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Nano Express

# One-photon scattering by an atomic chain in a two-mode resonator: cyclic conditions

Andrii S Sizhuk1* and Stanislav M Yezhov12

Author Affiliations

1 Department of Radiophysics, Kyiv Taras Shevchenko National University, Acad. Glushkova Avenue 4-g, Kyiv 03022, Ukraine

2 Department of Physics, Kyiv Taras Shevchenko National University, Acad. Glushkova Avenue 2, Kyiv 03022, Ukraine

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Nanoscale Research Letters 2014, 9:203  doi:10.1186/1556-276X-9-203

 Received: 5 December 2013 Accepted: 3 April 2014 Published: 1 May 2014

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

### Abstract

In this work, a chain of N identical two-level atoms coupled with a quantized electromagnetic field, initially prepared via a single-photon Fock state, is investigated. The N-particle state amplitude of the system is calculated for several space configurations of the atoms in the Weisskopf-Wigner approximation. It was shown that the space configuration of an atomic chain, the total number of atoms, and even the available volume for the field modes define the behavior of the system state amplitude with time. Applying the condition of ‘cyclic bonds’, presented in this work, to the elaborated theory allows to describe the system time evolution, practically, for any space configuration.

##### Keywords:
Atoms; Fock state; Weisskopf-Wigner approximation

### Background

The collective absorption (emission) of photons by an ensemble of identical atoms ‘provides valuable insights into the many-body physics of photons and atoms’ (quoted from [1]). Taking into account the quantization of electromagnetic field, many fundamental and interesting properties of the coupled systems of atoms and field are revealed. For example, when the average distances between atoms are much less than the ‘resonant transition’ wavelength of emitted (absorbed) light, the cooperative coupling leads to a substantial radiative shift of the transition energy and significant change in decay rate of the ensemble state. The latter was revealed through the various theoretical (for example, some relatively modern researches in [2-5]) and experimental investigations (see starting, for example, from [6,7] to the modern applications like described in [8] and impressively effective experimental realizations as in [1]). Some peculiar behavior in spontaneous emission is proper even in a system of atoms which can have a relative distance larger than the emission wavelength (see, for instance, [9]), and initially, only one atom or one-photon state is excited, as discovered in this paper.

In the present paper, a system (chain) of N identical two-level non-interacting atoms, prepared ‘via a single-photon Fock state’ in the one- or two-mode resonator, is investigated. The main goal of the paper is to obtain the information about the state of electromagnetic field and atomic system (chain) in a Weisskopf-Wigner approximation (see [10] chapter 6, page 206 and some comments in [11]). The calculations of the state amplitudes of the atomic system are made for several approximations in resonator (cavity) characteristics and for several types of space configurations of the atoms in the chain.

In this work, we study the case, in which the distances between atoms are quite large, so that the average distances between atoms are greater or in the same order than the ‘resonant transition’ wavelength. Therefore, we prepare an ensemble of N two-level atoms initially in ground state, and a single mode of the radiation field is excited in a ‘Fock’ state (so called one-photon state). This is the case of a purely monochromatic wave with zero line width under the consideration. A laser output in single mode operation can approximate this situation due to its high degree of monochromaticity (small line width) for instance. The mode of electromagnetic field is specified completely by giving its wave vectors k0 with atomic transition frequency ω = c|k0| and its polarization j (j = 1, 2).

The main feature, differentiating our research from others in this domain, is the developed direct and consistent solution to the N-particle equations, describing the time evolution of the N atomic probability state amplitudes. Besides, in certain sense, we explained the nature of the widely used Weisskopf-Wigner approximation that was not found in the reviewed by us scientific literature.

The goal of this paper can be formulated as an attempt to propose an adapted and simple in practical use theory, for example in the highly applied nanoscale physics. The proposed theoretical material requires corresponding experimental verification. As an idea of an application, the model system can be realized on atomic (developing the method proposed in [1] for the nuclei of 57Fe in certain composites, but this time for a visible region), chains of trapped ions (like in [8]), and molecular structures for further developing such techniques like FRET (described for instance in [12]), atomic chains like carbyne loops (for example, [13]), and microhole array synthesized by femtosecond laser radiation (see [14], for an instance).

Let us first provide below some general theoretical premises. More detailed derivations of the corresponding mathematical model can be found in [11].

### Methods

#### The equations of motion for the state amplitudes

We have assumed that the atomic energy levels have no linewidth, so that, only if ω = ω k 0 = Δ E ab / , the atoms can be able to absorb a photon. Obviously, this is an unrealistic case since it is impossible to have a completely monochromatic wave. In addition, for the case of the Fock initial state, in which we measured the energy precisely of the mode, the average electric field will be zero. In the forth of the law of energy conservation, an emitted photon will correspond to the same frequency ω k 0 (we can say it will occur with a high probability after a quite long time interval if the system has a damping).

Therefore, consider a collection of N identical atoms, at positions r1,…,rα,…,rN, coupled to a one mode electromagnetic (EM) field. Each atom α = 1..N is assumed to have only the two states |aα and |bα, separated by energy E α = E a α - E b α = ω . In the dipole approximation, the closed conservative system of identical atoms with the electromagnetic field in a cavity can be described by the Hamiltonian consisting of free atoms and electromagnetic field items with dipole-field coupling between the atoms and the electromagnetic field modes.

Inasmuch as at the initial time moment t = 0 all atoms α = 1..N of the ensemble are in the ground state |bα and EM field is in Fock state | 1 k 0 (that presents one photon with the wave vector k0), we look for a solution of the corresponding Schrödinger equation in the interacting picture in the following form:

Ψ = α = 1 N β α ( t ) | b 1 b 2 a α b N 0 + k , j γ k , j ( t ) | b 1 b 2 b N 1 k , j (1)

with the initial conditions:

β α ( 0 ) = 0 , γ k , j ( 0 ) = δ k , k 0 , (2)

where δ k , k 0 is Kroneker’s delta symbol. δ k , k 0 = 1 if k = k0, and δ k , k 0 = 0 if k k0. βα (t) (α = 1..N) and γk,j (t) (j = 1, 2) are the αth atom excited state amplitude with the others in the ground states and excited Fock field state amplitude of the jth polarization with the wave vector k, accordingly.

Then, the corresponding Schrödinger equation in the interacting picture yields the following system of equations:

β ̇ α ( t ) = i k , j g α ( k , j ) γ k , j ( t ) exp ( - i ( ν k - ω ) t + i kr α ) ; (3)

γ ̇ k , j ( t ) = i δ = 1 N g δ ( k , j ) β δ ( t ) exp ( i ( ν k - ω ) t - i kr δ ) , (4)

where

g α ( k , j ) = ν k 2 ε 0 V α · e k , j , (5)

where

α · e k , j = | α | cos θ k , j . (6)

Here, θk,j is the angle between dipole transition vector α (more accurately, non-diagonal dipole matrix element) and the jth unit polarization vector ek,j (j = 1,2 and ek,j · k= 0). V is the available by the system of atoms and field space volume. The frequencies νk correspond to the modes with the module of the wave vectors k equal to |k|.

Therefore, substituting Equation 4 into 3 and differentiating one more time, after applying the Weisskopf-Wigner approximation (details in [11]), we can derive the following system of evolution equations:

d 2 d t 2 β α ( t ) = - δ = 1 N β δ ( t ) Φ α δ - 2 D α d dt β α ( t ) , (7)

where

Φ α δ = j , | k | = k 0 g α ( k , j ) g δ ( k , j ) exp [ i k ( r α - r δ ) ] . (8)

And the decay rates Dα in the approximation can be estimated by the formula:

D α = 1 2 8 π 3 2 π 2 ε 0 1 2 π c 3 | | 2 ω 3 = 1 2 1 3 π ε 0 c 3 | | 2 ω 3 . (9)

The coefficient Dα (α = 1..N) describes the respective rate of decay for αth atom excited state. Note, that the ‘non-resonant’ items for the particle with distinguished from α indexes were disregarded in here in an assumption of quite large interatomic distances (see details in [11]).

### Results and discussion

#### An atomic chain with cyclically distanced atoms

α N sin ( kr α ) cos ( kr α ) = 0

Next, we try to make the calculations, using here the particular case of space configuration for the system atoms field. Below, for simplicity, only one polarized mode (j = 1) of the resonant field modes is taken into account with the common parameters gα and α for α = 1..N :

g α ( k , j ) = g α = g > 0 (10)

and

α · e k , j = | | (11)

for |k| = k0. In other words, the space angle distribution for the components Φαδis disregarded here, assuming the direction of the transition dipole moment α for any atom in the system coincides with the photon polarization in absorbing or emitting a resonant photon.

Then, from the system of Equation 7, in the case of a cavity with two resonant modes k = ± k0 and identical atoms with Dα D for α = 1..N, one derives that

d 2 d t 2 β α ( t ) = - 2 g 2 δ = 1 N β δ ( t ) cos ( k ( r α - r δ ) ) - 2 D d dt β α ( t ) . (12)

Using the notation

B c ( t ) = α = 1 N β α ( t ) cos kr α (13)

and the ‘cyclic’ condition α N sin ( kr α ) cos ( kr α ) = 0 yields the following relatively simple linear differential equation:

d 2 d t 2 B c ( t ) = - 2 g 2 α cos 2 ( kr α ) B c ( t ) - 2 D d dt B c ( t ) . (14)

Therefore, taking into account the initial conditions βδ (0) = 0 for α = 1..N, the solution of the above equation is as follows:

B c = α β α ( t ) cos ( kr α ) = C exp Ω 2 + t - exp Ω 2 - t , (15)

where Ω 2 = g 2 α cos 2 ( kr α ) and

Ω 2 ± = - D ± D 2 - Ω 2 2 . (16)

By analogy,

B s = α β α ( t ) sin ( kr α ) = C exp Ω 2 + t - exp Ω 2 - t , (17)

where Ω 2 = g 2 α sin 2 ( kr α ) and

Ω 2 ± = - D ± D 2 - Ω 2 2 . (18)

It is easy to see, that Ω 2 2 + Ω 2 2 = 2 g 2 N .

The field probability amplitudes can be obtained using the subsystem of Equation 4 of the full ‘conservative’ system of Equations 3 and 4. Therefore, substituting (15) and (17) into the Equation 4, and then taking into account the restrictions βα (0) = 0 for α = 1..N, we obtain that

γ k ( t ) = 2 ig C Ω 2 + f Ω 2 + , t - Ω 2 - f Ω 2 - , t - i C Ω 2 + f Ω 2 + , t - Ω 2 - f Ω 2 - , t + 1 ; (19)

and

γ - k ( t ) = 2 ig C Ω 2 + f Ω 2 + , t - Ω 2 - f Ω 2 - , t + i C Ω 2 + f Ω 2 + , t - Ω 2 - f Ω 2 - , t . (20)

where

f Ω , t = exp Ω t 2 Ω 2 sinh Ωt 2 . (21)

Note, here, we neglected the possible space angle distribution for the direction of the resonant wave vector k.

Inasmuch as cos(k (rα - rδ)) = cos (krα) cos (krδ) + sin (krα) sin (krδ), then, after substitution of the found superpositions (15) and (17) into the initial Equation 12, we derive the following integrable differential equation:

d 2 d t 2 β α ( t ) + 2 D d dt β α ( t ) = - 2 g 2 cos kr α B c ( t ) + sin kr α B s ( t ) . (22)

Integrating the left and right sides of the equation above (22) over time yields

d dt β α ( t ) + 2 D β α ( t ) = T α ( t ) , (23)

where

T α ( t ) = - 2 g 2 cos kr α F c ( t ) + sin kr α F s ( t ) + d dt β α ( 0 ) + 2 D β α ( 0 ) , (24)

and

F c , s ( t ) = 0 t B c , s ( t ) dt . (25)

According to the definition of the functions Fc,s (t)

F c ( t ) = C 1 Ω 2 + exp Ω 2 + t - 1 - 1 Ω 2 - exp Ω 2 - t - 1 ; (26)

and

F s ( t ) = C 1 Ω 2 + exp Ω 2 + t - 1 - 1 Ω 2 - exp Ω 2 - t - 1 . (27)

The solution of such linear first order differential equation, like (23), has the form:

β α ( t ) = 1 exp 2 Dt T α ( t ) exp 2 Dt dt . (28)

The integration in the last expression can be performed, yielding

T α ( t ) e 2 Dt dt = - 2 g 2 cos kr α C 1 Ω 2 + 1 Ω 2 + + 2 D e ( 2 D + Ω 2 + ) t - 1 - 1 2 D e 2 Dt - 1 - 1 Ω 2 - 1 Ω 2 - + 2 D e ( 2 D + Ω 2 - ) t - 1 - 1 2 D e 2 Dt - 1 + sin kr α C × 1 Ω 2 + 1 Ω 2 + + 2 D e ( 2 D + Ω 2 + ) t - 1 - 1 2 D e 2 Dt - 1 - 1 Ω 2 - 1 Ω 2 - + 2 D e ( 2 D + Ω 2 - ) t - 1 - 1 2 D e 2 Dt - 1 + d dt β α ( 0 ) + 2 D β α ( 0 ) 1 2 D e 2 Dt - 1 + C 0 . (29)

Therefore,

β α ( t ) = = - 2 g 2 cos kr α C H Ω 2 + , D , t - H Ω 2 - , D , t + sin kr α C H Ω 2 + , D , t - H Ω 2 - , D , t + d dt β α ( 0 ) + 2 D β α ( 0 ) 1 2 D 1 - e - 2 Dt + C 0 e - 2 Dt , (30)

where

H Ω , D , t = 1 Ω 1 Ω + 2 D e Ωt - e - 2 Dt - 1 2 D 1 - e - 2 Dt . (31)

The initial condition βα (0) = 0, for α = 1..N, sets the coefficient C0 equals 0. The initial time derivative d dt β α ( 0 ) can be determined, for example, if the system of Equation 3 from the initial ‘conservative’ full system of Equations 3 and 4 is chosen as a basis at the time moment t = 0. Then, the initial condition for the field state amplitude γk (0) = 1, where k = k0, sets the time derivative d dt β α ( 0 ) to the following expression:

d dt β α ( 0 ) = ig e kr α γ k ( 0 ) + e - kr α γ - k ( 0 ) = ig cos kr α + i sin kr α . (32)

Now, the question arises how to choose correctly the coefficients C and C. First of all, the choice has to satisfy the limitations on the probability amplitude, yielding the corresponding probability limited above by unit (the sum of all the modules squared of the introduced amplitudes equals unit probability). Secondly, the solution with the coefficients have to be consistent with the model decay (damping).

We observe that, formally, when the real part of the variable Ω is a negative quantity, that is Re (Ω) < 0, the introduced functions H and f have the following limits for quite long time intervals:

lim t H Ω , D , t = - 1 2 D Ω , when Re Ω < 0 ; (33)

lim t f Ω , t = - 1 2 Ω 2 , when Re Ω < 0 . (34)

Then,

lim t γ k ( t ) = ig C 1 Ω 2 - - 1 Ω 2 + - i C 1 Ω 2 - - 1 Ω 2 + + 1 ; (35)

lim t γ - k ( t ) = ig C 1 Ω 2 - - 1 Ω 2 + + i C 1 Ω 2 - - 1 Ω 2 + ; (36)

lim t β α ( t ) = - 1 D g 2 C 1 Ω 2 - - 1 Ω 2 + cos kr α + C 1 Ω 2 - - 1 Ω 2 + sin kr α + 1 2 D d dt β α ( 0 ) . (37)

As for an open system, in our case, it should be expected for a quite long time interval the total electromagnetic energy of the atoms-field system to be emitted into the subsystem causing the state damping. Therefore, let us define the coefficients C and Cin the following manner:

C = i Ω 2 - Ω 2 + 2 g Ω 2 + - Ω 2 - (38)

and

C = - Ω 2 - Ω 2 + 2 g Ω 2 + - Ω 2 - . (39)

Then, after substitution into the expressions for the time limits, one derive the logical finale of the system evolution:

lim t γ k ( t ) = 0 ; (40)

lim t γ - k ( t ) = 0 ; (41)

lim t β α ( t ) = 0 . (42)

The possible space configurations of the atomic system, satisfying the condition of ‘circularity’, can be easily found. For example, the set s3a1 (the notation ‘s3a1’ is just introduced here): kr 1 k · r 1 = π 6 , kr 2 = 2 π 3 , and kr3 = π. As an instance, it can also be the set s3a2: kr 1 = 2 π 3 , kr 2 = 3 π 2 , and kr 3 = π 6 + 2 π . A space configuration for five atoms can be represented, for example, by such set like s5a1: kr 1 k · r 1 = 2 π 3 , kr 2 = 3 π 2 , kr 3 = 5 π 2 , kr 4 = 7 π 2 , and kr 5 = 19 π 6 . Specifically, a combination, such like kr i = ± π n + o i π n + 2 i π with i = 1..N, n > 2, and o i π n < < π n , can generate the necessary magnitudes of the characteristic system frequencies Ω2 and Ω 2 (that, actually, are the corresponding Rabi frequencies), comparable with the given magnitude of the decay coefficient D.

Below we depict the atomic system behavior in the several introduced above configurations. Note, that the cited thereby Rabi frequencies were calculated in the SI system of units with the following notations: 1.05457 × 1 0 - 34 J sec/rad ; the electric permittivity of free space ε0 ≈ 8.8542 × 10-12 F/m; the speed of light in free space c = 299792458 m/sec; resonant wavelength close to the D2-line of a sodium atom λD ≈ 589.29 × 10-9 m; corresponding circular (in radians per second) resonant frequency ω res = 2 π c λ D 0.101747 × 1 0 16 π rad/sec ; non-diagonal so called ‘transition’ dipole matrix element (in the same order as for the D2-line transition, that is about 1 Debye) ρex = 1 × 3.33564 × 10-30 C m. For instance, if the available for the system of atoms and field volume has the value equal to V = 0.001 m3, then g = ρ ex ω res 2 ε 0 V 77.8597 π rad/sec .

Assume, for example, the available volume V = 10-13 m3 is somehow filled by the set s3a1 with D ≈ 107 rad/sec, initially coupled with one-photon Fock state. Then, g = ρ ex ω res 2 ε 0 V 77.8597 × 1 0 5 π rad/sec , Ω 2 155.7195 × 1 0 5 π rad/sec , and Ω 2 77.8597 × 1 0 5 2 π rad/sec . The corresponding graphs for probability to find each atom in the excited state are shown in Figure 1.

Figure 1. Time evolution of |βα (t)|2. V = 10-13 m3. Atoms are arranged in the set s3a1 with D ≈ 107 rad/sec. The bold solid line represents the atom with the space phase kr1 = π/6, the dot line is for the space phase kr2 = 2π/3, and the thin solid line corresponds to kr3 = π.

Let us see what happens when the available volume is increased by one order. This yields V = 10-12 m3 with the same three atoms (D ≈ 107 rad/sec) of the configuration s3a1. Then, g 24.6214 × 1 0 5 π rad/sec ; Ω 2 49.2428 × 1 0 5 π rad/sec and Ω 2 24.6214 × 1 0 5 2 π rad/sec . The corresponding graphs for each atom excited state probability are depicted in Figure 2.

Figure 2. Atom excited state probability |βα (t)|2. V = 10-12 m3. Atoms are arranged in the set s3a1 with D ≈ 107 rad/sec. The bold solid line represents the atom with the space phase kr1 = π/6, the dot line is for the space phase kr2 = 2π/3, and the thin solid line corresponds to kr3 = π.

Suppose now that the available volume is V = 10-13 m3, somehow filled by the set s5a1 with D ≈ 107 rad/sec initially coupled with one-photon Fock state. Then, g = ρ ex ω res 2 ε 0 V 77.8597 × 1 0 5 π rad/sec ; Ω 2 77.8597 × 1 0 5 2 π rad/sec , and Ω 2 155.7195 × 1 0 5 2 π rad/sec . The corresponding graphs for each atom excited state probability are shown in Figure 3.

Figure 3. Atomic excitation probability |βα (t)|2 as a function of time. V = 10-13 m3. Atoms are arranged in the set s5a1 with D ≈ 107 rad/sec. The bold solid line represents the atom with the space phase kr1 = 2π/3, the dot line is for the space phase kr5 = 19π/6, and the thin solid line corresponds to kr3 = 5π/2.

And again, let us see what happens when the available volume is increased by one order. This yields V = 10-12 m3 with the same five atoms (D ≈ 107 rad/sec) of the configuration s5a1. Then, g 24.6214 × 1 0 5 π rad/sec ; Ω 2 24.6214 × 1 0 5 2 π rad/sec , and Ω 2 49.2428 × 1 0 5 2 π rad/sec . The corresponding graph is in Figure 4. Note that the graphs Figures 3 and 4 of excited state probabilities are for the chosen three atoms with the following phases: kr 1 = 2 π 3 , kr 3 = 5 π 2 , and kr 5 = 19 π 6 .

Figure 4. Probability |βα (t)|2. V = 10-12 m3. Atoms are arranged in the set s5a1 with D ≈ 107 rad/sec. The bold solid line represents the atom with the space phase kr1= 2π/3, the dot line is for the space phase kr5 = 19π/6, and the thin solid line corresponds to kr3 = 5π/2.

As it was supposed in the derivative of the differential equations with the damping items such like (12) (see the details in the work [11], the available volume V for the system of atoms and field defines the ‘available’ modes for the electromagnetic field. The value of volume V can determine one of the inequalities D < Ω2 and D > Ω2 ( D < Ω 2 and D > Ω 2 ), therefore defining the character of the system relaxation. Such fundamental system property was illustrated in the figures. It is interesting to note that increasing the system volume V, therefore increases the ‘available’ number of quantized field modes, the maximum probability to find an atom in its excited state decreases. Other interesting feature, shown in the proposed graphs, is the different character of relaxation for each excited atom. The latter depends, as shown here, on the space phase krα, where α = 1..N.

On this note, therefore, let our narration to come to the following conclusions, in short.

### Conclusions

Thus, in this work, we investigated a chain of N identical two-level long distanced atoms prepared ‘via a single-photon Fock state’. The functional dependence of the atomic state amplitudes on a space configuration and time is derived in the Weiskopf-Wiegner approximation.

It was shown that in increasing the system volume V, the maximum value of probability to find an atom in its excited state decreases. The feature can be experimentally investigated at the proposed nanoscale limit for the space configuration of atoms.

Hence, the Weiskopf-Wiegner approximation was revealed through the provided application to the many-body system at the nanoscale limit for the atomic space phases. The found solution (30) cannot be counted as a particular one, or as a limit of such, for the initial systems of Equations 3 and 4 that represent only a closed conservative system of atoms and an electromagnetic field. Thus, we can say that the model described in this work, besides the atoms and the electromagnetic field, implicitly contains a third participant guaranteeing a total system relaxation with time. It is interesting to note here that the ‘complete’ decay of the system excitations was strongly imposed by the choice of the coefficients C (38) and C (39).

The methods, described in this work, of solving the system of linear differential equations can be applied even for more general situations when the boundary ‘circular’ conditions are not satisfied. In certain cases, the problem allows to extend the system by adding a subsystem of non-sufficient number of atoms (or an atom when the expression α N sin ( kr α ) cos ( kr α ) < 1 is actual) without influencing the main characteristics under our interest. Besides, it is interesting to investigate the dependence of value of the mentioned construction α N sin ( kr α ) cos ( kr α ) on the location of the origin of a coordinate system.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

SAS performed the calculations and analysis of the results. YSM analyzed the methods of investigation and drafted the manuscript. Both authors read and approved the final manuscript.

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# Time Shifting Property of Laplace Transform

## Laplace Transform

The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.

Mathematically, if $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a time domain function, then its Laplace transform is defined as −

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}\:\:\:\:\:\:...(1)}$$

Equation (1) gives the bilateral Laplace transform of the function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as,

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}\:\:\:\:\:\:...(2)}$$

## Time Shifting Property of Laplace Transform

Statement - The time shifting property of Laplace transform states that a shift of t0 in time domain corresponds to the multiplication by a complex exponential $\mathit{e}^{-\mathit{st_{\mathrm{0}}}}$ in s-domain. Therefore, if

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{s}\right)}}$$

Then, according the time shifting property,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{e^{\mathit{-st_{\mathrm{0}}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}}$$

Proof

By the definition of the Laplace transform, we have,

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}}$$

If

$$\mathrm{\mathit{t}\to \mathrm{\left( \mathit{t-t_{\mathrm{0}}}\right )}}$$

Then

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\right]}\:\mathrm{=}\:\int_{\mathrm{0} }^{\infty }\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}}$$

Substituting $\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}$ = u in RHS of the above equation. Then,

$$\mathrm{\mathit{t}\:\mathrm{=}\:\mathrm{\left(\mathit{u\mathrm{+}t_{\mathrm{0}}}\right)}\:\mathrm{and}\:\mathit{dt}\:\mathrm{=}\:\mathit{du}}$$

$$\mathrm{\therefore\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\right]}\:\mathrm{=}\:\int_{\mathrm{0} }^{\infty }\mathit{x}\mathrm{\left(\mathit{u}\right)}\mathit{e^{-s \mathrm{\left( \mathit{u+t_{0}} \right )}}\:\mathit{du}} \:\mathrm{=}\:\mathit{e^{\mathit{-st_{\mathrm{0}}}}\int_{\mathrm{0}}^{\infty}\mathit{x}\mathrm{\left(\mathit{u}\right)}\mathit{e^{-su}}\:\mathit{du}}}$$

$$\mathrm{\Rightarrow \mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\right]}\:\mathrm{=}\:\mathit{e^{-st_{\mathrm{0}}}}\int_{\mathit{-\infty }}^{\infty}\mathit{x}\mathrm{\left(\mathit{u}\right)}\mathit{e^{-su}}\:\mathit{du}\:\mathrm{=}\:\mathit{e^{-st_{\mathrm{0}}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}$$

$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{e^{\mathit{-st_{\mathrm{0}}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}}$$

Similarly, if $\mathit{t} \to \mathrm{\left(\mathit{t\:\mathrm{+}\:t_{\mathrm{0}}}\right)}$ Then,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t\:\mathrm{+}\:t_{\mathrm{0}}}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{e^{\mathit{st_{\mathrm{0}}}}\mathit{X}\mathrm{\left(\mathit{s}\right)}}}$$

Hence, it proves that a time shift of t0 corresponds to the multiplication by a complex exponential $\mathit{e}^{-\mathit{st_{\mathrm{0}}}}$ in the s-domain.

## Numerical Example

Using time shifting property of Laplace transform, find the Laplace transform of the signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left ( \mathit{t-\mathrm{5}} \right )}$

Solution

The given signal is,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left ( \mathit{t-\mathrm{5}}\right)}}$$

The Laplace transform of the unit step function is given by,

$$\mathrm{\mathit{L}\mathrm{\left [\mathit{u}\mathrm{\left(\mathit{t}\right)}\right ]}\:\mathrm{=}\:\frac{1}{\mathit{s}}}$$

Hence, by using the time shifting property of LT $\mathrm{\left [ i.e.,\mathit{x}\mathrm{\left(\mathit{t-t_{\mathrm{0}}}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{e}^{-\mathit{st_{\mathrm{0}}}} \mathit{X}\mathrm{\left(\mathit{s}\right)}\right ]}$, we obtain,

$$\mathrm{\mathit{L}\mathrm{\left [\mathit{u}\mathrm{\left(\mathit{t-\mathrm{5}}\right)}\right ]}\:\mathrm{=}\:\mathit{e^{-\mathrm{5\mathit{s}}}}\mathit{L}\mathrm{\left [\mathit{u}\mathrm{\left(\mathit{t}\right)}\right ]}\:\mathrm{=}\:\frac{\mathit{e^{-\mathrm{5\mathit{s}}}}}{\mathit{s}}}$$

format: plain


text:
## What alternative notation of subscripts/superscripts that are also treated as SYMBOLS in Mathematica?

5

2

I tried to avoid using the subscript $A_0$ in the Module, and get an error with $A[0]$.

In[119]:= Module[{A[0] = 1},A[0]]

During evaluation of In[119]:= Module::lvset: Local variable specification {A[0]=1} contains A[0]=1, which is an assignment to A[0]; only assignments to symbols are allowed. >>

Out[119]= Module[{A[0] = 1}, A[0]]

EDIT:

Alternatively, I tried to define "A0, A1, A2", but failed with command

Sum[Ai, {i, 3}]

I just want a Symbol of combining a letter with a postfix of number, e.g. Ai, A_i, A[i], A(i),... any notation that is treated as a Symbol in MMa would be okay.

EDIT2:

For example,

In[130]:= Module[{A0 = 1},Sum[Symbol["A" <> ToString[i]], {i, 0, 3}]]

Out[130]= A0 + A1 + A2 + A3

But I wish to replace $A0$ with $1$, and get 1+ A1 + A2 + A3

This would work when A0 defined globally.

In[132]:= Sum[Symbol["A" <> ToString[i]], {i, A0 = 1; 0, 3}]

Out[132]= 1 + A1 + A2 + A3

What about using the Notation package, in particular, Symbolize?

– Simon – 2012-06-10T06:29:54.360

10

Honestly I would suggest to not create symbols with Symbol or ToExpression for that use case. You can very well use Downvalues as you tried in the first place also for local variables, you just can't make those definitions within the local symbols list, e.g. this will work:

Module[{A},A[0] = 1; Sum[A[i],{i,0,3}]]

As explained in answers to other questions (e.g. here) localication with Module will actually create a "new" symbol (something like A$573) which will in this case leak from the Module (this is called lexical scoping, at least in Mathematica speak. It can be argued whether what Module does actually qualifies for what that term is used in general). These leaking symbols can be used as a feature, but I doubt it is what you want in this case. The following might be closer to what you actually had in mind, although it also can have subtleties when used in more complicated circumstances (see again e.g. here for more details and links): Block[{A}, A[0] = 1; Sum[A[i],{i,0,3}]] There is also Array which I think should be mentioned in that context: Module[{A}, A[0] = 1; Total@Array[A, 4, 0]] With In[207]:= Module[{A}, A[0] = 1; Sum[A[i], {i, 0, 3}]], I got this  Out[207]= 1 + A$2471[1] + A$2471[2] + A$2471[3]. Is this normal? How to remove "$2471" here? – Osiris Xu – 2012-06-12T00:33:48.150 @OsirisXu: yes, that is the expected behaviour. As I mentioned, typically the Block construct seems more appropriate in this context, but it all depends. What you see with Module is the local variables leaking. You'll find quite some answers on this site where this is explained and where this is used as a feature (search for Module,scoping,local variables). – Albert Retey – 2012-06-12T08:55:07.327 Nice. Thank you. – Osiris Xu – 2012-06-12T20:25:30.210 5 One option is to assemble symbols as strings and use Symbol: Sum[ Symbol["x" <> ToString[i]], {i, 5} ] x1 + x2 + x3 + x4 + x5 For use in Module you will need to inject this symbol, e.g. using With: With[{sym = Symbol["x" <> ToString[3]]}, Module[{sym = 7}, sym + 3] ] 10 Also be aware of Unique and its function: Unique["x"] x6 Judging from your second example Module may be the wrong tool for the job. Consider: Block[{A0 = 1}, Sum[Symbol["A" <> ToString[i]], {i, 0, 3}]] Sum[Symbol["A" <> ToString[i]], {i, 0, 3}] /. A0 -> 1 Sum[a[i], {i, 0, 3}] /. a[0] -> 1 1 + A1 + A2 + A3 1 + A1 + A2 + A3 1 + a[1] + a[2] + a[3] This is nice. But this notation is not that convenient to use. Any other alternatives? I just want a Symbol of combining a letter with a postfix of number, e.g. Ai, A_i, A[i], A(i),... any notation that is treated as a Symbol in MMa would be okay. – Osiris Xu – 2012-06-10T00:40:50.643 @OsirisXu could you give an example application for this? That will help me determine methods that may be applicable. – Mr.Wizard – 2012-06-10T00:42:00.937 For example, In[130]:= Module[{A0 = 1},Sum[Symbol["A" <> ToString[i]], {i, 0, 3}]] Out[130]= A0 + A1 + A2 + A3 But I wish to replace$A0$with$1\$, and get 1+ A1 + A2 + A3 – Osiris Xu – 2012-06-10T00:45:59.340

@OsirisXu I need more context than that. What operation or procedure are you doing? For that example would Sum[A[i], {i, 0, 3}] /. A[0] -> 1 be acceptable? – Mr.Wizard – 2012-06-10T00:49:09.933

4In other words, @Osiris, what are you really trying to do? – J. M.'s ennui – 2012-06-10T00:51:38.607

@Mr.Wizard Yes, that does the trick. Thanks a lot. – Osiris Xu – 2012-06-10T00:54:44.023

@OsirisXu okay, I added it to my answer. Notice the form with Block as well, but make sure you understand what Block does before you start using it. Generally I think you'll have better luck with Replace or ReplaceAll (/.) for what I believe you're doing. – Mr.Wizard – 2012-06-10T00:58:59.653


text:
1. ## Ordinary differential equations....

If { f1(x), f2(x) } is one set of two linearly independent solution on a<=x<=b and { g1(x), g2(x) } be another set of two linearly independent solution, then show that there exists a constant c not equal to 0 such that
W [ g1(x), g2(x) ] = c W [ f1, f2] (x)
for all a<=x<=b.

where,
W[] is wronskian, which is defined as

W [ f1(x), f2(x)]= f1 (x) * f2 ' (x) - f1' (x) * f2' (x)

(determinant involving f1, f2 and f1' , f2' )

f1 ' (x) is derivative of f1(x) i.e., d( f1(x))/dx

2. There is Liouville's theorem, which states that the Wronskian $W$ of two independent solutions of the ODE $y''+ay'+by=0$ satisfies $W'(t)={\rm e}^{-\int_{t_0}^{t} a(u)du}W(t_0)$. Apply this for $W_1=W(f_1,f_2)$ and $W_2=W(g_1,g_2)$, divide by parts, and integrate.

3. Originally Posted by niranjan
If { f1(x), f2(x) } is one set of two linearly independent solution on a<=x<=b and { g1(x), g2(x) } be another set of two linearly independent solution, then show that there exists a constant c not equal to 0 such that
W [ g1(x), g2(x) ] = c W [ f1, f2] (x)
for all a<=x<=b.

where,
W[] is wronskian, which is defined as

W [ f1(x), f2(x)]= f1 (x) * f2 ' (x) - f1' (x) * f2' (x)

(determinant involving f1, f2 and f1' , f2' )

f1 ' (x) is derivative of f1(x) i.e., d( f1(x))/dx
Are we to assume that this is a second order, linear, homogeneous differential equation? If not, it is not true. If yes, it would have been helpful to say so.

4. Are we to assume that this is a second order, linear, homogeneous differential equation?

Υeap, that's the only (college/uni course maybe?) setting
where the question makes sense.


text:
# For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$

For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$

My try :

Left hand side :
\begin{align} \sum_{d|p^k}\sigma (d) &= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) + \cdots + \sigma(p^k) \\ &= \dfrac{p^{0+1}-1}{p-1} + \dfrac{p^{1+1}-1}{p-1} + \cdots + \dfrac{p^{k+1}-1}{p-1} \\&= \dfrac{1}{p-1}\left( (p + p^2 + \cdots + p^{k+1}) - (k+1)\right) \\&= \dfrac{1}{p-1}\left(\dfrac{p(p^{k+1}-1)}{p-1}- (k+1)\right) \\ \end{align}

I'm not sure if I am in right path; this doesn't look simple. Any help ?

• Reduce to the case of prime powers (as it looks like you've already done), then compute both sides of $n = p^k$. You don't need to compute a closed form for the left side directly; just count how many times each $p^r$ appears in the sum. Aug 25, 2014 at 21:16

Define the functions $N$ and $1$ by $N(n)=n$ and $1(n)=1$ for all $n \in \mathbb{N}.$

An equivalent, but easier to write, solution is to note that the function $\sum_{d|n}\sigma(d)$ is the Dirichlet convolution $\sigma\ast 1$ and thus the definition $\sigma=N*1$ implies that $$\sigma*1=(N*1)*1=N*(1*1).$$ Now notice that $1*1$ is the classical divisor function $\tau$ and we therefore get $$\sigma*1=N*\tau,$$ which provides the required answer.

You can reduce to prime powers to get the job done, but that doesn't get at the heart of the equality.

$$\sum_{d\mid n}\sigma(d)=\sum_{d\mid n}\sum_{r\mid d}r=\sum_{r\mid d\mid n}r=\cdots$$

Can you continue? For each $r\mid n$, how many times is $r$ a summand of the above sum?

• Exactly! I have felt the same when working as all I was doing was just algebra. If I am understanding correctly, it seems in your method $n$ need not be a single prime power($p^k$) ? Aug 25, 2014 at 21:21
• and $r$ appears $\tau(d)*\tau(n)$ times i guess, not so sure sorry Aug 25, 2014 at 21:22
• Indeed, $n$ is an arbitrary natural number here. We sum an $r$ for each instance of $r\mid d\mid n$, in other words the number of $r$s is equal to the number of $d$s for which $r\mid d\mid n$, which is equals the number of divisors of $n/r$... can you proceed further from here? What does the sum look like at this point? Aug 25, 2014 at 21:28
• i see it now xD so the multiplier for each $r$ is $\tau(n/r)$ in the sum; for example, $r=n$ appears only once because $\tau(n/n) = 1$ Aug 25, 2014 at 21:43
• change of variables ($r\leftrightarrow\frac{n}{r}$): $~\sum\limits_{r\mid n}r\,\tau(\frac{n}{r})=\sum\limits_{r\mid n}\frac{n}{r}\tau(r)$ Aug 25, 2014 at 21:51

looking at this intuitively, firstly we note that: $$\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d) \\ = \sum_{d|n} d\tau(n/d)$$ so now we are summing the divisors $d$ of $n$, each divisor being counted with multiplicity $\tau(n/d)$. so you just have to persuade yourself that this multiplicity is appropriate.

but this is evident if we look at a particular $d$, since for any integer $m$ we have $$dm | n \Leftarrow \Rightarrow m | n/d$$ in words there is a 1-1 correspondence between (a) multiples of $d$ which divide $n$, and (b) divisors of $n/d$


text:
How to rotate a procedural material

I am faking sci-fi towers with windows on each floor by using a procedural texture, but I can't get it to be horizontal instead of vertical.

This is the node setup right now:

This question seems to be about the same thing but for one, I can't read the nodes (which is why I squished the node layout horizontally so all the print is visible in the screenshot), and two, my best stab at copying it doesn't seem to do anything.

So how do I rotate the stripes 90 degrees so they form neat horizontal bands around the axis of the tower? When I change the numbers in the Rotation column of the Mapping node, nothing happens.

• You can always right click on image and select something like 'open image in new tab' (Chrome), or just check here: i.stack.imgur.com/8haQ4.png – cgslav Jul 22 '17 at 19:21
• You can't "Rotate Materials" you can only rotate textures. Anyway you already got it, just rotate the texture coordinates in your Vector Mapping node, maybe 90 degrees in the Y axis will do, depending on your object orientation – Duarte Farrajota Ramos Jul 22 '17 at 23:00
• @LukeD Thanks - i even know that but it had slipped my mind. Unfortunately, the setup from the similar question still doesn't help. – kim holder Jul 24 '17 at 16:03
• @DuarteFarrajotaRamos I have tried repeatedly to do that, and no change in those numbers affects the orientation of the texture at all. However, increasing the Z scale yields a result so similar i'll move on for now. – kim holder Jul 24 '17 at 16:07
• Could you upload .blend file? I'll try it. – cgslav Jul 24 '17 at 17:15

2 Answers

There are several issues with your model that prevent proper results.

1. You have inverted normal in your model - Unrelated to texturing but can cause issues, fix it by ticking Flip in the Screw modifier
2. Your curve object has unnaplied transformations (more importantly scale) and has an offcentered origin point, both of which will negatively affect your texture coordinates and make them harder to work with. Apply scale and properly place the center relative to the object.
3. You inconsistently use Vector type of mapping in Vector Mapping nodes for textures, all should be set to Texture
4. You use duplicate Texture Coordinates Vector mapping and Wave texture nodes which are not necessary
5. Your shader node setup also seems far too convoluted, and looks like it could be simplified.

Here is my proposed node setup to achieve an identical result. Not sure exactly what you final result should look like, you may need to adjust the color ramp node to suit your liking. The Modulo node will control the tiling size.

I experimented with the Rotation values in the Mapping node and found that the bands are horizontal when Y rotation is 45d and X rotation is -35.26d .

-35.26 corresponds to arcsin( -1 / sqrt(3) ) which happens to correspond to the angle between the Z axis and the vector [1,1,1] . I suspect this is because the wave texture uses x+y+z as input into its wave function.

• Then there is something going on with Blender on my system, i guess. Doing that still leaves me with diagonal stripes. The only thing that affects stripe orientation for me is the scale of the Z axis. i.imgur.com/H62pMIQ.png – kim holder Jul 24 '17 at 18:27


text:
Standard Error of prediction for Logistic Sigmoid function

(previously: Finding the prediction interval for logistic regression)

Update 2:

This paper describes what I am looking to implement.

Update:

• Based upon some comments I am renaming this question. I am asking for the std error to calculate prediction bands on a Logistic sigmoid function, which is fit using regression not the regression predictions, which are binary.

• Predictions are being made using scipy.special.expit like so: prediction = [expit(x*beta + alpha) for x in a_pred]. The results look like this:

Based on Kerby Shedden's answer and this I now have to methods of calcualting the SE for the prediction interval:

$$SE = \sqrt(xSx^T)$$

and

$$SE = \sqrt(MSE + xSx^T)$$

However, is it legitimate to caluate the MSE based on residuals from binary data as shown:

TL;DR

there are two type of SE(Standard error) - More Detail

How do I calculate the SE for a predicted indivdual (not the mean) put into a logistic regression model statsmodels.discrete.discrete_model.Logit

Like this for linear regression in statsmodels or this in R.

I can extract model parameter variance, covariance and std dev but that is it....the model does not return std error of predictions - how do I calculate these?

Full

I am trying to fit a prediction interval for logitistic regression model. I am using statsmodels although I am happy hear answers using another package.

My procedure so far:

Fit the model to data df:

log_mdl = statsmodels.discrete.discrete_model.Logit.from_formula ("hit ~ a",df).fit()

The models parameters returned:

                           Logit Regression Results
==============================================================================
Dep. Variable:                    hit   No. Observations:                  200
Model:                          Logit   Df Residuals:                      198
Method:                           MLE   Df Model:                            1
Date:                Tue, 10 Mar 2020   Pseudo R-squ.:                  0.6209
Time:                        14:02:57   Log-Likelihood:                -45.308
converged:                       True   LL-Null:                       -119.52
Covariance Type:            nonrobust   LLR p-value:                 3.823e-34
==============================================================================
coef    std err          z      P>|z|      [0.025      0.975]
------------------------------------------------------------------------------
Intercept     -4.0571      0.532     -7.624      0.000      -5.100      -3.014
a              0.0990      0.015      6.657      0.000       0.070       0.128
==============================================================================

then I make some predictions for values a_pred so I can plot the line:

 log_pred = [expit(x* log_mdl.params[1] + log_MLE.params[0]) for x in a_pred]

I now wish to find the prediction interval for each prediction.

To do this I need the SE (Standard Error) of each prediction:

However there are two type of SE - More detail:

Ideally there would be a method like this for linear regression in statsmodels or this in R.

I have found this for 95% Confidence interval for of the true logit, which is the same as the martic operation in this. Which is apparently what R returns as the SE for a prediction.

However I am not sure if this is the SE of the predicted mean (Confidence interval) or the SE of a predicted individual (Prediction interval).

I have also found this in which there are two method of calcualting the CI, one for the funct (i assume mean) and another for an observation (I assume this means a single prediction). This reflects what is said here, that there are two SEs one for the mean and another for a single prediction which includes the variation of the signal not just the accuracy of the estimated mean.

How can I find the SE of a predicted individual?

• The response variable for Logit is binary. The prediction interval is the set {0, 1} unless either one as probability larger than the confidence level, and then it would be just one point. I don't see how prediction intervals make much sense for a binary variable. Mar 10, 2020 at 18:14
• Aplogies, I am making probability predictions like so: expit(x*beta+alpha) where alpha and beta are taken from the logistic model. I have updated the question. Mar 10, 2020 at 19:38
• A subject is either in one category or the other, so a prediction interval with probabilities does not make sense to me. Unless the prediction interval contains the closed interval $[0,1]$, the prediction interval is assured to exclude the correct value.
– Dave
Mar 10, 2020 at 21:07
• This may be of interest. Mar 10, 2020 at 22:26
• Does this answer your question? How are the standard errors computed for the fitted values from a logistic regression? Mar 14, 2020 at 11:38

Something like below should work:

x = log_mdl.model.exog
c = log_mdl.cov_params()
vcov = np.dot(x, np.dot(c, x.T))
se = np.sqrt(np.diag(vcov))

If x is very large there are faster ways to do this that only compute the diagonal of vcov, i.e.

v = (x * np.dot(x, c)).sum(1)
se = np.sqrt(v)

But this is less transparent.

For terminology, I think you could refer to this as the standard error for the logit probabilities.

• I have tried this, SE = sqrt( X C Xt). it yields very close confidence bands - depending on if there is a lot of data - this seems to relate only to the confidence of the predicted mean, not the interval of 95% prediction based upon the spread of the data and the accuracy of the mean. I found an alternative definition: SE = sqrt( MSE + (X C Xt)), where MSE is the mean squared error. This seems like it would account for data variability but I how do I find the MSE for a logistic sigmoid fit? Is it legitimate to find the residuals based upon binary values? Mar 11, 2020 at 9:35
• These confidence bands correctly capture the uncertainty in the probabilities, or more specifically the log odds for P(Y=1 | X=x). When the sample size is large and the number of variables is small or moderate, then these standard errors will be fairly small. This does not capture the predictive uncertainty in a future observation. There is no sensible way to talk about a predive standard deviation for logistic regression becaue what you are predicting is a category label, not a numeric value. Mar 12, 2020 at 11:59

The usual way to assess predictive uncertainty for logistic regression is using cross-validation. If you want something that sort of looks like the OLS prediction interval, you can use something like below:

import numpy as np                                                                                import statsmodels.api as sm                                                                                import pandas as pd                                                                                from scipy.stats.distributions import norm                                                                                n, p = 200, 3
# Generate data
x = np.random.normal(size=(n, p))                                                                                par = np.r_[1, 0, -1]
lpr = np.dot(x, par)
pr = 1 / (1 + np.exp(-lpr))
y = (np.random.uniform(size=n) < pr).astype(np.int)

# Fit a model
model = sm.GLM(y, x, family=sm.families.Binomial())
result = model.fit()

# Calculate the accuracy.
c = result.cov_params()
lpr_hat = np.dot(x, result.params)
pr_hat = 1 / (1 + np.exp(-lpr_hat))
sd = np.sqrt(np.diag(np.dot(x, np.dot(c, x.T))))
q = pr_hat*(1 - norm.cdf(-lpr_hat/sd)) + (1 - pr_hat)*norm.cdf(-lpr_hat/sd)

The bottom five lines are a plug-in estimate of the probability that the observed value and predicted value are the same, for each case.


text:
# Continuous Monitoring of Distributed Data Streams over a Time-based Sliding Window

Abstract : The past decade has witnessed many interesting algorithms for maintaining statistics over a data stream. This paper initiates a theoretical study of algorithms for monitoring distributed data streams over a time-based sliding window (which contains a variable number of items and possibly out-of-order items). The concern is how to minimize the communication between individual streams and the root, while allowing the root, at any time, to be able to report the global statistics of all streams within a given error bound. This paper presents communication-efficient algorithms for three classical statistics, namely, basic counting, frequent items and quantiles. The worst-case communication cost over a window is $O(\frac{k} {\epsilon} \log \frac{\epsilon N}{k})$ bits for basic counting and $O(\frac{k}{\epsilon} \log \frac{N}{k})$ words for the remainings, where $k$ is the number of distributed data streams, $N$ is the total number of items in the streams that arrive or expire in the window, and $\epsilon < 1$ is the desired error bound. Matching and nearly matching lower bounds are also obtained.
Keywords :
Type de document :
Communication dans un congrès
Jean-Yves Marion and Thomas Schwentick. 27th International Symposium on Theoretical Aspects of Computer Science - STACS 2010, Mar 2010, Nancy, France. pp.179-190, 2010, Proceedings of the 27th Annual Symposium on the Theoretical Aspects of Computer Science

Littérature citée [29 références]

https://hal.inria.fr/inria-00456120
Contributeur : Publications Loria <>
Soumis le : vendredi 12 février 2010 - 10:07:44
Dernière modification le : lundi 20 novembre 2017 - 15:14:01
Document(s) archivé(s) le : vendredi 18 juin 2010 - 20:19:43

### Fichier

chan.pdf
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### Identifiants

• HAL Id : inria-00456120, version 1

### Citation

Ho-Leung Chan, Tak-Wah Lam, Lap-Kei Lee, Hing-Fung Ting. Continuous Monitoring of Distributed Data Streams over a Time-based Sliding Window. Jean-Yves Marion and Thomas Schwentick. 27th International Symposium on Theoretical Aspects of Computer Science - STACS 2010, Mar 2010, Nancy, France. pp.179-190, 2010, Proceedings of the 27th Annual Symposium on the Theoretical Aspects of Computer Science. 〈inria-00456120〉

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text:
# Christmas Sentences

I've received an early Christmas gift with a combination lock and a letter containing a poem and some sentences. There are several typos: I don't believe they're intentional, but they might be. I have not yet solved it.

Who and When is all you need
One letter each,

EIGHTH
Renowned Curator Jacques Sauniere staggered through the valuted archway of the museum's Grand Gallery

TENTH
It was five o' clock on a winter's morning Syria

THIRD
It was a bright cold day in April, and the clocks were striking thirteen.

FIFTH
"Christmas won't be Christmas without any presents,"
grumbled Jo, lying on the rug.

The combination lock doesn't have every letter.

1. RSTLNBDMJP
2. LHAEIOUYRT
3. RSTLNACDEO
4. DHKYRSTLNE

These are the same for every MasterLock, so they likely don't have anything to do with the actual solution, just whether a solution could be valid.

I've already figured this much out:

Every sentence is the opening sentence of a book. The first (Renowned curator...) is from The Da Vinci Code, the second (It was five o' clock...) is from Murder on the Orient Express, the third (It was a bright cold day...) is from 1984, and the fourth (Christmas won't be Christmas...) is from Little Women.

The following codes are incorrect:
* NHNL (nth letter of each title)
* CHNL (nth letter of each title, with no The in Da Vinci Code)
* NIOS (nth letter of each author's name)

• Is this an original puzzle of yours or is it from another source? We require attribution for puzzles, we like to be sure its creator gets credit. Welcome to PSE. – Thomas Markov Dec 18 '19 at 13:57
• @ThomasMarkov I got this as an early Christmas present. I don't know where the gifter got the puzzle, but it seems to be original. – leo60228 Dec 18 '19 at 13:59
• Like a postcard or something? – Thomas Markov Dec 18 '19 at 14:01
• Yes. There's precedent on the site for original puzzles from IRL friends: puzzling.stackexchange.com/questions/20825/… – leo60228 Dec 18 '19 at 14:02
• A realization that could hopefully help others-- the numbers in the titles all relate to the numbers from the book. As @P1Storius pointed out, the prologue in The Da Vinci Code (clue 1) states that it was 10:46 PM. Clue 2 starts with "Tenth," and could relate to it. Clue 2 includes 5 o'clock-- clue 4 has fifth. Clue 3 is from 1984 -- Clue 1 says 8th. Clue 4 is about the March family (3rd month). Clue 3 says third. This might give us the order of an already-tested solution? – NegativeFriction Dec 18 '19 at 17:36

New attempt, I think I have it this time.

It's either SUTH, SUTY, or SUTT

The "crack" to this was realizing that

Each clue's corresponds to another clue if you look at the time. If you read the start of the Da Vinci Code, it begins at 10:48 (tenth). Murder on The Orient Express starts at 5 AM (fifth). The clocks were striking 13-- in 1984 (Eighth). Little Women is about Josephine March. March is the third month (third). The Murder on the Orient Express one is trickier.

From this, we can determine that the order of the clues is not simply top-to-bottom.

But here's the other kicker-- we've got a mystery of who.

I believe that who is a character who was centrally dead in the book.

Consider the following to support this:

The Da Vinci code opens with the murder of Jacques Sauniere. In Little Women, Beth March dies. In Murder on the Orient Express, we have two dead bodies-- Daisy Armstrong and Ratchett (the titular murdered individual). In 1984, Winston is widely accepted to have died figuratively (though not literally).

Now, let's look at how we can use these individuals to solve this.

If we look at the nth letter of the main character's name in the clue that corresponds to the given number, we get a number on each dial. For example: 1984 is about Winston Smith. The eighth letter of his name is "S." The Da Vinci Code is about Jacques Sauniere. The tenth letter of his name is U. The third letter of Beth March's name is s. Murder on the Orient Express is a bit tricky, but I have 3 possibilities. 1) The H in Ratchett. 2) The Y in Daisy Armstrong. 3) The T in Armstrong.

Let me know if one of those 3 works. I'm overly invested in this riddle.

• None of those worked. The chain fell off, so I have my present (some candy), but the lock is still locked. :( – leo60228 Dec 18 '19 at 20:27
• well, drat. If and when you get the correct answer/logic, please let me know. I was so darned certain I had this one.... – NegativeFriction Dec 19 '19 at 13:01

Could it be

SLER?

This is based on taking the nth letter of the character or time that appears in the riddle. I had to do a little bit of fudging to make elements work, such as

making Jacques Sauniere's name not count spaces, even though other names count them, and changing "the clock strikes 13" to the clock striking 1-- since a 12 hour clock would strike one, and "I" wasn't an option in the combination lock, and updating Jo's name to include her last name, Jo March.)

Interesting riddle. I'll be interested to know the accurate answer if mine's incorrect-- and it feels like it might be, given the amount of fudging I had to do.

• Just tried it, this is incorrect. :( – leo60228 Dec 18 '19 at 14:17
• I figured-- thought I'd give it my best crack! Here's a thought-- maybe it's the nth letter of the character in the timeth sentence? IE, if book reads "It was 3 o'clock. I like oranges. Sam picked flowers" and the riddle said "second," we'd look at the 3rd sentence (3 o'clock) of the book, and the second letter of the character ("Sam")? Trying to figure out that "Who and When" thing... – NegativeFriction Dec 18 '19 at 14:20

I feel like this is wrong because it's too obvious, but what about

MTTR

These letters are the

EIGHTH, TENTH, THIRD, and FIFTH in the letter lists you've provided.

This is very likely not correct, since

I bet those letters are on dials, so there's no point of reference for which should be considered first.

• Yeah, they're on dials. – leo60228 Dec 18 '19 at 20:27

Regarding the clue

Who and When is all you need

Is it perhaps

NIOS

Because these correspond to

The author names: Dan Brown, Agatha Christy, George Orwell and Louisa May Alcott
The numbers indicate the Nth letter of their names

What I don't understand yet is what the other clue means:

Who and When is all you need
The first three books have a clear time of the day (and in some cases day of the year) that they take place:

Da Vinci Code: Prologue ; Louvre Museum, Paris 10:46 P.M. ; Renowned curator Jacques Sauniere staggered (...) Murder on the Orient Express: It was five o' clock (...) 1984: It was a bright cold day in April, and the clocks were striking thirteen. [which I interpret as 13:00, or 1:00 p.m.] Little women: No time clue in the beginning of the book

• Per the OP's hint, your suggestion is not correct; they have already tried it. As to time-- The book Little Women follows the March family. That may indicate the number 3 somehow, as March is the third month in the Julian calendar? – NegativeFriction Dec 18 '19 at 17:30
• Oh my, how did I miss that... will have to give it a bit more thought – P1storius Dec 18 '19 at 17:46
• I wrote a comment on the overall post, but you caused me to realize something that may be relevant-- The number in each heading corresponds to a number from one of the excerpts. Clue 1 says eighth-- one clue is about 1984. Clue 2 says tenth, and Da Vinci Code starts at 10:46. Clue 3 says third, and March is the third month (family name is March in Little Women). Clue 4 says fifth, and Orient Express starts at 5 o'clock in the morning. Not sure how they relate, but I don't think that that crossover is by accident. – NegativeFriction Dec 18 '19 at 17:49


text:
# Impulse response for an LTI system

I'm new to signal processing and working my way through a textbook. There is an exercise where a causal LTI system is given that responds to a rectangular pulse.

I have an exercise where a causal LTI system responds for a rectangular pulse

$$s(t) = \Pi(t) = \operatorname{rect}(t)$$

with a triangular pulse.

$$g(t) = \Lambda(2t) = \operatorname{tri}(2t)$$

One task is to find out what the impulse response for this system is. The solution is

$$h(t) = \frac{d}{dt} h_{\epsilon}(t)$$

because $$\delta(t) = \frac{d}{dt} \epsilon(t)$$

where $$\epsilon(t)$$ is the unit step function, $$\delta(t)$$ the unit impulse function.

and looks like this

I don't really understand the solution and how it turned to an endlessly oscillating rect pulse. Unfortunately, the textbook doesn't explain its approach. So, why does the output signal looks like this?

EDIT:

Translation: A causal LTI system responds to a rectangular pulse $$s(t) = \operatorname{rect}(t)$$ with $$g(t) = \Lambda(2t)$$

a) What is the response for $$s_1(t) = \operatorname{rect}(\frac{t-1}{2})$$?

b) What is the unit step response?

c) What is the impulse response?

And the solutions for a) - c) from the textbook

• Please define all symbols you are using. – Hilmar Apr 18 at 13:46
• Done. Sorry, I thought the symbols were uniformly defined in signal processing. – Brain Damage Apr 18 at 14:00
• did i hit the definitions well enough, @Hilmar? – robert bristow-johnson Apr 18 at 17:08
• Well, they are in my experience -- but you're not using them. Either your book or your professor is a bit idiopathic in their choice of symbols, or you're coming at this from a completely different starting point (theoretical math?). – TimWescott Apr 18 at 17:10
• At any rate -- things still aren't clear, and Robert and I both think we're experienced hands. Could you take a picture of the problem statement from the book, and post it here? Or do your best to quote it word-by-word? – TimWescott Apr 18 at 17:11

EDIT : Based on RBJ's comment that the indicated the system cannot possibly be an LTI, due to the fact that the output spectrum contains frequencies that do not exist at the input spectrum..., I've looked at the input-output spectrums, and came to the following conclusion :

Given the input:

$$s(t) = \begin{cases} { 1 ~~~,~~~-0.5 < t < 0.5 \\ 0~~~,~~~\text{otherwise} }\end{cases} \tag{1}$$

and the output: $$g(t) = \begin{cases} { 2t+1 ~~~,~~~ -0.5 < t < 0 \\ 1-2t ~~~,~~~ ~~~~~~~~0 < t < 0.5 }\end{cases} \tag{2}$$

The corresponding spectrums (apart from linear scale factors !) are :

$$S(\omega) = \frac{ \sin(\omega /2) }{ \omega} \tag{3}$$

and

$$G(\omega) = \left( \frac{ \sin(\omega/4) }{ \omega} \right)^2 \tag{4}$$

Now, the input spectrum contains all frequencies except the zeros of the sine function which occur at $$\omega = 2\pi k$$, whereas the output spectrum contains all frequencies except those $$\omega = 4\pi k$$. Therefore, RBJ is correct in his claim that this particular input-output pair violates the LTI principle of output cannot contain frequencies that do not exist at the input.

However, in my opinion, redefining the output triangular-pulse duration as $$\hat{g}(t) = \begin{cases} { 2t+1 ~~~,~~~ -1 < t < 0 \\ 1-2t ~~~,~~~ ~~~~~0 < t < 1 }\end{cases} \tag{5}$$

eliminates the associated problem, as the output spectrum becomes:

$$\hat{G}(\omega) = \left( \frac{ \sin(\omega/2) }{ \omega} \right)^2 \tag{6}$$

which includes all frequencies except at those $$\omega = 2\pi k$$. Thus the missing frequencies at the input are also missing at the output.

Hence, I would like to assume that there was a typo at the duration of the triangular-pulse output $$g(t)$$, and that otherwise the analysis based on LTI remains valid... Thanks to RBJ for the reminder... EDIT-1-ENDS

EDIT-2 : However, now happens another problem. If the system is to remain LTI (so that subsequent analysis remains valid), then $$\hat{g}(t)$$ (the redefined $$g(t)$$) should be used in the remaning progress. But then the observed waveforms in the figures will be different from what's being drawn. Most notably, the step-response obtained in part-b, becomes a flat output (a unit-step with finite length triangular tails), and finally the impulse response found as step-response's derivative will be a single rectangular-pulse at the origin (and one more at positive infinity) and not a periodic waveform.

So I think the problem should be completely redefined to make any sense...

EDIT-2-ENDS

Your system is stated to be causal and LTI. And an example response for a single rectangular-pulse input $$s(t)$$ is provided to be a single triangular-pulse output $$g(t)$$:

$$g(t) = \mathcal{T}\{s(t)\} = s(t) \star h(t)$$

where $$h(t)$$ is the impulse response of the LTI system, and $$\star$$ denotes convolution.

The followings are easily deduced from the properties of LTI systems:

a-) Observe that $$s_1(t)$$ is given by a sum of two shifted $$s(t)$$ pulses side by side as:

$$s_1(t) = s(t-0.5) + s(t-1.5)$$

then from the LTI property, conclude the output as:

$$g_1(t) = \mathcal{T}\{s_1(t)\} = \mathcal{T}\{s(t-0.5) + s(t-1.5)\} = g(t-0.5) + g(t-1.5)$$

the output for $$s_1(t)$$ consists of two triangular pulses repeated (and the origin is shifted).

b-) Now, instead of two pulses, consider an infinity of such rectangular pulses shifted to locate side by side. Then the output will also be infinity of triangular pulses located side by side, as given by the figures in the case b:

$$\epsilon(t) = \sum s(t -0.5-n) \implies h_{\epsilon}(t) = \mathcal{T}\{ \epsilon(t) \}$$ \begin{align} h_{\epsilon}(t) &= \mathcal{T}\{ \sum s(t -0.5-n) \} \\ &= \sum \mathcal{T}\{s(t -0.5-n) \} \\ &= \sum g(t-0.5-n) \end{align}

incidentally, the signal $$\epsilon(t)$$ is a unit-step function, and the corresponding response $$h_{\epsilon}(t)$$ is known as step-response of an LTI system.

c-) then the impulse response of the LTI system can be found based on the relation of step-reponse to impulse-response. Namely, if $$h_{\epsilon}(t)$$ is the step-response of an LTI system, then the impulse response $$h(t)$$ is :

$$h(t) = \frac{ d h_{\epsilon}(t)} {dt}$$

you can see from the figure in c-) that the impulse response $$h(t)$$ is indeed the differentiation of the step-response; i.e, the triangular pulses are converted into bipolar rectangular pulses, and they are repeated indefinetely at the rate of triangular pulses in the step-response. Hence, the figure in case-c.

or it could as well be directly verified as the differentitation property of convolution. Actually this property is a consequence of the differentiation property of the convolution operator...

• Fat, once you look at Aufgabe 1.1 a), you can tell that the particular system is not linear at all. it is a multiplier with $g(t)$ as the output, the periodic triangle wave is one input, and $s(t)$ is the other input. – robert bristow-johnson Apr 18 at 20:12
• okay, i'm wrong. i see how it could be LTI where a $s(t)=\Pi(t)$ is the input and a single $g(t)=\Lambda(2t)$ is the output. there might be a zero divided by zero issue where the $S(f)=\operatorname{sinc}(f)$ spectrum of the input becomes a $G(f)=\frac12 \operatorname{sinc}^2(\frac{f}2)$ spectrum in the output. gonna be a very strange frequency response. I cannot see how the length of the output (one unit of time) can be the same as the length of the input (also one unit of time). how does the LTI system not ring out a little longer? – robert bristow-johnson Apr 18 at 20:36
• Actually, there are frequencies of the input spectrum with zero amplitude that somehow get multiplied by the frequency response and become non-zero. so this is not a possible LTI system because the output contains frequency components that do not exist in the input. – robert bristow-johnson Apr 18 at 20:41
• i am refraining from a -1. but the answer is wrong. – robert bristow-johnson Apr 18 at 20:44
• @robertbristow-johnson see the edit please – Fat32 Apr 18 at 21:25


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# Finding the zeroes using Chebyshev polynomials

Use the zeroes of $\bar{T}_3$ and transformations of the given interval to construct an interpolating polynomial of degree 2 for $f(x)={ 1\over x}$ over the interval $[1,3]$

My biggest issue is finding the zeroes. How exactly do I do that? I have been using the formula $$\frac{1}{2}\left[(b-a)\cos\left(\frac{N+\frac{1}{2}-n}{ N}\cdot\pi\right) +a+b\right]$$ Where $[a,b]$ is the interval, $N$ is the degree and for $n=1,2,3,\ldots, N.$ $$\bar{x_k}=\frac{1}{2}\left[(3-1)\cos\left(\frac{2+\frac{1}{2}-1}{2}\cdot\pi\right) +1+3\right]=1.29289322$$ This is of course wrong. If anyone is familiar with this topic then please guide me.

The final answer is $P(x)=.3489153-.1744576(x-2.866025)+.1538462(x-2.866025)(x-2)$.

-
Those cannot be the zeroes. If $x=2.866025$ then $P(x) = .3489153$... Ah nevermind, you're looking for the zeros of a different polynomial. The question didn't render properly on my phone –  Arkamis Oct 21 '12 at 17:55
hmmm I am quite confused then. So how would you solve this then? –  math101 Oct 21 '12 at 17:59
Well actually it would be the zeroes of the $\bar{T_3}$ not of the interpolating polynomial –  math101 Oct 21 '12 at 18:03
I think I have it figured out. The answer has been factored for some reason –  math101 Oct 21 '12 at 18:11

The zeroes of $T_3(x)=4x^3-3x$ are $0,\pm\frac{\sqrt 3}{2}$. The first idea is to apply the mapping from the interval $[-1,1]$ (the natural domain for the Chebyshev polynomials) into your desired domain $[1,3]$. It can be easily seen that the required mapping is $f(x)=x+2$, and thus you are to interpolate $\frac1{x}$ over the points $2,2\pm\frac{\sqrt 3}{2}$. You can now apply any of a number of methods (Lagrange, Newton, Neville-Aitken) for constructing the quadratic polynomial interpolant. That part I leave to you.


text:
# Why (Generalized) Complex Sinusoids are Important

## GUIDE: Mathematics of the Discrete Fourier Transform (DFT). Why (Generalized) Complex Sinusoids are Important

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## Why (Generalized) Complex Sinusoids are Important

As a preview of things to come, note that one signal5.10 is projected onto another signal using an inner product. The inner product computes the coefficient of projection5.11 of onto . If (a sampled, unit-amplitude, zero-phase, complex sinusoid), then the inner product computes the Discrete Fourier Transform (DFT), provided the frequencies are chosen to be . For the DFT, the inner product is specifically

Another commonly used case is the Discrete Time Fourier Transform(DTFT) which is like the DFT, except that the transform accepts an infinite number of samples instead of only . In this case, frequency is continuous, and

The DTFT is what you get in the limit as the number of samples in the DFT approaches infinity. The lower limit of summation remains zero because we are assuming all signals are zero for negative time. This means we are working with unilateral Fourier transforms. There are also corresponding bilateral transforms for which the lower summation limit is .

If, more generally, (a sampled complex sinusoid withexponential growth or decay), then the inner product becomes

and this is the definition of the transform. It is a generalization of the DTFT: The DTFT equals the transform evaluated on the unit circle in the plane. In principle, the transform can also be recovered from the DTFT by means of “analytic continuation” from the unit circle to the entire plane (subject to mathematical disclaimers which are unnecessary in practical applications since they are always finite).

Why have a tranform when it seems to contain no more information than the DTFT? It is useful to generalize from the unit circle (where the DFT and DTFT live) to the entire complex plane (the transform’s domain) for a number of reasons. First, it allows transformation of growingfunctions of time such as unstable exponentials; the only limitation on growth is that it cannot be faster than exponential. Secondly, the transform has a deeper algebraic structure over the complex plane as a whole than it does only over the unit circle. For example, the transform of any finite signal is simply a polynomial in . As such, it can be fully characterized (up to a constant scale factor) by itszeros in the plane. Similarly, the transform of an exponential can be characterized by a single point of the transform (the point which generates the exponential); since the transform goes to infinity at that point, it is called a pole of the transform. More generally, the transform of any generalized complex sinusoidis simply a pole located at the point which generates the sinusoid. Poles and zeros are used extensively in the analysis of recursivedigital filters. On the most general level, every finite-order, linear, time-invariant, discrete-time system is fully specified (up to a scale factor) by its poles and zeros in the plane.

In the continuous-time case, we have the Fourier transformwhich projects onto the continuous-time sinusoids defined by , and the appropriate inner product is

Finally, the Laplace transform is the continuous-time counterpart of the transform, and it projects signals onto exponentially growing or decaying complex sinusoids:

The Fourier transform equals the Laplace transform evaluated along the “ axis” in the plane, i.e., along the line , for which . Also, the Laplace transform is obtainable from the Fourier transform via analytic continuation. The usefulness of the Laplace transform relative to the Fourier transform is exactly analogous to that of the transform outlined above.

<< Previous page  TOC  INDEX  Next page >>


text:
1. ## Cubic equation

I've been stuck on this problem for a while, after starting C1 AS level maths a few weeks ago. Here is the question:

Curve C => y = ax³ - x² + x - 5, a is some constant.
Point P => (-1/2, y)
Tangent to C at P is parallel to y = (7x - 1)/2

Find a.

Any help will be appreciated, thanks.

2. ## Re: Cubic equation

Originally Posted by b150
I've been stuck on this problem for a while, after starting C1 AS level maths a few weeks ago. Here is the question:

Curve C => y = ax³ - x² + x - 5, a is some constant.
Point P => (-1/2, y)
Tangent to C at P is parallel to y = (7x - 1)/2

Find a.

Any help will be appreciated, thanks.
Parellel to y = (7/2)x - 1/2 means the slope of the tangent (at x = -1/2) is 7/2
So if f(x) = ax^3 -x^2 + x - 5, then f ' (-1/2) = ...

Can you take it from here?

3. ## Re: Cubic equation

Originally Posted by b150
I've been stuck on this problem for a while, after starting C1 AS level maths a few weeks ago. Here is the question:

Curve C => y = ax³ - x² + x - 5, a is some constant.
Point P => (-1/2, y)
Tangent to C at P is parallel to y = (7x - 1)/2

Find a.

Any help will be appreciated, thanks.
The tangent of C is $f'\left(-\dfrac{1}{2}\right)$ and since parallel lines have the same gradient $f'\left(-\dfrac{1}{2}\right) = \dfrac{7}{2}$

Solve for a

edit: too slow

4. ## Re: Cubic equation

This looks like a calculus question.

Do you know how to find the derivative of this cubic function?

Do you know what a derivative tells about a function?

5. ## Re: Cubic equation

Originally Posted by e^(i*pi)
...

edit: too slow
...but much better looking!

6. ## Re: Cubic equation

haha I realised how to do it like 5 minutes after I posted :P Thanks anyway.

7. ## Re: Cubic equation

Originally Posted by TheChaz
...but much better looking!
I don't see how you can say that e^(i*pi) is much better looking ! He doesn't even show a picture of himself.

... Oh! You mean his post was much better looking? Yes, it does look good -- LaTeX does that.


text:
Basic College Mathematics (10th Edition)

$21\frac{1}{8}-20\frac{1}{4}=21.125-20.25=0.875=\frac{7}{8}$ inches
To change the fractions to decimals, first we need to write $\frac{1}{8}$ and $\frac{1}{4}$ as decimals. We do this by dividing: $~~~~0.125$ $8\overline{)1.000}$ $~~~~~~~\underline{8}$ $~~~~~~~20$ $~~~~~~~\underline{16}$ $~~~~~~~~~~40$ $~~~~~~~~~~\underline{40}$ $~~~~~~~~~~~~~0$ $~~~~0.25$ $4\overline{)1.00}$ $~~~~~~~\underline{8}$ $~~~~~~~~20$ $~~~~~~~~\underline{20}$ $~~~~~~~~~~~0$ Therefore, $21\frac{1}{8}=21.125$ and $20\frac{1}{4}=20.25$. We find the difference: $21.125-20.25=0.875$ inches We convert this to a fraction: $0.875=\frac{875}{1000}=\frac{7}{8}$ This matches our answer to Exercise 34. We see that the fraction and decimal methods yielded equivalent answers, although it was simpler to work in fractions.


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G = C2×Q8⋊F5order 320 = 26·5

Direct product of C2 and Q8⋊F5

Series: Derived Chief Lower central Upper central

 Derived series C1 — C20 — C2×Q8⋊F5
 Chief series C1 — C5 — C10 — D10 — C4×D5 — C4⋊F5 — C2×C4⋊F5 — C2×Q8⋊F5
 Lower central C5 — C10 — C20 — C2×Q8⋊F5
 Upper central C1 — C22 — C2×C4 — C2×Q8

Generators and relations for C2×Q8⋊F5
G = < a,b,c,d,e | a2=b4=d5=e4=1, c2=b2, ab=ba, ac=ca, ad=da, ae=ea, cbc-1=ebe-1=b-1, bd=db, cd=dc, ece-1=b-1c, ede-1=d3 >

Subgroups: 650 in 162 conjugacy classes, 60 normal (32 characteristic)
C1, C2, C2 [×2], C2 [×4], C4 [×2], C4 [×8], C22, C22 [×6], C5, C8 [×2], C2×C4, C2×C4 [×15], Q8 [×2], Q8 [×8], C23, D5 [×2], D5 [×2], C10, C10 [×2], C4⋊C4 [×3], C2×C8 [×4], C22×C4 [×3], C2×Q8, C2×Q8 [×8], Dic5 [×2], Dic5 [×2], C20 [×2], C20 [×2], F5 [×2], D10 [×2], D10 [×4], C2×C10, Q8⋊C4 [×4], C2×C4⋊C4, C22×C8, C22×Q8, C5⋊C8 [×2], Dic10 [×2], Dic10 [×5], C4×D5 [×4], C4×D5 [×4], C2×Dic5, C2×Dic5, C2×C20, C2×C20, C5×Q8 [×2], C5×Q8, C2×F5 [×4], C22×D5, C2×Q8⋊C4, D5⋊C8 [×2], D5⋊C8, C4⋊F5 [×2], C4⋊F5, C2×C5⋊C8, C2×Dic10, C2×Dic10, C2×C4×D5, C2×C4×D5, Q8×D5 [×4], Q8×D5 [×2], Q8×C10, C22×F5, Q8⋊F5 [×4], C2×D5⋊C8, C2×C4⋊F5, C2×Q8×D5, C2×Q8⋊F5
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], C2×C4 [×6], D4 [×4], C23, C22⋊C4 [×4], SD16 [×2], Q16 [×2], C22×C4, C2×D4 [×2], F5, Q8⋊C4 [×4], C2×C22⋊C4, C2×SD16, C2×Q16, C2×F5 [×3], C2×Q8⋊C4, C22⋊F5 [×2], C22×F5, Q8⋊F5 [×2], C2×C22⋊F5, C2×Q8⋊F5

Smallest permutation representation of C2×Q8⋊F5
On 80 points
Generators in S80
(1 44)(2 45)(3 41)(4 42)(5 43)(6 46)(7 47)(8 48)(9 49)(10 50)(11 51)(12 52)(13 53)(14 54)(15 55)(16 56)(17 57)(18 58)(19 59)(20 60)(21 61)(22 62)(23 63)(24 64)(25 65)(26 66)(27 67)(28 68)(29 69)(30 70)(31 71)(32 72)(33 73)(34 74)(35 75)(36 76)(37 77)(38 78)(39 79)(40 80)
(1 19 9 14)(2 20 10 15)(3 16 6 11)(4 17 7 12)(5 18 8 13)(21 31 26 36)(22 32 27 37)(23 33 28 38)(24 34 29 39)(25 35 30 40)(41 56 46 51)(42 57 47 52)(43 58 48 53)(44 59 49 54)(45 60 50 55)(61 71 66 76)(62 72 67 77)(63 73 68 78)(64 74 69 79)(65 75 70 80)
(1 69 9 64)(2 70 10 65)(3 66 6 61)(4 67 7 62)(5 68 8 63)(11 76 16 71)(12 77 17 72)(13 78 18 73)(14 79 19 74)(15 80 20 75)(21 41 26 46)(22 42 27 47)(23 43 28 48)(24 44 29 49)(25 45 30 50)(31 51 36 56)(32 52 37 57)(33 53 38 58)(34 54 39 59)(35 55 40 60)
(1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)(21 22 23 24 25)(26 27 28 29 30)(31 32 33 34 35)(36 37 38 39 40)(41 42 43 44 45)(46 47 48 49 50)(51 52 53 54 55)(56 57 58 59 60)(61 62 63 64 65)(66 67 68 69 70)(71 72 73 74 75)(76 77 78 79 80)
(2 3 5 4)(6 8 7 10)(11 18 12 20)(13 17 15 16)(14 19)(21 33 22 35)(23 32 25 31)(24 34)(26 38 27 40)(28 37 30 36)(29 39)(41 43 42 45)(46 48 47 50)(51 58 52 60)(53 57 55 56)(54 59)(61 73 62 75)(63 72 65 71)(64 74)(66 78 67 80)(68 77 70 76)(69 79)

G:=sub<Sym(80)| (1,44)(2,45)(3,41)(4,42)(5,43)(6,46)(7,47)(8,48)(9,49)(10,50)(11,51)(12,52)(13,53)(14,54)(15,55)(16,56)(17,57)(18,58)(19,59)(20,60)(21,61)(22,62)(23,63)(24,64)(25,65)(26,66)(27,67)(28,68)(29,69)(30,70)(31,71)(32,72)(33,73)(34,74)(35,75)(36,76)(37,77)(38,78)(39,79)(40,80), (1,19,9,14)(2,20,10,15)(3,16,6,11)(4,17,7,12)(5,18,8,13)(21,31,26,36)(22,32,27,37)(23,33,28,38)(24,34,29,39)(25,35,30,40)(41,56,46,51)(42,57,47,52)(43,58,48,53)(44,59,49,54)(45,60,50,55)(61,71,66,76)(62,72,67,77)(63,73,68,78)(64,74,69,79)(65,75,70,80), (1,69,9,64)(2,70,10,65)(3,66,6,61)(4,67,7,62)(5,68,8,63)(11,76,16,71)(12,77,17,72)(13,78,18,73)(14,79,19,74)(15,80,20,75)(21,41,26,46)(22,42,27,47)(23,43,28,48)(24,44,29,49)(25,45,30,50)(31,51,36,56)(32,52,37,57)(33,53,38,58)(34,54,39,59)(35,55,40,60), (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30)(31,32,33,34,35)(36,37,38,39,40)(41,42,43,44,45)(46,47,48,49,50)(51,52,53,54,55)(56,57,58,59,60)(61,62,63,64,65)(66,67,68,69,70)(71,72,73,74,75)(76,77,78,79,80), (2,3,5,4)(6,8,7,10)(11,18,12,20)(13,17,15,16)(14,19)(21,33,22,35)(23,32,25,31)(24,34)(26,38,27,40)(28,37,30,36)(29,39)(41,43,42,45)(46,48,47,50)(51,58,52,60)(53,57,55,56)(54,59)(61,73,62,75)(63,72,65,71)(64,74)(66,78,67,80)(68,77,70,76)(69,79)>;

G:=Group( (1,44)(2,45)(3,41)(4,42)(5,43)(6,46)(7,47)(8,48)(9,49)(10,50)(11,51)(12,52)(13,53)(14,54)(15,55)(16,56)(17,57)(18,58)(19,59)(20,60)(21,61)(22,62)(23,63)(24,64)(25,65)(26,66)(27,67)(28,68)(29,69)(30,70)(31,71)(32,72)(33,73)(34,74)(35,75)(36,76)(37,77)(38,78)(39,79)(40,80), (1,19,9,14)(2,20,10,15)(3,16,6,11)(4,17,7,12)(5,18,8,13)(21,31,26,36)(22,32,27,37)(23,33,28,38)(24,34,29,39)(25,35,30,40)(41,56,46,51)(42,57,47,52)(43,58,48,53)(44,59,49,54)(45,60,50,55)(61,71,66,76)(62,72,67,77)(63,73,68,78)(64,74,69,79)(65,75,70,80), (1,69,9,64)(2,70,10,65)(3,66,6,61)(4,67,7,62)(5,68,8,63)(11,76,16,71)(12,77,17,72)(13,78,18,73)(14,79,19,74)(15,80,20,75)(21,41,26,46)(22,42,27,47)(23,43,28,48)(24,44,29,49)(25,45,30,50)(31,51,36,56)(32,52,37,57)(33,53,38,58)(34,54,39,59)(35,55,40,60), (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30)(31,32,33,34,35)(36,37,38,39,40)(41,42,43,44,45)(46,47,48,49,50)(51,52,53,54,55)(56,57,58,59,60)(61,62,63,64,65)(66,67,68,69,70)(71,72,73,74,75)(76,77,78,79,80), (2,3,5,4)(6,8,7,10)(11,18,12,20)(13,17,15,16)(14,19)(21,33,22,35)(23,32,25,31)(24,34)(26,38,27,40)(28,37,30,36)(29,39)(41,43,42,45)(46,48,47,50)(51,58,52,60)(53,57,55,56)(54,59)(61,73,62,75)(63,72,65,71)(64,74)(66,78,67,80)(68,77,70,76)(69,79) );

G=PermutationGroup([(1,44),(2,45),(3,41),(4,42),(5,43),(6,46),(7,47),(8,48),(9,49),(10,50),(11,51),(12,52),(13,53),(14,54),(15,55),(16,56),(17,57),(18,58),(19,59),(20,60),(21,61),(22,62),(23,63),(24,64),(25,65),(26,66),(27,67),(28,68),(29,69),(30,70),(31,71),(32,72),(33,73),(34,74),(35,75),(36,76),(37,77),(38,78),(39,79),(40,80)], [(1,19,9,14),(2,20,10,15),(3,16,6,11),(4,17,7,12),(5,18,8,13),(21,31,26,36),(22,32,27,37),(23,33,28,38),(24,34,29,39),(25,35,30,40),(41,56,46,51),(42,57,47,52),(43,58,48,53),(44,59,49,54),(45,60,50,55),(61,71,66,76),(62,72,67,77),(63,73,68,78),(64,74,69,79),(65,75,70,80)], [(1,69,9,64),(2,70,10,65),(3,66,6,61),(4,67,7,62),(5,68,8,63),(11,76,16,71),(12,77,17,72),(13,78,18,73),(14,79,19,74),(15,80,20,75),(21,41,26,46),(22,42,27,47),(23,43,28,48),(24,44,29,49),(25,45,30,50),(31,51,36,56),(32,52,37,57),(33,53,38,58),(34,54,39,59),(35,55,40,60)], [(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25),(26,27,28,29,30),(31,32,33,34,35),(36,37,38,39,40),(41,42,43,44,45),(46,47,48,49,50),(51,52,53,54,55),(56,57,58,59,60),(61,62,63,64,65),(66,67,68,69,70),(71,72,73,74,75),(76,77,78,79,80)], [(2,3,5,4),(6,8,7,10),(11,18,12,20),(13,17,15,16),(14,19),(21,33,22,35),(23,32,25,31),(24,34),(26,38,27,40),(28,37,30,36),(29,39),(41,43,42,45),(46,48,47,50),(51,58,52,60),(53,57,55,56),(54,59),(61,73,62,75),(63,72,65,71),(64,74),(66,78,67,80),(68,77,70,76),(69,79)])

38 conjugacy classes

 class 1 2A 2B 2C 2D 2E 2F 2G 4A 4B 4C 4D 4E 4F 4G ··· 4L 5 8A ··· 8H 10A 10B 10C 20A ··· 20F order 1 2 2 2 2 2 2 2 4 4 4 4 4 4 4 ··· 4 5 8 ··· 8 10 10 10 20 ··· 20 size 1 1 1 1 5 5 5 5 2 2 4 4 10 10 20 ··· 20 4 10 ··· 10 4 4 4 8 ··· 8

38 irreducible representations

 dim 1 1 1 1 1 1 1 1 2 2 2 2 2 4 4 4 4 4 8 type + + + + + + + + - + + + + + - image C1 C2 C2 C2 C2 C4 C4 C4 D4 D4 D4 SD16 Q16 F5 C2×F5 C2×F5 C22⋊F5 C22⋊F5 Q8⋊F5 kernel C2×Q8⋊F5 Q8⋊F5 C2×D5⋊C8 C2×C4⋊F5 C2×Q8×D5 C2×Dic10 Q8×D5 Q8×C10 C4×D5 C2×Dic5 C22×D5 D10 D10 C2×Q8 C2×C4 Q8 C4 C22 C2 # reps 1 4 1 1 1 2 4 2 2 1 1 4 4 1 1 2 2 2 2

Matrix representation of C2×Q8⋊F5 in GL6(𝔽41)

 1 0 0 0 0 0 0 1 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40
,
 1 39 0 0 0 0 1 40 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1
,
 30 11 0 0 0 0 15 11 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40
,
 1 0 0 0 0 0 0 1 0 0 0 0 0 0 40 1 0 0 0 0 40 0 1 0 0 0 40 0 0 1 0 0 40 0 0 0
,
 9 0 0 0 0 0 9 32 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0

G:=sub<GL(6,GF(41))| [1,0,0,0,0,0,0,1,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40],[1,1,0,0,0,0,39,40,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1],[30,15,0,0,0,0,11,11,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40],[1,0,0,0,0,0,0,1,0,0,0,0,0,0,40,40,40,40,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0],[9,9,0,0,0,0,0,32,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,0] >;

C2×Q8⋊F5 in GAP, Magma, Sage, TeX

C_2\times Q_8\rtimes F_5
% in TeX

G:=Group("C2xQ8:F5");
// GroupNames label

G:=SmallGroup(320,1119);
// by ID

G=gap.SmallGroup(320,1119);
# by ID

G:=PCGroup([7,-2,-2,-2,-2,-2,-2,-5,56,422,184,1684,438,102,6278,1595]);
// Polycyclic

G:=Group<a,b,c,d,e|a^2=b^4=d^5=e^4=1,c^2=b^2,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,c*b*c^-1=e*b*e^-1=b^-1,b*d=d*b,c*d=d*c,e*c*e^-1=b^-1*c,e*d*e^-1=d^3>;
// generators/relations

׿
×
𝔽


text:
# Far point

In visual perception, the far point is the farthest point at which an object can be placed (along the optical axis of the eye) for its image to be focused on the retina within the eye's accommodation. It is sometimes described as the farthest point from the eye at which images are clear. The other limit of eye's accommodation is the near point.

For an unaccommodated emmetropic eye, the far point is at infinity, but for the sake of practicality, infinity is considered to be 6 m because the accommodation change from 6 m to infinity is negligible. See visual acuity or snellen chart for details about 6/6 (m) or 20/20 (ft) vision.

For an unaccommodated myopic eye, the far point is closer than 6 m. It depends upon the refractive error of the person's eye .

For an unaccommodated hypermetropic eye, incident light must be converging before entering the eye so as to focus on the retina. In this case (the hypermetropic eye) the focus point is behind the retina in virtual space, rather than on the retina screen.

Sometimes far point is given in diopters, which refers to the inverse of the distance stated above (see Simple myopia). For example, an individual that can see clearly out to 50 cm would have a far point of ${\displaystyle {\frac {1}{50\ {\text{cm}}}}=2\ {\text{diopters}}}$.

## Vision correction

A corrective lens can be used to correct myopia by imaging an object at infinity onto a virtual image at the patient's far point. According to the thin lens formula the required optical power P is

${\displaystyle P\approx {\frac {1}{\infty }}-{\frac {1}{{\mathit {F}}P}}=-{\frac {1}{{\mathit {F}}P}}}$,[1]

where FP is the distance to the patient's far point. P is negative, because a diverging lens is required.

This calculation can be improved by taking into account the distance between the spectacle lens and the human eye, which is usually about 1.5 cm:

${\displaystyle P=-{\frac {1}{{\mathit {FP}}-0.015\;{\text{m}}}}}$.

For example, if a person has FP = 30 cm, then the optical power needed is P = −3.51 diopters where one diopter is the reciprocal of one meter.

## References

1. "Vision Correction | Physics". courses.lumenlearning.com. Retrieved 2019-12-05.


text:
# Inverse 4-parameters Cumulative Weibull Distribution

May someone write the inverse of the FOUR-parameters Weibull cumulative distribution (percent point function), please? I am referring to the 4-parameters formula appearing in Sigmaplot.

• There probably is no formula for the inverse. May 31 at 19:55
• Existing for the three-parameters distribution, I wonder if it also exists for the four-parameters one. Jun 1 at 6:29


text:
# nLab proof relevance

foundations

## Foundational axioms

foundational axiom

## Removing axioms

#### Constructivism, Realizability, Computability

intuitionistic mathematics

# Contents

## Idea

In constructive mathematics, proof relevance refers to the concept that mathematical proofs are mathematical objects themselves and that besides just knowing that a proposition has some proof it is relevant to remember the way a given proof is constructed (and to remember possibly different proofs of the same proposition).

This idea is formalized notably by the idea of propositions as types in type theory where a proof is an actual term (the proof term) of some type.

For instance logical disjunction “A or B” in type theory (in particular in homotopy type theory) may be exhibited by the sum type $A + B$ of the types representing the induvidual propositions, and a proof/term of $A + B$ carries in it the information of whether it proved $A$ or $B$ to (hence) prove their disjunction. Similarly, a proof of $C \to (A + B)$ will include the information of which proofs of $C$ lead to proofs of $A$ and which proofs of $C$ lead to proofs of $B$.

## References

### Prehistory

In

the following complaint about mathematical proof (in section 12 Historical and mathematical proof of the Preface) might be read as being a complaint about the traditional non-constructive concept of proof and about the traditional lack of proof relevance:

All the same, while proof is essential in the case of mathematical knowledge, it still does not have the significance and nature of being a moment in the result itself; the proof is over when we get the result, and has disappeared. The process of mathematical proof does not belong to the object; it is a function that takes place outside the matter in hand.

footnote 42: Mathematical truths are not thought to be known unless proved true. Their demonstrations are not, however, kept as parts of what they prove, but are only our subjective means towards knowing the latter. In philosophy, however, consequences always form part of the essence made manifest in them, which returns to itself in such expressions.

See also earlier conceptions of proofs expressing the ‘cause’ of a theorem, where proofs by contradiction in particular were taken generally to fail. Such an idea goes back to Aristotle for whom a proper answer to the question “Why is the angle in a semicircle a right-angle?” gives its cause.

• Paolo Mancosu, Philosophy of Mathematics and Mathematical Practice in the Seventeenth Century, OUP, 1996.

Last revised on July 19, 2017 at 08:59:07. See the history of this page for a list of all contributions to it.


text:
# How to Make 3D Text and Logos Using Cinema 4D Renderer in After Effects

Learn how to convert 2D text and shapes into 3D with After Effects. Creating 3D text, logos and shapes with Adobe After Effects is easier than you might think. In this tutorial we will learn how to use the Cinema 4D renderer that comes with the latest version of After Effects, and how we can use it to extrude text and shape layers to create your very own 3D logo.

## What You'll Learn in This Tutorial

In this tutorial we'll show you how to activate and use the Cinema 4D renderer to create a 3D project within Adobe After Effects. You'll learn:

• How to convert 2D text and shapes into 3D
• How to activate Cinema 4D renderer so that you can access the 3D tools available in After Effects
• How to use Lighting
• How to navigate around a 3D scene in Adobe After Effects

## How to Activate Cinema 4D Renderer

First let's create a new composition in After Effects. Take care to look at the settings and make sure that they are suitable for your project. For this example we will be using the following:

• Width: 1920 px
• Height: 1090 px
• Frame Rate: 30 fps

Next let's create a text layer by using the Text Tool (Control + T) and type in some text that you want to create.

Decide what font you want to use for the text. For this example we will be using RNS Sanz which you can find from the Envato Elements website.

You can also add some color to the text at this stage by using the Character Panel and choosing a specific fill color that you want to use.

Click on the small cube icon next to the text layer to turn it into a 3D layer. This will allow the layer to be manipulated in three dimensions and will also add some extra options to the bottom right of the preview panel.

And by clicking on the Renderer Dropdown Box you can choose between Classic 3D and Cinema 4D. Classic 3D is the traditional After Effects renderer whereas Cinema 4D will give you more advanced options and arguably a better final look to the project. So let's choose Cinema 4D.

Another way to activate the Cinema 4D renderer is by going into the Composition Settings and choosing the 3D Renderer tab.

From here you can choose the 3D renderer you want to use, and also see the details of what sort of functions and enabled or disabled when using them.

## How to Create 3D Text

Once the Cinema 4D Renderer has been activated, you can go into the text options in the layer panel. Under the transform options, you'll notice that there are some new geometry and material options available to use.

Under the Geometry options, increase the Extrusion Depth value to create some depth to the text and make it 3D.

To make the 3D effect clearer, let's add a light layer to the project (Layer > New > Light).

For this example we'll be using the following light settings:

• Light Type: Point
• Color: FFFFFF
• Intensity: 127%
• Falloff: Smooth
• Falloff Distance: 969

With the light now applied to the project, the 3D effect should now appear much stronger than before, with some lighting and shadow effects that help enhance the look of the text.

Next choose Angular for the bevel style and increase the value to your liking. For this example we'll be using a value of 5.

Once applied you'll notice that an angular bevel has been applied to the outside of your text.

## How to Convert a Flat Logo to 3D

For a flat logo, the process is very similar. First you need to import your logo. For this example we are using a vector logo created in Adobe Illustrator.

Click and drag it into your composition and then use the transform options to scale and position it in place.

Once you are happy with the way that it looks, right click on the logo layer and choose Create > Create Shapes from Vector Layer.

This will create a new shape layer. Click on the small cube icon next to the shape layer to turn it into a 3D layer.

Then under the Geometry options, you should find the same type of options that were available to you with the text layer. Adjust the Extrude and Bevel options to your liking so that they match the look of your 3D text.

## How to Add a Camera and Light

Select the light layer and in the viewport you'll notice that it will be represented by a 3D gizmo. It's a little tool that has three handles or arrows that let you position the object or rotate the object in three dimensions (you can adjust the X, Y, and Z dimensions).

In fact, if you open up the object options and take a look at the transform properties, you'll see that a lot of these now have a third dimension that we can adjust. Adjust the position of the light to your liking so that the light effects both the logo and the text.

Next let's throw in an after effects camera. You can do this by right clicking down in a blank area and going to New > Camera.

Select a 35 mm preset and click okay.

From there you can use the different camera controls to manipulate the scene and get a view that you want. Simply press the C key on the keyboard to go through each of the camera controls which include:

• Orbit
• Pan
• Dolly

## Congratulations

Congratulations! And that's how you can make convert 2D text and logos to 3D in Adobe After Effects. Now that you've learned the basics, check out some of the other tutorials we have on Adobe After Effects below.

I hope you've found this tutorial useful and I'll see you next time on Envato Tuts+!