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# How does electricity produce magnetisim? [duplicate]

I am in class 10 and I have just read the electromagnetism chapter. They always keep on saying that electricity produces magnetism! but how!? We know electrons move, but how do they produce something that can exert a force on something metallic far away from those electrons? I have tried to read about it on wikipedia but I am not able to understand the complicated terms! Please answer in such a way that it will remove all my confusions!

• How? Because that is a fundamental property of charges! If they move, they generate a magnetic field. In many ways, there is no more deep explanation. If you understand special relativity, you can transform the electric field from the rest frame of the particles to the laboratory frame, and see that a magnetic field arises, but I guess, you will not find that very illuminating. – Sebastian Riese Jun 3 '15 at 16:17
• I donnot find it even illuminating! – Aaryan Dewan Jun 3 '15 at 16:21
• physics.stackexchange.com/questions/184646/… – HolgerFiedler Jun 3 '15 at 16:35
• Op, there exist several descriptions of this system at various levels of complexity...to help give an appropriate answer, it would help to know a few things: are you familiar with calculus, or have an intuition for the concept of a field? How has electricity been described in class 10 thus far? – anon01 Oct 12 '15 at 16:42

I'll attempt to "explain" it as simply as possible but without cheating ("not simpler"). This is indeed not an answer to the "how" does something happen, but rather how physicists describe it.

• in physics, one calls a scalar field a function $\psi:\mathbb{R}^4 \rightarrow \mathbb{R}$ from $\mathbb{R}^4$ to $\mathbb{R}$, (where $\mathbb{R}^4$ is the mathematical object corresponding to "space-time", i.e. a set of points whose position is given by 4 numbers, the first coordinate being the time t, and the 3 others the (x,y,z) coordinates). One could also consider a function from space $\mathbb{R}^3$ only, to $\mathbb{R}^3$

This notion allows to describe for example the evolution the temperature in a room: a temperature scalar field $\psi$ would associate to each point $P\in\mathbb{R}^4$ with "temporal position" $t_P$ and spatial position $(x_P,y_P,z_P)$ the value denoted $\psi(t_P,x_P,y_P,z_P)$ of the temperature at that precise time and spatial position.

• In eletromagnetism one needs the notion of vector field: to each point in space-time is associated a vector:

1- the eletric field $\vec{\mathbf{E}}: \mathbb{R}^4 \rightarrow \mathbb{R}^3$ describes the force that is exerted on a charged particule with charge $q$ at space-time position $(x_P,y_P,z_P)$ via the following formula $$\vec{\mathbf{F}} =q \vec{\mathbf{E}}(t_P,x_P,y_P,z_P)$$ $\vec{\mathbf{F}}$ is a vector, it is the mathematical description of the force exerted on a charged particule at position $(x_P,y_P,z_P)$ in presence of the field $\vec{\mathbf{E}}$. The vector $\vec{\mathbf{F}}\in\mathbb{R}^3$ (belongs to $\mathbb{R}^3$), one indeed needs 3 numbers to describe the direction and intensity of a force.

2- The magnetic field $\vec{\mathbf{B}}: \mathbb{R}^4 \rightarrow \mathbb{R}^3$ describes another force that is exerted on charged particles (or also on particles with a magnetic momentum). In the case of charged particles, the formula is given by $$\vec{\mathbf{F}} =q \vec{\mathbf{v}}\wedge \vec{\mathbf{B}}(t_P,x_P,y_P,z_P)$$ where $\vec{\mathbf{v}}$ is the speed of the particle and $\vec{\mathbf{v}}\wedge \vec{\mathbf{B}}$ (also denoted $\vec{\mathbf{v}}\times \vec{\mathbf{B}}$) is some vector defined by some formula which depends on $\vec{\mathbf{v}}$ and $\vec{\mathbf{B}}$, but one can just remember that it is perpendicular to both $\vec{\mathbf{v}}$ and $\vec{\mathbf{B}}$. In particular this formula says that if the particle does not move, the magnetic field has no effect on it.

To summarize, if you are a charged particle with charge $q$ at position $(x_P,y_P,z_P)$ and at that point in space and time $t_P$ there is an electric field with value $\vec{\mathbf{E}}(t_P,x_P,y_P,z_P)$ and a magnetic field with value $\vec{\mathbf{B}}(t_P,x_P,y_P,z_P)$ then you are submitted to the so-called Lorentz force $$\vec{\mathbf{F}} =q \left( \vec{\mathbf{E}}(t_P,x_P,y_P,z_P)+ \vec{\mathbf{v}}\wedge \vec{\mathbf{B}}(t_P,x_P,y_P,z_P)\right)$$

There is a distinct phenomenon: an electron or any charged particle also produces an electric field which is given by the Coulomb law (previously the electric field was a given data, we did not care about how it was produced) $$\vec{\mathbf{E}}_{produced} = \pm \frac{q}{\left\lVert\vec{\mathbf{r}}\right\rVert} \vec{\mathbf{r}}$$ (I'm not sure which sign) where $q$ is the charge, and $\vec{\mathbf{r}}$ is a vector giving the relative position at which we want to know the value of $\vec{\mathbf{E}}$ with respect to the position of the charged particle. $\left\lVert\vec{\mathbf{r}}\right\rVert$ is the norm of a vector, i.e. its length. A more complicated formula gives the magnetic field created by an moving charge.

So an eletric field is different from a magnetic field as to how it affects a charged particle, however there is a relation between the two. All the quantities mentionned depend on the referential, exactly like the notion of speed: if you observe an object, you will not measure the same speed depending on whether you are in a train or fixed at some place. The thing I'll skip now are the arguments, the reasoning that led to the following relation, (wikipedia) between the eletric and magnetic field $\vec{\mathbf{E}}_1,\vec{\mathbf{B}}_1$ in a first referential, and those $\vec{\mathbf{E}}_2,\vec{\mathbf{B}}_2$ in a second: (argument in a few words: "symmetry"of the Maxwell equations) $$\vec{\mathbf{E}}_2 = \gamma \left( \vec{\mathbf{E}}_1 + \vec{\mathbf{v}} \wedge \vec{\mathbf{B}}_1\right ) - \left ({\gamma-1} \right ) \left( \frac{\vec{\mathbf{E}}_1 \cdot \vec{\mathbf{v}}}{\lVert\vec{\mathbf{v}}\rVert} \right) \vec{\mathbf{v}}$$

$$\vec{\mathbf{B}}_2 = \gamma \left( \vec{\mathbf{B}}_1 - \frac {\vec{\mathbf{v}} \wedge \vec{\mathbf{E}}_1}{c^2} \right ) - \left ({\gamma-1} \right ) \left( \frac{\vec{\mathbf{B}}_1 \cdot \vec{\mathbf{v}}}{\lVert\vec{\mathbf{v}}\rVert} \right) \vec{\mathbf{v}}$$ (where $\vec{\mathbf{v}}$ now denotes the speed of the second referential relative to the first one (i.e not the same $\vec{\mathbf{v}}$ as before) and $\gamma$ is some function of $\vec{\mathbf{v}}$.) It is based on this formula that people say that eletric and magnetic fields are the same thing. E.g. even if $\vec{\mathbf{B}}_1=0$, but $\vec{\mathbf{E}}_1, \vec{\mathbf{v}} \neq 0$ then $\vec{\mathbf{B}}_2\neq 0$, in words even if in a first referential there were no magnetic field but there is an eletric field, then in a second referential that is moving you will observe a magnetic field.

• I am 15 years old :-P – Aaryan Dewan Jun 4 '15 at 9:39
• you can always read it again in a few years... – Noix07 Jun 4 '15 at 9:52
• This is not the most appropriate level to answer this question - as is clear by the wording. – anon01 Oct 12 '15 at 16:35

There is neither electricity alone nor magnetism alone produced by it. There exists an inseparable electromagnetic field produced by (moving) electric charges, in mutual interaction with them, and the splitting to its "electric" and "magnetic" parts depends upon the motion of the observer.The concept of electricity alone and magnetism that produces is due to educational reasons, since there is no way to understand this during your school years (except the case to be a genious). In your next years such questions will be the motivation to meet : (1) Maxwell with his famous equations of electrodynamics, a unification of "electricity" and "magnetism" and (2) Einstein with his theory of Special Relativity and his ideas, unification of the 3-dimensional space and 1-dimensional time to the 4-dimensional space-time.

But don't be disappointed by all these. These ideas must not prevent you from studying the trees :electrostatics (Coulomb's Law etc), magneto-statics (Biot and Savart Law, Faraday's Law etc) and others, in order to see later the forest of Classical Electrodynamics (Maxwell's Equations etc) inextricably coupled with the theory of Special Relativity.

Don't forget, Ladies and Sirs : we try to answer to a 15 years old boy !!!

• This is the best answer so far: no one else has made the IMO crucial point that the two phenomena belong to one, indivisible whole. – Selene Routley Jun 5 '15 at 6:54
• A moving electric charge does not emit light. – Jiminion Jun 5 '15 at 13:50
• There is a difference between a magnetic field and an e-m field. A horseshoe magnet is not emitting and e-m field. – Jiminion Jun 5 '15 at 14:14

It is an observed result that a moving electric charge can produce a magnetic force. As to the 'how', that's a bit troublesome. At this level, one has to accept some things as axiomatic.

As to the force at a distance, this is also true of ordinary electric potential. So magnetic force at a difference should be no more troublesome than electric force at a distance.

It not produced from, it is the exact same force that do difference thing. Magnetic force is like a torque coiling around direction of electricity. And so if electric charge run in circle then the magnetic that coiled around will be merged in one direction, produce a strong magnet

Still the force is actually came from attractive and repulsive force of electric charge

There are youtube video explain about how electric charge force being magnet related to relativity theory of Einstien

I'd just like to add to diracpaul's excellent answer, which to summarize, makes the point that both electricity and magnetism belong to an inseparable whole: you can't treat one as a more fundamental phenomenon producing the other.

Probably the most compelling reason for looking at things in this "indivisible" way is special relativity, wherein we accept the postulate that there is no way of telling whether one is in uniform motion relative to something else without looking outside our laboratory. This is essentially as described in Galileo's famous allegory of Salviati's Ship, and it was taken up again very earnestly by Einstein after having been abandoned in the 19th century because 19th century scientists before the Michelson Morley experiment believed in the luminiferous aether (and thus one could in principle always measure one's motion relative to the aether).

So, if Galileo's postulate truly does hold, then we cannot treat electricity and magnetism as one "causing" the other because different observers in different, relatively moving (at a constant relative velocity) will measure different relative amounts of "electricity" and "magnetism" in an electromagnetic field. It is the way Einstein made sense of certain generator and motor thought experiments that seemed to prefer a particular reference frame if one insists on thinking of electricity producing magnetism or the other way around.

Electricity and magnetism are united in a geometrical object called a 2-tensor which you likely won't have met yet and these objects transform very differently when one shifts one's reference frame from the three component vectors that are commonly used in electrostatic and magnetostatic formulas you have probably seen. The formulas you have most likely seen may give the impression of electricity producing magnetism, but they are not accurate when changing reference frames (without some really messy accounting).


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Linear Algebra : Row Operations and Equivalent Systems

Example Questions

← Previous 1

Example Question #1 : Row Operations And Equivalent Systems

Which matrix is equivalent to

Explanation:

By adding  times row one to row two we get

Example Question #2 : Row Operations And Equivalent Systems

Which matrix is equivalent to

Explanation:

By adding row one to row two we get

Example Question #3 : Row Operations And Equivalent Systems

Which matrix is equivalent to

Explanation:

We add row two to row one to get

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

format: plain


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Applied Exercises 8.115

$\Delta U=q+w$

Semi Yoon
Posts: 60
Joined: Fri Sep 28, 2018 12:27 am

Applied Exercises 8.115

An experimental automobile burns hydrogen for fuel. At the beginning of a test drive, the rigid 30.0-L tank was filled with hydrogen at 16.0 atm and 298 K. At the end of the drive, the temperature of the tank was still 298 K, but its pressure was 4.0 atm. (a) How many moles of H2 were burned during the drive? (b) How much heat, in kilojoules, was given off by the combustion of that amount of hydrogen?

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# Response of Linear System to Stochastic Process

Somehow I am getting the variance{u(n)} equal to '0' !! This is the case when I take the coefficient 'a' as real. As it is not mentioned in the question I need to find the solution to this question for a complex 'a' also. I would be glad I anybody could help me solve it. I waold also like to clear my concepts about the complex coefficients, as in how does the Z and inverse Z-transform takes place when 'a' is complex. This is a back problem of the second chapter in the book - Adaptive Filters: Theory and Application by Farhang-Boroujeny

Consider the following LTI system with impulse response $h[n]$

$$\{v[n]\} \longrightarrow \boxed{H(z)} \longrightarrow \{u[n]\}$$

From the analysis of LTI systems with WSS random inputs, the following results deduced. Given the input $v[n]$ which is zero mean white WSS random process, its ACS (Auto-Correlation Sequence) sequence is: $$\phi_{vv}[m] = \sigma_v^2 \delta[m]$$ where $\sigma_v^2$ variance of the input.

Then the output ACS is given by: $$\phi_{uu}[m] = h[m] \star h[-m]^{*} \star \phi_{vv}[m]$$

Similary the input and output PSD (Power Spectral Densities) are related by: $$S_{uu}(w) = H(w) H(w)^{*} S_{vv}(w) = |H(w)|^2 S_{vv}(w)$$ which follows from the relation that PSD of a WSS RP is the DTFT of its ACS; i.e., $$S_{uu}(w) =\sum_{m=-\infty}^{\infty} \phi_{uu}[m] e^{-j w m} \longleftrightarrow \phi_{uu}[m] = \frac{1}{2\pi} \int_{-\pi}^{\pi} S_{uu}(w) e^{j w m} dw$$

Then the variance of the output is given by the following: $$\sigma_{uu}^2 = \phi_{uu}[0] = \frac{1}{2\pi} \int_{-\pi}^{\pi} S_{uu}(w)dw$$

Now using the given information $H(z) = \frac{1}{1-a z^{-1}}$ and $\phi_{vv}[m] = 1 \delta[m]$ you can compute the variance in either way:

$$\sigma_u^2 = \left( h[m] \star h[-m]^{*} \star \delta[m] \right) |_{m=0}$$ or $$\sigma_u^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |\frac{1}{1-a e^{-jw}}|^2 dw$$

For example chosing the first approach with $h[n] = a^n u[n]$ yields: \begin{align} \sigma_u^2 &= \left( h[m] \star h[-m]^{*} \star \delta[m] \right) |_{m=0} \\ \sigma_u^2 &= \left( h[m] \star h[-m]^{*} \right) |_{m=0} \\ \sigma_u^2 &= \sum_{k=-\infty}^{\infty} h[m-k]h[-k]^{*} |_{m=0} \\ \sigma_u^2 &= \sum_{k=-\infty}^{\infty} h[-k]h[-k]^{*} \\ \sigma_u^2 &= \sum_{k=-\infty}^{\infty} a^{-k} u[-k] (a^{-k})^{*} u[-k]\\ \end{align}

$$\sigma_u^2 = \sum_{k=-\infty}^{0} (a a^{*})^{-k} = \sum_{k=-\infty}^{0} |a|^{-2k} = \sum_{k=0}^{\infty} (|a|^{2})^k = \frac{1}{1-|a|^2}$$

• Thank you very much @Fat32. I really appreciate your efforts. I am trying to figure out every little detail of this. Will let you know for sure if I have trouble understanding anything. – Copernicus Sep 15 '17 at 3:56
• @Rohan have you finished your analysis? – Fat32 Sep 17 '17 at 21:47
• Yes, and it is so apparent to me now. Thank you very much @Fat32. – Copernicus Sep 17 '17 at 22:26
• @Rohan I'm glad I could help. Please note that for the last infinite sum to converge we should have $|a| < 1$, as gven in the question. – Fat32 Sep 17 '17 at 22:44
• I was not considering the time reversal property, I took only the delay ($x[n − k] <=> z^−kX(z)$) into account. I dont know why it didnt work!! – Copernicus Sep 17 '17 at 22:53


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# How To Import Two Classes With The Same Name in JavaScript

Sometimes in working, we want to import two classes with the same name in JavaScript. This article will show you two ways to do so by using the “as” keyword. It can use both during import and export. Let’s get started.

## Import two classes with the same name in JavaScript

Below are two ways to import two classes with the same name in JavaScript.

### Use the “as” keyword in import

When there are two classes with the same name to import, we can use the “as” keyword to rename one or both classes. Consider the code for the two classes below. We have two different files, but the inside contains the same class name.

// Same class name
export class Student {
constructor(name, school) {
this.name = name;
this.school = school;
}
}
// Same class name
export class Student {
this.age = age;
}
}

As you can see, these two classes have the same name. We will import these two classes as following:

import { Student } from './student1.js';

// Use the keyword "as" to change Student to Student2
import { Student as Student2 } from './student2.js';

const student1 = new Student('John', 'Harvard');
console.log(student1.name);

// Use a new class name to create the object
const student2 = new Student2(17, 'Wall Street');
console.log(student2.age);

Output:

John
17

We use the keyword “as” to change to Student2.

Using the “as” keyword, we rename the class right from the time of import and then use the new name in the code.

### Use the “as” keyword in export

Another way is to rename it right from the time of export, so when we import, we only need to import the renamed class. In the example below, we will rename the class of student 2.

class Student {
this.age = age;
}
}

// Use the keyword "as" to change Student to Student2
export {
Student as Student2
};

Using the “as” keyword while exporting, we changed Student to Student2 and imported it.

import { Student } from './student1.js';

// Use new name to import
import { Student2 } from './student2.js';

const student1 = new Student('John', 'Harvard');
console.log(student1.name);

const student2 = new Student2(17, 'wall street');
console.log(student2.age);

Output:

John
17

So when importing student2, we need to import the changed class name.

## Summary

This article shows us two common ways to import two classes with the same name in JavaScript. You can use both methods at the same time or use one of them. We often use the keyword “as” at import because it is easier to observe. Hope you succeed.

Maybe you are interested:


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Help with approximation of the length of a curve in $\mathbb{R}^n$

I would like to have some hints on this exercise since I struggle to begin.

Let $$c : [a,b] \rightarrow \mathbb{R}^n$$ be a curve such that

$$l(c) := \sup_{a

I need to show that: for all $$\epsilon>0$$, there exist a $$\delta>0$$ such that

$$\Big| l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \| \Big| < \epsilon$$

for all partition $$P=\{ a=t_0 with $$\delta(P)<\delta$$, where $$\delta(P)= \max_{k=0,...,n-1} \{ t_{k+1}-t_k \}$$.

• There's lots of information packed in this problem, lots of little facts to exploit. Have you attempted to use of any of these facts? For example the function $c$ is uniformly continuous. Also, if $P'$ is a refinement of $P$ then $\Sigma_{P'} \ge \Sigma_P$ (I use $\Sigma_P$ as shorthand for the summation expression in your question); in other words, refining the partition moves the approximation closer to the supremum $l(c)$. – Lee Mosher Apr 30 at 15:46
Let's proceed by "reductio ad absurdum" i.e. by contradiction: suppose that there exists an $$\epsilon_o>0$$ such that, for every $$\delta>0$$ the following inequality holds $$\Big| l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \| \Big| \ge \epsilon_0>0\label{1}\tag{1}$$ for all partitions $$P=\{ a=t_0 of the interval $$[a,b]$$ (note that the can chose the unrestricted set of partitions since by the contradiction hypothesis, \eqref{1} holds for all $$\delta >0$$).
Now, since $$l(c) := \sup_{a we have that $$\begin{split} l(c)\ge \sum_{i=0}^{k-1} \| c(t_{i+1})-c(t_i) \| &\: \text{ for all paritions P of }[a,b]\\ &\Updownarrow\\ \end{split}$$ $$\Big| l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \| \Big|= l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \|.\label{2} \tag{2}$$ Therefore for all paritions $$P$$ of $$[a,b]$$ we have that $$\begin{split} l(c)\;- &\sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \|\ge \epsilon_o \\ l(c)>l(c)-\epsilon_o\ge&\sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \|>0 \end{split}\quad \text{ for all paritions P of }[a,b]\label{3}\tag{3}$$ Thus, there exists an upper limit for all the sums $$\sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \|$$ which is strictly less than $$l(c)$$ and this contradicts the fact that $$l(c)$$ is defined as the supremum i.e. the minimum of the set of all upper limits.
Note that the proof offered is non constructive as all proofs by contradiction. This means that, given an arbitrary $$\epsilon>0$$, we are not able to estimate the maximum partition dimension $$\delta$$ for which the inequality $$\Big| l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \| \Big| < \epsilon$$ holds by using the steps of the proof. Perhaps one could try an constructive proof by exploiting the fact that, as noted by Lee Mosher in his comment to the question, a partition finer than a given one produces a larger value of the associated sum: however, it seem to me that this conjectured constructive way would be quite hard to follow.


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# How to calculate air density?

For my data, I have temperature(F), atmospheric pressure and dew point.

I wanted to get a rough estimate of the air density, using all three of these.

In addition, how would I get an even rougher estimate using just temp and dew?

• Use the ideal gas law for air density, given atmospheric pressure and temperature. If you only have dew point temperature and air temperature, you can't get an estimate of air density because the vapor pressure of water vapor is independent of air pressure. – David White Dec 7 '16 at 23:14
• Ok, so I read up on ideal gas law, and I couldn't find any simply formulas in regards to adding dew. – Danny W Dec 8 '16 at 0:19
• An ideal gas has a particle density determined by temperature and pressure. Density, though, depends on gas particle WEIGHT, and H2O is a lighter molecule than O2 or N2. – Whit3rd Dec 8 '16 at 1:06
• @DannyW, you (or I) may be missing a fine point here. For a "rough" estimate, ignore the amount of water vapor in the air, if you are talking about ambient temperature. If the temperature is not ambient, please specify conditions that are somewhat more specific. – David White Dec 8 '16 at 3:27
• What about simply calculating the densities by the universal gas formula, and adding them? – peterh says reinstate Monica Dec 8 '16 at 4:08

Parameters which you have are temperature, atmospheric pressure and dew point. Parameters which are need for calculating the air density are temperature, atmospheric pressure, relative humidity and saturated vapor pressure.

In this case, the relative humidity is necessary to be calculated from the dew point.

The relative humidity can be obtained by the ratio of the saturated water vapor amount $s(t0)$, $s(t)$ at the dew point $t0$ and the temperature $t$. Namely, the relative humidity $Rh$ can be expressed as follows.

$$Rh = \frac{s(t0)}{s(t)} \times 100$$

$s(t)$ can be obtained from the equation of state of water vapor.

$$s(t) = \frac{217 Ps}{t + 273.15}$$

, where the saturated water vapor pressure $Ps$ [Pa] can be obtained from Tetens's formula.

$$Ps = 611 \times 10^{7.5 t / (t + 237.3)}$$

Here, the relative humidity could be obtained. As a next step, it is calculated the air density.

The air density can be obtained from Jones's formula. Jones's paper is F. E. Jones, “The air density equation and the transfer of the mass unit”, J. Res. Natl. Bur. Stand. 83, 1978, pp. 419-428.

The air density $\rho$ is

$$\rho = \frac{0.0034848}{t + 273.15} (P - 0.0037960 \cdot Rh \cdot Ps)$$

, where $t$ [Celsius] and $P$ [Pa] are temperature and atmospheric pressure, respectively. The unit of air density $\rho$ is [kg/m$^3$].

The unit of temperature which used here is Celsius. So if you want to use Fahrenheit as a unit of temperature, please convert it. If my explain is difficult to understand, I apologize. Because my English is poor.

If you want to quickly check the above calculation, you can confirm it using following AWK command. The input values for "echo" are atmospheric pressure, temperature and dew point, respectively.

$echo '1013.25 25 14' | awk '{ps = 611 * 10^(7.5 *$2 /($2 + 237.3))} {ps0 = 611 * 10^(7.5 *$3 /($3 + 237.3))} {st = 217 * ps / ($2 + 273.15)} {st0 = 217 * ps0 / ($3 + 273.15)} {rh = 100 * st0 / st} {ro = ($1 * 10^2 - 0.003796 * rh * ps) * 0.0034848 / (\$2 + 273.15)} END{print "\nAir density is " ro " [kg/m^3]";}'

When atmospheric pressure, temperature and dew point are 1013.25 hPa, 25 degrees C and 14 degrees C, the air density is 1.17693 [kg/m^3].


text:
1. Other
2. Other
3. propositional equation help im having trouble proving this is p...

# Question: propositional equation help im having trouble proving this is p...

###### Question details

Propositional equation help... I'm having trouble proving this!

Is (p ⇒ q) ∧ (q ⇒ r) equivalent to (p ⇒ r)? Explain


text:
# Convolution and the square root of the Fourier Transform

The Fourier transform is a linear transform. At least in the discrete, finite dimensional case, it can be represented as a unitary matrix, meaning that it has a nontrivial square root (although not necessarily a unique one). These square roots have the property that if you apply the transformation twice, you get the Fourier transform.

What can be said about these linear transformations? In particular I'd be interested in learning the following two things:

1. What do the columns of the matrix in question look like? (Basis functions)
2. Is there an equivalent of the convolution theorem for this transform? What do pointwise multiplication and convolution become in the transformed domain?
3. Does this have an extension to continuous function spaces, or to distributions, as does the Fourier transform?

## 2 Answers

To take it one step further than DisintegratingByParts's answer:

The Fourier transform $\mathcal F$ is unitary on $L^2({\mathbb R})$ with spectrum $\{1,i,-1,-i\}$. This means that $\mathcal F$ effectively behaves like a finite-dimensional diagonal operator. In particular, (basically) any square root of $\mathcal F$ can be written as a polynomial of degree $3$ in $\mathcal F$ itself. These polynomials can be deduced by Lagrange interpolation and equal

$$p(x) = \pm \frac{1}{4} (x+1)(x^2+1) \pm \frac{1+i}{4 i \sqrt{2}} (x^2-1)(x+i) \pm \frac{i}{4} (x-1)(x^2+1) \pm \frac{-1+i}{4 i \sqrt{2}} (x^2-1)(x-i).$$

So the answers are:

1. On $\mathbb R$, the basis functions are Hermite functions like DisintegratingByParts mentioned. In higher dimensions, $L^2(\mathbb R^d)$ can be decomposed into invariant subspaces whose angular parts are given by spherical harmonics. The book of Stein and Weiss has a very nice exposition.

2. Generally convolution, pointwise multiplication, differentiaion, and multiplication by monomials will not transform into anything nice simply because the different terms in $p({\mathcal F})$ will transform differently. (It's easy to check that in none of the cases do coefficients magically vanish to produce unexpected symmetry).

3. Because it's a polynomial in $\mathcal F$, it rather directly extends to distributions.

The Fourier transform $\mathcal{F} : L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ is a unitary operator that is diagonalized by the orthonormal basis of Hermite functions: $$h_n(x) = (-1)^n(2^n n!\sqrt{\pi})^{-1/2}e^{x^2/2}\frac{d^n}{dx^n} e^{-x^2}.$$ Specifically, $$(\mathcal{F}h_n)(k) = (-i)^n h_n(k).$$ From this you can define a square root: $$\sqrt{\mathcal{F}}f = \sum_{n=0}^{\infty}(-i)^{n/2}\langle f,h_n\rangle h_n,$$ where $(-i)^{n/2}$ is either square root of $(-i)^n$, and $\langle f,h_n\rangle = \int_{-\infty}^{\infty}f(t)h_n(t)dt$ is the $L^2$ inner product of $f$ and $h_n$. I don't know the properties of such square roots.


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# These roots have roots

Algebra Level 5

Find the number of ordered pairs of real numbers $(x,y)$ such that

$\begin{cases} x = \sqrt{y+1000} - \sqrt{y+100} \\ y = \sqrt{x+ 1000 } - \sqrt{ x +100} .\\ \end{cases}$

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# Equation of a Straight Line

Equation of a Straight Line
Go back to  'Coordinate Geometry'

Do you know that the equation of a line is really great to describe all sort of situations around you?

For example, Harry has just turned 18 and his parents gifted him a car on his birthday.

The car has 27000 miles in it and every year he will travel 12,000 miles to his college.

We can represent it in the form of an equation of line as $$y=27000x+12000$$.

In this mini-lesson, we will explore the world of the equation of a straight line. We will try to understand the general equation of a line, straight line formula, the way of finding the equation of a straight line and discover other interesting aspects of it. You can check out the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page.

We don't want you to worry, as we, at Cuemath, understand how this topic can be confusing for you.

We are here to help you understand this concept in a clear way.

## What Is the Equation of a Straight Line?

### Straight Line: Definition

A straight line is a figure formed when two points $$A (x_1 , y_1)$$ and $$B (x_2 , y_2)$$ are connected with minimum distance between them, and both the ends extended to infinity.

### Straight Line Formula

The standard form of a linear equation with variables $$x$$ and $$y$$ is:

 $ax + by = c$

where $$a, b, c$$ are constants and $$x, y$$ are variables.

## The Slope of a Straight Line

When you draw a line in XY-plane, have you observed that the line forms two angles with $$x$$ axis which are supplementary?

Suppose a line $$l$$ makes an angle of $$\theta$$ with a positive direction of $$x$$ axis.

The angle $$\theta$$ is called the inclination of the line and $$\tan{\theta}$$ is called the slope of the line.

Note that $$x$$-axis has the slope 0. In fact, all lines parallel to $$x$$-axis have the slope 0.

Also, the slope of all the vertical lines including $$y$$-axis is not defined.

### Slope Formula

Suppose we have a line joining two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ having inclination of $$\theta$$.

Then, the slope $$m$$ of the line is given by:

 $$m=\dfrac{y_2-y_1}{x_2-x_1}$$

## How To Find the Equation of a Straight Line?

### Using General Form

The general form of a straight line is given by $$ax+by=c$$.

Let's transform the equation $$y=2x-1$$ in the general form.

Add 1 to both sides of the equation.

So, we obtain the general form as $$2x-y=1$$.

Can you transform $$y=-\dfrac{x}{2}+7$$ in the general form?

### Using Standard Form

What if you are given just two points $$(x_1, y_1$$) and $$(x_2, y_2)$$ and you have to find the equation of a straight line joining these two points?

Let us take a point $$(x,y)$$ lying between $$(x_1, y_1$$) and $$(x_2, y_2)$$ on the line.

Notice that these three points are collinear.

So, we can write

\begin{align}\dfrac{y-y_1}{x-x_1}&=\dfrac{y_2-y_1}{x_2-x_1}\\y-y_1&=\left(\dfrac{y_2-y_1}{x_2-x_1}\right)(x_2-x_1)\end{align}

The equation of the standard form is:

 $$y-y_1=\left(\dfrac{y_2-y_1}{x_2-x_1}\right)(x_2-x_1)$$

### Using Point-slope Form

The equation of a line whose slope is $$m$$ and which passes through a point $$(x_1, y_1$$) is found using the point-slope form.

The equation of the point-slope form is:

 $${y-y_1=m(x-x_1)}$$

where $$(x, y)$$ is a random point on the line.

Let us see how to find the point-slope form.

We will derive this formula using the equation for the slope of a line.

Let us consider a line whose slope is $$m$$.

Let us assume that $$(x_1, y_1$$) is a known point on the line.

Let $$(x,y)$$ be any other random point on the line whose coordinates are not known.

We know that the equation for the slope of a line is:

\begin{aligned}\text{Slope }&= \dfrac{ \text{Difference in y-coordinates}}{ \text{Difference in x-coordinates}}\\[0.2cm]m&= \dfrac{y-y_1}{x-x_1} \end{aligned}

Multiplying both sides by $$(x-x_1)$$,

$m(x-x_1) = y-y_1$

This can be written as,

$y-y_1=m(x-x_1)$

Hence the point-slope formula is proved.

### Using Slope-intercept Form

Now, suppose a line is known to you with its slope and an intercept on one of the axes.

Say, a line intersects the $$y$$-axis at the point $$(0,c)$$.

Using the point-slope form we have $$y-c=m(x-0)$$ or $$y=mx+c$$.

 $y=mx+c$

where $$c$$ is the $$y$$ intercept.

Similarly, if $$d$$ is the $$x$$ intercept, then slope-intercept form is:

 $y=m(x-d)$

## Equations of Straight Line on Graph

The graph of a linear equation in one variable $$x$$ forms a vertical line parallel to $$y$$-axis and vice-versa.

The graph of a linear equation in two variables $$x$$ and $$y$$ forms a straight line.

Explore the graphs of equations of a straight line using the simulation below.

Important Notes
1. The equation of a straight line is also called a linear equation.

2. If the product of slopes of two straight lines is -1, then lines are perpendicular to each other.

3. If two straight lines are parallel to each other, then they have the same slope.

## Solved Examples

 Example 1

Carlos will begin attending classes for the sixth grade from next week.

He suddenly realized that he doesn't have either a notebook or a pen.

He goes to a shop to purchase them and realizes that the cost of a notebook is twice more than $5 for a pen. Represent this above information using the equation of a straight line. Solution Let's assume the following. The cost of a pen =$$$x$$

The cost of a notebook = \$$$y$$

According to the given information,

 $$\therefore$$ The equation of straight line is $$y=2x +5$$.
 Example 2

Look at the population and year graph.

Can you determine the population in the year 2010 using the slope of the line AB?

Solution

The line AB passes through $$A(1985, 92)$$ and $$B(1995, 97)$$.

Let $$x_1=1985, y_1=92, x_2=1995$$ and $$y_2=97$$.

Using the slope equation we can calculate the slope of the line AB.

\begin{align}\text{Slope of AB}&=\dfrac{y_2-y_1}{x_2-x_1}\\&=\dfrac{1995-1985}{97-92}\\&=\dfrac{1}{2}\end{align}

Let the population in the year 2010 be $$z$$.

Then, the line AB must pass through $$C(2010,z)$$.

\begin{align}\text{Slope of AB}&=\text{Slope of BC}\\\dfrac{1}{2}&=\dfrac{z-97}{2010-1995}\\\dfrac{15}{2}&=z-97\\z&=104.5 \end{align}

 $$\therefore$$ Population in year 2010 is 104.5 millions.
 Example 3

Jolly's teacher asked her to find the equation of a line passing through two points as shown on the blackboard.

Can you help her?

Solution

The given two points are:

\begin{aligned} (x_1, y_1)&=(1,3)\\[0.2cm] (x_2,y_2)&=(-2, 4) \end{aligned}

To use the point-slope form, we first need to find the slope of the line.

The slope of the line is found using the slope equation.

\begin{aligned} m &= \dfrac{y_2-y_1}{x_2-x_1}\\[0.2cm]&= \dfrac{4-3}{-2-1}\\[0.2cm]&= \dfrac{1}{-3}\\[0.2cm]&=- \dfrac{1}{3} \end{aligned} The equation of the line is found using the point-slope form. \begin{aligned} y-y_1&=m(x-x_1)\\[0.2cm]y-3&=- \dfrac{1}{3}(x-1)\\[0.2cm] y-3&=- \dfrac{1}{3} x+ \dfrac{1}{3}\\[0.2cm] \text{Adding 3 }& \text{ on both sides},\\[0.2cm] y&= - \dfrac{1}{3}x+ \dfrac{10}{3} \end{aligned}

To express this in the standard form, first, we multiply both sides by $$3$$

$3y=-x+10$

Adding $$x$$ on both sides,

$x+3y=10$

 $$\therefore$$ Equation of the given line in the standard form is $$x+3y=10$$.
 Example 4

Determine the equation of a line parallel to the $$x$$-axis and passing through the point (1, 3).

Solution

The line parallel to $$x$$ axis has 0 slope.

So, $$m=0$$.

Let $$x_1=1$$ and $$y_1=3$$.

By point-slope form, the equation of the line is $${y-y_1=m(x-x_1)}$$

\begin{align}y-3&=0\times(x-1)\\y-3&=0\\y&=3\end{align}

 $$\therefore$$ The equation of the line is $$y=3$$.

Think Tank
1. Which of the following graphs may represent the equation $${y+2= -2(x-1)}$$?

## Interactive Questions

Here are a few activities for you to practice.

## Let's Summarize

We hope you enjoyed learning about the Equation of a Straight Line with the examples and practice questions. Now, you will be able to easily solve problems on the concentric circles.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

## FAQs on Equation of a Straight Line

### 1. What is a straight line in math?

A straight line is a figure formed when two points $$A (x_1 , y_1)$$ and $$B (x_2 , y_2)$$ are connected with minimum distance between them, and both the ends extended to infinity.

### 2. Who invented the equation of a straight line?

René Descartes invented the equation of a straight line.

### 3. What is the formula for the equation of a straight line?

Generally, we represent a straight line using the formula: $$ax + by = c$$, where $$x$$ and $$y$$ are variables.

### 4. What is the y-intercept formula?

If a line intersects the $$y$$-axis at the point $$(0,c)$$, then $$c$$ is the $$y$$-intercept.

### 5. How do you find the slope and y-intercept of an equation?

The slope $$m$$ of a line joining two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is given by $$m=\dfrac{y_2-y_1}{x_2-x_1}$$.

The $$y$$ intercept of a straight line $$ax + by = c$$ is found by substituting the $$x$$ as 0.

### 6. How do you graph a slope-intercept equation?

The slope-intercept form is given by $$y=mx+c$$. It is plotted on graph by joining the two points satisfying the equation.

### 7. What is the y-intercept in y = -2x?

The $$y$$-intercept in $$y=-2x$$ is 0.

### 8. How do you write an equation for a vertical and horizontal line?

The equation of a horizontal line passing through $$(a,b)$$ is of the form $$y=b$$.

The equation of a vertical line passing through $$(a,b)$$ is of the form $$x=a$$.

### 9. How do you find the equation of a straight line when given two points?

The equation of a straight line joining two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is given by $$(y-y_1)=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$$.

### 10. How do you find the equation of a line from a pair of straight lines?

A pair of straight lines gives us two equations of straight lines.

More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus


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Calculate expectation of Cumulative distribution function of a normal distribution

I have to calculate the expectation of the Cumulative Distribution Function of a normally distributed random variable X, which has variance 1 and mean 0. I calculated the integral of the CDF (taken as an RV distributed on [0,1] and X and got 0 as the answer. I am not sure if I've done this right. Is there a better way of solving this question?

1 Answer

The expectation of the cumulative distribution function is independent of the distribution. The cumulative distribution function is uniformly distributed over $[0,1]$, so its expectation is $\frac12$.

• For a continuously distributed variable this works. Not otherwise: for example a constant r.v. will give $1$ instead of $1/2$.
– Ian
Apr 17, 2016 at 15:34


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# PlanetMath graphics sandbox

## Primary tabs

Type of Math Object:
Data Structure
Major Section:
Reference
Groups audience:

## Mathematics Subject Classification

### picture request

Since this entry seems to be a hub of graphics activity, I'm
posting my request here.

I am working on an entry on basic methods of enumerative
combinatorics. Right now I'm working on the principle of
inclusion-exclusion. In its most basic form, this says that
$|S| + |T| = |S\cup T| + |S\cap T|.$
Intuitively this is true because if you mark each element of
$S$ with a daub of blue paint and then mark each element of
$T$ with a daub of red paint, then each element of $S\cup T$
will have at least one daub of paint on it, red or blue, and
each element of the intersection $S\cap T$ will have two daubs
of paint on it.

What I would like to include to illustrate this is a Venn diagram
showing the intersection of two circles, one marked $S$ and
one marked $T$. The components $S\setminus T$ and $T\setminus S$
each should have an x inside, while the intersection should have
two xs.

Is anyone willing or able to contribute such a picture?


text:
preprekomW

2020-12-03

A rectangular coil of wire, 22cm by 35 cm and carrying a current of 1.4A, is oriented with the plane of its loop perpendicular to a uniform 1.5T magnetic field.
a) calculate the net force and torque which the magnetic field exerts on the coil.
b)the coil is rotated through a 30 degree angle about the y axis, the left side coming out of the plane. Calculate the net force and torque which the magnetic field now exerts on the coil

Expert


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# Can there be a universe with different mathematics?

I've this friend who asked me to ask this to some respectable mathematician or physicist. So, I decided to post this here.

Can there be universe with different laws of mathematics?

I tried to tell him that mathematics is how we define it.

He asked me if 2+2 could be 3. I do not have a clue what that means, so I think he wants to know if the Laws of Thought could be violated.

I said that even if it is, we cannot think of it so why bother?

The subsequent discussion was very cryptic for me but in conclusion he asked me to post it here.

4 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
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Note: you must add a full line of space before and after lists for them to show up correctly
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# up as a code block.

print "hello world"
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# up as a code block.

print "hello world"
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Remember to wrap math in $ ... $ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

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This is always an interesting subject. Theoretical physicists and cosmologists imagine there might be "different universes", because "laws of physics might be different in each". The analogy with mathematics is that it has definitions and axioms, which can "be different", so we already have "different abstract universes of mathematics". But it doesn't stop us from discovering and understanding those different abstract universes of mathematics, which is currently easier for us to do now than it is for us to discover other "different universes". In fact, we're able to imagine such different universes and their different laws of physics through the power of mathematics.

A related question, not synonymous with yours posted here, is whether or not we could someday run across an alien civilization that uses mathematics that contradicts ours. But if any two peoples, whether they're presently on Earth, or one of them is in the past or in the future or alien, at least agree to definitions and axioms of a given branch of mathematics, they are going to arrive at the same conclusions. Our mathematics may differ because it's possible that we're starting with different definitions and axioms---but nothing precludes us from working out consequences of such different definitions and axioms right here and now! Ready to try your hand at imagining some alien mathematics?

- 4 years, 3 months ago

So it's your contention that, if we can agree to a common set of axioms, then we and the aliens will necessarily always reach the same conclusions. This seems like the mathematical version of the premise to "Chance and Necessity" by Jacques Monod. If we and the aliens by chance have been working with the same set of axioms prior to "first contact", then by necessity none of our conclusions will contradict theirs. Would we not also have to have been working with the same set of rules of inference, or are these not in play?

I am also wondering if all the elements of the (short) list of "Laws of Thought" are challenged by quantum logic. I can see that being the case for the Laws of Non-Contradiction and the Excluded Middle, but the Law of Identity would still seem to be on solid footing. You would know better than I would, though, so I'm just curious what your thoughts are.

- 4 years, 3 months ago

"If we can agree to a common set of axioms, then we and the aliens will necessarily always reach the same conclusions" depends on both of us agreeing to a kind of a mathematics that delivers repeatable results. It's just one kind of formalism, one that follows, say, the classic three laws of logic as you've listed. But it's hardly the only formalism possible, and quantum logic hints at many more. In a way, as art imitates life, and often conversely, physical reality follows mathematical reality, and conversely as well. We're familiar with Newtonian physics, which is repeatable and deterministic, and yet there's the reality of quantum physics, which are not, but nevertheless there exists a kind of a logic (or madness) behind quantum physics. It's quite possible we could meet with advanced aliens that prefers thinking at the quantum level (because in many ways it's actually more efficient in delivering results), so that we'd get the perception that they must be engaged in very intelligent abstract thought, and yet such abstract thought appear to be very alien to our own "kind of mathematics".

But, as I keep saying here, the power and the beauty of mathematics is that there's nothing to PRECLUDE us from imagining and developing such "alien mathematics"!

- 4 years, 3 months ago

I'd imagine that any alien species that "visits" us will have already mastered the interplay between mathematical and physical reality, as well as the ability to think at the quantum level, so I guess we should just get on with it and prepare for that possibility. :)

- 4 years, 3 months ago

Yeah, that's right, we should just get on with it, and start thinking alien mathematics instead of waiting for aliens to come here and baffle us with it. If aliens can think it over there, we can think it here.

- 4 years, 3 months ago

Well, in a different universe, it IS a theory that basic physics could be different. Who knows if the other universes even have sophisticated life forms like us that do math and stuff? Also, the idea of multiple universes is a theory. To test it, they would have to find a black hole (difficult) or make a wormhole (super difficult). Personally, I believe that wormholes just have dead stars and infinite density. Well, hmm, infinite density? How do objects get sucked in then? How much is infinity in that case?

- 4 years, 3 months ago

Depends. What does he mean by "3"? Or by "2"? For example 2 is defined as the successor of 1, which is why 1+1=2. If 3 is defined as the successor of 2, then we will have 2+1 = 3. But, if instead, 4 was defined as the successor of 2, and 3 is defined as the successor of 4, then yes we do have 2+2 = 3.

Alternatively, $2 + 2 \equiv 3 \pmod{1}$. So yes, there are universes (even on earth), where we allow for that arithmetic.

Staff - 4 years, 3 months ago

Could there be an universe where the laws of thought do not hold?

Staff - 4 years, 3 months ago

We're almost already doing that right now in the absolutely weird and often incomprehensible field of quantum mechanics. Quantum mechanics has already stirred vigorous re-examination of the foundations of logic, i.e., it's shaking up our notions about "laws of thought". If you want to see some otherworldly "laws of thought", check out quantum logic.

- 4 years, 3 months ago

Maybe there is another type of maths where you can divide by zero....

- 4 years, 3 months ago

Wheel theory!

Staff - 4 years, 3 months ago

Someone reinvented the wheel!? :) I hadn't heard of Wheel Theory before, so thanks for mentioning it. It's a bit weird, but apparently it serves as a better model for the way a computer actually does arithmetic, with "apparently" being the operative word here. Of course, as per our discussion, it doesn't need to be useful, it just needs to be. :D

- 4 years, 3 months ago

Well, some people believe that other universes have different physics from ours, so maybe. Though, scientist do not really believe the different physics theory because there is no proof to support it.

- 1 year, 1 month ago

The ultimate goal of mathematics is to eliminate any need for intelligent thought.

- 4 years, 3 months ago

I STRONGLY and VOCIFEROUSLY disagree with that view!

- 4 years, 3 months ago

I second Michael on this one; see my comments to Bobbym None below.

- 4 years, 3 months ago

Maybe, but over here that is not what I mean by mathematics.

Staff - 4 years, 3 months ago

Is the axiom of choice correct?

- 4 years, 3 months ago

It's an axiom, which means it's up to you if you want to assume it to be true. You can equally assume it to be false, and thus define a different branch of mathematics. The classic example is the Parallel postulate, of which there are at least 3 variations, 1) given a line and through a given point not on it, only 1 line can be parallel to the first 2) ditto, but no lines can be parallel, 3) ditto, but many lines can be parallel, the 3 variations which have led to 1) Euclidean geometry, 2) Riemann geometry, and 3) Lobachevskian geometry. All of them "not incorrect".

- 4 years, 3 months ago

Good examples! I just wanted to bring up the often controversial axiom of choice to make the point about there being no absolute mathematics, and that different paradigms of mathematics already exist simultaneously, each with their own unique set of axioms.

- 4 years, 3 months ago

Oh, well, that was a really good choice you made, the [infamous] Axiom of Choice. It's almost as bad as having to make an whole axiom in response to the simple question, "Can I pick?" An universe where you aren't even allowed to pick anything at all would be a strange one, all right.

- 4 years, 3 months ago

@Arron Kau , @Calvin Lin , @bobbym none : I'd be obliged to have your opinion here.

Staff - 4 years, 3 months ago

"The ultimate goal of mathematics is to eliminate any need for intelligent thought." Whitehead may have been prophetic in that view. The Discretists led by Zeilberger and us computational boys could not agree more. The formalists think that mathematics is just a game with rules invented by humans for humans.

- 4 years, 3 months ago

Mathematics has no "goal". It has axioms and rules of our choosing, (see Michael's comments above), can be used as a prosaic tool or an artist's paint brush, be a salvation for compulsive minds, and more, but it certainly has no overarching goal. Philosophy has goals but no rules, physics has both goals and (some) rules, and mathematics has rules but no goals. AI researchers and advocates may share Whitehead's view, but I doubt Gödel or von Neumann did. To this end, a quote from the latter: "The whole system [i.e., "pure" mathematics] seems to function without any direction, without any reference to usefulness, and without any desire to do things which are useful." The fact that so much of this "uselessness" eventually finds a use is besides the point; just because something ends up being useful doesn't mean that there was any initial objective to be so.

- 4 years, 3 months ago

Nicely put, Brian!

In trying to verify the real origin of the quote, "The ultimate goal of mathematics is to eliminate all need for intelligent thought", attributed to Alfred North Whitehead, I've accidentally found a nice PDF book (free!) that seems like an excellent reference book for problems typically found in Brilliant

Concrete Mathematics

I still can't find out if Alfred is actually the author of that quote. But the context of it has to do with "mechanization of mathematics", i.e. the project of putting it all into a computer in algorithmic form, so that the "drudgery of mathematical computation", including proofs, is eliminated, and we all just go ask Wolfram Alpha. I'm sure Stephen Wolfram would just like that.

- 4 years, 3 months ago

The real origin of the quote and the sensation is my status.

Only kidding, I found it in A=B

Staff - 4 years, 3 months ago

Interesting book, and free is always good. :) I've just read the preface so far, and I find the philosophy behind it quite refreshing. As for the Whitehead quote, I agree that, if indeed he did say it, then there must have been some additional context. He was a deep thinker, so taking this quote at face value would do a disservice to his legacy.

- 4 years, 3 months ago

Intelligence is not supposed to be lost while doing mathematics, I am sure. But the more mechanical the results, the better it is.

Over here, we are picturing mathematics as an activity rather than a structure. I do not mean that mathematics is supposed to take over intelligence but think of it, the more mechanical the results the cooler it is. Would you rather prove a theorem that is much generalised or would you want it to have five cases? Which do you think would be more elegant?

It is not possible to define intelligence but let's accept equating it with raw processing power for a second. Which algorithm would a mathematician want to develop more? An O(n^2) algorithm or an O(n log(n)) algorithm?

Think for a second what if the quartic formula was as easy as the quadratic formula. Would you call the simplifier a good mathematician?

Isn't a mathematician with a general solution with a problem far better than a mathematician who can solve a few cases and gives you an algorithm that is time consuming and have to make you think how to optimise it? E.g, compare Fermat's Little theorem and Euler generalisation.

And oh, take the popular example of Wolfram|Alpha. Many people on Brilliant (not me) look down upon its users believing that it is of low intelligence to use it. But wouldn't you consider yourself a great mathematician, if you could program your own Wolfram|Alpha?

Staff - 4 years, 3 months ago

You're opening up another subject, the matter of "mechanization of mathematics". My take on that is that you could be the a genius with a hand shovel, but the job is going to the guy that knows how to operate a backhoe. Few people today waste time doing long multiplications and divisions by hand, when we've got calculators. Computers and math software enormously multiplies our ability to do math--and thus should free us to take our abstract thinking to an even higher level. So, that quote is simply badly worded. The truth should be closer to, "The ultimate goal of mechanization of mathematics is to allow intelligent thought to expand."

Even though Alfred Whitehead spent much of his life in trying to develop the "definite axiomatization of mathematics" (and failed grandly!), I have my doubts that he actually said that quote. It doesn't quite sound like him.

- 4 years, 3 months ago

$Strange\;Behavior \xrightarrow[]{\text{Willing to eliminate strange behavior}}Mathematics \; + \;Intelligent\; Thought \\ \\ Mathematics \; + \; Intelligent\; Thought \xrightarrow{\text{Willing to eliminate intelligent thought}}Better \; Mathematics \; + \; More\; Intelligent\; Thought \\ \\ Better \; Mathematics \; + \; More\; Intelligent\; Thought \xrightarrow{\text{Willing to mechanise mathematics}} Mechanical \; Tools \; + \; More\; Creative\; Thought \;+ \;Even\; Better\; Mathematics$

Though mathematics actually encourages more intelligent and creative thought, it is driven by the inspiration to eliminate the need of it from wherever possible.

Staff - 4 years, 3 months ago

I think it feeds on itself. While physicists and mathematicians loathe unexplained strange behavior, as soon as they resolve strange behavior so that it's no longer strange, they go look for more. Kind of like how entomologists are fascinated by the weirdest bugs they can find---even if it takes them halfway around the world. I know I'm guilty of that, I look for the weird and the strange in physics and mathematics, and then complain that it makes no sense, while being secretly excited to have found such things.

- 4 years, 3 months ago

Improving existing tools is never a bad thing, whether it's in mathematics, carpentry or any creative/productive venture. It helps to lessen the drudgery and more fully access our potential. Making new tools that make several existing tools redundant can be considered an even greater achievement, (although it may put a few noses out of joint). Creating "tools", if you can call them that, which have no clear and present use might be seen as foolish by the general public, but that is often what happens in mathematics. These are the "What ifs", and they've played, and will continue to play, a crucial role in our epistemological and technological evolution, but they were prompted by curiosity rather than necessity. Whatever allows us to reach our creative potential is a good thing, and that's what mathematics has always done and will continue to do, so I don't know where Whitehead was coming from when he said, that, (if indeed he did). Perhaps the quote had some additional context that we are not privy to, or perhaps he was just having a no-good, very bad day. :(

- 4 years, 3 months ago

Staff - 4 years, 3 months ago

NO!

Or, that's my initial response.

Then, YES!

Either way, I will put this on my "to do when you have room to breathe" - list. Thanks!

- 4 years, 3 months ago


text:
Warning: Use of undefined constant REQUEST_URI - assumed 'REQUEST_URI' (this will throw an Error in a future version of PHP) in /home/rajivsharma9/public_html/satchamp.in/wp-content/themes/Divi/functions.php on line 74
Question 18 - SAT CHAMP
Select Page

What is the domain of the function defined by

f(x) = \sqrt [ 3 ]{ \begin{matrix} { 16x }^{ 2 }- & 1 \end{matrix} }\ \ \ \ ?

(A) -1/4 ≤ x ≤ 1/4

(B) -1/4 ≤ x or x  ≥ 1/4

(C) x ≥ 0

(D) All Real Numbers

Solution:

Choice D is correct.

The radicand in a cube root can either be positive of negative.

Therefore, there is no restriction on the value of x.

So, the domain of the function defined by

f(x) = \sqrt [ 3 ]{ \begin{matrix} { 16x }^{ 2 }- & 1 \end{matrix} }\ \ \ \

is the set of all real numbers.

It is different from radicand in a square root.

If the problem would have asked for the domain of

f(x) = \sqrt [ ]{ \begin{matrix} { 16x }^{ 2 }- & 1 \end{matrix} }\ \ \ \ ,

Then

\ { \begin{matrix} { 16x }^{ 2 }- & 1 \end{matrix} }\ \ \ \

should be greater than or equal to zero.

For example,

\sqrt [ 3 ]{ -27 } = -3

However,

\sqrt{ -16 }

is not defined for any real number.


text:
I'm would like to achieve something like Object A will jump to position A in parabolic way. For example:

I am able to make the object jump but i do not know how to calculate to make the object jump to the precise location.

• do you need real physics or fake physics ? – Raxvan Jul 30 '14 at 14:01
• What is the different between real physics and fake physics? which one is better? – tekminewe Jul 30 '14 at 15:17
• "real" physics use Newton's laws of motion and take into account gravity, mass, and other forces. "Fake" physics is just a function that you write out of your head that looks good, something like f(x) = sin( pow(x,0.5f) * Pi);. Consider having point A and B and you want to jump ObjectA from A to B => ObjectA.position = lerp(A,B,Time) + Vector.Up * f(Time) and Time is from 0 to 1. No approach is better than the other, "fake" physics can look better or worse , it all depends on your imagination – Raxvan Jul 30 '14 at 15:35
• Thank you for your explaination. Then how can I achieve that in real physics? – tekminewe Jul 31 '14 at 5:29

SpriteKit uses SKAction objects to accomplish most of its functionality. What you're seeking is SKAction's followPath:duration: action.

The following accomplishes this:

    UIBezierPath *path = [UIBezierPath bezierPath];
[path moveToPoint:CGPointZero];


text:
The MathNet Korea
Information Center for Mathematical Science

### 논문검색

Information Center for Mathematical Science

#### 논문검색

Journal of Differential Equations
( Vol. 78 NO.2 / (1989))
The Singular Limit Dynamics of Semilinear Damped Wave Equations

Pages. 262-307
Abstract Introduction 1. The Equations and Some Preliminaries 2. Setting of the Problem and Nain Results 3. Modifying the Equation Far from the Attractor 4. Plan of the Proof 5. Spaces of Curves with Exponentially Weighted Norms 6. Resolution of the Nonhomogeneous Linear Equations in Exponentially Weighted Spaces 7. The Mappings $T_\varepsilon$: A Contraction Theorem with Parameters 8. Applying the Contraction Theorem 9. Further Regularity, the Time Derivatives, and the Manifolds $M_\varepsilon$ 10. Proofs of Lemmata 6.1 and 6.2


text:
# Hide WSTP application from taskbar

I just updated a small MathLink program that I made a while ago, and recompiled it using the new WSTP system (Mathematica 11.2).

The new version runs fine, but now this external WSTP program is shown in the taskbar (on Windows 10) once it's called via Install[], which was not the case with the old version that I created a while ago with the old MathLink system. On one of my computers, a (frozen) console window is shown when I click on the taskbar item, but on another system nothing happens when I click on that taskbar item...

Interestingly, this even happens for the pre-compiled WSTP programs supplied with Mathematica (for example the addtwo program described on the Install[] help page).

I build my program using VS 2017 using the instructions in the Mathematica documentation. I already tried this instruction to force the linker to use the "Windows" type instead of "Console", but it didn't change anything...

Any ideas on how one can get rid of these WSTP programs showing up in the taskbar?

• If you follow the structure of the examples that come with Mathematica, the button will appear in the taskbar. This has always been the case. It has nothing to do with MathLink/WSTP updates. Perhaps try comparing your old program with the new one and see where the difference is. I would suspect the MLInitializeIcon part, but I am not very familiar with Windows anymore. If you want to compile in console mode, you'd use the main() function instead of WinMain, but then you'd get a console window, even worse than a taskbar button. – Szabolcs Feb 21 '18 at 22:46
• Ok - didn't know that this was always the case for the standard programs! My old program used the main() function, but there was no console window... If I use the main() function in my updated version, I get a console window... so not sure where the problem is... I compiled the old version with MinGW though, so maybe using VS is the "problem"! – mjayvizzle Feb 21 '18 at 23:01

if (!WSInitializeIcon(hinstCurrent, nCmdShow)) return 1;

from the WinMain function in the standard WSTP C/C++ source file works, and no taskbar item is shown.
(Not sure why my old version, which used the main() function and was compiled with MinGW, didn't generate a console window or taskbar item though).


text:
01 Sep 2019

# Introduction

In order to process an analog signal such as sound or light with a digital computer it is necessary to convert it to a digital signal by taking measurements at discrete intervals. For example, consider the process of sending sound to another person via cellphone. The analog sound signal is first converted to an analog electric signal by the microphone. The electric signal is then sampled in order to produce a digital signal which can be processed by the phone. On the receiving end, the digital signal is converted back into an analog one by the speaker. Sending the signal to and from cell towers require additional conversions between the digital signal and an analog radio wave.

Digital cameras are also a form of sampling since we are measuring light intensities at discrete pixels.

Since sampling only records a small part of the analog signal, it makes sense to worry that some information is lost in the process. Even a small degradation would be catastrophic for communications since after many sampling steps these errors would compound on each other and produce noticeable distortions.

This is where the sampling theorem comes to the rescue. According to the theorem, as long as the analog signal has a bounded number of frequencies (this bound is called the bandwidth) it is possible to exactly reconstruct the signal using only a finite number of samples. In addition, given the bandwidth, the theorem specifies the number of samples per second (called the sampling rate) needed to achieve this reconstruction. This optimal sampling rate is sometimes called the Nyquist rate.

In the remainder of this post we will show why the sampling theorem isn’t all that surprising using some basic linear algebra. We will also derive an efficient algorithm for reconstructing a continuous signal from a sequence of samples and implement it with a few lines of python. Finally, we’ll see what happens when if we use a sampling rate that is lower than the Nyquist rate. One example of such under-sampling occurs when you take a picture of a pattern of lines whose frequency is too high for the cameras sampling rate. In such cases the image contains new patterns at unexpected lower frequencies. You can also see a version of this effect by quickly scrolling the image at the top of this page up and down unless your screen has a way higher frame rate than mine. This is a special case of a phenomenon called aliasing which we will analyze in detail.

# Sampling at a Glance

In this section we’ll go into a bit more technical detail in order to state the sampling theorem more rigorously and show why it makes sense.

$f(t) = a_0 + a_1 \cos(t) + b_1 \sin(t) + a_2 \cos(2t) + b_2 \sin(2t)$

You may recognize this as a Fourier expansion of $f$ and we’ll go into more detail about this later. For now it suffices to note that $f$ is determined by the 5 dimensional vector $\mathbf{F} = [a_0, a_1, a_2, b_1, b_2]^T$.

Suppose we evaluate $f$ at $t_0$ and get the measurement $f(t_0)$. This imposes a linear constraint on $\mathbf{F}$:

$a_0 + \cos(t_0)a_1 + \sin(t_0)b_1 + \cos(2t_0)a_2 + \sin(2t_0)b_2 = f(t_0)$

Since $\mathbf{F}$ has 5 variables, it stands to reason that we can calculate it by measuring $f$ at 5 points. This is the key idea behind the sampling theorem. Constraining $f$ to the above form is another way of saying that the bandwidth of $f$ is $N = 2$ and the sampling theorem says that $f$ can be reconstructed from $2N + 1 = 5$ samples.

This method of reconstructing a signal from a sequence of samples is helpful for understanding why $2N+1$ samples are sufficient, but solving a linear system with $2N+1$ variables is not feasible in practice. For example, the human ear can detect frequencies of up to about 20kHz which means that when processing sound we must take around 40 thousand samples per second. Since solving a general linear system of equations has a complexity of $\mathrm{O}(n^3)$, the method outlined above is clearly not practical.

In the following sections we will develop a bit of Fourier theory and show that the above system of equations has a very special form which can efficiently be solved by a Fast Fourier Transform with a complexity of just $\mathrm{O}(n\log(n))$.

# The Fourier Series

Let $f(t): \mathbb{R} \rightarrow \mathbb{R}$ be a continuous periodic function with period $T$. This means that for each $t \in \mathbb{R}$, $f(t + T) = f(t)$. An example of such a function is the humble $\cos(\frac{2\pi}{T}t)$. More generally, for each $n$ the function $\cos(n\frac{2\pi}{T}t)$ has period $T$. Furthermore, we say that $\cos(n\frac{2\pi}{T}t)$ has a frequency of $n$ because it repeats itself $n$ times on each interval of length $T$.

It turns out that any continuous function with period $T$ can be approximated arbitrarily well by a function of the form:

$f(t) \sim a_0 + \sum_{n=0}^N\left(a_n\cos\left(n\frac{2\pi}{T}t\right) + b_n\sin\left(n\frac{2\pi}{T}t\right)\right)$

for some $N$. If this approximation is exact, we say that the function has bandwidth $N$. This is another way of saying that all of the frequencies appearing in the function are between $0$ and $N$.

For the purposes of this post, and for signal processing in general, it will be easier to use an alternative but equivalent expression for functions with bandwidth $N$ using the complex exponential function rather than trigonometric functions. Recall that by the Euler identity,

$e^{i\theta} = \cos(\theta) + i \sin(\theta)$

From this together with the identities $\cos(-t) = \cos(t)$ and $\sin(-t) = -\sin(t)$ it is easy to deduce:

\begin{align*} \cos(\theta) &= \frac{1}{2}(e^{i\theta} + e^{-i\theta}) \\ \sin(\theta) &= -\frac{i}{2}(e^{i\theta} - e^{-i\theta}) \end{align*}

Therefore, we can represent a function with bandwidth $N$ by

$f(t) = \sum_{n=-N}^{N} F_n e^{n \frac{2\pi i}{T}t}$

for some complex numbers $F_{-N}, \dots, F_0, \dots, F_N \in \mathbb{C}$. We call the series $F_n$ the Fourier series of $f(t)$. Note that this series has $2N+1$ elements. Our goal in the next section will be to understand how to efficiently calculate the coefficients $F_n$ based on $2N+1$ samples of $f(t)$.

# The Discrete Fourier Transform

In this section we’ll derive the discrete Fourier transform which converts a sequence of $2N + 1$ samples of a function $f(t)$ with period $T$ and bandwidth $N$ into the Fourier series $F_{-N}, \dots, F_N$.

Since $f(t)$ has period $T$, it suffices to take samples in the interval $[0, T]$. We will assume that the samples are taken from points that are evenly spaced starting at $0$. Explicitly, the space between consecutive samples is $T_s := T / (2N + 1)$ and we sample at the points:

$t_k = kT_s$

For $k = 0, \dots , 2N$. Here is an example of a function with period $T = 2\pi$ and bandwidth $N = 4$ together with the $2N+1 = 9$ samples:

Let’s try to express the value of the sample taken at the $k$-th point in terms of the Fourier series of $f(t)$. Using the Fourier expansion of $f(t)$ we have

\begin{align*} f(t_k) &= \sum_{n=-N}^{N} F_n e^{n \frac{2\pi i}{T}t_k} = \sum_{n=-N}^{N} F_n e^{n \frac{2\pi i}{T} k\frac{T}{2N+1}} \\ &= \sum_{n=-N}^{N} F_n e^{\frac{2\pi i}{2N + 1}nk} \end{align*}

We can simplify this expression by introducing the $2N + 1$-th root of unity $W_{2N+1}$ which is defined to be

$W_{2N+1} := e^{-\frac{2\pi i}{2N + 1}}$

Our expression for $f(t_k)$ can now be expressed as

$f(t_k) = \sum_{n=-N}^{N} F_n W_{2N+1}^{-kn}$

Note that this looks very similar to the expression for multiplication by a matrix whose value at $(k, n)$ is $W_{2N+1}^{-kn}$. However, the indexing from $-N$ to $N$ in the sum is a bit awkward so we will reindex using the change of variable $m = n + N$. The expression now becomes

$f(t_k) = \sum_{m=0}^{2N} F_{m-N} W_{2N+1}^{-k(m - N)}$

and multiplying both sides by $W_{2N+1}^{-kN}$ gives

$$\label{linear-eq} W_{2N+1}^{-kN}f(t_k) = \sum_{m=0}^{2N} F_{m-N} W_{2N+1}^{-km}$$

At last, the sum on the right is indeed the expression for the $k$-th row of the multiplication of a matrix and a vector which we can write out explicitly as follows

$$\label{fft-matrix} \begin{bmatrix} f(t_0) \\ W_{2N+1}^{-N}f(t_1) \\ W_{2N+1}^{-2N}f(t_2) \\ \vdots \\ W_{2N+1}^{-2N^2}f(t_{2N}) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & W_{2N+1}^{-1} & W_{2N+1}^{-2} & \cdots & W_{2N+1}^{-2N} \\ 1 & W_{2N+1}^{-2} & W_{2N+1}^{-4} & \cdots & W_{2N+1}^{-2 \cdot 2N} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & W_{2N+1}^{-2N} & W_{2N+1}^{-2N \cdot 2} & \cdots & W_{2N+1}^{-2N \cdot 2N} \end{bmatrix} \begin{bmatrix} F_{-N} \\ F_{-N + 1} \\\\ F_{-N + 2} \\ \vdots \\ F_N \end{bmatrix}$$

Let’s denote this matrix by $M_{2N + 1}$. It is an interesting exercise in the properties of the root of unity $W_{2N+1}$ to show that the columns of $M_{2N + 1}$ are orthogonal and that they all have a norm of $2N + 1$. From this it immediately follows that

$M_{2N + 1}^{-1} = \frac{1}{2N + 1} M_{2N+1}^*$

where $M_{2N+1}^*$ denotes the conjugate transpose. By multiplying both sides of \ref{fft-matrix} with $M_{2N+1}^*$ and using the fact that the complex conjugate of $W_{2N+1}^{-1}$ is $W_{2N+1}$ we arrive at the following key equality:

\begin{equation*} \begin{bmatrix} F_{-N} \\ F_{-N + 1} \\ F_{-N + 2} \\ \vdots \\ F_N \end{bmatrix} = \frac{1}{2N + 1} \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & W_{2N+1}^{1} & W_{2N+1}^{2} & \cdots & W_{2N+1}^{2N} \\ 1 & W_{2N+1}^{2} & W_{2N+1}^{4} & \cdots & W_{2N+1}^{2 \cdot 2N} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & W_{2N+1}^{2N} & W_{2N+1}^{2N \cdot 2} & \cdots & W_{2N+1}^{2N \cdot 2N} \end{bmatrix} \begin{bmatrix} f(t_0) \\ W_{2N+1}^{-N}f(t_1) \\ W_{2N+1}^{-2N}f(t_2) \\ \vdots \\ W_{2N+1}^{-2N^2}f(t_{2N}) \end{bmatrix} \end{equation*}

The matrix appearing in the above equality is known as the discrete Fourier transform which is denoted by DFT. So we can rewrite the equality more compactly as:

$$\label{dft} \begin{bmatrix} F_{-N} \\ F_{-N + 1} \\ F_{-N + 2} \\ \vdots \\ F_N \end{bmatrix} = \frac{1}{2N+1} \mathrm{DFT} \left( \begin{bmatrix} f(t_0) \\ W_{2N+1}^{-N}f(t_1) \\ W_{2N+1}^{-2N}f(t_2) \\ \vdots \\ W_{2N+1}^{-2N^2}f(t_{2N}) \end{bmatrix} \right)$$

The upshot of all this is that we now have an explicit expression for computing the Fourier series of the function $f(t)$ based on a list of $2N+1$ samples.

Now that we have the algorithm in hand, let’s turn to the runtime. Since the DFT is the multiplication of a matrix by a vector, the naive implementation would have complexity $\mathrm{O}(N^2)$ which is better than our original back of the envelope estimation of $\mathrm{O}(N^3)$ from the introduction, but still not fast enough for serious signal processing.

Luckily, there are some interesting symmetries in the DFT matrix which allow us to evaluate the transform with a divide and conquer approach that is appropriately called the fast Fourier transform (FFT) and which has a complexity of $\mathrm{O}(N\log(N))$. It is important to note that applying the FFT to a vector gives the same result as multiplying the vector with the DFT matrix, it is just a faster implementation of that multiplication.

The FFT is the workhorse of modern signal processing and has fast implementations in many libraries and even in hardware. We’ll work through an example using Python in the next section.

# A Working Example

In this section we’ll put the theory of the previous sections into practice using Python. For this example we’ll consider the following function which has period $T = 2\pi$ and bandwidth $N = 2$:

$f(t) = \cos(t) + \sin(2t)$

We would like to reconstruct $f$ based only on samples taken at $2N+1 = 5$ evenly spaced points $t_0, \dots, t_4$ in the interval $[0, 2\pi]$. Here is a plot of the function on the interval $[0, 2\pi]$ together with the 5 sampled points

In the following code snippet we create an array containing the samples $f(t_i)$ and then use equation \ref{dft} from the previous section to compute the Fourier series coefficients $F_n$:

import numpy as np

# Setup
# -----
N = 2  # The bandwidth of f(t)
T = 2*np.pi  # The period of f(t)

# The space between consecutive samples
T_s = T / (2*N + 1)

# The 2N+1'th root of unity
W = np.exp(-(2*np.pi*1j)/(2*N + 1))

# Generate the samples
# --------------------
# Define the function
def f(t):
return np.cos(t) + np.sin(2*t)

# Generate the values t_0, ..., t_{2N}
t = np.array([n*T_s for n in range(2*N + 1)])

# Evaluate f(t_i) for i = 0, ..., 2N
ft = f(t)

# Reconstruct the coefficients of the Fourier
# series of f(t):  F_{-2}, F_{-1}, F_0, F_1, F_2
# using equation (3) above.
# ----------------------------------------------
scaled_ft = np.array([W**(-i*N) * ft[i]
for i in range(2*N + 1)])

# Apply the fast Fourier transform
F = np.fft.fft(scaled_ft) / (2*N + 1)
print("The Fourier series of f(t) is: ", F)

The output of this program is:

$F = [\frac{i}{2}, \frac{1}{2}, 0, \frac{1}{2}, -\frac{i}{2}]$

Let’s verify this by plugging these coefficients into the definition of the Fourier series and check that we get $f(t)$:

\begin{align*} \sum_{n=-2}^2 F_n e^{n \cdot it} &= \frac{i}{2}e^{-2 \cdot it} + \frac{1}{2}e^{-it} + 0 + \frac{1}{2}e^{it} -\frac{i}{2}e^{2 \cdot it} \\ &= \frac{1}{2}(e^{it} + e^{-it}) - \frac{i}{2}(e^{2\cdot it} - e^{-2\cdot it}) \\ &= \cos(t) + sin(2t) = f(t) \end{align*}

# Aliasing

In this final section we’ll take a look at what happens if we try to use $2N + 1$ samples to reconstruct a function whose bandwidth is greater than $N$. This is known as under-sampling since we are using fewer points than the theory allows. As we will see, a common symptom of under-sampling is that the reconstructed function has spurious low frequency components. This effect is known as aliasing.

A classic instance of aliasing is known as the wagon wheel effect. The name comes from phenomenon in which fast turning wheels in old (and even recent) movies seem to be slowly spinning backwards. To see the relationship with sampling, suppose for example that the wheel is rotating in the clockwise direction at a rate of $N$ revolutions per second. Consider the function $f(t) = \sin(-N \cdot 2\pi t)$ which describes the height of a fixed point on the wheel as a function of time. This function has a period of $1$ and a bandwidth of $N$. We can think of the process of taking a video of the wheel as measuring evenly spaced samples of $f(t)$ - one sample per frame. According to the sampling theorem, we will only be able to reconstruct the true motion of the wheel if the camera is operating at a frame rate of at least $2N+1$ frames per second. If the frame-rate is too low, the aliasing effect will introduce incorrect low frequency components which in this case manifest as a wheel which is slowly turning in the opposite direction.

We’ll now get into the technical details and understand exactly how aliasing happens. To make the equations nicer, suppose for simplicity that we are sampling a function with the following Fourier series:

$f(t) = \sum_{n=-N}^{N} F_n e^{n \frac{2\pi i}{T}t} + \sum_{n=N+1}^{3N+1} F_n e^{n \frac{2\pi i}{T}t}$

In other words, in addition to the $2N+1$ non zero coefficients of a function with bandwidth $N$, there are an additional $2N+1$ non zero coefficients to the right of $N$.

Now, suppose we naively sample this function at only $2N+1$ points as if it had bandwidth $N$ and try to measure the Fourier coefficients $F_{-N}, \dots, F_N$. Similar to our derivation of the DFT above, the value of the sample $f(t_k)$ is equal to

\begin{align*} W_{2N+1}^{-kN}f(t_k) &= \sum_{m=0}^{2N} F_{m-N} W_{2N+1}^{-km} + \sum_{m=2N+1}^{4N+1} F_{m-N} W_{2N+1}^{-km} = \\ &= \sum_{m=0}^{2N} F_{m-N} W_{2N+1}^{-km} + \sum_{m=0}^{2N} F_{m+N+1} W_{2N+1}^{-k(m + 2N + 1)} \\ &= \sum_{m=0}^{2N} (F_{m-N} + F_{m+N+1}) W_{2N+1}^{-km} \end{align*}

Note the similarity of this to equation \ref{linear-eq} in our original derivation of the DFT . Similarly to there we can conclude that

\begin{equation*} \begin{bmatrix} F_{-N} + F_{N+1} \\ F_{-N + 1} + F_{N+2} \\ F_{-N + 2} + F_{N+3} \\ \vdots \\ F_N + F_{3N + 1} \end{bmatrix} = \frac{1}{N} \mathrm{DFT} \left( \begin{bmatrix} f(t_0) \\ W_{2N+1}^{-N}f(t_1) \\ W_{2N+1}^{-2N}f(t_2) \\ \vdots \\ W_{2N+1}^{-2N^2}f(t_{2N}) \end{bmatrix} \right) \end{equation*}

This means that if we run the DFT algorithm under the assumption that the bandwidth of $f(t)$ is $N$, our estimate of the coefficient $F_n$ for $-N \leq n \leq N$ will be $$F'_n = F_n + F_{n + (2N + 1)}$$. In other words, our estimates of the lower frequency components $F_{-N}, \dots, F_N$ are clouded (“aliased”) by the higher frequency components $F_{N+1}, \dots, F_{3N+1}$.

The general pattern in the case of a function $f(t)$ with arbitrarily high bandwidth is that is that our estimate of $F_n$ for $-N \leq n \leq N$ will be:

$$\label{aliasing-sum} F'_n = \sum_{m=-\infty}^{\infty} F_{n + m(2N + 1)}$$

Note that even though the sum is written as infinite, in practice all but finitely many coefficients are zero since we are assuming that $f(t)$ has finite bandwidth.

In particular, even if the original signal $f(t)$ didn’t have any low frequencies at all, if we try to reconstruct if from too few samples , the reconstructed function may have low frequency components which are contributed from the higher frequencies of the original $f(t)$.

Let’s now apply this theory to a simple “wagon wheel” setup. Suppose we have a wheel that is spinning clockwise at a rate of $4$ revolutions per second. If we choose some point on the wheel, it’s height relative to the center of the wheel as a function of time is:

$f(t) = \sin(-4\cdot 2\pi t)$

This function has a period of $T = 1$ and a bandwidth of $N = 4$. Its Fourier series is:

$f(t) = -\frac{i}{2}e^{-4\cdot 2\pi i \cdot t} + \frac{i}{2}e^{4\cdot 2\pi i \cdot t}$

So all of it’s Fourier coefficients are zero except for $F_{-4} = -\frac{i}{2}$ and $F_{4} = \frac{i}{2}$.

By the sampling theorem, if we wanted to perfectly reconstruct this function we’d need to sample it at a rate of $2N + 1 = 9$ samples per second.

Suppose we sample this function at only $2\cdot 2 + 1 = 5$ samples per second corresponding to an incorrect bandwidth of $N’ = 2$:

According to equation \ref{aliasing-sum} and the Fourier coefficients of $f(t)$ we calculated above, if we apply the DFT algorithm to these samples we’ll recover the following (incorrect) Fourier coefficients which to avoid confusion will be denoted by $G_n$:

\begin{align*} G_{-2} &= F_{-2} = 0 \\ G_{-1} &= F_{-1} + F_{-1 + 5} = 0 + \frac{i}{2} = \frac{i}{2} \\ G_{0} &= F_{0} = 0 \\ G_{1} &= F_{1} + F_{1 - 5} = 0 - \frac{i}{2} = -\frac{i}{2} \\ G_{2} &= F_2 = 0 \end{align*}

Plugging these coefficients into the Fourier series expression we see that the function we have measured is:

$g(t) = G_{-1}e^{-2\pi i \cdot t} + G_1 e^{2\pi i \cdot t} = \frac{i}{2} e^{-2\pi i \cdot t} + \frac{i}{2} e^{2\pi i \cdot t} = \sin(2\pi \cdot t)$

This corresponds to a wheel rotating counter-clockwise (“backwards”) with a rate of just one revolution per second. Here is a plot of the original function $f(t)$ together with the samples and the reconstructed function $g(t)$:

As we can see, the DFT algorithm has indeed found a function $g(t)$ with bandwidth $N’=2$ that passes through all of the $2N’+1 = 5$ samples but unfortunately this is not equal to $f(t)$ which has a higher bandwidth of $N = 4$.


text: Simple conditions for matrices to be bounded operators on $l_p$ The two theorems proved yield simple yet reasonably general conditions for triangular matrices to be bounded operators on $l_p$. The theorems are applied to N\"orlund and weighted mean matrices. Keywords:Triangular matrices, NÃ¶rlund matrices, weighted means, operators, on $l_p$.Categories:47B37, 47A30, 40G05


text:
Operations on Formal Power Series

Правка en1, от rng_58, 2017-12-17 20:07:08

You are given the first n (or n + 1 if necessary) terms of a former power series P(x) = c0 + c1x + c2x2 + .... What operations can be performed efficiently?

• Obviously, P(x) + Q(x), P(x) - Q(x), P'(x), , kP(x) for a given constant k, can be done in O(n).

• P(x)Q(x) can be done in O(nlogn) by FFT.

• can be done in O(nlogn): Link, check problem E

• can be done in O(nlogn): Link, check problem E

• exp(P(x)) can be done in O(nlogn): Link, check Figure 1, left

• Open: Can we do more complicated operations like P(Q(x)), P(x)1 / k, sin(P(x)), arcsin(P(x)), etc.? Are there other important operations?

• Probably a bit related to the computation of : when we are given two big decimal number x and y, can we compute x / y?

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en1 rng_58 2017-12-17 20:07:08 1157 Initial revision (published)


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Search: All articles in the CJM digital archive with keyword unitary group

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1. CJM 2008 (vol 60 pp. 1067)

Kariyama, Kazutoshi
 On Types for Unramified \$p\$-Adic Unitary Groups Let \$F\$ be a non-archimedean local field of residue characteristic neither 2 nor 3 equipped with a galois involution with fixed field \$F_0\$, and let \$G\$ be a symplectic group over \$F\$ or an unramified unitary group over \$F_0\$. Following the methods of Bushnell--Kutzko for \$\GL(N,F)\$, we define an analogue of a simple type attached to a certain skew simple stratum, and realize a type in \$G\$. In particular, we obtain an irreducible supercuspidal representation of \$G\$ like \$\GL(N,F)\$. Keywords:\$p\$-adic unitary group, type, supercuspidal representation, Hecke algebraCategories:22E50, 22D99

2. CJM 2002 (vol 54 pp. 634)

Weber, Eric
 Frames and Single Wavelets for Unitary Groups We consider a unitary representation of a discrete countable abelian group on a separable Hilbert space which is associated to a cyclic generalized frame multiresolution analysis. We extend Robertson's theorem to apply to frames generated by the action of the group. Within this setup we use Stone's theorem and the theory of projection valued measures to analyze wandering frame collections. This yields a functional analytic method of constructing a wavelet from a generalized frame multi\-resolution analysis in terms of the frame scaling vectors. We then explicitly apply our results to the action of the integers given by translations on \$L^2({\mathbb R})\$. Keywords:wavelet, multiresolution analysis, unitary group representation, frameCategories:42C40, 43A25, 42C15, 46N99