text stringlengths 1 1.11k | source dict |
|---|---|
And another way:
logLine[min_, max_] := Module[{lines, labels},
lines = Line[{{Log[#], -1}, {Log[#], 1}}] & /@ Range[min, max];
labels = Text[#, {Log[#], 1.7}] & /@ Range[min, max];
Graphics[{
labels,
Line[{{Log[min], 0}, {Log[max], 0}}],
lines
}, AspectRatio -> 1/10
]
]
We have then that logLine[1,12] yields
To plot an arbitrary range we could use the following function:
logLineRange[range_] := Module[{lines, labels},
lines = Line[{{Log[#], -1}, {Log[#], 1}}] & /@ range;
labels = Text[#, {Log[#], 1.7}] & /@ range;
Graphics[{
labels,
Line[{{Log[Min[range]], 0}, {Log[Max[range]], 0}}],
lines
}, AspectRatio -> 1/10
]
]
Having defined that function, we can then do this:
logLineRange[{1.07, 1.15, 1.20, 1.29, 1.38, 1.44}] | {
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"url": "http://mathematica.stackexchange.com/questions/31470/how-to-plot-logarithmic-scales/31471"
} |
model, we give an optimal algorithm, i. The biased coin has a 75% chance of landing on heads. Why is this so? Isn't the unbiased variance. The probability is 0. If we had some weird biased coin that had both sides being head, then the flipping of this coin would be a deterministic process since we would always get a head every flip. Summing these for the total expected number of flips is p/p + (1-p)/(1-p) = 2. Biased coins. " It's a tiny difference, of course, but it's good to know if want to up your chances of winning a bet!. I got a question on the coin flip project. often denoted by uppercase letters, often X, Y, and Z. The ratio of successful events A = 210 to total number of possible combinations of sample space S = 1024 is the probability of 6 heads in 10 coin tosses. 1) is positive half of the time. The most biased coin problem asks how many total coin flips are required to identify a "heavy" coin from an infinite bag containing both "heavy" coins with mean $\theta_1 \in (0,1)$, | {
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"url": "http://zynh.rdswatersports.it/biased-coin-flip.html"
} |
The case $Y_{k,2}$ is symmetrical to the case $Y_{k,1}$. Thus by applying a similar argument, there is an open set containing the point $x$ that is disjoint from the set $Y_{k,2}$.
Now consider the case $Y_{k,\varnothing}$. If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,\varnothing}$, then we are done. So assume $B_k(x) \cap Y_{k,\varnothing} \ne \varnothing$. Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,\varnothing}$. Note that $t_1>x_1$ and $t_2 > x_2$. Now consider the following open set:
$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_1
The set $G$ is an open set containing the point $x$. We claim that $G \cap Y_{k,\varnothing}=\varnothing$. Suppose $g \in G \cap Y_{k,\varnothing}$. Then $x_1 and $x_2. Consider the following set:
$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2 | {
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geometric-optics
What should be different in case of magnification from the case of focusing?
How can these actions be combined? In essence you are describing making a microscope: in the simplest case that is a two-lens system, where the magnification comes about from having a lens near the object (the objective lens) close to the focal distance of that lens, and a second lens (the ocular) collecting that light and focusing it for you (onto your eye, or onto a screen / sensor etc).
The basic equation for a microscope can be found at http://www.schoolphysics.co.uk/age16-19/Optics/Optical%20instruments/text/Microscope_/index.html
In your case, you need to move the ocular lens so that you get a real image instead of a virtual image - in other words, you have to put the objective lens closer to the object than its focal length. | {
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ros2, ros2-control, joint-trajectory-controller, ros2-controllers
I hope this was not too confusing and I really hope someone has some ideas, this has cost me many restless nights. after fiddling around for weeks I have found the solution to my problem!
The problem is not due to the holding position error but that I used velocity as a control interface which does not work in the aforementioned setup.
It worked when i changed it to:
ur_joint_trajectory_controller:
ros__parameters:
joints:
- UR_shoulder_pan_joint
- UR_shoulder_lift_joint
- UR_elbow_joint
- UR_wrist_1_joint
- UR_wrist_2_joint
- UR_wrist_3_joint
command_interfaces:
- velocity
state_interfaces:
- position
- velocity
state_publish_rate: 100.0
action_monitor_rate: 20.0
open_loop_control: true | {
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} |
electricity, electrical-resistance, electrical-engineering
$$
\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} = 0
$$
and since we are assuming no $z$ or $\phi$ dependence, this becomes
$$
\frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) = 0 \quad \Rightarrow \quad r \frac{\partial V}{\partial r} = C \quad \Rightarrow \quad V(r) = C \ln r + D,
$$
where $C$ and $D$ are constants of integration. These latter constants will be determined by the boundary conditions; if we require $V(r_o) = 0$ and $V(r_i) = V_0$, we get
$$
C \ln r_o + D = 0 \quad \text{and} \quad C \ln r_i + D = V_0,
$$
which can be solved to yield
$$
C = \frac{1}{\ln(r_i/r_o)} \quad \text{and} \quad D = - \frac{\ln r_o}{\ln(r_i/r_o)}.
$$
Plugging this back in, we get the solution for $V$:
$$
V(r) = V_0 \frac{\ln (r/r_o)}{\ln (r_i/r_o)}
$$
Now that we've done this, the rest is more straightforward. The electric field is
$$ | {
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ros
Title: hector exploration error msgs
Hi, All,
A while ago I asked a question on using move base and hector exploration together here. Now, I get some error messages when I run SLAM, hector exploration, movebase together | {
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electricity
Title: Light bulbs with voltage source or current source When we connect 2 light bulbs to a voltage source in parallel, the equivalent resistance of the bulbs is R/2 (1/Req = 1/R + 1/R). If we connect these lights in serial we have Req = 2R;
I= V/Req is valid for both circuits, and P=I*V (with V being the same for circuit 1 and 2)
This means the parallel circuit produces more light.
How does this change for a circuit with a current source instead of a voltage source?
Which of these 2 is brightest with a current source? The quick answer is: The series connection produces more light. This can be determined with simple reasoning as follows:
(1) The current is fixed and
(2) Series connected elements have the same current through them | {
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astrophysics, velocity, integration, galaxies
The first integral just uses the standard change of variables to polar coordinates. If $f(r)$ is some function which does not depend on $\theta$ (just as the DF only depends on $v_t$) we have
$$\int \mathrm{d}^2\boldsymbol{x} f(r)=\int_0^{2\pi} \mathrm{d}\theta \int_0^\infty\mathrm{d}r \,r f(r)=2\pi\int_0^\infty\mathrm{d}r \,f(r)$$
which explains the first result.
The second is similar. Firstly we have
$$\int \mathrm{d}x \,x^2\int \mathrm{d}y \,f(r)=\int \mathrm{d}^2\boldsymbol{x}\,x^2 f(r)=\int_0^{2\pi} \mathrm{d}\theta \int_0^\infty\mathrm{d}r \,r x^2f(r)$$
However, by symmetry (as $f$ depends only on $r$) this is equal to
$$\int \mathrm{d}x \int \mathrm{d}y \,y^2f(r)=\int \mathrm{d}^2\boldsymbol{x}\,y^2 f(r)=\int_0^{2\pi} \mathrm{d}\theta \int_0^\infty\mathrm{d}r \,r y^2f(r).$$
Equating the two gives | {
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Oh, is that right? So what happens if I take a point there? Oh yeah, I remember what to do. So let's start at that point on that ellipse. And those were the levels sets f equal constant.
So what's the first search direction? What direction do I move from x0 y0? Do I move along the ellipse? Absolutely not, because along the ellipse f is constant. The gradient direction is perpendicular to the ellipse.
So I move perpendicular to the ellipse. And when do I stop? Pretty soon, because very soon I'm going back up again.
I haven't practiced with this curve. But I know-- and time is up, thank God. So what do I know is going to happen? And by Friday we'll make it happen? So what do we see for the curve, the track of the-- it's say it?
AUDIENCE: Zigzag.
GILBERT STRANG: It's a zigzag, yeah. We would like to get here, but we're not aimed here at all. So we zig, zig, zig zag, and very slowly approach that point. And how slowly? With that multiplier, 1 minus b over 1 plus b. | {
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"url": "https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-22-gradient-descent-downhill-to-a-minimum/"
} |
sorting
}
public boolean done() {
return (unknown.size() == 0);
}
public Compare copy() {
Compare res = new Compare(this.id);
res.greater = copy(greater);
res.less = copy(less);
res.unknown = copy(unknown);
return res;
}
public ArrayList<Integer> copy(ArrayList<Integer> c) {
ArrayList<Integer> res = new ArrayList<Integer>();
for (int i = 0; c.size() > i; i++) {
res.add(c.get(i));
}
return res;
}
public void compare(Compare other, boolean greater, Compare[] items) { //true if greater than other
if(greater) {
this.greater.add(other.id);
other.less.add(this.id);
}
else {
this.less.add(other.id);
other.greater.add(this.id);
}
this.update_unknown(items);
other.update_unknown(items);
}
public void all_greater(Compare[] items) {
for (int i = 0; greater.size() > i; i++) {
ArrayList<Integer> g =items[greater.get(i)].greater;
if(g != null) { | {
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"tags": "sorting",
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} |
computer-architecture, education, reference-question
Title: How does a computer work? I have been a computer nerd for many many years. I can program in quite a few languages, and I can even build them. I sat down with a buddy the other day and asked how a computer actually takes electricity and does something with it, and we just couldnt figure it out, and Google wasn't much help either.
I mean, how does a computer take a constant flow of electricity and turn it into 1's and 0's and then actually do something with those 1's and 0's like turn a light on for 15 seconds?
I understand gates (AND, OR, NOR, NAND, NOT) and a little about diodes, resistors and transistors, but I figured this would be the perfect place to have it explained in true laymens terms! | {
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statistical-mechanics, temperature, renormalization, ising-model, spin-models
$$ \beta H(\sigma_i;J') = -J'\sum_i^{N/3} \sigma_i\sigma_{i+1} + \frac{N}{3}\cdot\mathrm{const.} $$
where $$ J' = \mathrm{tanh}^{-1}[(\mathrm{tanh} \, J)^3].$$
Up to a constant (which cancels out when we divide by the partition function), the Hamiltonian has almost the same form as the initial one. If we now send $N/3 \rightarrow N$ (which I guess we can do without harm in the limit $N\rightarrow \infty$), maybe rescale the lattice, and forget about the constant, we recover precisely the first Hamiltonian, but with a different coupling constant. | {
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"tags": "statistical-mechanics, temperature, renormalization, ising-model, spin-models",
"url": null
} |
magnetic-fields, fourier-transform, resonance, nuclear-magnetic-resonance
Title: Why do the matrices in k-space and the final MRI image have to match? Naturally, there is no correspondence pixel-to-pixel between Fourier space (k-space) and the final 2D image of an MRI - k-space stores the Fourier coefficients, hence each pixel in it affects the whole final image.
Further, there are tons of examples online in which cutting off part of k-space degrades the final image, but only to some extent.
Yet, in general MRI spatial encoding of the image takes place by dephasing the precessing hydrogen atoms across a selected slice as many time in each direction as the number of pixels we want to acquire in the final 2D image. I'm aware that there are methods to acquire half of k-space, for example, to save time, but I'm focusing on the math for the more general method.
So the question is why do we need to match the number of frequency and phase encoding steps (points in Fourier space) during the acquisition of the matrix of the final image? | {
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statistics, distance, self-study
Title: How do I test a difference between two proportions representing fatality rate for Covid 19 in Philippines and World (except Philippines)? I'm trying to analyse if the fatality rate from my country (A third world country) vary significantly from the world's fatality rate.
So I'd basically have two samples, labeled (Philippines) and (World excluding the Philippines) then i can compute the fatality rate for the 2 groups.
Does Mcnemar's test apply here for me to check if fatality rate in the Philippines is higher, or do you have any suggestions? Thanks It is not a case of paired nominal data. Hence, Mc Nemar's test can not be applied to check whether there is a higher fatality rate in Philippines ?. THE fatality rate is given for Philippines and world ( excluding Philippines ). As defined, it is expressed as proportion. Therefore, t-test/z-test shall be appropriate given that you meet other conditions such as sample-size. | {
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Thanks.
Try to construct your solution from this figure
cross posting
5. ## Re: Running Around a Track
I'm sorry. I'm not drawing anything from that figure.
6. ## Re: Running Around a Track
another method
assume track is 1200ft B rate 1200/78=15.38 fps A rate 1200/91= 13.19fps
after B runs 1 lap A is 1029ft ahead of B 91 * 13.19 = 1029 ft
Rate of closure = 15.38-13.19 = 2.2 fps
closure time 1029/2.2 = 467.7 sec
total elapsed time = 467.7 +78= 555.7 sec
7. ## Re: Running Around a Track
Originally Posted by bjhopper
total elapsed time = 467.7 +78= 555.7 sec
typo; 545.7 ; 546 really: LCM(91,78)
These are quite simple: have the faster runner start behind the slower,
the distance behind being the track length; how long does it take to catch up?
8. ## Re: Running Around a Track
Hello Wilmer,
Long 'time no see.How are you?
Soroban' s equation gives elapsed time until runners meet. 546 sec exactly
My answer 555.7 sec.There is no typo error.
Can you show what I did wrong? | {
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rosdep, ros-electric
[sudo] password for joq:
Reading package lists... Done
Building dependency tree
Reading state information... Done
python-qt4 is already the newest version.
python-qt4-dev is already the newest version.
python-sip-dev is already the newest version.
libgenrunner-dev is already the newest version.
libpyside-dev is already the newest version.
libshiboken-dev is already the newest version.
python-pyside.qtcore is already the newest version.
python-pyside.qtgui is already the newest version.
python-qt4-gl is already the newest version.
shiboken is already the newest version.
0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.
successfully installed python-qt-bindings
rosdep python-qt-bindings failed check-presence-script after installation
rosdep install ERROR:
failed to install python-qt-bindings | {
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electrolysis
If salt water is electrolyzed, the first reactions on the electrodes are :
$$\ce{Cathode : 2 H2O + 2 e- -> 2 OH- + H2}$$
$$\ce{Anode : 2 Cl- -> Cl2 + 2 e-}$$
As a consequence, $\ce{H2}$ and $\ce{Cl2}$ are produced on the electrodes, and $\ce{OH-}$ is emitted in the solution at the cathode. But $\ce{Cl2}$ is partially soluble in water, and goes into solution at the anode. After a while the negative ions $\ce{OH-}$ migrate to the anode (like $\ce{Cl-}$) where they meet the dissolved $\ce{Cl2}$ molecules and react according to :
$$\ce{Cl2 + 2 OH- -> Cl- + ClO- + H2O}$$ No free chlorine gas $\ce{Cl2}$ is produced any more at the anode, as it is consumed to create the hypochlorite ion $\ce{ClO-}$ The obtained solution is called "bleach". Now if this bleach is slightly heated, the hypochlorite ions disproportionate according to : $$\ce{3 ClO- -> ClO3^- + 2 Cl-}$$ So the solution contains now three negative ions containing chlorine atoms : $\ce{Cl-, ClO-, ClO3^-}$, instead of one ! | {
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python, computational-geometry
def compute_direction(common_point, first_endpoint, second_endpoint):
first_vector = tuple(map(operator.sub, first_endpoint, common_point))
second_vector = tuple(map(operator.sub, second_endpoint, common_point))
return first_vector[0]*second_vector[1] - first_vector[1]*second_vector[0]
def does_line_segments_intersect(first_segment, second_segment):
d1 = compute_direction(first_segment[0], first_segment[1], second_segment[0])
d2 = compute_direction(first_segment[0], first_segment[1], second_segment[1])
d3 = compute_direction(second_segment[0], second_segment[1], first_segment[0])
d4 = compute_direction(second_segment[0], second_segment[1], first_segment[1])
if d1*d2 < 0 and d3*d4 < 0:
return True
pass | {
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quantum-field-theory, renormalization, regularization
Is that true? to me, it seems better to define renormalization as a collection of techniques for adjusting the theory to obtain physical results. I'll explain. According to Wilson's renormalization group, a quantum field theory always inherently has a cutoff parameter, so in any case integrals should be done only up to the cutoff, so there are no infinite quantities. Yet the results are still not consistent with observation if you don't renormalize the calculations (e.g. using counterterms).
Am I correct? Is it true that the usual presentation of renormalization as a tool for removing divergences is a misinterpretation of the true purpose of it? You're totally right. The Wikipedia definition of the renormalization is obsolete i.e. it refers to the interpretation of these techniques that was believed prior to the discovery of the Renormalization Group. | {
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ros, gazebo, laser, node, ros-indigo
Originally posted by nouf on ROS Answers with karma: 71 on 2015-04-10
Post score: 0
May be this gazebo tutorials will help I guess . There they mentions about adding lasers in URDf model.
<gazebo reference="your_link_name">
<plugin name="your_link_laser_controller" filename="libgazebo_ros_laser.so">
... plugin parameters ...
</plugin>
</gazebo>
Originally posted by SVS with karma: 233 on 2015-04-13
This answer was ACCEPTED on the original site
Post score: 1 | {
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• Please check my Edit 2. – A. Ray May 27 '16 at 3:22 | {
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it cannot, explain why not. x��=]��q��+�ͷIv��Y)?ز�r$;6EGvU�"E��;Ӣh��I���n v��K-�+q�b ��n�ݘ�o6b�j#�o.�k}���7W~��0��ӻ�/#���������$����t%�W ��� For the function f shown below, determine we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0.If not, explain why not. We can see its geometric meaning as follows: \Rolle’s theorem" by Harp is licensed under CC BY-SA 2.5 Theorem 1.2. Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). Question 0.1 State and prove Rolles Theorem (Rolles Theorem) Let f be a continuous real valued function de ned on some interval [a;b] & di erentiable on all (a;b). f c ( ) 0 . The Common Sense Explanation. stream After 5.5 hours, the plan arrives at its destination. Get help with your Rolle's theorem | {
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general-relativity, gravity
$$u^\alpha=\left(\left(1-\frac{2GM}{r}\right)^{-1/2},0,0,0\right)$$
and they would define the total energy (rest mass plus kinetic, but not including potential energy) as
$$T = -g_{ab}u^a p^b = \frac{m}{1-\frac{2GM}{r}} \approx m + \frac{GMm}{r}.$$
To derive this, we used the fact that $E=m$ is conserved, which means that $p^t = \dfrac{m}{\left( 1-\frac{2GM}{r}\right)^{1/2}}$. If we continue to assign the potential energy $V = \frac{GMm}{r}$, then we get
$$E=T-V = m + K-V = m \implies K=V $$
So in some sense the relations you postulated hold when there is a well-defined "effective mass" i.e. Killing energy, but in a general spacetime with no timelike Killing vector, you won't be able to make a sensible definition of such a thing. | {
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function in find derivatives of radical function essentially involves converting the radical i.e! Obtain when we find the first three derivatives but not really sure what to do after.. Therefore we use the definition of derivative to calculate the derivative of a function using definition. … derivative of a radical and prove a formula for the nth derivative to do after that power. Stays the same regardless of the derivative of a radical a point is the slope of,... Are multiplying, we rewrote the radical in terms of u\displaystyle { }! Fundamental: a radical difference radical in terms of u\displaystyle { u } u differentiates _ ( x_+4x_+7 and... The nth derivative constant rule, sum rule, power rule the of! Math, please use our google custom search here liberal party function involve... An exponent and using the definition of derivative to calculate the derivative of radical! Having trouble finding the derivative of a function by applying the power rule and thus its derivative is also.. | {
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ros, gazebo-ros
Title: Spawning via gazebo_ros spawn_entity seems to not put /tf into namespace
Hello,
I want to spawn a robot dynamically via
ros2 run gazebo_ros spawn_entity.py -entity robot1 -x 0 -y 0 -z 0 -file ~/turtlebot3_ws/install/turtlebot3_gazebo/share/turtlebot3_gazebo/models/turtlebot3_waffle_pi/model.sdf -robot_namespace robot1
but when I do so, the /tf topic seems to be outside of the namespace. I did an echo of it with 2 robots running, but I could not distinguish data from the header of the messages.
Many documentation and tutorials only cover the general purpose, but I was not able to get any information wether /tf should be /robot1/tf so far.
Do you know if this is ok, or how I can remap etc.?
Edit:
Because its the same .sdf file, and /tf only transforms relations for the robots parts, this could be ok, right?
Thank you very much
Originally posted by Zeckurbo on ROS Answers with karma: 105 on 2022-08-30
Post score: 0
It should look similar to this: | {
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python, performance
And you can write some simple tests like this :
tests_cases = ([], [1], [1, 2], [1, 2, 3], [3, 2, 1], [2, 2], [2, 2, 2])
for lst in tests_cases:
for i in range(math.factorial(len(lst))):
a = list(from_rank(i, lst))
b = list(from_rank2(i, lst))
if a != b:
print(i, a, b) | {
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11. Oct 31, 2011
lurflurf
sin'(0)=1 is only true in radians, it can be taken as the definition of radians.
sin'(0)=pi/Arccos(-1)
As I said above the sine and cosine in another scaling are easily expressible in terms of the usual one, but that does not make them equal. Some confusion might be resulting from thinking of degrees as a way of expressing a number versus as being a different number. Like if we said
60 degrees=π/3
or
degrees=π/180
by sine in degrees we mean a function where
(60,√3/2)
not
(π/3,√3/2) | {
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python, python-3.x, gui, raspberry-pi
#now that loop has finished, print the final count
return spin """
def startMeasure(self): #this is the function triggered to start the measure of speed
c=calculations()
m=measure()
v=c.speed(m.hydroReel(),c.spinCounter(m.measurementTime()))
v=round(v,4)
l=[m.hydroDepth(),v]
app.setLabel("e5",v)
app.hideButton("Start measure")
file1=''
def insertMeasure(self): #this function inserts all the data into a csv file
global file1,mis_0,l
m=measure()
myFile = open(m.fileName(),'a')
with myFile:
writer = csv.writer(myFile,lineterminator='\n')
if file1!=m.fileName():
firstLine = ["Vertical", "Edge distance", "Depth"]+["HydroDepth","Speed"]*5
writer.writerow(firstLine)
file1=m.fileName()
mis_0=float(m.edgeDist())
depth=float(app.getEntry("e4"))
v=app.getLabel("e5")
l.append(depth)
l.append(v) | {
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ros, ros-melodic
that is perfectly possible, but then I would have expected the call to publish(..) to appear in the body of the service callback. Not right after initialisation. | {
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homework-and-exercises, special-relativity, metric-tensor, coordinate-systems, observers
=\sqrt{1- |v|^2}dt.
$$
Thus the elapsed proper time (the Minkowski analogue of arc-length) along the curve is
$$
\tau(t) = \int^t_0 \sqrt{1- |v(t')|^2}dt'
$$
This relates the $t$ parameter to the $
\tau$ parameter. As the curve is everywhere timelike, the map is invertible and $t$ can be recovered from $\tau$.
Any stuff on differential forms and open sets is just overthinking.
Would you have the same problem with the formula
$$
s = \int \sqrt{1+ \left(\frac {dx}{dz}\right)^2 + \left(\frac {dy}{dz}\right)^2 } dz
$$
computing the arc-length of a nearly vertically-directed Euclidean curve $(x(z),y(z),z)$? | {
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c++, c++11
I also think I can't use std::map since retrieval is \$O(ln)\$.
Do you think the use of shread_ptr inside the deque for memory management is a good design choice?
I used sharad_ptr so that I don't have to do the manual delete of the allocated memory. I also make the class store the shared_ptr rather than a copy of the student because the Student class may contain other information that might be costly to copy. Also a student can be in more then one class ( This way instead of creating multiple copy shared_ptr will only increase the count - Here also I am not sure is it really necessary to increase count because since there will only be one copy I was also thinking about passing by const ref so that we dont have the overhead of increasing the count itself) | {
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First, by the rearrangement inequality, the sum on the right side is maximized when $x_i$'s are arranged in ascending order. Thus, it suffices to show the case with $x_1 \le x_2 \le \cdots \le x_n$. In other words, we wish to show $$(1+a_n)(x_1 + \cdots + x_n)^2 \ge 2 \, n \, x_1 \, (a_1 \, x_1 + \cdots + a_n \, x_n).$$
Second, suppose $x_2 > x_1$, we can replace $x_n$ by $x_n + x_2 - x_1$, and then replace $x_2$ by $x_1$. This would leave sum $x_1 + \cdots + x_n$ unchanged, while increasing the sum on the right-hand side. We can also do this for $x_3, \dots, x_{n-1}$. This means we only need to show the case with $x_1 = x_2 = \cdots = x_{n-1} \le x_n$. Define $\Delta \equiv x_n - x_1$, we only need to show that \begin{align} (1+a_n)(n \, x_1 + \Delta)^2 \ge 2 \, n \, x_1 \, (a_1 \, x_1 + \cdots + a_n \, x_1 + a_n \, \Delta) \\ = 2 \, n \, x_1 \, \left[ \frac{1}{2}(1 + a_n) \, n \, x_1 + a_n \, \Delta\right]. \end{align} | {
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beginner, vba, excel
End Sub | {
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php, programming-challenge, time-limit-exceeded
$chunk1 = array_slice($a, 0, $d);
$chunk2 = array_slice($a, $d);
return array_merge($chunk2, $chunk1);
}
For this input:
$a = range(1, 100000);
$n = count($a);
$d = 500;
… I've measured these times on my local machine*:
original: 1.8130s
optimized: 0.4294
rewritten: 0.0054s
Exponential expression vs. bitwise Operators
Your program has a flaw. It won't calculate the result correctly for large array sizes, because of this:
$n <= 10 ^ 5
^ is bitwise Xor and not pow:
pow(10, 5);
10 ** 5;
Try to avoid globals
I can see, that n is an input parameter that is not part of the function's given signature. However, I would try to avoid globals and get the value manually, if needed:
$n = count($a);
You can read more about this here:
PHP global in functions
Are global variables in PHP considered bad practice? If so, why?
* macOS 10.13, I7 2.5 GHz, 16GB RAM, MAMP PHP 7.2.1 | {
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python, python-3.x, object-oriented, generator, elasticsearch
No need for usage() to call sys.exit(). After usage() is called, and returns normally, execution reaches the end of the file, which if this is the main program file, will naturally end the program. Of course, if this is not the main program file, the main guard would have not run either method, the execution would reach the end of the file completing the importation of the file as a module in another program.
Stop Writing Classes
See "Stop Writing Classes" for a PyCon talk by Jack Diederich.
A class with no instance data members probably shouldn't be a class. Neither ESWriter nor CWLogs have any instance data members.
A class with no constructor and only one public method to call shouldn't be a class. Neither ESWriter nor CWLogs have a constructor. Both have a single public method, called immediately after constructing a class instance, so the instance is not even saved.
These should not be classes.
Private name mangling | {
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ros, compressedimage
Run an image_transport/republish node that will subscribe to the CompressedImage topic, publish it as an Image topic, and subscribe to that.
Look at the format field of the CompressedImage message to determine the compression type, then use an appropriate Python library to manually decompress the image data and extract the height and width.
Originally posted by preed with karma: 88 on 2017-09-29
This answer was ACCEPTED on the original site
Post score: 2
Original comments
Comment by Younès on 2017-09-30:
I can't use republish since it is not recognized by my Ubuntu 16.04. How could I resolve this problem?
Comment by preed on 2017-10-02:
It does exist in 16.04, although you might not have it installed. Try sudo apt install ros-kinetic-image-transport.
Comment by Younès on 2017-10-02:
It was already installed in my Ubuntu. I used the command rosrun image_transport republish compressed in:=/irat_red/camera/image image_transport:=compressed raw out:=/irat_red/camera/image
It works | {
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organic-chemistry, reaction-mechanism, aromatic-compounds, organic-reduction
Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry. Chemistry is an experimental science supported by computational methods. Qualitative explanations for reaction mechanisms follow as a result. More on that issue later.The mechanism of the Birch reduction has been investigated by Zimmerman and Wang.1 What are the experimental results? The Birch reduction is first order in substrate, electrons and alcohol. Therefore, the rate-limiting step is the protonation of the radical anion 2 of anisole 1. Based on the observation that the dianion of anthracene is more basic than its radical anion, 2 the authors designed an experiment utilizing 2% d1-t-BuOH as the alcohol source. The argument is that the radical anion, which is less basic than the anion, will be more discriminating,3 favoring protonation rather than deuteration (isotope effect). The more basic anion, 4 or 7, which is less discriminating, will give rise to a | {
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ros, frame, moveit, tf2, transform
[...] and can you tell me why the arrow link
is starting from 2 Joint?
Sorry, I do not fully understand this question. The position and orientation of pm_slot2 frame depends on the base-frame. The base frame is the world frame in the fanuc world. I think the yellow line shows this dependence.
I do no transformation. I just set the pose reference frame to pm_slot2 and define a pose-msg (not a poseStamped-msg) with the coordinates in the pm_slot2 frame (0,0,0) (x,y,z) / (0,0,0,1) (quaternions). For the position (x,y,z) this is correct, but I think the orientation (quaternions) is wrong.
Thank you!
Finally, I found the solution. I had to transform my target to the pm_slot2 frame.
It was just one line of code: target = tubehandling.tfl.transformPose('/base', target)
Originally posted by nicob with karma: 37 on 2021-12-09
This answer was ACCEPTED on the original site
Post score: 0
Original comments
Comment by Ranjit Kathiriya on 2021-12-09:\ | {
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astrophysics, thermodynamics, soft-question
Now imagine an extremely high-density material such as a neutron star or black hole. I suspect it's not possible to describe the internal energy of (or anything about) an individual atom inside a black hole, since it's impossible to inspect beyond the event horizon, and since I assume that atoms don't exist in an distinguishable form there. But presumably the atoms had energy when they became part of the thing. How is that energy accounted for? The internal energy of a black hole is just its mass. You can measure the mass of a black hole by its gravitational effects on outside bodies, and then extrapolate an equivalent energy using $E=mc^{2}$. | {
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moveit, ros-kinetic
<< " set_payload(" << req.payload << ", [" << req.center_of_gravity.x << ", " << req.center_of_gravity.y
^
/home/shu/catkin_ws/src/Universal_Robots_ROS_Driver/ur_robot_driver/src/ros/hardware_interface.cpp:395:28: error: ‘ur_msgs::SetPayload::Request {aka struct ur_msgs::SetPayloadRequest_<std::allocator<void> >}’ has no member named ‘center_of_gravity’
<< ", " << req.center_of_gravity.z << "])" << std::endl
^
/home/shu/catkin_ws/src/Universal_Robots_ROS_Driver/ur_robot_driver/src/ros/hardware_interface.cpp: In lambda function:
/home/shu/catkin_ws/src/Universal_Robots_ROS_Driver/ur_robot_driver/src/ros/hardware_interface.cpp:394:63: error: ‘ur_msgs::SetPayload::Request {aka struct ur_msgs::SetPayloadRequest_<std::allocator<void> >}’ has no member named ‘center_of_gravity’ | {
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javascript, calculator, ecmascript-6
Try to use regex more
Right here you have a couple of regexes:
.replace('$', '').replace(',', '')
instead you could simplify this to:
.replace(/[$,]/g, '');
this will also be global and will remove all occurrences.
String templates
You're using string templates which is awesome but you can still make some more things string templates:
(6450 * 24).toString() + '.00'
can become...
`${6450 * 24}.00` | {
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turing-machines, undecidability, decision-problem
Algorithm $A$ could run forever, and so one could not tell whether $M$ performs at least 4 steps on some input string. The problem sounds to me undecidable. However I can build a complementary algorithm of $A$ (i.e. $A'$) and I can build a new algorithm $B$ using $A$ and $A'$ in parallel. In this case the problem sounds to me decidable. Your problem is decidable. If $M$ always executes less than 5 steps, then it never sees more than the first 4 symbols of its input. Hence it suffices to run $M$ on all inputs of length at most 4. | {
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IRRATIONAL NUMBERS
1. An irrational number is any real number that cannot be expressed as a ratio of integers.
2. The square root of any positive integer is either an integer or an irrational number. So, $$\sqrt{x}=\sqrt{integer}$$ cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{17}$$, ...).
This week's PS question
This week's DS Question
Theory on Number Properties: math-number-theory-88376.html
DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59
Please share your number properties tips below and get kudos point. Thank you.
_________________
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Joined: 22 Feb 2012
Posts: 11
Followers: 0
Kudos [?]: 2 [0], given: 259 | {
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Now to prove that the Paley construction actually gives a Hadamard design.
In our setting q=4m-1 so r=2m-1 is odd and -1 is not a square.
Corollary (i) says |S|=2m-1 so our blocks are of the right size.
To prove it's a 2-design, we've seen already that we must show that
if a,b are distinct elements of k we must count solutions s,s' in S of
s-s'=a-b. Let c be the nonzero field element a-b. By part (i) of
the Proposition, the count is 1/4 the number of solutions in k* of
x^2-y^2=c. Now there are q-1 solutions in k, because we have
x^2-y^2=(x-y)(x+y) and there are q-1 factorizations, each of which
yields unique (x,y) because 2 is invertible in k. Of those we must
throw out solutions with x=0 or y=0. The number of x=0 solutions is
2 or 0 according as -c is or is not a square; for y=0 the count is
2 or 0 according as +c is or is not a square. But by (ii) exactly
one of these two happens. Hence the number of solutions of x^2-y^2=c | {
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meteorology, tropical-cyclone, geoengineering
Title: Ways, Through Science, to Counteract Hurricane Forces? I know nothing significant about meteorology and definitely not a science guy so this will probably seem naive...
But today I am wondering if there is any existing science or methods a government (with enough funds) could employ, through science, to lessen the force of a hurricane?
I vaguely remember something about them being affected by air temperatures, so couldn't something be done to somehow inject heat or refrigeration into the area? Or something to counteract against it's speed increases?
What approaches would one consider in such an endeavor? It might be possible but hasn't been done in practice
One idea is that smoke could be introduced into the hurricane and, because the smoke is heavier than air, the wind speeds could be reduced. | {
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There is a partial answer, which corresponds to the value $$k=2.$$
Firstly, since $$\det A^k= {\det}^k A = \underbrace{111\dots11}_k\det A,$$ then the easiest case of the solution is $$\det A=0,\tag1$$ as in the given example.
Let us consider the possible dimensions $$n$$ of the matrix $$A.$$
$$\color{brown}{\textbf{Case n=1.}}$$
The case is trivial, it does not correspond with the task statement.
Also, the equation $$a^k = \underbrace{111\cdot11}_k$$ has not solutions.
This fact excludes solutions in the form $$A=aE,$$ where $$\;E\;$$ is an arbitrary unit matrix (or transformed unit matrix).
$$\color{brown}{\textbf{Case n=2.}}$$
The equation $$\begin{pmatrix} a & b \\ c & d\end{pmatrix}^2 = 11\begin{pmatrix} a & b \\ c & d\end{pmatrix},$$ or $$\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & bc+d^2\end{pmatrix}^2 = \begin{pmatrix} 11a & 11b \\ 11c & 11d\end{pmatrix},$$ $$\begin{cases} a+d=11\\ bc = ad, \end{cases}$$ | {
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python, regex
It is called like this:
if __name__ == '__main__':
fileextensions = ['\.doc', '\.o', '\.out', '\.c', '\.h']
try:
currentdir = os.getcwd()
except OSError:
print 'Error occured while getting current directory'
sys.exit(1)
for myfiles in os.listdir(currentdir):
matchextname(fileextensions, myfiles)
Could you please review the code and suggest if there is any better way of doing this, or share any other comments related to errors/exception handling which are missing - or anything else in terms of logic? I think the way to code this, that most clearly indicates what you are doing, is to use os.path.splitext to get the extension, and then look it up in a set of extensions:
import os.path
extensions = set('.doc .o .out .c .h'.split())
_, ext = os.path.splitext(filename)
if ext in extensions:
print(filename)
A couple of other comments on your code: | {
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ros, node, topic
Original comments
Comment by Kenn on 2017-07-17:
The typical ROS way to do this is to represent each source as a separate topic and subscribe to both topics.
I definitely didn't know this but still is it more efficient that the other way? Considering you have to call two functions and synchronize them, subscribing to one topic sounds easier.
Comment by Kenn on 2017-07-17:
and that is what is stuck in my mind for now, so I have to solve it somehow. I thought ROS specific approach to this would be to use services but as I said I wanna fully understand nodes. Do you have any idea how I can do it with two nodes and one topic? And thank you for the help.
Comment by ahendrix on 2017-07-17:
Each publisher sends a complete message; not part of a message. If you look at the data that you receive when you try to use a single topic, you'll see that each message only has half of the data filled in.
Comment by Kenn on 2017-07-18: | {
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velodyne
Can any one Plz Help me with this ????
Thank you :)
Originally posted by Nikka on ROS Answers with karma: 3 on 2016-10-06
Post score: 0 | {
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books, summation
Title: Closed-form expression for the sums of the rows of the Trinomial triangle There is a question in the second chapter of the book that I'm unable to solve, and unfortunately algorist.com does not provide a rigorous enough solution, or maybe I can't quite understand it.
Here is the question: | {
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The Ohio State University linear algebra 2568 exam problem. How to Tell If Two Functions Are Inverses, How to Tell if a Function Has an Inverse Function (One-to-One). rewrite it as x = an expression involving y, i.e. So we know the inverse function f-1 (y) of a function f(x) must give as output the number we should input in f to get y back. The inverse graphed alone is as follows. Step 2: Make the function invertible by restricting the domain. This is the currently selected item. So, how do we check to see if two functions are inverses of each other? Inverse function calculator helps in computing the inverse value of any function that is given as input. No packages or subscriptions, pay only for the time you need. This page explores the derivatives of invertible functions. See the answer. We can determine if a function has an inverse function if a value of y corresponds to only one value of x. To show that the function is invertible we have to check first that the function is One to | {
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machine-learning, cross-validation, class-imbalance, sampling
Imagine the problem of cancer detection, where your original dataset is unbalanced: 10% of the patients have cancer y=1 and the remaining 90% don't y=0. If you train a classifier which is prone to error on unbalanced datasets (such as an Artificial Neural Network), you may end up predicting always the majority class: y=0.
If you oversample to a new distribution, let's say 50/50, your classifier is expected to increase the performance, specially on the positive class. Nonetheless, to measure the performance on real data, which is by itself skewed, measure on oversampled may not be the best choice.
Thus, if you are optimizing the hyperparameters or choosing from a set of classifiers, cross-validating with oversampled data may provide you with a different perspective on the classifier's ability to predict on both classes with equal importance. Nonetheless, if you are estimating the real-life prediction capability, I would not advise you to oversample such validation data! | {
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optics, fourier-transform, crystals, photonics
Actually no. Waves with momentum exactly equal to the band edge are not traveling waves, but are known as standing waves. Read on to see why. If you had chosen another frequency above the gap, the direction would not necessarily be the same because this is a folded band diagram. You'd need to add $0.5$ to whatever value was indicated on the abscissa to account for the extra momentum from a neighboring Brillouin zone, which give you the total momentum at that frequency. The higher order bands actually are extensions from neighboring Brillouin zones. E.g., a frequency of $0.55\frac{\omega a}{2 \pi c}$ indicates $~\frac{2}{3} \Gamma M = \frac{1}{3}\vec{b_2}$, which can't be the total momentum, because that is momentum at $~0.28\frac{\omega a}{2 \pi c}$ also, and these are not the same modes. There is an extra component of either $-\frac{1}{3}\vec{b_2}$ from the neighboring $\Gamma M$ zone, or there is a component in the $M K$ zone, which would mean a direction off axis from the reciprocal | {
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training
Title: Running multiple times of a model is for model randomness or data randomness? When a paper report the average and std of a model on a dataset, it means that they have changed the split of training and test sets and run the model multiple times or they just run the model on constant splitting multiple times to find the randomness of model? Which one is more common? Usually, we report mean and variance for k-fold crossvalidation and similar techniques. We run the model multiple times in different data, but that can be applied to the second case too as, for example, a Neural Network can be initialized with random weights multiple times for test in the model.
If the paper does not explicitly say that they re-splited the dataset, then is more likely that this variance comes from random initialization. You should read the experimental procedures used and try to determine all sources of variance for papers results. | {
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c#, interface, exception, compiler
private static Type BuildType(Type interfaceType)
{
var assemblyName = new AssemblyName($"DynamicAssembly_{Guid.NewGuid():N}");
var assemblyBuilder = AppDomain.CurrentDomain.DefineDynamicAssembly(assemblyName, AssemblyBuilderAccess.Run);
var moduleBuilder = assemblyBuilder.DefineDynamicModule("DynamicModule");
var typeName = $"{RemoveInterfacePrefix(interfaceType.Name)}_{Guid.NewGuid():N}";
var typeBuilder = moduleBuilder.DefineType(typeName, TypeAttributes.Public);
typeBuilder.AddInterfaceImplementation(interfaceType);
var properties = interfaceType.GetProperties(BindingFlags.Instance | BindingFlags.Public);
BuildProperties(typeBuilder, properties);
return typeBuilder.CreateType();
string RemoveInterfacePrefix(string name) => Regex.Replace(name, "^I", string.Empty);
} | {
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43. myininaya Group Title
so what is f' first?
44. sam0004 Group Title
-8x/(x^4+1)
45. sam0004 Group Title
X=-1 and x=1
46. myininaya Group Title
(x^2+1)^2 does not equal x^4+1
47. myininaya Group Title
just leave is at (x^2+1)^2
48. sam0004 Group Title
Increasing from negative infinity to 0 and decreasing from 0 to positive infinity
49. myininaya Group Title
also x^2+1 is always positive so f' and f exist everywhere so you just need to find when f'=0 which is when -8x=0 which is when x=?
50. myininaya Group Title
ok yep
51. sam0004 Group Title
F(x)=3x^2-12x=5 over the closed interval 0,3. Find absolute extrema.
52. sam0004 Group Title
Do i again begin with taking the derivative?
53. myininaya Group Title
see if there any critical numbers between 0 and 3 first
54. myininaya Group Title
which means you need to find f' which all means you need find when f'=0 this looks like you meant to write a polynomial so we don't need to worry when f' dne | {
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quantum-state, measurement, density-matrix, probability
From here, I think that the difference between $\rho$ and $\sigma$ should play a part, but I'm not sure how to proceed. Is my idea correct?
Help would be appreciated. Yes, you can use $\rho - \sigma$ to construct a measurement where the measurement statistics differ for $\rho$ compared to $\sigma$. The issue is that $\rho - \sigma$ is not necessarily positive, while a measurement needs to consist of positive operators (no negative eigenvalues) that sum to the identity.
Here is a place to start: Consider the Hermitian operator $\Lambda = (\rho - \sigma)$ and then take its spectral decomposition to be $\Lambda = A - B$ where $A$ and $B$ are both positive operators. You can write these operators in terms of the eigenvalues/eigenvectors of $\Lambda$:
\begin{align}
A &= \sum_{i: \lambda_i > 0} \lambda_i |\lambda_i\rangle \langle \lambda_i| \\
B &= \sum_{i: \lambda_i < 0} (-\lambda_i) |\lambda_i\rangle \langle \lambda_i|
\end{align} | {
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algorithms, python, breadth-first-search, maze
I leave it as an exercise for you to draw this with pen & paper. Can you take it from here?
Note that the duplicate (3, 4) is version 2 is suspicious. Can you guess where it comes from?
I would like to know if there is a "canonical" version which is considered best practice.
You typically want the invariant that everything in the queue is already put in the seen queue. As a formula, this amounts to saying $$\text{work queue}\ \subseteq\ \text{seen queue}.$$
Otherwise, you could pick the tip of the work queue, say item $i_1$, go through its neighbors to then add its neighbor $i_2$ to the work queue. However, $i_2$ was already part of the work queue before (perhaps just after $i_1$!), but you couldn't know since you didn't keep the seen queue synchronized.
Hence, implementing this invariant amounts in your code to
putting start onto the seen queue
queue = collections.deque([[start]])
seen_or_visited = set([start]) | {
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model-checking, software-verification, transition-systems
The states of $\texttt{TS}$ are all possible combinations of values of variable $x$ and register $r$ (no output variable $y$) in $\texttt{SC}$.
The label of each state in $\texttt{TS}$ consists of all the variables (including the register) which are evaluated true given the valuations in $\texttt{SC}$.
For example, $\{x, r, y \}$ for state $(x = 1, r = 1)$ means that all variables are true according to $\lambda_y = \lnot (x \oplus r)$ and $\delta_r = x \lor r$ given $x = 1, r = 1$.
The transitions (arcs with arrow) between states in $\texttt{TS}$ result from the functions $\lambda_y$ and $\delta_r$.
For example, $(x = 0,r = 1) \to (x = 0,r = 1)$ if the next input bit equals 0.
For more details and a general recipe for modeling sequential hardware circuits as transition systems, please refer to the book "Principles of Model Checking" (Section 2.1.2) by Christel Baier and Joost-Pieter Katoen. The MIT Press. | {
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integer ratios. When we divide two integer numbers we will have an equation that looks like the following: Sometimes, we are only interested in what the remainder is when we divide A by B. It's often represented with vertical bars, so |-5| = |5| = 5, for example. Solving Modulus Inequalities. Definition of modulus in the Definitions. Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members. Geometric representation of complex numbers. Basic arithmetic is, of course, built-in to PowerShell. It also offers good support for object-oriented programming, functional programming, and data-driven programming. Example: 10 mod 3 = 1 (because 10 divided by 3 = 3 with remainder 1). But –1 will not work, and neither will +1, because they're too close to zero. If a is not congruent to b modulo m, we write a 6 b( mod m). In this case we have z 1 = 6 + 3 i and z 2 = 10 + 8 i. Modular arithmetic (sometimes called clock arithmetic) is a system of | {
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neural-networks, deep-learning, convolutional-neural-networks, comparison
Title: Can a vanilla neural network theoretically achieve the same performance as CNN? I perfectly understand that CNN takes into account the local dependency of each pixel to the nearby pixels. In addition, CNNs are spatially invariant which means that they are able to detect the same feature anywhere in the image. These qualities are useful in image classification problems given the nature of the problem.
How does a vanilla neural net exactly falls short on these properties? Am I right in claiming that a vanilla neural net has to learn a given feature in every part of the image? This is different than how a CNN does it, which learns the feature once and then detects it anywhere in the image.
How about local pixel dependancy? Why can't a vanilla neural network learn the local dependency by relating one pixel to its neighbors in the 1D input? | {
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machine-learning, machine-learning-model, convolution, image-size, image
Title: inner workings of Mobile-net resolution multiplier - what does it do? i have a question concerning the way Mobile Net's resolution parameter works. From the article itself and from the blog posts on the topic (1, 2) I wasn't able to find an answer to my question. It is said that “the resolution multiplier changes the input dimensions of the image, reducing the internal representation at every layer”. Does that mean it (the parameter):
A) breaks down each image into a, for example, respective 224x224 bits of pixels sub-images, over which the network runs its convolution?
OR
B) takes one single 224x224 slice of pixels from, for example, the middle of the image, over which the network runs its convolutions?
OR
C) Squishes the original image into a, for example 224x224 pixel representation of itself, over which the network runs its convolutions? I contacted the authors of the mobile net architecture and papers and they provided me with the following answer: | {
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mineralogy
All of those tiny fist-sized samples are probably pretty much different types of minerals. So then to scale that to buckets or bins and tractors full of each type, I have never seen.
So it seems that these small shops get the minerals in small quantities from somewhere other than a Home-Depot--like place. That's what I'm wondering about (the first part of the question), where they get the minerals from. I have heard that these people actually go out rock hunting and find them themselves. I am not wondering where specifically they go and find each different type (I have no specific minerals I am interested in), just generally wondering if they are actually going out into the fields with a chisel and a brush and digging stuff up, putting them in their pockets, and then taking them home and cleaning them up.
If that's not the case, I wonder what they actually do to get them (at a high level). | {
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Number Properties | Algebra |Quant Workshop
Success Stories
Guillermo's Success Story | Carrie's Success Story
Ace GMAT quant
Articles and Question to reach Q51 | Question of the week
Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
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Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com | {
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polygon. As a demonstration of this, drag any vertex towards the center of the polygon. The Exterior angle of regular polygon formula explained with examples. The sum of the exterior angles of convex polygons is 360°. Some of the worksheets for this concept are Interior and exterior angles of polygons, Interior angles of polygons and multiple choices, 6 polygons and angles, Infinite geometry, Work 1 revised convex polygons, 15 polygons mep y8 practice book b, 4 the exterior angle theorem, Mathematics linear 1ma0 angles polygons. Simply enter one of the three pieces of information! For instance, in an equilateral triangle, the exterior angle is not 360° - 60° = 300°, as if we were rotating from one side all the way around the vertex to the other side. Calculate angles of regular and irregular polygons and create tessellations and tiling patterns. If a polygon has 5 sides, it will have 5 interior angles. The formula for calculating the size of an exterior angle is: exterior angle of a | {
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beginner, game, haskell, snake-game
Right-hand sides composed solely of an if-statement are good candidates for writing guards.
stepSnake snake dir
| validCoord tail' && not (tail' `elem` toList snake') = Just (head, snake'', tail')
| otherwise = Nothing
This is a very dense function though, I think it might be easier to write if you wrap the version using a Seq snake around a primitive list version. Your snakes will presumably never grow so long as to actually cause a user noticeable slowdown between steps due to list processing, so you might think about tossing out Seq altogether.
stepSnake :: Snake -> Direction -> Maybe (Coord, Snake, Coord)
stepSnake snake dir = fmap (second fromList) (stepSnakeList (toList snake) dir)
where second f (a, b, c) = (a, f b, c) | {
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c++, error-handling, iterator, assembly, serialization
// figure out number of bytes for this opcode
if (std::distance(it, end) > opbytes) {
it = end;
// throw or return 0
}
// disassemble the instruction thoroughly
std::advance(it, opbytes);
return opbytes;
} | {
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and matrix-matrix multiplication will be outlined. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. This topic is collectively known as matrix algebra. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. While off diagonal elements are zero. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Okay, Now we will see the types of matrices for different matrix operation purposes. The scalar matrix is basically a square matrix, whose all off-diagonal | {
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kinetics, enzymes
Let $K_B$ be the binding equilibrium constant for species B, then $25K_B$ is that for A. The species [ES] for A is then at free energy $\Delta G^0_A=-RTln(25K_B)$ and B at $\Delta G^0_B=-RTln(K_B)$, which means that the activation energy (from [ES] to its transition state towards products) is greater for A by the difference between these values which is $RTln(25)$. The ratio of rate constants is the ratio of Arrhenius equations each of which we assume has the form $k=A_0exp(-(E_{TS}+\Delta G)/RT)$ (where $A_0$ is the pre-exponential term assumed equal). Most terms cancel giving the ratio of rate constants as 25 with B being faster. | {
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• I don't think this is a general formula. – true blue anil Oct 17 '16 at 15:28
• @trueblueanil it is the general formula (look at my answer), I just did not see any explanation on how SkeletonBow arrived there. – Thanassis Oct 17 '16 at 15:31
• Oh, I misread the question as the $8$ sided die scoring higher. Drat, can't up vote now, so sorry ! – true blue anil Oct 17 '16 at 15:33
• @trueblueanil no problem. I saw your answer (which you deleted), and figured that you thought it was for numbers higher than or equal to the ones on the 8-die. – Skeleton Bow Oct 17 '16 at 15:37
• @Thanassis to be honest, it was quite simple: I drew it out on a grid and realized that every time you add a number to the smaller $m$, the number of possibilities grew by $m-1$. This meant that I needed to find the total sum of the first $m-1$ integers and then divide that by the total number of squares on the grid, which would be $m\cdot n$ and then got the simplified result. – Skeleton Bow Oct 17 '16 at 15:40 | {
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ros, navigation, tf-broadcaster, robot-localization, transform
process_noise_covariance: [0.05, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0.05, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0.06, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0.03, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0.03, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0.06, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0.025, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0.025, 0, 0, 0, 0, 0, 0, 0, | {
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clustering, unsupervised-learning, kernel, rbf
The numbers of clusters (4): try it out with different numbers of clusters, and you will find more trivial results
The positions of observations: move #0 to the left and you may find results that make more sense
The low number of observations: #6 is not considered as an outlier, due to its vicinity to the #5, #8, #9 triplet. Indeed, the more points there are in an area, the largest its influence is. This is also how #4 and #7 can be considered part of the (#5, #6, #8, #9) cluster, rather than the (#2, #3) cluster. Try removing #9 and see how the results suddenly make sense.
The chosen kernel (not tried on my side, but it is likely that the results may vary a lot if you increase the gamma value or change the kernel shape)
TL;DR: your case is an example of how spectral clustering can go wrong with low number of observations and dimensions. | {
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quantum-mechanics, quantum-information, quantum-computer
$$
$$
K^6 = IZZIIZZ
$$
The stabilizer set establishes valid codewords for a state if the equation $$s\left|\psi\right\rangle=\left|\psi\right\rangle,\;\;\;\forall s \in S \;\;\;\;\; (1)$$
is satisfied. That means $\left|\psi\right\rangle$ is a +1 eigenstate of $s$.
We then consider a practical example of the usage of these stabilizers | {
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graph-isomorphism
Title: On lattice and code isomorphism We know deciding isomorphism between lattices or codes is difficult if the presentation is through arbitrary bases. What if the presentation of the lattice is through minimum bases? Likewise the corresponding problem for codes? The reduction from graph isomorphism to linear code isomorphism (Petrank and Roth '97) has the property that the vectors used in the reduction are precisely the lowest-weight vectors, having weight 5, while all other vectors have weight at least 6. So even when given by minimum-weight vectors, Linear Code Equivalence is still GI-hard. | {
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deep-learning, reinforcement-learning
Questions:
Should one rely on the B2 state to iterate over possible actions from this state (next state) to get an approximation of highest reward (max Q)?
Then, why do we store the A1 and move-to-B2 information at all in the replay buffer?
Or I am wrong and we just use the A1 and iterate over possible actions (including that to B2) to get the max Q?
Edit: I think I have found an answer ). We need to store previous state (A1) and action (move to B2) in order to create the state-action distribution, which will be met with the expected long-term reward distribution, that we get after the next state routine. Right? The TD Target (for learning update) for using $\hat{q}(s,a)$ neural network in Q-learning is:
$$r + \text{max}_{a'} \hat{q}(s',a')$$
In order to calculate this, you need a starting state $s$, the action taken form that state $a$, and the resulting reward $r$ and state $s'$. | {
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vehicles such as bikes car a. Both a magnitude and direction. with a velocity v is define momentum and its si unit product of its of. Term impulse, and Sports term impulse, and state its SI units are still in... ( P ) is the SI unit of momentum is kilogram meters per second points ) 1/ t d f! Its inertia ) 1/ P=mv ) v 2 â v 1 ) define momentum and it 's unit! T d P f = d t d P f = m ( v 2 â 1! Previous Year Narendra Awasthi MS Chauhan 's SI unit ⦠Write its SI units:. Kilogram meters per second -1 ; the SI unit kg m/s, or equivalently, N s is! For measuring time is the product of units of mass and velocity ( P=mv ) quantity... Mass the total quantity of motion possessed by an object of mass and velocity 1.5... Define intensity of radiation on the basis of photon picture of light however, momentum is vector! You have to measure velocity in SI unit of mass and velocity of any object rolling a! 2011, 2014 ) or define one newton of force is newton quantities obtained..., you have to | {
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$x=\dfrac{25}{29} \pm \dfrac5{29} \sqrt{373}$
Take these x-values and plug them into the equation of the line to get the y-coordinates of the points of intersection.
EDIT: Use the attached sketch to control your results. | {
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gravity, string-theory, branes, compactification, kaluza-klein
This latter question is considered far more serious and has received far more attention.
The models with large extra dimensions explain the existence of the hierarchy by saying it is somewhat illusory, and the true Planck scale is much closer to the Fermi scale than we think. But I suppose this might also explain the stability of the hierarchy: in this scenario, the heavier scales of physics just aren't much heavier than the Fermi scale, so their effects need not be disruptive.
For other models of physics, such as those with small extra dimensions, the hierarchy is still large, so the stability problem has to be resolved some other way. Until 2012, the standard expectation was that new physics just above the Fermi scale, such as supersymmetry or compositeness, neutralized the heavy virtual effects, and thereby stabilized the hierarchy. | {
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machine-learning, ensemble-modeling
Title: Learning Algorithm that decide which model gives better results for each testing instance Is their any existing Ensemble technique which uses subset of training data to predict which algorithm is better for predicting each instance of testing data?
Let's say we have N sized training set and K sized testing set in which a particular attribute needed to be predicted using the training set. But there are hundreds of algorithms and ways we can use. We can divide training set into two parts and train each model with first half and decide test on second half. Based on characteristics, we can decide which algorithm to use for real test cases (K sized set). As an example lets say dataset have an attribute named "temperature". Particular algorithm may work well when temperature is higher than 100 Celsius. We can then classify all the 100 degree or above instances to particular class. Then final prediction will be done based on that by with that model class trained with all N sized data. | {
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all data available at the kth step. GaussianMatrix[r, {n1, n2}] gives a matrix formed from the n1\[Null]^th derivative of the Gaussian with respect to rows and the n2\[Null]^th derivative with respect to columns. The basis used in the present study involves multiple Gaussian kernels. Because SAS/IML is a matrix language, you can evaluate the Gaussian kernel on a grid of integer coordinates (x,y) where x is an integer in the range [1,5] and y is in the range. Recently, Micchelli and Pontil (2005) have shown that, for a general class of kernel-based classification meth-ods, the associated optimal kernel selection problems are in fact convex problems. The Gaussian function at scales s=. When I first learned about Gaussian processes (GPs), I was given a definition that was similar to the one by (Rasmussen & Williams, 2006):. Lin The linear kernel. We describe an approach to kernel selection in Support Vector Machines (SVMs) driven by the Gram matrix. the Gram matrix is bounded using | {
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kinematics
Title: Tricycle kinematics
I am learning Robotics. I came across the following equations for tricycle kinematics. I don't understand why the velocity along the $y$-axis is $0$ in the robot reference frame. Shouldn't it be $V_s\sin\alpha$? Or else, the robot will not turn right? The $y$-axis in the tricycle's reference frame is parallel to the rear axle of the tricycle. The two wheels on that axle impose a constraint on the tricycle's motion when that motion has been constrained to no slippage (as specified in the lower left bubble in your graphic where it states we assume no slippage for the expression $V_y(t)=0$). Because the back wheels are presumably perpendicular to their axle (the tricycle's $y$-axis), they will prevent all motion parallel to this axle unless the wheels are allowed to slide sideways. Remember that wheels cannot roll sideways, they can only roll perpendicular to their axle!
Edit: | {
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acid-base, corrosion
Title: Does 99.8% acetic acid cause severe skin burns like formic acid? I've so far handled only concentrated fuming nitric, sulfuric & hydrochloric acids from the inorganic side. Even during an accidental spill on the hand I did say only the $\ce{HNO3}$ affects the skin by turning it yellow and the top layer sometimes peels off.
But in the organic acid side I've had once experience in the past long long ago with formic acid and it was pretty nasty. A really pungent smell and if it gets on the hand then immediately the top layer of skin will peel off. Worse than there inorganic ones. Is 99.8% acetic acid this strong? | {
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string-theory, black-holes, quantum-gravity, ads-cft, holographic-principle
Title: Has the black hole information loss paradox been settled? This question was triggered by a comment of Peter Shor's (he is a skeptic, it seems.) I thought that the holographic principle and AdS/CFT dealt with that, and was enough for Hawking to give John Preskill a baseball encyclopedia; but Kip Thorne is holding out, as are others. Why? It is a matter of opinion, largely dependent on what you mean by settled: | {
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
### Show Tags
26 Feb 2015, 21:43
Hi All,
Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeating "pattern" of the units digits.
Here's another way to organize the information.
We're given [(2222)^333][(3333)^222]
We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]
(2222)(3333) = a big number that ends in a 6
Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ([(2222)(3333)]^222) will be a gigantic number that ends in a 6. | {
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electromagnetism, forces, electrostatics, electric-fields
1 The Abraham-Lorentz force complicates this statement a bit, but if we're assuming that the effects of radiation are negligible, then so are the effects of the Abraham-Lorentz force. | {
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cell-biology, cell-membrane, cellular-respiration, mitochondria, definitions
This is a very interesting topic and is best understood with an explanation of the JC-1 Dye.
Explanation
Mitochondrial membrane potential changes - depending on cellular events.
The dye 'JC-1' is able to indicate mitochondrial membrane potential in different cells. | {
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$$\left(y^2 - 1\left)\frac{y}{y^2}\right.\right. = \frac{y^2 - 1}{y} = y - \frac{1}{y} = \sqrt{x + 1} - \frac{1}{\sqrt{x + 1}}$$
This means that the "oblique" asymptote is $y = \sqrt{x + 1}$.
• $y = \sqrt{x + 1}$ is a curvilinear asymptote. It is not oblique since an oblique asymptote is a line. – N. F. Taussig Sep 30 '14 at 9:32
• @N.F.Taussig Hence why I put quotes around the term oblique...this function asymptotes to the function $f(x) = \sqrt{x}$ as I stated--it doesn't exactly asymptote to that, but it essentially does. – Jared Sep 30 '14 at 9:32
• I knew you understood the question. I just wanted to make sure that the person who posted the question understood why you put the quotes there. – N. F. Taussig Sep 30 '14 at 9:33 | {
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• Yes, it is correct. – Timbuc Mar 28 '15 at 16:24
• The $\frac{\text{first~term}}{1-r}$ formula comes from $(1-r)(1+r+r^2+\ldots + r^n) = 1-r^{n+1}$ (multiply out to see why). If $|r|<1$ then $r^{n+1} \to 0$ as $n\to \infty$ giving the formula $1+r+r^2 + \ldots = \frac{1}{1-r}$. See for example this and this – Winther Mar 28 '15 at 16:27
• mathworld.wolfram.com/GeometricSeries.html – Poppy Mar 28 '15 at 16:29
• @Winther So, as far as I understood from the last statement from the WolframMathWorld webpage you sent me, if the sum started from $k=1$, instead of $k=0$, instead of $1$ and the numerator, we could have $r$, the ratio, so we would have the formula $\frac{r}{1 - r}$? – nbro Mar 28 '15 at 16:37 | {
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java
Title: Data structure limited by range For a utility class, I've created a particular implementation of a list where the list's items can be limited to a specific range, using the google guava Range utility. Items in the list must accordingly implement Comparable. When an item is added or retrieved from the list, they will be within the specified range.
Note: I'm using this class to store commits/revisions from git or svn repositories, then users can limit information to a specific range. I know I could simply filter out information, but I decided to go all out and see if I could turn it into a learning experience
I have two different implementations of this:
Accept any items, but hide them based on the range. The range could then be modified, and previously hidden items could be exposed.
Only accept items within the range, and when the range is updated, all items outside of the range are discarded. | {
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python, text-mining, apache-hadoop, k-means, distance
Second, you might consider a different feature modeling approach, rather than tf-idf. Nothing against tf-idf--it works fine for many problems--but I like to start with binary feature modeling, unless I have experimental evidence showing a more complex approach leads to better results. That said, it's possible that k-means could respond strangely to the switch from a floating-point feature space to a binary one. It's certainly an easily-testable hypothesis!
Finally, you might look at the expected class distribution in your data set. Are all classes equally likely? If not, you may get better results from either a sampling approach, or using a different distance metric. k-means is known to respond poorly in skewed class situations, so this is something to consider as well! There is probably research available in your specific domain describing how others have handled this situation. | {
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meteorology, atmosphere-modelling, visualization
If iice=0:
qprc=qr qrain=qr and qclw=qc if T>0C
qcld=qc =0 =0 if T<0C
qsnow=qs and qclice=qc if T<0C
=0 =0 if T>0C
If iice=1:
qprc=qr+qs qrain=qr and qclw=qc
qcld=qc+qi qsnow=qs and qclice=qc
Independent of the above definitions, the scheme can use different
assumptions of the state of hydrometeors:
meth='d': qprc is all frozen if T<0, liquid if T>0
meth='b': Bocchieri scheme used to determine whether qprc
is rain or snow. A temperature assumption is used to
determine whether qcld is liquid or frozen.
meth='r': Uses the four mixing ratios qrain, qsnow, qclw,
and qclice
The routine uses the following
expressions for extinction coefficient, beta (in km**-1),
with C being the mass concentration (in g/m**3): | {
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PRACTICE PROBLEMS: For problems 1 & 2, use sigma notation to write the Macluarin series for the given. it is a polynomial, or its sum is an entire transcendental function, which is regular in the entire place and which possesses an essential singular point at infinity. 15, we say that the radius of convergence is zero and that the radius of convergence is infinity for case (iii). Centre, Radius, And Interval Of Convergence Of Derivative Power Series. So the radius of convergence of this series is actually 1, because x goes from 0 up to 1, and then from 0 down to 1. Then, and. Power series don't have to be centered at 0. 2 Radius of Convergence Radius of Convergence There are exactly three possibilities for a power series: P a kxk. The radius of convergence r is a nonnegative real number or ∞ such that the series converges if. 3 of the text, one can integrate or differentiate one power series to get another. We can find R generally by ratio test. 9 Problem 1E. n=0 n P 1 1 n (b) (z/a) , | {
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quantum-field-theory, quantum-electrodynamics, polarization, scattering-cross-section
Title: Soft bremsstrahlung calculation in Peskin and Schreoder On page 183 of Peskin and Schroeder, we have the following scattering cross section
$$\tag{6.23}d\sigma(p\rightarrow p'+\gamma)=d\sigma(p\rightarrow p')\cdot \int\,\frac{d^3k}{(2\pi)^3}\frac{1}{2k}\sum_{\lambda=1,2}e^2\bigg|\frac{p'\cdot\epsilon^{(\lambda)}}{p'\cdot k}-\frac{p\cdot\epsilon^{(\lambda)}}{p\cdot k}\bigg|$$
My question is that $\frac{1}{2k}$ makes no sense because $k$ is a 4 vector. Should it be $\frac{1}{2k^0}$ instead?
Then Peskin and Schroeder claims that the differential probability of radiating a photon with momentum $k$, given that the electron scatters from $p$ to $p'$ is
$$\tag{6.24}d(\text{prob})=\frac{d^3k}{(2\pi)^3}\sum_{\lambda}\frac{e^2}{2k}\bigg|\boldsymbol{\epsilon_\lambda}\cdot\Big(\frac{\boldsymbol{p'}}{p'\cdot k}-\frac{\boldsymbol{p}}{p\cdot k}\Big)\bigg|^2$$ | {
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assembly
section .text
global main ; provide "main" entry point for glibc
extern printf, scanf ; these will be linked in from glibc
main:
push rbx ; will be used to keep sum (r12-r15 not used/modified)
push r12 ; will be used as counter
sub rsp,8 ; input buffer in stack + stack 16B alignment fixed | {
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astronomy, orbital-motion, moon, tidal-effect
Apparent size
If $R_m$ is the radius of the Moon and $D_m$ its geocentric distance (that is, the distance between the centre of the Moon and the centre of the Earth), then its geocentric angular diameter is
$$
\delta = 2\arcsin\left(\frac{R_m}{D_m}\right).
$$
Obviously, changing both $R_m$ and $D_m$ by a factor 2 wouldn't change the geocentric angular size. However, we're not observing the Moon from the centre of the Earth, but from a particular location on the surface on the Earth: when the Moon is above the horizon, we're slightly closer to it than the geocentric distance. In particular, if the Moon is directly overhead, it extends an angle
$$
\delta' = 2\arcsin\left(\frac{R_m}{D_m-R_e}\right),
$$
where $R_e$ is the radius of the Earth. So if we change $R_m$ and $D_m$ by a factor 2, we'd measure an angle
$$
\delta'' = 2\arcsin\left(\frac{2R_m}{2D_m-R_e}\right) < \delta',
$$
in other words, the Moon would actually appear slightly smaller from a particular location.
Diurnal parallax | {
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# Is there an irrational number arbitrarily close to another irrational number?
I know that there is a rational number arbitrarily close to an irrational, due to the density of real number. But what about an irrational number? Thanks!
Yes, consider $\alpha + \frac{1}{n}$ where $\alpha$ is irrational and $n$ is an integer. $\alpha + \frac{1}{n}$ is also irrational and can be made arbitrarily close to $\alpha$ by choosing $n$ to be sufficiently large.
• Thank you. Stupid question but can one define something like the number closest to an irrational? For example, the question "which number is closest to a given irrational number is rational or irrational?" makes no sense right? just making sure. – ANANDA PADHMANABHAN S Jan 21 '16 at 5:25
• Irrationals can differ in how well they can be approximated by rationals, check this out: mathworld.wolfram.com/IrrationalityMeasure.html – Dan Brumleve Jan 21 '16 at 5:29 | {
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interference, spectroscopy
Now, the thing that really worries me is that from 630 nm the tungsten source (cyan) was measured to be of higher intensity than the tungsten and the deuterium source!
I don't know how to explain this result. A friend of mine told me that saturation could be a reason but I don't believe this since it is odd to think that if I provide a pixel with more photons it will count less, i.e. the slope of the curve of counted photons as a function of incident photons is negative. | {
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gas-laws, concentration, heat
Title: How would one calculate the Heat Capacity Ratio for a multiple compound Gas? So I'm attempting to calculate the heat capacity ratio (known also as adiabatic index) of a gas with several concentrations of different compounds such as N2, H2O, CO2, CH4 etc. And while I have an idea (I show below) I haven't actually found any confirmation that this is how the ratios work. Perhaps there is some special way the gases interact changing the heat capacity ratio of the entire gas that someone knows of?
What I'm assuming I should do is take the ratios of each and multiply them by the percentage of the weights within the gas. For example if my gas was 20% CO2 and 80% N2, and CO2 had a ratio of K=1.2, and N2 had a ratio of K=1.33 (just as placeholder values), then my calculation would be
Total ratio of my gas : (0.2)(1.2) + (0.8)(1.33) = 1.304 | {
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quantum-mechanics, crystals
My first question arises here: Why can we assume that the kinetic energy term is invariant under translation?
Regardless, assuming it is invariant under translation for the purpose of this question, we define a translation operator, $\hat{T}$, such that, $\hat{T}f(r)=f(r+T)$, and that $\hat{T}$ can be represented by a function $T(n_1,n_2,n_3)$. Consequently, there are an infinite number translation operators, indexed by $n_1,\,n_2$, and $n_3$. We may hence define a second translation operator, $\hat{T}'$, indexed by $n_1',\,n_2'$, and $n_3'$, such that $\hat{T}\hat{T}'=\hat{T}'\hat{T}$, and therefore, that $[\hat{T},\hat{T'}]=0$.
The book goes on to define $\hat{T}''=\hat{T}\hat{T}'$, corresponding to the translation vector, $T''$, indexed by $n_1+n_1',\,n_2+n_2'$, and $n_3+n_3'$. | {
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