSDW performs better. However, it should be noted that this conclusion is limited in scope to control inputs generated from virtual passive dynamic walking which does not have knowledge of actuator fault. A controller which considers actuator fault could make APPPDW also better in terms of global stability.
We considered the case of biped dynamic walking in the presence of one ankle actuator fault. It was shown that biped walking in the presence of one ankle actuator fault can still lead to successful walking gait cycle. Two cases were analyzed: APUSDW and APPPSDW. Both were found to give stable walking although the controlled had no knowledge of actuator fault. However, for virtual passive dynamic walking base control algorithm considered, the APUSDW is shown to have better global stability. We wish to further investigate stability properties with other controllers which consider actuator fault in generating control inputs to the walker.
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