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### Wave Diffraction by Two Strips with Fractional Boundary Conditions
Vasil Tabatadze1, 2, *, Kamil Kara¸cuha1, and Eldar I. Veliyev1, 3
Abstract—In this article, a solution of the plane wave diffraction problem by two axisymmetric strips with different dimensions is considered. Fractional boundary conditions are required on the surface of each strip. Several cases of strip’s dimension, configurations, and fractional orders are considered, and numerical results are obtained. The near electric field distribution, Total Radar Cross Section frequency characteristics, and the Poynting vector distribution in the vicinity of these strips are calculated and illustrated. For the fractional order 0.5, the solution is found analytically.
1. INTRODUCTION
In the last decades, the tools of the fractional calculus which use fractional operators have an increasing number of applications in physics and electrodynamics. The interest in fractional operators is related to the fact that fractional operators broaden the notion of the initial operator. If the initial operator is related to the description of some canonical cases, the fractional operators with non-integer order can describe a new problem with new properties. Therefore, fractional operators generalize the class of problems studied in different areas of the science [1–3].
Applications of fractional operators in electrodynamics were developed by Engheta and then, Veliev and Ivakhnychenko [1–5]. Engheta proposed a fractional paradigm which states that if the function and its first derivative describe two canonical states of the electromagnetic field, then the fractional derivative describes an intermediate state of the field between the canonical states [1]. Veliev and Ivakhnychenko developed this idea further and considered the generalization of boundary conditions. In this article, there are introduced fractional boundary conditions (FBC) which correspond to intermediate boundary conditions between the perfect electric conductor and perfect magnetic conductor [6]. These boundary conditions describe a new material which was studied in the previous works [7–11]. In these works, electromagnetic scattering problems were considered such as the diffraction by one strip, two strips with the same length and a half-plane with fractional boundary condition where the theoretical part is formulated, and for one strip case, some numerical results are also obtained. Some other authors also have publications in this direction and made contributions in fractional calculus development in electrodynamics [12–15].
This study more deeply investigates the properties of such a new material which satisfies the fractional boundary condition. The theoretical description of electromagnetic plane wave scattering by the two strips with different dimensions is given in the theoretical section. After that, the total electric field distribution and Total Radar Cross Section (TRCS) are investigated and obtain numerical results. In the general case, the solution of the problem is reduced to a solution of a system of linear algebraic equations (SLAE) which can be obtained with a given accuracy. However, for the values of the fractional order (FO) α= 0.5, the solution is obtained in an analytical form which is the advantage
Received 25 June 2019, Accepted 29 August 2019, Scheduled 16 September 2019
* Corresponding author: Vasil Tabatadze ([email protected]).
1 Informatics Institute of Istanbul Technical University, Maslak, Istanbul 34469, Turkey. 2 Tbilisi State University, 1 Chavchavadze
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of the fractional derivative method (FDM) [7–9]. In results of the numerical simulation part, TRCS shows the presence of resonances which are associated with the excitation of quasi-self-oscillations of a double strip resonator.
The method represented in the paper is an efficient and new fundamental approach for diffraction problems because the method requires one integral equation and one current density for each strip, whereas by using impedance boundary condition, two current densities namely, electric and magnetic current densities, and two integral equations for each strip are needed [7].
2. THEORETICAL PART
In this problem, there are two axisymmetric strips with different widths and infinite length in z-axis located at y = ±l planes. The widths of the strips are 2a1 and 2a2, respectively. In Figure 1, the
geometry of the problem is given.
Figure 1. The geometry of the problem.
As an incident field, there is anE-polarized electromagnetic plane wave. The electric field has only a z component and can be written as Ezi =e−ik(xcosθ+ysinθ) wherek is the wavenumber, andθ is the
angle of incidence. A total electric field is given asEz =Ezi+EzswhereEzsstands for the total scattered electric field by both strips. The total electric fieldEz should satisfy the fractional boundary condition (FBC) on each strip. The fractional boundary condition is given in Eq. (1).
Dkyα Ez(x, y) = 0, y → ±l, x∈(−ai, ai), i= 1,2 (1) where α is a fractional order (FO); the fractional derivative is given in Eq. (1); Dkyα is determined by the Riemann-Liouville equation [4, 5, 16] which has the form as Eq. (2).
yf(y) = 1 Γ (1−α)
d dy
y
−∞
f(t)
(y−t)αdt, (2)
where, Γ(1−α) is Gamma Function, andα∈(0,1).
When α= 0, the boundary condition corresponds to the perfect electric conductor (PEC). When α= 1, it stands for the perfect magnetic conductor (PMC). For FO between 0< α <1, FBC corresponds to the intermediate case between PEC and PMC. Previous studies show that when fractional order is in this range, the strip boundary corresponds to the material with a pure imaginary impedance which is called fractional impedance. The relation between the fractional order and fractional impedance is given with the following formula,ηα=sintan
π
2α
for any incidence angle [7, 8]. From this expression, it
is easy to get the next formula α= 1 ln
1−ηαsinθ 1+ηαsinθ
[4, 7, 9].
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be solved [8, 9]. In order to formulate the problem, first, time dependency and field expressions need to be defined. Here, throughout formulation, the time dependency is e−iωt, and it is omitted everywhere. The scattered electric field Ezs(xy) is the sum of two parts which are the scattered fields by the first and second strips, respectively. Total scattered electric field isEzs(x, y) =Es1
z (x, y) +Ezs2(xy), and each part can be found by Eq. (3) [9].
Esjz (x, y) =
−∞
fj1−αxGαx−x, yjdx, j = 1,2, y1 =y−l, y2=y+l, (3)
where,
Gαx−x, y=i 1 4πD
α ky −∞ eik
q(x−x)+|y|√1−q2
(1−q2)α−21dα.
Here,f1−α(x) is the fractional density of surface current on the strip. In Eq. (3),j= 1,2 corresponds to the upper or lower strip, respectively, and(x−x, y) is the fractional Green’s function which is, in the two-dimensional case,(x−x, y) =4iDkyα H0(1)(k (x−x)2+y2) [5, 7, 8]. In the expression,H
(1) 0 (z)
is the Hankel function of the first kind and zero order. Note that in Eq. (3), the plane wave expansion of Hankel function is used. Then, Eq. (4) is achieved by representing fj1−α(x) with its corresponding Fourier transformFj1−α(q) [9, 10].
Ezsj(x, yj) =−ie
±iπ
2α 4π
−∞F
1−α
j (q)eik
xq+|yj|√1−q2
1−q2α−21 dq (4)
wherey1 =y−l,y2 =y+l,j= 1,2, and
Fj1−α(q) =
1
1
˜
fj1−α(ξ)e−iεqξdξ,
˜
fj1−α(ξ) = 2επj
−∞
Fj1−α(q)eiεjqξdq,
˜
fj1−α(ξ) = afj1−α(ξ), j =kaj, ξ = ax j.
Equations (3) and (4) are chosen so that the total electric field Ez satisfies Helmholtz equation, and also, the scattered electric fieldEzs satisfies Sommerfeld radiation condition at the infinity [7, 9].
After applying the FBC on the surface of each strip and multiplying both sides of the integral equation (4) by aia
ie
−ikxτdx, the integral equation system is as Eq. (5) where the upper sign stands
fori= 1, and, for the lower sign,i= 2 needs to be taken into account. Different signs in the exponent (e∓iklsinθ) correspond to the upper and lower strips, respectively. Keep in mind that (±i), which comes from the fractional boundary condition, is given in Eq. (1). Here, (+) sign corresponds to the upper part of each strip, and (−) sign stands for the lower part of each strip.
−∞F
1−α
i (q)
sin (kai(q−τ)) (q−τ)
1−q2α−21 dq=−i4π(±i)αe∓iklsinθsinαθsin(kai(τ + cosθ)) (τ + cosθ)
−∞F
1−α
j (q)
sin (kaı(q−τ)) (q−τ) e
ik2l√1−q21
q2α−12 dq, i, j= 1,2, i=j (5)
In order to solve the set of integral equations (IE) in Eq. (5), the IE is reduced to the system of linear algebraic equation by presenting the unknown ζnαi as series of the Gegenbauer polynomials given in Eq. (6) [9–11]. Here, the Gegenbauer polynomials series for the normalized current density are used with the weighting function (1−ξi)α−12 in order to satisfy well-known Meixner’s edge condition [17].
˜
fi1−α(ξi) = (1−ξi)α−12
n=0
ζnαiC α n(ξi)
(4)
The corresponding Fourier transform is shown in Eq. (7) [9–11].
Fi1−α(q) = 2π Γ (α+ 1)
n=0
(−i)nαiβnαJn+α(iq)
(2iq)α , (7)
where,
i=kai, βnα = Γ (Γ (nn+ 2+ 1)α), i= 1,2.
The weighting function (1−ξi)α−12 in Eq. (6) is chosen so that it satisfies edge conditions [17]. After Eq. (7) is put in Eq. (5), then multiplied by Jk+ατα(1τ) and Jk+ταα(2τ) for corresponding integral
equations i= 1 and i= 2, respectively, the overall result for the upper and lower strips is as Eq. (8). Note that IE in Eq. (5) is integrated with parameter τ between (−∞,∞) in order to get Eq. (8).
n=0
(−i)nαiβnαCknα(i,i)+
i j
α
n=0
(−i)nαjβnαCknα(i,j)=γki,α, i, j= 1,2, i=j (8)
Here,
Cknα(i,i) =
−∞
Jn+α(iq)Jk+α(iq)
1−q2α−12 q2α dq,
Cknα(i,j) =
−∞
Jn+α(jq)Jk+α(iq)eik2l
1−q2
1−q2α−12
q2α dq, i=j,
γki,α = −i2(±i)α(2i)αΓ (α+ 1)e∓iklsinθtanαθ(1)kJk+α(icosθ).
It can be shown that the system of linear algebraic equation (SLAE) (8) can be reduced to SLAE of the Fredholm type of the second kind [9]. Then, the coefficients ζnαj can be found with any desired accuracy (within the machine precision) by using the truncation of SLAE (8). Note that, for the fractional order α = 0.5, the solution can be found analytically [8, 9]. After applying the boundary condition for FOα= 0.5, analytically, Fourier transform of the current density in the upper and lower strips can be found by using Eq. (5), and the result is given in Eq. (9).
Fi0.5(τ) =
−i4eiπ4 sinθ
sin(kai(τ + cosθ)) (τ + cosθ) e
∓iklsinθsin(kaj(τ+ cosθ))
(τ+ cosθ) e
±iklsinθ
eik2l
1−τ2
1−eik4l√1−τ2 (9)
wherei, j= 1,2,i=j.
Having expression for Fi1−α in Eq. (7) and Fi0.5 in Eq. (9), the radiation pattern of the scattered field in the far zone can be found by using Eq. (10). Using the stationary phase method for ka→ ∞, the scattered electric field gets the next form:
Ezs(x, y) =A(kr) Φα(φ). (10)
Here,
A(kr) =
2 πkre
ikr−iπ
4 , and Φα = Φα1 (φ) + Φα2(φ),
where,
Φα1 (φ) = −i 4e
±iπα
2 F11−α(cosφ) (sinαφ)e−iklsinφ,
Φα2 (φ) = −i 4e
±iπα
2 F21−α(cosφ) (sinαφ)e+iklsinφ.
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field in the far zone. The radiation pattern (RP) Φα(φ) expression in Eq. (10) is used for Total Radar Cross Section calculation [18]. The definition of the Total Radar Cross Section is given in (11).
σt=
2π
0
|Φα|2 (11)
As it has already been mentioned, the total electric fieldEz has only az-component, and the magnetic field hasxandy componentsHx andHy, respectively. The Poynting vector S hasx andy components and is calculated by using the formula in Eq. (12) [19].
S = 1
2Re [E×H
], S
x =12ReEzHy∗, Sy = 12Re [EzHx∗] (12)
Here, () denotes the complex conjugate.
3. RESULTS OF NUMERICAL SIMULATIONS
Based on the described mathematical algorithm, the program package was created in MatLab, and several cases of strips dimension and fractional orders were considered. The near field distribution was calculated, and the Total Radar Cross Section dependence on the wavenumber was constructed. The results are given below.
First, the case in which the strips have the same length is considered. Figure 2 shows the family of graphs depending on different fractional orders. Total Radar Cross Section (σT) has dependency on = ka for different Fractional orders (α = 0.5,0.75,1) (normal incidence). As seen, 2 strips have resonance frequencies for all values of the fractional orders which is investigated in this article. There exists an analytical expression to find the resonant frequency for such a quasi-resonator when the strips have the same dimensions. For the resonant frequency, the analytical expression gives kl πn2 (n = 1,2,3, ...) [19]. For the first resonance (n = 1), the value of kl is equal to kl π2. For n = 2, kl π. Our results in Total Radar Cross Section by using fractional derivative approach correspond to the similar outcome with analytical results. In Figures 2, 3, and 5, the wavenumber (k) is changed while the widths of the strips (a1 =a2=a= 1) are fixed, and they coincide to the half of the distance
between the strips (l=a1 =a2 =a). The resonant values ofka, which in these cases, are equal to kl,
are close to the analytical values. The first resonance is atka≈1.9, and the second one is atka≈3.4. The deviation from analytical values is because the analytical formulation is derived for the case of the strip which has much larger width than the wavelength (a λ= 2kπ) [19].
Figure 3 shows a family of graphs corresponding to different angles of incidence (θ = π6432) at α = 1. For θ = π642, the resonances at ka 1.9 and ka 3.4 are preserved, but for θ = π3, it disappears.
Figure 2. Total Radar Cross Section σT when a=a1 =a2= 1,l= 1,θ= π2 forα= 0.5, 0.75, 1.
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If the distance between the strips is decreased (l= 0.5), the first resonance atka≈1.9 practically disappears for all angles of incidence because in this case, the widths of the strips are twice of the distance between the strips (2l) (a = a1 = a2 = 2l). The analytical value of the first resonance is no
longer at π2, but it is atπ. Therefore, a resonance appears at ka≈3.4 which is close to π. This result is shown in Figure 4. It is also considered the case when the fractional order is α = 0.5, and then, the corresponding family of graphs is constructed. This is shown in Figure 5. As seen for this case, resonance values are higher. In the previous case (Figure 3 andα= 1), first resonance values are up to 5, and the second resonance values are reaching 3. On the other hand, forα= 0.5 (Figure 5), the first resonance values are reaching 6, and the second ones are up to 18. Figure 6 represents the total electric field Ez distribution. This field corresponds to the resonant value (ka1 = 3.4). As seen between the
strips, there exist high field values, and these high field values correspond to standing wave. Figure 7 gives the total field Ez distribution at the first resonance, (ka1 = 1.9). The field maximum in Figure 7
is higher than the field maximum in Figure 6 (approximately 6 versus 4) as expected from Figure 3. The amplitudes of the total electric fields are given for different configurations in Figures 6–12.
Figure 8 shows the total electric fieldEz distribution at a non-resonant frequency (ka1 = 1.5). For
this case, high field values are not observed between the strips. After that, the total electric field Ez distribution for intermediate fractional orders is considered. Figure 9 shows the total electric field Ez distribution at the resonance frequency when α = 0.5. The field maximum is 9 which is higher than that in the case of Figure 6. Figure 10 shows the total electric field Ez distribution for α = 0.75. In this case, the field maximum reaches 10.
Figure 4. Total Radar Cross Section σT when a = a1 = a2 = 1, l = 0.5, α = 1 and θ =
π
6432.
Figure 5. Total Radar Cross Section σT when a = a1 = a2 = 1, l = 1, α = 0.5 and θ =
π
6432.
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Figure 7. Total electric fieldEz,a1 =a2 = 1,l= 1,α= 1, ka1= 1.9,θ= π2.
Figure 8. Total fieldEz,a1 =a2= 1, l= 1, α= 1,ka1 = 1.5,θ= π2.
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Figure 10. Total electric fieldEz,a1=a2 = 1, l= 1, α= 0.75,ka1= 3.4, θ= π2.
Figure 11. Total electric fieldEz,a1=a2 = 1, l= 1, α= 0.75,ka1= 3.4, θ= π3.
Figure 11 shows the result for the incidence angle θ= π3 instead of θ = π2. As seen, in this case, there exists a resonant field between the strips, but it has a smaller amplitude with respect to the normal incidence case as given in Figure 10 (the maximum amplitude is 4 versus 10). When the incident angle is changed and taken asθ= π4, the results are as given in Figure 12.
For this case, the resonance disappears. The high field values and standing waves between the strips do not exist.
Now, the investigation is done when the strip’s dimensions are different: a1 = 3 and a2 = 4.
Figure 13 shows the Total Radar Cross Section dependence on =ka. As seen, high resonances are observed at ka14.8 and 9.6. The corresponding wavenumber values (k) arek≈1.6 and k≈3.2,
respectively. Because the half of the distance between the strips (l) is equal to 1, the corresponding values ofklarekl= 1.6 andkl= 3.2, respectively. This means that by increasing the dimensions of the strips, the resonance frequencies approach analytical values given in the formula kl≈ πn2 (n= 1,2, ...) compared to Figure 2 [19]. The Maximum value of Total Radar Cross Section value reaches 60 at the second resonanceka1 = 9.6 and the fractional orderα= 0.25. If the incidence angle is taken as θ= π4,
the resonance maximum is shifted, broadened, and decreased. This is seen in Figure 14.
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Figure 12. Total field Ez,a1 =a2 = 1,l= 1,α= 0.75, ka1 = 3.4,θ= π4.
Figure 13. Total Radar Cross Section σT when a=a1 = 3,a2 = 4,l= 1,θ= π2.
Figure 14. Total Radar Cross Section σT when a=a1 = 3,a2 = 4,l= 1,θ= π4.
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Figure 16. Total electric fieldEz,a1= 3, a2= 4, l= 1, α= 0.75,ka1 = 9.5,θ= π2.
Figure 17. Total electric fieldEz,a1= 3, a2= 4, l= 1, α= 0.75,ka1 = 9,θ= π3.
(11)
for different configurations.
Now, the total electric field Ez distribution is calculated atα = 0.75 and compared to Figure 10. The corresponding total fieldEz distribution is given in Figure 16.
The total electric field Ez distribution on this figure is similar to Figure 10. The only difference is the higher field amplitudes in this case. In Figure 10, the maximum amplitude is approximately 10. On the other hand, in Figure 16, the maximum amplitude is approximately 90.
If the incidence angle is changed and taken asθ= π3, the total electric fieldEz distribution will get the next form which is shown in Figure 17.
As seen, there is still a resonance, but the amplitude of the field between the strips is much smaller, and it is approximately 4. By decreasing the incidence angle, the resonance becomes weaker.
Figure 18 represents the total electric field Ez distribution for the fractional orderα= 0.01. This is very close to the perfect electric conductor. The maximum field value is 3.5 which is smaller than α= 0.25 and α= 0.75 cases considered above.
Figure 19. Poynting vector distribution a1 = 1,
a2= 1, l= 0.5, α= 1,ka1 = 1,θ= π2.
Figure 20. Poynting vector distribution a1 = 1,
a2= 1, l= 0.5, α= 0.01,ka1= 1, θ= π2.
Figure 21. Poynting vector distribution a1 = 1,
a2= 1, l= 0.5, α= 0.5,ka1= 1, θ= π2.
Figure 22. Poynting vector distribution a1 = 1,
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Distribution of the Poynting vector given in Eq. (13) is also calculated for different cases. Figure 19 and Figure 20 show the Poynting vector distribution for α = 1 andα = 0.01, respectively. In the case of Figure 19, diffraction problem corresponds to the PMC case, and the energy penetrates through the strips. On the other hand, Figure 20 stands for PEC case at a non-resonant case (ka1 = 1), so energy
flows around the strips and could not penetrate between the strips which creates a shadow region. In other words, in the region between the strips, the Poynting vectors have smaller length.
Figure 21 and Figure 22 show the Poynting vector distribution for α = 0.5 and α = 0.75, respectively. In Figure 21, the energy goes around the first strip but penetrates through the second strip from both sides. In the case of Figure 22, it is close to PMC due to having fractional order α = 0.75, so the flow of the energy is similar to Figure 19.
The Poynting vector distribution is also calculated when strip dimensions and the distance between them are different. Figure 23 and Figure 24 show the Poynting vector distribution whena1= 3, a2 = 4
forα= 1 and α= 0.01, respectively. In Figures 23 and 24, it is seen that the energy cannot penetrate between the strips.
Figure 23. Poynting vector distribution a1 = 3,
a2= 4, l= 1,α= 1,ka1 = 3,θ= π2.
Figure 24. Poynting vector distribution a1 = 3,
a2= 4, l= 1, α= 0.01,ka1 = 3,θ= π2.
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Finally, Figure 25 shows the Poynting vector distribution when a1 = 3, a2 = 4 for α = 0.5. Here,
vortexes are seen on the upper part.
4. CONCLUSION
In this article, a new mathematical method for the solution of the diffraction problem by the two axisymmetric strips with different widths and fractional boundary conditions is considered. The formulation for plane wave diffraction by the double strips with different sizes is done, and for the FO α= 0.5 case, an analytical expression is found. Numerical experiments are conducted for different dimensions of the strips, and Total Radar Cross Section frequency characteristics show that for fractional order between 0 and 1, the resonance peaks are increased, and we get a resonator with higher values of Total Radar Cross Section. The total electric field Ez distribution confirms the existence of high field values between the plates.
ACKNOWLEDGMENT
The authors of this article would like to express the gratitude to Prof. Dr. Nader Engheta and Prof. Dr. Ertu˘grul Kara¸cuha for their useful discussions.
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References
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Turkish exports data is used to determine the structural change in Turkish foreign trade because there is not a significant change occurred in Turkish imports between | 7,981 | 26,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.90876 |
http://www.physicsforums.com/showthread.php?s=b3ea40159cce81ba91a60f9036fd4bad&p=4814805 | 1,411,134,623,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657131376.7/warc/CC-MAIN-20140914011211-00207-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 735,410,657 | 9,924 | # Do AM waves have symmetrical amplitude?
by ChromeBit
Tags: amplitude, symmetrical, waves
P: 19 I've been finding that many videos online depict AM waves (the modulated wave that is transmitted from a circuit/received by your aerial before demodulation) differently. Some show the waves as having symmetric amplitude so at a point (p) on the wave, the voltage is equal to +a and -a. Overall the wave has a voltage of zero (I'm not sure if I've used the correct term here can someone advise me?); I've heard this is why crystal radios have a diode - because one half of the wave must be removed to avoid the wave cancelling out. However, I've seen other sources which don't show an identical wave. Can anyone help me?
Sci Advisor PF Gold P: 2,085 A wave cannot simultaneously have amplitudes +a and -a. The voltage oscillates rapidly at the carrier frequency f0 (which is 1000 kHz for "AM station 100"), while the audio appears as a slow modulation of the carrier amplitude. Thus, when the modulation has the value a, the actual signal oscillates between +a and -a every 1/f0 seconds.
Mentor P: 12,015 In addition to what marcusl said, remember that the AM transmission also has sidebands which will vary in frequency and amplitude as well.
P: 759
Do AM waves have symmetrical amplitude?
Quote by ChromeBit I've heard this is why crystal radios have a diode - because one half of the wave must be removed to avoid the wave cancelling out.
Yes, that is one way to think of it. Once the + or - side is removed you can then filter the rf out and recover the original audio.
This image shows an rf signal modulated by a single audio tone. You cannot see the individual rf waves in this image because the sweep frequency of the oscilloscope is too slow to resolve it. But you can clearly see the audio signal increasing and decreasing it's amplitude. Thus the term "amplitude modulation".
Quote by ChromeBit However, I've seen other sources which don't show an identical wave. Can anyone help me?
Is This the other way you have seen it? This image shows the frequency spectrum of the same am signal. It's just another way of viewing the same thing. The horizontal axis is the frequency and the vertical axis is the amplitude. The peak in the center is the carrier and the two peaks on the left and right of it are the sidebands produced by the audio modulation. The difference in frequency between the carrier and the two sidebands are the same as the single tone audio modulation. The crystal diode in the receiver mixes the carrier with the sidebands to produce a difference and a sum frequency, of which the difference frequency is the original single tone audio.
PF Gold P: 82 ... When you see equal opposing amplitudes, you are actually looking at an "envelope" AM (amplitude modulated) waveform such as TurtleMeister has shown in his top pic above. As he mentions, the actual RF fine-lined high frequency dosn't show up here, inside the modulation envelope, because the CRT scope "drawing" resolution is too poor. In AM, the broadcast transmitter takes the coarse low-frequency audio envelope wave you see directly from a microphone (or such) and increases, or decreases, (modulates) the envelope at this corresponding audio frequency. The antenna "pulses" it into the air. Since the top and bottom seem to cancel out (average out) at audio frequencies, they are cut in half (decoded) by a diode and the high frequency half "RF filler" discarded. The remaining envelope-half rises and falls from zero, drives a speaker, and you hear the sound wave only. Since the modulation process is a pulsing voltage, pulses from lightening often interfere. Some crude depictions might show the envelope already cut in half, or even just the audio, and claim it is the AM signal, so keep this in mind. FM (frequency modulation) works similar except the envelope is a set of continuously straight (instead of wavy) lines above and below the contained high frequency RF. In this case, one would need to see the fine RF lines in between the outer envelope limits to notice that the inner high speed waveform spreads (audio freq lowers), or contracts tighter (audio freq goes up) to recognize from the picture that a lower modulated frequency has been imposed upon the RF as an audio signal. Since the decoding receiver does not respond to pulses or voltage spikes (just frequency variations), lightening, and other sparks, have very little effect making a nice quiet background for the signal. EDIT: From wikipedia http://en.wikipedia.org/wiki/Amplitude_modulation You can see that "AM waves do not have symmetrical amplitude" in that each RF wave alternates up, then down. Wes ...
P: 19 Thanks everyone, this cleared nearly everything up!
Thanks
PF Gold
P: 12,269
Quote by ChromeBit Thanks everyone, this cleared nearly everything up!
Read as much as you can about this. There are many really half baked descriptions and 'explanations' to be found on the Web. There is a mix of theoretical (ideal) explanations and practical examples where things are far from ideal. One thing you should bear in mind is that it is very rate, in practice, for an AM signal to be carrying just a simple tone so the envelope may look any shape.
P: 18 Perhaps you are seeing something being confused with SSB (single side-band) signalling, though that is a kind of FM rather than AM. No, the signals you get from the air must have a zero DC component, which means they are symmetrical, and rectification (what used to be called "detection") is required in order to receive them. DC cannot be transformed, nor can it be transmitted. | 1,223 | 5,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2014-41 | latest | en | 0.964148 |
http://www.csm.ornl.gov/comp_chemistry/qm/quantummechanics/source/fock.f | 1,495,648,347,000,000,000 | text/plain | crawl-data/CC-MAIN-2017-22/segments/1495463607849.21/warc/CC-MAIN-20170524173007-20170524193007-00195.warc.gz | 469,189,668 | 2,196 | program fock c ****************** Program FOCK ****************** c This program is designed to read in the LCAO-MO c coefficient matrix, the one- and two-electron AO c integrals and to form a CLOSED SHELL Fock matrix c (i.e., a Fock matrix for species with all doubly c occupied orbitals). c c Copyright (C) 1992 Jeff Nichols and Jack Simons c c ************************************************** implicit real (a-h,o-z) include "limits.h" logical yesno character*30 dfname, fname, dscrpt c c Set core needed to: c 3 square arrays of order maxorb c 1 array of dimension canonical(maxorb**4) c total = 3*mxosq + mxo4 c dimension pcore(3*mxosq + mxo4) dimension ipntr(10) itotal = 3*mxosq + mxo4 + 10 call modinfo('fock ') call shwmem('fock ',itotal) call goon 10 continue write(*,*) write(*,*)' Input the number of orbitals in your system: ' read(*,*)iorb if(iorb.gt.maxorb)then write(*,1010)maxorb goto 10 endif ipntr(1) = 1 ipntr(2) = ipntr(1) + iorb*iorb ipntr(3) = ipntr(2) + iorb*iorb ipntr(4) = ipntr(3) + iorb*iorb iorb2 = iorb*(iorb+1)/2 iorb4 = iorb2*(iorb2+1)/2 ipntr(5) = ipntr(4) + iorb4 c c Open the MO-AO coefficient file. c dscrpt = 'MO-AO coefficients' ids = 18 dfname = 'mocoefs.dat' idf = 11 fname = ' ' ifn = 30 iflnum = 31 call rfile(fname,ifn,dscrpt,ids,dfname,idf,iflnum,yesno) if(yesno)then call rdmatd(fname,ifn,dscrpt,ids,pcore(ipntr(1)),iorb, & iorb,iflnum) c c Close MO-AO coefficients file. c close(unit=31) else call rdmatu(dscrpt,ids,pcore(ipntr(1)),iorb,iorb) endif call wtmatu(dscrpt,ids,pcore(ipntr(1)),iorb,iorb) c c Open the integral file. c dscrpt = '1e- and 2e- integrals' ids = 21 dfname = 'ao_integrals.dat' idf = 16 fname = ' ' ifn = 30 iflnum = 30 call rfile(fname,ifn,dscrpt,ids,dfname,idf,iflnum,yesno) if(.not.yesno)then write(*,*)' You must have the AO integrals available ' write(*,*)' on disk ... this can be done by running ' write(*,*)' program INTEGRAL. ' call qmexit endif call driver(pcore(ipntr(1)),pcore(ipntr(2)),pcore(ipntr(3)), & pcore(ipntr(4)),iorb,iorb2,iorb4) call qmexit stop 1010 format(' Maximum number of orbitals is ',I2,'. ',/, & ' Please try again. ') end subroutine driver(c,p,f,g,n,n2,n4) implicit real (a-h,o-z) logical yesno character*30 dfname, fname, dscrpt dimension c(n,n),p(n,n),f(n,n),g(n4) c c Get the 1e- integrals. c call get1e(g,n,n2) c c Put the 1e- Hamiltonian directly into the Fock matrix. c ij = 0 do 2 i = 1,n do 1 j = 1,i ij = ij + 1 f(i,j) = g(ij) f(j,i) = f(i,j) 1 continue 2 continue c c Get the 2e- integrals. c call get2e(g,n,n4) c c Close 1e- and 2e- integrals file. c close(unit=30) 5 continue c c Determine the number of closed shell orbitals. c write(*,*)' Please input the number of doubly occupied ' write(*,*)' orbitals. ' read(*,*)ndocc write(*,1000)ndocc if(ndocc.gt.n)then write(*,*)' The number of doubly occupied orbitals cannot ' write(*,*)' exceed the total number of orbitals. Please ' write(*,*)' try again. ' goto 5 endif c c Calculate the charge bond order matrix P(MR,MS). c do 30 mr = 1,n do 20 ms = 1,n p(mr,ms) = 0.0e0 do 10 k = 1,ndocc p(mr,ms) = p(mr,ms) + c(mr,k)*c(ms,k) 10 continue 20 continue 30 continue c c Write out the charge bond order matrix. c write(*,*)' The charge bond order matrix: ' call matout(p,n,n,n) c c Form the CLOSED SHELL Fock matrix. c do 70 mu = 1,n do 60 nu = 1,n do 50 mr = 1,n do 40 ms = 1,n icx1 = icanon(mu,mr,nu,ms) icx2 = icanon(mu,mr,ms,nu) f(mu,nu) = f(mu,nu) + & p(mr,ms)*(2.0e0*g(icx1) - g(icx2)) 40 continue 50 continue 60 continue 70 continue dscrpt = 'Fock matrix' ids = 11 call wtmatu(dscrpt,ids,f,n,n) dfname = 'fock.out' idf = 8 fname = ' ' ifn = 30 iflnum = 32 call wfile(fname,ifn,dscrpt,ids,dfname,idf,iflnum,yesno) if(yesno)then call wtmatd(fname,ifn,dscrpt,ids,f,n,n, & iflnum) c c Close Fock matrix file. c close(unit=32) endif return 1000 format(' There is ',I2,' doubly occupied orbital(s). ') end | 1,431 | 3,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-22 | longest | en | 0.554687 |
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``` If a=3,b=4,c=5 and a+b+c=0 then angle between a and b is
```
7 years ago
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``` Dear pallavi
|a|=3,|b|=4,|c|=5
a+b+c=0
a+b =-c
|a+b| =|-c|
|a+b|2 =|-c|2
|a|2 +|b|2 +2a.b =|c|2
9 + 16 +2a.b= 25
a.b=0
so a and b are perpendicular
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7 years ago
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## Basic Construction 2
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In $\mathbb{R}^3$ under the standard inner product, the length of
$$\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$$
is 1. However, under the innner product
$\left&space;\langle&space;\overrightarrow{x},&space;\overrightarrow{y}&space;\right&space;\rangle&space;=&space;a&space;x_1&space;y_1&space;+&space;x_2&space;y_2&space;+&space;x_3&space;y_3$
where a>0, the length becomes
$\left&space;\|&space;e_1&space;\right&space;\|&space;=&space;\sqrt{\left&space;\langle&space;e_1,&space;e_1&space;\right&space;\rangle}&space;=&space;\sqrt{&space;a&space;\times&space;1&space;\times&space;1+0\times&space;0+0\times&space;0}&space;=&space;\sqrt{a}$
It is obvious to me that under the standard inner product the unit vector has length of 1; also I know the above calculation of the length under the non-standard inner product is following the definition of length. However, how would one visualize the length of the unit vector being $\sqrt{a}$ under the non-standard inner product defined above ?
Similarly, I have problem visualizing the fact that
$$\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$$
and
$$\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$$
are not orthogonal to each other under the non-standard inner product.
• For the first part, you could perhaps tihink of a tape measure along the $x$-axis marked in both metres and feet? For the second part, those two vectors are orthogonal under the particular non-standard metric you have given. Feb 10, 2021 at 22:17
• Thanks for your comment. Those two are indeed orthogonal - it was my mistake. I was meant to write some other metric. Also you mentioned metres and feet - it starts to make a little sense to me now. Feb 10, 2021 at 22:43
While vectors under norm generated by standard inner product can be visualized on $$n$$ dimensional sphere, the non-standard products leads to $$n$$ dimensional ellipsoids.
You can rewrite the inner product as $$x^TAy$$ where $$A$$ is positive definite matrix. For setting $$A = I$$ you have standard inner product. An expression $$x^TAy$$ is so-called quadratic form which can be visualised as an ellipsoid (under assumption that $$A$$ is positive definite which is also necessary to fulfil axioms of inner product).
In your example (I switched to 2D only but this can be generalized for $$n$$ dimensions) vector $$\begin{pmatrix} 1 & 0 \end {pmatrix}$$ is unit vector which lays on x axis, vector $$\begin{pmatrix} 0 & 1 \end {pmatrix}$$ layes on y axis and ends of these vectors lays on unit sphere with centre in point (0,0). Under your non-standard inner product, vector $$\begin{pmatrix} 1 & 0 \end {pmatrix}$$ lays again on x axis and also on major axis of ellipse, vector $$\begin{pmatrix} 0 & 1 \end {pmatrix}$$ lays again on y axis and also on minor axis of ellipse. The ellipse is centered in point (0,0). A length of the ellipse major axis is $$\sqrt{2}$$ and minor axis has length 1. | 835 | 2,959 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-27 | latest | en | 0.856393 |
https://www.convertunits.com/info/kilojoule | 1,701,757,361,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00008.warc.gz | 809,826,151 | 12,932 | ## Measurement unit: kilojoule
Full name: kilojoule
Plural form: kilojoules
Symbol: kJ
Category type: energy
Scale factor: 1000
## SI unit: joule
The SI derived unit for energy is the joule.
1 joule is equal to 0.001 kilojoule.
## Convert kilojoule to another unit
Convert kilojoule to
Valid units must be of the energy type.
You can use this form to select from known units:
Convert kilojoule to
## Definition: Kilojoule
The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3.
So 1 kilojoule = 103 joules.
The definition of a joule is as follows:
The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889). | 237 | 819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-50 | latest | en | 0.773085 |
http://mathhelpforum.com/calculus/145166-compute-10th-derivative-print.html | 1,527,252,404,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867092.48/warc/CC-MAIN-20180525121739-20180525141739-00428.warc.gz | 185,374,510 | 3,130 | # Compute the 10th derivative of ...
• May 17th 2010, 11:22 AM
s3a
Compute the 10th derivative of ...
The question is attached. I know the power series representation of cos(x) and therefore can get cos(6x^2)/x^2 but I don't know what to do about the -1.
Any help would be greatly appreciated!
• May 17th 2010, 12:23 PM
Random Variable
Start with the known Maclaurin series of $\displaystyle \cos x$
$\displaystyle \cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + ...$
then $\displaystyle \cos 6x^{2} = 1 - \frac{(6x^{2})^{2}}{2!} + \frac{(6x^{2})^{4}}{4!} - \frac{(6x^{2})^{6}}{6!} + ... = 1 - \frac{6^{2}x^{4}}{2!} + \frac{6^{4}x^{8}}{4!} - \frac{6^{6}x^{12}}{6!} + ...$
and $\displaystyle \cos 6x^{2} -1 = -\frac{6^{2}x^{4}}{2!} + \frac{6^{4}x^{8}}{4!} - \frac{6^{6}x^{12}}{6!} + ...$
finally $\displaystyle \frac{\cos 6x^{2}-1}{x^{2}} = -\frac{6^{2}x^{2}}{2!} + \frac{6^{4}x^{6}}{4!} - \frac{6^{6}x^{10}}{6!} + ...$
For a Macluarin series, the coefficent of the $\displaystyle x^{10}$ term is $\displaystyle \frac{f^{(10)}(0)}{10!}$
The coefficient of the $\displaystyle x^{10}$ term in the above Macluarin series is $\displaystyle -\frac{6^{6}}{6!}$
so $\displaystyle \frac{f^{(10)}(0)}{10!} = -\frac{6^{6}}{6!}$
EDIT: and $\displaystyle f^{(10)}(0) = - \frac{6^{6} 10!}{6!} = - 235146240$ (Surprised)
• May 17th 2010, 12:48 PM
s3a
How do you get from the before-last step to the last one? (From the one that says "so" to the one with the emoticon)
• May 17th 2010, 12:53 PM
lilaziz1
the 10! came from the exponent on $\displaystyle \frac{6^{6}x^{10}}{6!}$. If you were to take the derivative of that 10 times, you would get $\displaystyle \frac{6^{6} 10!}{6!}$
• May 17th 2010, 01:00 PM
Random Variable
Quote:
Originally Posted by s3a
How do you get from the before-last step to the last one? (From the one that says "so" to the one with the emoticon)
It should be $\displaystyle f^{(10)} (0) = - \frac{6^{6} 10!}{6!}$
• May 17th 2010, 01:04 PM
s3a
Oh that 10! was also tricky but I get it now! Thanks! | 833 | 2,041 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.82156 |
https://brainmass.com/physics/power/conservation-of-energy-86848 | 1,527,301,733,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00115.warc.gz | 520,606,122 | 19,242 | Explore BrainMass
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# Conservation of Energy
1. An object of mass 0.23 kg is initially at the origin and is acted on by the sole force F=(0.50 N)i. After a certain amount of time, the object is at a position r=(0.80 m)i. What is the changed in the object's kinetic energy?
2. 74 gram ball is tossed straight up in the air, rises to a maximum point, then falls back until it is .6 m below the position of the hand that tossed it up. A second ball is simply dropped from the same hand position and also lands 0.60 m below that position. What is the net work done by gravity in the two cases?
3. The net force acting on a particle depends on the position of the particle on the x-axis according to the relation F=Fo +Cx, w here Fo=5N and C=-2N/m. The particle is initially at rest at the point x=0 m when the force begins to act. (a) Calculate the work done by the force when the particle reaches x values of 1, 2, 3, and 4 m. (b) Determine any positions (other than at x=0 m) where the work done is zero. (c) Is the force conservative?
4. The maximum power of a particular horse is 1 hp. With what speed can this horse pull a sled on level ground if the weight of the sled with its load is 5000 N and the coefficient of kinetic friction is Mk = 0.03? What is the maximum speed on a 5-degree upward incline?
#### Solution Preview
Answer is in the attached file as well. Thank You.
--------------------------------------------------------
1. An object of mass 0.23 kg is initially at the origin and is acted on by the sole force F=(0.50 N)i. After a certain amount of time, the object is at a position r = (0.80 m)i. What is the change in the object's kinetic energy?
Change in energy = change in kinetic energy (since the object moves in the same level it's potential energy remains same)
Change in energy = change in kinetic energy = force * distance = 0.50 * 0.80 = 0.4 J
2. 74 gram ball is tossed straight up in the air, rises to a maximum point, then falls back until it is .6 m below the position of the hand that tossed it up. A second ball is simply dropped from the same hand position and also lands 0.60 m below that position. ...
#### Solution Summary
Four questions on Mechanics have been answered.
\$2.19 | 560 | 2,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-22 | latest | en | 0.914421 |
https://physics.stackexchange.com/questions/771939/deriving-probability-in-bells-inequality | 1,726,113,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00406.warc.gz | 421,208,298 | 41,215 | # Deriving probability in Bell's inequality
Measuring the spin with an arbitrary angle $$\theta$$ with respect to the plane xz we obtain the spin operator (multiplying Pauli matrices per the projection $$(\cos\theta,\sin\theta)$$)
$$\hat{\mathrm{S}}_{\hat{n}}=\sin(\theta)\hat{\mathrm{S}}_x+\cos(\theta)\hat{\mathrm{S}}_y=\frac{\hbar}{2}\left[\begin{array}{cc} \cos (\theta)& \sin(\theta) \\ \sin (\theta)& -\cos(\theta) \end{array}\right]$$
and when diagonalized, thr eigenvectors and eigenvalues found are
$$\left[\begin{array}{l} \cos(\theta/2)\\ \sin(\theta/2) \end{array}\right],\quad\left[\begin{array}{c} -\sin(\theta/2)\\ \cos(\theta/2) \end{array}\right]$$
you can write the up component from the z axis basis to the arbitrary direction basis:
$$\left[\begin{array}{l} 1 \\ 0 \end{array}\right]=\cos(\theta/2)\left[\begin{array}{l} \cos (\theta/2) \\ \sin (\theta/2) \end{array}\right]-\sin(\theta/2)\left[\begin{array}{c} -\sin (\theta/2) \\ \cos (\theta/2) \end{array}\right]$$
and the down:
$$\left[\begin{array}{l} 0 \\ 1 \end{array}\right]=\sin(\theta/2)\left[\begin{array}{l} \cos (\theta/2) \\ \sin (\theta/2) \end{array}\right]+\cos(\theta/2)\left[\begin{array}{c} -\sin (\theta/2) \\ \cos (\theta/2) \end{array}\right]$$
In Bell type experiments you measure first with an angle $$\theta$$ the first particle and $$\theta'$$ the second. Substituting $$\theta$$ for $$\theta'$$ everything is the same.
The state of the 2 particles needs to be up-down or down-up for conservation of angular momentum:
$$|\psi\rangle=\frac{1}{\sqrt{2}}(|ud\rangle_z-|du\rangle_z)$$
So the probability of for instance up-up (up in each particle in each direction set by each angle) is:
$$P=|_{\theta}\langle u|_{\theta'}\langle u|\psi\rangle|^2=\ldots=\frac{1}{2}|\sin(\theta/2)\cos(\theta'/2)-\sin(\theta'/2)\cos(\theta/2)|^2=\frac{1}{2}\sin^2(\Delta/2)$$
with $$\Delta=\theta'-\theta$$.
But my question is: if the initial supersposition had a change in the pase and we had a plus sign, i.e.,
$$|\psi\rangle=\frac{1}{\sqrt{2}}(|ud\rangle_z+|du\rangle_z)$$
we would obtain a sum of angles, not a difference. And the result the books show is a difference of angles, so you need the initial state with a minus. How is this possible? How is it that the phase is so important? Am I missing something?
The phase is important! If you change the phase in the initial state, you no longer have the property that the state takes the form $$|ud\rangle+|du\rangle$$ in any basis. It is only the singlet state (i.e., the one with the minus sign) that has this property.
It is still possible to show that any state with any relative phase $$|ud\rangle+\exp(i\phi)|du\rangle$$ is maximally entangled, it's just that the correlations you expect from the pairs of measurements will be different.
• Thanks, but I still can't understand why the singlet state is the only taking the form $|ud\rangle+|du\rangle$ in any basis. Sure I need more background. Could you please provide a reference? Commented Jul 14, 2023 at 7:07
• @David that is a calculation you can do! Write your up and down eigenstates in your $\theta$ basis, as you have done, construct $|ud\rangle_\theta-|du\rangle_\theta$, and show that it is equal to the construction in your original basis $|ud\rangle_z-|du\rangle_z$. Much better than a reference Commented Jul 14, 2023 at 13:39
• @David similar answer physics.stackexchange.com/a/175940/291677. Then further reading physics.stackexchange.com/a/421274/291677 Commented Jul 14, 2023 at 13:41 | 1,074 | 3,510 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-38 | latest | en | 0.609795 |
http://korsika.ning.com/profiles/blogs/how-great-thou-art-free-lyrics-3 | 1,725,873,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00407.warc.gz | 16,236,642 | 12,004 | Korsika
La Tramontane - Ferienhaus direkt am Meer
# How Great Thou Art Free Lyrics
- Multiple synoptic symbols can be sent, but only one METAR code per symbol. METAR codes are decoded to give the mean wind speed, max temp, min temp, and pressure. Pressure is encoded as minus and plus values with 1 unit increments between them.
- Wind speeds are encoded in a form similar to the standard WMO FM 10X synoptic format. Negative wind speeds are assigned the -1 index. Positive wind speeds are assigned the +1 index. Intermediate wind speeds are assigned the 0 index. Wind speed is always given relative to a specified 10 knot running speed.
- The mss code is given on a per symbol basis and only one is given. It can be easily decoded by subtracting the -1 index from the 10 index and interpreting the negative result as the minus value.
- The max temp is given as the plus of the high temp and low temp. This is encoded as +/- 30, +/- 15, +/-10, and +/- 5 for 30 degrees, 15 degrees, 10 degrees, and 5 degrees, respectively. For example, if the high temp is 70 degrees and the low temp is 55 degrees, then the max temp is 25.
- The min temp is given as the minus of the high temp and low temp. This is encoded as +/- 30, +/- 15, +/-10, and +/- 5 for 30 degrees, 15 degrees, 10 degrees, and 5 degrees, respectively. For example, if the high temp is 70 degrees and the low temp is 55 degrees, then the min temp is -35.
- The pressure is encoded as plus and minus pressure symbols, with 1 unit of pressure between them. This is encoded as the sequence "+0, -1, 0, 0,..., 0, -1, 0, +1, +2, +3,..., +x, -x" where x equals the pressure of the symbol (minus symbol is indicated by 0 and plus symbol by 1, and pressure levels are enumerated from 0 to (2^((n-1)/2))-1).
- METAR codes are decoded to give the station name, the mean wind speed, max temp, min temp, and pressure. Pressure is encoded as minus and plus values with 1 unit of 45cee15e9a
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Turn on keyboard macros (repeat keys) and keymapping.
Each time a key is pressed, it is repeated a specified number of times.
KEYMACRO also lets you:
Set the number of repeats per key.
Activate or deactivate macros for different keys.
Activate or deactivate macros based on whether the keyboard is asleep or active.
Save and load macros from a text file.
Install:
Copy the file "macro.txt" into your
~/.config/KEYMACRO/ folder.
Source:
* THIS FILE IS PROVIDED UNDER THE TERMS OF THE ECLIPSE PUBLIC LICENSE
* ("AGREEMENT"). ANY USE, REPRODUCTION OR DISTRIBUTION OF THIS FILE
* CONSTITUTES RECIPIENTS ACCEPTANCE OF THE AGREEMENT.
* You can obtain a current copy of the Eclipse Public License from
*
*
* Contributors:
* Kris De Volder
******************************************************************************/
package org.mockserver.serialization;
import java.io.IOException;
/**
* @author Kris De Volder
*/
public interface OutputSerializer {
public void write(T value) throws IOException;
}
Q:
SQL Server Query to do a Inner Join and return the first record in the select statement
I need help writing an inner join to return the result of
select * from table where column = 'value'
and also return the first record of the select statement (so that if a tie exists, the first row returned is used)
A:
SELECT TOP 1 *
FROM table
WHERE column = 'value'
ORDER BY
ROW_NUMBER() OVER (ORDER BY [PK] DESC)
A number
https://clic.umoncton.ca/d2l/lms/blog/view_userentry.d2l?ownerId=20...
https://celebitems.es/en/blog/69_bill-gates-en.html | 1,074 | 4,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.790737 |
https://root.cern.ch/doc/master/ScanBuilder_8cxx_source.html | 1,675,151,478,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00811.warc.gz | 538,435,345 | 7,022 | ROOT Reference Guide
ScanBuilder.cxx
Go to the documentation of this file.
1// @(#)root/minuit2:$Id$
2// Authors: M. Winkler, F. James, L. Moneta, A. Zsenei 2003-2005
3
4/**********************************************************************
5 * *
6 * Copyright (c) 2005 LCG ROOT Math team, CERN/PH-SFT *
7 * *
8 **********************************************************************/
9
10#include "Minuit2/ScanBuilder.h"
13#include "Minuit2/MinimumSeed.h"
15#include "Minuit2/MnFcn.h"
16#include <cmath>
17
18namespace ROOT {
19
20namespace Minuit2 {
21
23 const MnStrategy &, unsigned int, double) const
24{
25 // find the function minimum performing a parameter scan (using MnParameterScan class)
26 // function gradient is not used
28 MnUserParameterState upst(seed.State(), mfcn.Up(), seed.Trafo());
29 MnParameterScan scan(mfcn.Fcn(), upst.Parameters(), seed.Fval());
30 double amin = scan.Fval();
31 unsigned int n = seed.Trafo().VariableParameters();
32 MnAlgebraicVector dirin(n);
33 for (unsigned int i = 0; i < n; i++) {
34 unsigned int ext = seed.Trafo().ExtOfInt(i);
35 scan(ext);
36 if (scan.Fval() < amin) {
37 amin = scan.Fval();
38 x(i) = seed.Trafo().Ext2int(ext, scan.Parameters().Value(ext));
39 }
40 dirin(i) = std::sqrt(2. * mfcn.Up() * seed.Error().InvHessian()(i, i));
41 }
42
43 MinimumParameters mp(x, dirin, amin);
44 MinimumState st(mp, 0., mfcn.NumOfCalls());
45
46 return FunctionMinimum(seed, std::vector<MinimumState>(1, st), mfcn.Up());
47}
48
49} // namespace Minuit2
50
51} // namespace ROOT
class holding the full result of the minimization; both internal and external (MnUserParameterState) ...
const MnAlgebraicSymMatrix & InvHessian() const
Definition: MinimumError.h:50
const MnAlgebraicVector & Vec() const
const MinimumError & Error() const
Definition: MinimumSeed.h:30
const MnUserTransformation & Trafo() const
Definition: MinimumSeed.h:32
const MinimumParameters & Parameters() const
Definition: MinimumSeed.h:29
const MinimumState & State() const
Definition: MinimumSeed.h:28
MinimumState keeps the information (position, Gradient, 2nd deriv, etc) after one minimization step (...
Definition: MinimumState.h:27
Wrapper class to FCNBase interface used internally by Minuit.
Definition: MnFcn.h:30
double Up() const
Definition: MnFcn.cxx:39
const FCNBase & Fcn() const
Definition: MnFcn.h:47
unsigned int NumOfCalls() const
Definition: MnFcn.h:39
Scans the values of FCN as a function of one Parameter and retains the best function and Parameter va...
const MnUserParameters & Parameters() const
API class for defining three levels of strategies: low (0), medium (1), high (>=2); acts on: Migrad (...
Definition: MnStrategy.h:27
class which holds the external user and/or internal Minuit representation of the parameters and error...
const MnUserParameters & Parameters() const
double Value(unsigned int) const
double Ext2int(unsigned int, double) const
unsigned int ExtOfInt(unsigned int internal) const
FunctionMinimum Minimum(const MnFcn &, const GradientCalculator &, const MinimumSeed &, const MnStrategy &, unsigned int, double) const override
Definition: ScanBuilder.cxx:22
Double_t x[n]
Definition: legend1.C:17
const Int_t n
Definition: legend1.C:16
VecExpr< UnaryOp< Sqrt< T >, VecExpr< A, T, D >, T >, T, D > sqrt(const VecExpr< A, T, D > &rhs)
This file contains a specialised ROOT message handler to test for diagnostic in unit tests. | 912 | 3,393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-06 | latest | en | 0.284207 |
https://nbviewer.jupyter.org/github/sgsinclair/alta/blob/master/ipynb/utilities/SimpleSentimentAnalysis.ipynb | 1,603,962,104,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107903419.77/warc/CC-MAIN-20201029065424-20201029095424-00411.warc.gz | 437,054,043 | 6,160 | # Simple Sentiment Analysis¶
This notebook shows how to analyze a collection of passages like Tweets for sentiment.
This is based on Neal Caron's An introduction to text analysis with Python, Part 1.
This Notebook shows how to analyze one tweet.
### Setting up our data¶
Here we will define the data to test our positive and negative dictionaries.
In [6]:
theTweet = "No food is good food. Ha. I'm on a diet and the food is awful and lame."
positive_words=['awesome','good','nice','super','fun','delightful']
In [7]:
type(positive_words)
Out[7]:
list
### Tokenizing the text¶
Now we will tokenize the text.
In [8]:
import re
theTokens = re.findall(r'\b\w[\w-]*\b', theTweet.lower())
print(theTokens[:10])
['no', 'food', 'is', 'good', 'food', 'ha', 'i', 'm', 'on', 'a']
### Calculating postive words¶
Now we will count the number of positive words.
In [14]:
numPosWords = 0
for banana in theTokens:
if banana in positive_words:
numPosWords += 1
print(numPosWords)
1
### Calculating negative words¶
Now we will count the number of negative words.
In [10]:
numNegWords = 0
for word in theTokens:
if word in negative_words:
numNegWords += 1
print(numNegWords)
2
In [18]:
v1 = "0"
v2 = 0
v3 = str(v2)
v1 == v3
Out[18]:
True
### Calculating percentages¶
Now we calculate the percentages of postive and negative.
In [11]:
numWords = len(theTokens)
percntPos = numPosWords / numWords
percntNeg = numNegWords / numWords
print("Positive: " + "{:.0%}".format(percntPos) + " Negative: " + "{:.0%}".format(percntNeg))
Positive: 6% Negative: 11%
### Deciding if it is postive or negative¶
We are going assume that a simple majority will define if the Tweet is positive or negative.
In [12]:
if numPosWords > numNegWords:
print("Positive " + str(numPosWords) + ":" + str(numNegWords))
elif numNegWords > numPosWords:
print("Negative " + str(numPosWords) + ":" + str(numNegWords))
elif numNegWords == numPosWords:
print("Neither " + str(numPosWords) + ":" + str(numNegWords))
print()
Negative 1:2
## Next Steps¶
Let's try another utility example, this time looking at more Complex Sentiment Analysis.
CC BY-SA From The Art of Literary Text Analysis by Stéfan Sinclair & Geoffrey Rockwell. Edited and revised by Melissa Mony.
Created August 8, 2014 (Jupyter 4.2.1)
In [ ]: | 656 | 2,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-45 | latest | en | 0.591089 |
https://encyclopediaofmath.org/index.php?title=Catalan_surface&printable=yes | 1,723,689,656,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00445.warc.gz | 180,077,539 | 5,726 | # Catalan surface
A ruled surface whose rectilinear generators are all parallel to the same plane. Its line of restriction (cf. Ruled surface) is planar. The position vector of a Catalan surface is $r=\rho(u)+vl(u)$, where $l''(u)\neq0$, $(l,l',l'')=0$. If all the generators of a Catalan surface intersect the same straight line, then the surface is a conoid. | 96 | 361 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.774827 |
http://www.pyramydair.com/blog/2005/06/more-about-sighting-in-how-to.html | 1,397,909,189,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00553-ip-10-147-4-33.ec2.internal.warc.gz | 628,126,097 | 19,350 | Wednesday, June 08, 2005
More about sighting-in: How to determine the two intersection points
By B.B. Pelletier
The following question was a comment to the June 1 post, At what range should you zero your scope?
Question:
How do you figure out what the two intersections are? Is there a formula?
Just shoot & see
No doubt, a formula could be constructed, but I do it the easy way. Simply aim at a specific point at a known distance, shoot and see where the pellet impacts relative to the aim point. That's where I came up with the distances I gave in my June 1 post.
Each pellet has a different trajectory
My distance figures in that posting were not exact. They change slightly because each type of pellet has a unique ballistic flight. If it has higher drag, it slows down faster and the trajectory is more pronounced. The figures I gave were for a domed pellet, like the Crosman Premier, which is pretty standard for field use.
If you shoot wadcutters or hollowpoints that have a sharp shoulder, such as the RWS Super-H-Point, your trajectory will be more pronounced because the pellet is less aerodynamic. The difference will be small at close range but will increase rapidly as the distance passes 30 yards. That's why wadcutters and hollowpoints are not as good for long-range shooting (unless you take the time to learn their performance characteristics well).
Twenty yards is a common zero distance
The initial point of intersection I gave in the posting (20 yards) is based on seven years of competitive experience shooting field target. I had to learn the performance of a great many pellets (all .177, but that makes little difference to what we are doing). Although there are a LOT of variables, I soon noticed that nearly every shooter had a first zero point of around 20 yards. When I tried it myself, I discovered why.
What a precision target shooter wants is the most forgiving zero possible. One where, if the yardage is not exactly what the shooter estimated, the difference in pellet impact point is very small. If you zero your gun so the first impact point is 20 yards, you'll get another 10 to 20 yards of flat trajectory and any apparent rise or fall of the pellet is less than one pellet diameter. The first intersection is at 20 yards and the second is at 30 yards (a 10-yard flat spot) with a gun shooting around 725-750 f.p.s. When you get it up to 950 f.p.s., the second point will be all the way out to 40 yards. If you go even faster, you'll have a flatter trajectory but also get blown groups and leading. It's not worth it.
You CAN stretch out the first point of intersection to 30 yards, if you like. All it takes is a scope adjustment. BUT, your flat spot will be MUCH shorter than when you were zeroed at 20 yards. It may only be 5 yards long. Zero at 40 yards for the first point, and the flat spot MAY be yards long. Get it?
What's really happening is that the pellet is dropping the moment it leaves the muzzle. You have about 30 to 40 yards to play with before the trajectory is a real downward slide. By zeroing at 20 yards, you get a nice long flat spot and can still adjust your scope or hold-over for distances outside that range.
Will zeroing at 5 feet give me a really long flat trajectory?
At this point you may be wondering why I'm saying 20 yards is a good zero point. Why not zero at 5 FEET and enjoy as much of that flat trajectory as possible? If your scope looked straight through the center of your barrel, you could do that. Because you have to mount your scope high above the barrel, you can't make it work that way. The separation of the scope axis and bore axis introduces parallax that has to be accounted for when you sight-in.
This is already a long post, so I'm going to end it here and continue this discussion on another day. I hope I've answered some of your questions. Please feel free to post comments or additional questions!
At June 10, 2005 12:17 PM, Anonymous said...
What is the practical range for an airgun.
What is the "effective" range for an airgun used for hunting.
Is twenty or 30 yards what everyone shoots at?
At June 10, 2005 12:33 PM, B.B. Pelletier said...
Well, your question is a VERY deep one, and I think it deserves a better answer than I can give her.
Watch this week's blog for my answer.
And, thanks,
B.B.
At June 15, 2005 2:49 PM, Anonymous said...
I'm planning on deer hunting w/a 6 mm PCP loaded w/70gr Hollowpoints. I have accuracy down to a 1/4 inch @ 100 yds.
Am I nuts?
At June 15, 2005 7:38 PM, B.B. Pelletier said...
For deer I like to see 500 foot-pounds on target and no .243 air rifle I know of comes close to that.
B.B.
At June 15, 2005 11:37 PM, Anonymous said...
I'm thinking thats NO problem, I forget all the stats on what he told me. I do know I shot duck decoys @ 200 to 250 yds w/it.(Rangefinder said 250, guy w/scope said parallex was 225)Either way w/the accuracy & what it does to smaller targets.....I say it'll drop a deer in its tracks.
I'll check w/manufacturer on the ft/pounds, but I've seen some amazing results w/these things......They are handmade by a very talented machinist friend of a friend.
The 30 caliber goes through 2 by 4's sideways. I realize this shot was at an unmeasured distance but, its probably 75 yards.
When I get the ft/# numbers I'll get back to you.
Thanks
At February 10, 2006 9:19 PM, Anonymous said...
I have never tried a scope of any kind, but am interested in trying one on a pistol. What is the difference between a pistol scope and a rifle scope? Can a rifle scope be used on a pistol? I am assuming a scope marketed specifically as a pistol scope is meant to be viewed at arms length, but can't imagine what that must be like.
At February 15, 2006 8:13 AM, B.B. Pelletier said...
The main difference between pistol and rifle scopes is the eye relief. You have about 20 inches with a pistol scope but around three with a rifle scope. That allows you to extend your arms.
Pistol scopes are lower powered and they tend to cost about twice what comparable rifle scope do.
B.B.
At May 19, 2007 1:35 PM, Anonymous said...
if i sight my springer (Crosman Storm Xt) at about 35 feet how much further can i still hit targets accuratly ??
At May 19, 2007 2:52 PM, B.B. Pelletier said...
35 feet is 12 yards. That's a poor distance at which to sight in, but I hear this kind of question all the time because that's all the distance you have where it is convenient to shoot. So this is what you do. Sight in to strike one inch BELOW your aim point at 35 feet. That should put you very close to dead on at 20-30 yards. If you sight in at 35 feet, you will be off at most other distances.
B.B.
At May 22, 2007 3:47 PM, Anonymous said...
so if i sight at 35 ft and follow your directions i should be pretty close on target at 20 to 30 yards, but if i shoot a target at 35 feet it will mean that i will shoot and hit 1 inch lower right????
At May 22, 2007 5:04 PM, B.B. Pelletier said...
That's correct. Now you may be off by half an inch, but it should be pretty close if yoiu do it this way.
B.B.
At May 27, 2007 5:46 PM, Anonymous said...
thank you for your advice. one question can u use remington oil or reel magic to lube springers breech???
At May 28, 2007 5:56 AM, B.B. Pelletier said...
Rem Oil contains mineral spirits and is therefore not to be used in compression chambers. I cannot find technical information about Reel Magic and therefore cannot recommend its use in compression chambers.
B.B.
At June 14, 2007 4:54 PM, Anonymous said...
i finally measured the space i have and it is actually 50-55 feet so should i still do the same as if i have 30-35 feet??
At June 14, 2007 7:00 PM, B.B. Pelletier said...
No. The difference in distance at this range is critical. So use 30 feet and you're done.
B.B.
At July 08, 2007 6:56 PM, Anonymous said...
i used rem oil on my gun what should i do ( it's compression chambered
At July 09, 2007 6:36 AM, B.B. Pelletier said...
I don't understand your question. What does "compression chambered" mean and how does that relate to Rem Oil?
B.B.
At July 09, 2007 11:35 PM, Anonymous said...
i used oil for pumps and put some in the compressing chamber of my air rifle will this affect it .. the oil is Bell and Gossett pump lubricant...
At July 10, 2007 6:03 AM, B.B. Pelletier said...
You didn't mention what rifle you shoot and it makes a difference. A Benjamin 392 probably won't mind, but I wouldn't use a non-airgun oil in a Air Arms S410 or a Career 707.
B.B.
At July 10, 2007 2:49 PM, Anonymous said...
i am having some troubles when sighting in. i get close grouping for about 3 to 4 shots and then they start to wander off is it the gun or the scope??
At July 10, 2007 2:59 PM, B.B. Pelletier said...
can you tell me a little more? Like gun, scope, mount, range at which this is happening? What pellet are you using? Is the scope adjusted to the limit of either knob?
Are you using the artillery hold?
B.B.
At July 10, 2007 4:52 PM, Anonymous said...
gun is crosman quest ammo is crosman pointed the range is 50 feet i get it to sight in small groups and then they just go all over the place i'm getting frustrated and the scope is a powerline 3-9x32 scope...
At July 10, 2007 5:05 PM, B.B. Pelletier said...
Hang in there!
Is the scope adjusted to the limit either to the right or up high?
Are you holding the rifle on the open flat of your hand - not touching the stock with your fingertips? Are you holding the stock just forward of the triggerguard? Do you allow the rifle to recoil as much as it can?
Are you breathing and relaxing before you align the crosshairs? By aligning them before you relax you are sertting the rifle up to throw the shots high and to the right, if you are right-handed.
It is very frustrating to learn to shoot a breakbarrel air rifle. But the technique I just described is the only way. Shooting off cloth or sandbags is a recipe for failure.
B.B.
At July 10, 2007 7:57 PM, Anonymous said...
i leave the stock on a pillow to stability....i let the gun recoil as if i am shooting it normally. so i should not use a pillow or anything right??????i tend to shoot more to the left..
At July 10, 2007 8:22 PM, Anonymous said...
I AM SHOOTING TO MUCH TO THE LEFT..I AM HOLDING THE STOCK WITH MY HAND AND FINGERS AROUND THE STOCK... I DONT LET THE GUN RECOIL AS MUCH AS IT CAN.... AND I HAVE BEEN RESTING MY HAND ON A LPILLOW FOR STABILITY... I AM SUPPOSED TO USE A PILLOW OR SOMTHING TO HELP WITH SHAKING?????I GET ALOT OF SCATTERD SHOTS AND THEN I GET A CLOSE SHOT AND THEN THEY SCATTER AGAIN I AM ABOUT TO GIVE UP BECAUSE IT MIGHT BE THE GUN OR SCOPE.......
At July 11, 2007 5:47 AM, B.B. Pelletier said...
You need to learn the artillery hold for AN Y spring piston air rifle. They all require it. And breakbarrels are the worst.
The shakinjg is something you need to deal with. It's one of the things shooters nbeed to overcome.
Cloth will not help you. You have to rest a recoiling spring rifle on the FLAY of your open palm.
I can't find any blogs I have done on this hold, so I will do a new one for you. Tomorrow.
B.B.
At July 11, 2007 2:49 PM, Anonymous said...
i really think that it is not the way i hold the gun because i can get the gun shooting where i want it and then it shoots all over the place i am probaly gonna stick with my open sights.......or buy a new scope for the third time the gun itself shoots okay bu not the best when shooting with open sights...
At January 14, 2008 1:17 PM, Anonymous said...
B.B,
If i zero my scope at 20 yards,about how many inches under will a pellet hit its target at 100yards (shooting at 900 fps).
Is there any formula?
At January 14, 2008 2:50 PM, B.B. Pelletier said...
You'll hit about 3-4 feet low at 100 yards. Try it on a large piece of cardboard, knowing the pellet will drop.
B.B.
At January 14, 2008 3:23 PM, Anonymous said...
Wow..it's about 1 meter...that is big...more than i expected really lol
In the rotating elevation adjustment ,it is said : 1 click,1/4'' at 100 yards....what does that mean?
About how many clicks will i have to turn to get my shots higher to zero at 100 yards?...is it about 160 clicks? Lets say for every click,the shot will go up 1/4 inch for 100 yard distance(0.6 centimeters)....I have about 1 meter to catch up(100 centimeters).Then for 100 centimeters,it will be 100/0.6=166 clicks....
Is that correct? Does the rotating adjustment rotate that many??
Thanx for clearing things :)
At January 14, 2008 5:36 PM, B.B. Pelletier said...
The scope marking mean what you think they mean.
Most scopes have not enough vertical adjustment to shoot 100 yards. You need an adjustable scope mount for the rest.
But instead of talking about it, why not test it yourself? Make an aim point about 3 feet above a large piece of cardboard and see where the pellet hits. If you hit the cardboard (it should be a piece at least a meter square) mark your hit and see how far the pellet dropped.
Guessing, which is what I am doing, only goes so far.
Count the clicks in your scope. That's what I do and they vary widely.
B.B.
At January 15, 2008 7:55 AM, Anonymous said...
B.B,
Yes i will try it myself of course,but right now the scope i bought was broke(i told you about it).And it is under repair.The guy said he will make it stronger from the inside so that the crosshair wont rotate anymore...and ALSO this time i will buy a scope stop that has a recoil absorber and i will buy a double mount system so that i can fit the scope stop in between those 2 mounts.
Do you think this will prevent my scope (Gamo sporter 3-9x 40) from breaking again?
Thanx,and sorry for asking that many questions,but i hope you dont't mind ...
At January 15, 2008 8:49 AM, B.B. Pelletier said...
That's a question I can't answer. It depends on what the rebuilder does. But since he says he can do it, it sounds like he knows what he's doing. Many scope repair places don't even acknowledge the difference between airgun stresses and those from firearms.
B.B.
At January 15, 2008 4:52 PM, Anonymous said...
B.B,
I hope the Recoil absorber of the stop scope will greatly reduce the recoil on the scope....
At July 04, 2008 10:03 PM, Anonymous said...
Hi,
I recently purchased my first air rifle, a Gamo Big Cat 1200. I followed your tips for sighting in at 35 ft, using Gamo PBA Raptor pellets. I then decided to see how other pellets performed and my problems began. I tried RWS Superpoint Field Line 8.2 grains, RWS Superdome 8.3 gr and RWS Wad cutters and all hit from four to six inches lower than the Raptors and I ran out of clicks on my scope before I got close to the center. Is there something wrong with my initial sighting in? What do you recommend?
Thanks,
Ed
At July 05, 2008 6:58 AM, B.B. Pelletier said...
Ed,
You have barrel droop. Have you read our discussions abut it? It's very common on breakbarrels.
To correct your scope you either need to shim under the rear ring or you need to use an adjustable mount.
The Raptors hit higher because they are going so much faster. But you will want to use the other pellet because they are more accurate, so you do have to correct the scope.
B.B.
At July 07, 2008 7:52 PM, Anonymous said...
Thanks for the quick reply. I've quickly read a few of your discussions and did add the 35mm piece "shim" in the rear of the scope mount. I was able to bring down the number of clicks on my scope and got the RWS Superdomes on target at 20 yds. That is ½ “ high at 20 yds. However, when shooting at 15 and 10 yds things got very confusing, and really need your help. I don’t wish to clog up this forum with my problems, so if you wish to write me privately my email address is eforteza@prtc.net. So you know, I’m new at air rifles but been shooting all my life from mortars, M-14, M16 to bows, so I know a little about shooting. However, this really has me scratching my head. As I mentioned before, I’m shooting a Gamo Big Cat with two different pellets, Gamo PBS Raptors and RWS Superdomes. Here were the results with the Superdomes and I did it twice just to make sure. The RWS Superdomes; at 20 yds were hitting ½ high at 20 yds from aim point. When I moved to 15 yds. they were hitting down ½ “ and right 1 ½” from center. When I moved in to 10 yds. they were hitting down 2” and to the right ½”. It doesn’t make sense to me how they went to the right and down, and looped around from 3:30 o'clock to 5:30 o'clock as I got closer. Hope all this makes sense to you. I contacted Gamo and they're willing to check it out under warranty, but I wanted to see if I can figure it out first before spending on shipping charges. Thanks again, Ed from Puerto Rico
At July 08, 2008 8:48 AM, B.B. Pelletier said...
Ed,
There are two possibilities here, a scope that's grossly out of alignment or a spiraling pellet. I suspect the latter, because as you go farther from the muzzle, the dispersion changes.
Spiraling pellets are due to gross instability. That pellet, the RWS Superdome, is wrong for your rifle. If you plot all the hits on a single piece of paper you will see the spiral. From that, you can even predict where the pellet will strike at a different range.
Don't expect a Raptor to be accurate at any range.
The Superdome is too narrow for your bore. You need a fatter pellet. Try some Gamo pellets and try some Crosman Premiers.
Incidentally, centerfire bullets also spiral, though they do it over much longer distances.
By the way, I was a 4.2" mortar platoon leader, so we share some common experiences.
B.B.
At July 08, 2008 12:38 PM, Anonymous said...
B.B.;
81 MM, PLT-SGT 0369, 3rdMarDiv, Okinawa, 1973.
Are you suggesting the Crosman Premiers 7.9 gr dome lights? You mentioned as I go farther from the muzzle, the dispersion changes. What do you mean? In my case the groups move as I get closer. Haven’t tried it longer, until I get it correct from 20 on in. I noticed the information did not type correctly in my previous comment, so to confirm, at 20 yds the hits are half inch high, at 15 yds the group moves right two inches and down one inch; and at 10 yds the group is half inch right and two inches low. At 15 way right and at 10 way low.
I will contact Pyramyd about ordering the pellets and hopefully get on target.
ED
At July 08, 2008 1:53 PM, B.B. Pelletier said...
Ed,
Mortar platoon leader and later company commander of Combat Support Company, 2nd Battalion, 81st Armor, 1st Armored Division, Erlangen, Germany, 1974-1977.
You hadn't mentioned the caliber of your rifle before, so I couldn't be specific, but now that I know, also get some 10.2-grain JSB Exact pellets. I think among all of those I have mentioned this spiraling tendency will go away.
As for the dispersion, I have actually watched spiraling pellets out to 50 yards and they don't disperse all the time. Sometimes they just rotate around a center at the same distance all the way. Other times they do disperse. You should be able to put a target at 18 yards and predict where it will be hit, based on the other targets.
B.B.
At July 08, 2008 3:24 PM, Anonymous said...
Waiting list until November for these :(
At July 08, 2008 4:43 PM, B.B. Pelletier said...
Yes, and when they arrive they will evaporate in two weeks.
B.B.
At July 27, 2008 8:41 PM, Anonymous said...
Hi B.B.:
Follow up to our last. I ordered a box of Crosman Premiers 7.9 (JSB were on back order) and the results are better. I set my zero at 12 yds and my groups at 10 yds are 1/2 low and at 20 yds 1/2 high, but they are vertical. Need to see how the group does at longer distances. Semper Fi!
At January 01, 2009 11:47 AM, Anonymous said...
i know this dont really go with this forum but i got a beeman rs2 sportsman series break barrel for christmas.when i took it out i mounted the scope and sighted it in at 15 yards,but it was shooting 2 in. groups?so i started playing wit it and after about 200 rounds i was hitting pepsi bottles at 85 yards aiming at the top of the bottle and my groups improved at 20 yards i was hitting the same hole.after about 600 rounds i oiled it and was shooting fine for 50 or so rounds and then it just started goin crazy. i went from shooting half inch 3 shot groups at 15 yd. to 3 in. groups using same pellet (crosman premiers)at same distance and its getting aggravating,please help!
At January 01, 2009 1:57 PM, B.B. Pelletier said...
I think I know what's wrong. You are shooting Crosman Premiers, which are wonderfully accurate, BUT, they do lead the barrel.
Here is what I want you to do. I want you to clean your barrel. I really want you to use a new brass brush loaded with JB Non-Embedding Bore Cleaning Compound. Please read this report and clean your gun exactly as I describe there:
http://www.pyramydair.com/blog/2005/11/is-your-airgun-barrel-really-clean.html
I have talked hundreds of shooters through a similar problem and I'm pretty sure it will work for you. You can buy the JB compound at a good gun store if you want to speed up the process. Don't use Hoppes No. 9 or any other solvent cleaners -- only use JB paste that most benchrest shooter clean their barrels with.
Then get back to me on the current day's blog (whenever that is) and tell me your results. I will then tell you how to continue to keep shooting the same pellets and not have to clean the barrel again.
http://www.pyramydair.com/blog/
B.B.
At January 13, 2009 1:54 PM, Anonymous said...
thank you for the information,it helped ALOT.i went back to shooting 3 shot one hole groups at 20 yards!
At January 13, 2009 1:59 PM, kevin said...
Anonymous,
Great news that your barrel cleaning improved accuracy!
Here is a link to take you to current/active dialogue:
http://www.pyramydair.com/blog/
Look forward to hearing from you.
kevin
At January 13, 2009 1:59 PM, B.B. Pelletier said...
You can do me a favor in return. Would you mind commenting about your accuracy problem on the current day's blog and tell them what you did to fix it? A lot of people don't yet believe this works.
B.B.
At April 21, 2009 12:17 AM, Anonymous said...
B.B.
I'm a 34yr old pilot that hasn't fired a rifle down range since my 10th grade JROTC class. I recently purchased a Crosman Quest 1000 to take me back to the happy place I recall the range was for me. Many things you talked about helped me thru the 1st 50rds I've fire, is their any advice for do's and don'ts with my new rifle.
Thanks,
Wings.
At April 21, 2009 5:42 AM, Mr B. said...
Hi Wings,
Welcome to the wonderful world of air guns. You posted to a blog that was written in 2005. Come join us at http://pyramydair.com/blog/ and ask your question on our current blog.
There's a group of folks there that will be happy to answer your questions.
Mr B.
At April 21, 2009 7:23 AM, Anonymous said...
Tom am I missing something? It seems to me that the separation distance between bore and scope has to come into play here. For example on my condor with rings and a tri-rail there's quite a gap there, meaning I have to point the scope looking more down to get on-sight at 20 yards. With that much of an "X" (the difference of the scope looking down, or the bore pointing up) and assuming 950 fps using 21.1 Gr. .22 Cal. Beeman Kodiaks. Do I still want to use 20 yards? I'm looking for no more than 1/2 an inch above line of sight. Thanks From SavageSam
At April 21, 2009 7:47 AM, B.B. Pelletier said...
SavageSam,
Keep the 20 yards! It's too handy not to use.
The scope separation will THEN affect where the SECOND impact point is - and that you determine on the range.
In other words, don't think about it too much. But if you DON'T use the 20 yards with normal velocities (800-950 f.p.s.) you will have a screwy setup.
B.B.
At April 21, 2009 8:48 AM, Anonymous said...
Thank you VERY MUCH for the quick reply. I'm hopeing to get out today before it gets too hot. From SavageSam.
At July 04, 2009 10:40 AM, Anonymous said...
I think this discussion is all about scope on a rifle which is at least 1 inch above .
If I use default factory open sight , do I still need to follow this sight-in procedure ?
thanks !
At July 04, 2009 1:34 PM, B.B. Pelletier said...
Here is an easy way to understand this problem and to answer your question with any gun. If you put a target against the muzzle and sighted on the same target with the sight, where would you hit, relative to where the sights are looking?
You would hit below the point of aim, of course.
The discussion in this report is the range at which you want the aim point and the impact point to coincide. And it works for open sights as well as scopes.
As you point out, scopes will always be higher than open sights, but both are higher than the bore.
B.B. | 6,517 | 24,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2014-15 | latest | en | 0.967036 |
https://codegolf.stackexchange.com/questions/233116/how-to-golf-a-fork-in-haskell | 1,721,657,076,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00672.warc.gz | 157,392,572 | 42,188 | # How to golf a fork in Haskell?
I'm trying to make a pointfree function f that can act like APL's dyadic forks/Husk's §, so f g h i a b equals g (h a) (i b), where g is a binary function, h and i are unary functions, and a and b are integers (I'm only dealing with integers and not other types right now). I converted the function to pointfree in the following steps, but it's 26 bytes with a lot of parentheses and periods and even a flip, so I'd like to know how to golf it.
f g h i a b=g(h a)(i b)
f g h i=(.i).g.h
f g h=(.g.h).(.)
f g=(.(.)).(flip(.)).(g.)
f=((.(.)).).((flip(.)).).(.)
Try it online!
• Plugging into pointfree.io gives an 20-byter ((flip.((.).)).).(.). Commented Aug 4, 2021 at 23:21
• @Bubbler ...Oh. Can you put that as an answer, with an explanation of how it works? It'd be very helpful.
– user
Commented Aug 4, 2021 at 23:22
• The reason I wrote it in a comment is that I don't know how it works :P Btw, should the arguments be ordered exactly like that? Commented Aug 4, 2021 at 23:22
# 18 bytes
(flip.).(.).((.).)
Try it online!
Since your function has no argument repetition and no argument deletion, it essentially becomes a BC calculus golf (in Haskell terms, a golf using just B=(.) and C=flip). I will use B and C combinators and convert to Haskell code later.
\g h i a b -> g (h a) (i b)
\g h i a -> B (g (h a)) i
\g h i -> C (\a -> B (g (h a))) i
\g h -> C (\a -> B (g (h a)))
-- choice 1
\g h -> C (B B (B g h))
-- choice 1.1
\g -> B C (B (B B) (B g))
B C.B (B B).B
(flip.).(((.).).).(.) -- 21 bytes
-- choice 1.2
\g -> B (B C (B B)) (B g)
B (B C (B B)).B
((flip.((.).)).).(.) -- 20 bytes; pointfree.io
-- choice 2
\g h -> C (B (B B g) h)
\g -> B C (B (B B g))
B C.B.B B
(flip.).(.).((.).) -- 18 bytes; shortest
As a bonus, if you can swap i and a in the definition of f, it becomes an ordered LC and allows a very short form using just (.):
\g h a i b -> g (h a) (i b)
\g h a i -> B (g (h a)) i
\g h a -> B (g (h a))
\g h -> B (B B g) h
\g -> B (B B g)
B B (B B)
(.).((.).) -- 10 bytes
• Excellent answer. It seems like a mechanical process of enumeration. Why is pointfree.io not able to find the shortest one? Is the original insight that the problem was equivalent to a BC calculus golf the stumbling block? Commented Aug 5, 2021 at 0:27
• @Jonah The primary reason is that a.b.c can be both (a.b).c and a.(b.c), which give different results at the end. I'm pretty sure pointfree.io uses only one of the possible abstractions to get some correct result. Commented Aug 5, 2021 at 0:34
• what does "BC calculus" mean? Where can I find more info to learn about it? Commented Aug 5, 2021 at 17:23
• B and C are combinators, i.e. essentially functions that are used in a system with no explicit variables. Asking for a point-free definition of a function is more or less asking for a definition in terms of combinators. BC calculus is simply the system you get if you allow combinators B and C (+ applications) only.
– zale
Commented Aug 5, 2021 at 20:36
• @NooneAtAll To add to zale's comment, BC calculus is equivalent to linear logic/linear lambda calculus where each argument is used exactly once in some order. Commented Aug 6, 2021 at 0:02 | 1,037 | 3,202 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-30 | latest | en | 0.760408 |
http://www.trickyriddles.com/riddle/4746-Five-is-Equaled-to%3F | 1,718,492,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00043.warc.gz | 60,114,146 | 4,384 | ### Riddle: Five is Equaled to?
• Category : Logic
• Rating: 3.74
• Print Riddle
Can you solve this riddle?
If 1=5
2=25
3=125
4=3125
5= ? | 58 | 140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-26 | latest | en | 0.481635 |
http://www.early-retirement.org/forums/f28/calculate-number-portfolio-buckets-69405.html | 1,527,212,648,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866917.70/warc/CC-MAIN-20180525004413-20180525024413-00279.warc.gz | 364,397,435 | 29,167 | Calculate Number / Portfolio Buckets
11-25-2013, 06:38 PM #1 Recycles dryer sheets Join Date: Aug 2013 Posts: 349 Calculate Number / Portfolio Buckets I'm working on calculating my "number" and doing so from the bottom up using buckets & time periods. Here are the buckets. ELE - Essential Living Expenses - portfolio goal is a very low risk 0% real return (no risk, keep pace with inflation.) DLE - Desired Living Expenses - portfolio goal is conservative 2% real return (low to moderate risk beat inflation by 2%.) LL - Legacy & Luxury - portfolio goal is aggressive growth using various strategies to maximize total return. Here are the time periods (we are 47/49). ER - Early Retirement (10.5 years) - Covers period from now until DW can access her IRAs (DW 49 to 59.5). RR - Regular Retirement (12 years) - Covers period after Early Retirement until Social Security can be accessed (DW 59.5 to 71.5). SR - Subsidized Retirement (16.5 years) - Covers period after Regular Retirement until old age (DW 71.5 to 88). OA - Old age (12 years) - Covers period of reduced income need until departure. (DW 88 to 100) ::: numbers adjusted for ease of explanation ::: For our ELE bucket we need 2.415M based on the following.ER - 80k * 10.5 years = 840k RR - 70k * 12.0 years = 840k SR - 30k * 16.5 years = 495k OA - 20k * 12.0 years = 240k For our DLE bucket we need 1.125M based on the following.ER - 30k * 10.5 years = 315k RR - 30k * 12.0 years = 360k SR - 20k * 16.5 years = 330k OA - 10k * 12.0 years = 120k From this our number 3.54M (plus social security) to cover a 51 year retirement with very conservative investments (matching inflation on essentials and return 2% real for non-essentials.) Money over and above the 3.54M can go into a LL portfolio and withdrawn as desired. Regarding the account types, the first 10.5 years is being withdrawn from a taxable account. From there on out the money comes from 90% ROTH and 10% IRA. Now for the portfolios. ELE - Goal: Keep Pace with Inflation - Low Risk VTSMX - Total Stock Market - 13.70% - \$330.75k VGTSX - Total International Stock - 5.87% - \$141.75k VBMFX - Total Bond Market - 64.35% - \$1,554.00k VTIBX - Total International Bond - 16.09% - \$388.5k DLE - Goal: 2% Real Return - Moderate Risk VTSMX - Total Stock Market - 40.88% - \$459.90k VGTSX - Total International Stock - 17.52% - \$197.10k VBMFX - Total Bond Market - 33.28% - \$374.40k VTIBX - Total International Bond - 8.32% - \$93.60. Aggregating the two portfolios gives the following. VTSMX - Total Stock Market - 22.33% - \$790.65k VGTSX - Total International Stock - 9.57% - \$338.85k VBMFX - Total Bond Market - 54.47% - \$1,928.40k VTIBX - Total International Bond - 13.62% - \$482.10k These portfolios came from Vanguard. I created a portfolio for each time period and aggregated them by goal and then again overall using this tool. https://personal.vanguard.com/us/fun...recommendation I could really use some help / suggestions here since I have limited experience. __________________
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11-25-2013, 06:56 PM #2 Moderator Emeritus Join Date: Apr 2011 Location: The Woodlands, TX Posts: 7,853 Not sure what the questions are. Whos is working and why only figure DW requirements? Not nearly enough information. __________________
11-25-2013, 06:59 PM #3
Give me a museum and I'll fill it. (Picasso)
Give me a forum ...
Join Date: Jul 2008
Posts: 20,728
Quote:
Originally Posted by aja8888 ...why only figure DW requirements?
Are there other requirements?
__________________
"Old age is the most unexpected of all things that can happen to a man" -- Leon Trotsky
11-25-2013, 07:10 PM #4
Recycles dryer sheets
Join Date: Aug 2013
Posts: 349
Quote:
Originally Posted by NW-Bound Are there other requirements?
Exactly. :-) DW is 18 months older than I and has a family history of longer life. I calculated everything based on her age to 100 and us getting social security when I hit 70. She'll get spousal social security. The income numbers are for the household, we don't separate anything, it's all our pile.
I guess my questions are . . .
Does this approach make sense given Bernstein's idea of having won the game and taking minimal risk for the portion of your portfolio needed to cover essential expenses?
Do the portfolios that Vanguard recommended make sense to match inflation and to return a 2% real return?
Are there better ways of doing so?
11-25-2013, 08:21 PM #5
Give me a museum and I'll fill it. (Picasso)
Give me a forum ...
Join Date: Feb 2013
Posts: 5,578
Quote:
Originally Posted by aim-high Does this approach make sense given Bernstein's idea of having won the game and taking minimal risk for the portion of your portfolio needed to cover essential expenses? Do the portfolios that Vanguard recommended make sense to match inflation and to return a 2% real return? Are there better ways of doing so?
This is Bernstein's recent thinking on safe assets for essential expenses -
"You want to end up with a portfolio that matches your liabilities, meaning the amount you'll need to spend in retirement. The rule of thumb I came up with, based on annuity payouts and spending patterns late in life, is that you should save 20 to 25 times your residual living expenses -- that is, the yearly shortfall you have to make up after Social Security and any pension.
This portfolio should be in safe assets: Treasury Inflation-Protected Securities, annuities, or even short-term bonds.
Anything above that, you can invest in risky assets."
The worst retirement investing mistake - Sep. 4, 2012
Here is another article on the sequence of return risk potential flaw in the balanced stock / bond portfolios for essential retirement expenses -
http://money.msn.com/mutual-fund/the...your-portfolio
11-25-2013, 08:23 PM #6 Thinks s/he gets paid by the post Join Date: Oct 2010 Posts: 4,171 I like your way of thinking about your retirement. More than like, think it's pretty awesome, actually. If I had something to add, I'd add it, but I'm afraid I'm not smart enough. Posting simply because I want to see a checkmark in the list and can see what smarter people say.
11-25-2013, 08:28 PM #7 Thinks s/he gets paid by the post Join Date: Jul 2011 Location: Rio Rancho Posts: 1,527 I'm having trouble with the concept of your plan. The way I read your plan conceptually is that it is layered over time, where the bottom layer is ELE, next above that is DLE, and the top layer is LL. Sort of like Maslow's hierarchy of needs. Is that correct? or did you mean the concept of 'chunks' of time rather than layers. The chunks would align with the periods of time you have layed out (ER, RR, SR, OA). Or do you use both chunks and layers? edit: I have my RE plan laid out in the chunks, but with your two lower layers in each chunk. each bucket is one chunk instead of a bucket per layer. __________________ "We live the lives we lead because of the thoughts we think" Michael O’Neill
11-25-2013, 08:40 PM #8
Recycles dryer sheets
Join Date: Aug 2013
Posts: 349
Quote:
Originally Posted by daylatedollarshort This is Bernstein's recent thinking on safe assets for essential expenses - "You want to end up with a portfolio that matches your liabilities, meaning the amount you'll need to spend in retirement. The rule of thumb I came up with, based on annuity payouts and spending patterns late in life, is that you should save 20 to 25 times your residual living expenses -- that is, the yearly shortfall you have to make up after Social Security and any pension. This portfolio should be in safe assets: Treasury Inflation-Protected Securities, annuities, or even short-term bonds. Anything above that, you can invest in risky assets." The worst retirement investing mistake - Sep. 4, 2012
Thanks, for the quote / link. He pegs retirement age at 65ish. Going 18 years earlier than that means I'll need 35 to 40 times residual living expenses. Also, it looks like even my conservative portfolio is too risky based on the assets he's talking about.
It doesn't make sense to annuitize my portfolio at my age. (Of course, Bernstein would say anyone retiring at my age is crazy.) I don't see it as retiring, I see it as doing work I like to do without worrying about remuneration.
Quote:
Interest rates usually more than keep up with inflation. It's true that real yields right now are historically low, but as a student of financial history I have to believe that's not going to last forever.
Who thought they'd be this low for as long as they have been? How long can the Fed keep "risk free" at a negative real return.
11-25-2013, 08:48 PM #9
Recycles dryer sheets
Join Date: Aug 2013
Posts: 349
Quote:
Originally Posted by timo2 I'm having trouble with the concept of your plan. The way I read your plan conceptually is that it is layered over time, where the bottom layer is ELE, next above that is DLE, and the top layer is LL. Sort of like Maslow's hierarchy of needs. Is that correct? or did you mean the concept of 'chunks' of time rather than layers. The chunks would align with the periods of time you have layed out (ER, RR, SR, OA). Or do you use both chunks and layers? edit: I have my RE plan laid out in the chunks, but with your two lower layers in each chunk. each bucket is one chunk instead of a bucket per layer.
Both chunks and layers.
Using Maslow as you described, the base is ELE, then DLE, then LL at the top. But I divide the ELE up by time periods because the annual ELE changes over a 51 year retirement.
The first period I'm at higher expenses, kids still in the home, and no access to withdraw money form my IRA.
The second period the kids are gone and I can tap the IRA.
The third period I we are older, reduced expenses some, and we're getting Social Security.
The fourth period we're really old and aren't taking trips to Europe.
Add all those periods up and I get my overall ELE. I do the same for my DLE. That is my number. Anything over that number goes into the LL investment pool.
I just realized, with the goal of a 2% real on my DLE, I'll have to adjust those numbers down some to account for the compounded return.
11-25-2013, 08:55 PM #10 Give me a museum and I'll fill it. (Picasso)Give me a forum ... Join Date: Feb 2013 Posts: 5,578 If you want 2% real, 30 year TIPS are close at ~1.50%. Do you have access to stable value funds? Our SV returns have actually been not too bad in real terms since inflation is so low.
11-25-2013, 11:01 PM #11 Dryer sheet wannabe Join Date: Feb 2005 Posts: 17 If all you're trying to do is to keep up with inflation with the ELE portfolio, I'd stay entirely with TIPS/IBonds. You could buy TIPS maturing each year with the specific required income. I'd think that is the lowest risk for someone who has the funds. You might be interested in listening to Zvi Bodie. Also, this bogleheads thread talks about how to build a TIPS ladder that generates an inflation adjusted income for each of the retirement years: Bogleheads • View topic - TIPS Ladder Spreadsheets in General & Two in Particular
11-26-2013, 06:38 AM #12
Recycles dryer sheets
Join Date: Aug 2013
Posts: 349
Quote:
Originally Posted by zakenjanei If all you're trying to do is to keep up with inflation with the ELE portfolio, I'd stay entirely with TIPS/IBonds. You could buy TIPS maturing each year with the specific required income. I'd think that is the lowest risk for someone who has the funds. You might be interested in listening to Zvi Bodie. Also, this bogleheads thread talks about how to build a TIPS ladder that generates an inflation adjusted income for each of the retirement years: Bogleheads • View topic - TIPS Ladder Spreadsheets in General & Two in Particular
Thanks! This is from one of the posts on the bogleheads thread.
Quote:
That sounds about what I'd like to do with my essential living expenses.
11-26-2013, 07:38 AM #13
Recycles dryer sheets
Join Date: Jan 2013
Posts: 117
Quote:
Originally Posted by aim-high Exactly. :-) DW is 18 months older than I and has a family history of longer life. I calculated everything based on her age to 100 and us getting social security when I hit 70. She'll get spousal social security. The income numbers are for the household, we don't separate anything, it's all our pile.
If she has a longer life expectancy than you, make sure you run the numbers assuming there is only one SS and much higher taxes when you die. This can have a big impact on plans.
11-26-2013, 07:48 AM #14 Thinks s/he gets paid by the post Join Date: Oct 2006 Posts: 3,961 I'm pretty analytical, so I like your double-layered analysis. I'm a big fan of splitting expenses into "needs" and "wants" like you have done. Nice work. When you add all this up, you're stock/bond AA is 32/68, and your first year withdrawal rate is 3.1%. That's plenty conservative, which seems to be your intent. It sounds like you've got some money left over for the LL category. Unless you plan to spend all of it in the first couple years, that's another safety net. The step down from RR to SR is pretty large, \$50k. Presumably some SS and some belief that your expenses will go down. Some people here would be pretty cautious about SS. Some will try to plan for the possibility that the last phase of your life will be "dependent" and you may be paying other people to do things for you that you're accustomed to doing yourself.
11-26-2013, 07:59 AM #15 Give me a museum and I'll fill it. (Picasso)Give me a forum ... Join Date: Nov 2010 Location: Vermont & Sarasota, FL Posts: 17,873 I'm actually not that keen on the approach and prefer a single integrated approach rather than chunking it up the way you have. I do have changes to spending as we age in my base plan which it appears that you have. The resultant AA is too conservative given your young ages IMO. Even the Vanguard Target Retirement 2010 fund is 40/60 and the Target Retirement 2020 fund is ~60/40 so your 30/70 seems too conservative to me. I'm older than you and target 60/40 but I concede that we each have different levels of comfort with absorbing investment risk. __________________ If something cannot endure laughter.... it cannot endure. Patience is the art of concealing your impatience. Slow and steady wins the race. Retired Jan 2012 at age 56...60/35/5 AA
11-26-2013, 08:38 AM #16 Recycles dryer sheets Join Date: Aug 2013 Posts: 349 RetiringAt55 - Thanks. Good thinking. I was using both my SS and my DW's spousal SS. I'll remove the spousal from the plan. I don't think there will be a tax impact though because the income will be 90% from a Roth IRA. Independent - correct assumption on the decrease from RR to SR. I was thinking we'd get 36k in SS benefits plus reduced expenses. That will be adjusted (see above) and I'll need to revisit the idea of paying for things we do ourselves, especially in the OA phase. pb4uski - it feels too conservative to me also, especially given the potential of 50 years without producing income. We will have another 10x or so in the LL portfolio. I'm not opposed to risk. We took a lot of it to get here - much more than I realized until I read this forum. Perhaps I've been scared too far the other way. However, I've made it, lost it, and made it again. I don't want to risk losing it again. I'm in the process now of divesting from highly concentrated positions in some rather illiquid investments.
11-26-2013, 08:57 AM #17
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Quote:
Originally Posted by pb4uski I'm actually not that keen on the approach and prefer a single integrated approach rather than chunking it up the way you have. I do have changes to spending as we age in my base plan which it appears that you have. The resultant AA is too conservative given your young ages IMO. Even the Vanguard Target Retirement 2010 fund is 40/60 and the Target Retirement 2020 fund is ~60/40 so your 30/70 seems too conservative to me. I'm older than you and target 60/40 but I concede that we each have different levels of comfort with absorbing investment risk.
It depends on your risk level and retirement expenses. If you have 40 years of retirement expenses in savings + LTC covered, all you really need is to keep up with inflation. Or if you can live off other income streams, like part time work, pensions and/or SS, then you don't really need to take any portfolio risks other than maybe keep up with inflation.
11-26-2013, 01:57 PM #19
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Quote:
Originally Posted by Svensk Anga I considered buying some of the previously issued 20 and 30 year TIPS on the secondary market but decided against it. These can sell for substantially more than par, depending on accumulated inflation adjustments since issue. The government is obligated to pay back par plus inflation at maturity. If we get a bout of deflation (as in Japan recently), one could lose the current premium over par. Better I think to park the money in something shorter term and then buy the TIPS at auction every year to fill out the rungs where a cheap secondary market TIPS is not available. See: Treasury Inflation-Protected Securities (TIPS) - Markets Data Center - WSJ.com Anything over 1000 in the last column is at some risk. The TIPS maturing in the 2040's and 2029 look fairly reasonable. TIPS may not earn enough interest to provide a real return after taxes. In your case, with so much Roth space, you could count on a real return.
We have been doing the same. We sold our initial round of TIPS when it looked like yields had bottomed out. I couldn't bear to see all those 20 - 40% total returns disappear.
Now that yields are rising we are now re-buying the positive yield, longer maturities again in chunks at auction.
I have bought TIPS on the secondary market at times in the past. I would do that again if yields went over 2%. I think the prospect of 20+ years of deflation is not impossible, but slim odds compared to a fairly risk free 2+% real return. I bought some on the secondary market when yields jumped on deflation fears in the past and in hindsight I wish I had bought much more than I did at the time.
11-26-2013, 03:49 PM #20 Give me a museum and I'll fill it. (Picasso)Give me a forum ... Join Date: May 2006 Location: west coast, hi there! Posts: 5,990 Just a comment on the recent emphasis on matching liabilities and bond maturities. This has come about after the 2008-2009 disaster. Before it wasn't nearly as popular. One should pay attention to historical real rates. I'd not want to purchase TIPS myself until they approach historical real returns (2.3% for 10 years, maybe 2.4% for 20 years). Also one does not want to fund a ladder all at once as then one is initially locking in only today's rates. __________________
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests) | 4,978 | 19,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-22 | latest | en | 0.881395 |
https://exercism.io/tracks/java/exercises/rail-fence-cipher/solutions/2e5953a66b1a41478691c01736aab6ae | 1,575,685,013,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540491871.35/warc/CC-MAIN-20191207005439-20191207033439-00209.warc.gz | 363,801,112 | 7,561 | # trevans1's solution
## to Rail Fence Cipher in the Java Track
Published at Jul 13 2018 · 0 comments
Instructions
Test suite
Solution
#### Note:
This solution was written on an old version of Exercism. The tests below might not correspond to the solution code, and the exercise may have changed since this code was written.
Implement encoding and decoding for the rail fence cipher.
The Rail Fence cipher is a form of transposition cipher that gets its name from the way in which it's encoded. It was already used by the ancient Greeks.
In the Rail Fence cipher, the message is written downwards on successive "rails" of an imaginary fence, then moving up when we get to the bottom (like a zig-zag). Finally the message is then read off in rows.
For example, using three "rails" and the message "WE ARE DISCOVERED FLEE AT ONCE", the cipherer writes out:
``````W . . . E . . . C . . . R . . . L . . . T . . . E
. E . R . D . S . O . E . E . F . E . A . O . C .
. . A . . . I . . . V . . . D . . . E . . . N . .
``````
``````WECRLTEERDSOEEFEAOCAIVDEN
``````
To decrypt a message you take the zig-zag shape and fill the ciphertext along the rows.
``````? . . . ? . . . ? . . . ? . . . ? . . . ? . . . ?
. ? . ? . ? . ? . ? . ? . ? . ? . ? . ? . ? . ? .
. . ? . . . ? . . . ? . . . ? . . . ? . . . ? . .
``````
The first row has seven spots that can be filled with "WECRLTE".
``````W . . . E . . . C . . . R . . . L . . . T . . . E
. ? . ? . ? . ? . ? . ? . ? . ? . ? . ? . ? . ? .
. . ? . . . ? . . . ? . . . ? . . . ? . . . ? . .
``````
Now the 2nd row takes "ERDSOEEFEAOC".
``````W . . . E . . . C . . . R . . . L . . . T . . . E
. E . R . D . S . O . E . E . F . E . A . O . C .
. . ? . . . ? . . . ? . . . ? . . . ? . . . ? . .
``````
Leaving "AIVDEN" for the last row.
``````W . . . E . . . C . . . R . . . L . . . T . . . E
. E . R . D . S . O . E . E . F . E . A . O . C .
. . A . . . I . . . V . . . D . . . E . . . N . .
``````
If you now read along the zig-zag shape you can read the original message.
# Running the tests
You can run all the tests for an exercise by entering
``````\$ gradle test
``````
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### RailFenceCipherTest.java
``````import org.junit.Assert;
import org.junit.Ignore;
import org.junit.Test;
public class RailFenceCipherTest {
private RailFenceCipher railFenceCipher;
@Test
public void encodeWithTwoRails() {
railFenceCipher = new RailFenceCipher(2);
Assert.assertEquals("XXXXXXXXXOOOOOOOOO",
railFenceCipher.getEncryptedData("XOXOXOXOXOXOXOXOXO"));
}
@Ignore("Remove to run test")
@Test
public void encodeWithThreeRails() {
railFenceCipher = new RailFenceCipher(3);
Assert.assertEquals("WECRLTEERDSOEEFEAOCAIVDEN",
railFenceCipher.getEncryptedData("WEAREDISCOVEREDFLEEATONCE"));
}
@Ignore("Remove to run test")
@Test
public void encodeWithEndingInTheMiddle() {
railFenceCipher = new RailFenceCipher(4);
Assert.assertEquals("ESXIEECSR",
railFenceCipher.getEncryptedData("EXERCISES"));
}
@Ignore("Remove to run test")
@Test
public void decodeWithThreeRails() {
railFenceCipher = new RailFenceCipher(3);
Assert.assertEquals("THEDEVILISINTHEDETAILS",
railFenceCipher.getDecryptedData("TEITELHDVLSNHDTISEIIEA"));
}
@Ignore("Remove to run test")
@Test
public void decodeWithFiveRails() {
railFenceCipher = new RailFenceCipher(5);
Assert.assertEquals("EXERCISMISAWESOME",
railFenceCipher.getDecryptedData("EIEXMSMESAORIWSCE"));
}
@Ignore("Remove to run test")
@Test
public void decodeWithSixRails() {
railFenceCipher = new RailFenceCipher(6);
Assert.assertEquals("112358132134558914423337761098715972584418167651094617711286",
railFenceCipher.getDecryptedData("133714114238148966225439541018335470986172518171757571896261"));
}
}``````
``````import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
//this was an absolute nightmare. Can't wait to see all the one-line answers to this...
public class RailFenceCipher {
int numberOfRails;
int sequenceLength; //how many characters before rail is back to 0
public RailFenceCipher(int rails){
numberOfRails = rails;
sequenceLength = 2*rails - 2;
}
//I think I could use a stream for this but this way is easier for me to process mentally
public String getEncryptedData(String input){
StringBuilder[] sba = new StringBuilder[numberOfRails]; //StringBuilder for each rail
for(int i = 0; i < numberOfRails; i++){
sba[i] = new StringBuilder();
}
//Split the string into rails
for(int i = 0; i < input.length(); i++){
if(i%sequenceLength > numberOfRails-1) { //going back up the rails
sba[sequenceLength - (i%sequenceLength)].append(input.charAt(i));
}
else { //going down the rails
sba[i%sequenceLength].append(input.charAt(i));
}
}
//concatenate the rails
for(int i = 1; i < sba.length; i++){
sba[0].append(sba[i]);
}
return sba[0].toString();
}
//processing this mentally was a challenge, period
public String getDecryptedData(String input){
int numSequences = (int)Math.ceil((double)input.length() / sequenceLength);
//char array for each sequence of characters
char[][] sequences = new char[numSequences][sequenceLength];
//split characters into sequences
int currentSequence = 0;
int currentRail = 0;
int remainingSequences = numSequences; //decrement this when the last, partial sequence is filled
boolean firstCharacter = true; //all rails except the top and bottom have 2 characters
int sequenceCounter = 0; //used to find place in sequence string
for(int i = 0; i < input.length(); i++){
//put the character in the array at the right place
if(!firstCharacter){
sequences[currentSequence][sequenceLength - sequenceCounter] = input.charAt(i);
}
else{
sequences[currentSequence][sequenceCounter] = input.charAt(i);
}
//update state
if(!firstCharacter || currentRail == 0 || currentRail == numberOfRails-1){
firstCharacter = true;
//decrement remainingSequences if necessary
if((currentSequence*sequences[0].length) + (sequenceCounter+1) >= input.length()){
remainingSequences = numSequences -1;
}
currentSequence++;
}
else {
//decrement remainingSequences if necessary
if((currentSequence*sequences[0].length) + (sequenceLength-sequenceCounter) >= input.length()){
remainingSequences = numSequences -1;
currentSequence++;
}
else {
firstCharacter = false;
}
}
//hit finished sequence, start next rail
if(currentSequence >= remainingSequences){
currentSequence = 0;
currentRail++;
sequenceCounter++;
}
}
//concatenate the sequences
StringBuilder result = new StringBuilder();
for(char[] sequence : sequences){
result.append(sequence);
}
return result.toString().trim(); //trim off the uninitialized characters
}
}`````` | 1,894 | 6,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-51 | latest | en | 0.722143 |
https://www.careerlauncher.com/center-microsite/blog.jsp?id=guqq3vLNb84%3D | 1,701,543,939,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00334.warc.gz | 790,482,413 | 5,091 | ## Crack CAT 2023: Tips to Crack TITA Questions
The Common Admission Test (CAT) is a highly competitive exam that opens doors to prestigious management programs in India. To crack CAT 2023, it is essential to prepare meticulously, especially for the challenging TITA (Type In The Answer) questions. This blog post aims to provide valuable tips and strategies to help you excel in solving TITA questions and maximize your overall CAT performance.
1. Understand the Format: First and foremost, familiarize yourself with the format of TITA questions in CAT. Unlike multiple-choice questions, TITA questions require you to type in the answer directly. These questions test your conceptual understanding and problem-solving skills without any answer options to guide you.
2. Strengthen Your Fundamentals: To tackle TITA questions effectively, it is crucial to have a strong foundation in the relevant topics. Review and revise the fundamental concepts across different sections, such as Quantitative Aptitude, Verbal Ability, Data Interpretation, and Logical Reasoning. Solidifying your understanding of core concepts will enhance your problem-solving abilities.
3. Practice Time Management: Time management is vital in CAT, especially when solving TITA questions. Develop a strategy to allocate time wisely for each question, considering their difficulty level. Set a time limit for each TITA question and practice solving them within that timeframe during mock tests and practice sessions. This practice will help you build speed and accuracy.
4. Enhance Calculation Speed: Since TITA questions require you to compute and derive the answers, improving your calculation speed can significantly boost your performance. Practice mental calculations, learn shortcut techniques for arithmetic operations, and enhance your speed in calculations involving fractions, percentages, and decimals. This will save valuable time during the exam.
5. Read the Question Carefully: When encountering a TITA question, read it carefully to understand the requirement and identify the key information. Pay attention to the units mentioned, data provided, and any specific instructions. Analyze the question from different angles to determine the approach that will lead you to the correct answer.
6. Estimate and Approximate: In some cases, exact calculations may not be necessary to arrive at the answer. Develop the skill of estimation and approximation to quickly narrow down the options. Round off numbers, simplify expressions, and eliminate answer choices that are clearly far from the estimated value. This strategy can save time and help you make educated guesses.
7. Review Your Answers: Since TITA questions do not offer answer options, it is crucial to review your answers for accuracy and precision. Double-check the calculations, ensure that you have provided the answer in the correct format (decimal, fraction, percentage, etc.), and verify that you have answered all parts of the question, if applicable.
8. Practice, Practice, Practice: Consistent practice is the key to mastering TITA questions. Solve a wide range of TITA questions from previous CAT papers and other relevant study materials. Familiarize yourself with different question patterns and difficulty levels. Regular practice will improve your problem-solving skills and build confidence in tackling TITA questions.
Conclusion: With dedicated preparation and the right strategies, you can excel in solving TITA questions in CAT 2023. Understand the format, strengthen your fundamentals, practice time management, enhance calculation speed, read questions carefully, estimate and approximate, review your answers, and engage in regular practice. By following these tips, you will be well-prepared to crack CAT 2023 and achieve your goal of pursuing a management program of your choice. | 709 | 3,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-50 | latest | en | 0.896311 |
https://stacks.math.columbia.edu/tag/080Q | 1,723,376,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00508.warc.gz | 435,645,139 | 6,477 | Lemma 37.16.7. Let $f : X \to S$ be a morphism of schemes of finite presentation. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $x \in X$ with image $s \in S$. If $\mathcal{F}$ is flat at $x$ over $S$ and $(\mathcal{F}_ s)_ x$ is a flat $\mathcal{O}_{X_ s, x}$-module, then $\mathcal{F}$ is finite free in a neighbourhood of $x$.
Proof. If $\mathcal{F}_ x \otimes \kappa (x)$ is zero, then $\mathcal{F}_ x = 0$ by Nakayama's lemma (Algebra, Lemma 10.20.1) and hence $\mathcal{F}$ is zero in a neighbourhood of $x$ (Modules, Lemma 17.9.5) and the lemma holds. Thus we may assume $\mathcal{F}_ x \otimes \kappa (x)$ is not zero and we see that Theorem 37.16.2 applies with $f = \text{id} : X \to X$. We conclude that $\mathcal{F}_ x$ is flat over $\mathcal{O}_{X, x}$. Hence $\mathcal{F}_ x$ is free, see Algebra, Lemma 10.78.5 for example. Choose an open neighbourhood $x \in U \subset X$ and sections $s_1, \ldots , s_ r \in \mathcal{F}(U)$ which map to a basis in $\mathcal{F}_ x$. The corresponding map $\psi : \mathcal{O}_ U^{\oplus r} \to \mathcal{F}|_ U$ is surjective after shrinking $U$ (Modules, Lemma 17.9.5). Then $\mathop{\mathrm{Ker}}(\psi )$ is of finite type (see Modules, Lemma 17.11.3) and $\mathop{\mathrm{Ker}}(\psi )_ x = 0$. Whence after shrinking $U$ once more $\psi$ is an isomorphism. $\square$
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• 2 comment(s) on Section 37.16: Critère de platitude par fibres
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 557 | 1,623 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.643462 |
https://byjus.com/question-answer/arun-is-now-half-as-old-as-his-father-twelve-years-ago-the-fathers-age/ | 1,721,633,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517833.34/warc/CC-MAIN-20240722064532-20240722094532-00755.warc.gz | 126,620,286 | 22,776 | 1
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Question
# Arun is now half as old as his father. Twelve years ago the fathers age was three times as old as Arun. Find their present ages
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Solution
## Let Aruns age be x years now.Then his fathers age =2x years12 years ago, Aruns age was (x−12) years and his fathers age was (2x−12) years.Given that, (2x−12)=3(x−12)2x−12=3x−3636−12=3x−2xx=24Therefore, Arun's present age =24 years.His father's present age 2(24)=48 years.Verification:Arun's ageFather'a ageNow : 244812 years ago24−12=1248−12=3636=3 (Arun's age)=3(12)=36
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https://irawen.blogspot.com/2017/10/single-link-list.html | 1,716,527,003,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00114.warc.gz | 278,784,131 | 42,841 | ## Recent Post
In this list each and every nodes are linked with one another in only one direction from head node to tail node.Here each node has been divided into two parts-
1)Information/Data part- It holds any type of value which may be either numeric or string.
Pointer start points first/head node link part of each node points to the next node.last nodes link part is NULL,if contains null in the link part.it means it is last node of the list.
struct node
{
data_type variable_name;
}
eg:-
struct node
{
int info;
}
Singly linked list insertion in an empty list
start=null
struct node
{
int info;
}
struct node *tmp;
tmp=(struct node *)malloc (sizeof(struct node))
tmp--->info=10;
tmp--->info=info;
start=tmp
Singly linked list insertion at the beginning
first check node is empty or not
struct node
{
int info;
}
struct node *tmp;
tmp=(struct node *)malloc(sizeof(struct node))
tmp--->info=1;
start=tmp;
struct node *P;
P=start;
while(P!=null)
{
print P--->info;
}
Singly linked list insertion at the end-
struct node
{
int info;
struct node *tmp;
tmp=(struct node *)malloc(sizeof(struct node));
struct node *P;
P=start;
while(P!=null)
Insertion at nth position in singly linked list-
P=start
for(i=1;i<position-1&&P!=null;i++)
• Deletion of the only node
• Deletion at the end (last node)
• Deletion of the nth node
Deletion of the only node-
struct node *tmp;
tmp=start;
start=null;
free(tmp);
Deletion at the beginning of first node
struct node *tmp;
tmp=start;
start=start--->;
free(tmp);
Deletion at the end node (last node)
struct node *P;
P=start; | 405 | 1,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.681521 |
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Question From class 11 Chapter SOLUTION AND PROPERTIES OF TRIANGLE
# In , if where all symbols have their usual meaning, then is an acute angled triangle is an obtuse angled triangle is right angled which is not isosceles is isosceles which is not right angled
If then the triangle is right angled (b) obtuse angled (c)isosceles (d) equilateral
3:46
The points are the vertices of (A) an equilaterla triangle (B) an isosceles rilghat angled triangle (C) a scalene triangle (D) an isosceles triangle which is not righat angled
3:50
ABC is a right isosceles triangle, right angled at B. <br> Then = _________
1:49
In any triangle ABC if , then the triangle is(A) Right angled (C) Isosceles (B) Equilateral (D) None of these
1:48
If then is (i) Isosceles (ii) Right Angled (iii) Equilateral (iv) Right angled isosceles
3:31
If are two non-collinear vectors and a, b, and c represent the sides of a satisfying then is (where is perpendicular to the plane of ) a. an acute-angled triangle b. an obtuse-angled triangle c. a right-angled triangle d. a scalene triangle
1:02
ABC is an isosceles right angled triangle whose angle C is right angled. Prove that .
2:21
If and have a common root where a, b and c are sides of a triangle , then
2:36
ABC is a isosceles right angled triangle, right angled at C. prove that
2:34
In a triangle the altitude from is not less than andthe altitude from is not less than . The triangle is right angled (b) isosceles obtuse angled (d) equilateral
2:52
ABC is an isosceles right triangle, right-angled at . Prove that:
1:57
is an isosceles right triangle right-angled at . Prove that .
1:24
Show that points A(3,-1) , B(-1 , 2) and C (6,3) form an isosceles right -angled triangle when joined .
2:07
Is the triangle, whose vertices are , a right-angled triangle, an acute-angled triangle or an obtuse-angled triangle?
1:03
The points , and are the vertices of (A) an obtuse-angled triangle (B) an acute-angled triangle (C) a right-angled triangle (D) none of these
2:44
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Microconcepts | 952 | 3,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-16 | latest | en | 0.777355 |
http://dict.cnki.net/h_53881975000.html | 1,576,231,096,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540553486.23/warc/CC-MAIN-20191213094833-20191213122833-00308.warc.gz | 40,988,079 | 8,200 | 全文文献 工具书 数字 学术定义 翻译助手 学术趋势 更多
fire allocations 的翻译结果: 查询用时：0.157秒
在分类学科中查询 所有学科 航空航天科学与工程 更多类别查询
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fire allocations 火力分配(4)
火力分配
Optimal Partitions and Global Optimization for Aiming Points in Fire Allocations 火力分配的最优剖分与投掷点的全局优化 短句来源 Modeling and Solution of Scheduling Problem and Fire Allocations for Multi-targets BVR Attacking 超视距多目标攻击排序及火力分配建模与解算 短句来源 Aiming at the key problem in BVRAC(Beyond Visual Range Air Combat):scheduling of multi-targets attacking and fire allocations,a synthetical dominant index method is proposed to estimate the operational effectiveness. 针对未来超视距空战条件下的多目标攻击排序和制导武器火力分配问题,提出了一种用以评估超视距空战作战效能的综合优势指数法; 短句来源 An application example in two-to-eight assignment is given,its calculation is implemented by Lindo60.Finally,a nonlinear programming model for fire allocations in one-to-four is proposed and Lingo5.0 software is utilized. 以2对8攻击排序为例,利用求解线性规划软件Lindo6.0进行解算。 最后,建立了1对4攻击火力分配的非线性规划模型,并利用求解非线性规划软件Lingo5.0进行解算。 短句来源
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In this paper, a method for optimization of thefire effect is described, taking into comprehensive consider ationof fire accuracy, fire area, target assignment and fire allocation, Amathematical model is established and simplification of the modelreduces the calculation to a great extent. The objective function is thethreat from the enemy and the constraints are a limited number of theshell, In the light of specific conditions, reduced gradient method isselected, which further reduses the calculation.... In this paper, a method for optimization of thefire effect is described, taking into comprehensive consider ationof fire accuracy, fire area, target assignment and fire allocation, Amathematical model is established and simplification of the modelreduces the calculation to a great extent. The objective function is thethreat from the enemy and the constraints are a limited number of theshell, In the light of specific conditions, reduced gradient method isselected, which further reduses the calculation. By calculation of thepractical example, a desired result is obtaiaed.The model and the alg-orithm given in this paper may be used to optimize the scheme for ope-rating ang commanding the artillery troop. 本文综合考虑我炮兵射击误差,炮弹威力,射击幅员,目标性质,目标分配和火力分配等因素,在弹药量一定的条件下,把敌目标对我的威胁程度最小化。建立了模型,并将模型简化,使问题维数降低,计算量大幅度减小。求解方法结合了问题的特点,选用了简化梯度法使计算量进一步减小。对实例进行了计算,达到了预期的结果。文中所提出的模型与算法可用于炮兵作战指挥方案的优化。 A method is proposed in[1]for fire allocations based on an optimization model.Two problems remain to be solved:(1) how to compute an optimal partition for allocating the firepower to a few groups of targets;and(2)how to compute the aiming points of the firepower so that a globally optimal damage is achieved.In this paper,the problem(1)is completely solved via a dynamical programming method.We modify the Alienor-Gabriel algorithm to solve the problem(2).Some computational results are presented to compare... A method is proposed in[1]for fire allocations based on an optimization model.Two problems remain to be solved:(1) how to compute an optimal partition for allocating the firepower to a few groups of targets;and(2)how to compute the aiming points of the firepower so that a globally optimal damage is achieved.In this paper,the problem(1)is completely solved via a dynamical programming method.We modify the Alienor-Gabriel algorithm to solve the problem(2).Some computational results are presented to compare with those in[5]. 火力分配的最优剖分与投掷点的全局优化李彦君，顾基发，汪寿阳（中国科学院系统科学研究所，北京１０００８０）ＯｐｔｉｍａｌＰａｒｔｉｔｉｏｎｓａｎｄＧｌｏｂａｌＯｐｔｉｍｉｚａｔｉｏｎｆｏｒＡｉｍｉｎｇＰｏｉｎｔｓｉｎＦｉｒｅＡｌｌｏｃａｔｉｏｎｓＹａｎ... Lanchester combat theory and differential games are combined to es-tablish a new model to dynamically describe the infiuence of performance of informationwar system on optimal fire allocation in this paper.In our new model each warring sideis supported by an information war system. At the end of this paper some properties ofoptimal fire allocation strategy are derived and illustrated in tactical sense。 基于兰彻斯特作战理论，提出了一个新的微分对策模型来研究在交战双方均有信息战系统协助作战的条件下的最优火力分配策略。又运用微分对策理论对该模型进行分析和求解，并对所得到的结论作出符合战术意义的解释。 << 更多相关文摘
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# Increase/Decrease Value of variable/cell using scroll bar?
Hi,
is there a way to control the value of a variable or a cell value by
using a scroll bar?
I have 3 stacked bar graphs that compare the value "Amount Invested"
with "Net Income"
of 3 different strategies.
>From the worksheet holding the graph I would like to be able to
increase/decrease the value of "Amount Invested", this value will
affect the "Net Income", and view the results.
ie I have a cell say B2, this is the data series for the top part of
one of the bar graphs "Amount Invested".
I would like to be able to increase/decrease this value by 1% for each
1% move in either direction and be able to watch the graph change as I
do this.
The upper limit for the value would be detirmined by the available
income.
When the value in B2 is calculated I would like the sheet to
automatically calculate the max amount that can be contributed and work
out what percentage B2 is of that notional maximum and put the scroll
bar in the relevant position to represent the current value.
Is this possible?
Bernie
## Related Results
### Macro to increase the value of a cell each time its run
macro to increase the value of a cell each time its run
thats pretty much what i need ! LOL
thanks everyone in advance
### Is it possible to increase the value of all cells in one row on the same number?
Is it possible to increase the value of all cells in one row on the same number, for example on 10?
Is it possible to do in in the MS Word table also ?
### Increase the value of a cell by a percentage
How do i increase the value of a cell by a percentage and keep it in the
existing cell. I have a spreadsheet that has payment values in for customers.
I want to increase this amount by .08%. I have a mailmerge that creates
letters based on the amount in that cell that is why I need to keep it in the
same cell. thanks
### Use vlook to show the value of a cell using a formula
I would like to use the VLOOK formula to show the value of a cell that is
using a formula. For example. I have entered data on Sheet 1 in cell b2. On
Sheet 2 cell b2 is a formula =IF(ISBLANK(Sheet1!b2)," ",Sheet1!b2). Since
there is data on Sheet 1, data appears on Sheet 2. On Sheet 3 I want to use
a VLOOK command that will find the value on Sheet 2 in the A column and show
the value of Sheet 2 b2.(I am using ranges, I have just simplified this for
explanation purposes) I would like to avoid referencing Sheet 1 from Sheet 3
if possible. Can this be done?
David
### How to increase the value of the cell by some percent
I am trying to increase the value of cells by certain percentage. example i have a worksheet with some data in cells C6 to F20 with different numbers. i want to increase all the cells by 5% , is it possible?.
### How can I get the value of a cell to increase in steps?
I want a formula where in a logic statement, I can automatically increase the
value of a cell by one.
### Increase the starting value of a cell by an amount each day
I apologize if this is addressed in another post. I've been searching for this answer in a few different forums including this one and none of the answers seemed to cover exactly what I was trying to do (or I just didn't understand it). Hopefully, someone can help me through here
I keep a running account of my hours at work though a spreadsheet. I make sure the hours I am tracking in this excel sheet match up with the hours I have billed in another system.
The long and the short of it is, I have a cell with a value that I input and I'm trying to get it to increase every work day by 8.
So I put in a starting value.. say 40.. and every weekday, it increases that cell value by 8. Then I make sure the hours I have tracked always match that cell.
Example:
On 12/7 the value of the cell is 40
On 12/8 the value of the cell becomes 48
and so on..
Maybe there is a better way to do this or there is a template that does something like what I'm talking about?
Aside from this work tracking "solution" I am fumbling through, I would like to know the formula to increase the value of a cell by a number each day anyway..
### Text value of a cell
how can I return the text value of a cell useing"sumif" formula
### How to set the value of a cell using its row and column numbers?
If I have the row and column number s of a cell (calculated in one part of
the worksheet) how can I set the value of this cell to the value I want.
This is usally done in Access by setvalue for a control. so you send data to
the place you want. Is this feasible in Excel and how?
### Assigning the previous value of a cell to a variable
Hi
I've been using the following code which assigns the value in the cell that has been modified to a variable 'Result'.
```
VB:
Range)
Dim RowNo As Long, ColNo As Long, Result As String
RowNo = Target.Row
ColNo = Target.Column
Result = Cells(RowNo, ColNo).Value
End Sub
If you like these VB formatting tags please consider sponsoring the author in support of injured Royal Marines
```
This may seem impossible, but I also need to assign the previous value of that cell to a variable.
Is there a way of doing this? It's a long shot....
Cheers
Johno
### How to assign negative value using Scroll Bar in Worksheet
Excel 2003 SP3
Win XP Pro SP2
I use a scroll bar assigned to a cell in order to vary the value of this cell.
The range of the variable can be given in the "Properties" of the scroll
bar. For positive values this works very fine. As soon as I go towards
negative values, the displayed value in the cell jumps to 65'XXX (2 at the
power of something I suppose). How can I get there negative values? If I
itroduce them manually in the assigned cell using the keyboard, it works fine.
### How do i update the values of variables without using the Calculate Sub
My question is: "can i have values changed live on my spreadsheet by a vba function such as -Range("O19").Value = 0- when a timer in cell A1 reaches certain values without any user interaction?"
I have a spreadsheet which has various formulas that are affected by a timer in cell A1.
When certain conditions are met i have a vba routine that will run an external programme dependant on the condition. The routine is part of a Private Sub Worksheet_Calculate() script but because i have 78 different if-then routines to run different versions of the software the whole thing slows down so much that by the time a 'time-flag' has been reached the spreadsheet has missed it.
I am now trying to use the routine under Private Sub Worksheet_SelectionChange(ByVal Target As Range) but i find that the variables that my vba script send to the spreadsheet are not updated until some key movement or mouse click is made in the spreadsheet itself which is not acceptable and that when my timer in cell A1=0 the script to run the external software doesn't fire.
I have 78 of these mini routines for values of P28 of 200 to 278
```If
Range("p28").Value = 202 Then
Range("o19").Value = 0
Range("O15").Value = 0
RetVal = Shell("C:Program FilesWorkspace Macro Pro 6.5Workspace Macro Pro.exe 2-1st&3rd.wksp/u", 1)
End If```
Can anyone offer any advice ASAP 'coz this is driving me crazy.
### Increase value of cell with each new sheet
I was wondering if it was possible to increase the value of a cell each time a new sheet is created within a workbook. For example, I created a series of events with reoccurring dates (2nd through the 15th), on the next sheet I continued the dates (16th through the 29th). If another user creates a new sheet in the future can the copied sheet then continue the patter automatically (30th and on) without having to change the first date manually?
I would like to use this process for other functions as well such as number the pay periods and so on.
Thanks.
### VBA: Referencing a worksheet using the value of a cell
Helllo,
I need to figure out how to reference a worksheet in a MS excel workbook using the value of a cell. For example, instead of having the line - Sheets("Test Macro Sheet").Select I want to be able to type "Test Macro Sheet" into the cell in a different worksheet and have it referenced that way.
Can anyone help with this?
Thank you.
### Plotting in Excel with a y-axis that changes with the value of a cell
Hi, I am kind of new to vba. I am trying to do basically the same as
Andy Pope did on this thread: [FONT=Arial][SIZE=3]VBA To Create Charts/Graphs From Non Continuous Ranges[/SIZE][/FONT]
, i have an excel sheet full with data and I want to plot some graphs but must depend on the variable value of a cell. The value will always be introduced in the same cell. If A3 has a value of 2, then column C should be selected (1+2 = 3 = column"C") and plotted vs. column A (x-axis data)
The value of column A will always be plotted vs. the variable columns that depend on the value of the cell. The data range would begin in row 4 because row 3 is being used to introduce the variable on the cell. This macro/code should be able to run every time that a new value is introduced to the cell. The already plotted graphs should not be deleted.
I was trying to do something like this but it doesnt work!
```
VB:
Range
i = Range("A3").Value
Dim sheetName As String
sheetName = "DataSource"
Dim WSD As Worksheet
Set WSD = Worksheets(sheetName)
finalRow = Range("A65536").End(xlUp).Row
Dim dataString As String
dataString = "R4C1" & ":R4C" & (1+i)
Set rngChtData = WSD.Range(dataString)
Set rngChtXVal = WSD.Range("\$A\$1:\$A\$finalRow")
If you like these VB formatting tags please consider sponsoring the author in support of injured Royal Marines
```
I appreciate your help!!
### Use Value Of Active Cell To Name Row
I'm trying to take the value of a cell and use the value as a name for the row.
If cell a1 has value = June. I want to change the name of row 1 to June.
I'm not sure what I'm doing wrong with the following code.
Sub Name_a_row()
'
'
Dim TheName As String
Dim RowNum As Integer
TheName = ActiveCell.Value
RowNum = ActiveCell.Row
End Sub
Thanks for the help.
### Using visible value of a cell for validation
Dear All,
This is most likely a simple problem.
I have a sheet where the user must input a date (dd-mm-yyyy). Based on this date, an adjacent cell displays the name of the weekday (I do this by copying the value of the original date, and displaying the cell as "dddd".
I would very much like the value of this cell to be the name of the weekday, as I have another cell which checks to see if the inputted date, is (for example) a tuesday. I do this by using:
```
VB:
(CELLA="tuesday";"TRUE";"FALSE")
If you like these VB formatting tags please consider sponsoring the author in support of injured Royal Marines
```
This doesn't work though.
I had hoped that there was a copy/paste option to only copy the visible data of a cell, but no.
Does anyone have any tips on how to do this? I have searched for something similar, but was unable to find anything.
cheers,
Peter
PS. I included a small example, it's very simplified, maybe it's excessive, but it might help explaining what's going wrong.
### Finding max array value of variable cell range
Hi All,
I'm currently stuck in trying to find out to get a max array value
from a variable cell range,
the data is divided in 5 collums, and the variable cell range should
be dependent upon the first column. the maximum value should be
available from column c to e. the first colums has blanks in between
of variable spaces. I need to get the maximal amount of the array
based upon the id code in first column (including the row of the id
code, excluding the row of the second id code)...
example:
A B C D E
12341 data data data
12341
42343
23432
does anyone had a clue how to fix my problem (offset, dmax or
anything) ?? I really don't know how to fix this problem,
best regards
### Increase the value in a cell simply by using the arrow keys?
I would like to be able to increase the value in a cell by 1 by using the up arrow key. Obviously, the default action in Excel is to move up to the next cell. Is there any way to change this, perhaps with a plug in?
### Using the Value of a Cell to Reference a Worksheet
Anyone know how I can use the value of a cell to determine what worksheet I collect data from?
History:
I have one tab that is basically a calendar that lists the days as 1-31. For each day, I am completing numerous calculations based on the data associated with that day.
I also have 31 tabs. One for each day. I input the daily data on the respective day.
I do not want to have to change the formulas for each month and well, as much as I would like the 1st to always be on the same day, it just isn't happening. So come Feb I will need to change the weekday that the 1st falls on (the rest of the days will change automatically). this is why I can't just use =SUMIF('1'!\$H:\$H,"MILE",'1'!\$S:\$S). I can't hard code '1' as the worksheet, I need it to reference the cell that contains the date, say H3 for the sake of an example.
I tried something like:
=sumif((value(H3))!\$H:\$H,"MILE",(value(H3))!\$S:\$S)
I need the worksheet name in the formula to be dynamic.
Any suggestions?
I need it to function in Excel 2003 since it needs to run at work.
### If value of cell +1, then value of another cell -1
I am creating a spreadsheet/inventory count and wish to create a formula. First, we have general merchandise, and then we have pre-packaged kits which contain multiple items together, ready to ship and counted as 1 "pre-packed" unit. So if cells B1:B10 contained the values of how many individual parts/items we currently have, and then B11 was the number of prepacked units, is there a formula so that when the value of B11 is increased (lets say by 1), then the values of B1-B10 decrease by 1? I want it to decrease the values of B1-10 by the amount that the value of B11 increases. So if I increase the value of B11 by 22, all the mentioned cells would decrease by 22 consequently.
Thank you very much in advance, and I apologize for any confusion or obvious ineptitude at Excel!
### VBA - Read Values of Many Cells
Excel 2003 > I am developing a comprehensive VBA function that needs to read the value of many cells in a worksheet. I don't want to Name all those cells because then I will need to pass too many variables to the function. I can use the following set of objects and properties ...
```... but that's not desirable either because the cell locations are then hard coded. Can someone suggest another solution?
```
### In vba what command is used to determine if a particular cell on a particular sheet changed? some kind of event? how to get the old and new value of the cell?
in vba what command is used to determine if a particular cell on a
particular sheet changed? some kind of event? how to get the old and new
value of the cell?
### Increase value of a cell in VBA
I wish to increase the value of cell A1 by 1. How do I do this ?
No luck finding an answer? You could always try Google. | 3,657 | 15,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2015-32 | longest | en | 0.93019 |
https://scholaron.com/homework-answers/let-fx-y--x2--2163361 | 1,638,571,390,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362919.65/warc/CC-MAIN-20211203212721-20211204002721-00184.warc.gz | 544,866,585 | 20,329 | /
/
/
Let f(x, y) = x^2 + 2xy + 5y^2 + 2x + 10y - 3. At which point(s) does f(x, y) have possible maximum
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# Question : Let f(x, y) = x^2 + 2xy + 5y^2 + 2x + 10y - 3. At which point(s) does f(x, y) have possible maximum : 2163361
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) Let f(x, y) = x2 + 2xy + 5y2 + 2x + 10y - 3. At which point(s) does f(x, y) have possible maximum/minimum values?
A) (-1, 17) and (0, 5)
B) (1, 0)
C) (-1, 0) and (0, 1)
D) (0, -1)
E) none of these
2) Let f(x, y) = 5x2 - 5y2 + 2xy + 34x + 38y + 12. At which point does f(x, y) have a possible maximum or minimum value?
A) (4, 3)
B) (4, -3)
C) (-4, 3)
D) (-4, -3)
3) Let f(x, y) = yex + xy2. At which point does f(x, y) have a possible maximum or minimum value?
A) ((√(2)/2), - (√(2)e√(2)/2/2))
B) ((1/2), -e1/2)
C) (0, 0)
D) ((1/2), - (e1/2/2))
E) none of these
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
4) Find all points (x, y) where f(x, y) = 3xy - x2 - y2 - 2x - y + 3 has a possible relative maximum or minimum.
Enter your answer as just (a, b) where a, b are reduced fractions of form (c/d).
5) Find all points (x, y) where f(x, y) = x2 - 2y2 + 4x - 6y + 8 has a possible relative maximum or minimum.
Enter your answer as just (a, b) where a, b are either integers or reduced fractions of form (c/d).
6) Find all points (x, y) where f(x, y) = 2x2 + 2y3 - x - 6y + 14 has a possible relative maximum or minimum.
Enter your answer exactly as just (a, b), (c, d) with b > d and where a, b, c, d are either integers or reduced fractions of form (e/f).
7) Find all points (x, y) where f(x, y) = x3 - y2 - 3x + y + 5 has a possible relative maximum or minimum.
Enter your answer exactly as just (a, b), (c, d) with a > c and where a, b, c, d are either integers or reduced fractions of form (e/f).
8) Find all points (x, y) where f(x, y) = x2 + y3 - 6y2 + 6x - 15y has a possible relative maximum or minimum.
Enter your answer exactly as just (a, b), (c, d) with b < d and where a, b, c, d are all integers.
9) Find all points (x, y) where f(x, y) = x2 + xy + y2 - x - y + 2 has a possible relative maximum or minimum.
Enter your answer exactly as just (a, b) where a, b are reduced fractions of form (c/d) or integers.
10) Find all points (x, y) where f(x, y) = xy - 2x2 + x - 4y + 1 has a possible relative maximum or minimum.
Enter your answer exactly as just (a, b) where a, b are either reduced fractions of form (c/d) or integers.
11) Let f(x, y, z) = x2y + (x/z). Find (∂f/∂z).
Enter your answer as a polynomial in x in standard form (unlabeled).
12) Let f(x, y, z) = ex^2 + y^2 + z^2. Find (∂f/∂z).
Enter your answer as a polynomial in ex^2 + y^2 + z^2 in standard form (unlabeled).
13) Let f(x, y) = √(x2 + y2). Compute (∂f/∂x) at (3, 4).
Enter just a reduced fraction (a/b).
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
14) Let f(x, y) = x4 - y2 + 3x2 + 2y - 7. The first partial derivatives of f(x, y) are zero at the points (0, 1) and (-1, 1). Use the second derivative test to determine the nature of f(x, y) at each of these points.
A) (0,1) neither relative maximum nor minimum, (-1, 1) maximum
B) (-1, 1) relative maximum, (0, 1) neither relative maximum nor minimum
C) (0, 1) relative maximum, (-1, 1) relative minimum
D) (0, 1) no conclusion possible, (-1, 1) minimum
E) none of these
15) Find all points (x, y) where f(x, y) = x2 + xy + y2 - 3x + 2 has a possible relative maximum or minimum. Use the second-derivative test to determine, if possible, the nature of f(x, y) at each of these points.
A) (-2, 1) is a relative maximum
B) (-2, 1) is a relative minimum
C) (2, -1) is a relative maximum
D) (2, -1) is a relative minimum
16) Find all points (x, y) where f(x, y) = x3 - 12y + y2 has a possible relative maximum or minimum. Use the second-derivative test to determine, if possible, the nature of f(x, y) at each of these points.
A) (0, 0) gives a relative maximum
B) Test Inconclusive
C) (0, 6) gives a relative maximum
D) (0, 0) gives a relative minimum
17) Find all points (x, y) where f(x, y) = (1/x) + xy - (1/y) has a possible relative maximum or minimum. Use the second-derivative test to determine, if possible, the nature of f(x, y) at each of these points.
A) (1, 1) gives a relative minimum point
B) Neither a relative maximum nor a relative minimum at (-1, 1)
C) (-1, 1) gives a relative minimum point
D) (-1, 1) gives a relative maximum point
## Solution 5 (1 Ratings )
Solved
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Test: Relations And Functions - 1
25 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Relations And Functions - 1
Description
This mock test of Test: Relations And Functions - 1 for JEE helps you for every JEE entrance exam. This contains 25 Multiple Choice Questions for JEE Test: Relations And Functions - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Relations And Functions - 1 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Relations And Functions - 1 exercise for a better result in the exam. You can find other Test: Relations And Functions - 1 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1
If R is a relation from a non – empty set A to a non – empty set B, then
Solution:
Let A and B be two sets. Then a relation R from set A to set B is a subset of A × B. Thus, R is a relation from A to B ⇔ R ⊆ A × B.
QUESTION: 2
The range of the function f(x) = 7-x Px-3 is
Solution:
Here, 0 ≤ x- 3 ≤ 7 - x
⇒0 ≤ x - 3 and x - 3 ≤ 7 - x
By solvation, we will get 3 ≤ x ≤ 5
So x = 3,4,5 find the values of 7-x Px - 3 by substituting the values of x
QUESTION: 3
Let R be the relation over the set of straight lines of a plane such that l1 R l2 ⇔ l1 ⊥ l2. Then, R is
Solution:
To be reflexive, a line must be perpendicular to itself, but which is not true. So, R is not reflexive
For symmetric, if l1 R l2 ⇒ l1 ⊥ l2.
⇒ l2 ⊥ l1 ⇒ l1 R l2 hence symmetric
For transitive, if l1 R l2 and l2 R l3
⇒ l1 R l2 and l2 R l3 does not imply that l1 ⊥ l3 hence not transitive.
QUESTION: 4
The diagram given below shows that
Solution:
Because, the element b in the domain A has no image in the co-domain B.
QUESTION: 5
Which of the following is an even function?
Solution:
Because, f(- x) = f(x) is the necessary condition for a function to be an even function, which is only satisfied by x2+ sin2x .
QUESTION: 6
The binary relation S = Φ (empty set) on set A = {1, 2, 3} is
Solution:
Reflexive : A relation is reflexive if every element of set is paired with itself. Here none of the element of A is paired with themselves, so S is not reflexive.
Symmetric : This property says that if there is a pair (a, b) in S, then there must be a pair (b, a) in S. Since there is no pair here in S, this is trivially true, so S is symmetric.
Transitive : This says that if there are pairs (a, b) and (b, c) in S, then there must be pair (a,c) in S. Again, this condition is trivially true, so S is transitive.
QUESTION: 7
The void relation (a subset of A x A) on a non empty set A is:
Solution:
The relation { } ⊂ A x A on a is surely not reflexive. However, neither symmetry nor transitivity is contradicted. So { } is a transitive and symmetry relation on A.
QUESTION: 8
A relation R in a set A is called reflexive,
Solution:
A relation R on a non empty set A is said to be reflexive if fx Rx for all x ∈ R, Therefore, R is reflexive.
QUESTION: 9
The domain of the function f = {(1, 3), (3, 5), (2, 6)} is
Solution:
The domain in ordered pair (x,y) is represented by x-coordinate. Therefore, the domain of the given function is given by : {1, 3, 2}.
QUESTION: 10
The domain of the function
Solution:
x - 1 ≥ 0 and 6 – x ≥ 0 ⇒ 1 ≤ x ≤ 6.
QUESTION: 11
Let R be the relation on N defined as x R y if x + 2 y = 8. The domain of R is
Solution:
As x R y if x + 2y = 8, therefore, domain of the relation R is given by x = 8 – 2y ∈ N.
When y = 1,
⇒ x = 6 ,when y = 2,
⇒ x = 4, when y = 3,
⇒ x = 2.
therefore domain is {2, 4, 6}.
QUESTION: 12
If n ≥ 2, then the number of onto mappings or surjections that can be defined from {1, 2, 3, 4, ……….., n} onto {1, 2} is
Solution:
The number of onto functions that can be defined from a finite set A containing n elements onto a finite set B containing 2 elements = 2− 2.
QUESTION: 13
A relation R in a set A is called symmetric, if
Solution:
A relation R on a non empty set A is said to be symmetric if fx Ry ⇔ yRx, for all x , y ∈ R .
QUESTION: 14
The range of is
Solution:
We have ,
Therefore, range of f(x) is {-1}.
QUESTION: 15
The function f(x) = sin x2 is
Solution:
For even function: f(-x) = f(x) ,
therefore, f(− x)
= sin (− x)2 = sin x2 = f(x).
QUESTION: 16
Which of the following is not an equivalence relation on I, the set of integers ; x, y
Solution:
If R is a relation defined by xRy : ifx ⩽ y, then R is reflexive and transitive But, it is not symmetric. Hence, R is not an equivalence relation.
QUESTION: 17
If A = {1, 2, 3}, then the relation R = {(1, 2), (2, 3), (1, 3) in A is
Solution:
A relation R on a non empty set A is said to be transitive if fxRy and y Rz ⇒ xRz, for all x ∈ R. Here, (1, 2) and (2, 3) belongs to R implies that (1, 3) belongs to R.
QUESTION: 18
A relation R in a set A is called transitive, if
Solution:
A relation R on a non empty set A is said to be transitive if fx Ry and yRz ⇒ x Rz, for all x ∈ R.
QUESTION: 19
The range of the function f(x) =|x−1| is
Solution:
We have, f(x) = |x−1|, which always gives non-negative values of f(x) for all x ∈ R.Therefore range of the given function is all non-negative real numbers i.e. [0,∞).
QUESTION: 20
The range of the function is
Solution:
As the denominator of the function is a modulus function i.e.
QUESTION: 21
Let A = {a, b, c} and R = {(a, a), (b, b), (c, c), (b, c)} be a relation on A. Here, R is
Solution:
Correct Answer :- b
Explanation:- A = {a, b, c} and R = {(a, a), (b, b), (c, c), (b, c)}
Any relation R is reflexive if fx Rx for all x ∈ R. Here ,(a, a), (b, b), (c, c) ∈ R. Therefore , R is reflexive.
For the transitive, in the relation R there should be (a,c)
Hence it is not transitive.
QUESTION: 22
A relation R from C to R is defined by x Ry iff |x| = y. Which of the following is correct?
Solution:
QUESTION: 23
A relation R in a set A is said to be an equivalence relation if
Solution:
A relation R on a non empty set A is said to be reflexive iff xRx for all x ∈ R . .
A relation R on a non empty set A is said to be symmetric if fx Ry ⇔ y Rx, for all x , y ∈ R .
A relation R on a non empty set A is said to be transitive if fx Ry and y Rz ⇒ x Rz, for all x ∈ R.
An equivalence relation satisfies all these three properties.
QUESTION: 24
Let f: R → R be a mapping such that f(x) = . Then f is
Solution:
QUESTION: 25
Which of the following is a polynomial function?
Solution:
A polynomial function has all exponents as integral whole numbers. | 2,073 | 6,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-04 | latest | en | 0.863817 |
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## Revision of A way of getting proper normal subgroups of small index from Sun, 19/04/2009 - 00:58
### Quick description
Given a group, it can be useful to know that there is a proper normal subgroup of small index, where 'small' is defined in terms of some finite invariant of the group. To obtain such a result, it suffices to show that the group has a non-trivial action on a sufficiently small finite set. These may arise from the internal structure of the group.
An action of a group on a finite set is nothing more or less than a homomorphism from to , the group of permutations of . As has order , the image of under this homomorphism has order at most ; in fact, it has order dividing , by Lagrange's theorem. We thus obtain a normal subgroup of index dividing as the kernel of this homomorphism. The kernel is a proper normal subgroup if and only if the action of on is non-trivial.
### Prerequisites
Basic group theory
### Example 1
Let be a group with a subgroup such that the index of in is finite. Then there is a normal subgroup of contained in of index dividing .
Proof: We have an action of by right multiplication on the set of right cosets of , that is, a homomorphism from to . Now let be the kernel of ; this is automatically a normal subgroup. The cardinality of is precisely the index of in , so , and hence divides . The subset of that fixes the coset is precisely itself, so is contained in . Hence is the normal subgroup we require.
Indeed, it can be seen that the given above is in fact the largest normal subgroup of to be contained in , known as the \emph{core} of in , in the sense that it contains all normal subgroups of that are contained in . This is because the stabiliser of the coset is the conjugate of , and given any such that is normal in , we have . So acts trivially on the right cosets of , and is thus contained in .
### Example 2
We say a subgroup of a group is if for every automorphism of . This is a stronger property than normality, and is useful primarily for the following reason: if is a subgroup of , and is a characteristic subgroup of , then every automorphism of that sends to itself induces an automorphism of , and hence also sends to itself. In particular, if is normal in then is normal in , and if is characteristic in then is characteristic in (so the property of being a characteristic subgroup is transitive).
In addition, it can be said that 'any subgroup uniquely specified by a family of characteristic subgroups is itself characteristic', even if was chosen arbitrarily. This is because any automorphism of will preserve element-wise, so the specification of is unchanged by applying the automorphism.
Let be a group with normal (characteristic) subgroups and , such that contains with finite, and such that the commutator of and is not contained in . Then has a proper normal subgroup of index dividing .
Proof: Since and are normal in , we have an action of by conjugation on the quotient . Since is not contained in (which is the same as saying that is not in the centre of ), this action is non-trivial. By mapping onto its quotient and composing this with the action of on , we get a non-trivial homomorphism from to . The kernel of is therefore a proper normal (characteristic) subgroup of of index at most . If and are characteristic, then is characteristic, since it has been specified uniquely by our choice of and .
If more is known about the structure of , then the bound on the index of the proper normal subgroup can be improved. For instance, if it is known that all equivalence classes of elements of under the action of its automorphism group have size at most , then must act on a single equivalence class non-trivially, giving a proper normal (characteristic) subgroup of index dividing .
For an example of how to use this in infinite group theory, suppose is a group with a characteristic subgroup of finite index that does not contain the derived subgroup of (this ensures that is non-abelian, so itself acts non-trivially on by conjugation). Now suppose is any group containing as a normal (characteristic) subgroup, but of arbitrary index. Then the argument above applies, and we obtain a proper normal (characteristic) subgroup of of finite index. Moreover, if has a descending sequence of characteristic subgroups of finite index such that each factor is non-abelian, then we obtain a corresponding strictly descending sequence of normal (characteristic) subgroups of .
### This looks nice, but I for
This looks nice, but I for one will not understand it without some help with basic definitions. Here are the ones I don't know: non-central section (or any kind of section for that matter), derived subgroup. I sort of half remember what a characteristic subgroup is but wouldn't mind being reminded.
### Normal subgroups
Here's a simple result in the same spirit: if H is a subgroup of G of index n, then there is a normal subgroup K of G contained in H of index dividing n!. Proof: G acts by left multiplication on the coset space G/H, which has cardinality n. There are only n! possible ways any group element can act here, so the set of elements which act trivially is a subgroup of index dividing n!. But one easily checks that this subgroup is a normal subgroup that is contained in H.
### Normal subgroups
For me, it seems unnatural to say this without mentioning the symmetric group. I would give the proof as follows. The action of on is the same thing as a homomorphism . If is the kernel of this homomorphism then is isomorphic to a subgroup of and so of order dividing . It's clear that is contained in (indeed, it's precisely the intersection of with its conjugates).
### I've gone ahead and written
I've gone ahead and written up this example as a way of trying to get started an article on proving results by letting a group act on a finite set. I think it's a nice one. (I came across it last year when giving supervisions on a first Cambridge course in group theory: the lecturer had set it as a question.)
### Some basic definitions
Tim wrote:
"I for one will not understand it without some help with basic definitions. Here are the ones I don't know: non-central section (or any kind of section for that matter), derived subgroup. I sort of half remember what a characteristic subgroup is but wouldn't mind being reminded."
Let G be a group.
A SECTION of G is simply a homomorphic image of a subgroup of G (or phrased differently if there is a subgroup H
### I see that you too have
I see that you too have written up the example that Terence Tao suggested. I think this is absolutely fine: the example can now be found in two genuinely distinct ways: one by somebody who wants to find a normal subgroup of small index (who will naturally arrive at this article) and the other by somebody who wants to understand why people are interested in group actions and how they manage to use them to solve problems that are not ostensibly about actions (who will be led to the article on proving results by letting a group act on a finite set. In general, I am all in favour of the same example popping up in more than one place, and wouldn't even be unduly disturbed if it was simply cut and pasted from one article to another, though that has not happened in this case.
### I've rewritten the article in
I've rewritten the article in a way that hopefully assumes less background of the reader, and also included Terence's example. This is indeed a very good example of what I am talking about. Is it a bad thing to have it written up here and in Tim's article, or should there just be a link to the most detailed instance of an example (Tim's in this case)? | 1,666 | 7,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-34 | latest | en | 0.929034 |
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COMP171 Hashing Hashing … Again, a (dynamic) set of elements in which we do ‘search’, ‘insert’, and ‘delete’ Linear ones: lists, stacks, queues, … Nonlinear ones: trees, graphs (relations between elements are explicit)
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## Hashing
### Hashing …
• Again, a (dynamic) set of elements in which we do ‘search’, ‘insert’, and ‘delete’
• Linear ones: lists, stacks, queues, …
• Nonlinear ones: trees, graphs (relations between elements are explicit)
• Now for the case ‘relation is not important’, but want to be ‘efficient’ for searching (like in a dictionary)!
• Generalizing an ordinary array,
• An array is a direct-address table
• A set of N keys, compute the index, then use an array of size N
• Key k at k -> direct address, now key k at h(k) -> hashing
• Basic operation is in O(1)!
• To ‘hash’ (is to ‘chop into pieces’ or to ‘mince’), is to make a ‘map’ or a ‘transform’ …
### Hash Table
• Hash table is a data structure that support
• Finds, insertions, deletions (deletions may be unnecessary in some applications)
• The implementation of hash tables is called hashing
• A technique which allows the executions of above operations in constant average time
• Tree operations that requires any ordering information among elements are not supported
• findMin and findMax
• Successor and predecessor
• Report data within a given range
• List out the data in order
### General Idea
• The ideal hash table data structure is an array of some fixed size, containing the items
• A search is performed based on key
• Each key is mapped into some position in the range 0 to TableSize-1
• The mapping is called hash function
Data item
A hash table
### Unrealistic Solution
• Each position (slot) corresponds to a key in the universe of keys
• T[k] corresponds to an element with key k
• If the set contains no element with key k, then T[k]=NULL
### Unrealistic Solution
• Insertion, deletion and finds all take O(1) (worst-case) time
• Problem: waste too much space if the universe is too large compared with the actual number of elements to be stored
• E.g. student IDs are 8-digit integers, so the universe size is 108, but we only have about 7000 students
### Hashing
Usually, m << N
h(Ki) = an integer in [0, …, m-1] called the hash value of Ki
The keys are assumed to be natural numbers, if they are not, they can always be converted or interpreted in natural numbers.
### Example Applications
• Compilers use hash tables (symbol table) to keep track of declared variables.
• On-line spell checkers. After prehashing the entire dictionary, one can check each word in constant time and print out the misspelled word in order of their appearance in the document.
• Useful in applications when the input keys come in sorted order. This is a bad case for binary search tree. AVL tree and B+-tree are harder to implement and they are not necessarily more efficient.
### Hash Function
• With hashing, an element of key k is stored in T[h(k)]
• h: hash function
• maps the universe U of keys into the slots of a hash table T[0,1,...,m-1]
• an element of key k hashes to slot h(k)
• h(k) is the hash value of key k
### Collision
• Problem: collision
• two keys may hash to the same slot
• can we ensure that any two distinct keys get different cells?
• No, if N>m, where m is the size of the hash table
• Task 1: Design a good hash function
• that is fast to compute and
• can minimize the number of collisions
• Task 2: Design a method to resolve the collisions when they occur
### Design Hash Function
• A simple and reasonable strategy: h(k) = k mod m
• e.g. m=12, k=100, h(k)=4
• Requires only a single division operation (quite fast)
• Certain values of m should be avoided
• e.g. if m=2p, then h(k) is just the p lowest-order bits of k; the hash function does not depend on all the bits
• Similarly, if the keys are decimal numbers, should not set m to be a power of 10
• It’s a good practice to set the table size m to be a prime number
• Good values for m: primes not too close to exact powers of 2
• e.g. the hash table is to hold 2000 numbers, and we don’t mind an average of 3 numbers being hashed to the same entry
• choose m=701
### Deal with String-type Keys
• Can the keys be strings?
• Most hash functions assume that the keys are natural numbers
• if keys are not natural numbers, a way must be found to interpret them as natural numbers
• Method 1: Add up the ASCII values of the characters in the string
• Problems:
• Different permutations of the same set of characters would have the same hash value
• If the table size is large, the keys are not distribute well. e.g. Suppose m=10007 and all the keys are eight or fewer characters long. Since ASCII value <= 127, the hash function can only assume values between 0 and 127*8=1016
a,…,z and space
272
• Method 2
• If the first 3 characters are random and the table size is 10,0007 => a reasonably equitable distribution
• Problem
• English is not random
• Only 28 percent of the table can actually be hashed to (assuming a table size of 10,007)
• Method 3
• computes
• involves all characters in the key and be expected to distribute well
### Collision Handling: (1) Separate Chaining
• Lilke ‘equivalent classes’ or clock numbers in math
• Instead of a hash table, we use a table of linked list
• keep a linked list of keys that hash to the same value
Keys:
Set of squares
Hash function:
h(K) = K mod 10
### Separate Chaining Operations
• To insert a key K
• Compute h(K) to determine which list to traverse
• If T[h(K)] contains a null pointer, initiatize this entry to point to a linked list that contains K alone.
• If T[h(K)] is a non-empty list, we add K at the beginning of this list.
• To delete a key K
• compute h(K), then search for K within the list at T[h(K)]. Delete K if it is found.
### Separate Chaining Features
• Assume that we will be storing n keys. Then we should make m the next larger prime number. If the hash function works well, the number of keys in each linked list will be a small constant.
• Therefore, we expect that each search, insertion, and deletion can be done in constant time.
• Disadvantage: Memory allocation in linked list manipulation will slow down the program.
• Instead of following pointers, compute the sequence of slots to be examined
• Open addressing: relocate the key K to be inserted if it collides with an existing key.
• We store K at an entry different from T[h(K)].
• Two issues arise
• what is the relocation scheme?
• how to search for K later?
• Three common methods for resolving a collision in open addressing
• Linear probing
• Double hashing
• To insert a key K, compute h0(K). If T[h0(K)] is empty, insert it there. If collision occurs, probe alternative cell h1(K), h2(K), .... until an empty cell is found.
• hi(K) = (hash(K) + f(i)) mod m, with f(0) = 0
• f: collision resolution strategy
### Linear Probing
• f(i) =i
• cells are probed sequentially (with wrap-around)
• hi(K) = (hash(K) + i) mod m
• Insertion:
• Let K be the new key to be inserted, compute hash(K)
• For i = 0 to m-1
• compute L = ( hash(K) + I ) mod m
• T[L] is empty, then we put K there and stop.
• If we cannot find an empty entry to put K, it means that the table is full and we should report an error.
### Linear Probing Example
• hi(K) = (hash(K) + i) mod m
• E.g, inserting keys 89, 18, 49, 58, 69 with hash(K)=K mod 10
To insert 58, probe T[8], T[9], T[0], T[1]
To insert 69, probe T[9], T[0], T[1], T[2]
### Primary Clustering
• We call a block of contiguously occupied table entries a cluster
• On the average, when we insert a new key K, we may hit the middle of a cluster. Therefore, the time to insert K would be proportional to half the size of a cluster. That is, the larger the cluster, the slower the performance.
• Linear probing has the following disadvantages:
• Once h(K) falls into a cluster, this cluster will definitely grow in size by one. Thus, this may worsen the performance of insertion in the future.
• If two clusters are only separated by one entry, then inserting one key into a cluster can merge the two clusters together. Thus, the cluster size can increase drastically by a single insertion. This means that the performance of insertion can deteriorate drastically after a single insertion.
• Large clusters are easy targets for collisions.
• f(i) = i2
• hi(K) = ( hash(K) + i2 ) mod m
• E.g., inserting keys 89, 18, 49, 58, 69 withhash(K) = K mod 10
To insert 58, probe T[8], T[9], T[(8+4) mod 10]
To insert 69, probe T[9], T[(9+1) mod 10], T[(9+4) mod 10]
• Two keys with different home positions will have different probe sequences
• e.g. m=101, h(k1)=30, h(k2)=29
• probe sequence for k1: 30,30+1, 30+4, 30+9
• probe sequence for k2: 29, 29+1, 29+4, 29+9
• If the table size is prime, then a new key can always be inserted if the table is at least half empty (see proof in text book)
• Secondary clustering
• Keys that hash to the same home position will probe the same alternative cells
• Simulation results suggest that it generally causes less than an extra half probe per search
• To avoid secondary clustering, the probe sequence need to be a function of the original key value, not the home position
### Double Hashing
• To alleviate the problem of clustering, the sequence of probes for a key should be independent of its primary position => use two hash functions: hash() and hash2()
• f(i) = i * hash2(K)
• E.g. hash2(K) = R - (K mod R), with R is a prime smaller than m
### Double Hashing Example
• hi(K) = ( hash(K) + f(i) ) mod m; hash(K) = K mod m
• f(i) = i * hash2(K); hash2(K) = R - (K mod R),
• Example: m=10, R = 7 and insert keys 89, 18, 49, 58, 69
To insert 49, hash2(49)=7, 2nd probe is T[(9+7) mod 10]
To insert 58, hash2(58)=5, 2nd probe is T[(8+5) mod 10]
To insert 69, hash2(69)=1, 2nd probe is T[(9+1) mod 10]
### Choice of hash2()
• Hash2() must never evaluate to zero
• For any key K, hash2(K) must be relatively prime to the table size m. Otherwise, we will only be able to examine a fraction of the table entries.
• E.g.,if hash(K) = 0 and hash2(K) = m/2, then we can only examine the entries T[0], T[m/2], and nothing else!
• One solution is to make m prime, and choose R to be a prime smaller than m, and set
hash2(K) = R – (K mod R)
• Quadratic probing, however, does not require the use of a second hash function
• likely to be simpler and faster in practice
• Actual deletion cannot be performed in open addressing hash tables
• otherwise this will isolate records further down the probe sequence
• Solution: Add an extra bit to each table entry, and mark a deleted slot by storing a special value DELETED (tombstone)
### Perfect hashing
• Two-level hashing scheme
• The first level is the same as with ‘chaining’
• Make a secondary hash table with an associated hash function h_j, instead of making a list of the keys hashing to the same slot | 3,060 | 11,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-17 | longest | en | 0.847399 |
https://www.jiskha.com/questions/365580/a-deli-offers-its-cheese-sandwich-with-various-combinations-of-mayonnaise-lettuce | 1,618,671,935,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00202.warc.gz | 939,038,718 | 5,130 | # math for educators
A deli offers its cheese sandwich with various combinations of mayonnaise, lettuce, tomatoes, pickles, and sprouts. 11 types of cheese are available. How many different cheese sandwiches are possible?
1. 👍
2. 👎
3. 👁
1. Each of the ingredients are in two states, with, or without. So
number=2^5 * 12 assuming only one cheese, or no cheese is a choice
1. 👍
2. 👎
👤
bobpursley
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a 14- gram serving of mayonnaise contains 11 grams of fat. what percen tof the mayonnaise, to the nearest tenth of a percent, is fat? | 822 | 3,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-17 | latest | en | 0.904744 |
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2009-08-11 03:15:20
farah345
Guest
### nice fact
Take any 4-digit number, and its reverse, and subtract the two. For example,
7694-4967=2727
Now add all the digits:
2+7+2+7=18
1+8=9
The answer will always be 9. Can someone prove how?
## #2 2009-08-11 03:29:25
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: nice fact
Hi Farah;
represent the 2 numbers like this.
As you can see the expression on the right hand side is divisible by 9.
Any number that is divisible by nine when you add its digits they will sum to 9.
Last edited by bobbym (2009-08-11 03:30:47)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #3 2009-08-11 03:33:10
Kurre
Member
Registered: 2006-07-18
Posts: 280
### Re: nice fact
Il give you some tips. First prove that the first number (4-digit minus reverse) is divisible by 9. We know (or you can prove that too) that a number is divisible by 9 iff its sum of digits is divisible by 9. So when we first add the digits it must be a number divisible by 9. What possible numbers are there?
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## #4 2009-08-11 09:35:02
Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791
### Re: nice fact
farah345 wrote:
The answer will always be 9. Can someone prove how?
1111-1111 = 0
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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## #5 2009-08-11 10:21:52
soroban
Member
Registered: 2007-03-09
Posts: 452
. .
. .
.
Offline
## #6 2009-08-11 10:34:34
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: nice fact
Hi Farah;
It also works on 5 digit numbers:
Again the right hands side is divisible by 9.
Last edited by bobbym (2009-08-11 10:35:59)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #7 2009-08-11 22:41:49
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,698
### Re: nice fact
Hi soroban,
soroban wrote:
. .
. .
.
I tested this in Excel...which found that it is true for 3645 of the 9000 numbers in the range 1000 to 9999.
I didn't try to weed out multiples of 1111 (or anything else).
Found by Excel:
5-digit results
10890: 3645 times
10989: 640 times
Example: 8991 - 1998 = 6993: 6993 + 3996 = 10989
I haven't tried to work out why, but of the numbers that I checked the middle two digits were always 99.
4-digit results
9999: 2880 times
1170, 1251, 1332, 1413, 1494, 1575, 1656, 1737, 1818 & 1998: a total of 3815 times
3-digit results
261, 342, 423, 504, 585, 666, 747 & 828: a total of 648 times
2-digit results
99: 162
1-digit results
Zero: 90
The digit-sum of the above multiple-digit numbers always reduces to 9: eg,
(a) 10989's digit-sum is 27, whose digit-sum is 9
(b) 9999's digit sum is 36, whose digit-sum is 9
(c) 747's digit sum is 18, whose digit-sum is 9
Last edited by phrontister (2009-08-11 23:37:37)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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## #8 2014-05-03 01:24:58
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,965
Website
### Re: nice fact
Cool Fact: The lastest kernel is kernel 3.14
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
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## #9 2014-05-03 04:05:35
ShivamS
Member
Registered: 2011-02-07
Posts: 3,648
### Re: nice fact
I hear it doesn't make much of a difference.
Offline
## #10 2014-05-03 04:28:55
Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,965
Website
### Re: nice fact
no. not much. but it is the pi'th kernel
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
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#101
Picture a four-dimensional space (x, y, z, t). Picture “a single point” (let’s call it (x, y, z, 0). Is that single point within the framework of the 4-D space? Why yes… yes it is. QED.
It was energy. We can’t get to what it was prior to that, if anything. The universe expands at the point of the Big Bang (and has continued to expand). That doesn’t mean that it didn’t exist at the time of the energy->matter transformation. At that point, we have physical extension (if only at a point).
#102
No, I am not putting any limitation on God. I am just explaining the state of affair.
You see. You use before and yet insist that God created time.
#103
Apparently there is a big debate among scientists at the moment about the nature of time. I don’t think you and I are going to solve it.
I don’t think it makes sense to speak in terms of God existing in time, time isn’t a thing in it’s own right.
#104
God’s being is independent of any temporal state of affairs. God is before time in a way similar to the way a locomotive car is before the caboose, or where a stick pushing a rock is before the rock in order of dependency, insofar as the continued movement of the rock is dependent upon the stick and not vice versa. In such a way, “before” isn’t always a temporal description.
#105
I agree but the OP doesn’t seem to be satisfied with that logic.
#106
Picture a single point. Is that single point within the framework of 4-D space? Two points are required to draw a line, so there are no lines that can be drawn to make axes for 4-D space. We cannot picture a 4-D space within a universe of a single point. QED.
When the point expands (the bang in big bang) it extends in multiple dimensions including time. This sets the framework which we can label with (x,y,z,t). But there is no reason to think that framework existed before the Big Bang. If you work within a theological assumption of creation ex nihilo, there is every reason to think there was nothing before creation.
Same for 11-D space, as well.
#107
It was probably JUST before Genesis 1.1, if I had to guess.
To your point, time is hardly an objective thing, thus your assumption in the question might be unreasonable.
#108
Can a have an n-dimensional space in a single point? I think the answer is ‘yes’. It’s an interesting boundary condition for the math!
I’m not picturing anything existing prior to the existence of the energy which gave rise to the Big Bang. However, at the point (in time, as it were, wouldn’t you say? that would be t=0) at which the energy existed and the Bang occurred, we would already be in a ‘framework’, even if that framework was infinitesimally small. (And, to tell the truth, that’s the real issue here: it’s not really a ‘point’, but an infinitesimally small ‘space’.)
#109
In Big Bang cosmology, the Planck epoch or Planck era is the earliest stage of the Big Bang, before the time passed was equal to the Planck time , t P, or approximately 10−43 seconds. There is no currently available physical theory to describe such short times, and it is not clear in what sense the concept of time is meaningful for values smaller than the Planck time. Wikipedia sv Planck Epoch
The Planck Epoch is the period we are discussing, which is effectively so small that it is equivalent to a single point. When we speak of a singularity, we mean a point of an infinitely dense universe. (Infinitely dense implies the reciprocal of zero volume = single point) The planck constants allow for the gap between the limit of a sequence and the actual sequence, but not by giving enough spacetime to measure.
Once the Planck Epoch has passed (10^-43 sec) it is possible to project axes onto the universe. Before that time the “framework” is imaginary, wholly outside the universe.
#110
Philosophically speaking, however, if you’ve already started the clock, then you’ve already got a t-axis.
#111
The statement is misleading and does not give the complete truth.
#112
And if you have not started the clock, you do not have a t axis.
What is so hard about this? The Big Bang is the moment spacetime starts, the moment the universe is created from nothing. It is a singularity, where things change. Time comes into existence with the Big Bang, not before it. The clock starts, but there is no time axis until 10^-43 seconds have passed.
If you do not believe in the Big Bang, say so. If you don’t think it was a singularity, what do you think. Just do not keep asserting that time was around before, because that defies the definition of the Big Bang.
#113
No. Time was in existence when the angels were created. They had their free will and some disobeyed committing sin. Mortal sin requires sufficient reflection for there to be full consent of the will. And sufficient reflection presupposes at least a short duration of time.
#114
You’re supposing angels think and need to process and reflect on knowledge like we do. If Thomas Aquinas is correct about the angelic intellect. They don’t.
#115
Please explain how an angel would commit a sin of disobedience.
#116
Do they choose, at the moment of gaining new knowledge with no need for reflection or processing, to order their will towards God or against? They do not have brains or material-based consciousness like we do which is what requires us to process, think, and makes us all “wishy washy,” going back and forth. They don’t even have passible emotions like we do for that reason.
#117
Thought is necessarily linear, so yes, your deity is subjected to time.
#118
Completely and spectacularly wrong!
Why would thought need to be linear?
Why would the linearity of thought make God subject to time?
What does a deity subject to time mean? How do you define deity?
#119
We’re talking about the foundational principle of reality here, that which makes reality real, that which can’t properly be said “He is a thing” but only simply “He Is,” Subsistent Existence in itself. What makes you think that this “thinks” at all?
#120
You’re thinking about some type of mega powerful patron of a people.
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Author : Prasant Pandey
Updated On : June 25, 2024
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##### Overview: The quantitative ability section plays an essential role in the IPMAT exam because it holds more weight than the verbal ability section. Those who have studied Mathematics in class 12 may feel that this section is easy. But there are a few students from a commerce or arts background who might feel that the quantitative section is challenging, and scoring in this is tough because of the numerous topics to be completed. We will discuss what type of preparation is required for commerce/arts students to crack the quantitative section of this exam. Get complete insights on scoring the highest marks in this section.
In the Integrated Program in Management Aptitude Test, you will be mainly tested on quantitative aptitude and verbal ability.
As a commerce or art student, you may be confused about preparing for the quantitative section as you don't have a mathematics subject in your 11th and 12th standards.
Remember, the difficulty level of the entrance exam varies between easy to moderate levels. Hence, you can easily crack the exam with a high score with regular practice and preparation strategy.
Our experts at Super grads have compiled a few recommendations to assist you in better understanding how commerce or art students should prepare for the IPMAT.
### IPMAT Preparation Strategy for Commerce or Arts Students
IPM preparation should mix brilliant work and hard work like any aptitude test. Aspirants from any background can easily crack the exam on the first attempt by following a proper study plan and preparation strategy.
The IPMAT entrance exam has only a few questions on probability, polynomials, algebraic equations, etc. So, just focusing on fundamental concepts up to your class 10 will help you clear this section.
Therefore, practicing at least one to two IPMAT Question Papers weekly will help improve your time management skills and speed and know different techniques to solve Maths questions quickly in the exam.
Start brushing up on your grammar skills and vocabulary for the verbal ability section. Also, speed plays a vital role in qualifying for the entrance exam.
### How Can a Commerce/Arts Student Start Preparing for the IPMAT Exam 2025?
Usually, aspirants with a Maths background require three to four months to prepare for the Integrated Program in Management Aptitude Test.
When it comes to commerce/art background, you must require at least six months for exam preparation.
However, the preparation level differs from one aspirant to another depending on your current abilities, how many hours you study daily, your understanding of concepts, etc.
Therefore, it is essential to keep to following points in mind while preparing an IPMAT Study Plan & Preparation Strategy 2025.
### List all Topics You Need to Study!
Perhaps the first step to preparing for the quantitative section is creating a study schedule that includes a list of all subjects and topics per the.
Putting your obligations on paper will help you better understand what you really have to do and which topics you need to focus on more.
If you assign a tough concept from Arithmetic Problems, choose an easy topic, Modern maths. This way, you can build interest to study more and not get bored.
### Know Weaknesses and Strengths
During the preparation, you need to know your weaknesses and strengths. It will help build strong areas and focus on the weaker areas to work hard.
• Start the preparation with the tough ones, and next go to an easier one.
• Try to solve the IPMAT Maths Questions with one or more methods. By doing this, your accuracy level increases in solving different types of questions.
### Develop a Schedule
Based on the time available, you can schedule your study sessions. Add your study sessions to your calendar to ensure you have time set aside expressly for studying.
Plan out which topic you will study on which day to ensure that you devote enough time to each topic.
For example, Monday to Wednesday can be set aside for the topics like number systems, algebra, and geometry. At the same time, Thursday to Saturday can be devoted to time & work, percentage, Simple & Compound Interest, probability, Profit Loss & Discount, and Modern Maths.
If your schedule is busy, you may have to adjust your study plan, but you must learn at least one quantitative topic in one day.
### Take Breaks and Schedule Time for Other Activities Too!
If your schedule includes long hour study sessions, you must ensure to take breaks every so often to stretch, hydrate, and rest your mind. This way, you can keep your brain fresh and help prevent you from feeling overwhelmed.
To achieve daily targets, your mind will be more receptive during time devoted to studying. If you study for long hours, you will get discouraged and be tempted to give up. You should schedule time for nonacademic activities, such as exercise, hobbies, and socializing with other students.
### IPMAT Syllabus for Commerce/Arts Students 2025
Before starting your preparation, you must go through the detailed syllabus, as it will help you get an idea about the topics you need to focus on more per their weightage.
Check out the table below to know the IPMAT Syllabus for Commerce Students.
IPMAT Quant Topics IPMAT Verbal Topics Algebra & Arithmetic Reading Comprehension Number System Paragraph Completion Set Theory Word Jumble Data Interpretation Word Usage Geometry Vocabulary and Analogy Percentage, Profit, and Loss Para Jumbles Time, Speed Fill in the Blanks Distance, Time, and Work Grammar Mensuration Tenses
### Best Books for IPMAT Preparation
In the quant section, the questions are designed to test your problem-solving ability skills. To understand basic concepts of Mathematics, you can refer to the 10th and 12th NCERT books.
R.D. Sharma is one of the best-suggested books for Quant preparation as it explains every topic in-depth and includes many questions.
For English, you can refer to the following IPMAT preparation books:
• Word Power Made Easy by Norman Lewis
• English Grammar and Composition by Wren and Martin
## Subject-wise Preparation Strategy for IPMAT
There are three sections in the IPMAT exam, i.e., Quantitative Ability (MCQ), Quantitative Ability (SA), and Verbal Ability.
As per the previous year's trends, 60% of the question paper comprises quant questions, and 40% includes verbal questions.
The following are some of the best IPMAT Preparation Tips 2025 that will help you score well in the upcoming entrance exam.
Following these tips will help you crack the exam with a high score.
### How should Commerce or Arts Students prepare for IPMAT Quantitative Ability?
In the quant section, the questions are designed to test your fundamentals rather than testing your computational efficiency. So, get comfortable with simple concepts like percentages, time & speed, problems on ages, simple interest, ratios & proportions, mixtures, etc.
Also, you should cover permutation & combinations and probability topics, as most of the time, you can expect maximum questions from these topics. Getting acquainted with these topics would definitely help you get a fantastic score in the exam.
The following are some of the simple tips that will help enhance your IPMAT Maths Preparation.
• Prepare a list of topics in which you are strong and weak. This helps in planning your preparation and studying topics accordingly.
• Try regularly revising all the important formulas related to the abovementioned topics.
• Start and complete your favorite topics first and then move to complex topics.
• Practice as many questions as possible from the previous year's papers to understand your preparation levels.
### How should Commerce or Arts Students prepare for IPMAT Verbal Ability?
In the section, the questions are straightforward and test your overall expertise in the English language and capability to use contextual vocabulary.
The following are some of the important tips to enhance your IPMAT verbal ability preparation.
• Firstly, habitually read the newspaper for at least 45 minutes daily.
• Try to learn 10-15 new words daily and note down the meaning and example of each word you find unfamiliar or difficult to remember. This is one of the best ways to improve your vocabulary.
• For reading comprehension, identify the paragraph's tone; it helps answer questions very quickly.
• Do not stick to one style of reading. Try and explore different types. In this way, you can remember the new words quickly.
• The difficulty level of each question varies in this section. Therefore, analyzing previous year's papers will help you understand the type of questions and the difficulty level.
Download Free IPMAT Study Material
Fill your details
Frequently Asked Questions
How to Prepare for IPMAT without coaching?
What are the tips for Commerce Students to prepare for Maths section?
What is teh IPMAT syllabus for commerce students?
What are the best IPMAT Maths Preparation Tips for Commerce/Arts students?
Can a Non-Maths students crack IPMAT exam on the first attempt?
Whaich is the best study material for IPMAT quantitative ability section?
# How should Commerce or Arts Students prepare for IPMAT 2025?
Author : Prasant Pandey
June 25, 2024
SHARE
##### Overview: The quantitative ability section plays an essential role in the IPMAT exam because it holds more weight than the verbal ability section. Those who have studied Mathematics in class 12 may feel that this section is easy. But there are a few students from a commerce or arts background who might feel that the quantitative section is challenging, and scoring in this is tough because of the numerous topics to be completed. We will discuss what type of preparation is required for commerce/arts students to crack the quantitative section of this exam. Get complete insights on scoring the highest marks in this section.
In the Integrated Program in Management Aptitude Test, you will be mainly tested on quantitative aptitude and verbal ability.
As a commerce or art student, you may be confused about preparing for the quantitative section as you don't have a mathematics subject in your 11th and 12th standards.
Remember, the difficulty level of the entrance exam varies between easy to moderate levels. Hence, you can easily crack the exam with a high score with regular practice and preparation strategy.
Our experts at Super grads have compiled a few recommendations to assist you in better understanding how commerce or art students should prepare for the IPMAT.
### IPMAT Preparation Strategy for Commerce or Arts Students
IPM preparation should mix brilliant work and hard work like any aptitude test. Aspirants from any background can easily crack the exam on the first attempt by following a proper study plan and preparation strategy.
The IPMAT entrance exam has only a few questions on probability, polynomials, algebraic equations, etc. So, just focusing on fundamental concepts up to your class 10 will help you clear this section.
Therefore, practicing at least one to two IPMAT Question Papers weekly will help improve your time management skills and speed and know different techniques to solve Maths questions quickly in the exam.
Start brushing up on your grammar skills and vocabulary for the verbal ability section. Also, speed plays a vital role in qualifying for the entrance exam.
### How Can a Commerce/Arts Student Start Preparing for the IPMAT Exam 2025?
Usually, aspirants with a Maths background require three to four months to prepare for the Integrated Program in Management Aptitude Test.
When it comes to commerce/art background, you must require at least six months for exam preparation.
However, the preparation level differs from one aspirant to another depending on your current abilities, how many hours you study daily, your understanding of concepts, etc.
Therefore, it is essential to keep to following points in mind while preparing an IPMAT Study Plan & Preparation Strategy 2025.
### List all Topics You Need to Study!
Perhaps the first step to preparing for the quantitative section is creating a study schedule that includes a list of all subjects and topics per the.
Putting your obligations on paper will help you better understand what you really have to do and which topics you need to focus on more.
If you assign a tough concept from Arithmetic Problems, choose an easy topic, Modern maths. This way, you can build interest to study more and not get bored.
### Know Weaknesses and Strengths
During the preparation, you need to know your weaknesses and strengths. It will help build strong areas and focus on the weaker areas to work hard.
• Start the preparation with the tough ones, and next go to an easier one.
• Try to solve the IPMAT Maths Questions with one or more methods. By doing this, your accuracy level increases in solving different types of questions.
### Develop a Schedule
Based on the time available, you can schedule your study sessions. Add your study sessions to your calendar to ensure you have time set aside expressly for studying.
Plan out which topic you will study on which day to ensure that you devote enough time to each topic.
For example, Monday to Wednesday can be set aside for the topics like number systems, algebra, and geometry. At the same time, Thursday to Saturday can be devoted to time & work, percentage, Simple & Compound Interest, probability, Profit Loss & Discount, and Modern Maths.
If your schedule is busy, you may have to adjust your study plan, but you must learn at least one quantitative topic in one day.
### Take Breaks and Schedule Time for Other Activities Too!
If your schedule includes long hour study sessions, you must ensure to take breaks every so often to stretch, hydrate, and rest your mind. This way, you can keep your brain fresh and help prevent you from feeling overwhelmed.
To achieve daily targets, your mind will be more receptive during time devoted to studying. If you study for long hours, you will get discouraged and be tempted to give up. You should schedule time for nonacademic activities, such as exercise, hobbies, and socializing with other students.
### IPMAT Syllabus for Commerce/Arts Students 2025
Before starting your preparation, you must go through the detailed syllabus, as it will help you get an idea about the topics you need to focus on more per their weightage.
Check out the table below to know the IPMAT Syllabus for Commerce Students.
IPMAT Quant Topics IPMAT Verbal Topics Algebra & Arithmetic Reading Comprehension Number System Paragraph Completion Set Theory Word Jumble Data Interpretation Word Usage Geometry Vocabulary and Analogy Percentage, Profit, and Loss Para Jumbles Time, Speed Fill in the Blanks Distance, Time, and Work Grammar Mensuration Tenses
### Best Books for IPMAT Preparation
In the quant section, the questions are designed to test your problem-solving ability skills. To understand basic concepts of Mathematics, you can refer to the 10th and 12th NCERT books.
R.D. Sharma is one of the best-suggested books for Quant preparation as it explains every topic in-depth and includes many questions.
For English, you can refer to the following IPMAT preparation books:
• Word Power Made Easy by Norman Lewis
• English Grammar and Composition by Wren and Martin
## Subject-wise Preparation Strategy for IPMAT
There are three sections in the IPMAT exam, i.e., Quantitative Ability (MCQ), Quantitative Ability (SA), and Verbal Ability.
As per the previous year's trends, 60% of the question paper comprises quant questions, and 40% includes verbal questions.
The following are some of the best IPMAT Preparation Tips 2025 that will help you score well in the upcoming entrance exam.
Following these tips will help you crack the exam with a high score.
### How should Commerce or Arts Students prepare for IPMAT Quantitative Ability?
In the quant section, the questions are designed to test your fundamentals rather than testing your computational efficiency. So, get comfortable with simple concepts like percentages, time & speed, problems on ages, simple interest, ratios & proportions, mixtures, etc.
Also, you should cover permutation & combinations and probability topics, as most of the time, you can expect maximum questions from these topics. Getting acquainted with these topics would definitely help you get a fantastic score in the exam.
The following are some of the simple tips that will help enhance your IPMAT Maths Preparation.
• Prepare a list of topics in which you are strong and weak. This helps in planning your preparation and studying topics accordingly.
• Try regularly revising all the important formulas related to the abovementioned topics.
• Start and complete your favorite topics first and then move to complex topics.
• Practice as many questions as possible from the previous year's papers to understand your preparation levels.
### How should Commerce or Arts Students prepare for IPMAT Verbal Ability?
In the section, the questions are straightforward and test your overall expertise in the English language and capability to use contextual vocabulary.
The following are some of the important tips to enhance your IPMAT verbal ability preparation.
• Firstly, habitually read the newspaper for at least 45 minutes daily.
• Try to learn 10-15 new words daily and note down the meaning and example of each word you find unfamiliar or difficult to remember. This is one of the best ways to improve your vocabulary.
• For reading comprehension, identify the paragraph's tone; it helps answer questions very quickly.
• Do not stick to one style of reading. Try and explore different types. In this way, you can remember the new words quickly.
• The difficulty level of each question varies in this section. Therefore, analyzing previous year's papers will help you understand the type of questions and the difficulty level.
Download Free IPMAT Study Material
Fill your details
Frequently Asked Questions
How to Prepare for IPMAT without coaching?
What are the tips for Commerce Students to prepare for Maths section?
What is teh IPMAT syllabus for commerce students?
What are the best IPMAT Maths Preparation Tips for Commerce/Arts students?
Can a Non-Maths students crack IPMAT exam on the first attempt?
Whaich is the best study material for IPMAT quantitative ability section?
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Aug 2, 2006
We are working on a spreadsheet that would project what our labor cost would be for next week. I need some help in figuring out how to calculate overtime when an employee reaches 40 hours.
## Net To Gross Calculation - Payroll
Jul 7, 2007
I have created a spreadsheet to work out deductions from gross weekly pay to give a net amount. I would like to be able to get from a net amount back to a gross amount but am unsure of how to do this.
Deductions are tax and national insurance. A person has a tax code for example 885 meaning they can earn up to 8850 a year before being taxed of 8850/52 a week so anything after 8850 is taxable. There are two tax rates (10% )and (18%). The band for 10% is 10500 per year or 10500/52=201.92 a week. This means any taxable pay above 201.92 is taxable at 18%.
To work out National insurance deductions for an employee, weekly gross amount- 100 x 10% giving the National insurance to be deducted. So Gross - Tax - National Insurance = Net amount.
Attached is a spreadsheet I created to work it out. Firstly does this look okay? Secondly is it possible to get back to the gross amount given ONLY a net amount and a tax code?
## Overtime Calculation
Nov 12, 2009
=IF(a9>40,(a9-40*1.5))
Obviously this is not correct because the result is FALSE.
## Calculation Of Overtime
Feb 1, 2008
I use Excel 2007, and I need help with an overtime calculator. It pertains to a specific wage order, which has three basic principles:
• Any hours over 16 in one day are double-time. (2x)
• Any hours over 40 in a week are time-and-a-half (1.5x)
• Any hours over 48 in a week are double-time. (2x)
I worked 5 hours on a Monday, 18 hours on a Tuesday, 18 hours on a Wednesday, and 13 hours on a Thursday. (I work in a residential group home, so 24 hour shifts are common). That totals 54 hours, and the correct overtime breakdown should be:
• 40 regular hours.
• 8 hours at time-and-a-half, and
• 6 hours of double time.
I’m using the following formulas: ...
## Overtime Calculation On Timesheet
Apr 12, 2008
" =(C2 >D2)*MEDIAN(0,D2-1/4,1/2)+MAX(0,MIN(3/4,D2+(C2 >D2))-MAX(1/4,C2)) "
approach to sort out Day/Night Hours. Its bomb proof!
A new situation demands overtime payments......start and finish time can be any time day or night (crap job!), overtime is payable after 8 hours. Thus I have day (0600-1800) standard rate, day (0600-1800) overtime rate, night (1800-0600) standard rate, night (1800-0600) overtime rate.
So, starting at 1400 and finishing at 0100 give 4 hours day std + 4 hours std night + 3 hours night o/time; whereas starting at 0200 and finishing at 1300 gives 4 hours std night + 4 hours day std + 3 hours day o/time.
I'm using Excel 2003 and 2007 so use the Excel 97-2003 format.
## Overtime Calculation For Time Worked Over 8 Hr Per Day Or Over 40 Hours Per Week?
Aug 4, 2013
My overtime pays is anything exceed over 8 hours per day or over 40 hours per week. Right now I can only calculate overtime by either over 8 hr/day or over 40 hr/ week. I need a way to combine both.
## Calculation Of Shift Allowance Affected By Overtime And Overnight Shifts
Apr 21, 2014
I have an Excel sheet where users enter shift start and finish times (normal Excel time format) - for example: A1 might be 18:00 and A2 might be 06:00 for an overnight shift from 6pm to 6am.
All I need to calculate from these times are the number of hours to which an allowance applies, under the following conditions.
The allowance is paid for all hours worked between 18:00 and 06:00.After 10 hours, a shift becomes overtime and no shift allowance is paid.Shifts are regularly worked overnight (i.e. past midnight into the following morning)
Example scenarios include:03:00 - 15:00 would pay 3 hours of shift allowance09:00 - 21:00 would pay 1 hour of the allowance (as the shift becomes overtime from 19:00)03:00 - 21:00 would pay 3 hours19:00 to 08:00 would pay 10 hours (as the shift becomes overtime from 05:00 the following morning)22:00 to 10:00 would pay 8 hours
And so on.
I feel like I’ve got most of the pieces of the jigsaw, but I can’t put them together! I’ve got the following formulae working out bits of what I think I need:
[Code] ........
The above works out the number of hours before 6am and after 6pm respectively (which I can then SUM), and I believe also accounts for overnight shifts. This obviously doesn't include the more-than-10-hours criteria yet.
The larger formula now looks like this:
[Code] ......
…but this still doesn’t work properly! The sections referring to 1.25 were my attempt to get the shift allowance to stop if the past-midnight shift continues past 06:00, but I don't think it works properly.
I also know I’m probably using *24 and /24 more than I need to, but that’s partly so I’ve got a better grasp of what the formula is doing.
Once this is working, I'm happy using an IF… >10 formula to prevent the total number of hours of shift allowance being more than 10. However, I’m really struggling to find one single formula that will factor in shifts that might start before 6am and finish after 6pm (i.e. 05:00 – 19:00, which should pay one hour), shifts that might go past midnight and possibly past 06:00 the next day, and so on.
Lastly – not to try and complicate things further – there is an optional cell elsewhere, say A3, where a user enters ‘Y’ if the individual takes an unpaid 30 minute break at some point during the shift. This is to be deducted from whatever type of hours are being paid at the end of the shift. For example, if a shift is from 01:00 - 11:00 with the break, it would pay 5 hours with a shift allowance and 4.5 without. If the shift was 13:00 - 23:00, it would pay 5 without the allowance and 4.5 with the allowance. Is there a practical way of doing this, or does this become much more complicated?
## Excel 2003 :: Time Card / Sheet For Semi-monthly Pay Period With Overtime Calculation
Nov 19, 2012
I'm working on a dynamic payroll spreadsheet that will automatically calculate the overtime worked in a week. Right now, I'm running into a snag. My issue is with the formula in Column R. Right now, as shown below, it is doing the calculation based on regular hours minus 40 to determine the OT time. The snag is very messy and it lay in this: while the row by row calculations for total overtime worked for the week is correct, the sum at the bottom is very much off. I need an accurate method to sum the hours of overtime for the given column.
Here are the guidelines for the pay periods and overtime:
1. The pay periods for the month go from the 1st to the 15th and the 16th to EOM (End of Month). This means that the pay period could end on any given day of the week. More on this in a moment.
2. A work week is defined as Sunday to Saturday.
3. Overtime is calculated based on the rule of anything over 40 hours in a given work week.
4. Holiday hours worked do not count towards the 40 hour mark in granting overtime since Holiday pay is automatically overtime.
If it were just a matter of a bi-weekly (every 2 weeks) pay period, I would simply state =IF(weekday(DATE)=7,Hours_Worked - 40,0), and tag a SUM(range) at the bottom. Unfortunately, with it being a semi-monthly (twice a month), the end of the pay period could be a Wednesday, so a reference to day of the week won't work unless the formula can dynamically determine which set of data to evaluate.
I'm completely willing to toss out the current method of determining overtime. This is the calculations sheet that references a cleanly formatted and designed time card on a tab called "Time Card", so this isn't the full workbook. In fact, once the whole thing is done, this calculation sheet will be hidden.
Columns M and N (which are formula referenced in Column P) are basic End - Start calculations and were hidden to simplify the display as well as the number of formulas displayed.
Column L (formula referenced) is a Yes/No display for if the date in question is holiday pay.
Excel 2003
H
I
O
P
Q
R
S
1
Start Work
Time Out
Day Count
[code].....
## Overtime And Double Overtime Calculations?
May 21, 2014
calculate overtime in California. I found a few previous posts, but none that meet all of the requirements. Employees get overtime in CA for:
More than 8 hours are worked in a day (up to 12)
More than 40 hours are worked in a week
Hours worked on the 7th consecutive day (up to 8)
Employees get double overtime for:
More than 12 hours are worked in a day
More than 8 hours worked on the 7th consecutive day
To eliminate the the 7th day issues, I am just using helper cells for hours worked on the 7th day of the work week. So far, what I have is what I found in a previous post:
=MAX(0,SUM(A13:A26)-40-SUMIF(A13:A26,">8")+8*COUNTIF(Daily Total Hours,">8"))+SUMIF(A13:A26,">8")-8*COUNTIF(A13:A26,">8")
This will calculate the hours of overtime over 40 in a week and 8 in a day, but will not differentiate between hours 8-12 and hours 12-?
## Payroll Save As PDF Using Macro?
Jun 17, 2014
I have a problem in saving multiple pdfs from excel. I am posting a sheet here. First sheet has the employee details. Second sheet have the template for payslip. My aim is to save pdfs of templates of all the employee in the list using a macro.
currently i am doing it as
1. In sheet 2, I lookup each employee using the serial number of employees at the top left corner cell of the template.
2. Once the details are updated in sheet 2 i save it as pdf in my desktop.
The problem with recording macro and looping is that the pdfs are overwritting and the last pdf alone is saved. i could not find how to change the pdf name for every loop.
## Payroll :: Calculating These Formulas?
Jan 12, 2007
I am having a little trouble with IF. My pest control sales people are paid on straight commission. In addition to paying them a sliding percentage of the total contract value I pay them 50% of everything above \$100 they charge on the initial service.
calculating these formulas?
Attached is the payroll document.
## Function To Calculate Payroll
Nov 18, 2008
I am trying to calculate payroll. I have 2 columns where regular hours and overtime hours are manually inputed. They are then multiplied by their respective pay per hour columns to come up with regular and overtime pay. The next column adds these to get total gross pay. That gross pay is then multiplied by the FICA and FICA Med factors to figure those taxes(2 different columns). I then have a column the adds all deductions to get total deductions(Fed,FICA,FICAMed,St). The last column subtracts total deductions from the gross pay column to get net income. My problem is the net pay column is \$.01 off sometimes.
I think what is happening is I obviously have all columns in dollar amounts with 2 decimal points. Some function columns have multiple decimal points in the answer and then are only showing the 2 decimal points. When those columns are used in the next equation, instead of using the dollar amount that is showing with 2 decimal points, it is remembering the multiple decimal points. This is resulting in being a penny off when I get to the end. How can I get the equations to use what actually shows in the columns(2 decimal points) instead of remembering multiple decimal places?
## How To Lookup Payroll Records
Mar 7, 2014
Each payweek I receive about 600 records some with duplicat names if an ee works overtime. I then look at previous payroll and extract a given account code. Because some ee's work overtime the rows do not always synchronize in count or names.
Any BEST lookup function? Or BEST tool? Is access a more logical option?
## Hyperlink To 53rd Payroll Week
Jul 30, 2013
Sso if you've done payroll before, you may have run into the 53rd instance of a payroll week. Today I've been editing my payroll template to compensate for this occurrence and I've almost got it complete but of all things to mess up on me, a hyperlink is the problem.
I have a legend where I can see all of the payroll week periods, the amount paid, PTO used, etc. And ALL of my payroll sheets are named "Payroll #1", "Payroll #2" ... "Payroll #52" and have a cell on my legend that hyperlinks to cell \$A\$1 of the corresponding sheet. But when I went to add in this 53rd week, the hyperlink has decided not to work. It will work if I name it anything else, and it will work if I take the # out of the name it will work, but I want to have the # in the sheet name because all other 52 sheets are named thusly and I would rather fix this one link than go back and replace 52 other working links and risk breaking something else.
Here is an example of one of my working links:
Screen Shot 2013-07-30 at 5.41.42 PM.png
I created the Payroll #53 the same way:
Screen Shot 2013-07-30 at 5.42.50 PM.png
But it always resets to this as soon as I try using the link:
Screen Shot 2013-07-30 at 5.42.15 PM.png
And pops up this error:
Screen Shot 2013-07-30 at 5.47.35 PM.png
I do have a VBA Macro that hides both that sheet, and the row referencing Payroll #53 on the legend by using a checkbox. I'm not sure why the # sign is making everything so difficult when there are 52 other hyperlinks that use it also. I tried deleting my macro to see if that was part of the problem but it still refuses to properly create the hyperlink. I also tried rebooting to see if that was part of the problem but it wasn't.
As a side note, this particular cell will not link to any page with a # in the name.
Screen Shot 2013-07-30 at 5.45.51 PM.png
I tried creating the hyperlink on a Windows based machine (since Microsoft is native to the OS after all) and it worked. So it must be a bug or something in the Mac version.
## Profit / Loss Payroll And Cost Of Supplies
Mar 6, 2013
I have multiple departments I need to keep track of multiple numbers with. For example, I need to see the cost of payroll for last fiscal year, vs this fiscal year and have excel show that as a gain (bad) or shrink (good).
I need to track sales this FY vs last FY, obviously, a gain is good and shrink is bad.
Finally, I need to track d&d goods this FY vs last FY.
Where I'm getting held up is, for example last year's payroll was 12000 and this year's is 12500 but next months is 11000 this FY vs 13000. How do I get excel to figure out that payroll decreasing is good and not just do simple subtraction or negative addition thereby allowing me to just plug the numbers in and excel figure out if what I'm throwing at the cell is bad or good comparing to last year?
Also, if I could figure out the if/then to show positive numbers green and negative numbers red that would be quite awesome too.
## Desperately Seeking Solution - Payroll Tax Withholding
Feb 1, 2007
I need to determine federal tax withholding from gross pay.
If gross amount (for S-0 = single no dependents) is
Over.................But NOT Over............Then Subtract.........And multiply
\$0.00...............\$195.00.....................\$5 1.......................10%
\$195.00............\$645.00.....................\$99 .......................15%
\$645.00............\$1,482.00..................\$351 ......................20%
\$1,482.00..........\$3,131.00..................\$447 ......................25%
\$3,131.00..........\$6,763.00..................\$849 .....................30%
If gross is \$400 then what are the total taxes taken out? Could someone please provide a formula?
I tried this one: =IF(M10
## Payroll Extract To Work Only On Certain Pay Period Dates.
Feb 9, 2007
I currently have an Excel payroll extract that populates a start date and end date via the calender control 11 user form.
What I have found out is I need to inject some sanity to this application. I can not have users select days that are outside of a pay period.
The users should only be able to select the first or the 16th of the month for a pay period start. Then they can only select the 15th or the 28th/31st for the pay period end.
What I am trying to do is have them select the month and the pay period start date, then the end date would automaticly be selected. But I don't want to have to create a bunch of loops to counter for the differing month end dates or leap years.
## Creating Payroll Spreadsheet With Dropdown List Linking To Values
Dec 3, 2013
I am trying to attempt to create a payroll spreadsheet with certain aspects and with multiple worksheets. so what im trying to accomplish is this
1/ on my main worksheet would have the payroll template there would be 2 columns (1) would be "routes" (2) would be the dollar value of that route.
2/ i want to be able to pick from a drop down list in colume (1) which will have various routes such as A,
B, C, D, E etc
3/ once i have selected a route in colum (1) i want to have the dollar value that is associated to that route to show up in column (2)
ex/ route A worth \$1
route B worth \$2
create a dropdown list but thats about it, i have not been able to link any values together.
## Payroll Spreadsheet - Highlight Cells That Have Hours Entered For Vacation
Jun 18, 2014
I have a payroll spreadsheet and I want to highlight cells that have hours entered for vacation. For example, in cell E4 I put the job name which is "VACATION" and in cells G4 through M4 I put the hours in for each day. (G4 is Monday, H4 is Tuesday, etc.)
I tried to put a conditional formatting using the following formula ="IF(E4=""VACATION"")" then I chose the fill color however it is not working.
## How To Calculate Overtime
Sep 5, 2008
I am trying to formulate a formula that will calculate overtime hours worked.
Now standard hours are 17:30pm - 20:45pm. Anything outside these hours are overtime. If the start time is 18:00pm then the person is still paid from 17:30pm @ standard rate regardless.
Now I am trying to work out a formula that will cover hrs outside of the standard hrs AND hrs unworked but paid for.
see attached! September tab {blue highlighted cells}
## Calculating Overtime
Apr 3, 2008
I'm looking to calculate OT wages when they happen vs only at the End of Week totals. ie... if the employee hits 40 hours midshift on a Wed, I want to calculate what the total dollars would be for Wed.... a few hours at regular time plus what ever hours above 40 at time and a half.
## Sum Overtime Hours
Mar 19, 2009
I have a report given to me formatted as general. These are overtime hours for 5000+ associates. The time is shown as 4.52 being 4 hours and 52 minutes.
If I sum 4.52, 5.1, .18... I get 9.8 when in fact it is 10hrs 20m. I need this to display as 10.2
In fact I have done it in the past but lets just say im ready for the weekend.
## Calculating The Overtime
Jan 15, 2010
We are guaranteed (right now) a 10 hour day. (we are on 4 - 10 hour shifts). So if we work 9 hours say on Monday, we get 9 hours of straight time and an hour of short work week (approx 80% of pay). Now if we work 11 hours on Tuesday (which they can do unfortunately) I get 11 hours of straight time and no overtime. We have to make up for the short work week hour. So a less complicated explanation would be if I dont work more than 40 hours per week, no overtime no matter what I actually worked per day. Seems pretty simple but what I want and need to do is to calculate it per day.
Mon 10 Hours
Tues 11 Hours
Wed 9.5 Hours
Thurs 10 hours
I should get:
Mon 0 overtime
Tues .5 overtime
Wed 0 overtime
Thurs 0 overtime
I got it to the point where If the day where we get short work week is first and overtime after that, it works. But if we work overtime first then get short work week later in the week, it wont calculate it. I know why it wont work now but I dont know how to make it work. LOL Here's a link to the file.
## Calculate Overtime
Jan 9, 2007
I have a time sheet for my employee's that I need to calculate their overtime in 1.5x and 2x rates.
Their overtime totals are done in individule columns from D33 to S33. The first 4 hours per day are charged at 1.5x and anything over that is 2x. I want to show the 1.5x in one box and the 2x in another. I do believe that I need two formulas one in each of the boxes where the final totals would go.
Here's an example, in columns D37 to D41 the employee has worked 12, 14, 9, 16 and 14.5 hours. so that's 17 hours @ 1.5x and 8.5 @ 2x.
## Counting Overtime With A Twist
Feb 5, 2010
Each employee has a different plan time each day of the week.
Mon - 9 Tues - 9 Wed - 7 Thur - 8 Fr - 7
Overtime is really counted after 40
There are 5 sheets for each day of the week mon-fri lets say sheet1..2...3...etc, on each sheet there is a column a with plan and column b with actual hrs. What I'm looking for is a summary sheet for each employee to see where they trend in OT after each day so...
person1 works 10 hrs on mon with a 9 hour plan thus 1 hr trending OT. On the summary page person1 would now show 1 hrs of OT. Now if they would work 9 hrs on tues it would still show 1 hr of OT. On wed however they worked 6 hrs so now the summary page would 0.
## Employee Timesheet With Overtime
Nov 2, 2008
i am creating a weekly time sheet for my company.the problem that i have is when the persons time reaches 40 hours, the time needs to be calculated in the overtime field. this is really tough for me when the person reaches 40 hours in the middle of the work day. I cant figure it out. i have attached the spreadsheet if you would like to look.
## Calculate Overtime Hours
Aug 6, 2008
I am trying to get my time sheet to work out right but for some reason the formulas are a little more then what I can do. My time sheet is set up on a weekly bases. I have a regular time line, an overtime line and a total time line. an example I have is I work 12 hours a day I need an 8 to show up in the regular hours, 4 in the overtime and 12 in the total hours.
## Calculating Overtime In Pivot Table
Nov 30, 2010
How to report Overtime in a pivot table. Apparently this is more difficult than it seems. Please take a look at the cross posting at [URL]....
I have a sample file there
## IF Formula With Array - Calculating Overtime?
Dec 15, 2013
I have 3 basic job categories... each of those categories start overtime at a different hour.
So, if employee A is a dock worker, he starts overtime at 25 hours. If employee A is an office worker, he starts overtime at 40 hours and if he is a driver, he starts overtime at 55 hours. So based on that info, I'm wanting my spreadsheet to figure out how many hours each employee has left for the week.
The 2nd part question is how many hours per day is left for the week. Rather than making a separate tab for each day of the week, I'd rather the spreadsheet know what day of the week it is and divide accordingly.
## Sum Formula - Calculating Overtime Hours
Feb 20, 2014
Example: In cells A1:A10 random number between 0 & say 20, need to sum ABOVE 8 = (calculating overtime hours)
E.g.
A1 = 0
A2 = 8
A3 = 8
A4 = 10 (giving 2)
A5 = 12 (giving 4)
A6 = 5
A7 = 13.5 (giving 5.5)
A8 = 8
A9 = 0
A10 = 16 (giving 8)
A11 = (Total overtime above 8 hours) 2+4+5.5+8 = 19.5
Need to be able to increase rows and drag across. | 5,980 | 23,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-26 | latest | en | 0.947204 |
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# Chapter_7_Kim_F.pdf
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Chapter 7 Frequency Response 1
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Objective Discuss the general frequency response characteristics of amplifiers of amplifiers. Derive the system transfer functions Develop the Bode diagrams of the magnitude and phase of the transfer functions. Analyze the frequency response of transistor circuits with capacitors with capacitors. Determine the Miller effect and Miller capacitance. Determine the high-frequency response of basic Determine the high frequency response of basic transistor circuit configurations. 2
7.1 Amplifier Frequency Response So far, it is assumed in AC circuit analysis that: analysis that: (1) The coupling and the bypass capacitor are short (2) The parasitic load and transistor capacitance are open. In reality, gain factor including voltage, current, transconductance are a function of frequency. We need to investigate the effects f it i ll f 3 of capacitors in all frequency range.
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7.1 Amplifier Frequency Response Band Width Why this kind of shape? (1) f<f L Coupling and bypass capacitor is not a perfect short. (2) f L <f<f H Coupling and bypass capacitor are short. Load and TR capacitors are open (3) f>f H All are getting short. Ex) Audio : 20Hz to 20Khz
7.1.1. Equivalent Circuit Depending on frequency range, the equivalent circuit changes. (1) f < f L (Low frequency range) C li d b it t h t - Coupling and bypass capacitor are not short. - Load and TR capacitance is ignored. (2) f L < f < f H (Mid frequency range) Co pling and b pass capacitor are short - Coupling and bypass capacitor are short. - Load and TR capacitors are open (3) f > f H (High frequency range) Coupling and bypass capacitor are short - Coupling and bypass capacitor are short. - Load and TR capacitors are not open, getting short. Assumption: f L << f H 5
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7.2. System Transfer Function -Transfer Function- ) ( ) ( s Y H h(t) or H(s) x(t) y(t) X(s) Y(s) ) ( s X s = ) ( ) ( ) ( s X s H s Y = Name of function Expression Voltage TF C t TF ) ( / ) ( s Vi s Vo ) ( / ) ( s Ii s Io Current TF Transresistance TF T d TF ) ( / ) ( s Ii s Vo ) ( / ) ( 6 Transconductance TF s Vi s Io | 615 | 2,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-30 | longest | en | 0.862144 |
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# NOMINAL function
This article describes the formula syntax and usage of the NOMINAL function in Microsoft Excel.
## Description
Returns the nominal annual interest rate, given the effective rate and the number of compounding periods per year.
## Syntax
NOMINAL(effect_rate, npery)
The NOMINAL function syntax has the following arguments:
• Effect_rate Required. The effective interest rate.
• Npery Required. The number of compounding periods per year.
## Remarks
• Npery is truncated to an integer.
• If either argument is nonnumeric, NOMINAL returns the #VALUE! error value.
• If effect_rate ≤ 0 or if npery < 1, NOMINAL returns the #NUM! error value.
• NOMINAL (effect_rate,npery) is related to EFFECT(nominal_rate,npery) through effective_rate=(1+(nominal_rate/npery))*npery -1.
• The relationship between NOMINAL and EFFECT is shown in the following equation:
## Example
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Data Description 0.053543 Effective interest rate 4 Number of compounding periods per year Formula Description Result =NOMINAL(A2,A3) Nominal interest rate with the terms above 0.05250032 | 314 | 1,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-35 | latest | en | 0.71203 |
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Member
i fuked up the easiset question q 10 my first reaction was a buudhism which i put down ,but then to be a sick kunt and prove it to myself mathematically i done the number in 96-number in 91 instead of working out the percentage in the rush i put the answer 2 be b ffs im really pissed off
well 7 out of 10
bex
Active Member
I got...
1a
2c
3b
4a
5a
6c
7c
8d
9c
10b
Buddhism has the greatest rate of membership increase?? How did you guys get that?
josie_is_slut
Member
number 10 was easy
ok christianity and judasim, u can easily see its increased by a bit
i doubled buddism in 1991 and doubled islam in 1991
buddishm had a 60 thousand more than 1996 wen doubled
and islam had a 80 thousand more than in 1996 wen doubled
therefore buddism increased at a faster rate becoz wen the figure is doubled, and the closer it is to 1996, the faster it increases
just look at the fractions, kinda hard to explain, and do anyting to keep them both in ratio, double em, triple em, square em wateva so u can see a bigger differencce between the figures relative to each other
Killah33
Member
Why the hell was a Maths question in there.....
i mean seriously
bex
Active Member
ehh what a dumb question!!
Sarah J
Member
yeah. q10 bugged me. Why the hell would they put a maths question in there? It didn't occur to me that I'd be doing maths in a religion exam so I didn't bring a calculator.:mad1:
schwang_thang
Member
well it wasnt really a maths qstn
all u had 2 do is look at the 1996 figures. both r approx 200 000, so all u had 2 do is look at the 1991 figures and find out which had the lesser figure, which wud mean that it increased at a greater rate in the same time period.
Killah33
Member
yeh but dude
it was a MATHS question
sure it was easy, but CAMAN
WaTs 2 CuM AhEd
Member
how can the last m.c answer be A. isnt the answer christianity.......................
kaseita
Member
no
rate of increase. christianity went up (or down. can't remember the figure) very little proportionally.
Sarah
Member
i think q10 was one of those general knowledge religion questions.
i tried to work it out mathematically but i couldn't. i think ur just meant to know that buddhism is fastest growing religion in australia, islam fastest growing in world and christianity isn't growing very fast
i'm just testing
Fi
Member
Anyway, i agree with sarah, we were meant to know the answer without having to look at the stats.
At first i thought it was a maths question, but then remembered i had read it somewhere
SploJo
New Member
Youre spose to just KNOW Buddhism is the fastest growing. We learnt it in our course, using a calculator i think just screwed a lot of people up because it asked for RATE not by how many.
As for the Uniting Church q. its Presbyterian im pretty sure- a little late in adding my 2cents but whoever argued the 1/2 joined half didn compared to 3/4 of congregationalists is correct. We also learnt that--not from our teachers though, from independent research.
The MC was pretty good all up, some VERY tricky questions mainly 4 and 7 for me.
Anyways all the best in SOR. Keep in mind that if you didnt go as good as you wanted to---dont worry not THAT many ppl made it into band6 last year not sure whether that helps but i thought id share anyways...
Well have a KILLER time on schoolies!!!
bex
Active Member
dont worry not THAT many ppl made it into band6 last year
Yeah! a grand total of 2!
DC10
Member
seriously? 2 ppl?!
i thought it's like 48 ppl getting band 6 last year!
studies of religion is such a death trap LOL
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Chapter 2 Class 10 Polynomials
Serial order wise
Transcript
Question 4 If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3, find other zeroes. Let p(x) = x4 − 6x3 − 26x2 + 138 x − 35 Given roots are 2 ± √3 Since both are factors Since x = 2 + √𝟑 is a zero, x − (2 + √3) is a factor x − 2 – √3 is a factor Since x = 2 – √𝟑 is a zero, x − (2 – √3) is a factor x − 2 + √3 is a factor Hence, (x − 2 − √𝟑) × (x − 2 + √𝟑) is also a factor. ( (x − 2) − √3) × ( (x − 2) + √3) is also a factor. (x − 2)2 − (√3)2 x2 + 22 − 4x − 3 x2 + 4 − 4x − 3 x2 − 4x + 1 ∴ x2 − 4x + 1 is a factor of p(x) Now by dividing p(x) by (x2 − 4x + 1) We can find out other factors 𝑥2−4𝑥+ 1 − 𝟐𝒙 𝒙4 − 6𝒙𝟑 − 26𝒙^𝟐+ 138x − 35 Now, we find zeroes of x2 − 2x − 35 x2 – 2x − 35 = 0 We find the zeroes using Splitting the middle term method x2 − 7x + 5x − 35 = 0 x (x −7) + 5 (x − 7) = 0 (x + 5) (x − 7) = 0 So, x = –5 and x = 7 are zeros Therefore, the zeroes of p(x) are 2 + √𝟑 , 2 − √𝟑 , −5 and 7 Splitting the middle term method We need to find two numbers whose Sum = −2 & Product = −35
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Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. | 577 | 1,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.802336 |
https://www.analyzemath.com/inversefunction/inverse_function.html | 1,721,920,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00740.warc.gz | 538,841,322 | 32,541 | # Inverse Function
First, the definition and properties of inverse function are reviewed. Then the graphs of of one to one functions functions and their inverses are invetsigated graphically.
More tutorials on finding inverse functions are included.
## Definition, properties of inverse functions
### 1 - Definition
Let function f be defined as a set of ordered pairs as follows:
f = { (-3 , 0) , (-1 , 1) , (0 , 2) , (1 , 4) , (5 , 3)}
The inverse of function f is defined by interchanging the components (a , b) of the ordered pairs defining function f into ordered pairs of the form (b , a). Let g be the inverse of function f; g is then given by
g = {(0 , - 3) , (1 , - 1) , (2 , 0) , (4 , 1) , (3 , 5)}
The plots of the set of ordered pairs of function f and its inverse g are shown below.
Function f and its inverse g are reflection of each other on the line y = x.
### 2 - Inverse Function Notation
The inverse function, denoted f -1, of a one-to-one function f is defined as
f -1(x) = {(y,x) | such that y = f(x)}
Note: The -1 in f
-1 must not be confused with a power.
If function f is not a one-to-one then it does not have an inverse. More discussions on one to one functions will follow later.
### 3 - Domain and Range of a Function and its Inverse
Let function f be defined by the set of ordered pairs as follows:
f = {(1,0),(4,5),(6,9)}
According the definition above, the inverse function of f is given by:
f -1 = {(0,1),(5,4),(9,6)}
The domain of f is equal to the range of f -1 and the range of f is equal to the domain of f -1.
### 4 - Composition of a Function and its Inverse
• f(f -1(0)) = f(1) = 0
• f(f -1(5)) = f(4) = 5
• f(f -1(9)) = f(6) = 9
and
• f -1(f(1)) = f -1(0) = 1
• f -1(f(4)) = f -1(5) = 4
• f -1(f(6)) = f -1(9) = 6
and in general
f(f
-1(x)) = x , x in the domain of f -1
f
-1(f(x)) = x , x in the domain of f
### 5 - What is the inverse function needed for?
In some situations we now the output of a function and we need to find the input and that is where the inverse function is used.
Example: Find x such that 0 < x < π/2 and
sin(x) = 0.2
x = arcsin(0.2) , here arcsin is the inverse of sin(x).
### 6 - Which functions have an inverse function (invertible functions) ?
Let function f be defined as a set of ordered pairs as follows:
f = { (-3 , 0) , (-1 , 2) , (0 , 2) , (1 , 4) , (5 , 3)}
The inverse of function f is obtained by interchanging the components (a , b) of the ordered pairs defining function f into ordered pairs of the form (b , a). Let g be the inverse of function f; g is then given by
g = {(0 , - 3) , (2 , - 1) , (2 , 0) , (4 , 1) , (3 , 5)}
Below are shown the Venn diagrams of function f and its inverse g and we note that g is
not a function (input 2 has two outputs -1 and 0). We say that function f is not invertible (does not have an inverse) because it is not a one-to-one functions or only one to one functions have an inverse. Below are shown the graphs of f and its inverse g and we note again that g(2) = 0 and g(2) = - 1 and a vertical line would pass by both points (2 , 0) and (2 , -1) and therefore g is not a function. More on one-to-one functions.
### 7 - Important properties of a function and its inverse
1) The domain of f -1 is the range of f
2) The range of f
-1 is the domain of f
3) (f
-1o f)(x) = x for x in the domain of f
4) (f
o f -1)(x) = x for x in the domain of f -1
5) The graphs of f and f
-1 are reflection of each other on the line y = x
More on the
properties of inverse functions is included in this website.
### 8 - Graphical investigation of the properties of the inverse functions
The graphs of three functions (in blue) and their inverses (in red) are shown below. examine the graphs and answer the questions below.
### 9 - Questions
1) For each function, select some points on the graph of f and determine their coordinates, then find their inverses on the graph of the inverse function.
2) Why are the graphs of a function and its corresponding inverse a reflection of each other on the line y = x?
3) For each function, find its domain and range and deduce the domain and range of the corresponding inverse then verify your results graphically.
## Answers to the Above Questions
1) If (a,b) is a point on the graph of f then point (b,a) is a point on the graph of f-1
a) F(x) = x
2 , x ≥ 0.
points on the graph of f: (0,0) , (1,1) , (2,4)
corresponding points on f
-1(x) : (0,0), (1,1) and (4,2)
b) g(x) = x
3.
points on the graph of g: (-1,-1) , (0 , 0) , (1,1) , (2,8)
corresponding points on g
-1(x) : (-1,-1) , (0,0), (1,1) and (8,2)
c) h(x) = ln(x).
points on the graph of h: (1,0) , (4 , 1.4)
corresponding points on h
-1(x) : (0,1) , (1.4,4)
2) From the definition of the inverse function given above, if point (a,b) is on the graph of function f then point (b,a) is a point on the graph of f
-1. But points (a,b) and (b,a) are reflection of each other on the line y = x and therefore the whole graph of a function and its inverse are reflection of each other on the line y = x.
3) a) f(x) = x
2 , x ≥ 0
The domain of f is given by the interval: [0 , + ∞)
The range of f is given by the interval: [0 , + ∞)
One of the properties of a function and its inverse is that the domain of f is the range of f
-1 and the range of f is the domain of f-1. Hence the domain of f-1 is given by the interval [0 , + ∞)
and the range of f
-1 is given by the interval [0 , + ∞).
b) g(x) = x
3
The domain of g is given by the interval: (- ∞ , + ∞)
The range of g is given by the interval: (- ∞ , + ∞)
Hence The domain of g
-1 is given by the interval: (- ∞ , + ∞)
The range of g
-1 is given by the interval: (- ∞ , + ∞)
c) h(x) = ln(x)
The domain of h is given by the interval: (0 , + ∞)
The range of h is given by the interval: (- ∞ , + ∞)
Hence
The domain of h
-1 is given by the interval: (- ∞ , + ∞)
The range of h
-1 is given by the interval: (0 , + ∞)
### 10 - Exercises
Exercise 1:
a) Find the domain and range of function f defined by
f = {(-4,2),(-3,1),(0,5),(2,6)}
b) Find the inverse function of f and its domain and range.
Exercise 2:
Which of these functions do not have an inverse?
f = {(-1,2),(-3,1),(0,2),(5,6)}
g = {(-3,0),(-1,1),(0,5),(2,6)}
h = {(2,2),(3,1),(6,5),(7,1)}
### Answers to Above Exercises
Exercise 1:
a) domain of f = {-4,-3,0,2} and range of f = {2,1,5,6}
b) inverse of f = {(2,-4),(1,-3),(5,0),(6,2)}
domain of inverse of f = {2,1,5,6} and range of inverse of f = {-4,-3,0,2}.
Exercise 2:
Functions f and h are not one to one functions and therefore do not have inverses.
### More References and links to the Inverse Functions
Find inverse of exponential functions
Inverse Functions Graphing
Applications and Use of the Inverse Functions
Find the inverse of a Function - Questions
Find the Inverse Function (1) - Tutorial.
Find the Inverse Function (2) - Tutorial
Definition of the Inverse Function - Interactive Tutorial
Find Inverse Of Cube Root Functions.
Find Inverse Of Square Root Functions.
Find Inverse Of Logarithmic Functions.
Find Inverse Of Exponential Functions. Properties of Inverse Functions
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## Boat And Stream Quiz for IBPS , SBI , RBI: Quiz – 2
Boat And Stream plays a significant role in Quantitative Aptitude Section of banking exams such as IBPS, SBI and RBI PO and Clerk. You will get at least 1-2 questions from Boat and Stream in one of IBPS, SBI and RBI PO & clerk exam. So, aspirants should focus on Boat And Stream questions in detail. Here, we are providing you with the Boat and Stream questions quiz with the detailed solution so that you can easily prepare for Boat and Stream questions. We are providing here all-important latest pattern-based questions and Previous Year Questions of Boat And Stream of various Government Exam like IBPS, SBI, and RBI PO and Clerk exam. This Boat And Stream quiz we are providing is free. Attempt this Boat and Stream quiz to practice important questions with answers and solutions. And score better in IBPS, SBI and RBI PO and Clerk exam.
Boat And Stream Quiz to improve your Quantitative Aptitude for SBI Po & SBI clerk exam, IBPS PO & IBPS Clerk exam, IBPS RRB PO and assistant exam, LIC AAO, LIC Assistant and other competitive exam.
1. If the speed of boat in still water is five times to speed of current and boat cover 12 km downstream in 2 hours. Then in how many time boats will cover 100 km in upstream.
(a) 12 hr
(b) 18 hr
(c) 25 hr
(d) 28 hr
(e) 22 hr
2. The time taken by a boat to row 12 km against the stream is same as time taken by boat to row 21 km with the stream. If speed of boat in still water is 5.5 kmph then find speed of boat is how much percent more than speed of stream?
3. Speed of current is 75% of speed of boat in still water and a boat covers 42 km downstream and 21 km in upstream in 9 hours. Find speed of current?
(a) 9km/hr
(b) 8km/hr
(c) 6km/hr
(d) 4km/hr
(e) 12km/hr
4. The sum of upstream and downstream speed of a boat is 36 km/h and if the boat travels 90 km downstream in 4 hr 30 min, then find the time taken by boat to travel 256 km upstream?
(a) 16 hr
(b) 12 hr
(c) 15 hr
(d) 10 hr
(e) 8 hr
5. River flows from B to A (1000 km) with speed of 2 km/hr. Two boats P and Q, both having speed in still water of 10 km/h starts from A and B respectively and meets at point C. Find the ratio of distance AC and CB.
(a) 3 : 4
(b) 1 : 1
(c) 2 : 3
(d) 4 : 3
(e) None of these
6. A boatman can cover a river of 60 km length and came back at its initial point in 4.5 hrs. If speed of boat is thrice than that of the speed of stream then find the speed of boat?
(a) 10
(b) 30
(c) 20
(d) 60
(e) 25
7. If speed of boat in upstream is double than the speed of current and speed of boat in still water is 27 km/hr. Then find the time taken by boat to travel 54 km downstream? (in hour)
(a) 1.5
(b) 1.8
(c) 2.5
(d) 1.2
(e) 2
8. A boat in upstream covers 110 km in same time as it covers 140 km in downstream. Find speed of stream is what percent of the speed of boat in still water.
(a) 12%
(b) 16%
(c) 15%
(d) 10%
(e) 20%
9. A man can row 27 km upstream and 56 km downstream in 17 hours. If speed of boat in still water is 150% more than speed of current, find speed of boat while travelling upstream?
(a) 5 kmph
(b) 3 kmph
(c) 2 kmph
(d) 8 kmph
(e) 7 kmph
10. If the sum of downstream and upstream speed of a boat is 12 kmph and ratio between speed of boat in sill water and upstream speed of boat is 3 : 2. Find the time require to cover 50 km by the boat if it cover half distance upstream and half downstream ?
Boat & Stream Quiz- 2 PDF
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http://www.astro.umd.edu/~ricotti/NEWWEB/teaching/ASTR415/InClassExamples/NR3/index_by_section.htm | 1,511,251,908,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806327.92/warc/CC-MAIN-20171121074123-20171121094123-00627.warc.gz | 335,551,170 | 7,685 | ## Index of Code Files by Book Section
Note: Book section links are active for subscribers to Numerical Recipes Electronic, or as manual references to the hardcopy book. Webnote links are active for anyone. You must be connected to the Internet for these links to work. Links to files (ending in .h) are local to Numerical Recipes Code and do not require an Internet connection.
1.0 Introduction
calendar.h calendar routines
1.4 Vector and Matrix Objects
nr3.h defines Numerical Recipes classes used by all routines
2.1 Gauss-Jordan Elimination
gaussj.h Gauss-Jordan decomposition
2.3 LU Decomposition and Its Applications
ludcmp.h LU decomposition and related
2.4 Tridiagonal and Band-Diagonal Systems of Equations
tridag.h solve tridiagonal linear systems
banded.h routines for banded matrices
2.5 Iterative Improvement of a Solution to Linear Equations
ludcmp.h LU decomposition and related
2.6 Singular Value Decomposition
svd.h singular value decomposition
2.7 Sparse Linear Systems
tridag.h solve tridiagonal linear systems
sparse.h sparse matrix routines
linbcg.h solution of sparse linear system by biconjugate gradient method
asolve.h biconjugate gradient sparse linear solver (example)
2.8 Vandermonde Matrices and Toeplitz Matrices
vander.h solve Vandermonde linear systems
toeplz.h solve Toepltz linear systems
2.9 Cholesky Decomposition
cholesky.h Cholesky decomposition
2.10 QR Decomposition
qrdcmp.h QR decomposition and related
3.1 Preliminaries: Searching an Ordered Table
interp_1d.h interpolation routines for one dimension
interp_linear.h linear interpolation
3.2 Polynomial Interpolation and Extrapolation
interp_1d.h interpolation routines for one dimension
3.3 Cubic Spline Interpolation
interp_1d.h interpolation routines for one dimension
3.4 Rational Function Interpolation and Extrapolation
interp_1d.h interpolation routines for one dimension
3.5 Coefficients of the Interpolating Polynomial
polcoef.h polynomial coefficients from polynomial values
3.6 Interpolation on a Grid in Multidimensions
interp_2d.h interpolation routines for two dimensions
3.7 Interpolation on Scattered Data in Multidimensions
interp_rbf.h interpolation by radial basis functions
krig.h fitting or interpolation by kriging
interp_curve.h interpolate along a curve
3.8 Laplace Interpolation
interp_laplace.h Laplace interpolation on a grid
4.2 Elementary Algorithms
4.3 Romberg Integration
4.4 Improper Integrals
4.6 Gaussian Quadratures and Orthogonal Polynomials
gauss_wgts.h weights for Gaussian and related quadratures
4.8 Multidimensional Integrals
5.1 Polynomials and Rational Functions
poly.h operations on polynomials
5.3 Series and Their Convergence
series.h routines for summing a series
levex.h sample program for Levin transformation
5.7 Numerical Derivatives
dfridr.h numerical derivative by Ridders' method
5.8 Chebyshev Approximation
chebyshev.h Chebyshev approximation and related
5.9 Derivatives or Integrals of a Chebyshev-Approximated Function
chebyshev.h Chebyshev approximation and related
5.10 Polynomial Approximation from Chebyshev Coefficients
chebyshev.h Chebyshev approximation and related
pcshft.h shift a polynomial
5.11 Economization of Power Series
pcshft.h shift a polynomial
chebyshev.h Chebyshev approximation and related
5.13 Rational Chebyshev Approximation
ratlsq.h rational function fitting
6.1 Gamma, Beta, and Related Functions
gamma.h gamma function and related
6.2 Incomplete Gamma Function
incgammabeta.h incomplete gamma and beta functions
erf.h error function and related distributions
6.3 Exponential Integrals
expint.h exponential integral
6.4 Incomplete Beta Function
incgammabeta.h incomplete gamma and beta functions
6.5 Bessel Functions of Integer Order
bessel.h Bessel functions of integer order
6.6 Bessel Functions of Fractional Order
besselfrac.h Bessel functions of fractional order
6.7 Spherical Harmonics
plegendre.h Legendre functions and spherical harmonics
6.8 Fresnel Integrals, Cosine and Sine Integrals
frenel.h Fresnel integrals
cisi.h cosine and sine integrals
6.9 Dawson's Integral
dawson.h Dawson's integral
6.10 Generalized Fermi-Dirac Integrals
fermi.h Fermi-Dirac integrals
6.11 Inverse of the Function x log(x)
ksdist.h Kolmogorov-Smirnov distribution functions
6.12 Elliptic Integrals and Jacobian Elliptic Functions
elliptint.h elliptic integrals
6.13 Hypergeometric Functions
hypgeo.h hypergeometric function
6.14 Statistical Functions
erf.h error function and related distributions
distributions.h probability distributions and related
incgammabeta.h incomplete gamma and beta functions
ksdist.h Kolmogorov-Smirnov distribution functions
7.1 Uniform Deviates
ran.h random number generators
7.2 Completely Hashing a Large Array
hashall.h hash a large array
7.3 Deviates from Other Distributions
deviates.h random deviates from various distributions
7.4 Multivariate Normal Deviates
multinormaldev.h multivariate normal deviates
7.5 Linear Feedback Shift Registers
primpolytest.h test whether a polynomial is primitive
7.6 Hash Tables and Hash Memories
hash.h routines for hash tables and hash memory
7.7 Simple Monte Carlo Integration
mcintegrate.h Monte Carlo integration
7.8 Quasi- (that is, Sub-) Random Sequences
sobseq.h Sobol quasi-random sequence
8.1 Straight Insertion and Shell's Method
sort.h sorting and related routines
8.2 Quicksort
sort.h sorting and related routines
8.3 Heapsort
sort.h sorting and related routines
8.4 Indexing and Ranking
sort.h sorting and related routines
8.5 Selecting the Mth Largest
sort.h sorting and related routines
iqagent.h estimate distribution function from a sample
8.6 Determination of Equivalence Classes
eclass.h find equivalence classes
9.0 Introduction
scrsho.h plot a function interactively
9.1 Bracketing and Bisection
roots.h routines for root finding in one dimension
9.2 Secant Method, False Position Method, and Ridders' Method
roots.h routines for root finding in one dimension
9.3 Van Wijngaarden-Dekker-Brent Method
roots.h routines for root finding in one dimension
9.4 Newton-Raphson Method Using Derivative
roots.h routines for root finding in one dimension
9.5 Roots of Polynomials
roots_poly.h roots of a polynomial
zrhqr.h polynomial roots by eigenvalue method
qroot.h Bairstow's method for complex root
9.6 Newton-Raphson Method for Nonlinear Systems of Equations
mnewt.h multidimensional Newton root finding
9.7 Globally Convergent Methods for Nonlinear Systems of Equations
roots_multidim.h routines for root finding in multidimensions
10.1 Initially Bracketing a Minimum
mins.h minimization routines
10.2 Golden Section Search in One Dimension
mins.h minimization routines
10.3 Parabolic Interpolation and Brent's Method
mins.h minimization routines
10.4 One-Dimensional Search with First Derivatives
mins.h minimization routines
10.5 Downhill Simplex Method in Multidimensions
amoeba.h downhill simplex minimization
10.6 Line Methods in Multidimensions
mins_ndim.h minimization routines in multidimensions
10.7 Direction Set (Powell's) Methods in Multidimensions
mins_ndim.h minimization routines in multidimensions
10.8 Conjugate Gradient Methods in Multidimensions
mins_ndim.h minimization routines in multidimensions
10.9 Quasi-Newton or Variable Metric Methods in Multidimensions
quasinewton.h quasi-Newton or DFP multidimensional minimization
10.11 Linear Programming: Interior-Point Methods
interior.h linear programming by interior point method
10.13 Dynamic Programming
dynpro.h dynamic programming
stringalign.h Needleman-Wunsch string alignment
11.1 Jacobi Transformations of a Symmetric Matrix
eigen_sym.h routines for eigenproblems on symmetric matrices
11.2 Real Symmetric Matrices
eigen_sym.h routines for eigenproblems on symmetric matrices
11.3 Reduction of a Symmetric Matrix to Tridiagonal Form
eigen_sym.h routines for eigenproblems on symmetric matrices
11.4 Eigenvalues and Eigenvectors of a Tridiagonal Matrix
eigen_sym.h routines for eigenproblems on symmetric matrices
11.6 Real Nonsymmetric Matrices
eigen_unsym.h routines for eigenproblems on nonsymmetric matrices
12.2 Fast Fourier Transform (FFT)
fourier.h FFT routines in one dimension
12.3 FFT of Real Functions
fourier.h FFT routines in one dimension
12.4 Fast Sine and Cosine Transforms
fourier.h FFT routines in one dimension
12.5 FFT in Two or More Dimensions
fourier_ndim.h FFT routines in multidimensions
12.6 Fourier Transforms of Real Data in Two and Three Dimensions
fourier_ndim.h FFT routines in multidimensions
rlft3_sharpen.h example using FFT to sharpen an image
13.1 Convolution and Deconvolution Using the FFT
convlv.h data convolution
13.2 Correlation and Autocorrelation Using the FFT
correl.h data correlation
13.4 Power Spectrum Estimation Using the FFT
spectrum.h spectral estimation using FFTs
13.6 Linear Prediction and Linear Predictive Coding
linpredict.h linear prediction
13.7 Maximum Entropy (All-Poles) Method
linpredict.h linear prediction
13.8 Spectral Analysis of Unevenly Sampled Data
period.h Lomb periodogram for spectral analysis
fasper.h fast evaluation of Lomb periodogram
13.9 Computing Fourier Integrals Using the FFT
dftintegrate.h Fourier integrals using FFT
13.10 Wavelet Transforms
wavelet.h wavelet routines
14.1 Moments of a Distribution: Mean, Variance, Skewness
moment.h calculate moments of samples
14.2 Do Two Distributions Have the Same Means or Variances?
stattests.h various statistical tests
moment.h calculate moments of samples
14.3 Are Two Distributions Different?
stattests.h various statistical tests
kstests.h Kolmogorov-Smirnov tests in one dimension
14.4 Contingency Table Analysis of Two Distributions
stattests.h various statistical tests
14.5 Linear Correlation
stattests.h various statistical tests
14.6 Nonparametric or Rank Correlation
stattests.h various statistical tests
14.8 Do Two-Dimensional Distributions Differ?
kstests_2d.h Kolmogorov-Smirnov tests in two dimensions
quadvl.h used in two-dimensional K-S test
14.9 Savitzky-Golay Smoothing Filters
savgol.h Savitzky-Golay filters
15.2 Fitting Data to a Straight Line
fitab.h fit a straight line
15.4 General Linear Least Squares
fitlin.h general linear fit
fitsvd.h general linear fit using SVD
fit_examples.h examples for fitting
15.5 Nonlinear Models
fitmrq.h Levenberg-Marquardt nonlinear fitting
fit_examples.h examples for fitting
15.7 Robust Estimation
fitmed.h fit a line minimizing absolute deviation
15.8 Markov Chain Monte Carlo
mcmc.h Markov chain Monte Carlo routines
16.1 Gaussian Mixture Models and k-Means Clustering
gaumixmod.h Gaussian mixture models
kmeans.h k-means classification
16.3 Markov Models and Hidden Markov Modeling
markovgen.h Markov model generator
hmm.h hidden Markov models
16.4 Hierarchical Clustering by Phylogenetic Trees
phylo.h phylogenetic tree algorithms
16.5 Support Vector Machines
svm.h support vector machines
17.0 Introduction
odeint.h ordinary differential equations
stepper.h base class for ODE steppers
17.1 Runge-Kutta Method
rk4.h Runge-Kutta integration (pedagogical example)
17.2 Adaptive Stepsize Control for Runge-Kutta
stepperdopr5.h Dormand-Prince fifth-order stepper
17.3 Richardson Extrapolation and the Bulirsch-Stoer Method
stepperbs.h Bulirsch-Stoer stepper base class
17.4 Second-Order Conservative Equations
stepperstoerm.h Bulirsch-Stoer stepper, second-order equations
17.5 Stiff Sets of Equations
stepperross.h fourth-order stiffly stable Rosenbrock stepper
17.7 Stochastic Simulation of Chemical Reaction Networks
stochsim.h stochastic simulation of ODEs
18.1 The Shooting Method
shoot.h two point boundary problem by shooting
18.2 Shooting to a Fitting Point
shootf.h two point boundary problem by shooting to a fitting point
18.4 A Worked Example: Spheroidal Harmonics
sfroid.h spheroidal harmonics example using relaxation method
difeq.h example solving two point boundary problem by relaxation
sphoot.h spheroidal harmonics example using shooting
sphfpt.h spheroidal harmonics example using shooting to a fitting point
19.1 Fredholm Equations of the Second Kind
fred2.h linear Fredholm equations of the second kind
19.2 Volterra Equations
voltra.h Volterra integral equations
19.3 Integral Equations with Singular Kernels
fred_singular.h singular Fredholm equations
20.5 Relaxation Methods for Boundary Value Problems
sor.h successive over-relaxation
20.6 Multigrid Methods for Boundary Value Problems
mglin.h full multigrid method (FMG) for linear equation
mgfas.h multigrid method full approximation storage
20.7 Spectral Methods
weights.h differentiation matrix for spectral methods
21.1 Points and Boxes
pointbox.h geometric point and box
kdtree.h KD-tree routines
21.2 KD Trees and Nearest-Neighbor Finding
kdtree.h KD-tree routines
21.3 Triangles in Two and Three Dimensions
circumcircle.h circumcircle of three points
21.4 Lines, Line Segments, and Polygons
polygon.h geometric polygon
21.6 Triangulation and Delaunay Triangulation
delaunay.h Delaunay triangulation
21.7 Applications of Delaunay Triangulation
delaunay.h Delaunay triangulation
voronoi.h construct Voronoi diagram
21.8 Quadtrees and Octrees: Storing Geometrical Objects
sphcirc.h geometrical circle or sphere
qotree.h QO-tree routines
22.1 Plotting Simple Graphs
psplotexample.h generate a PostScript plot (example)
22.3 Gray Codes
igray.h Gray code
22.4 Cyclic Redundancy and Other Checksums
icrc.h cyclic redundancy checksum
decchk.h decimal digit error check
22.5 Huffman Coding and Compression of Data
huffcode.h Huffman compression
22.6 Arithmetic Coding
arithcode.h compression by arithmetic coding
22.7 Arithmetic at Arbitrary Precision
mparith.h multiprecision arithmetic
Webnote.3 Implementation of Stiel
stiel.h Stieltjes procedure for quadrature weights
Webnote.9 Complete VEGAS Code
rebin.h used by vegas.h
Webnote.10 Complete miser Code
ranpt.h used by miser.h
Webnote.11 Code Listing for selip
selip.h select Mth largest in place
Webnote.12 Routine Implementing the Simplex Method
simplex.h linear programming by simplex method
Webnote.14 Code Implementation for the Traveling Salesman Problem
anneal.h simulated annealing
Webnote.15 Code for Minimization with Simulated Annealing
amebsa.h downhill simplex minimization with simulated annealing
Webnote.18 Code for External or Memory-Local Fourier Transform
fourfs.h FFT on external storage
Webnote.19 Code Listing fitexy
fitexy.h fit a line with errors in both x and y
Webnote.20 Routine Implementing Eighth-order Runge-Kutta
stepperdopr853.h Dormand-Prince eighth-order stepper
Webnote.24 StepperSie Implementation
steppersie.h semi-implicit extrapolation stepper
Webnote.25 Solvde Implementation
solvde.h relaxation solution of ODEs
Webnote.26 Code for PSpage and PSplot
psplot.h generate a PostScript plot
Webnote.27 Code for machar
machar.h probe machine characteristics | 3,685 | 14,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-47 | latest | en | 0.676193 |
http://www.google.com/patents/US6754284?dq=6,073,142 | 1,496,101,683,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613135.2/warc/CC-MAIN-20170529223110-20170530003110-00343.warc.gz | 636,307,278 | 18,101 | ## Patents
Publication number US6754284 B2 Publication type Grant Application number US 10/225,982 Publication date Jun 22, 2004 Filing date Aug 22, 2002 Priority date Aug 22, 2002 Fee status Lapsed Also published as Publication number 10225982, 225982, US 6754284 B2, US 6754284B2, US-B2-6754284, US6754284 B2, US6754284B2 Inventors Original Assignee Motorola, Inc. Export Citation Patent Citations (7), Referenced by (2), Classifications (10), Legal Events (6) External Links:
Method and apparatus for minima enlargement
US 6754284 B2
Abstract
A modulator (102) modulates a plurality of symbols to create a modulated signal. A transmitter is coupled to the modulator (102). The transmitter transmits the modulated signal if the modulated signal is above a threshold. A minima enlarger (104) is coupled to the modulator (102). The minima enlarger (104) computes an adjustment to the plurality of symbols if the modulated signal is below the threshold. A summer (106) is coupled to the modulator (102) and the minima enlarger (104). The summer (106) adds the adjustment to the plurality of symbols if the modulated signal is below the threshold.
Images(2)
Claims(17)
We claim:
1. A method comprising the steps of:
receiving a plurality of symbols;
modulating the plurality of symbols to create a first modulated signal;
comparing the first modulated signal to a threshold;
if the first modulated signal is above the threshold, transmitting the first modulated signal; and
if the first modulated signal is below the threshold,
modulating the adjusted plurality of symbols to create a second modulated signal, and
transmitting the second modulated signal if the second modulated signal exceeds the threshold; otherwise, repeating the steps of computing, adding and the second step of modulating.
2. The method of claim 1 wherein the adjustment is computed with the following equation: $\Delta \left(k,m\right)=C·W\left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)·g\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]·{}^{j\left({\phi }_{p}-\frac{2\pi }{M}{k\left(p\right)}_{M}\right)}$
wherein:
k=frequency axis;
m=time axis;
C=a constant that ensures that minima is sufficiently enlarged;
W(k,m)=a matrix;
p=output sample;
D=filter decimation rate;
I=filter interpolation rate;
g=pulse shaping filter;
φp=phase; and
M=number of subchannels.
3. The method of claim 1 wherein the threshold is predetermined.
4. The method of claim 1 wherein the threshold is dynamically adjusted.
5. The method of claim 1 wherein the plurality of symbols are divided in time and frequency.
6. The method of claim 5 wherein the step of modulating comprises at least one of mixing, pulse shaping, and rate changing the plurality of symbols.
7. A method comprising the steps of:
receiving a plurality of symbols;
mixing the plurality of symbols using an inverse fast Fourier transform to create a plurality of mixed symbols;
pulse shaping the plurality of mixed symbols to create a first modulated signal;
comparing the first modulated signal to a threshold;
if the first modulated signal is above the threshold, transmitting the first modulated signal; and
if the first modulated signal is below the threshold,
computing an adjustment to the plurality of mixed symbols,
adding the adjustment to the plurality of mixed symbols to create an adjusted plurality of mixed symbols,
pulse shaping the adjusted plurality of mixed symbols to create a second modulated signal, and
transmitting the second modulated signal if the second modulated signal exceeds the threshold; otherwise, repeating the steps of computing, adding and the second step of pulse shaping.
8. The method of claim 7 further comprising the step of changing a sampling rate of at least one of the first modulated signal and the second modulated signal.
9. The method of claim 7 wherein the adjustment is computed with the following equation: ${\delta }_{r}\left(n,m\right)=C·{}^{{\mathrm{j\phi }}_{p}}·g\left[\mathrm{mI}+{\left(\mathrm{nD}\right)}_{I}\right]{w}_{r}\left({\left(n-p\right)}_{M},⌊\frac{\mathrm{nD}}{I}⌋-m\right)$
wherein:
n=time sample index
m=time axis;
C=a constant that ensures that minima is sufficiently enlarged;
p=output sample;
D=filter decimation rate;
I=filter interpolation rate;
g=pulse shaping filter;
φp=phase; and
M=number of subchannels.
10. The method of claim 7 wherein the threshold is predetermined.
11. The method of claim 7 wherein the threshold is dynamically adjusted.
12. The method of claim 7 wherein the plurality of symbols are divided in time and frequency.
13. An apparatus comprising:
a modulator for modulating a plurality of symbols to create a modulated signal;
a transmitter, coupled to the modulator, for transmitting the modulated signal if the modulated signal is above a threshold;
a minima enlarger, coupled to the modulator, for computing an adjustment to the plurality of symbols if the modulated signal is below the threshold; and
a summer, coupled to the modulator and the minima enlarger, for adding the adjustment to the plurality of symbols if the modulated signal is below the threshold.
14. The apparatus of claim 13 wherein an output of the modulator is coupled to the input of the minima enlarger, an output of the minima enlarger is coupled to the input of the summer, and the output of the summer is coupled to the input of the modulator.
15. The apparatus of claim 14 wherein an output of the modulator is coupled to the transmitter.
16. The apparatus of claim 13 wherein a comparator is integrated within the modulator.
17. The apparatus of claim 13 wherein the modulator is a frequency division multiplexing modulator.
Description
FIELD OF THE INVENTION
The present invention relates generally to a method and apparatus for minima enlargement.
BACKGROUND OF THE INVENTION
Wireless telecommunication systems sometimes are divided into a series of cell areas covering a service area. Each cell area has a transmitting base station using an operating frequency set comprising a plurality of radio channels to communicate with mobile subscribers. Each channel represents an information signal at a particular frequency carrier or band.
In many instances it is advantageous to combine these channels for transmission purposes. The channels can all be combined by a broadband signal combiner into a multi-subchannel signal at lower power levels and then amplified by a single linear amplifier (or its equivalent, a plurality of linear amplifiers in parallel, each amplifying a reduced power version of the same multi-carrier sub-channel signal) to raise the multi-subchannel signal to an appropriate transmit power level.
As data rate requirements rise, the symbol rate necessary in forthcoming protocols will cause the modulation bandwidth to exceed the coherence bandwidth of the channel. This requires an expensive equalizer at the receiver to compensate for intersymbol interference created by the time dispersion in a multipath channel. Another application of multi-subchannel technology is to split the single, high symbol rate modulation into a plurality of lower rate sub-channels that each has a low enough modulation bandwidth to avoid the need for an adaptive equalizer.
Highly linear multi-subchannel modulations have large dynamic ranges where the minimum voltage can approach −50 dB or lower from the signal mean. This will essentially take the linear amplifier used to increase the power of the signal prior to transmission down to idle bias current and cause a spike in the load impedance with unpredictable results. In addition, with recent supply modulation techniques used to increase the linear amplifier efficiency, if the supply voltage approaches zero, the linear amplifier could have problems with phase, noise and stability.
In the past, floor clamp circuits have been employed to prevent the voltage from dropping below a specified threshold. This results in frequency domain splatter and a reduction in linearity of the signal. Furthermore, different types of signals can tolerate different amounts of distortion and thus different amounts of compression. Prior-art techniques have not attempted to tailor symbols on multiple subchannels, including empty subchannels, in a controlled, individualized manner to improve the minima to average power ratio.
Clearly then, a need exists for an improved method and apparatus for minima enlargement.
BRIEF DESCRIPTION OF THE FIGURES
A preferred embodiment of the invention is now described, by way of example only, with reference to the accompanying figures in which:
FIG. 1 illustrates a block diagram of a transmit modem in accordance with the present invention; and
FIG. 2 illustrates a pictorial representation of minima enlargement in accordance with the present invention.
DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT
It will be appreciated that for simplicity and clarity of illustration, elements shown in the figures have not necessarily been drawn to scale. For example, the dimensions of some of the elements are exaggerated relative to each other. Further, where considered appropriate, reference numerals have been repeated among the figures to indicate corresponding elements.
The present invention discloses an improved method and apparatus for minima enlargement in a modulator. The present invention allows control of the compression in accordance with the type of information being sent and will tailor symbols on multiple subchannels, including empty subchannels, in a controlled, individualized manner to improve the minima to average power ratio.
Turning to the figures, FIG. 1 illustrates a block diagram of the transmit modem with the iterative minima enlarger. As illustrated, the block diagram comprises an unmodulated symbols input 100, a frequency division multiplexing (“FDM”) modulator 102, a minima enlarger 104, and a summation component 106. The unmodulated symbols, X(k,m) 100, is the complex baseband symbol matrix with k selecting the frequency axis, and m selecting the time axis. The FDM modulator 102 modulates symbols according to the following equation: $\begin{array}{cc}x\left(n\right)=\sum _{m=0}^{{N}_{s}-1}\text{ }g\left[\mathrm{mI}+{\left(\mathrm{nD}\right)}_{I}\right]\sum _{k=0}^{M-1}\text{ }X\left(k,⌊\frac{\mathrm{nD}}{I}⌋-m\right){}^{j\frac{2\pi }{M}{k\left(n\right)}_{M}}& \left(1\right)\end{array}$
where,
x(n)=the nth sample of the modulated output;
g=the pulse shaping filter;
Ns=the pulse shaping filter length (in units of symbol periods);
I=the filter interpolation rate;
D=the filter decimation rate; and
M=the number of subchannels.
The inner sum of equation (1) performs the mixing of the subchannels, and the outer sum of equation (1) performs the pulse shaping and rate change as described in greater detail in U.S. Pat. No. 6,134,268, titled “Apparatus for Performing a Non-Integer Sampling Rate Change in a Multichannel Polyphase Filter,” which is herein incorporated by reference. It is known in the art that equation (1) can be efficiently implemented using an inverse fast Fourier transform (“IFFT”) to perform the mixing as shown in equations (2) and (3) below: $\begin{array}{cc}x\left(n\right)=\sum _{m=0}^{{N}_{s}-1}\text{ }g\left[\mathrm{mI}+{\left(\mathrm{nD}\right)}_{I}\right]{x}_{r}\left({\left(n\right)}_{M},⌊\frac{\mathrm{nD}}{I}⌋-m\right)& \left(2\right)\end{array}$
where xr(n,m) is given by: $\begin{array}{cc}{x}_{r}\left(n,m\right)=M·{-1}^{}\left\{X\left(k,m\right)\right\}=M·\frac{1}{M}\sum _{k=0}^{M-1}\text{ }X\left(k,m\right){}^{j\frac{2\pi }{M}\mathrm{kn}}& \left(3\right)\end{array}$
It is important to note that the present invention is not limited to any particular method for performing FDM modulation, but rather, it should be obvious to those skilled in the art that the present invention can utilize any of the many available methods of performing FDM modulation and still remain within the spirit and scope of the present invention.
The minima enlarger 104 detects signal minima and computes an adjustment matrix that when combined with the unmodulated symbols, X(k,m), 100 enlarges the signal above the minima threshold, Mp. The signal minima is defined as any output sample whose magnitude is less than a specified minima threshold, Mp. The minima threshold, Mp, can either be predetermined or dynamically adjusted based on system needs. When the signal drops below the minima, the problems discussed in the background manifest themselves. The adjustment matrix of the present invention, however, brings the signal above the threshold and avoids the aforementioned problems.
The summation component combines the unmodulated symbols, X(k,m), 100 with an adjustment computed by the minima enlarger component 104.
Let us now focus the discussion on the minima enlarger component 104. When the minima enlarger component 104 detects a minima at output sample p with phase of φp, as described by the following equation,
|x(p)|<M p withx(p)=φp (4)
it computes an additive symbol adjustment to X(k,m) called Δ(k,m) such that passing X′(k,m)=X(k,m)+Δ(k,m) through the FDM modulator would result in x′(p)≧Mp, where x′(p) is given by the following equation: $\begin{array}{cc}\begin{array}{c}{x}^{\prime }\left(p\right)=\text{ }\sum _{m=0}^{{N}_{s}-1}\text{ }g\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]\sum _{k=0}^{M-1}\text{ }{X}^{\prime }\left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right){}^{j\frac{2\pi }{M}{k\left(p\right)}_{M}}\\ =\text{ }\sum _{m=0}^{{N}_{s}-1}\text{ }g\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]\sum _{k=0}^{M-1}\left\{X\left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)+\Delta \left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)\right\}{}^{j\frac{2\pi }{M}{k\left(p\right)}_{M}}\end{array}& \left(5\right)\end{array}$
Noting that the system is linear, the contribution of Δ(k,m) to x′(p) can be treated separately: $\begin{array}{cc}{x}_{\Delta }\left(p\right)=\sum _{m=0}^{{N}_{s}-1}\text{ }g\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]\sum _{k=0}^{M-1}\text{ }\Delta \left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right){}^{j\frac{2\pi }{M}{k\left(p\right)}_{M}}& \left(6\right)\end{array}$
The minima enlarger 104 must find the Δ(k,m) such that:
|x(p)+x Δ(p)|≧M p (7)
In practice, this may be accomplished by adjusting the minima such that |x′(p)|=Mp·α, with α>1. That is:
|x(p)+x Δ(p)=M p ·α (8)
FIG. 2 pictorially illustrates minima enlargement. First, it should be noted that the magnitude of x(p) 200 is close to the origin, and hence the output of the linear amplifier is no longer reliable. Second, it should also be noted that φp 202, the phase of xΔ(p) 204, is the same as the phase of x(p)200. Making all contributions towards xΔ(p) 204 coherent in this direction minimizes the magnitude.
Working from equation (6): $\begin{array}{cc}\angle \text{ }{x}_{\Delta }\left(p\right)=\angle \sum _{m=0}^{{N}_{s}-1}\text{ }g\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]\sum _{k=0}^{M-1}\Delta \left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right){}^{j\frac{2\pi }{M}{k\left(p\right)}_{M}}={\phi }_{p}& \left(9\right)\\ {\phi }_{p}-\frac{2\pi }{M}{k\left(p\right)}_{M}=\sum _{m=0}^{{N}_{s}-1}\text{ }\sum _{k=0}^{M-1}\text{ }\mathrm{\angle \Delta }\left(k,\text{ }⌊\frac{\mathrm{pD}}{I}⌋-m\right)& \left(10\right)\end{array}$
The optimal solution to this is to require each component of the double summation to have the necessary phase: $\begin{array}{cc}\mathrm{\angle \Delta }\left(k,\text{ }⌊\frac{\mathrm{pD}}{I}⌋-m\right)={\phi }_{p}-\frac{2\pi }{M}{k\left(p\right)}_{M}\text{ }\mathrm{for}\text{ }0\le k\le M-1\text{ }\mathrm{and}\text{ }0\le m\le {N}_{s}-1& \left(11\right)\end{array}$
Now consider the magnitude xΔ(p) 204. The total symbol error introduced is given by: $\begin{array}{cc}\sum _{m=0}^{{N}_{s}-1}\text{ }\sum _{k=0}^{M-1}\text{ }\Delta \left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)& \left(12\right)\end{array}$
In order to minimize this quantity, examine the magnitude of xΔ(p) from equation (1): $\begin{array}{cc}\text{ }{x}_{\Delta }\left(p\right)=\sum _{m=0}^{{N}_{s}-1}\text{ }g\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]\sum _{k=0}^{M-1}\Delta \left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right){}^{j\frac{2\pi }{M}{k\left(p\right)}_{M}}& \left(13\right)\\ =\sum _{m=0}^{{N}_{s}-1}\text{ }g\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]\sum _{k=0}^{M-1}\Delta \left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)& \left(14\right)\end{array}$
The pulse-shaping filter is acting as a weighting function to the symbol adjustment matrix. Moving symbols that align with large filter coefficients have a greater impact on the magnitude than others. The pulse-shaping filter coefficients themselves can be used to weight the changes.
As a final component to Δ(k,m), it is noted that the various symbol types have different importance. For example, in most applications pilot and synchronization symbols should be altered much less than data symbols. Define W(km) as a matrix the same dimension as X(k,m) that contains weights for each entry in X(k,m). The complete equation for Δ(km) is given by: $\begin{array}{cc}\Delta \left(k,m\right)=C·W\left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)·g\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]·{}^{j\left({\phi }_{p}-\frac{2\pi }{M}{k\left(p\right)}_{M}\right)}& \left(15\right)\end{array}$
The constant C is in place to ensure that the minima is sufficiently enlarged according to equation (8), after which, the linear amplifier will be operating in a region where the output is reliable:
|x Δ(p)+x(p)|=M p ·αM p ·α−|x(p)|=x Δ(p) (16)
$\begin{array}{cc}{M}_{p}·a-x\left(p\right)=\sum _{m=0}^{{N}_{s}-1}\text{ }g\left(\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right)\sum _{k=0}^{M-1}\Delta \left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right){}^{j\frac{2\pi }{M}\mathrm{kn}}& \left(17\right)\\ =\sum _{m=0}^{{N}_{s}-1}\text{ }g\left(\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right)\sum _{k=0}^{M-1}C·W\left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)·g\left(\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right)·{}^{j\left({\phi }_{p}-\frac{2\pi }{M}k\right)}{}^{j\frac{2\pi }{M}\mathrm{kn}}& \left(18\right)\\ =\sum _{m=0}^{{N}_{s}-1}\text{ }{g}^{2}\left(\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right)\sum _{k=0}^{M-1}C·W\left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)& \left(19\right)\end{array}$
Pulling C to the front, noting that |g2|=g2 and that W(k,m) is positive and real: $\begin{array}{cc}{M}_{p}·a-x\left(p\right)=C\sum _{m=0}^{{N}_{s}-1}\text{ }{g}^{2}\left(\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right)\sum _{k=0}^{M-1}W\left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)& \left(20\right)\\ C=\frac{{M}_{p}·a-x\left(p\right)}{\sum _{m=0}^{{N}_{s}-1}\text{ }{g}^{2}\left[\mathrm{mI}+{\left(\mathrm{pD}\right)}_{I}\right]\sum _{k=0}^{M-1}W\left(k,⌊\frac{\mathrm{pD}}{I}⌋-m\right)}& \left(21\right)\end{array}$
Computing Δ(k,m) in this manner requires the re-computation of the IFFT for xr(n,m) in equations (2) and (3). It is more efficient to solve for x′r(n,m), the IFFT of Δ(k,m). Since the system is linear, the component from the symbol adjustment matrix can be computed and summed with xr(n,m):
x′ r(n,m)=x r(n,m)+δr(n,m) (22)
where, $\begin{array}{cc}{\delta }_{r}\left(n,m\right)=M·{-1}^{}\left\{\Delta \left(k,m\right)\right\}=M·\frac{1}{M}\sum _{k=0}^{M-1}\text{ }\Delta \left(k,m\right){}^{j\frac{2\pi }{M}\mathrm{kn}}& \left(23\right)\end{array}$
Substituting for Δ(k,m) and using the pointer updates for k and m: $\begin{array}{cc}{\delta }_{r}\left(n,m\right)=\sum _{k=0}^{M-1}\text{ }C·W\left(k,⌊\frac{\mathrm{nD}}{I}⌋-m\right)·g\left[\mathrm{mI}+{\left(\mathrm{nD}\right)}_{I}\right]{}^{j\left({\phi }_{p}-\frac{2\pi }{M}{k\left(p\right)}_{M}\right)}{}^{j\frac{2\pi }{M}\mathrm{kn}}& \left(24\right)\\ =C·{}^{j\text{ }{\phi }_{p}}\sum _{k=0}^{M-1}\text{ }W\left(k,⌊\frac{\mathrm{nD}}{I}⌋-m\right)·g\left[\mathrm{mI}+{\left(\mathrm{nD}\right)}_{I}\right]{}^{j\left(\frac{2\pi }{M}{k\left(n-p\right)}_{M}\right)}& \left(25\right)\\ =C·{}^{j\text{ }{\phi }_{p}}·g\left[\mathrm{mI}+{\left(\mathrm{nD}\right)}_{I}\right]\sum _{k=0}^{M-1}W\left(k,⌊\frac{\mathrm{nD}}{I}⌋-m\right){}^{j\left(\frac{2\pi }{M}{k\left(n-p\right)}_{M}\right)}& \left(26\right)\\ {\delta }_{r}\left(n,m\right)=C·{}^{j\text{ }{\phi }_{p}}·g\left[\mathrm{mI}+{\left(\mathrm{nD}\right)}_{I}\right]{w}_{r}\left({\left(n-p\right)}_{M},⌊\frac{\mathrm{nD}}{I}⌋-m\right)& \left(27\right)\end{array}$
Where wr(n,m) is defined as the one-dimensional IFFT of W(k,m) along the frequency axis: $\begin{array}{cc}{w}_{r}\left(n,m\right)=M·{-1}^{}\left\{W\left(k,m\right)\right\}=M·\sum _{k=0}^{M-1}\text{ }W\left(k,m\right){}^{j\left(\frac{2\pi }{M}\mathrm{kn}\right)}& \left(28\right)\end{array}$
Thus, in a first embodiment, the present invention provides a method and apparatus for receiving a plurality of symbols, and modulating the plurality of symbols to create a first modulated signal. The first modulated signal is then compared to a threshold. If the first modulated signal is above the threshold, the first modulated signal is transmitted. If the first modulated signal, however, is below the threshold, an adjustment to the received plurality of symbols is computed and added to the received plurality of symbols to create an adjusted plurality of symbols. The adjusted plurality of symbols is then modulated to create a second modulated signal. The second modulated signal is transmitted if it exceeds the threshold; otherwise, the steps of computing, adding and the second step of modulating are repeated until the modulated signal exceeds the threshold.
In an alternative embodiment, the present invention provides a method and apparatus for receiving a plurality of symbols, mixing the plurality of symbols using an inverse fast Fourier transform to create a plurality of mixed symbols, and pulse shaping the plurality of mixed symbols to create a first modulated signal. The first modulated signal is then compared to a threshold. If the first modulated signal is above the threshold, the first modulated signal is transmitted. If the first modulated signal is below the threshold, an adjustment to the plurality of mixed symbols is computed and added to the plurality of mixed symbols to create an adjusted plurality of mixed symbols. Pulse shaping is performed on the adjusted plurality of mixed symbols to create a second modulated signal. As in the first embodiment, the second modulated signal is transmitted if the second modulated signal exceeds the threshold; otherwise, the steps of computing, adding and the second step of pulse shaping are repeated until the modulated signal exceeds the threshold.
While the invention has been described in conjunction with specific embodiments thereof, additional advantages and modifications will readily occur to those skilled in the art. The invention, in its broader aspects, is therefore not limited to the specific details, representative apparatus, and illustrative examples shown and described. Various alterations, modifications and variations will be apparent to those skilled in the art in light of the foregoing description. Thus, it should be understood that the invention is not limited by the foregoing description, but embraces all such alterations, modifications and variations in accordance with the spirit and scope of the appended claims.
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Referenced by
Citing PatentFiling datePublication dateApplicantTitle
US8306486 *Jul 23, 2009Nov 6, 2012Panasonic CorporationMethods and apparatus for reducing the average-to-minimum power ratio of communications signals in communications transmitters
US20110021165 *Jul 23, 2009Jan 27, 2011Panasonic CorporationMethods and apparatus for reducing the average-to-minimum power ratio of communications signals in communications transmitters
Classifications
U.S. Classification375/295, 370/343, 375/260
International ClassificationH04L5/06
Cooperative ClassificationH04L27/2627, H04L5/06, H04L27/2614
European ClassificationH04L5/06, H04L27/26M3A, H04L27/26M2
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Effective date: 20120622 | 8,421 | 25,798 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-22 | latest | en | 0.770919 |
http://mo-kisha.ru/applications-of-single-phase-squirrel-cage-induction-motor.html | 1,569,033,327,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574182.31/warc/CC-MAIN-20190921022342-20190921044342-00550.warc.gz | 121,308,336 | 7,929 | # Applications of single phase squirrel cage induction motor, what is squirrel cage induction motor
Hence, at that moment the rotor experiences zero tangential force hence the rotor decelerates for the moment. Since the rotor has a very low resistance, the starting torque was very poor. Synchronous speed is the speed of rotation of the magnetic field in a rotary machine, and it depends upon the frequency and number poles of the machine. Transactions of the American Institute of Electrical Engineers.
This induces a current in the rotor winding, which produces its own magnetic field. Notice that the single wire loop is connected to a resistor and together they form a closed loop. In both cases, they are isolated from their shafts.
## Three Phase Induction Motor
The squirrel cage winding of a synchronous machine will generally be smaller than for an induction machine of similar rating. This motor is one phase since it is so small. Bearings are provided as the base for rotor motion, and the bearings keep the smooth rotation of the motor. After giving the supply, if the rotor is made to rotate in either direction by an external force, then the motor will start to run. The intrinsic performance of this machine matches well with the nature of wind, having a variable speed and not being in our control.
Polyphase motors have rotor bars shaped to give different speed-torque characteristics. At standstill, the rotor current is the same frequency as the stator current, and tends to travel at the outermost parts of the cage rotor bars by skin effect. It is simple, from the name itself we can understand that here, the induction process is involved.
In simple words, the induction motor which uses a squirrel cage rotor is called squirrel cage induction motor. In squirrel cage induction motor starting torque is very low. Thus the outer cage has a high resistance with low reactance to resistance ratio and the inner cage has low resistance but high reactance to resistance ratio. To withstand high currents, geilenkirchen singletreff the rotor structure is modified and more resembles a cage than a winding. Similar is the operation of the induction motor in parallel with a synchronous motor serving as a power factor compensator.
In effect the rotor is carried around with the magnetic field but at a slightly slower rate of rotation. It is made of thin laminations, separated by varnish insulation, to reduce eddy currents circulating in the core. The three types of capacitor motors are capacitor start, capacitor run, and capacitor start and run motors. Thus, the rotor will feel the torque that rotates the rotor in the direction of the rotating magnetic flux. From Wikipedia, the free encyclopedia.
These motors are well suited for those applications where there is a limitation in starting current. Squirrel cage induction motors are commonly used in many industrial applications. We saw Squirrel cage induction motor Applications.
## Working principle and types of an Induction Motor
Point to be noted that fan used in home appliances is a single phase induction motor which is inherently not self-starting. In a single-phase split-phase motor, reversal is achieved by changing the connection between the primary winding and the start circuit. That is why we use a capacitor in the single-phase induction motor. It has only one phase still it makes the rotor to rotate, so it is quite interesting. The rotor of the squirrel cage motor has no winding, and there is no need for the rotor to be electrically connected to any electricity.
• The method of changing the direction of rotation of an induction motor depends on whether it is a three-phase or single-phase machine.
• When current flows through the rotor circuit it generates rotor flux.
• The normal running windings within such a single-phase motor can cause the rotor to turn in either direction, so the starting circuit determines the operating direction.
• Some very small rotors which operate on the basis of eddy current, have their rotor as solid steel without any conductors.
This rotating magnetic field has a speed which is known as the synchronous speed. The capacitor of these motors are sometimes built onto the motor or located remotely away from the motor primarily making it easier to replace. When an alternating current is run through the stator windings, partnervermittlung elitepartner a rotating magnetic field is produced.
Here rotor core is laminated to avoid power losses from eddy current and hysteresis. For this to work the motor must see a reactive load, and either be connected to a grid supply or an arrangement of capacitors to provide excitation current. The construction of Stator of the induction motor is almost the same as other motors. All the necessary information in one place.
1. There are two types of induction motors.
2. Here you will know the Squirrel cage induction motor working principle, construction, application and advantages of the squirrel cage induction motor.
3. But the rotor construction differs with respect to the type which is specified above.
4. The winding on a stator is mounted in such a way that it provides low reluctance path for generated flux by A.
5. So here there is two fluxes, one is the stator flux and the other is the rotor flux, and the rotor flux will lag behind the stator flux.
## What is Squirrel Cage Induction Motor
The rotor flux will be lagging in respect of the stator flux. This rotor is made up from parallel aluminum or copper bars. The interaction between the stator's revolving magnetic field and the copper-pipe-rotor's induced magnetic field produces a torque and thus rotation. If the copper pipe is inserted inside the stator, there will be an induced current in the pipe, and this current will produce a magnetic field in the pipe. The windings are accessible through slip rings.
## Working & Applications
Actually, wandern single when we give supply to stator winding then the current starts to flow in the coil will produce a magnetic flux in the coil. Rotor conductor bars are short-circuited with two end rings. Or we can say in synchronous speed. These rotor currents generate their self-magnetic field which will interact with the field of the stator.
The motor rotor shape is a cylinder mounted on a shaft. The rotor is constructed with deep and narrow slots so as to obtain high reactance while starting. At a given power rating, lower speed requires a larger frame.
This is the working principle of both single and three phase induction motors. These motors consist of the split phase, shaded pole, and capacitor motors. There are end rings which are welded or electrically braced or even bolted at both ends of the rotor, thus maintaining electrical continuity. The stator poles are equipped with an additional winding in each corner called a shade winding as shown in fig. Induction motors are cheaper than synchronous motors.
## Induction Motor Working Principle Types & Definition
Since rotation at synchronous speed would result in no induced rotor current, an induction motor always operates slightly slower than synchronous speed. The interaction of the magnetic fields of currents in the stator and rotor produce a torque on the rotor. Because of that, zagreb the rotor will feel a torque which will make the rotor to rotate in the direction of the rotating magnetic field.
Various regulatory authorities in many countries have introduced and implemented legislation to encourage the manufacture and use of higher efficiency electric motors. To optimize the distribution of the magnetic field, windings are distributed in slots around the stator, with the magnetic field having the same number of north and south poles. It is almost same for any given synchronous motor or a generator.
Single phase capacitor motors are the next step in the family of single phase induction motors. The generating mode for induction motors is complicated by the need to excite the rotor, which begins with only residual magnetization. The loop ends are fixed to the slip rings, and two brushes make the connection between the slip rings and the external circuit. The moment rotor catches the rotating magnetic field the rotor current drops to zero as there is no more relative motion between the rotating magnetic field and rotor.
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http://r-tutorials.com/category/exercise-database/ | 1,696,475,699,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00449.warc.gz | 36,020,950 | 10,333 | Exercise Database
R Exercises – 71-80 – Loops (For Loop, Which Loop, Repeat Loop), If and Ifelse Statements in R
1. Simple ifelse statement Create the data frame ‘student.df’ with the data provided below: Use a simple ‘ifelse’ statement to add a new column ‘male.teen’ to the data frame. This is a boolean column, indicating T if the observation is a male younger than 20 years. 2. Double for loop Write a double for loop which prints
R Exercises – 61-70 – R String Manipulation | Working with ‘gsub’ and ‘regex’ | Regular Expressions in R
Required packages and datasets 1. ‘College’ dataset – Colleges in Texas a. Get familiar with the ‘college’ dataset and its row names. b. Get a vector with the college names (‘college.names’) which you will need in the further steps of this and the next exercises. c. Get a vector (‘texas.college’) which contains all colleges with ‘Texas’ in its name.
R Exercises – 51-60 – Data Pre-Processing with Data.Table
Required packages for the excises 1. ‘College’ dataset – Basic row manipulations a. Transform ‘College’ from ‘ISLR’ to data.table. Make sure to keep the University identifier. We will use this new data.table called ‘dtcollege’ throughout this block of exercises. b. Get familiar with the dataset and its variables. c. Extract rows 40 to 60 as a new data.table (‘mysubset’).
R Exercises – 41-50 – Working with Time Series Data
1. Simple time series plot on ‘non-ts’ data a. Get 200 random numbers and call the object ‘mydata’. Let’s set a seed of 14 for reproducibility. b. Get a time series plot without converting to class ‘ts’. c. Add ablines to the chart to indicate the horizontal boundaries of 0 and 1. 2. Working with ‘xts’ a. Get and load
R Exercises – 31-40 – Data Frame Manipulations
1. Working with the ‘mtcars’ dataset a. Get a histogram of the ‘mpg’ values of ‘mtcars’. Which bin contains the most observations? b. Are there more automatic (0) or manual (1) transmission-type cars in the dataset? Hint: ‘mtcars’ has 32 observations. c. Get a scatter plot of ‘hp’ vs ‘weight’. 2. Working with the ‘iris’ dataset
R Exercises – 21-30 – The Apply Family of Functions
1. Function ‘apply’ on a simple matrix: a. Get the following matrix of 5 rows and call it ‘mymatrix’ b. Get the mean of each row c. Get the mean of each column d. Sort the columns in ascending order 2. Using ‘lapply’ on a data.frame ‘mtcars’ a. Use three ‘apply’ family functions to get the minimum values
R Exercises for Beginners – 11-20 – Easy Functions
1. a. Write a function “myfun” of x to the power of its index position (x, x^2, x^3, …) b. Test the function with an x of 1:10 c. Enlarge the function “myfun” with a division through the index position (x, x^2 / 2, x^3 /3, …) 2. a. Write a simple moving average
R Exercises for Beginners – 1-10
Practicing is a crucial part of learning a new language. Statistical languages like R are no exception of that rule. Many of my students think the same and would love to see more exercises. Therefore, I decided to write an R exercise sheet for beginners and blog it over here. These R exercises are an
Famous and Very Useful Pre-Installed Exercise Datasets in R
As most of you surely know, R has many exercise datasets already installed. That simply means, as soon as you installed R Base, which includes the library ‘datasets’, you have ample opportunity to explore R with real world data frames. For me as course content creator those datasets help tremendously, because with them I can | 908 | 3,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.803121 |
http://medical-dictionary.thefreedictionary.com/densities | 1,539,980,359,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512434.71/warc/CC-MAIN-20181019191802-20181019213302-00226.warc.gz | 230,346,383 | 14,385 | # density
(redirected from densities)
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Related to densities: Population densities
## density
[den´sĭ-te]
1. the ratio of the mass of a substance to its volume.
2. the quality of being compact.
3. the quantity of matter in a given space.
4. the quantity of electricity in a given area, volume, or time.
5. the degree of film blackening in an area of a photograph or radiograph.
## den·si·ty (ρ),
(den'si-tē),
1. The compactness of a substance; the ratio of mass to unit volume, usually expressed as g/cm3 (kg/m3 in the SI).
2. The quantity of electricity on a given surface or in a given time per unit of volume.
3. radiologic physics the opacity to light of an exposed radiographic or photographic film; the darker the film, the greater will be the measured density.
4. clinical radiology a less exposed area on a film, corresponding to a region of greater x-ray attenuation (radiopacity) in the subject; the more light transmitted by the film, the greater the density of the subject will be; this is not actually the opposite of sense 3, because one concerns film density and the other subject density.
[L. densitas, fr. densus, thick]
## density
/den·si·ty/ (den´sit-e)
1. the quality of being compact or dense.
2. quantity per unit space, e.g., the mass of matter per unit volume. Symbol d.
3. the degree of darkening of exposed and processed photographic or x-ray film.
## density (D)
[den′sitē]
Etymology: L, densus, thick
1 the amount of mass of a substance in a given volume. The greater the mass in a given volume, the greater the density. See also mass, volume.
2 (in radiology) the degree of x-ray film blackening.
## density
The amount of a substance per unit volume Imaging
1. The compactness in a scan which reflects the type of tissues seen in CT and MR scans.
2. The amount of 'hard' or mineralized tissue in a plain film. See Bone mineral, Current density, Muscle fiber density, Spin density, Vapor density.
## den·si·ty
, pl. densities (dens'i-tē, -tēz)
1. The compactness of a substance; the ratio of mass to unit volume, usually expressed as g:cm3 (kg:m3 in SI).
2. The quantity of electricity on a given surface or in a given time per unit of volume.
3. radiologic physics The opacity to light of an exposed radiographic or photographic film; the darker the film, the greater the measured density.
4. clinical radiology A less-exposed area on a film, corresponding to a region of greater x-ray attenuation (radiopacity) in the subject; the more light transmitted by the film, the greater the density of the subject; this is not actually the opposite of the sense 3 definition, because one concerns film density and the other subject density.
[L. densitas, fr. densus, thick]
## density
the ratio of mass to volume. Measured in kilograms per cubic metre (kg.m-3).
## density
An indication of the compactness of a substance. It is expressed as the ratio of the mass of the substance to its unit volume. The common units are g/cm3 and kg/m3. This property is usually given by lens manufacturers, the greater the density of a material, the greater its weight, all other factors being equal.
## den·si·ty
, pl. densities (dens'i-tē, -tēz)
1. Compactness of a substance.
2. Quantity of electricity on a given surface or in a given time per unit of volume.
3. radiologic physics opacity to light of an exposed radiographic or photographic film; the darker the film, the greater the measured density.
4. clinical radiology a less exposed area on a film, corresponding to a region of greater x-ray attenuation (radiopacity) in the subject.
[L. densitas, fr. densus, thick]
## density,
n the concentration of matter, measured by mass per unit volume.
density, radiographic,
n the degree of darkening of exposed and processed photographic or radiographic film, expressed as the logarithm of the opacity of a given area of the film.
## density
1. the ratio of the mass of a substance to its volume.
2. the quality of being compact.
3. the quantity of matter in a given space.
4. the quantity of electricity in a given area, volume or time.
5. the degree of film blackening in an area of a photograph or radiograph.
population density
number of animals per unit of area; important in relation to the rate of spread of disease.
density sampling
## Patient discussion about density
Q. what are the sources for high density lipoprotein? I have heard that high density lipoprotein is good for heart. What differences does it make in heart’s health and what are the sources for high density lipoprotein?
A. Hi Liam, it is very important that we have high density lipoprotein (HDL) in our body. The fact is that the HDL is formed inside the body. They are known as good cholesterol as they are famous for their protection for heart against the heart diseases. It has been found that Vitamin B3 or Niacin consumption increases the count of HDL. It’s good to cut on the diet having more of saturated fats and oils, which increases the chances of heart attack.
More discussions about density
References in periodicals archive ?
Whether dead crow densities will be associated with the number of human cases in future years or other geographic areas is unknown.
The explanations for this pattern suggest three corollary hypotheses: virus "overwinters" in temperate rodent communities as persistent infections in older adult animals, which serve as a reservoir for reintroducing virus into susceptible young animals in the spring; spring antibody prevalence is a function of the population density (infectious and susceptible) the year before (the high fall population densities and higher spring antibody prevalence at the Colorado trapping webs in spring 1995 provide tentative support for this hypothesis); and deviations from typical environmental conditions alter the pattern of infection in potentially predictable directions.
The LGMR technology also allows for future increases in track densities by completely removing the servo tracks from the data side of the tape.
In this example we saw that the traditionally known facts, that compound viscosity can be altered by changes in plasticizer level or polymer type, and thereby affect sponge product densities.
A higher-output MP formulation, designated MP+++, was developed to handle the higher track densities of the DDS-4 format.
To increase areal densities in longitudinal recording, as well as increase overall storage capacity, the data bits on a disc must be made smaller and put closer together.
Women whose mammograms showed over 65 percent dense tissue developed breast cancer at a rate more than 400 percent higher than that of women with densities of less than 5 percent.
Another alternative to DMFC is the direct formic acid fuel cell (DFAFC) technology that has the potential to offer five to six times higher power densities.
Future I/O-intensive applications will require higher access densities than are indicated by the current development roadmaps.
Line of modular infrared heating devices (SpotIR, LineIR, StripIR, PanelIR, ChambIR, Hi-TempIR, and SimulateIR) produce heat flux densities ( -3 watts/[in.
for at least five years and compared their bone densities with those of age- and weight-matched women not on the hormone.
TOKYO -- New Devices Allow More Densities and Options to Stack with Other Semiconductors in Multi-Chip and System-in-Package Designs
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Open / Close | 1,654 | 7,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-43 | longest | en | 0.837347 |
http://www.unicode.org/mail-arch/unicode-ml/y2010-m07/0361.html | 1,511,521,999,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807650.44/warc/CC-MAIN-20171124104142-20171124124142-00144.warc.gz | 521,096,678 | 2,816 | Re: Reasonable to propose stability policy on numeric type = decimal
From: karl williamson (public@khwilliamson.com)
Date: Wed Jul 28 2010 - 23:33:39 CDT
• Next message: Martin J. Dürst: "Re: Reasonable to propose stability policy on numeric type = decimal"
Asmus Freytag wrote:
> On 7/25/2010 6:05 PM, Martin J. Dürst wrote:
>>
>>
>> On 2010/07/26 4:37, Asmus Freytag wrote:
>>
>>> PPS: a very hypothetical tough case would be a script where letters
>>> serve both as letters and as decimal place-value digits, and with modern
>>> living practice.
>>
>> Well, there actually is such a script, namely Han. The digits (一、
>> 二、三、四、五、六、七、八、九、〇) are used both as letters and as
>> decimal place-value digits, and they are scattered widely, and of
>> course there are is a lot of modern living practice.
> Martin,
>
> you found the hidden clue and solved it, first prize :)
>
> They do not show up as gc=Nd, nor as numeric types Digit or Decimal.
>
> The situation is worse than you indicate, because the same characters
> are also used as elements in a system that doesn't use place-value, but
> uses special characters to show powers of 10.
Is it the case that a sequence of just these characters, without any
intervening characters, and not adjacent to the special characters you
mention always mean a place-value decimal number?
>
> However, as I indicated in my original post, in situations like that,
> there are usually some changes in practice that took place. Much of the
> living modern practice in these countries involves ASCII digits. While
> the ideographic numbers are definitely still used in certain contexts,
> I've not seen them in input fields and would frankly doubt that they
> exist there. I would fully expect that they are supported as number
> format for output, at least in some implementations, and, of course,
> that input methods convert ASCII digits into them. In other words, I
> wonder whether automatic conversion goes only one-way for these numbers.
> I would suspect it, for the general case, but I don't actually know for
> sure.
>
> For someone in Karl's situation, it would be interesting to learn
> whether and to what extent he should bother supporting these numbers in
> his language extension.
>
> A./
>
This archive was generated by hypermail 2.1.5 : Wed Jul 28 2010 - 23:37:34 CDT | 585 | 2,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-47 | latest | en | 0.941685 |
https://www.jiskha.com/display.cgi?id=1270392491 | 1,503,074,451,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104704.64/warc/CC-MAIN-20170818160227-20170818180227-00369.warc.gz | 917,355,843 | 4,478 | Physics
posted by .
To reduce the frequency of a particle when it behaves like a wave, we need to ...
-Increase its mass
-Reduce its speed
-Increase its speed
-Hit it with low energy photons
-Hit it with high energy photons
Does the same wave theory apply? i.e. V = f(lambda)
• Physics -
When speaking of the wave velocity of a a particle, you have to distinguish between the group velocity of the "wave packet" of waves of different frequencies (which tends to move with the particle while spreading out due to the uncertainty principle) and the "phase velocity" of the different waves that make up the wave packet. The phase velocity exceeds the speed of light. A particles can be thought of as a wave packet consisting of "beats" in its Be Broglie waves.
The formula V = f(lambda) only applies if V is the phase velocity, not the particle velocity.
You CAN use the rule
h*frequency = energy
which implies that you have to increase the energy and the speed of the particle, in order to increase its wave frequency.
For a better explanation, see
http://en.wikipedia.org/wiki/Phase_velocity
• Physics -
So the best answer would be increasing the speed, rightÉ
• Physics -
Yes. While I'm at it, I misspelled De Broglie in my previous answer.
• Physics -
I thought to reduce the frequency of a particle, you would have to decrease the speed. no?
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Which of the following statements correctly describes the changes that occur in a sound wave, as it passes from cool air into warmer air? | 553 | 2,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-34 | latest | en | 0.920728 |
https://kr.mathworks.com/matlabcentral/cody/problems/17-find-all-elements-less-than-0-or-greater-than-10-and-replace-them-with-nan/solutions/1884171 | 1,586,550,752,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370511408.40/warc/CC-MAIN-20200410173109-20200410203609-00336.warc.gz | 549,492,405 | 15,782 | Cody
Problem 17. Find all elements less than 0 or greater than 10 and replace them with NaN
Solution 1884171
Submitted on 25 Jul 2019 by Sean Ferrell
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
x = [ 5 17 -20 99 3.4 2 8 -6 ]; y_correct = [ 5 NaN NaN NaN 3.4 2 8 NaN ]; assert(isequalwithequalnans(cleanUp(x),y_correct))
2 Pass
x = [ -2.80 -6.50 -12.60 4.00 2.20 0.20 -10.60 9.00]; y_correct = [ NaN NaN NaN 4.00 2.20 0.20 NaN 9.00] assert(isequalwithequalnans(cleanUp(x),y_correct))
y_correct = NaN NaN NaN 4.0000 2.2000 0.2000 NaN 9.0000 | 247 | 665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-16 | latest | en | 0.572633 |
http://math.stackexchange.com/questions/145066/how-to-find-out-what-changes-applied-to-integral | 1,469,660,921,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827080.38/warc/CC-MAIN-20160723071027-00235-ip-10-185-27-174.ec2.internal.warc.gz | 164,014,463 | 18,015 | # How to find out what changes applied to integral?
I have got such integral $$\int{\frac{\sqrt{x^2+1}}{x+2}dx}$$ and with Maple I got something like this: $$\int\frac{1}{2} + \frac{1+3u^2+4u^3}{-2u^2+2u^4-8u^3}du$$ And I want to know how to achive this changes.
I tried to use WolframAlpha, but there is scarier solution. This integral was for Gaussian quadrature method, so it's analytic solution is horrible.
-
Substituting $x=-8u+2u^2$ could get somewhat close to what you are expecting, but not exactly the expression you have. – Kirthi Raman May 14 '12 at 17:22
## 1 Answer
1. We can use the Euler substitution $t=\sqrt{x^{2}+1}-x$ to obtain a rational fraction in terms of $t$ $$\begin{eqnarray*} I =\int \frac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\frac{1}{2}\int \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }\mathrm{d}t. \end{eqnarray*}$$
2. Since the integrand is a rational fraction, we can expand it into partial fractions and integrate each fraction. $$\begin{equation*} \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }=1-\frac{1}{t^{2}}+ \frac{4}{t}+\frac{20}{t^{2}-4t-1}. \end{equation*}$$
Added. Detailed evaluation. From $t=\sqrt{x^{2}+1}-x$, we get $x=\dfrac{1-t^{2}}{2t}$ and $\dfrac{dx}{dt}=-\dfrac{t^{2}+1}{2t^{2}}$. So we have $$\begin{eqnarray*} I &=&\int \frac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\int \frac{t+\frac{1-t^{2}}{2t}}{\frac{1-t^{2}}{2t}+2}\left( -\frac{t^{2}+1}{2t^{2}}\right) \mathrm{d}t \\ &=&\frac{1}{2}\int \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) } \mathrm{d}t. \end{eqnarray*}$$
Expanding into partial fractions as above, we obtain $$\begin{eqnarray*} 2I &=&\int 1-\frac{1}{t^{2}}+\frac{4}{t}+\frac{20}{t^{2}-4t-1}\mathrm{d}t \\ &=&\int 1\mathrm{d}t-\int \frac{1}{t^{2}}\mathrm{d}t+4\int \frac{1}{t} \mathrm{d}t+20\int \frac{1}{t^{2}-4t-1}\mathrm{d}t \\ &=&t+\frac{1}{t}+4\ln \left\vert t\right\vert -2\sqrt{5}\ln \frac{\sqrt{5}t-2 \sqrt{5}+5}{5-\sqrt{5}t+2\sqrt{5}}+C \\ &=&\sqrt{x^{2}+1}-x+\frac{1}{\sqrt{x^{2}+1}-x}+4\ln \left( \sqrt{x^{2}+1} -x\right) \\ &&-2\sqrt{5}\ln \frac{\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) -2\sqrt{5}+5}{5-\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) +2\sqrt{5}}+C. \end{eqnarray*}$$
Therefore the given integral is $$\begin{eqnarray*} I &=&\frac{1}{2}\left( \sqrt{x^{2}+1}-x\right) +\frac{1}{2}\frac{1}{\sqrt{ x^{2}+1}-x}+2\ln \left( \sqrt{x^{2}+1}-x\right) \\ &&-\sqrt{5}\ln \frac{\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) -2\sqrt{5}+5}{5-\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) +2\sqrt{5}}+C. \end{eqnarray*}$$
- | 1,141 | 2,503 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2016-30 | latest | en | 0.582992 |
https://www.coursehero.com/file/6514396/The-Mean-Value-Theorem/ | 1,529,287,858,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859923.59/warc/CC-MAIN-20180618012148-20180618032148-00103.warc.gz | 776,876,377 | 100,171 | The Mean Value Theorem
# The Mean Value Theorem - The Mean Value Theorem In this...
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The Mean Value Theorem In this section we want to take a look at the Mean Value Theorem. In most traditional textbooks this section comes before the sections containing the First and Second Derivative Tests because many of the proofs in those sections need the Mean Value Theorem. However, we feel that from a logical point of view it’s better to put the Shape of a Graph sections right after the absolute extrema section. So, if you’ve been following the proofs from the previous two sections you’ve probably already read through this section. Before we get to the Mean Value Theorem we need to cover the following theorem. Rolle’s Theorem Suppose is a function that satisfies all of the following. 1. is continuous on the closed interval [ a,b ]. 2. is differentiable on the open interval ( a,b ). 3. Then there is a number c such that and . Or, in other words has a critical point in ( a,b ). To see the proof of Rolle’s Theorem see the Proofs From Derivative Applications section of the Extras chapter. Let’s take a look at a quick example that uses Rolle’s Theorem. Example 1 Show that has exactly one real root. Solution From basic Algebra principles we know that since is a 5 th degree polynomial it will have five roots. What we’re being asked to prove here is that only one of those 5 is a real number and the other 4 must be complex roots. First, we should show that it does have at least one real root. To do this note that
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and that and so we can see that . Now, because is a polynomial we know that it is continuous everywhere and so by the Intermediate Value Theorem there is a number c such that and . In other words has at least one real root. We now need to show that this is in fact the only real root. To do this we’ll use an argument that is called contradiction proof. What we’ll do is assume that has at least two real roots. This means that we can find real numbers a and b (there might be more, but all we need for this particular argument is two) such that . But if we do this then we know from Rolle’s Theorem that there must then be another number c such that .
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https://www.physicsforums.com/threads/trajectory-velocity-time-question.300673/ | 1,527,260,291,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867094.5/warc/CC-MAIN-20180525141236-20180525161236-00270.warc.gz | 801,062,385 | 15,937 | # Homework Help: Trajectory/velocity/time question
1. Mar 18, 2009
### Puchiko
1. The problem statement, all variables and given/known data
(I'm translating the problem from Slovak, so it might sound a bit choppy at times-my apologies)
There's a road. On the road, there's a moving vehicle carrying a tree log, and a tourist is walking alongside the vehicle. The tourist's velocity is 2 m/s, and he wants to know how long the log is. When he walks from the front end to the back end of the log (keep in mind the log is moving in the opposite direction) he counts 16 steps. When walking from the back end to the front end of the log, he counts 112 steps (the vehicle is now moving in the opposite direction than the tourist). The length of his step is 0.75 m.
a) What is the velocity of the vehicle?
b) What is the length of the log?
2. Relevant equations
s=t.v
s....trajectory
t....time
v....velocity
3. The attempt at a solution
I'm rather stumped. I converted the tourists steps:
It took him 12m and 6s to walk from the front to the back end of the vehicle.
It took him 84m and 42s to walk from the back to the front end of the vehicle.
And that's about as far as I got. I could calculate the vehicle velocity if I had the log length, just like I could calculate the log length if I had the velocity. But having neither?
I also deduced he must be moving quicker than the vehicle, because he managed to reach the front end from the back one.
Thanks for any help.
2. Mar 18, 2009
### LowlyPion
Welcome to PF.
It seems like the length of the log can be expressed as
T1*(Vp + Vt) = L
And T2*(Vp - Vt) = L
Where Vp is the velocity of the person, Vt the Velocity of the truck.
T1 and T2 you already figured looks like.
3. Mar 18, 2009
### Puchiko
So, I put together the equation by combining yours, and calculated the thing. My results are
a) The speed of the vehicle is 1,5 m/s.
b) The log is 21 m long.
Can anyone check my results? Thanks.
Also, though I get how you got the first equation (since they're moving in the same direction, it's just time multiplied by their combined speed), I don't understand how I'd put together the second one. Can you explain? Thankies, and thanks so much for your help-I thought I was never going to solve it!
4. Mar 18, 2009
### LowlyPion
That's what I get anyway.
In the one you are walking against the speed of the log. So in relative terms the end of the log is approaching the walker at the combined rates of their speeds.
In the other he is walking toward the front of the truck and the front end of the log at the rate of his speed less the speed that the truck is moving since it's moving in the same direction he is.
In both he walks the length of the log at the effective rate in the time given by his walking speed and length of stride.
5. Mar 18, 2009
### Puchiko
Thanks for all your help, it is appreciated. | 743 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.962505 |
https://metanumbers.com/24454 | 1,600,797,366,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00746.warc.gz | 548,615,559 | 7,433 | 24454
24,454 (twenty-four thousand four hundred fifty-four) is an even five-digits composite number following 24453 and preceding 24455. In scientific notation, it is written as 2.4454 × 104. The sum of its digits is 19. It has a total of 2 prime factors and 4 positive divisors. There are 12,226 positive integers (up to 24454) that are relatively prime to 24454.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 19
• Digital Root 1
Name
Short name 24 thousand 454 twenty-four thousand four hundred fifty-four
Notation
Scientific notation 2.4454 × 104 24.454 × 103
Prime Factorization of 24454
Prime Factorization 2 × 12227
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 24454 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 24,454 is 2 × 12227. Since it has a total of 2 prime factors, 24,454 is a composite number.
Divisors of 24454
1, 2, 12227, 24454
4 divisors
Even divisors 2 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 36684 Sum of all the positive divisors of n s(n) 12230 Sum of the proper positive divisors of n A(n) 9171 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 156.378 Returns the nth root of the product of n divisors H(n) 2.66645 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 24,454 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 24,454) is 36,684, the average is 9,171.
Other Arithmetic Functions (n = 24454)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 12226 Total number of positive integers not greater than n that are coprime to n λ(n) 12226 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2709 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 12,226 positive integers (less than 24,454) that are coprime with 24,454. And there are approximately 2,709 prime numbers less than or equal to 24,454.
Divisibility of 24454
m n mod m 2 3 4 5 6 7 8 9 0 1 2 4 4 3 6 1
The number 24,454 is divisible by 2.
Classification of 24454
• Arithmetic
• Semiprime
• Deficient
Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
Base conversion (24454)
Base System Value
2 Binary 101111110000110
3 Ternary 1020112201
4 Quaternary 11332012
5 Quinary 1240304
6 Senary 305114
8 Octal 57606
10 Decimal 24454
12 Duodecimal 1219a
20 Vigesimal 312e
36 Base36 iva
Basic calculations (n = 24454)
Multiplication
n×i
n×2 48908 73362 97816 122270
Division
ni
n⁄2 12227 8151.33 6113.5 4890.8
Exponentiation
ni
n2 597998116 14623445928664 357601746739549456 8744793114768942397024
Nth Root
i√n
2√n 156.378 29.0257 12.5051 7.54519
24454 as geometric shapes
Circle
Diameter 48908 153649 1.87867e+09
Sphere
Volume 6.12545e+13 7.51467e+09 153649
Square
Length = n
Perimeter 97816 5.97998e+08 34583.2
Cube
Length = n
Surface area 3.58799e+09 1.46234e+13 42355.6
Equilateral Triangle
Length = n
Perimeter 73362 2.58941e+08 21177.8
Triangular Pyramid
Length = n
Surface area 1.03576e+09 1.72339e+12 19966.6
Cryptographic Hash Functions
md5 0af03ed6bc08aa81017a8453e36cdfa9 391f093e1b800d434c69fce3b62c5b7672b9e3a7 4772641f101b4503096ec0c6b62d8619edd8e250cd7fb3df15ae7746a5b541e0 94e527b4986f70261d820e4c4301fd7b26819979815fffc21e411be3200aeb922d648d1a3857b6bed8d64fe0d27e6ec56351e8269c0884974077eea7716d4572 9d7ac8892dea52087a1ae715e7d774c36e562cdc | 1,425 | 4,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-40 | latest | en | 0.826469 |
panoradio-sdr.de | 1,726,191,945,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00233.warc.gz | 434,384,041 | 17,349 | September 13, 2024
# Correlation for Time Delay Analysis
Time delay analysis finds the delay (also called the “lag”) between two signals, that are shifted in time. It is the most important part of time-difference-of-arrival (TDOA) transmitter localization. This article presents an introduction to time delay analysis with application to TDOA localization with RTL-SDRs.
For delay analysis, correlation in the time domain is widely used. The correlation function plots the similarity between two signals for all possible lags \tau.
Corr(\tau) = \sum_{t=0}^{N-1}s_1(t)s_2(t+\tau)
The peak of the correlation function occurs at the lag with the best similarity between the two signals, i.e. the estimated delay.
\tau_{estimated} = \underset{\tau}{\arg\max} (Corr(\tau))
For robust determination of the delay, it is important to be able to read out the peak of the correlation function precisely. For that purpose, the correlation should have a distinct and outstanding peak (“good quality”). If, on the other hand, many peak are potential lags or a peak is distributed among several bins, accurate measurement of delay becomes difficult (“bad quality”).
The quality of the correlation is influenced by different parameters:
• The type of received RF signal (especially its bandwidth)
• The different properties of the receivers
• The different propagation paths between transmitter and the receivers
• The applied correlation method
## Which signals exhibit good correlation properties?
The quality of correlation is first of all largely dependent on the transmitted signal. If the designer has the freedom to choose the transmitted signal, he should use one with known good correlation properties such as a Barker code or maximum-length-sequences. In generic TDOA systems the designer has no influence on the transmitted signal. In either case it is useful to understand the influence of the signal type on correlation.
In general there is a strong connection between bandwidth and correlation quality. Signals with large bandwidth and noise-like waveforms (e.g. maximum-length-sequences) correlate much better than narrow-band signals, that exhibit strong periodicity, as depicted below.
In theory, the correspondence between correlation and bandwidth is established by the Wiener-Khinchin-Theorem. In essence, it says that the power spectrum of a signal equals the power spectrum of its autocorrelation function (autocorrelation = correlation of a signal with a time-delayed version of itself). The perfect (and desirable) correlation is a single sharp peak. The power spectrum of such a sharp peak is wideband and following the theorem, so is the power spectrum of the time signal.
It needs to be emphasized, that bandwidth here refers to the “short-time” bandwidth of the signal portion located in the correlation window and not the “general” bandwidth over a very long time. This is because only a comparatively short slice (or window) of the RF signal stream can be used for time delay analysis. Example: A FM signal may have a bandwidth of several 10 kHz in general (over a long period). The window used for correlation is 10,000 samples. Then the “short-time” bandwidth in this short window needs to be considered. The short-time bandwidth of a FM signal, however, is strongly dependent on the transmitted content (silence corresponds to a sine wave and thus a very narrow bandwidth).
The figure below shows the bandwidth and correlation of successive slices of an FM signal. Bandwidth and correlation largely vary. It is advantageous to analyze the received signals and perform correlation on those slices, that have large bandwidth only.
## Time delay analysis for complex IQ signals
Most radios, such as the RTL-SDR, deliver IQ samples at its output that can be interpreted as a complex signal, exhibiting a phase and amplitude information. In the following, s_1(t) is the complex IQ signal from one receiver and s_2(t) the complex IQ signal from the other receiver. Again the time delay between the signals is to be found using the correlation function.
For IQ signals, four different variants of the correlation function are considered here:
1. Complex correlation is the standard approach to calculate the correlation of complex signals: |\sum s_1(t) \cdot s_2^*(t+\tau)| with * denoting “complex conjugate”.
2. Real-valued correlation of the amplitude, or the absolute value (magnitude), of the signals \sum |s_1(t)| \cdot |s_2(t+\tau)|
3. Real-valued correlation of the phase of the signals \sum \measuredangle s_1(t) \cdot \measuredangle s_2(t+\tau) with \measuredangle s = \arg(s)
4. Real-valued correlation of the phase difference of the signals \sum [ \measuredangle s_1(t) - \measuredangle s_1(t-1)] \cdot [ \measuredangle s_2(t+\tau) - \measuredangle s_2(t-1+\tau)]
For any correlation method it is advantageous to remove the DC component of the input signals. (It mitigates the effect that the correlation function tends to have larger values at the center)
The applicability of the different methods is dependent on the modulation of the signal. The correlations based on phase and phase difference only respond to angular modulation (most used modulation types exhibit some component of angular modulation). Correlation of amplitude responds to amplitude modulation only, i.e. they hardly work on constant envelope modulation, such as FM. The method of complex correlation takes into account both amplitude and phase and is therefore suited for all modulation types.
However, modulation type is not the only criterion for choosing a specific correlation method. It is also dependent on different real-world flaws that will be considered below.
## The influence of phase, amplitude and frequency offset
In a TDOA system, two receivers ideally output two identical versions of the signal, having only a time-shift between them. In real-world systems, both signals also differ in other parameters, like amplitude, phase and frequency offset.
A constant amplitude offset may originate in different propagation paths or receiver gain settings. Fortunately, it does not impact time delay analysis.
Differences in phase and frequency may originate in separate receivers, when phase and frequency of local oscillators and sampling circuits are not exactly synchronized (which is difficult in when the receivers are not located in the same place). Especially the use of low-cost radios like the RTL-SDR practically limits synchronization capabilities. While usually a constant phase offset does impact time delay analysis, frequency offsets are of greater interest, because they influence the results of correlations using phase information.
Frequency Offset: For the analysis of frequency offset, a signal has been captured by an RTL-SDR sampling at 2 Msps. Then the correlation of this signal with a frequency shifted version of itself is analyzed to model a frequency offset in the receiver.
The influence of the frequency offset is strongly dependent on the used correlation method and the window size. The figures below show the four different correlations under frequency offset for a window size of 1,000,000 and 10,000 samples.
For a window size of 1,000,000 samples even tiny frequency offsets above some Hz make the complex correlation and correlation of phase useless. For shorter windows the tolerance on frequency offsets is greater. But even then frequency offsets need to be kept below some 100 Hz.
The simulations also show that the correlation of the amplitude and the correlation of the phase difference are robust against frequency offsets. The correlation of phase differences works well up to some 100 kHz. The correlation of amplitude values shows no impact at all, because it does not use phase information.
## Robustness to noise
In general, all correlation methods are very robust to noise, if the noise components in the two signals are independent to some extent (This is because two perfectly independent noise signals do not correlate at all). In practical systems, noise is not always completely independent and the correlation methods exhibit different robustness to noise. The figure above shows, that correlation of the phase difference is slightly less robust to noise than the other three correlation methods. This is supported by the general observation that the differentiation operation amplifies noise. This should be considered when dealing with very low SNRs.
## How to choose the window size?
The length of the window used for correlation analysis is a trade-off between several goals. The window length should be
1. large enough to provide a sufficient signal overlap, if the receivers cannot be triggered in time accurately
2. large enough to average signal differences others than time delay, such as noise and distortion originating in the different propagation path and physically different receivers
3. small enough to reduce the impact of frequency offset (unless correlation of the amplitude or the phase difference is used)
4. small enough to reduce the amount of required calculations and amount of data to be exchanged between the receivers
Comment on 1: A pair of internet connected RTL-SDR receivers can be simultaneously triggered with an accuracy of approximately some hundred milliseconds using SSH commands. This requires the window length to be larger than this time delay in order to capture a signal portion, that overlaps. For a trigger accuracy of 200 ms and a sample rate of 2 Msps, the window length must be at least 400,000 samples.
## Summary and Recommendations
Careful selection of the correlation method improves the robustness and avoids errors in time delay measurements for complex IQ signals.
In summary, complex correlation should be chosen, whenever the receivers are exactly synchronized in frequency. If a frequency offset cannot be avoided, the correlation of amplitude or the correlation of the differential phase should be considered. Correlation of the phase is not recommended.
In a practical TDOA system based on RTL-SDRs the correlation of amplitude and/or phase difference may be suited most.
In any case, the modulation type of the analyzed signal should be considered, especially for constant envelope modulation (FM) correlation of amplitude is not applicable.
The achievable resolution and measurement accuracy is mostly dependent on the bandwidth and modulation of the received signal, which cannot be influenced by the receiver. | 2,078 | 10,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.931475 |
https://sinaumedia.com/branch-of-physics-and-explanations/ | 1,669,659,548,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00410.warc.gz | 549,786,421 | 21,260 | # Complete 20+ Branch of Physics and Explanations
Posted on
In this article we will discuss about the branch of physics. There are various branches of physics such as Mechanics, Fluid Mechanics, Thermodynamics, Electronics, Geometric Optics, Quantum Physics, Geophysics and others. Check out the full review below.
Physics is a natural science which is divided into several branches. Each of these branches of physics has a different focus of study from one another. The division of physics at the same time proves that physics is a very broad science. This is in line with the fact that many facets of the universe can be studied separately.
## Branch of Physics
All branch of physics have played many roles in human life. Following are the important branches in physics:
### 1. Mechanics
Mechanics is a branch of physics, which is the study of motion and changes in the shape of objects caused by forces. Mechanics is a branch of science in the early or oldest period of all branches of physics. Rows of physicists who laid the foundations of mechanics include: Archimedes, Galileo Galilei, and Isaac Newton.
Mechanics is a branch of physics that is widely applied in everyday life. The most important problem in mechanics is how things move and what causes them. This problem was solved by Newton in his laws of motion. The branch of physics mechanics is further divided into two, namely kinematics and dynamics.
• Kinematics is a part of mechanics that studies the motion of matter without regard to the cause of the object’s motion.
• Dynamics is a branch of mechanics that studies the relationship between the motion of an object and the cause of motion, namely force.
### 2. Fluid Mechanics
The next branch in physics is fluid mechanics. Fluid mechanics comes from the words mechanics and fluid. Mechanics is the study of motion, while fluid is a substance that when a force is applied to it, the substance will change shape continuously because it is unable to withstand the force, no matter how small the force acts.
So, mechanics is the study of motion in fluids (liquids, gases and plasmas) and the forces acting on them. Fluid mechanics is a science that has long been applied in everyday life. Since time immemorial, without us realizing it, many human activities have been related to fluid mechanics.
Famous laws in fluid mechanics such as Archimedes’ Law, Pascal’s Law, and Bernoulli’s Law. The physics branch of fluid mechanics is further divided into two, namely fluid kinematics and fluid dynamics.
• Fluid kinematics is a branch of fluid mechanics that studies the behavior of fluids at rest or not in motion.
• Fluid dynamics is the branch of fluid mechanics that studies fluids in motion or flow.
### 3. Thermodynamics
Physics also has a branch called thermodynamics. In simple terms, thermodynamics is the study of energy transformations, changes in state and the equilibrium of a group of particles that make up gases, liquids and solids, especially those related to thermal or heat properties.
Thermodynamics is often defined as the study of energy changes from one form to another, especially changes from heat energy into other forms of energy. The physicist who is known as the father of thermodynamics is Sadi Carnot.
### 4. Electronics
Electronics is a branch of physics that studies the emitting behavior and effects of electrons, as well as the control of charged particles in vacuum, gases, and semiconductor materials.
In other words, electronics is a branch of science that deals with how to flow and control electrons and their behavior and effects when in a medium such as a vacuum, gas, semiconductor material and in a pie.
Electronics is further divided into several branches, including:
• Digital electronics are electronic systems that work with digital systems or discrete signals.
• Analog electronics are electronic systems that have a continuous signal.
• Microelectronics is an electronic device that uses the concept of integration (integrated circuit-IC).
• Integrated circuits (ICs) are Active Electronics components consisting of a combination of hundreds, thousands and even millions of transistors.
• Optoelectronics is an electronic component that uses the properties of light or the interaction of light with matter.
### 5. Geometric Optics
Geometric optics is a branch of physics that studies the reflecting and refracting properties of light. A well-known law in this branch of physics is Snell’s law. The basics in geometric optics are as follows:
• The path of light in a homogeneous medium is straight
• The rays have no effect on one another
• The trajectory of the rays can be reversed
• The trajectory of light obeys Snell’s laws of reflection and refraction
### 6. Physical Optics
Physical optics is a branch of physics that studies the properties of interference, diffraction, and polarization of light. In another sense, physical optics is the study of the physical properties of light as a wave. Light is polychromatic, meaning that it consists of many colors called the color spectrum.
### 7. Quantum Physics
Quantum physics is a branch of physics that studies the physical properties of matter on an atomic or sub atomic scale. The essence of quantum physics is uncertainty, predictions in quantum theory are more likely than certain. Quantum physics is a branch of physics that will invite us to see the world of physics from a different perspective. A point of view that sees matter not only as matter, but also as waves.
### 8. Medical Physics
Medical physics is a branch of physical science that uses the principles, methods, and philosophies of physics in practice and research for the prevention, diagnosis, and treatment of disease. Medical physics is needed to reduce the risk of devices that use radiation by implementing and developing quality assurance methods in terms of safety and feasibility of medical radiation equipment as a whole.
### 9. Biophysics
Biophysics is a branch of physics that studies physical phenomena in living things. Biophysics is a combination of biology and physics. In biophysics, biological processes are studied using physical methods. The method is applied to biological organisms and systems to gain insight into how they work
### 10. Geophysics
Geophysics is a branch of physics that studies the earth including the atmosphere that surrounds it, such as volcanic eruptions, earthquakes, landslides, weather changes, and others.
Geophysics is the knowledge of the physical properties of the earth and its application includes methods: seismic, gravitational, magnetic, electrical and radioactive.
Geophysics is a combination of physics and geography.
### 11. Environmental Physics
Environmental physics is a branch of physics that studies everything about the earth and air, as well as those related to the environment. Environmental physics is further divided into several branches, namely soil physics (surface and deep), water physics, air physics, ocean physics, earthquake physics, cloud physics, atmospheric physics, and weather physics.
### 12. Astrophysics
Astronomy is a branch of physics that studies space and everything in it. In other words, Astrophysics is the study of celestial bodies such as the moon, planets, stars, including the sun. Astrophysics is a combination of physics and astronomy.
### 13. Economic Physics
Economic physics is a branch of physics that studies economics using physical methods. Economic physics is a combination of physics and economics. The physics framework is used to solve problems that exist in economics.
Radiation Physics is the science of physics that studies any process by which energy moves through a medium or through space, and is eventually absorbed by another object.
### 15. Atmospheric Physics
Atmospheric physics is a branch of physics to study the atmosphere. Atmospheric physicists attempt to model Earth’s atmosphere and the atmospheres of other planets using fluid flow equations, chemical models, radiation budgets, and energy transfer processes in the atmosphere (and how these bind to other systems such as the oceans).
### 16. Meteorology
Meteorology or weather science is a branch of atmospheric science that includes atmospheric chemistry and atmospheric physics, with the main focus being on the science of weather forecasting.
### 17. Computational Physics
Computational physics is the study of numerical implementation of algorithms to solve problems in areas of physics where quantitative theory already exists. In history, computational physics was the first application of modern computer science in the field of science, and is now a subsection of computational science.
### 18. Physics of solids
Solid matter physics is the study of the physical properties and behavior of substances in the solid phase
### 19. Physics of condensed matter
Condensed matter physics is a field of physics that deals with the exploration and manipulation of phenomena and physical properties of solid and liquid matter based on the principles of quantum mechanics and statistical physics.
### 20. Nuclear Physics
Nuclear physics is the study of atomic nuclei, and the changes in atomic nuclei. Particle physics is the branch of physics that studies the basic particles that make up matter and radiation, and the interactions between them. Particle physics is also called high energy physics.
## Relationship of Physics with Other Sciences
Physics is a very fundamental science among all the Natural Sciences. For example, in Chemistry, the arrangement of molecules and practical ways of changing certain molecules into others using the method of applying the laws of Physics. Biology must also rely heavily on physics and chemistry to explain the processes that take place in living things.
The purpose of studying Physics is so that we can know the basic parts of objects and understand the interactions between objects, and be able to explain natural phenomena that occur. Although physics is divided into several fields, the laws of physics apply universally. A review of a phenomenon from a particular field of physics will obtain the same results when viewed from another field of physics.
In addition, the basic concepts of physics not only support the development of physics itself, but also support the development of other sciences and technology. Physics supports both pure and applied research. Geologists in their research use gravimetric, acoustic, electrical and mechanical methods. Modern equipment in hospitals applies the principles of physics and astronomers require spectrograph optics and radio engineering.
## Benefits of Studying Physics
It is something that we often hear about the statement of some people or maybe most people think that physics is difficult. They think that physics is a science full of formulas and high-level mathematical calculations. This results in not everyone being able to study physics well. Even physics is often considered as the scourge of learning. This situation is almost the same as the assumption in mathematics.
But despite all the assumptions above, there is something more important for us to think about. That is about what are the advantages of studying physics, and what are the disadvantages if we do not want to study physics. In fact, whether we realize it or not, essentially every human being needs knowledge and keeps up with technological developments in order to live this life in harmony. Where the development of technology is certainly an implication of the physical sciences that have been studied by experts who are experts in their fields.
Studying physics has many benefits. Starting from the beginning of studying physics, physics has been proven to be able to help facilitate humans in carrying out activities of daily life. There are several benefits of studying physics, including:
• Physics plays a big role in technological discoveries.
• Through physics can reveal the secrets of nature.
• Physics is at the forefront of technological developments.
• Physics as a basic science that has a role in the development of other sciences.
• Physics trains us to think logically and systematically. | 2,392 | 12,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-49 | longest | en | 0.942654 |
https://www.coursya.com/product/introduction-to-financial-engineering-and-risk-management/ | 1,721,077,006,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00030.warc.gz | 627,868,459 | 54,998 | # Introduction to Financial Engineering and Risk Management
### Description
Introduction to Financial Engineering and Risk Management course belongs to the Financial Engineering and Risk Management Specialization and it provides a fundamental introduction to fixed income securities, derivatives and the respective pricing models. The first module gives an overview of the prerequisite concepts and rules in probability and optimization. This will prepare learners with the mathematical fundamentals for the course. The second module includes concepts around fixed income securities and their derivative instruments. We will introduce present value (PV) computation on fixed income securities in an arbitrage free setting, followed by a brief discussion on term structure of interest rates. In the third module, learners will engage with swaps and options, and price them using the 1-period Binomial Model. The final module focuses on option pricing in a multi-period setting, using the Binomial and the Black-Scholes Models. Subsequently, the multi-period Binomial Model will be illustrated using American Options, Futures, Forwards and assets with dividends.
### What you will learn
Course Overview
Welcome to Financial Engineering and Risk Management
Pre-Requisite Materials
Welcome to Week 2! This week, we will cover mathematical foundations that are necessary for the study of future modules. In a nutshell, we will introduce probabilities and optimization. The theory of probability is the mathematical language to characterize uncertainties, e.g. how to describe the chances that the price of a particular stock will go up tomorrow. To make things precise, we need probabilities. Optimization is a set of toolkits that allow us to search for optimal solutions. For example, given a budget constraint, how do we maximize the profit? We need mathematical optimization. Financial engineers apply probabilistic models to capture the regularities of financial products, and apply optimization techniques to optimize their strategies. These mathematical toolkits will serve as a cornerstone for your financial engineering career.
Introduction to Basic Fixed Income Securities
Welcome to Week 3! This week, we officially embark on the journey of financial engineering and risk management. We will start with the fundamentals of financial engineering, i.e. the principles of pricing. In financial markets, given a financial product, how do we calculate its prices? These pricing principles will serve as the cornerstone of our future modules. We will also cover the basics of fixed income instruments, which serve as the building blocks of financial markets. If you get stuck on the quizzes, you should post on the Discussions to ask for help. (And if you finish early, I hope you’ll go there to help your fellow classmates as well.)
Introduction to Derivative Securities
Welcome to Week 4! This week, we will cover a new family of financial products: derivative securities. Derivative securities, as the name suggests, are financial products that derive their value from some underlying assets, such as interest rates or stocks. The prosperity of modern financial markets is due in large part to the wide variety of derivative securities on the markets such as forwards, futures, swaps, and options as we will introduce in this module. We will also introduce the 1-period binomial model, a simplified framework that allows us to calculate the prices of derivative securities. Despite its simplicity, 1-period binomial model is the building block of more powerful pricing models as we will find out in future modules. As always, if you get stuck on the quizzes, you should post on the Discussions to ask for help. (And if you finish early, I hope you’ll go there to help your fellow classmates as well.) | 701 | 3,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.856373 |
http://openstudy.com/updates/4f760f09e4b0ddcbb89d411a | 1,512,965,919,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00717.warc.gz | 212,395,529 | 9,993 | • anonymous
Let f(x)=k/(x^3+x) if x>=1, and f(x)=0 otherwise where k is const. (a) For what value of k is f a probability density function? (b) Find the mean of this density function. I'm having a hard time finding my limits of integration, I think it's 1 to infinity. But if that's the case I end up with: k[ln(x) - (1/2)ln(x^2+1)] from 1 to infinity which looks like k(infinity - infinity) - k[0-(1/2)ln(2)] and... i'm lost.
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
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Not the answer you are looking for? Search for more explanations. | 383 | 1,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-51 | latest | en | 0.454794 |
http://earthquake.usgs.gov/learn/kids/abc/parents/pp.html | 1,394,491,791,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011038777/warc/CC-MAIN-20140305091718-00064-ip-10-183-142-35.ec2.internal.warc.gz | 53,604,959 | 2,840 | anic Panic is one of the most significant causes of injury in earthquakes. Trying to run during the violent shaking of a large earthquake often leads to sprained ankles and broken bones. Learn now what to do during an earthquake, practice it now before you need it and you have much better chance of coming through the next earthquake unharmed. For parents, panic is particularly dangerous because your children will follow your lead. If you panic, so will they. Home earthquake drills can help you all stay calm in the next big earthquake. Find sturdy tables in your house and teach your children to go under them if they feel an earthquake. Make your children's bed safe (nothing to fall on it, consider bookcases, pictures and windows) and teach them to stay in bed if the earthquake hits while they are asleep. Once a month, at different times of the day, pretend an earthquake has started and practice duck and cover. You will all feel safer the next time we have a real earthquake.
Prepared Every earthquake shows us that preparing before the event can minimize your losses during the event. Several preparedness tips can be found through this Parent's Guide, but it is not complete. Preparing for the earthquake should include eliminating hazards in your house (bolting the house to its foundation, securing water heaters and gas dryers, using florist's putty to secure bric-a-brac and Velcro for computers and televisions), conducting home drills (see Panic) and maintaining supplies (see Electricity, also include water, medicine, tools). The American Red Cross and California Governor's Office of Emergency Services publish excellent guides to home preparedness.
P-waves Earthquakes produce three general types of waves (see Waves) to radiate energy. Two are body waves, which means that they travel through the body of the Earth and the other is surface waves, which means that they travel along the surface of the Earth. The two body waves are called P waves (for Primary) and S waves (for Secondary waves). P-waves are compressional waves while S waves are shear waves. Shear waves cannot travel through a fluid so P-waves are the only ones that travel through the Earth's core.
P-waves travel faster, but S waves are usually 2-3 times larger than the P wave. This leads to the characteristic shape of an earthquake on a seismogram with a small P wave followed by a larger S wave. Because the P wave is traveling faster, the time between the P and S wave increases away from the earthquake.
In fact, just like the time between seeing lightning and hearing thunder can be used to estimate the distance to the lightning, the time between the P and S wave can tell you how far away the earthquake is. Local rock type and the depth of the earthquake cause slight variations, but the number of seconds between the P and S wave times 5 is approximately the distance in miles to the earthquake. (Remember that some of that distance may be down into the Earth.)
This relationship between P and S waves also affects some of our perceptions of earthquakes. For instance, people will often report a sharp sound before the earthquake or that they "heard" the earthquake coming. They are not hearing something before the earthquake but are rather hearing the P wave at the beginning and then feeling the S wave. If they had counted the time between the two, they could have figured out how far away the earthquake was! The smaller P wave is also responsible for many of the reports that animals "knew" an earthquake was coming. With their better hearing, animals are more likely to feel and hear the P wave. The human thinks the animal knew before the earthquake happened when actually the animal felt the beginning of the earthquake that the less sensitive human missed.
Children's Pictures: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Parent's Guide: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z | 820 | 3,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-10 | longest | en | 0.961187 |
https://math.stackexchange.com/questions/1602082/how-to-derive-that-for-every-real-y-0-for-every-positive-real-z-neq-1-t/1604051 | 1,566,633,258,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00459.warc.gz | 539,173,680 | 32,493 | # How to derive that for every real $y > 0$, for every positive real $z \neq 1$, there is a $x \in \mathbb{R}$ such that $y=z^x$.
I am not sure on how to derive the following statement concerning the reals (that I think should be true).
For every real $y > 0$, for every positive real $z \neq 1$, there is a $x \in \mathbb{R}$ such that $y=z^x$.
QUESTIONS:
• Is this part of the definition of exponentiation, or it is a result that can be proven?
• If it can be proven, how (maybe by using the Archimedean Property)?
Thank you for your time.
PS: Few days ago I asked almost the same question, but a user pointed out (rightly) that the question I really wanted to ask is the one here.
• Actually, as always, there were typos in my question, and you pointed them. I really had in mind the exponential function, hence what you wrote. Hence, I will correct it, because I still don't see how to obtain the correct statement. – Kolmin Jan 6 '16 at 15:22
• So you're just asking why the exponential functions $f_z : \mathbf{R}^+ \to \mathbf{R}^+$, given by $f_z (x) = z^x$ is surjective. You can show more by producing an inverse, namely $g_z (x) = \log _ z (x)$. – Future Jan 6 '16 at 15:28
• The $x$ you want can be expressed as $\frac{\ln y}{\ln z}$. Exactly how one proves the result depends on the details of how one develops the theory of the exponential function and logarithm. – André Nicolas Jan 6 '16 at 15:33
• Thanks a lot both for your feedbacks! – Kolmin Jan 6 '16 at 15:39
• @T.S.L Shouldn't it be $f_z : \mathbb{R} \to \mathbb{R}*$? – Kolmin Jan 6 '16 at 15:40
Since the exponential function $\exp(t) = e^t$ strictly increases on $\mathbb{R}$, (Check this in your analysis textbook) its inverse, the logarithmic function $\log t$ on $\mathbb{R}^+$ is well-defined. If we take $x = \log y / \log z$, $z^x = e^{\log y} = y$ clearly.
You can use the fact that $\mathbb{R}$ has the least upper bound property. Let $E$ be the set of all $t$ such that $t^n \lneq x$, then if we can show that $E$ is nonempty and bounded above, we are guaranteed the existence of $y =sup$ $E$ in $\mathbb{R}$. From there, if we assume that $y^n \lneq x$ or $y^n \gneq x$ we should be able to obtain a contradiction (or else we would have produced a "gap" in the real line). The uniqueness part of the proof is much easier since if a number had two distinct real nth roots, then one would necessarily be larger than the other.
Also this is not a part of the definition of exponentiation. Note that we can take exponents in $\mathbb{Q}$ but the statement you asked about no longer holds. So this is indeed something that must be proven.
$$y = a^{\log_a(y)} \to z=a, x=\log_a (y)$$ | 800 | 2,672 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-35 | latest | en | 0.89413 |
https://stats.stackexchange.com/questions/435821/need-intercept-for-plm-fixed-effects-in-r?noredirect=1 | 1,701,379,310,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00500.warc.gz | 625,124,026 | 41,529 | # Need intercept for plm (Fixed effects) in R
I have a question of fixed effects in R. So I am trying to use plm function to find fixed effects like following:
> plm(PM25 ~ policy + 1, data=subset(part2, Delhi == 1), model="within",
> index=c("station_id", "date")) %>% summary()
The results I get:
> Oneway (individual) effect Within Model
>
> Call: plm(formula = PM25 ~ policy + 1, data = subset(part2, Delhi ==
> 1), model = "within", index = c("station_id", "date"))
>
> Unbalanced Panel: n = 7, T = 73-159, N = 992
>
> Residuals:
> Min. 1st Qu. Median 3rd Qu. Max.
> -193.7049 -54.6776 -9.6843 54.7431 318.4094
>
> Coefficients:
> Estimate Std. Error t-value Pr(>|t|)
> policy 76.8167 9.9787 7.6981 3.354e-14 ***
> ---
> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> Total Sum of Squares: 6846900
> Residual Sum of Squares: 6458000
> R-Squared: 0.056803
> F-statistic: 59.2603 on 1 and 984 DF, p-value: 3.3535e-14
I am wondering how can I find the intercept?
The within model you specify amounts to fitting a separate intercept for each unit in your panel data set, as in $$y_{it}=a_i+\beta x_{it}+u_{it}$$ See, e.g., these posts for some context:
Difference between fixed effects dummies and fixed effects estimator?
How exactly does a "random effects model" in econometrics relate to mixed models outside of econometrics?
Hence, you cannot additionally fit an overall intercept $$a$$, as you would then have perfect collinearity between the $$a_i$$ and $$a$$, as the sum of the $$a_i$$ is identically one.
You can retrieve these estimates as follows:
data("Grunfeld", package = "plm")
wi <- plm(inv ~ value + capital,
data = Grunfeld, model = "within", effect = "twoways")
fixef(wi)
One might of course leave out one of the intercepts. Something related to what you seek seems to be achieved by this command: https://rdrr.io/rforge/plm/man/within_intercept.html | 620 | 1,965 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | longest | en | 0.840405 |
https://www.plati.ru/itm/dhs-13-1-option-20-decisions-ryabushko-ap/2222597?lang=en-US | 1,529,942,080,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00512.warc.gz | 875,939,484 | 11,753 | # DHS 13.1 - Option 20. Decisions Ryabushko AP
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# Description
1. Provide a double integral as an iterated integral with the outer integration over x and external integration with respect to y, if D is given the above lines.
1.20. D: y ≤ 0, x2 = -y, x = √1 - y2
2. Calculate the double integral over the region D, limited to lines.
D: y = x2 - 1, y = 3
3. Calculate the double integral using polar coordinates.
4. Calculate the area of a plane region D, bounded set tench.
4.20. D: y = 4 - x2, y = x2 - 2x
5. Use double integrals in polar coordinates to calculate the flat area of the figure bounded by the lines indicated.
5.20. (X2 + y2) 3 = a2x2y2
6. Calculate the volume of the body bounded by a given surface.
6.20. z = 2x2 + y2, x + y = 1, x ≥ 0, y ≥ 0, z ≥ 0 | 296 | 863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-26 | latest | en | 0.698422 |
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# m03 #36
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m03 #36 [#permalink]
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26 Mar 2011, 21:37
Why the answer is A? If something travels at the same speed and at the same distance twice is it not the same number of hours? Thanks!
A swimmer makes a round trip up and down the river which takes her X hours. If the next day she swims the same distance with the same speed in still water, which takes her Y hours, which of the following statements is true?
• X>Y
• X<Y
• X=Y
• X=1/2*Y
• none of the above
Pick numbers and then check them against the options. Take 12 km as the distance traveled up/down the river, and assume the swimmer's speed to be 4 km/h; the current being 2 km/h, which means 6 km/h down the river and 2 km/h up the river. Going upriver takes 2 hours, return journey takes 6, thus a total of 8 hours. In still water, 24 km requires 6 hours. Thus and x=8 and y=6.
1. or True.
2. or False.
3. or False.
4. or False.
5. None of the above. False because A is true.
The correct answer is A.
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Re: m03 #36 [#permalink]
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26 Mar 2011, 23:06
Hello mate !
Read the stem as you read a CR. And read it critically - you may not have to do any of that math.
A swimmer makes a round trip up and down the river which takes her X hours. If the next day she swims the same distance with the same speed in still water, which takes her Y hours, which of the following statements is true?
See the italics. First italic implies she had to face resistance of the river one way. The second time she swam in still water, she did not face any resistance. So the text is alluding that Y is less than X.
gsagula wrote:
Why the answer is A? If something travels at the same speed and at the same distance twice is it not the same number of hours? Thanks!
No the concept of average speed tells me that the even when she traveled the same distance in both cases, her average speed was low the first time and it was high the second time reducing the time marginally. Hope that helps !
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Re: m03 #36 [#permalink]
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27 Mar 2011, 00:36
Thanks a lot Dude!
No the concept of average speed tells me that the even when she traveled the same distance in both cases, her average speed was low the first time and it was high the second time reducing the time marginally. ! Hope that helps
But it just states that the speed is the same in the first and second time. It doesn't mention any variation. It also states that the distance is the same. I can't see any reasonable explanation for variation in time using the premises stated in this question.
The fact that still water has nothing to do with this question is definitely true. It didn't mention if the speed is related to ground or water. I assume that it is ground.
distance/speed = time
Sorry, I didn't get it your explanation.
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Re: m03 #36 [#permalink]
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27 Mar 2011, 01:28
gsagula wrote:
Why the answer is A? If something travels at the same speed and at the same distance twice is it not the same number of hours? Thanks!
A swimmer makes a round trip up and down the river which takes her X hours. If the next day she swims the same distance with the same speed in still water, which takes her Y hours, which of the following statements is true?
• X>Y
• X<Y
• X=Y
• X=1/2*Y
• none of the above
Pick numbers and then check them against the options. Take 12 km as the distance traveled up/down the river, and assume the swimmer's speed to be 4 km/h; the current being 2 km/h, which means 6 km/h down the river and 2 km/h up the river. Going upriver takes 2 hours, return journey takes 6, thus a total of 8 hours. In still water, 24 km requires 6 hours. Thus and x=8 and y=6.
1. or True.
2. or False.
3. or False.
4. or False.
5. None of the above. False because A is true.
The correct answer is A.
Let the speed of the swimmer in still water be $$v_{still}$$
Let the speed of the stream be $$v_{stream}$$
Let the one way distance be $$D$$
Time going upstream
$$T_{upstream} = \frac{D}{v_{still}-v_{stream}}$$
$$T_{downstream} = \frac{D}{v_{still}+v_{stream}}$$
$$Total Time = X = T_{upstream}+T_{downstream} = \frac{D}{v_{still}-v_{stream}}+\frac{D}{v_{still}+v_{stream}}$$
$$X = \frac{D*v_{still}+D*v_{stream}+D*v_{still}-D*v_{stream}}{(v_{still})^2-(v_{stream})^2}$$
$$X = \frac{2D*v_{still}}{(v_{still})^2-(v_{stream})^2}$$
Likewise;
$$Y = \frac{2D*v_{still}}{(v_{still})^2-(v_{stream})^2}$$
But here;
$$v_{stream}=0$$
When $$v_{stream}=0$$, the denominator becomes maximum making the result to be minimum.
Thus, we can say that $$Y$$ will be minimum time taken by the swimmer at $$v_{stream}=0$$.
Hence;
$$X>Y$$
Ans: "A"
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Re: m03 #36 [#permalink]
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27 Mar 2011, 01:49
fluke wrote:
gsagula wrote:
Why the answer is A? If something travels at the same speed and at the same distance twice is it not the same number of hours? Thanks!
A swimmer makes a round trip up and down the river which takes her X hours. If the next day she swims the same distance with the same speed in still water, which takes her Y hours, which of the following statements is true?
• X>Y
• X<Y
• X=Y
• X=1/2*Y
• none of the above
Pick numbers and then check them against the options. Take 12 km as the distance traveled up/down the river, and assume the swimmer's speed to be 4 km/h; the current being 2 km/h, which means 6 km/h down the river and 2 km/h up the river. Going upriver takes 2 hours, return journey takes 6, thus a total of 8 hours. In still water, 24 km requires 6 hours. Thus and x=8 and y=6.
1. or True.
2. or False.
3. or False.
4. or False.
5. None of the above. False because A is true.
The correct answer is A.
Let the speed of the swimmer in still water be $$v_{still}$$
Let the speed of the stream be $$v_{stream}$$
Let the one way distance be $$D$$
Time going upstream
$$T_{upstream} = \frac{D}{v_{still}-v_{stream}}$$
$$T_{downstream} = \frac{D}{v_{still}+v_{stream}}$$
$$Total Time = X = T_{upstream}+T_{downstream} = \frac{D}{v_{still}-v_{stream}}+\frac{D}{v_{still}+v_{stream}}$$
$$X = \frac{D*v_{still}+D*v_{stream}+D*v_{still}-D*v_{stream}}{(v_{still})^2-(v_{stream})^2}$$
$$X = \frac{2D*v_{still}}{(v_{still})^2-(v_{stream})^2}$$
Likewise;
$$Y = \frac{2D*v_{still}}{(v_{still})^2-(v_{stream})^2}$$
But here;
$$v_{stream}=0$$
When $$v_{stream}=0$$, the denominator becomes maximum making the result to be minimum.
Thus, we can say that $$Y$$ will be minimum time taken by the swimmer at $$v_{stream}=0$$.
Hence;
$$X>Y$$
Ans: "A"
Thanks Dude!
What did the question mean was that in the second event she had constant speed? In this case it was inaccurately stated, because I can assume that she had the same speed of the day before.
What the question states is that speed is the same.
We don't have two speeds:
v_{still} v_{stream}
In other words, the question states that both events had the same distance and the same speed.
In this case it doesn't matter where she is swimming.
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Re: m03 #36 [#permalink]
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27 Mar 2011, 01:58
gsagula wrote:
Thanks Dude!
What the question states is that speed is the same.
We don't have two speeds:
v_{still} v_{stream}
In other words, the question states that both events had the same distance and the same speed.
In this case it doesn't matter where she is swimming.
Speed of the swimmer is the same but the speed of the stream vary.
$$v_{still}$$ is the speed of the swimmer, which I agree, is same in both instances.
$$v_{stream}$$ is the speed of the stream(flowing water) is different in both instances (in the first instance it has some speed more than 0 and in second, it is still(not flowing) and is 0)
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Re: m03 #36 [#permalink]
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27 Mar 2011, 02:22
fluke wrote:
gsagula wrote:
Thanks Dude!
What the question states is that speed is the same.
We don't have two speeds:
v_{still} v_{stream}
In other words, the question states that both events had the same distance and the same speed.
In this case it doesn't matter where she is swimming.
Speed of the swimmer is the same but the speed of the stream vary.
$$v_{still}$$ is the speed of the swimmer, which I agree, is same in both instances.
$$v_{stream}$$ is the speed of the stream(flowing water) is different in both instances (in the first instance it has some speed more than 0 and in second, it is still(not flowing) and is 0)
Ow... Thanks again!
So, I believe that this question is flaw because it's not speed, It's relative speed. The question should specify if the speed is in relationship to ground or water (search for Relative Velocity). One more thing, if she has resistance to go up, she will have this energy in her favor to go down, and in this case the average will remain exactly the same.
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27 Mar 2011, 02:26
Thank you very much fluke and gmat1220!
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Re: m03 #36 [#permalink]
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27 Mar 2011, 08:14
gsagula wrote:
Ow... Thanks again!
So, I believe that this question is flaw because it's not speed, It's relative speed. The question should specify if the speed is in relationship to ground or water (search for Relative Velocity). One more thing, if she has resistance to go up, she will have this energy in her favor to go down, and in this case the average will remain exactly the same.
Sorry, this inference is wrong. Lets says you travel from SF to NY with the wind and make 150mph and travel back to SF from NY at 100mph against the wind. The average speed is not (150 + 100)/2 = 125mph. It will be less than 125mph. Similarly, if the swimmer travels 50mph with the river and back 30mph against the river. Her average speed is less than (50+30)/2 i.e. 40mph. Try this when your cruising in your car next time.
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Re: m03 #36 [#permalink]
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27 Mar 2011, 09:38
gmat1220 wrote:
gsagula wrote:
Ow... Thanks again!
So, I believe that this question is flaw because it's not speed, It's relative speed. The question should specify if the speed is in relationship to ground or water (search for Relative Velocity). One more thing, if she has resistance to go up, she will have this energy in her favor to go down, and in this case the average will remain exactly the same.
Sorry, this inference is wrong. Lets says you travel from SF to NY with the wind and make 150mph and travel back to SF from NY at 100mph against the wind. The average speed is not (150 + 100)/2 = 125mph. It will be less than 125mph. Similarly, if the swimmer travels 50mph with the river and back 30mph against the river. Her average speed is less than (50+30)/2 i.e. 40mph. Try this when your cruising in your car next time.
Right, it is Total Distance/Total Time, however the question states that "speed" is the same, and it kills the argument.
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Re: m03 #36 [#permalink] 27 Mar 2011, 09:38
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# m03 #36
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3,748 | 13,112 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-47 | latest | en | 0.899725 |
https://expertmcqs.com/ecat-mathematics-numbers-systems/ | 1,603,954,710,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107903419.77/warc/CC-MAIN-20201029065424-20201029095424-00279.warc.gz | 323,948,106 | 14,450 | # ECAT – Mathematics (Numbers Systems)
The following questions are from the basics of numbers systems, rational and irrational numbers, complex numbers and their properties.
## Mathematics-Number Systems Sample Questions
1. If a = b and b = c, then a = c. This property is called ?
(A) Reflexive
(B) Transitive
(C) Trichotomy
(D) Symmetric
2. If a > b and b > a, then ?
(A) a = b
(B) a ≠ b
(C) cannot be evaluated
(D) impossible
3. π (pi) denotes the ratio of the __________ to the __________ .
(A) area of a circle, circumference of the circle
(B) area of a circle, length of its diameter
(C) circumference of a circle, length of its radius
(D) circumference of a circle, length of its diameter
4. The value of π (pi) is a/an __________ .
(A) rational number
(B) irrational number
(C) integer
(D) complex number
5. If x2 + 1 = 0, then x = ?
(A) 1
(B) i
(C) ± 1
(D) ± i
1. B
2. A
3. D
4. B
5. D
6. The product of complex numbers (4,3) and (5,−6) is ?
(A) (18,3)
(B) (18,−3)
(C) (38,9)
(D) (38,−9)
7. (2,6) ÷ (4,2) = ?
(A) (1,1)
(B) (1,2)
(C) (2,1)
(D) (2,2)
8. (−i)−19 = ?
(A) 1
(B) −1
(C) i
(D) −i
9. The multiplicative identity of a complex number is ?
(A) (0,0)
(B) (1,0)
(C) (0,1)
(D) (1,1)
10. $\sqrt{-9}\times\sqrt{-9}=?$
(A) 9
(B) −9
(C) 9i
(D) −9i | 488 | 1,268 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-45 | latest | en | 0.721128 |
https://discuss.leetcode.com/topic/38118/sharing-my-20ms-c-solution-doing-paint | 1,516,475,818,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00188.warc.gz | 659,942,607 | 10,049 | # Sharing my 20ms C++ solution (doing paint)
• ``````class Solution {
private:
void paint(vector<vector<char>>& grid, vector<vector<bool>>& visited, int i, int j)
{
int m = grid.size();
int n = grid[0].size();
if(grid[i][j]=='1')
{
visited[i][j] = true;
if(i-1>=0&&visited[i-1][j]==false&&grid[i-1][j]=='1')
paint(grid, visited, i-1, j);
if(i+1<m&&visited[i+1][j]==false&&grid[i+1][j]=='1')
paint(grid, visited, i+1, j);
if(j-1>=0&&visited[i][j-1]==false&&grid[i][j-1]=='1')
paint(grid, visited, i, j-1);
if(j+1<n&&visited[i][j+1]==false&&grid[i][j+1]=='1')
paint(grid, visited, i, j+1);
}
}
public:
int numIslands(vector<vector<char>>& grid) {
int m = grid.size();
if(m==0)
return 0;
int n = grid[0].size();
if(n==0)
return 0;
vector<vector<bool>> visited(m, vector<bool>(n, false));
int i, j, count=0;
for(i=0; i<m; i++)
for(j=0; j<n; j++)
{
if(visited[i][j]==false && grid[i][j]=='1')
{
paint(grid, visited, i, j);
count++;
}
}
return count;
}
};``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 368 | 1,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-05 | latest | en | 0.52006 |
http://math.stackexchange.com/questions/379978/find-all-functions-f-that-assign-a-real-number-fx-to-every-real-number-x/380000 | 1,469,308,559,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823670.44/warc/CC-MAIN-20160723071023-00187-ip-10-185-27-174.ec2.internal.warc.gz | 166,856,143 | 18,068 | # Find all functions $f$ that assign a real number $f(x)$ to every real number $x$ . . .
Find all functions $f$ that assign a real number $f(x)$ to every real number $x$ such that $$(x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)$$
I've tried subbing in heaps of values but I keep getting things like $f(0)=f(0)$ and other such useless results.
Any help would be hugely appreciated.
-
Where did this come from? It just feels like constant functions should be the only solutions. – Ross Millikan May 3 '13 at 5:09
No, $f(x) = a x + b$ is always a solution. – Robert Israel May 3 '13 at 5:13
It is from a maths olympiad I sat last year – John Marty May 3 '13 at 5:13
yes, I worked out that ax+c is always a solution for real a and c, but why is it the only one – John Marty May 3 '13 at 5:13
Put $y = 1$
$$(x+1)f(x) + f(1) = (x+1)f(1) + f(x^2)$$
Put $y = 0$
$$xf(x) + f(0) = xf(0) + f(x^2)$$
and subtract latter from former, we get
$$f(x) + f(1) - f(0) = x(f(1) - f(0)) + f(1)$$
and so
$$f(x) = x(f(1) - f(0)) + f(0)$$
Since any $f(x) = ax+b$ is a solution, you are done.
-
Partial solution:
1. Let $y=0, x=c(\neq 0)$: $cf(c)+f(0)=cf(0)+f(c^2)$
2. Let $y=0, x=-c$: $-cf(-c)+f(0)=-cf(0)+f(c^2)$.
Subtracting, we get $c(f(c)+f(-c))=c(2f(0))$, so $f(c)+f(-c)=2f(0)$ for all nonzero $c$.
More partial solution:
Rewrite as $f(x^2)-f(y^2)=(x+y)(f(x)-f(y))$, then divide both sides by $x^2-y^2$ to get $$\frac{f(x^2)-f(y^2)}{x^2-y^2}=\frac{f(x)-f(y)}{x-y}$$
If we knew that $f$ was continuous, then taking limits as $y\rightarrow x$ we find that $f'(x)=f'(x^2)$.
- | 629 | 1,562 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2016-30 | latest | en | 0.846983 |
https://republicofsouthossetia.org/question/which-of-the-following-statements-is-true-about-a-graph-that-is-perfectly-symmetric-such-as-the-16114811-15/ | 1,638,035,790,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00012.warc.gz | 575,755,178 | 13,161 | ## Which of the following statements is true about a graph that is perfectly symmetric, such as the one above? The mean is greater
Question
Which of the following statements is true about a graph that is perfectly symmetric, such as the one above?
The mean is greater than the median.
The mean is equal to the median.
The mean is less than the median.
in progress 0
1 week 2021-11-17T16:40:37+00:00 2 Answers 0 views 0
## Answers ( )
1. Answer:
it is B) The mean is equal to the median.
Step-by-step explanation:
2. Answer:B, A
Step-by-step explanation: | 145 | 565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-49 | latest | en | 0.931699 |
https://discuss.leetcode.com/topic/95755/iterative-solution-using-stack-no-trim | 1,516,667,841,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891546.92/warc/CC-MAIN-20180122232843-20180123012843-00137.warc.gz | 640,538,005 | 9,080 | # Iterative solution using Stack no trim()
• 1 Group every 3 digits and push to stack
2 pop groups from stack with weight = "Billion", "Million", or "Thousand"
3 Within each 3 digits group with weight = "hundred", "tens" and "single"
The less than 20 map is awesome !!!
``````private final String[] LESS_THAN_20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
private final String[] TENS = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
public String numberToWords(int num) {
Stack<int[]> threeDigitStack = new Stack<>();
int i = 0;
int[] threeDigit = new int[3];
String result = "";
if (num == 0)
return "Zero";
if (num < 0) {
result = "Nagetive ";
num = -num;
}
while(num > 0) {
int digit = num % 10;
num = num / 10;
threeDigit[i] = digit;
i++;
if(i == 3) {
threeDigitStack.push(threeDigit.clone());
i = 0;
Arrays.fill(threeDigit,0);
}
}
if(i != 0 && (threeDigit[0] != 0 || threeDigit[1] != 0 || threeDigit[2] != 0) )
threeDigitStack.push(threeDigit.clone());
boolean firstLoop = true;
while(!threeDigitStack.isEmpty()) {
String weight = "";
if(threeDigitStack.size() == 4) weight = " Billion";
else if(threeDigitStack.size() == 3) weight = " Million";
else if(threeDigitStack.size() == 2) weight = " Thousand";
threeDigit = threeDigitStack.pop();
int lessThan20 = threeDigit[1] * 10 + threeDigit[0];
int total = threeDigit[2] * 10 + lessThan20;
if(total != 0 && !firstLoop)
result += " ";
if(threeDigit[2] != 0) {
result += LESS_THAN_20[threeDigit[2]] + " " + "Hundred";
if(lessThan20 != 0)
result += " ";
}
if(lessThan20 != 0) {
if (lessThan20 < 20)
result += LESS_THAN_20[lessThan20];
else {
result += TENS[threeDigit[1]];
if(threeDigit[0] != 0)
result += " " + LESS_THAN_20[threeDigit[0]];
}
}
result += weight;
firstLoop = false;
}
return result.substring(0,result.length());
}``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 665 | 2,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-05 | latest | en | 0.484803 |
https://chathamtownfc.net/how-many-planes-of-symmetry-does-a-cube-have/ | 1,638,692,181,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363149.85/warc/CC-MAIN-20211205065810-20211205095810-00298.warc.gz | 223,859,064 | 6,417 | Click come see complete answer. Likewise, how many planes the symmetry walk a octahedron have?
The nine Planes that Symmetry the the Cube and/or Octahedron It includes two different types of symmetry planes: three of the planes are orthogonal to the 3 4-fold symmetry axes; each together plane is parallel to, and also halfway between, two opposite deals with of the cube; these 3 planes are mutually orthogonal.
You are watching: How many planes of symmetry does a cube have?
Secondly, how numerous planes the symmetry go a tetrahedron have? 6 airplane
Consequently, how many planes that symmetry does a square pyramid have?
4 plane
Is octahedral symmetrical?
In chemistry, octahedral molecular geometry defines the shape of compound with six atoms or groups of atoms or ligands symmetrically arranged roughly a central atom, defining the vertices of an octahedron.
Nouman IdollaProfessional
## Is a cube a continuous polyhedron?
Regular Polyhedron
There space five and only 5 of this figures, additionally called the Platonic Solids: the tetrahedron, cube, octahedron, dodecahedron, and also icosahedron.
Liqun Garcia De ReinaProfessional
## Is a cube symmetrical?
A cube is a 3 dimensional shape, not a two dimensional one. This method that it has actually planes that symmetry instead of lines. A Cube has actually nine planes of symmetry. 3 of these operation perpendicular to the deals with they intersect.
Dilenia PanhuisExplainer
## How many planes of symmetry are there in a cylinder?
They have actually planes. Only 2 dimensional figures can have actually lines the symmetry. A circle has a limitless lines the symmetry, surely a cylinder has actually infinite planes the symmetry.
Beverlee MouginExplainer
## What is an 8 face 3d form called?
In geometry, an octahedron (plural: octahedra) is a polyhedron v eight faces, twelve edges, and also six vertices. The ax is most generally used to describe the consistent octahedron, a Platonic solid composed of eight it is provided triangles, 4 of which satisfy at each vertex.
Chere UlichExplainer
## What is Euler"s formula in geometry?
Euler"s formula, one of two people of two vital mathematical theorems of Leonhard Euler. The very first is a topological invariance (see topology) relating the number of faces, vertices, and also edges of any polyhedron. The is created F + V = E + 2, whereby F is the variety of faces, V the number of vertices, and E the variety of edges.
Orietta KrohnsPundit
## Why room there just 5 Platonic solids?
The Greeks recognized that there space only 5 platonic solids. However why is this so? The an essential observation is that the interior angles the the polygons meeting at a peak of a polyhedron add to much less than 360 degrees.
Romualdo RiberoPundit
## How numerous sides does a icosahedron have?
30
Arundhati PlachnerPundit
## How countless planes of symmetry does a rectangle have?
Answer and Explanation: A rectangle has two lines that symmetry. Picture a continual rectangle: if you wrinkles it in the center of the shape vertically you deserve to see one heat of
Clair CaboPundit
## What is rotational symmetry in math?
Rotational symmetry is when things is rotated around a center point (turned) a number of degrees and also the object show up the same. The bespeak of symmetry is the variety of positions the thing looks the exact same in a 360-degree rotation.
Aridani AmezianPundit
## What execute you typical by aircraft of symmetry?
A plane that symmetry is an imagine plane the bisects a molecule into halves the are mirror pictures of every other. In 1, the upright plane that bisects the methyl group, the carbon atom, and the hydrogen atom bisects the molecule into halves that are mirror images of each other. Therefore, that is a plane of symmetry.
Sion InderwiesTeacher
## What is diagonal aircraft of symmetry?
A plane the cuts with a molecule in a method that photos of all the molecule"s features beyond the plane it seems to be ~ to produce an identical molecule is a mirror plane or a plane that reflection. If together a plane bisects the angle between a pair the rotational axis C2, we have a diagonal mirror plane σd.
Armiche MoratallaTeacher
## How many planes space in a rectangle-shaped prism?
A cube is a special instance of a rectangular prism, which method it has actually two congruent level bases and four level planes or faces. This means there room a complete of 6 level planes in a cube.
Bouzekri KetelhodtSupporter
## What is the facility of symmetry?
Noun. centre the symmetry (plural centres the symmetry) (mathematics) A point, within an item or figure, through which any straight line additionally passes through two point out on the leaf of the figure at the very same distance native the centre yet on the opposite sides.
Maynard FohtSupporter
## What walk it median to be congruent?
Congruent. Angles space congruent as soon as they room the very same size (in degrees or radians). Sides room congruent once they space the exact same length.
Amanda CaesarBeginner
## How many planes space on a cube?
The cube has actually nine the opposite planes. Three planes lie parallel to the side squares and go v the centre (picture). Six planes go through opposite edges and two body diagonals. They division the cube into prisms.
Yaco SpilkyBeginner
## How many planes that symmetry has actually the pyramid?
A rectangular based pyramid has 2 plane of symmetry. A square based pyramid has 4 planes of symmetry.
Shella KonnekerBeginner
## Has at the very least one line of symmetry?
One form that has at the very least one heat of symmetry is a rectangle. A rectangle has two lines the symmetry. Other instances of shapes that have actually at least one line of symmetry include a square, with four lines the symmetry, and also an equilateral triangle, with three lines of symmetry.
See more: How Do You Say Pretty Girl In French Flawlessly, You Are A Pretty Girl
Escolastica CarralonBeginner
## What is the contrary in chemistry?
Symmetry in organic Chemistry. A symmetry facet is a line, a plane or a point in or through an object, around which a rotation or reflection leaves the thing in an orientation indistinguishable from the original. | 1,413 | 6,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-49 | latest | en | 0.925481 |
https://uk.pinterest.com/explore/sample-statistics/ | 1,495,567,898,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607649.26/warc/CC-MAIN-20170523183134-20170523203134-00361.warc.gz | 834,377,673 | 64,067 | The Research Process - Qualitative Pathway - the downloadable worksheet is exceptional. Step by step on the methodology section. http://www.bcps.org/offices/lis/researchcourse/develop_writing_method_qualitative.html
Cross-curricular numeracy display. Children used 'Handa's Surprise' as a stimulus to sample different fruits from the text. They collected data in the form of bar charts to illustrate their preferences.
Air Fares #travel #agencies http://travels.remmont.com/air-fares-travel-agencies/ #airline ticket # About Air Fares This page presents inflation-adjusted and unadjusted average air fares since 1995. Averages are computed using data from the Bureau of Transportation Statistics’ Passenger Origin and Destination (O D) Survey, a 10% sample of all... Read moreThe post Air Fares #travel #agencies appeared first on Travels.
### The Bard by the Numbers
Shakespeare in Numbers//This would be fun to show my students when we read Hamlet later in the semester
Table 1: Symbols of Population Parameter and their Corresponding Sample Statistic
### Sampling (DS4): HSS.ID.A.2
The Mission: Students will explore samples and statistics using real world data. They will take multiple samples from a roughly normally distributed population and observe that the samples statistics are normally distributed with a mean close to the population mean. This mission is aligned with Common Core State Standards: HSS.ID.A.2. The fun stuff: Candies, NBA players and sodas! But especially candies--the kids will take samples of candies in order to build a roughly normal…
sampling statistics Gallery
### NASA Scientist Tells CIOs to Set Realistic Expectations for Big Data
“Any time you compute with data you collected, you are calculating sample statistics,” said Dr. Amy Braverman, principal statistician at the Jet Propulsion Laboratory. “It’s important to acknowledge there is uncertainty in the results”
An elevator has a maximum occupancy of 16 people and a weight limit of 2500 lbs. The average weight of the population is 150 lbs, with a standard deviation of 27 lbs, and the distribution of weight in the population is approximately normal. Suppose that a random sample of size 16 people is chosen. a) Is this a problem about means or proportions? If proportions, do we meet the c) What symbol do we use to represent the sample statistic? Statistics homework help
Pinterest | 507 | 2,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-22 | latest | en | 0.855829 |
http://brainden.com/forum/index.php/topic/15322-hard-to-resist/ | 1,412,056,862,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037662880.10/warc/CC-MAIN-20140930004102-00302-ip-10-234-18-248.ec2.internal.warc.gz | 22,387,942 | 18,593 | ## Welcome to BrainDen.com - Brain Teasers Forum
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# Hard to Resist
6 replies to this topic
### #1 mmiguel
mmiguel
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• Gender:Not Telling
Posted 03 September 2012 - 10:16 AM
A funky looking piece of electrical material has a length of 2 cm on the x-axis, which is the axis of current flow.
Assume a coordinate system where the object is sitting between x = -1 cm and x = 1 cm.
The cross-sectional shape of the object is circular, with radius varying as a function of x like so:
r(x) = 0.04 + 0.4*x^2+x^4
r(x) is in cm
A differential volume of this material has a resistance of R ohms in the the direction of current flow, where R varies with distance from the x-axis. If the distance of the differential element from the x-axis is p, then R(p) = 2*sqrt(p)
R(p) is in ohms
What is the total resistance of the whole object for current flowing along the x-axis?
Edited by mmiguel, 03 September 2012 - 10:20 AM.
• 0
### #2 mmiguel
mmiguel
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• 134 posts
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Posted 03 September 2012 - 11:01 PM
A funky looking piece of electrical material has a length of 2 cm on the x-axis, which is the axis of current flow.
Assume a coordinate system where the object is sitting between x = -1 cm and x = 1 cm.
The cross-sectional shape of the object is circular, with radius varying as a function of x like so:
r(x) = 0.04 + 0.4*x^2+x^4
r(x) is in cm
A differential volume of this material has a resistance of R ohms in the the direction of current flow, where R varies with distance from the x-axis. If the distance of the differential element from the x-axis is p, then R(p) = 2*sqrt(p)
R(p) is in ohms
What is the total resistance of the whole object for current flowing along the x-axis?
Let me know if you want any hints
• 0
### #3 bonanova
bonanova
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Posted 03 September 2012 - 11:36 PM
Spoiler for approach
• 0
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell
### #4 Rob_Gandy
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Posted 04 September 2012 - 10:21 PM
A funky looking piece of electrical material has a length of 2 cm on the x-axis, which is the axis of current flow.
Assume a coordinate system where the object is sitting between x = -1 cm and x = 1 cm.
The cross-sectional shape of the object is circular, with radius varying as a function of x like so:
r(x) = 0.04 + 0.4*x^2+x^4
r(x) is in cm
A differential volume of this material has a resistance of R ohms in the the direction of current flow, where R varies with distance from the x-axis. If the distance of the differential element from the x-axis is p, then R(p) = 2*sqrt(p)
R(p) is in ohms
What is the total resistance of the whole object for current flowing along the x-axis?
Spoiler for approach
Spoiler for So...
• 0
### #5 mmiguel
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Posted 05 September 2012 - 04:47 AM
Spoiler for So...
p is radius from x axis
R(p) is resistance in ohms as a function of radius from x-axis
• 0
### #6 bonanova
bonanova
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Posted 06 September 2012 - 06:56 AM
Wikipedia has a helpful elementary discussion of resistance [unit = Ohm] of an object as it relates to the resistivity [unit = Ohm-cm] of a material. The easy relationship is that the resistance in Ohms of a 1-cm cube of material is equal numerically to its resistivity in Ohm-cm. Resistance = resistivity x length / area in compatible units.
It thus makes sense to speak of a material having a resistivity that is [or, as in the case of the OP, is not] constant. Thus, in the formula R(p) = 2*sqrt(p), R refers to the resistivity of the material. After you've specified the material's shape, i.e. created an object, it's meaningful to speak of resistance.
Again, the distinction is that resistivity is a property of a material, independent of its shape; resistance is a property of an object. Resistance depends on the object material's resistivity and on the object's shape.
Spoiler for So ...
• 0
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell
### #7 mmiguel
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Posted 06 September 2012 - 07:41 AM
Wikipedia has a helpful elementary discussion of resistance [unit = Ohm] of an object as it relates to the resistivity [unit = Ohm-cm] of a material. The easy relationship is that the resistance in Ohms of a 1-cm cube of material is equal numerically to its resistivity in Ohm-cm. Resistance = resistivity x length / area in compatible units.
It thus makes sense to speak of a material having a resistivity that is [or, as in the case of the OP, is not] constant. Thus, in the formula R(p) = 2*sqrt(p), R refers to the resistivity of the material. After you've specified the material's shape, i.e. created an object, it's meaningful to speak of resistance.
Again, the distinction is that resistivity is a property of a material, independent of its shape; resistance is a property of an object. Resistance depends on the object material's resistivity and on the object's shape.
Spoiler for So ...
I was attempting to recreate a problem from an EE final I had 1.5 years ago based on memory, but I realized after I submitted that I had some minor problems in the statement. One which causes the actual calculation to be slightly more difficult than I anticipated (harder integral), and now.... units.
I would argue that the differential volume elements have shape and are objects, but this would not be helpful, because integration involves more than just summing stuff together - it also involves a change in units due to multiplication of differential elements of whatever units belong to the variable of integration. Thus if you intend to use something as part of an integrand, such that after integration it becomes what you originally had, what's in the integrand is actually a density of the original quantity vs. being portions of the original quantity.
Spoiler for
In essence, bonanova is correct.
Edited by mmiguel, 06 September 2012 - 07:48 AM.
• 0
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0 members, 0 guests, 0 anonymous users | 1,806 | 7,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2014-41 | longest | en | 0.919234 |
http://oeis.org/A297322 | 1,563,646,932,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526560.40/warc/CC-MAIN-20190720173623-20190720195623-00060.warc.gz | 118,165,062 | 3,728 | This site is supported by donations to The OEIS Foundation.
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A297322 a(n) = [x^n] Product_{k>=1} (1 + k*x^k)^n. 7
1, 1, 5, 28, 137, 726, 3896, 21071, 115089, 633007, 3500740, 19448573, 108458924, 606787572, 3404112479, 19142919543, 107874784017, 609021410570, 3443952349385, 19503777943838, 110599636109572, 627924447630011, 3568885868192419, 20304321490356084 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS Robert Israel, Table of n, a(n) for n = 0..1000 FORMULA a(n) = A297321(n,n). MAPLE f:= proc(n) local k; coeff(series(mul((1+k*x^k)^n, k=1..n), x, n+1), x, n); end proc: map(f, [\$0..30]); # Robert Israel, Dec 28 2017 MATHEMATICA Table[SeriesCoefficient[Product[(1 + k x^k)^n, {k, 1, n}], {x, 0, n}], {n, 0, 24}] CROSSREFS Main diagonal of A297321. Cf. A297324, A297326, A297329. Sequence in context: A258628 A054148 A272319 * A037591 A270922 A037682 Adjacent sequences: A297319 A297320 A297321 * A297323 A297324 A297325 KEYWORD nonn AUTHOR Ilya Gutkovskiy, Dec 28 2017 STATUS approved
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Last modified July 20 14:21 EDT 2019. Contains 325185 sequences. (Running on oeis4.) | 518 | 1,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-30 | latest | en | 0.530868 |
http://mathforcollege.com/nm/mtl/che/03nle/index.html | 1,493,207,854,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121305.61/warc/CC-MAIN-20170423031201-00644-ip-10-145-167-34.ec2.internal.warc.gz | 237,753,725 | 8,506 | Transforming Numerical Methods Education for the STEM Undergraduate MOBILE | VIDEOS | BLOG | YOUTUBE | TWITTER | COMMENTS | ANALYTICS | ABOUT | CONTACT | COURSE WEBSITES | BOOKS | MATH FOR COLLEGE |
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Pre-Requisites for Learning Quadratic Equations [PDF] [DOC]
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PHYSICAL PROBLEM
The float in a commode submerges to a certain depth in water. Using Newton's third law of motion and Archimedes principle, this problem of finding the depth to which the float submerges results in a nonlinear equation. [PDF] [DOC]
NUMERICAL METHOD
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How does Bisection Method Work? [PDF] [DOC]
Background of Bisection Method [YOUTUBE 9:04]
Algorithm of Bisection Method [YOUTUBE 9:47]
Example of Bisection Method [YOUTUBE 9:53]
A Power Point Presentation on Bisection Method [PDF] [PPT]
Worksheet of Bisection Method [PDF] [MFILE]
Convergence Worksheet of Bisection Method [PDF] [MFILE]
Pitfall: Slow Convergence of Bisection Method Worksheet [PDF] [MFILE]
Test Your Knowledge of Bisection Method [HTML] [FLASH] [PDF] [DOC]
Pre-Requisites for Learning Newton-Raphson Method [PDF] [DOC]
Objectives of Newton-Raphson Method [PDF] [DOC]
How does Newton-Raphson Method Work? [PDF] [DOC]
Derivation of Newton-Raphson Method [YOUTUBE 8:24]
Example for Newton-Raphson Method [YOUTUBE 10:06]
Advantages & Drawbacks for Newton-Raphson Method: Part 1 of 2 [YOUTUBE 7:09]
Advantages & Drawbacks for Newton-Raphson Method: Part 2 of 2 [YOUTUBE 4:43]
Derivation from Taylor Series of Newton-Raphson Method [YOUTUBE 7:56]
Supercomputers have No Divide Unit - A Newton-Raphson Method Approach [YOUTUBE 10:14]
Supercomputers have No Divide Unit - Example [YOUTUBE 5:23]
Finding Square Root of a Number - A Newton-Raphson Method Approach [YOUTUBE 6:34]
Finding Square Root of a Number - Example [YOUTUBE 7:03]
A Power Point Presentation on Newton-Raphson Method [PDF] [PPT]
Worksheet of Newton-Raphson Method [PDF] [MFILE]
Convergence Worksheet of Newton-Raphson Method [PDF] [MFILE]
Pitfall: Division by Zero in Newton-Raphson Method Worksheet [PDF] [MFILE]
Pitfall: Slow Convergence at Inflection Points in Newton-Raphson Method Worksheet [PDF] [MFILE]
Pitfall: Root Jumps over Several Roots in Newton-Raphson Method Worksheet [PDF] [MFILE]
Pitfall: Roots Oscillates Around Local Maxima and Minima in Newton-Raphson Method Worksheet [PDF] [MFILE]
Test Your Knowledge of Newton-Raphson Method [HTML] [FLASH] [PDF] [DOC]
Pre-Requisites for Secant Method [PDF] [DOC]
Objectives of Secant Method [PDF] [DOC]
How does secant Method Work? [PDF] [DOC]
Derivation of Secant Method: Approach 1 of 2 [YOUTUBE 5:58]
Derivation of Secant Method: Approach 2 of 2 [YOUTUBE 6:59]
Algorithm of Secant Method [YOUTUBE 5:16]
Example of Secant Method [YOUTUBE 8:16]
A Power Point Presentation on Secant Method [PDF] [PPT]
Worksheet of Secant Method [PDF] [MFILE]
Convergence Worksheet of Secant Method [PDF] [MFILE]
Pitfall: Division by Zero in Secant Method Worksheet [PDF] [MFILE]
Pitfall: Root Jumps over Several Roots in Secant Method Worksheet [PDF] [MFILE]
Test your knowledge of secant method [HTML] [FLASH] [PDF] [DOC]
ANECDOTES
Quest for Cubic Equation Solution [PDF] [DOC]
Newton [HTML] [PDF] [DOC]
Raphson [HTML] [PDF] [DOC]
Galois [HTML] [PDF] [DOC]
AUDIENCE | AWARDS | PEOPLE | TRACKS | DISSEMINATION | PUBLICATIONS Copyrights: University of South Florida, 4202 E Fowler Ave, Tampa, FL 33620-5350. All Rights Reserved. Questions, suggestions or comments, contact mathforcollege@yahoo.com. This material is based upon work supported by the National Science Foundation under Grant# 0126793, 0341468, 0717624, 0836981, , 0836805. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation. Other sponsors include Matlab, MathCAD, USF, FAMU and MSOE. Based on a work at http://nm.mathforcollege.com. Holistic Numerical Methods licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. ANALYTICS | 1,268 | 4,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-17 | longest | en | 0.722585 |
http://oeis.org/A068551 | 1,575,852,776,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540515344.59/warc/CC-MAIN-20191208230118-20191209014118-00280.warc.gz | 104,024,393 | 5,785 | This site is supported by donations to The OEIS Foundation.
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A068551 a(n) = 4^n - binomial(2*n,n). 4
0, 2, 10, 44, 186, 772, 3172, 12952, 52666, 213524, 863820, 3488872, 14073060, 56708264, 228318856, 918624304, 3693886906, 14846262964, 59644341436, 239532643144, 961665098956, 3859788636664, 15488087080696, 62135313450064 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Number of rooted two-face n-edge maps in the plane (planar with a distinguished outside face). - Valery A. Liskovets, Mar 17 2005 Total number of returns to the x axis in all lattice paths using steps (1,1) and (1,-1) from the origin to (2n,0). Cf. A108747. - Geoffrey Critzer, Jan 30 2012 Total depth of all leaves in all binary trees on 2n+1 nodes. - Marko Riedel, Sep 10 2016 REFERENCES H. W. Gould, Combinatorial Identities, Morgantown, WV, 1972. p. 32. Hojoo Lee, Posting to Number Theory List, Feb 18 2002. V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..175 Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.8. Guo-Niu Han, Enumeration of Standard Puzzles Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy] Marko R. Riedel, Math.Stackexchange.com, Average depth of a leaf in a binary tree V. A. Liskovets and T. R. Walsh, Counting unrooted maps on the plane, Advances in Applied Math., 36, No.4 (2006), 364-387. FORMULA G.f.: 1/(1-4*x)-1/sqrt(1-4*x) = C(x)*2*x/(1-4*x) where C(x) = g.f. for Catalan numbers A000108. a(n) = sum(k>=1, binomial(2*m-2*k, m-k) * binomial(2*k, k) ). a(n+1) = 4*a(n) + 2*C(n) where C(n) = Catalan numbers. Conjecture: n*a(n)+2*(3-4*n)*a(n-1)+8*(2*n-3)*a(n-2)=0. - R. J. Mathar, Apr 01 2012 Recurrence (an alternative): n*a(n) = 2^9*(2*n-9)*a(n-5) + 2^8*(18-5*n)*a(n-4) + 2^6*(10*n-27)*a(n-3) + 2^5*(9-5*n)*a(n-2) + 2*(10*n-9)*a(n-1), n>=5. - Fung Lam, Mar 22 2014 Asymptotics: a(n) ~ 2^(2*n)*(1-1/sqrt(n*Pi)). - Fung Lam, Mar 22 2014 E.g.f.: (exp(2*x) - BesselI(0,2*x))*exp(2*x). - Ilya Gutkovskiy, Sep 10 2016 MAPLE A068551:=n->4^n - binomial(2*n, n): seq(A068551(n), n=0..30); # Wesley Ivan Hurt, Mar 22 2014 MATHEMATICA nn=20; c=(1-(1-4x)^(1/2))/(2x); D[CoefficientList[ Series[ 1/(1-2y x c), {x, 0, nn}], x], y]/.y->1 (* Geoffrey Critzer, Jan 30 2012 *) PROG (PARI) a(n)=if(n<0, 0, 4^n-binomial(2*n, n)) (MAGMA) [4^n - Binomial(2*n, n): n in [0..35]]; // Vincenzo Librandi, Jun 07 2011 (PARI) x='x+O('x^100); concat(0, Vec(1/(1-4*x)-1/sqrt(1-4*x))) \\ Altug Alkan, Dec 29 2015 CROSSREFS a(n) = 2*A000346(n-1) for n>0. a(n) = A045621(2n). Cf. A000984, A005470. Sequence in context: A080069 A243965 A218780 * A099919 A100397 A084059 Adjacent sequences: A068548 A068549 A068550 * A068552 A068553 A068554 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Mar 23 2002 STATUS approved
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Last modified December 8 18:37 EST 2019. Contains 329865 sequences. (Running on oeis4.) | 1,408 | 3,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-51 | latest | en | 0.755715 |
http://oeis.org/A046117 | 1,481,423,091,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00254-ip-10-31-129-80.ec2.internal.warc.gz | 199,176,008 | 3,943 | This site is supported by donations to The OEIS Foundation.
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A046117 Values of p+6 such that p and p+6 are both prime. 55
11, 13, 17, 19, 23, 29, 37, 43, 47, 53, 59, 67, 73, 79, 89, 103, 107, 109, 113, 137, 157, 163, 173, 179, 197, 199, 229, 233, 239, 257, 263, 269, 277, 283, 313, 317, 337, 353, 359, 373, 379, 389, 439, 449, 463, 467, 509, 547, 563, 569, 577, 593, 599, 607, 613 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS T. D. Noe, Table of n, a(n) for n=1..1000 Eric Weisstein's World of Mathematics, Sexy Primes. FORMULA a(n) = A087695(n) + 3. MATHEMATICA q=6; a={}; Do[p1=Prime[n]; p2=p1+q; If[PrimeQ[p2], AppendTo[a, p2]], {n, 7^2}]; a "and/or" Select[Prime[Range[3, 7^2]], PrimeQ[ # ]&&PrimeQ[ #-6]&] (* Vladimir Joseph Stephan Orlovsky, Aug 07 2008 *) PROG (PARI) forprime(p=2, 1e4, if(isprime(p-6), print1(p", "))) \\ Charles R Greathouse IV, Jul 15 2011 CROSSREFS Cf. A023201, A046117. Cf. A098428. Cf. A087695. Sequence in context: A244555 A052293 A038842 * A240900 A091923 A050719 Adjacent sequences: A046114 A046115 A046116 * A046118 A046119 A046120 KEYWORD nonn AUTHOR STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 612 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2016-50 | longest | en | 0.599955 |
https://roughtheory.org/category/family/ | 1,680,117,645,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00133.warc.gz | 545,011,119 | 25,437 | # Rough Theory
Theory In The Rough
## Congratulations to Nate
For those who haven’t seen, Nate’s daughter has arrived!
## Are We Human? Or Are We Dancer?
So I was driving somewhere with my five-year-old son on Saturday night. The car paused at a stoplight long enough for him to look out the window at a large crowd of people, dressed for dancing and waiting for a tram to the city. Most were female, which caused my son to remark: “There are lots of women there.” And then, after a short pause: “And only one human.”
“Only one human?” I asked, guessing what he meant but wanting to make sure.
“Only one human – that guy,” he pointed.
“Only one man, you mean.”
“Only one human,” he insisted.
“No – the women are human too,” I found myself wondering how, exactly, I ended up needing to defend this point.
“No they aren’t!” he said in the tone he usually uses when I’m saying something deliberately absurd.
“Yes they are. ‘Human’ includes both men and women – it’s a bigger term than either.”
“Noooo…” he said, sceptical.
“Yes!” I said, with attempted enthusiasm.
He gave me a “taking it under advisement” look, but seems, on the whole, to remain unconvinced…
## Intersecting Interests
There are certain kinds of family interactions that I sort of wonder: how often does this happen in other households?
My five-year old son has been playing tonight with some whiteboard markers and an eraser I brought home from university. A few minutes ago, he plonked his whiteboard down beside me and asked: “Mummy? Can you please draw me a Venn diagram?”
Uh… sure?
So I drew two circles with some overlapping space. Next question: “Do you like puppies?”
Uh… yes?
“I like puppies too! How do you draw a puppy?”
I drew a puppy – having been directed to a marginal space on the whiteboard, not close to the Venn diagram. My son then painstakingly drew a puppy inside the intersecting space. He proceeded to ask whether I liked other things. If I liked them, and he didn’t, they went into the non-intersecting bit of my circle alone. If he liked them, and I didn’t, they went into the non-intersecting bit of his circle.
We did have a conversation about Venn diagrams a few days ago. I just hadn’t been expecting that the concept would percolate, and come up again in quite this form…
## Talking to Myself
I have been editing the introductory chapter of the thesis (I’ll put this online with the others soon). This morning, editing was taking place with my son bouncing up and down all around me on the couch, chanting various things, clambering up my back, and generally doing the sorts of things kids do when adults are visibly concentrating on other things. At some point, he plonked down beside me to read the text over my shoulder – out loud. This resulted in an extended period of having the text read out to me just as I typed it in, which… had a bit of a surreal effect on the editing process. At some point, he paused, confused, and asked: “Where is the other person?”
“What other person?” I asked, confused myself.
“Who are you talking to? Is it Jessica?” he wanted to know.
I suddenly realised he thought I was chatting with someone else online, and tried to explain, “No no – I’m not chatting with anyone. I’m writing. There’s no one else there.”
He mulled over this for a while, looking increasingly puzzled, and then, in an uncertain, quavering voice, slowly asked, “So… are you? Are you talking to yourself?”
Hmm… Good question…
## Writing from Home
A very small sample of interactions with my son today: Read more of this post
## Up the Water Spout
I’m one of those people who wanders around the house, obsessively turning lights off. My obsession competes with the opposing impulses of someone who, disliking things that go bump in the night, tends to leave them on. The delicate and ever-shifting balance of power in this ongoing conflict is currently being upset by a third party: a huntsman spider, who has chosen to share our toilet with us these past few days. It’s a selective creature: it hasn’t yet introduced itself to me. It’s apparently not quite as large as the creature in this photo – although it certainly looms large enough in the imaginations of members of the household. I somewhat suspect that, even once it is no longer with us, the ghost of its presence will continue to haunt our discussion over whether lights in disused portions of the house are best left on…
## Making It Last
Like all young children, my son enjoys repetitious games. He’ll ask: “Again?” And then: “Again?” On and on until I finally announce a countdown – a number of times we can play the game, before we hit the final time, and then have to stop. Around this time last year, I mentioned on the blog the first moment when it began to occur to him that perhaps there might be some way to evade the dreaded “Last time!” decree. That effort didn’t work, but this initial failure was no reason to stop trying to circumvent the system… The current strategy involves waiting for the “Last time!” – and then making a break for it. So, we might be playing ball, for example, and get to the last toss – and, instead of actually taking his “last time”, he’ll run off with the ball and shut himself in a room.
This doesn’t, however, mean that the adult with whom he’s been playing is free to go off and leave the game unfinished. If my son walks out, and – as normally happens – finds the adult has given up after a few minutes, and wandered off to do something else, this is the occasion for tears and indignation: “I want my last time!!!” So back to the game he and the adult will go, preparing for the final ball toss. But once again he makes a break for it, and ends up hiding with the ball in his room.
The desire, apparently, is that the moment should be frozen – the game must last forever – eternally suspended, hanging incomplete – as long as my son doesn’t effect the “last time” by enacting his side of the event.
## Autodidact
One of the nice things about living and working centrally in Melbourne, is that you rarely really need a car. Read more of this post
## Some Lesser-Known Benefits of Higher Education
I just dropped my son off at his childcare centre, and had a nice conversation with the woman who heads the teaching team in his room. I’m very happy with the centre and the staff – not least because they’ve dealt extremely well my son’s rather… non-institutional personality, allowing him an unusual amount of flexibility to drift around within their schedules and routines. Their tolerance is paired, though, with a fair amount of bemusement, and it’s not unusual for staff to pull me aside to share stories about my son’s strange combination of politeness and intractability (I’ve overheard staff joking with one another, describing the phrase “no thank you” as “classic Lyle”). He seems to be perceived as having a positive temperament, but staff seem genuinely puzzled, given this, by his desire to go off and do his own thing – as though politeness ought to correlate with instant compliance or desire for conformity… Thus is the stuff of parent-teacher conferences made…
This morning, the familiar conversation around these things took an unexpected turn: “So… what’s your son’s sign?”
Thinking I must have misheard: “His… what?”
“His astrological sign?”
“Uh… I have no idea…”
“That’s okay – what’s his birthdate?” I provided this, and then received his sign in return. I tend to respond to this kind of thing with a sort of extreme blankness, which for me signifies that I don’t really want to get into a discussion with someone about what they’ve just said, as I’m concerned that they’d find my reaction offensive, and I don’t think the issue is important enough to justify providing offence. This blank reaction, though, is often interpreted in strange ways by other people. In this case, the interpretation, apparently, was that I was struck speechless by how impressive it should be that they should be able to deduce the sign from the birth date. They blushed, and then tried to reassure, “I know – don’t worry – I can only do this because I studied it at university. Helps me with understanding the kids’ personalities.” I’m not sure I find this reassuring…
(Just a side point, from an immigrant’s perspective: astrology and other forms of new age spirituality or practice (often in instrumentalised form, as practice of manipulation or at least prediction of external events) come up startlingly often, in my experience, in professional settings in Melbourne. Every workplace I’ve been in here – the university is no exception – has quite casual, apparently sincere, discussion around new age themes, often by people who are quite scathing in their opinions of mainstream religion. And I’m not just talking about watercooler discussion or chats over coffee – I’m talking about discussion introduced into staff meetings or other formal contexts. Not that everyone or even the majority of people in a workplace participate – but there is no visible public disapprobation to airing these perspectives in a professional setting. I don’t know that I have a question here – more a sort of expression of… anthropological curiosity: what gives? What’s with the strange combination of reflexive scepticism toward older, established faiths, and the receptivity to demonstrably rather recent new age beliefs? Or have I just had profoundly atypical experiences, leading to a kind of strange new age bias in my selection of workplaces?)
## When I Upgrade, I Want to Be…
I’ve been doing a lot of writing recently, mostly on a laptop that I cart around and perch precariously on my knees while I sit in various ergonomically-dubious positions. Today, my son walked up, wanting to sit on my lap. He expressed this by saying: “Could I be a laptop now, please?”
ouch, ouch, ouch… | 2,172 | 9,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.974815 |
https://de.maplesoft.com/support/help/maple/view.aspx?path=DifferentialGeometry%2FGroupActions%2FLiesThirdTheorem | 1,656,808,326,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00285.warc.gz | 249,817,791 | 42,315 | LiesThirdTheorem - Maple Help
GroupActions[LiesThirdTheorem] - find a Lie algebra of pointwise independent vector fields with prescribed structure equations (solvable algebras only)
Calling Sequences
LiesThirdTheorem(Alg, M, option)
LiesThirdTheorem(A, M)
Parameters
Alg - a Maple name or string, the name of an initialized Lie algebra $\mathrm{𝔤}$
M - a Maple name or string, the name of an initialized manifold with the same dimension as that of $\mathrm{𝔤}$
option - with output = "forms" the dual 1-forms (Maurer-Cartan forms) are returned
A - a list of square matrices, defining a matrix Lie algebra
Description
• Let g be an $n-$dimensional Lie algebra with structure constants $C$. Then Lie's Third Theorem (see, for example, Flanders, page 108) asserts that there is, at least locally, a Lie algebra of n pointwise independent vector fields $\mathrm{Γ}$ on an $n$-dimensional manifold with structure constants $C$.
• The command LiesThirdTheorem(Alg, M) produces a globally defined Lie algebra of vector fields $\mathrm{Γ}$ in the special case that is solvable. More general cases will be handled in subsequent versions of the DifferentialGeometry package.
• The command LiesThirdTheorem(A, M) produces a globally defined matrix of 1-forms (Maurer-Cartan forms) in the special case that the list of matrices A defines a solvable Lie algebra.
• The command LiesThirdTheorem is part of the DifferentialGeometry:-GroupActions package. It can be used in the form LiesThirdTheorem(...) only after executing the commands with(DifferentialGeometry) and with(GroupActions), but can always be used by executing DifferentialGeometry:-GroupActions:-LiesThirdTheorem(...).
Examples
> $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$$\mathrm{with}\left(\mathrm{LieAlgebras}\right):$$\mathrm{with}\left(\mathrm{GroupActions}\right):$$\mathrm{with}\left(\mathrm{Library}\right):$
Example 1.
We obtain a Lie algebra from the DifferentialGeometry library using the Retrieve command and initialize it.
> $L≔\mathrm{Retrieve}\left("Winternitz",1,\left[4,4\right],\mathrm{Alg1}\right)$
${L}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{+}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}{+}{\mathrm{e2}}\right]$ (2.1)
> $\mathrm{DGsetup}\left(L\right):$
We define a manifold M of dimension 4 (the same dimension as the Lie algebra).
Alg1 > $\mathrm{DGsetup}\left(\left[x,y,z,w\right],\mathrm{M1}\right)$
${\mathrm{frame name: M1}}$ (2.2)
M1 > $\mathrm{Γ1}≔\mathrm{LiesThirdTheorem}\left(\mathrm{Alg1},\mathrm{M1}\right)$
${\mathrm{Γ1}}{:=}\left[{\mathrm{D_x}}{,}{\mathrm{D_y}}{,}{\mathrm{D_z}}{,}\left({x}{+}{y}\right){}{\mathrm{D_x}}{+}\left({y}{+}{z}\right){}{\mathrm{D_y}}{+}{z}{}{\mathrm{D_z}}{+}{\mathrm{D_w}}\right]$ (2.3)
M1 > $\mathrm{Ω1}≔\mathrm{LiesThirdTheorem}\left(\mathrm{Alg1},\mathrm{M1},\mathrm{output}="forms"\right)$
${\mathrm{Ω1}}{:=}\left[{\mathrm{dx}}{-}\left({x}{+}{y}\right){}{\mathrm{dw}}{,}{\mathrm{dy}}{-}\left({y}{+}{z}\right){}{\mathrm{dw}}{,}{-}{\mathrm{dw}}{}{z}{+}{\mathrm{dz}}{,}{\mathrm{dw}}\right]$ (2.4)
We calculate the structure equations for the Lie algebra of vector fields Gamma1 and check that these structure equations coincide with those for Alg1.
M1 > $\mathrm{LieAlgebraData}\left(\mathrm{Γ1},\mathrm{Alg1a}\right)$
$\left[\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e2}}{+}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}{+}{\mathrm{e2}}\right]$ (2.5)
Example 2.
We re-work the previous example in a more complicated basis. In this basis the adjoint representation is not upper triangular, in which case LiesThirdTheorem first calls the program SolvableRepresentation to find a basis for the algebra in which the adjoint representation is upper triangular. (Remark: It is almost always useful, when working with solvable algebras, to transform to a basis where the adjoint representation is upper triangular.)
M1 > $\mathrm{L2}≔\mathrm{LieAlgebraData}\left(\left[\mathrm{e4},\mathrm{e2}-\mathrm{e4},\mathrm{e3},\mathrm{e1}+\mathrm{e3}\right],\mathrm{Alg2}\right)$
${\mathrm{L2}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e2}}\right]{=}{-}{\mathrm{e4}}{+}{\mathrm{e3}}{-}{\mathrm{e2}}{-}{\mathrm{e1}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e3}}\right]{=}{-}{\mathrm{e3}}{-}{\mathrm{e2}}{-}{\mathrm{e1}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{-}{\mathrm{e4}}{-}{\mathrm{e2}}{-}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e3}}{+}{\mathrm{e2}}{+}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e4}}{+}{\mathrm{e2}}{+}{\mathrm{e1}}\right]$ (2.6)
Alg2 > $\mathrm{DGsetup}\left(\mathrm{L2}\right)$
${\mathrm{Lie algebra: Alg2}}$ (2.7)
Alg2 > $\mathrm{Γ2}≔\mathrm{LiesThirdTheorem}\left(\mathrm{Alg2},\mathrm{M1}\right)$
${\mathrm{Γ2}}{:=}\left[\left({x}{-}{y}\right){}{\mathrm{D_x}}{+}\left({y}{+}{z}\right){}{\mathrm{D_y}}{+}{z}{}{\mathrm{D_z}}{+}{\mathrm{D_w}}{,}{-}\left({x}{-}{y}\right){}{\mathrm{D_x}}{-}\left({-}{1}{+}{y}{+}{z}\right){}{\mathrm{D_y}}{-}{z}{}{\mathrm{D_z}}{-}{\mathrm{D_w}}{,}{\mathrm{D_z}}{,}{-}{\mathrm{D_x}}{+}{\mathrm{D_z}}\right]$ (2.8)
M1 > $\mathrm{LieAlgebraData}\left(\mathrm{Γ2},\mathrm{Alg2a}\right)$
$\left[\left[{\mathrm{e1}}{,}{\mathrm{e2}}\right]{=}{-}{\mathrm{e4}}{+}{\mathrm{e3}}{-}{\mathrm{e2}}{-}{\mathrm{e1}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e3}}\right]{=}{-}{\mathrm{e3}}{-}{\mathrm{e2}}{-}{\mathrm{e1}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{-}{\mathrm{e4}}{-}{\mathrm{e2}}{-}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e3}}{+}{\mathrm{e2}}{+}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e4}}{+}{\mathrm{e2}}{+}{\mathrm{e1}}\right]$ (2.9)
Example 3.
Here is an example where one of the adjoint matrices has complex eigenvalues. The Lie algebra contains parameters $p$ and $b$.
M1 > $\mathrm{L3}≔\mathrm{Retrieve}\left("Winternitz",1,\left[5,25\right],\mathrm{Alg3}\right)$
${\mathrm{L3}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{2}{}{\mathrm{_p}}{}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{\mathrm{_p}}{}{\mathrm{e2}}{+}{\mathrm{e3}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{\mathrm{_p}}{}{\mathrm{e3}}{-}{\mathrm{e2}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e5}}\right]{=}{\mathrm{_b}}{}{\mathrm{e4}}\right]$ (2.10)
M1 > $\mathrm{DGsetup}\left(\mathrm{L3}\right):$
Alg3 > $\mathrm{Adjoint}\left(\mathrm{e5}\right)$
$\left[\begin{array}{ccccc}{-}{2}{}{\mathrm{_p}}& {0}& {0}& {0}& {0}\\ {0}& {-}{\mathrm{_p}}& {1}& {0}& {0}\\ {0}& {-}{1}& {-}{\mathrm{_p}}& {0}& {0}\\ {0}& {0}& {0}& {-}{\mathrm{_b}}& {0}\\ {0}& {0}& {0}& {0}& {0}\end{array}\right]$ (2.11)
Alg3 > $\mathrm{DGsetup}\left(\left[x,y,z,u,v\right],\mathrm{M3}\right)$
${\mathrm{frame name: M3}}$ (2.12)
M3 > $\mathrm{Γ3}≔\mathrm{LiesThirdTheorem}\left(\mathrm{Alg3},\mathrm{M3}\right)$
${\mathrm{Γ3}}{:=}\left[{\mathrm{D_x}}{,}{-}{\mathrm{D_x}}{}{z}{+}{\mathrm{D_y}}{,}{\mathrm{D_z}}{,}{\mathrm{D_u}}{,}\left(\frac{{1}}{{2}}{}{{z}}^{{2}}{-}\frac{{1}}{{2}}{}{{y}}^{{2}}{+}{2}{}{\mathrm{_p}}{}{x}\right){}{\mathrm{D_x}}{+}\left({\mathrm{_p}}{}{y}{-}{z}\right){}{\mathrm{D_y}}{+}\left({\mathrm{_p}}{}{z}{+}{y}\right){}{\mathrm{D_z}}{+}{\mathrm{_b}}{}{u}{}{\mathrm{D_u}}{+}{\mathrm{D_v}}\right]$ (2.13)
M3 > $\mathrm{LieAlgebraData}\left(\mathrm{Γ3},\mathrm{Alg3a}\right)$
$\left[\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{2}{}{\mathrm{_p}}{}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{\mathrm{_p}}{}{\mathrm{e2}}{+}{\mathrm{e3}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{\mathrm{_p}}{}{\mathrm{e3}}{-}{\mathrm{e2}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e5}}\right]{=}{\mathrm{_b}}{}{\mathrm{e4}}\right]$ (2.14)
Example 4.
We calculate the Maurer-Cartan matrix of 1-forms for a solvable matrix algebra, namely the matrices defining the adjoint representation for Alg1 from Example 1.
> $A≔\mathrm{Adjoint}\left(\mathrm{Alg1}\right)$
${A}{:=}\left[\left[\begin{array}{rrrr}{0}& {0}& {0}& {1}\\ {0}& {0}& {0}& {0}\\ {0}& {0}& {0}& {0}\\ {0}& {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrrr}{0}& {0}& {0}& {1}\\ {0}& {0}& {0}& {1}\\ {0}& {0}& {0}& {0}\\ {0}& {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrrr}{0}& {0}& {0}& {0}\\ {0}& {0}& {0}& {1}\\ {0}& {0}& {0}& {1}\\ {0}& {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrrr}{-}{1}& {-}{1}& {0}& {0}\\ {0}& {-}{1}& {-}{1}& {0}\\ {0}& {0}& {-}{1}& {0}\\ {0}& {0}& {0}& {0}\end{array}\right]\right]$ (2.15)
> $\mathrm{MaurerCartan}≔\mathrm{LiesThirdTheorem}\left(A,\mathrm{M1}\right)$
${\mathrm{MaurerCartan}}{:=}\left[\begin{array}{cccc}{-}{\mathrm{dw}}& {-}{\mathrm{dw}}& {0}{}{\mathrm{dx}}& {\mathrm{dx}}{+}{\mathrm{dy}}{-}\left({2}{}{y}{+}{z}{+}{x}\right){}{\mathrm{dw}}\\ {0}{}{\mathrm{dx}}& {-}{\mathrm{dw}}& {-}{\mathrm{dw}}& {\mathrm{dy}}{+}{\mathrm{dz}}{-}\left({2}{}{z}{+}{y}\right){}{\mathrm{dw}}\\ {0}{}{\mathrm{dx}}& {0}{}{\mathrm{dx}}& {-}{\mathrm{dw}}& {-}{\mathrm{dw}}{}{z}{+}{\mathrm{dz}}\\ {0}{}{\mathrm{dx}}& {0}{}{\mathrm{dx}}& {0}{}{\mathrm{dx}}& {0}{}{\mathrm{dx}}\end{array}\right]$ (2.16)
Note that the elements of this matrix coincide with the appropriate linear combinations of the forms in the list ${\mathrm{Ω}}_{1}$ from Example 1.
Alg1 > $\mathrm{MaurerCartan}\left[1,4\right],\mathrm{Ω1}\left[1\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&plus\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{Ω1}\left[2\right]$
${\mathrm{dx}}{+}{\mathrm{dy}}{-}\left({2}{}{y}{+}{z}{+}{x}\right){}{\mathrm{dw}}{,}{\mathrm{dx}}{+}{\mathrm{dy}}{-}\left({2}{}{y}{+}{z}{+}{x}\right){}{\mathrm{dw}}$ (2.17)
Alg1 > $\mathrm{MaurerCartan}\left[2,4\right],\mathrm{Ω1}\left[2\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&plus\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{Ω1}\left[3\right]$
${\mathrm{dy}}{+}{\mathrm{dz}}{-}\left({2}{}{z}{+}{y}\right){}{\mathrm{dw}}{,}{\mathrm{dy}}{+}{\mathrm{dz}}{-}\left({2}{}{z}{+}{y}\right){}{\mathrm{dw}}$ (2.18)
Alg1 > $\mathrm{MaurerCartan}\left[3,4\right],\mathrm{Ω1}\left[3\right]$
${-}{\mathrm{dw}}{}{z}{+}{\mathrm{dz}}{,}{-}{\mathrm{dw}}{}{z}{+}{\mathrm{dz}}$ (2.19)
Alg1 > $\mathrm{MaurerCartan}\left[1,1\right],\left(-1\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&mult\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{Ω1}\left[4\right]$
${-}{\mathrm{dw}}{,}{-}{\mathrm{dw}}$ (2.20) | 4,345 | 10,635 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 57, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-27 | longest | en | 0.709644 |
http://forums.studentdoctor.net/archive/index.php/t-502475.html | 1,369,266,007,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702525329/warc/CC-MAIN-20130516110845-00016-ip-10-60-113-184.ec2.internal.warc.gz | 105,967,846 | 2,432 | Student Doctor Network Forums > Pre-Medical Forums > MCAT Discussions > MCAT Study Question Q&A > Logarithm question... PDA View Full Version : Logarithm question... tncekm03-06-2008, 08:35 PMOkay, so I'm trying to derive the Hendersen Hasselbach equation, and its been a LONG time since I've worked with logarithms, so I want to make sure I'm doing this right. I couldn't find an explicit agreement with this via google, so maybe someone here can verify that I've been doing this right (it gives me the right derivation) Log(A/B) = -Log(B/A) b/c Log(A/B) = LogA - LogB --> -Log(A/B) = LogB - LogA --> -Log(A/B) = Log(B/A) --> Log(A/B) = -Log(B/A), correct? Thx EDIT: Just realized I constructed the equation wrong, so I didn't even need this to get HH eqn. But, I'm still interested in whether or not this is a valid proof :D Chemist015703-06-2008, 09:02 PMYep. Basically, LOG(1/X) = -LOG(X) fileserver03-06-2008, 09:29 PMyes, that is correct. or you can do this: " Log(A/B) = -Log(B/A) b/c " = - log (b/a) = - log (b^1 a^-1) = log (b^1 a^-1)^-1 = log (b^-1 a^1) = log (a/b) iA-MD201303-06-2008, 10:16 PMlooks good :) tncekm03-06-2008, 10:16 PMAwesome, thx everyone. werd03-07-2008, 05:56 PMit's always bugged me that hendersen/hasslebach got this 'famous' equation named after them... all they did was rearrange the equillibrium equation. it wasn't even a complex rearrangement; it can be empirically derived in about 45 seconds using algebra. oh well. tncekm03-08-2008, 12:35 AMit's always bugged me that hendersen/hasslebach got this 'famous' equation named after them... all they did was rearrange the equillibrium equation. it wasn't even a complex rearrangement; it can be empirically derived in about 45 seconds using algebra. oh well. Yeah... I agree. If you want to talk about WEAK how about the rest mass energy equation: E=mc² Its only three letters! :D | 577 | 1,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2013-20 | latest | en | 0.899577 |
https://www.ashokcharan.com/Marketing-Analytics/~sd-sales-per-store.php | 1,719,011,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00786.warc.gz | 576,727,018 | 12,569 | # Sales and Distribution — Average Sales per Store
The average sales per store can be calculated by dividing the total sales volume by the number of stores distributing the product. The number of stores distributing the product is determined by multiplying the numeric distribution by the number of stores in the market universe. $$\text{Average Sales per Store}=\frac{\text{Sales Volume}}{\text{Number of stores distributing the product}}$$ $$\text{Number of stores distributing product = Numeric Distribution × # of stores in Universe}$$
For example, if the sales volume is 10,000 kg, the numeric distribution is 50%, and the number of stores in the market breakdown is 800, the average sales per store would be:
$$Average \,Sales = \frac{10,000}{0.5 × 800} = \frac{10,000}{400} = 25 \,kg\,per\,store$$
While this measure is simple and easy to understand, it can be misleading as it does not consider the quality of distribution. To illustrate this point, let us examine the following hypothetical data for Nescafe and Maxwell House.
Nescafe:
• Sold in 2,000 outlets,
• Average sales per store = 200 kg/month
Maxwell House:
• Sold in 100 outlets,
• Average sales per store = 300 kg/month
#### Exhibit 30.6 Average sales per store and rate of sales per store for Nescafe and Maxwell House.
Based on average sales per store alone, it may seem that Maxwell House is selling at a faster pace than Nescafe. Intuitively, this does not sound right. Maxwell House’s distribution is confined to only 100 outlets, suggesting that it is a relatively small brand.
Upon filtering the data and drilling down to the 100 stores handling Maxwell House, we find that Nescafe is actually selling 1,500 kg per month in those stores (Exhibit 30.6), which is far greater than the average sales per store for Maxwell House.
This demonstrates that the average sales per store metric fails to account for the size of stores. This is a significant omission because smaller brands tend to be distributed only in larger stores that carry a wide assortment. As a brand grows and its distribution expands, its average sales per store tends to decrease as it moves into smaller stores.
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It supports a collection of simulation and analysis platforms such as Promotions and Space Planner for optimizing space and promotions, Plannogram for populating shelves and merchandising, a Due To Analysis dashboard that decomposes brand sales into the factors driving sales, and a Promotion Evaluator to evaluate the volume, value and profit impact of promotion plans. | 736 | 3,593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-26 | latest | en | 0.878912 |
http://mathoverflow.net/questions/40866/finding-generalised-lyndon-words | 1,469,598,825,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825366.39/warc/CC-MAIN-20160723071025-00224-ip-10-185-27-174.ec2.internal.warc.gz | 160,998,964 | 14,557 | # Finding generalised Lyndon words
Let $\Sigma = \lbrace a_1, \ldots, a_n, A_1, \ldots A_n \rbrace$ (where $A_i = a_i^{-1}$) and $\prec$ be a total ordering on $\Sigma$.
Let $\Sigma^*$ be the set of all words (generated by the alphabet $\Sigma$) and $\prec^*$ be the total ordering on $\Sigma^*$ induced by $\prec$ (dictionary / lexicographical ordering).
Let $G$ be a finitely presented group which acts on $\Sigma^*$.
For $w \in \Sigma^*$, let $[w]$ denote the equivalence class of words under $G$ (i.e. $[w] = \text{Orb}_G(w)$).
Let $\text{First}_G(w)$ be the first element of $[w]$ under the total ordering $\prec^*$ (i.e. $\text{First}_G(w)$ is the unique element of $[w]$ s.t. $\forall v \in [w] \backslash \lbrace \text{First}_G(w) \rbrace$, $\text{First}_G(w) \prec^* v$). The naive way to determine $\text{First}_G(w)$ is to first generate $[w]$ and then determine the 'first' element of this set, however in general $[w]$ may be an infinite set.
In the case when $G = \langle \Sigma^* | \rangle$ and $g \in G$ acts on $\Sigma^*$ by $g :w \mapsto gwg^{-1}$, $[w]$ is the set of cyclic permutations of $w$ and $\text{First}_G(w)$ is the unique Lyndon word in $[w]$. In this particular case, duval's algorithm will determine $\text{First}_G(w)$ without having to generate all of $[w]$.
Is there an algorithm for determining $\text{First}_G(w)$ without first determining all the elements of $[w]$ for a general group $G$ acting on $\Sigma^*$?
Or alternatively
Is there a $\Sigma$ and $G$ such that $\forall n$, $\exists w \in \Sigma^*$ such that there is no sequence $g_1, g_2, \ldots g_m \in G$ such that $(g_m \circ \cdots \circ g_1)(w) = \text{First}_G(w)$ and $\forall p < m$, $\text{length}((g_p \circ \cdots \circ g_1)(w)) - \text{length}(w) < n$?
i.e. For any bound $n$ there is a word $w$ that must be made more than $n$ letters longer during any sequence of group actions that take it to it's 'first' word.
-
You consider only reduced words? The action of $G$ on $F^*$ should preserve what? – Mark Sapir Oct 2 '10 at 20:52
can you please explain, why is [w] is the set of all cyclic permutations of w ? I can see only one side inclusion. – GA316 Apr 15 at 5:29 | 697 | 2,186 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2016-30 | latest | en | 0.760017 |
https://discourse.codecombat.com/t/assigning-rvalue/7896 | 1,550,979,342,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249578748.86/warc/CC-MAIN-20190224023850-20190224045850-00339.warc.gz | 549,176,486 | 4,187 | # Assigning Rvalue
#1
Hi i am not able to find out the problem in my code
’ if 1 + 1 == 3: # ∆ Make this false.
hero.moveXY(5, 15) # Move to the first mines.
if 2 + 3 == 5: # ∆ Make this true.
hero.moveXY(15, 40) # Move to the first gem.
# != means “is not equal to”.
if 2 + 2 == 4: # ∆ Make this true.
hero.moveXY(25, 15) # Move to the second gem.
# < means “is less than”.
if 2 + 2 < 5: # ∆ Make this true.
enemy = hero.findNearestEnemy()
hero.attack(enemy)
if 2 < 1: # ∆ Make this false.
hero.moveXY(40, 55)
if True 3 = 0: # ∆ Make this false.
hero.moveXY(50, 10)
if False 2 + 3 = 9: # ∆ Make this true.
hero.moveXY(55, 25)`
#2
Please format your code by surrounding it top and bottom with triple backticks ( ` )
#3
``````if 1 + 1 == 3: # ∆ Make this false.
hero.moveXY(5, 15) # Move to the first mines.
if 2 + 3 == 5: # ∆ Make this true.
hero.moveXY(15, 40) # Move to the first gem.
# != means "is not equal to".
if 2 + 2 == 4: # ∆ Make this true.
hero.moveXY(25, 15) # Move to the second gem.
# < means "is less than".
if 2 + 2 < 5: # ∆ Make this true.
enemy = hero.findNearestEnemy()
hero.attack(enemy)
if 2 < 1: # ∆ Make this false.
hero.moveXY(40, 55)
if True 3 = 0: # ∆ Make this false.
hero.moveXY(50, 10)
if False 2 + 3 = 9: # ∆ Make this true.
hero.moveXY(55, 25)
``````
#4
``````if True 3 = 0: # ∆ Make this false.
hero.moveXY(50, 10)
``````
Is not properly formatted code.
`True` is a keyword, `3` will be rendered as a number, `=` is the `assignment-operator`, and `0` is another number.
The simplest way to make `True` false would be to write it as `False` instead.
Further the `rvalue` error is a result of trying to assign a variable inside of an `if-statement`.
`=` is the assignment operator. It assigns something to something else.
`==` is a comparison (equality) operator, and checks if two things are equal.
#5
i solved it thanks for trying to help i appreciated that
#6
And what are === for?
To check is [1,53,235] equal to [1,53,235] and to compare with null? | 704 | 2,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-09 | latest | en | 0.757272 |
https://homeworkandessays.com/assignment-2-fair-shares-answer-5-questions/ | 1,606,259,190,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177607.13/warc/CC-MAIN-20201124224124-20201125014124-00461.warc.gz | 335,153,498 | 9,285 | # Assignment 2: Fair Shares ANSWER 5 QUESTIONS
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Assignment 2: Fair Shares
The Center City Anuraphilic (frog lovers) society has fallen on hard times. Abraham, Bobby and Charlene are the only remaining members and each feels equally entitled to take possession of the society’s collection of live rare tropical frogs. The decision is made to use the method of sealed bids and fair shares to decide who will take possession of the entire collection and how much will be paid in compensation to the other members.
Abraham unseals his estimate of the value of the collection at \$12,000.00. Bobby’s estimate of the value of the collection is \$6,000.00. Charlene values the collection at \$9,000.00.
• Who receives the collection of frogs?
• What is each person’s fair share of the monetary value of the collection?
• Why is the monetary amount of each fair share different?
• How much money is owed to each of the two people who do not “win” the collection of frogs?
• In your opinion how “Fair” is the process described above?
Now pretending for a moment that you like frogs, we will insert you into the situation under special circumstances. Despite (or perhaps because of) your love of all things amphibious, you currently lack the funds to pay each of the others their probable fair share. You will not receive the collection, but wish to receive as much money as possible. You have no knowledge of the amounts in each of the sealed bids, but strongly suspect that Abraham will bid between \$10,000.00 and \$12,000.00.
• Given that you cannot afford to “win” the process, describe how you will go about deciding what to put down for your own estimate of the value of the collection.
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Order your paper today and save upto 15% with the discount code 15BEST | 446 | 1,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | longest | en | 0.937862 |
https://www.acmicpc.net/problem/13965 | 1,709,373,774,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475806.52/warc/CC-MAIN-20240302084508-20240302114508-00183.warc.gz | 619,541,174 | 9,189 | 시간 제한메모리 제한제출정답맞힌 사람정답 비율
2 초 512 MB33191756.667%
## 문제
Alice: How can it be?
Rabbit: Trust me Alice. It always takes the same time. When I go from my home up the road to Queen of Hearts' Castle, my watch counts nine hours. However, if I continue cown to Mad Hatter's house, my watch counts just two hours in total. Isn't that great?
Alice: How can it be Rabbit? The path is longer and you take a shorter time to do it? How can it be?
Rabbit: Trust me Alice! It is all recorded in my logbook. You can check it. All my trips are there...
Alice: Rabbit, I do not think it can help me...
Rabbit: Alice, no matter where you are, or where you want to go, or the track you choose, you'll be able to find how long it takes you.
Alice: Really?
Rabbit: For sure!
Poor Rabbit, poor Alice.
White Rabbit is helping Alice finding a quick way home through the Rabbit's hole with his holy logbook of trips. The problem lied in the chronometer of its bizarre pocket watch (it displays the hours from zero to 12), and the way the Rabbit counts the time with it: If a journey takes 14 hours (real time), seeing the pointer resting above number one, he assumes it took one hour long.
Given that the White Rabbit is telling the truth, can you help Alice finding how long the shortest path home takes, using the Rabbit's logbook of trips?
Your task is to find the shortest real time it takes for Alice to go from her present location to the Rabbit's hole. For each trip, the White Rabbit wrote down the trip time, the number of visited locations (not necessarily distinct) and the sequence in which they were visited. That sequence defines the trip because there is at most one direct track between any two locations in the Wonderland and it takes the same time both ways.
## 입력
The first line contains four integers N, A, R and T, where: N is the number of distinct lacaions; A identifies the place where Alice is lacated; R corresponds to the Rabbit's hole location; and T is the number of trips recorded in White Rabbit's logbook. All locations are identified by numbers from 1 to N. Each of the next T lines describes a trip logged with format d p a1 a2 ... ap, where d is the trip duration (according to White Rabbit), p is the number of locations and a1 a2 ... ap is the sequence of visited locations.
## 출력
An integer representing the shortest (real) time it takes for Alice to get home.
## 제한
• 2 ≤ N ≤ 200 Number of locations
• 1 ≤ T ≤ 500 Number of trips in the logbook
• 2 ≤ p ≤ 800 Number of (possibly repeated) locations in a trip
• 1 ≤ dij ≤ 12 Real time of the direct track between ai and aj (if it exists)
• There are at most 200 direct tracks
## 예제 입력 1
3 1 3 3
3 4 1 2 3 2
4 3 1 2 1
1 4 1 2 1 3
## 예제 출력 1
9
## 예제 입력 2
5 5 1 9
0 3 1 2 3
1 4 1 4 2 3
6 4 3 4 1 3
11 5 1 3 4 2 1
4 4 1 2 4 1
6 6 1 2 3 1 4 3
7 4 2 3 4 1
11 3 4 3 5
12 5 5 2 4 2 5
## 예제 출력 2
6 | 853 | 2,881 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-10 | latest | en | 0.950053 |
http://summerschoolurban.eu/kinetic-anatomy-behnke-pdf/ | 1,534,609,563,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213691.59/warc/CC-MAIN-20180818154147-20180818174147-00082.warc.gz | 385,340,337 | 4,280 | # Kinetic anatomy behnke pdf
Please forward this error screen to sharedip-10718044127. An animation showing the relationship between pressure and volume when mass and temperature are held constant. The absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system. P kinetic anatomy behnke pdf the pressure of the gas, V is the volume of the gas, and k is a constant.
The equation states that the product of pressure and volume is a constant for a given mass of confined gas and this holds as long as the temperature is constant. The equation shows that, as volume increases, the pressure of the gas decreases in proportion. Similarly, as volume decreases, the pressure of the gas increases. The law was named after chemist and physicist Robert Boyle, who published the original law in 1662. This relationship between pressure and volume was first noted by Richard Towneley and Henry Power in the seventeenth century. This law was the first physical law to be expressed in the form of an equation describing the dependence of two variable quantities.
For a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional. Or Boyle’s law is a gas law, stating that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If volume increases, then pressure decreases and vice versa, when temperature is held constant. Boyle’s law states that at constant temperature for a fixed mass, the absolute pressure and the volume of a gas are inversely proportional.
The law can also be stated in a slightly different manner, that the product of absolute pressure and volume is always constant. Most gases behave like ideal gases at moderate pressures and temperatures. The technology of the 17th century could not produce high pressures or low temperatures. Hence, the law was not likely to have deviations at the time of publication.
Stating that the pressure and volume of a gas have an inverse relationship, recommendations for Exercise Preparticipation Health Screening. Category: Group Exercise Combine fitness, accredited training organization. Psychology The health benefits of regular physical activity are well, mariotte’s essay “De la nature de l’air” was reviewed by the French Royal Academy of Sciences in 1679. Category: Mind Body Linda, the three gas laws in combination with Avogadro’s law can be generalized by the ideal gas law. Color Atlas of Human Anatomy, the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system.
Category: Online Modules; the technology of the 17th century could not produce high pressures or low temperatures. Grace and flow, this relationship between pressure and volume was first noted by Richard Towneley and Henry Power in the seventeenth century. Sur la nature de l’air, includes 248 page book and exam. Purchase and Download the HIIT vs Continuous Endurance Training article and complete 10 question exam.
Daniel Bernoulli in 1737-1738 derived Boyle’s law using Newton’s laws of motion with application on a molecular level. The debate between proponents of Energetics and Atomism led Boltzmann to write a book in 1898, which endured criticism up to his suicide in 1906. P denotes the pressure of the system. V denotes the volume of the gas. So long as temperature remains constant the same amount of energy given to the system persists throughout its operation and therefore, theoretically, the value of k will remain constant. Here P1 and V1 represent the original pressure and volume, respectively, and P2 and V2 represent the second pressure and volume. Boyle’s law, Charles’s law, and Gay-Lussac’s law form the combined gas law. | 787 | 3,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-34 | latest | en | 0.947363 |
http://authenticinquirymaths.blogspot.com.au/2012_08_01_archive.html | 1,527,327,526,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867416.82/warc/CC-MAIN-20180526092847-20180526112847-00284.warc.gz | 23,340,938 | 25,406 | ## ...or How I Broke My Arm Last Night
This brief post is dedicated to an enthusiastic and energetic young man who took up the idea to do something kinesthetic while learning times tables facts. This is a great strategy to combine left- and right-brain and get things really embedded in the memory.
The choice of action was trampolining.
Sadly, our young mathematician misjudged the jump and ended up with a broken arm.
So be careful out there. Those times tables can be dangerous.
## ...and how it relates to multiplication facts
Alright - we all know this remarkable feature of multiplication.
And we already know the formula because someone taught us back in high school:
area = base x height
But what about my team of Year 4 kids?
What links could they find between area and multiplication facts?
So we started with some coloured rectangles, carefully cut to be pretty close to whole centimetres.
Then, using our multilink cubes we worked out how many we would need to cover the shape.
Much discussion ensued using words such as "area", "length", "height", "centimetre", "compare", "larger", "smaller" etc
"Ah, yes! 3 rows of 6 cubes. So the area is 18 square centimetres!"
And then the moment of epiphany!
Like a crack in the dam wall, the realisation came that we don't need to cover the whole shape.
We can in fact use our knowledge of multiplication facts to simplify this task.
So instead of putting out 150 centicubes, I only need to know the length of the base and the height.
Viola! 10 x 15 = 150
The final step was to compare the areas of larger shapes.
Who had the biggest rectangle? How do we know? How can you prove it?
## So what?
This may not seem too spectacular for a lesson idea, but the lights really came on for a few individuals in the class. Sometimes what seems obvious to us adults really is remarkable to kids.
## Here's a great idea for multiplication facts
...and like any great idea it was stolen from another teacher (Mrs P-W, two rooms down) who found it on a web-site somewhere.
For the day, all kids in Year 4 were given a sticky label with a multiplication fact written on it - this became their new name for the day.
Here's 48 and 64....
...and 108 having a chat with 56....
...and 63 with 144!
We all had to use our new "names" for the whole day. It was a great way to get the kids to think multiplication and to remember a few of those harder facts.
Teachers also participated - I became Mr 42 (or 6 x 7). (Douglas Adams fans may see the significance)
## Tuesday, 21 August 2012
### Churchill Fellow 2012
Earlier in the year I applied for Winston Churchill Memorial Trust Fellowship to study the inquiry process in mathematics education.
To my surprise, my application was successful and in 2013 I will have the opportunity to get out there and see what is happening.
Receiving my Churchill Fellowship from the Governor General, the Honorable Quentin Bryce
## The Learning Journey
During April and part of May 2013 I will be visiting schools in New York, San Francisco, Tokyo and Singapore. I will also be attending the NCTM conference in Denver, Colorado.
I'm excited already - it should be lots of fun and a great experience. While I have made preliminary arrangements with several schools, if you happen to be in any of these cities and would like to arrange a visit, let me know and I'll see what we can organise.
## Special Thanks
I'd like to thank several people who helped me in the application process - Peter Sullivan of Monash University, Melbourne; Philip Heath, Principal of Radford College; my staff team on Year 4 whose enthusiasm and support really encouraged me; and my wife and family - thanks everyone!
## Want to apply for a Fellowship?
Interested? Are you an Australian citizen? (UK and NZ - you have your own associations)
Then look at the Winston Churchill Memorial Trust website - http://www.churchilltrust.com.au/
## Thursday, 16 August 2012
### Maths is...
Listening to the voice of the child
Something I want to do more in my class is to listen to how the kids are feeling about what they are doing. I make too many assumptions about what they think and feel and I make important judgments based on my own skewed views.
So what would the kids say when I asked them to complete the sentence....
## Maths is...
I got my class, and the one next door belonging to CapitanoAmazing (find him on Twitter or at http://4thgradebigquestions.blogspot.com.au/), to finish the sentence. I sorted through their responses and found lots of interesting and thoughtful comments.
Here is a selection that I have grouped into sets.
## Awesomely Positive Responses!
There was always going to be a few of these. Some kids are so positive about everything they do. Being so effusive doesn't make these comments any less important - they need to be acknowledged and celebrated along with the rest of the team.
## This is hard but I think I can do it
An interesting group of responses from kids who found that Maths is easy but who were prepared to take the challenge and have a go.
I really encourage kids to try a variety of strategies to reach a solution. This really helps them, and me, to reflect on the problem solving strategies being used and which might be most effective.
Maths is meant to be a challenge. We need to take kids from what they know to what they don't know - this is the essence of learning. I'm really glad that these guys identified the challenging nature of learning.
## Help! I think I'm drowning!
There is almost a sense of desperation in some of these comments, where the child can see that things aren't going in the right direction but they don't know how to turn it around.
How many kids are sitting in my room thinking that they are failing? Not just in Maths but in writing? spelling? reading?
Well, now I know about a few of them and guess what? Now you're on the radar I can do something about it.
This is formative assessment, isn't it?
And some of these responses made me sit back and think.
Maths is a sum that has one answer? Really?
Or Maths is the basic number operations? Isn't there anything else - just figuring out?
I do encourage the "fun" idea and the problem solving side of maths so I'm glad a few kids identified that as part of the process.
And "Power and Knowledge"! A very interesting concept to explore.
## Maths is Life!
And then there were those who saw the real-life meaning and application of Maths.
## Maths is everyday life.
And if that message is all that my kids hear this year, I won't be disappointed.
(Well, maybe a bit - but it's a really important lesson to learn.)
## What an awesome marathon!
...and what a great run from Uganda's Stephen Kiprotich! Inspirational!
And in Year 4 we were inspired to have a go and see how we would go.
How far could we run (as a team) in 2 hours, 8 minutes and 1 second?
## The Challenge
We have 98 kids in Year 4. If they all ran around the quad for about 1.30 minute we would do the equivalent of the marathon time run by Kiprotich.
So, we got organised and started the run, which went from 8.40am to 10.48am.
## The Results
• the kids ran for 2 hours, 8 minutes and 1 second
• ran a total of 326 laps
And how did that compare to the Olympic runners?
• we measured our lap of the quad to be 76m
• so our total distance was 326 x 76 = 24 776m - a bit over half a marathon!
• and our speed was 24.776km divided by 128 minutes multiplied by 60 minutes = 11.61375km/h
And then we calculated how fast Steve Kiprotich ran:
• 42.195km divided by 128 minutes multiplied by 60 minutes = 19.76953125km/h
So our Year 4 team had done a great job, run over half a marathon and done it in a speed that made us really appreciate how fast those guys in London were running!
## Seriously, how much of this did the kids do?
Seriously? Well, they did all the running.
And we led them through the calculations. A few would have got there with minimal help but it was interesting to explain how to do the calculations.
And I think one or two got the idea of the relationship between time, distance and speed.
And we all had some fun!
## What are those numbers?
To celebrate Sally Pearson’s great win in the 100m hurdles today, the teachers planned a special welcome. When the kids came in to the building this morning we had the following numbers written on the white board:
100
85
105
1235
868
13
10
Sally - an Australian hero!
“All these numbers relate to the 100m hurdles,” we told the kids
No other information was provided. Time for the kids to start thinking.
Ah! Now some lights start flicking on.
“100 is the 100m for the race!”
“And 1235 is the time Sally ran – 12.35 seconds!”
"And there's 10 hurdles in the race."
After some prompting we got to the last numbers:
13m from the start line to the first hurdle.
Each hurdle is 8.5m apart.
It's 10.5m from the last hurdle to the finish line.
And the hurdles are 83.8cm high. (Why the crazy height? Well, it used to be 2' 9" in the imperial system)
...and in this discussion we covered such mathematical concepts as:
- decimal points
- need to use correct units of measurement
- fractions
- length, time and speed
A great conversation to start the day!
So we got out some till paper (long rolls of paper from an old cash register) and held it at 83.8cm so the kids could have a go at a hurdle.
## Learning an Old Game - Fly!
As a bit of fun, we learnt a an old game that I used to play at school called "Fly".
Here's the rules:
• You have seven sticks and you place them about a foot apart.
• Then all the kids stand in a line and run through putting one foot in between each stick.
• The last person in the line is the 'Fly", they run through and then at the last stick they jump as far as they like, then they pick a stick to be moved to the spot where they landed.
• Eventually the spacings get bigger and bigger and harder and harder. A person gets out when they put more than one foot in between two sticks (ie they don't make the jump) or miss one.
• When 'Fly' gets out, the second last jumper becomes the new 'Fly'
## Year 6 Exhibition - what's that?!
Our school operates under the International Baccalaureate Primary Years Programme (PYP). A key part of this programme is the Year 6 Exhibition, a culminating activity designed to demonstrate the awareness and understanding of the students of what they have learnt over their time in the PYP.
It involves group work.
It elicits deeper learning.
It is a transdisciplinary inquiry.
So I'm keen to see how our Year 6 students are going to use their maths skills and knowledge to explore this year's concept of "Conflict".
## The Good News - I am a Mentor!
Each collaborative group finds or is assigned a mentor - someone who can help the group find their direction, clarify their questions and distill their message.
Fortunately for the group that I will be mentoring, I will have a special interest in how they explore the concept of "Conflict".
And, of course, how they use their maths skills as part of this inquiry.
## The Bad News - I am a Mentor!
So look out team - you have been warned! I'm looking for something more than a column graph and a table of numbers. I'm hoping to see some creative and insightful uses of maths skills to underpin this significant inquiry into a big, big concept.
I'll let you know how the team and I get along as the journey unfolds...
## Year 4 Was pumped Up About the 100m
- Usain Bolt, Asafa Powell, Yohan Blake, Justin Gatlin. What a race!
....so we thought we'd do some inquiry about the race and see what mathematical concepts we could uncover.
## We Headed for the Oval
On the 100m track, we spaced students out at 10m intervals holding signs saying 10m, 20m, 30m, etc.
At the first whistle, children started running to see how far they could get in 9.63 seconds. When the clock got to 9.63 seconds, I blew the whistle again and the runners stopped, checking how far they had got and looking to see how far Usain Bolt had got in the same time.
## Our Responses
So what did the kids have to say? And what mathematical concepts could they uncover?
"I ran 55m. That's 45m less than Usain Bolt."
"I almost got halfway."
"I got 64m before 9.63 seconds was up. I ran 6.6 m/sec. Usain Bolt ran 10.38 m/sec."
"I got between 60m and 65m."
"Usain Bolt would be twice as fast as me. In the Olympics I wouldn't even be second last!"
"I ran approximately 55m whuile Usain Bolt would have run 100m. 100m divided by 9.63 seconds is about 10.38. So Usain Bolt ran 10.38 metres every second. If I divide the metres by 1000 you get 0.01038 km. Then you multiply by 3600 seconds to get km per hour - that is about 37 km per hour. That was his average speed."
....and my favourite:
"I look down the track and murmur to myself - Usain Bolt is really fast!" | 3,023 | 12,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-22 | latest | en | 0.94812 |
https://studyres.com/doc/3786658/unit-4---four-operations---difference-multiple | 1,708,477,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00508.warc.gz | 575,523,483 | 8,307 | Download Unit 4 - Four Operations : Difference-Multiple
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```WHOLE NUMBERS
4
FOUR OPERATIONS
4.1 DIFFERENCE-MULTIPLE MODEL CONCEPT
THINKING MATHEMATICALLY
EXAMPLE n
The total of two numbers is 432.
The difference between the two numbers is
178.
What are the two numbers?
SOLUTION:
Ans: 2 equal units plus 178.
?
What do we need to find first?
Large N
432
178
?
432 – 178 = 254
2 units → 254
1 unit → 254 ÷ 2 = 127 (Small Number)
127 + 178 = 305 (Large Number)
The two numbers are 127 and 305.
32
MathExpress
Process Skills Level 46
Represent the smaller number with
1 unit and the large number with
(1 equal unit + 178).
How many units represent the total?
Draw a model.
Small N
We draw a model to help visualise
the problem.
Ans: The value of 2 equal units.
By subtracting the difference (178)
from the total (432), we get the
value of 2 equal units.
```
Related documents | 296 | 1,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-10 | latest | en | 0.819238 |
https://forum.arduino.cc/t/resistivity-measurment-using-four-point-probe-and-arduino/428419 | 1,643,386,015,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00204.warc.gz | 319,638,713 | 4,845 | # Resistivity measurment using four point probe and arduino
I want to build an arduino project to measure the resistivity of semiconductor sheet by using four point probe method. I understand how the probe works in principle. it is just four equally spaced linearly arranged pins, a const. current is supplied through external pair of pins and voltage across the middle two is measured. I want to do same using arduino, since professional equipment is expensive. I don't want it to be as accurate as pro gear, it is more of a proof of concept project. but I don't understand where to start from, like how would I control the cc power supply using arduino and how to read micro volt voltages. I want to collect data through labVIEW or Mathematica and plot it. I'd be really thankful if anyone points me in the right direction.
This may be of interest, You could add Arduino with a 16/24 bit ADC.
http://www.kerrywong.com/2011/08/14/accurate-milliohm-measurement/
I have seen designs use an LM317 to create a constant current of 100ma not sure how good they sare
Hi,
Interesting. I have seen this done in the lab but I was never involved..
This is a good perspective (LINK)
Some Arduino Constant Current discussion HERE
But, if you can measure the current, and do math, you don't really have to have a precisely-controlled constant current, right?
It's mathematically convenient to have a nice constant current. I measure the resistance of, say, a 20 amp circuit breaker by using my high current variable supply to put 10 amps through it and then separately measure the voltage across the terminals. 0.1 V = 0.01 ohms for example.
Let us know how this works out...
317 seems to work ok. Depends on a lot of factors but I would imagine pumping 100mA through some semiconductors would be a bad idea. | 411 | 1,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-05 | latest | en | 0.956831 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=47015 | 1,369,226,204,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701670866/warc/CC-MAIN-20130516105430-00039-ip-10-60-113-184.ec2.internal.warc.gz | 322,940,954 | 1,453 | Question 66276
wHAT ARE THE STEPS TO DRAWING A PIE CHART.....HOW DO YOU DRAW A PIE CHART OR CONSTRUCT A PIE CHART CAN YOU EXPLAIN THE STEPS
LET US TAKE THE FOLLOWING DATA ON GRADES OBTAINED BY STUDENTS IN A CLASS
A = 50
B= 100
C=50
TOTAL=200
MAKE A TABLE OF % .
A = 50*100/200=25
B=100*100/200=50
C=50*100/200=25
TOTAL = 100
NOW TAKING TOTAL AS 360 DEGREES CONVERT THE ABOVE PERCENTAGES IN TO DEGREES
A=360*25/100=90
B=360*50/100=180
C=360*25/100=90
TOTAL = 360 DEGREES.
NOW DRAW ANY CIRCLE TAKING A CONVENIENT RADIUS WITH O AS CENTER .
DRAW ANY LINE OP SAY WITH P ON THE CIRCUMFERENCE OF THE CIRCLE.SUPPOSE WE DRAW
IT HORIZONTALLY.
NOW TO SHOW A GRADE FROM OP MEASURE AT 'O' 90 DEGREES CORRESPONDING TO 90 DEGREES WE GOT FOR 'A' GRADE STUDENTS.. TO GET LINE OA WITH A ON CIRCUMFERENCE,SUCH THAT ANGLE AOP=90 DEG.
NOW TO SHOW B GRADE FROM OA MEASURE AT 'O' 180 DEGREES CORRESPONDING TO 180 DEGREES WE GOT FOR 'B' GRADE STUDENTS.. TO GET LINE OB WITH B ON CIRCUMFERENCE,SUCH THAT ANGLE AOB=180 DEG.
NOW TO SHOW C GRADE FROM OB MEASURE AT 'O' 90 DEGREES CORRESPONDING TO 90 DEGREES WE GOT FOR 'C' GRADE STUDENTS.. TO GET LINE OC WITH C ON CIRCUMFERENCE,SUCH THAT ANGLE BOC=90 DEG.
WHEN IT IS THE END OF PLOT , YOU WILL NOTICE THAT C COINCIDES WITH 'P'.
IF NOT ,EITHER PLOTTING IS NOT COMPLETE OR YOUR CALCULATION IS WRONG.
see the drawing below.
now shade different zones IN DIFFERENT COLORS OR HATCHING
POA , LABELING IT AS A GRADE | 496 | 1,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2013-20 | latest | en | 0.477486 |
https://profsonly.com/price-in-tl-demand-of-girne-lefkosa-demand-of-guzelyurt-magusal-karpaz-supply-for-whole/ | 1,695,793,725,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00334.warc.gz | 513,403,627 | 32,196 | Get help from the best in academic writing.
# Price (in TL) Demand of Girne/ Lefkoşa Demand of Güzelyurt/Mağusal Karpaz Supply for Whole
Question: Price (in TL) Demand of Girne/ Lefkoşa Demand of Güzelyurt/Mağusal Karpaz Supply for Whole TRNC 1000 0 0 900 80000 5000 3000 – 72000 800 10000 6000 64000 700 15000 9000 56000 600 20000 12000 48000 500 25000 15000 40000 400 30000 18000 32000 300 35000 21000 24000 200 40000 24000 16000 100 45000 27000 8000 0 50000 30000 0 c) Assume the restaurants aim toShow transcribed image text 100% (2 ratings)C) Equilibrium Price = 500 Let’s take two prices to set as Price Floor are 600 and 700 A price bottom is a government-or group- assessed price control or limit on how low a pri…View the full answerTranscribed image text: Price (in TL) Demand of Girne/ Lefkoşa Demand of Güzelyurt/Mağusal Karpaz Supply for Whole TRNC 1000 0 0 900 80000 5000 3000 – 72000 800 10000 6000 64000 700 15000 9000 56000 600 20000 12000 48000 500 25000 15000 40000 400 30000 18000 32000 300 35000 21000 24000 200 40000 24000 16000 100 45000 27000 8000 0 50000 30000 0 c) Assume the restaurants aim to organize and put a price floor in the market for dinners. Choose two price levels in which an effective price floor can be set. Calculate the excess supply at those points, and the effect of price floor on the producer surplus. Assume restaurants decide voluntarily decrease the quantity of meals they supply to the level of quantity demanded at price floor. Assume initially the quantity supplied of restaurants is equal to the quantity they are willing to supply at the price floor level and each restaurant reduces its quantity supplied equally. How much should each restaurant reduce its quantity supplied in narrentano
## Australian Employment Law Assignment | Homework For You admission college essay help
Your task is to find a news articles published in the last 12 months and discuss the employment law issues. The article must be discussing Australian employment law.
Here are some ideas of where to look:The Age, Financial Review and other major newspapers.Workplace ExpressABC .net.au Choose the articles which are most interesting to you. Discuss and explain the articles to someone who has no legal knowledge at all. Your approach should be to the same as your approach to Question 1 of this assignment. (See my hints above). Get Business Law homework help today
## Regular Square Pyramid Assignment | Assignment Help Services admission college essay help
Price (in TL) Demand of Girne/ Lefkoşa Demand of Güzelyurt/Mağusal Karpaz Supply for Whole
A regular square pyramid has base edge b , height h , and slant height l . Three expressions give the surface area, A , of this pyramid. Which expression does NOT?
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## Probability Distribution Assignment | Assignment Help Services admission college essay help
Complete parts (a) and (b) below
The number of dogs per household in a small town
Dogs 0 1 2 3 4 5
Probability 0.649 0.0218 0.087 0.027 0.013 0.006
(a) find the mean variance and standerd deviation of the probability distribution
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## Ethical, Legal and Regulatory Policy Issues Assignment | College Homework Help essay help
Table of Ethical, Legal and Regulatory Policy Issues
Topic
Definition of the Topic
(minimum 2 sentences for each, a definition in your own words)
How the topic impacts and will be addressed in the selection and use of an EHR system (minimum 4 sentences for each topic – 2 on impact and 2 on how addressed)
1
Safe Design
2
Meaningful Use
3
Quality Improvement
4
Data Accuracy
5
Data Accessibility
6
Data Comprehensiveness
7
Data Consistency
8
Privacy
9
Confidentiality
10
Security
11
Individually Identifiable Health Information
12
Protected Health Information
13
HIPAA Privacy Rule
14
HIPAA Security Rule
15
16
Authentication
17
Authorization
18
Encryption
19
Technical Safeguards
20
Healthcare Ethical Principles
A. Addressing the Most Difficult Issue – Consider each of the 20 issues listed and choose the one that you believe will be the most difficult for the Midtown Family Clinic to implement.
Explain why you selected the issue and what should be done to ensure it is properly addressed when the system is selected, implemented and used. Your response will be assessed on its applicability to the Case Study and how well you support your choice and explain what should be done. (A paragraph of five to six sentences.)
B. Summary – briefly summarize the content of this section and tie the information together for the reader. (3-4 sentences) Get Information Systems homework help today
## Operations Security Assignment | College Homework Help essay help
Reflect on the connection between knowledge or concepts from these courses and how those have been, or could be, applied in the work place. (Working as a IT Consultant)
Subjects : Operations Security
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## Information Systems Assignment | College Homework Help essay help
There should be no plagiarism for the paper an should not use any content spinning sites. The professor is very strict about this and if he feels suspension about the paper he will fail me in this subject. The paper should be at least 8 papers with APA formatting. Below are the guidelines for the Paper given by my professor.
For this assignment, you’ll create a lightweight white paper for a fictitious Initial Coin Offering (ICO). You will use your imagination to create a brand new, hopefully unique, ICO, and develop a lightweight white paper that presents your ICO to prospective investors. A full white paper takes considerable time and effort to create. For this assignment, you’ll only be asked to provide the most essential elements.
The goal is for you to be creative in your application of block chain technology and explain your ideas.
To start, read the following article: How to Write a Good White Paper for Your ICO – https://applicature.com/blog/token-offerings/write-good-white-paper-ico
Then, explore new and existing block chain projects in any domain that interests you. You can start with an Internet search for “block chain use cases” and “current block chain projects.” Identify a few that you find interesting and learn about each one. Then, think of a new idea of how you could apply block chain in a new and useful way.
Once you have your idea, create your white paper. Your paper should be in APA format, and have the following sections:
2) Abstract – Summary of what your whitepaper contains
3) Introduction – Introduce readers to the problem you will solve, the motivation to solve it, and how you’ll present your solution.
4) Problem/Market consideration – Explain the current situation (expand the problem from the introduction).
5) Solution – Describe your solution.
6) Summary – Close the sale. Get Information Systems homework help today
## Fictitious Initial Coin Offering Assignment | College Homework Help essay help
For this assignment, you’ll create a lightweight white paper for a fictitious Initial Coin Offering (ICO). You will use your imagination to create a brand new, hopefully unique, ICO, and develop a lightweight white paper that presents your ICO to prospective investors. A full white paper takes considerable time and effort to create.
For this assignment, you’ll only be asked to provide the most essential elements.
The goal is for you to be creative in your application of block chain technology and explain your ideas.
To start, read the following article: How to Write a Good White Paper for Your ICO – https://applicature.com/blog/token-offerings/write-good-white-paper-icon
Then, explore new and existing block chain projects in any domain that interests you. You can start with an Internet search for “block chain use cases” and “current block chain projects.” Identify a few that you find interesting and learn about each one. Then, think of a new idea of how you could apply block chain in a new and useful way.
Once you have your idea, create your white paper. Your paper should be in APA format, and have the following sections:
2) Abstract – Summary of what your whitepaper contains
3) Introduction – Introduce readers to the problem you will solve, the motivation to solve it, and how you’ll present your solution.
4) Problem/Market consideration – Explain the current situation (expand the problem from the introduction).
5) Solution – Describe your solution.
6) Summary – Close the sale. Get Information Systems homework help today
## GSM and CDMA Assignment | College Homework Help essay help
For this assignment, discuss a minimum of three mobile web applications that are familiar to you or that you use regularly.
Describe the differences between GSM and CDMA. Discuss how GSM and CDMA support the mobile web applications you identified. Get Information Systems homework help today
## Which of the Equations Assignment | Assignment Help Services essay help
How many pounds of cashew nuts worth 75 cents a pound must be mixed with 10 pounds of pecans worth \$1.50 a pound to make a mixture worth \$1.00 a pound? Which of the equations could be used to solve for p, the pounds of cashew nuts?
1) 75p + 150(10) = 100(p + 10)
2) 75(p + 10) + 150(10) = 100(p + 10)
3) 75p + 150 = 100(p + 10)
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## Faster Rate Assignment | Assignment Help Services essay help
The equation y = 1.6x represents the number of laps Henry can swim over time, where y is the number of laps and x is time in minutes. This table shows the number of laps Larry can swim over time. How many laps can each boy complete per minute, and who can swim laps at a faster rate? Select from the drop-down menu to correctly complete the statements.
Get Math homework help today | 2,281 | 9,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.785633 |
https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/s13660-017-1480-2 | 1,726,239,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00882.warc.gz | 297,178,999 | 84,545 | # The regularized CQ algorithm without a priori knowledge of operator norm for solving the split feasibility problem
## Abstract
The split feasibility problem (SFP) is finding a point $$x\in C$$ such that $$Ax\in Q$$, where C and Q are nonempty closed convex subsets of Hilbert spaces $$H_{1}$$ and $$H_{2}$$, and $$A:H_{1}\rightarrow H_{2}$$ is a bounded linear operator. Byrne’s CQ algorithm is an effective algorithm to solve the SFP, but it needs to compute $$\|A\|$$, and sometimes $$\|A\|$$ is difficult to work out. López introduced a choice of stepsize $$\lambda_{n}$$, $$\lambda_{n}=\frac{\rho_{n}f(x_{n})}{\|\nabla f(x_{n})\| ^{2}}$$, $$0<\rho_{n}<4$$. However, he only obtained weak convergence theorems. In order to overcome the drawbacks, in this paper, we first provide a regularized CQ algorithm without computing $$\|A\|$$ to find the minimum-norm solution of the SFP and then obtain a strong convergence theorem.
## 1 Introduction
Let $$H_{1}$$ and $$H_{2}$$ be real Hilbert spaces and let C and Q be nonempty closed convex subsets of $$H_{1}$$ and $$H_{2}$$, and let $$A:H_{1}\rightarrow H_{2}$$ be a bounded linear operator. Let $$\mathbb {N}$$ and $$\mathbb{R}$$ denote the sets of positive integers and real numbers.
In 1994, Censor and Elfving [1] came up with the split feasibility problem (SFP) in finite-dimensional Hilbert spaces. In infinite-dimensional Hilbert spaces, it can be formulated as
$$\textit{Find }x\in C\textit{ such that }Ax\in Q,$$
(1.1)
where C and Q are nonempty closed convex subsets of $$H_{1}$$ and $$H_{2}$$, and $$A:H_{1}\rightarrow H_{2}$$ is a bounded linear operator. Suppose that SFP (1.1) is solvable, and let S denote its solution set. The SFP is widely applied to signal processing, image reconstruction and biomedical engineering [24].
So far, some authors have studied SFP (1.1) [517]. Others have also found a lot of algorithms to study the split equality fixed point problem and the minimization problem [1820]. Byrne’s CQ algorithm is an effective method to solve SFP (1.1). A sequence $$\{x_{n}\}$$, generated by the formula
$$x_{n+1}=P_{C}\bigl(x_{n}-\lambda_{n}A^{*}(I-P_{Q})Ax_{n} \bigr), \quad\forall n\geq0,$$
(1.2)
where the parameters $$\lambda_{n}\in(0,\frac{2}{\|A\|^{2}})$$, $$P_{C}: H\rightarrow C$$, and $$P_{Q}: H \rightarrow Q$$, is a set of orthogonal projections.
As is well-known, Cencor and Elfving’s algorithm needs to compute $$A^{-1}$$, and Byrne’s CQ algorithm needs to compute $$\|A\|$$. However, they are difficult to calculate.
Consider the following convex minimization problem:
$$\min_{x\in C}f(x),$$
(1.3)
where
\begin{aligned}& f(x)=\frac{1}{2}\big\| (I-P_{Q})Ax\big\| ^{2}, \end{aligned}
(1.4)
\begin{aligned}& \nabla f(x)=A^{*}(I-P_{Q})Ax, \end{aligned}
(1.5)
$$f(x)$$ is differentiable and the gradient f is L-Lipschitz with $$L>0$$.
The gradient-projection algorithm [21] is the most effective method to solve (1.3). A sequence $$\{x_{n}\}$$ is generated by the recursive formula
$$x_{n+1}=P_{C}(I-\lambda_{n} \nabla f)x_{n}, \quad\forall n\geq0,$$
(1.6)
where the parameter $$\lambda_{n}\in(0,\frac{2}{L})$$. Then we know that Byrne’s CQ algorithm is a special case of the gradient-projection algorithm.
In Byrne’s CQ algorithm, $$\lambda_{n}$$ depends on the operator norm $$\| A\|$$. However, it is difficult to compute. In 2005, Yang [22] considered $$\lambda_{n}$$ as follows:
$$\lambda_{n}:=\frac{\rho_{n}}{\|\nabla f(x_{n})\|},$$
where $$\rho_{n}>0$$ and satisfies
$$\sum_{n=0}^{\infty}\rho_{n} = \infty,\qquad \sum_{n=0}^{\infty}\rho_{n}^{2} < \infty.$$
In 2012, López [23] introduced $$\lambda_{n}$$ as follows:
$$\lambda_{n}:=\frac{\rho_{n}f(x_{n})}{\|\nabla f(x_{n})\|^{2}},$$
where $$0<\rho_{n}<4$$. However, López’s algorithm only has weak convergence.
In 2013, Yao [24] introduced a self-adaptive method for the SFP and obtained a strong convergence theorem. However, the algorithm is difficult to work out.
In general, there are two types of algorithms to solve SFPs. One is the algorithm which depends on the norm of the operator. The other is the algorithm without a priori knowledge of the operator norm. The first type of algorithm needs to calculate $$\|A\|$$, but $$\|A\|$$ is not easy to work out. The second type of algorithm also has a drawback. It always has weak convergence. If we want to obtain strong convergence, we have to use the composited iterative method, but then the algorithm is difficult to calculate. In order to overcome the drawbacks, we propose a new regularized CQ algorithm without a priori knowledge of the operator norm to solve the SFP and we obtain a strong convergence theorem.
Consider the following regularized minimization problem:
$$\min_{x\in C}f_{\beta}(x):=f(x)+\frac{\beta}{2}\|x \|^{2},$$
(1.7)
where the regularization parameter $$\beta>0$$. A sequence $$\{x_{n}\}$$ is generated by the formula
$$x_{n+1}=P_{C}\bigl(I-\lambda_{n}(\nabla f+ \beta_{n}I)\bigr)x_{n}, \quad\forall n\geq0,$$
(1.8)
where $$\nabla f(x_{n})=A^{*}(I-P_{Q})Ax_{n}$$, $$0<\beta_{n}<1$$, and $$\lambda_{n}=\frac{\rho_{n}f(x_{n})}{\|\nabla f(x_{n})\|^{2}}$$, $$0<\rho _{n}<4$$. Then, under suitable conditions, the sequence $$\{x_{n}\}$$ generated by (1.8) converges strongly to a point $$z\in S$$, where $$z=P_{S}(0)$$ is the minimum-norm solution of SFP (1.1).
## 2 Preliminaries
In this part, we introduce some lemmas and some properties that are used in the rest of the paper. Throughout this paper, let $$H_{1}$$ and $$H_{2}$$ be real Hilbert spaces, $$A:H_{1}\rightarrow H_{2}$$ be a bounded linear operator and I be the identity operator on $$H_{1}$$ or $$H_{2}$$. If $$f:H\rightarrow\mathbb{R}$$ is a differentiable functional, then the gradient of f is denoted by f. We use the sign ‘→’ to denote strong convergence and use the sign ‘’ to denote weak convergence.
### Definition 2.1
See [25]
Let D be a nonempty subset of H, and let $$T:D\rightarrow H$$. Then T is firmly nonexpansive if
$$\|Tx-Ty\|^{2}+\big\| (I-T)x-(I-T)y\big\| ^{2}\leq\|x-y \|^{2}, \quad\forall x, y\in D.$$
### Lemma 2.2
See [26]
Let $$T:H\rightarrow H$$ be an operator. Then the following are equivalent:
1. (i)
T is firmly nonexpansive,
2. (ii)
$$I-T$$ is firmly nonexpansive,
3. (iii)
$$2T-I$$ is nonexpansive,
4. (iv)
$$\|Tx-Ty\|^{2}\leq\langle x-y, Tx-Ty \rangle$$, $$\forall x, y \in H$$,
5. (v)
$$0\leq\langle Tx-Ty, (I-T)x-(I-T)y\rangle$$.
Recall $$P_{C}: H\rightarrow C$$ is an orthogonal projection, where C is a nonempty closed convex subset of H. Then to each point $$x\in H$$, the unique point $$P_{C}x\in C$$ satisfies the following property:
$$\|x-P_{C}x\|=\inf_{y\in C}\|x-y\|=: d(x,C).$$
$$P_{C}$$ also has the following characteristics.
### Lemma 2.3
See [27]
For a given $$x\in H$$,
1. (i)
$$z=P_{C}x \Longleftrightarrow\langle x-z,z-y\rangle\geq0$$, $$\forall y\in C$$,
2. (ii)
$$z=P_{C}x \Longleftrightarrow\|x-z\|^{2}\leq\|x-y\|^{2}-\|y-z\| ^{2}$$, $$\forall y\in C$$,
3. (iii)
$$\langle P_{C}x-P_{C}y, x-y\rangle\geq\|P_{C}x-P_{C}y\|^{2}$$, $$\forall x,y\in H$$.
### Lemma 2.4
See [28]
Let f be given by (1.4). Then
1. (i)
f is convex and differential,
2. (ii)
$$\nabla f(x)=A^{*}(I-P_{Q})Ax$$, $$\forall x\in H$$,
3. (iii)
f is w-lsc on H,
4. (iv)
f is $$\|A\|^{2}$$-Lipschitz: $$\|\nabla f(x)-\nabla f(y)\|\leq\|A\|^{2}\|x-y\|$$, $$\forall x,y\in H$$.
### Lemma 2.5
See [29]
Let $$\{a_{n}\}$$ be a sequence of nonnegative real numbers such that
$$a_{n+1} \leq(1-\alpha_{n})a_{n} + \alpha_{n}\delta_{n}, \quad n \geq0,$$
where $$\{\alpha_{n}\}_{n=0}^{\infty}$$ is a sequence in $$(0,1)$$ and $$\{ \delta_{n}\}_{n=0}^{\infty}$$ is a sequence in $$\mathbb{R}$$ such that
1. (i)
$$\sum_{n=0}^{\infty}\alpha_{n} = \infty$$,
2. (ii)
$$\limsup_{n\rightarrow\infty}\delta_{n} \leq0$$ or $$\sum_{n=0}^{\infty}\alpha_{n}|\delta_{n}| < \infty$$.
Then $$\lim_{n\rightarrow\infty}a_{n} = 0$$.
### Lemma 2.6
See [30]
Let $$\{\gamma_{n}\}_{n\in\mathbb{N}}$$ be a sequence of real numbers such that there exists a subsequence $$\{\gamma_{n_{i}}\}_{i\in \mathbb{N}}$$ of $$\{\gamma_{n}\}_{n\in\mathbb{N}}$$ such that $$\gamma _{n_{i}}<\gamma_{n_{i}+1}$$ for all $$i\in\mathbb{N}$$. Then there exists a nondecreasing sequence $$\{m_{k}\}_{k\in\mathbb{N}}$$ of $$\mathbb{N}$$ such that $$\lim_{k\rightarrow\infty}m_{k}=\infty$$ and the following properties are satisfied by all (sufficiently large) numbers $$k\in \mathbb{N}$$:
$$\gamma_{m_{k}}\leq\gamma_{m_{k}+1},\qquad \gamma_{k}\leq \gamma_{m_{k}+1}.$$
In fact, $$m_{k}$$ is the largest number n in the set $$\{1,\ldots,k\}$$ such that the condition
$$\gamma_{n}\leq\gamma_{n+1}$$
holds.
## 3 Main results
In this paper, we always assume that $$f:H\rightarrow\mathbb{R}$$ is a real-valued convex function, where $$f(x)=\frac{1}{2}\|(I-P_{Q})Ax\|^{2}$$, the gradient $$\nabla f(x)=A^{*}(I-P_{Q})Ax$$, C and Q are nonempty closed convex subsets of real Hilbert spaces $$H_{1}$$ and $$H_{2}$$, and $$A:H_{1}\rightarrow H_{2}$$ is a bounded linear operator.
### Algorithm 3.1
Choose an initial guess $$x_{0}\in H$$ arbitrarily. Assume that the nth iterate $$x_{n}\in C$$ has been constructed and $$\nabla f(x_{n})\neq0$$. Then we calculate the $$(n+1)$$th iterate $$x_{n+1}$$ via the formula
$$x_{n+1}=P_{C}\bigl(x_{n}-\lambda_{n} \bigl(A^{*}(I-P_{Q})Ax_{n}+\beta_{n}x_{n} \bigr)\bigr), \quad\forall n\geq0,$$
(3.1)
where $$\lambda_{n}$$ is chosen as follows:
$$\lambda_{n}=\frac{\rho_{n}f(x_{n})}{\|\nabla f(x_{n})\|^{2}}$$
with $$0<\rho_{n}<4$$. If $$\nabla f(x_{n})= 0$$, then $$x_{n+1}=x_{n}$$ is a solution of SFP (1.1) and the iterative process stops. Otherwise, we set $$n:=n+1$$ and go to (3.1) to evaluate the next iterate $$x_{n+2}$$.
### Theorem 3.1
Suppose that $$S\neq\emptyset$$ and the parameters $$\{\beta_{n}\}$$ and $$\{\rho_{n}\}$$ satisfy the following conditions:
1. (i)
$$\{\beta_{n}\}\subset(0,1)$$, $$\lim_{n\rightarrow\infty}\beta _{n}=0$$, $$\sum_{n=1}^{\infty}\beta_{n} = \infty$$,
2. (ii)
$$\varepsilon\leq\rho_{n}\leq4-\varepsilon$$ for some $$\varepsilon>0$$ small enough.
Then the sequence $$\{x_{n}\}$$ generated by Algorithm 3.1 converges strongly to $$z\in S$$, where $${z=P_{S}(0)}$$.
### Proof
Let $$x^{*}\in S$$. Since minimization is an exactly fixed point of its projection mapping, we have $$x^{*}=P_{C}x^{*}$$ and $$Ax^{*}=P_{Q}Ax^{*}$$.
By (3.1) and the nonexpansivity of $$P_{C}$$, we derive
\begin{aligned}[b] \big\| x_{n+1}-x^{*}\big\| ^{2} ={}&\big\| P_{C}\bigl(x_{n}- \lambda_{n}\bigl(A^{*}(I-P_{Q})Ax_{n}+\beta _{n}x_{n}\bigr)\bigr)-P_{C}x^{*}\big\| ^{2} \\ \leq{}&\big\| (1-\lambda_{n}\beta_{n})x_{n}-\lambda _{n}A^{*}(I-P_{Q})Ax_{n}-x^{*}\big\| ^{2} \\ ={}&\bigg\| \lambda_{n}\beta_{n}\bigl(-x^{*}\bigr) +(1-\lambda_{n}\beta_{n}) \biggl(x_{n}- \frac{\lambda_{n}}{1-\lambda_{n}\beta _{n}}A^{*}(I-P_{Q})Ax_{n}-x^{*}\biggr) \bigg\| ^{2} \\ ={}&\lambda_{n}\beta_{n}\big\| x^{*}\big\| ^{2} +(1- \lambda_{n}\beta_{n})\bigg\| x_{n}-\frac{\lambda_{n}}{1-\lambda_{n}\beta _{n}}A^{*}(I-P_{Q})Ax_{n}-x^{*} \bigg\| ^{2} \\ & -\lambda_{n}\beta_{n}(1-\lambda_{n} \beta_{n})\bigg\| x_{n}-\frac{\lambda _{n}}{1-\lambda_{n}\beta_{n}}A^{*}(I-P_{Q})Ax_{n} \bigg\| ^{2} \\ \leq{}&\lambda_{n}\beta_{n}\big\| x^{*}\big\| ^{2} +(1-\lambda_{n}\beta_{n})\bigg\| x_{n}- \frac{\lambda_{n}}{1-\lambda _{n}\beta_{n}}A^{*}(I-P_{Q})Ax_{n}-x^{*}\bigg\| ^{2}. \end{aligned}
(3.2)
Since $$P_{Q}$$ is firmly nonexpansive, from Lemma 2.2, we deduce that $$I-P_{Q}$$ is also firmly nonexpansive. Hence, we have
\begin{aligned}[b] \bigl\langle A^{*}(I-P_{Q})Ax_{n}, x_{n}-x^{*}\bigr\rangle &=\bigl\langle (I-P_{Q})Ax_{n}, Ax_{n}-Ax^{*} \bigr\rangle \\ &=\bigl\langle (I-P_{Q})Ax_{n}-(I-P_{Q})Ax^{*}, Ax_{n}-Ax^{*}\bigr\rangle \\ &\geq\big\| (I-P_{Q})Ax_{n}\big\| ^{2} \\ &=2f(x_{n}). \end{aligned}
(3.3)
Note that $$\nabla f(x_{n})=A^{*}(I-P_{Q})Ax_{n}$$. From (3.3), we obtain
\begin{aligned} &\bigg\| x_{n}-\frac{\lambda_{n}}{1-\lambda_{n}\beta _{n}}A^{*}(I-P_{Q})Ax_{n}-x^{*} \bigg\| ^{2} \\ &\quad=\big\| x_{n}-x^{*}\big\| ^{2}+\frac{\lambda_{n}^{2}}{(1-\lambda_{n}\beta_{n})^{2}}\big\| A^{*}(I-P_{Q})Ax_{n}\big\| ^{2} -\frac{2\lambda_{n}}{1-\lambda_{n}\beta_{n}}\bigl\langle A^{*}(I-P_{Q})Ax_{n}, x_{n}-x^{*}\bigr\rangle \\ &\quad=\big\| x_{n}-x^{*}\big\| ^{2}+\frac{\lambda_{n}^{2}}{(1-\lambda_{n}\beta_{n})^{2}}\big\| \nabla f(x_{n})\big\| ^{2}-\frac{2\lambda_{n}}{1-\lambda_{n}\beta_{n}}\bigl\langle \nabla f(x_{n}), x_{n}-x^{*} \bigr\rangle \\ &\quad\leq\big\| x_{n}-x^{*}\big\| ^{2}+\frac{\lambda_{n}^{2}}{(1-\lambda_{n}\beta_{n})^{2}}\big\| \nabla f(x_{n})\big\| ^{2}-\frac{4\lambda_{n}}{1-\lambda_{n}\beta _{n}}f(x_{n}) \\ &\quad=\big\| x_{n}-x^{*}\big\| ^{2}+\frac{1}{(1-\lambda_{n}\beta_{n})^{2}}\cdot \frac{\rho _{n}^{2}f(x_{n})^{2}}{\|\nabla f(x_{n})\|^{4}}\cdot\big\| \nabla f(x_{n})\big\| ^{2} \\ &\qquad{} -\frac{4\rho_{n}f(x_{n})}{(1-\lambda_{n}\beta_{n})\|\nabla f(x_{n})\| ^{2}}\cdot f(x_{n}) \\ &\quad=\big\| x_{n}-x^{*}\big\| ^{2}+\frac{\rho_{n}^{2}f(x_{n})^{2}}{(1-\lambda_{n}\beta _{n})^{2}\|\nabla f(x_{n})\|^{2}}-\frac{4\rho_{n}f(x_{n})^{2}}{(1-\lambda _{n}\beta_{n})\|\nabla f(x_{n})\|^{2}} \\ &\quad=\big\| x_{n}-x^{*}\big\| ^{2} -\rho_{n}\biggl(4- \frac{\rho_{n}}{1-\lambda_{n}\beta_{n}}\biggr)\cdot\frac {f(x_{n})^{2}}{(1-\lambda_{n}\beta_{n})\|\nabla f(x_{n})\|^{2}}. \end{aligned}
(3.4)
By condition (ii), without loss of generality, we assume that $$(4-\frac{\rho_{n}}{1-\lambda_{n}\beta_{n}})>0$$ for all $$n\geq0$$. Thus from (3.2) and (3.4), we obtain
\begin{aligned}[b] \big\| x_{n+1}-x^{*}\big\| ^{2} \leq{}&\lambda_{n} \beta_{n}\big\| x^{*}\big\| ^{2}+(1-\lambda_{n} \beta_{n}) \biggl(\big\| x_{n}-x^{*}\big\| ^{2} \\ &-\rho_{n}\biggl(4-\frac{\rho_{n}}{1-\lambda_{n}\beta_{n}}\biggr)\cdot\frac {f(x_{n})^{2}}{(1-\lambda_{n}\beta_{n})\|\nabla f(x_{n})\|^{2}} \biggr) \\ ={}&\lambda_{n}\beta_{n}\big\| x^{*}\big\| ^{2}+(1- \lambda_{n}\beta_{n})\big\| x_{n}-x^{*}\big\| ^{2} \\ & -\rho_{n}\biggl(4-\frac{\rho_{n}}{1-\lambda_{n}\beta_{n}}\biggr)\cdot\frac {f(x_{n})^{2}}{\|\nabla f(x_{n})\|^{2}} \\ \leq{}&\lambda_{n}\beta_{n}\big\| x^{*}\big\| ^{2}+(1- \lambda_{n}\beta_{n})\big\| x_{n}-x^{*}\big\| ^{2} \\ \leq{}&\max\bigl\{ \big\| x^{*}\big\| ^{2}, \big\| x_{n}-x^{*}\big\| ^{2} \bigr\} . \end{aligned}
(3.5)
Hence, $$\{x_{n}\}$$ is bounded.
Let $$z=P_{S}0$$. From (3.5), we deduce
\begin{aligned}[b] 0 &\leq\rho_{n}\biggl(4-\frac{\rho_{n}}{1-\lambda_{n}\beta_{n}}\biggr)\frac {f(x_{n})^{2}}{\|\nabla f(x_{n})\|^{2}} \\ &\leq\lambda_{n}\beta_{n}\|z\|^{2}+(1- \lambda_{n}\beta_{n})\|x_{n}-z\| ^{2}- \|x_{n+1}-z\|^{2}. \end{aligned}
(3.6)
We consider the following two cases.
Case 1. One has $$\|x_{n+1}-z\|\leq\|x_{n}-z\|$$ for every $$n\geq n_{0}$$ large enough.
In this case, $$\lim_{n\rightarrow\infty}\|x_{n}-z\|$$ exists as finite and hence
$$\lim_{n\rightarrow\infty}\bigl(\|x_{n+1}-z\|-\|x_{n}-z\|\bigr)=0.$$
(3.7)
This, together with (3.6), implies that
$$\rho_{n}\biggl(4-\frac{\rho_{n}}{1-\lambda_{n}\beta_{n}}\biggr)\frac{f(x_{n})^{2}}{\| \nabla f(x_{n})\|^{2}} \rightarrow0.$$
Since $$\liminf_{n\rightarrow\infty}\rho_{n}(4-\frac{\rho_{n}}{1-\lambda _{n}\beta_{n}})\geq\varepsilon_{0}$$ (where $$\varepsilon_{0}>0$$ is a constant), we get
$$\frac{f(x_{n})^{2}}{\|\nabla f(x_{n})\|^{2}}\rightarrow0.$$
Noting that $$\|\nabla f(x_{n})\|^{2}$$ is bounded, we deduce immediately that
$$\lim_{n\rightarrow\infty}f(x_{n})=0.$$
(3.8)
Next, we prove that
$$\limsup_{n\rightarrow\infty}\langle-z, x_{n}-z\rangle\leq0.$$
(3.9)
Since $$\{x_{n}\}$$ is bounded, there exists a subsequence $$\{x_{n_{i}}\}$$ satisfying $$x_{n_{i}}\rightharpoonup\hat{z}$$ and
$$\limsup_{n\rightarrow\infty}\langle-z, x_{n}-z\rangle=\lim _{i\rightarrow\infty}\langle -z, x_{n_{i}}-z\rangle.$$
By the lower semicontinuity of f, we get
$$0\leq f(\hat{z})\leq\liminf_{i\rightarrow\infty}f(x_{n_{i}})=\lim _{n\rightarrow\infty}f(x_{n})=0.$$
So
$$f(\hat{z})=\frac{1}{2}\big\| (I-P_{Q})A\hat{z}\big\| ^{2}=0.$$
That is, is a minimizer of f, and $$\hat{z}\in S$$. Therefore
\begin{aligned}[b] \limsup_{n\rightarrow\infty}\langle-z, x_{n}-z\rangle &=\lim _{i\rightarrow\infty}\langle -z, x_{n_{i}}-z\rangle \\ &=\langle-z, \hat{z}-z\rangle \\ &\leq 0.\end{aligned}
(3.10)
Then we have
\begin{aligned} \|x_{n+1}-z\|^{2} ={}&\big\| P_{C}\bigl(x_{n}- \lambda_{n}A^{*}(I-P_{Q})Ax_{n}-\lambda_{n} \beta _{n}x_{n}\bigr)-P_{C}z\big\| ^{2} \\ \leq{}&\big\| (1-\lambda_{n}\beta_{n})x_{n}- \lambda_{n}A^{*}(I-P_{Q})Ax_{n}-z\big\| ^{2} \\ ={}&\bigg\| \lambda_{n}\beta_{n}(-z) +(1-\lambda_{n}\beta_{n}) \biggl(x_{n}- \frac{\lambda_{n}}{1-\lambda_{n}\beta _{n}}A^{*}(I-P_{Q})Ax_{n}-z\biggr)\bigg\| ^{2} \\ ={}&(\lambda_{n}\beta_{n})^{2}\|z\|^{2} \\ &+(1-\lambda_{n}\beta_{n})^{2}\bigg\| x_{n}- \frac{\lambda_{n}}{1-\lambda _{n}\beta_{n}}A^{*}(I-P_{Q})Ax_{n}-z\bigg\| ^{2} \\ &+2(1-\lambda_{n}\beta_{n})\lambda_{n} \beta_{n}\biggl\langle x_{n}-\frac {\lambda_{n}}{1-\lambda_{n}\beta_{n}}A^{*}(I-P_{Q})Ax_{n}-z,-z \biggr\rangle \\ \leq{}&(1-\lambda_{n}\beta_{n})^{2} \|x_{n}-z\|^{2}+(\lambda_{n}\beta_{n})^{2} \| z\|^{2} \\ & +2(1-\lambda_{n}\beta_{n})\lambda_{n} \beta_{n}\langle x_{n}-z, -z\rangle+2\lambda_{n}^{2} \beta_{n}\bigl\langle \nabla f(x_{n}), z\bigr\rangle \\ \leq{}&(1-\lambda_{n}\beta_{n})\|x_{n}-z \|^{2} \\ &+\lambda_{n}\beta_{n}\bigl(\lambda _{n}\beta_{n}\|z\|^{2}+2(1-\lambda_{n}\beta_{n})\langle x_{n}-z, -z \rangle+2\lambda_{n}\big\| \nabla f(x_{n})\big\| \cdot\|z\|\bigr). \end{aligned}
Note that $$\|\nabla f(x_{n})\|^{2}$$ is bounded, and that $$\lambda_{n}\| \nabla f(x_{n})\|=\frac{\rho_{n}f(x_{n})}{\|\nabla f(x_{n})\|^{2}}\cdot\| \nabla f(x_{n})\|$$. Thus $$\lambda_{n}\|\nabla f(x_{n})\|\rightarrow0$$ by (3.8). From Lemma 2.5, we deduce that
$$x_{n}\rightarrow z.$$
Case 2. There exists a subsequence $$\{\|x_{n_{j}}-z\|\}$$ of $$\{\| x_{n}-z\|\}$$ such that
$$\|x_{n_{j}}-z\|< \|x_{n_{j}+1}-z\| \quad\text{for all }j\geq1.$$
By Lemma 2.6, there exists a strictly nondecreasing sequence $$\{m_{k}\}$$ of positive integers such that $$\lim_{k\rightarrow\infty}m_{k}=+\infty$$ and the following properties are satisfied by all numbers $$k\in \mathbb{N}$$:
$$\|x_{m_{k}}-z\|\leq\|x_{m_{k}+1}-z\|,\qquad \|x_{k}-z\|\leq \|x_{m_{k}+1}-z\|.$$
(3.11)
We have
\begin{aligned} \|x_{n+1}-z\| ={}&\big\| P_{C}\bigl(x_{n}- \lambda_{n}A^{*}(I-P_{Q})Ax_{n}-\lambda_{n} \beta _{n}x_{n}\bigr)-P_{C}z\big\| \\ \leq{}&\big\| (1-\lambda_{n}\beta_{n})x_{n}- \lambda_{n}A^{*}(I-P_{Q})Ax_{n}-z\big\| \\ ={}&\bigg\| \lambda_{n}\beta_{n}(-z) +(1-\lambda_{n}\beta_{n}) \biggl(x_{n}- \frac{\lambda_{n}}{1-\lambda_{n}\beta _{n}}A^{*}(I-P_{Q})Ax_{n}-z\biggr)\bigg\| \\ \leq{}&\lambda_{n}\beta_{n}\big\| (-z)\big\| +(1-\lambda_{n} \beta_{n})\bigg\| x_{n}-\frac {\lambda_{n}}{1-\lambda_{n}\beta_{n}}A^{*}(I-P_{Q})Ax_{n}-z \bigg\| \\ \leq{}&\lambda_{n}\beta_{n}\|z\|+(1-\lambda_{n} \beta_{n})\|x_{n}-z\|. \end{aligned}
Consequently,
\begin{aligned} 0 &\leq\lim_{k\rightarrow\infty}\bigl(\|x_{m_{k}+1}-z\|-\|x_{m_{k}}-z \|\bigr) \\ &\leq\limsup_{n\rightarrow\infty}\bigl(\|x_{n+1}-z\|-\|x_{n}-z \|\bigr) \\ &\leq\limsup_{n\rightarrow\infty}\bigl(\lambda_{n} \beta_{n}\|z\|+(1-\lambda _{n}\beta_{n}) \|x_{n}-z\|-\|x_{n}-z\|\bigr) \\ &=\limsup_{n\rightarrow\infty}\lambda_{n}\beta_{n}\bigl(\|z \|-\|x_{n}-z\|\bigr) \\ &=0. \end{aligned}
Hence,
$$\lim_{k\rightarrow\infty}\bigl(\|x_{m_{k}+1}-z\|-\|x_{m_{k}}-z\|\bigr)=0.$$
(3.12)
By a similar argument to that of Case 1, we prove that
\begin{aligned}& \limsup_{k\rightarrow\infty}\langle-z, x_{m_{k}}-z \rangle\leq0, \\& \|x_{m_{k}+1}-z\|^{2}\leq(1-\lambda_{m_{k}} \beta_{m_{k}})\|x_{m_{k}}-z\| ^{2}+\lambda_{m_{k}} \beta_{m_{k}}\sigma_{m_{k}}, \end{aligned}
(3.13)
where
$$\sigma_{m_{k}}=\lambda_{m_{k}}\beta_{m_{k}}\|z \|^{2}+2(1-\lambda _{m_{k}}\beta_{m_{k}})\langle x_{m_{k}}-z, -z\rangle+2\lambda_{m_{k}}\big\| \nabla f(x_{m_{k}})\big\| \cdot\|z\|.$$
In particular, from (3.13), we get
$$\lambda_{m_{k}}\beta_{m_{k}}\|x_{m_{k}}-z\|^{2} \leq\|x_{m_{k}}-z\|^{2}-\| x_{m_{k}+1}-z\|^{2}+ \lambda_{m_{k}}\beta_{m_{k}}\sigma_{m_{k}}.$$
(3.14)
Since $$\|x_{m_{k}}-z\|\leq\|x_{m_{k}+1}-z\|$$, we deduce that
$$\|x_{m_{k}}-z\|^{2}-\|x_{m_{k}+1}-z\|^{2}\leq0.$$
Then, from (3.14), we have
$$\lambda_{m_{k}}\beta_{m_{k}}\|x_{m_{k}}-z\|^{2} \leq\lambda_{m_{k}}\beta _{m_{k}}\sigma_{m_{k}}.$$
Then
$$\limsup_{k\rightarrow\infty}\|x_{m_{k}}-z\|^{2}\leq\limsup _{k\rightarrow \infty}\sigma_{m_{k}}\leq0.$$
(3.15)
Then, from (3.12), we deduce that
$$\limsup_{k\rightarrow\infty}\|x_{m_{k}+1}-z\|=0.$$
(3.16)
Thus, from (3.11) and (3.16), we conclude that
$$\limsup_{k\rightarrow\infty}\|x_{k}-z\|\leq\limsup _{k\rightarrow\infty }\|x_{m_{k}+1}-z\|=0.$$
Therefore, $$x_{n}\rightarrow z$$. This completes the proof. □
## 4 Conclusion
Recently, the SFP has been studied extensively by many authors. However, some algorithms need to compute $$\|A\|$$, and this is not an easy thing to work out. Others do not need to compute $$\|A\|$$, but the algorithms always have weak convergence. If we want to obtain strong convergence theorems, the algorithms are complex and difficult to calculate. We try to get over the drawbacks. In this article, we use the regularized CQ algorithm without computing $$\|A\|$$ to find the minimum-norm solution of the SFP, where $$\lambda_{n}=\frac{\rho _{n}f(x_{n})}{\|\nabla f(x_{n})\|^{2}}$$, $$0<\rho_{n}<4$$. Then, under suitable conditions, the explicit strong convergence theorem is obtained.
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## Acknowledgements
The authors thank the referees for their helping comments, which notably improved the presentation of this paper. This work was supported by the Fundamental Research Funds for the Central Universities [grant number 3122017072]. MT was supported by the Foundation of Tianjin Key Laboratory for Advanced Signal Processing. H-FZ was supported in part by the Technology Innovation Funds of Civil Aviation University of China for Graduate (Y17-39).
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Tian, M., Zhang, HF. The regularized CQ algorithm without a priori knowledge of operator norm for solving the split feasibility problem. J Inequal Appl 2017, 207 (2017). https://doi.org/10.1186/s13660-017-1480-2 | 10,544 | 27,111 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.88147 |
https://www.bartleby.com/essay/Production-P3SM2ZAXH3GEY | 1,601,366,551,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401632671.79/warc/CC-MAIN-20200929060555-20200929090555-00182.warc.gz | 649,813,043 | 10,813 | Production
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1.1 Chuck Sox makes wooden boxes in which to ship | motorcycles. Chuck and his three employees invest a total of 40 | hours per day making the 120 boxes. | a) What is their productivity? | | Total hours invested in production=40 hours per day | | produced number of boxes in a day=120 | | | | Productivity per day is =units produced/input used | | productivity per day is =120 boxes/40 hours | | productivity per day = 3 boxes per day | | | | their producivity is 3 boxes per day | | | b) Chuck and his employees have discussed redesigning the process | to improve efficiency. If they can increase the rate to 125 | per day, what will be their new productivity? | | | | New productivity=Units…show more content…
| | | | | Labor productivity = Total units produced/Total Labor hours | | Labor productivity = 100 tires / 400 hours | | Labor productivity = 2.5 tires per hour hour | | | | b) What is the multifactor productivity for these tires at | Lakefront Manufacturing? | | | | | Labor is 400 hours * \$12.50= \$5,000 | | Raw materialis 20,000 lbs * \$1 =\$20,000 | | Energy =\$5,000 | | Capital Costs = \$10,000 | | Total multifactor cost per day =\$40,000 | | | | | Multifactor productivity per dollar = units produced/Total multifactor cost | | multifactor productivity per dollar = 1,000 tires / \$40,000 | | | | | Multifactor productivity per dollar = 0.025 tires per dollar | | | | c) What is the percent change in multifactor productivity if Fok | can reduce the energy bill by \$1,000 per day without cutting | production or changing any other inputs? P | | | | | Total multifactor productivity cost = \$40,000 | | minus reduction in energy cost = \$1,000 | | Total multifactor productivity cost = \$39,000 | | | | | Productivity per dollar = units produced/Total cost | | Productivity per dollar = 1,000/\$39,000 | | Productivity per dollar = 0.026 per labor dollar | | | | | Percentage change in productivity = 0.026-0.025/0.026 | | Percentage change in productivity = 0.038 or 3.85% | 1.13 Charles Lackey operates a bakery in Idaho Falls, Idaho. | Because of its excellent product and | 553 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-40 | latest | en | 0.806156 |
http://mathhelpforum.com/algebra/218849-how-many-intersection-two-sets.html | 1,480,798,749,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541134.5/warc/CC-MAIN-20161202170901-00391-ip-10-31-129-80.ec2.internal.warc.gz | 178,895,599 | 12,516 | # Thread: How many in the intersection of two sets
1. ## How many in the intersection of two sets
I'm seeking confirmation that the correct answer to this question is $0$, which is one of the choices in the multiple-choice version of the question, and why.
Q: A dance studio has $35$ students. A poll shows that $12$ study tap and $19$ study ballet. What is the minimum number of students who are studying both tap and ballet?
I'm not clear if the question's original wording (which I don't have handy) meant that $n$ students studying tap means " $12$ students who study only tap" or " $12$ students who are studying at least tap and possibly also ballet" (and the same ambiguous wording for ballet).
But note that $12+19$ is less than $35$ total students. So are there $4$ other students who might have been taking both ( $4$ was not answer choice).
Also, $12$ students was not an answer choice - as in there are $12$ students taking tap and all are also taking ballet, which has $5$ more students, not taking tap.
2. ## Re: How many in the intersection of two sets
I'm seeking confirmation.
that the correct answer to this question is $0$, which is one of the choices in the multiple-choice version of the question, and why.
Q: A dance studio has $35$ students. A poll shows that $12$ study tap and $19$ study ballet. What is the minimum number of students who are studying both tap and ballet?
I'm not clear if the question's original wording (which I don't have handy) meant that $n$ students studying tap means " $12$ students who study only tap" or " $12$ students who are studying at least tap and possibly also ballet" (and the same ambiguous wording for ballet).
But note that $12+19$ is less than $35$ total students. So are there $4$ other students who might have been taking both ( $4$ was not answer choice).
Also, $12$ students was not an answer choice - as in there are $12$ students taking tap and all are also taking ballet, which has $5$ more students, not taking tap.
Unless you can post the exact wording of the original, it is pointless to even try to help.
As posted, the numbers are inconstant, unless there are others qualifiers left out
For example: if 19 take only ballet and 12 take only tap, then this can be answered.
Try to find the exact wording.
3. ## Re: How many in the intersection of two sets
Q: A dance studio has $35$ students. A poll shows that $12$ study tap and $19$ study ballet. What is the minimum number of students who are studying both tap and ballet?
I see four possibilities: the numbers can refer to students that study (1) those disciplines and possibly something else or (2) those disciplines only, and (a) there are other disciplines besides tap and ballet or (b) there are no other disciplines.
(1a) The answer is 0 under the following scenario: 12 people study only tap. 19 people study only ballet and 4 people study salsa.
(1b) This variant is impossible.
(2a) The answer is 0 under the same scenario as in (1a).
The question as it is phrased strongly suggests (1a).
4. ## Re: How many in the intersection of two sets
Originally Posted by Plato
Unless you can post the exact wording of the original, it is pointless to even try to help.
As posted, the numbers are inconstant, unless there are others qualifiers left out
For example: if 19 take only ballet and 12 take only tap, then this can be answered.
Try to find the exact wording.
The original question's wording is ambiguous. That was my point. And it seems that it must be a typo or something.
But since you asked I went and looked for the exact wording:
In a dance school with 35 students, a poll shows that 12 are studying tap dance and 19 are studying ballet. What is the minimum number of students in the school who are studying both tap dance and ballet?
A. 0
B. 7
C. 9
D. 12
E. 31
5. ## Re: How many in the intersection of two sets
Originally Posted by emakarov
I see four possibilities: the numbers can refer to students that study (1) those disciplines and possibly something else or (2) those disciplines only, and (a) there are other disciplines besides tap and ballet or (b) there are no other disciplines.
(1a) The answer is 0 under the following scenario: 12 people study only tap. 19 people study only ballet and 4 people study salsa.
(1b) This variant is impossible.
(2a) The answer is 0 under the same scenario as in (1a).
The question as it is phrased strongly suggests (1a).
You defined scenario (1a) to mean that those numbers represent people who are studying that discipline and possibly others. So for (1a) 12 ppl study tap and ballet, 19 ppl study ballet (7 of them study only ballet), and 4 ppl study salsa. Answer 12 (which is a choice - contrary to what I said in the OP).
Scenario (2b) seems impossible: tap and ballet are the only disciplines and there are not even 35 students in those two classes.
6. ## Re: How many in the intersection of two sets
The original question's wording is ambiguous. That was my point. And it seems that it must be a typo or something.
But since you asked I went and looked for the exact wording:
In a dance school with 35 students, a poll shows that 12 are studying tap dance and 19 are studying ballet. What is the minimum number of students in the school who are studying both tap dance and ballet?
A. 0
B. 7
C. 9
D. 12
E. 31
The only slightly ambiguous thing I see about this question is whether there are other disciplines studied in the school. There is no ambiguity with respect to studying tap or ballet exclusively. If I say that I am studying computer science, it means just that. It does not mean that I am not studying anything else besides CS. Similarly, if it is said that 12 people are studying tap dance, it means that they are studying tap dance for sure, but possibly other disciplines as well.
Concerning other disciplines besides tap and ballet in the school, in the absence of any statement to the contrary it is natural to allow this possibility. Also, as post #3 says and as explained below, variant (1b) is impossible and is not one of the answer options. Therefore, the most natural interpretation is (1a).
Originally Posted by emakarov
I see four possibilities: the numbers can refer to students that study (1) those disciplines and possibly something else or (2) those disciplines only, and (a) there are other disciplines besides tap and ballet or (b) there are no other disciplines.
(1a) The answer is 0 under the following scenario: 12 people study only tap, 19 people study only ballet and 4 people study salsa.
(1b) This variant is impossible.
(2a) The answer is 0 under the same scenario as in (1a). | 1,580 | 6,639 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 31, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-50 | longest | en | 0.98072 |
https://www.cfd-online.com/Forums/main/81394-strcture-pressure-correction-equation-piso-stable-over-time.html | 1,713,311,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00031.warc.gz | 640,342,320 | 15,495 | # Strcture of pressure correction equation in PISO stable over time ?
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October 26, 2010, 05:09 Strcture of pressure correction equation in PISO stable over time ? #1 New Member Join Date: Feb 2010 Posts: 7 Rep Power: 16 Hello everybody, I have a problem in understanding why the things I'm doing work. I'm using the PISO algorithm to solve the Navier-Stokes equations. In there I have to solve the Poisson-like pressure equation which I'm solving with an algebraic multigrid solver. Unfortunatly my PISO algorithm does not use a real Poisson pressure equation like -/LAPLACE/ p= RHS, but it includes the flux a caused by this correction into the operator. This is, I have - /DIV/ a /DIV/ p = RHS. For the standard PISO with a real Poisson pressure equation it is pretty clear that the structure of the matrix resulting from discretising this will never change over the time steps. But how is this in my case ? I see that the matrix' structure doesn't really change so I can use the AMG coarsening once and once again without giving a new setup. But this is too heuristic in my eyes. So my question is: Can I argue from any physical point of view that the structure of my matrix stays as it is ? Does anyone have an idea or a hint where I could read about this ? KIndest regards, Sebi
October 26, 2010, 22:19 #2 Senior Member N/A Join Date: Mar 2009 Posts: 189 Rep Power: 17 PISO algorithm was a method proposed for pressure-velocity coupling of incompressible flows. There are many ways in which the algorithm has been used in the literature. Check the paper based on which the code was written. It might be a good starting point to understand. | 414 | 1,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.948668 |
http://slideplayer.com/slide/3191518/ | 1,529,806,634,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865995.86/warc/CC-MAIN-20180624005242-20180624025242-00066.warc.gz | 294,485,553 | 57,729 | # Bell work Do you think energy in the form of work would be needed to go towards the positive charge or towards the negative charge? Why?
## Presentation on theme: "Bell work Do you think energy in the form of work would be needed to go towards the positive charge or towards the negative charge? Why?"— Presentation transcript:
Bell work Do you think energy in the form of work would be needed to go towards the positive charge or towards the negative charge? Why?
Current Electricity, Lesson 1
Physics Coach Stephens
Electric Field & the Movement of Charge
Electric circuits allow us to: cook our food, light our homes, air-condition our work and living space, entertain ourselves with movies and music and even allows us to drive to work or school safely. We will explore the reasons for why charge flows through wires of electric circuits and the variables that affect the rate at which it flows. The means by which moving charge delivers electrical energy to appliances in order to operate them will be discussed in detail.
Understanding Non-Contact Force
You must understand the concept of how an electric field can influence charge within a circuit as it moves from one location to another. Electric force was described as a non-contact force in the last unit on Static Electricity. Two oppositely charged balloons can have an attractive affect on each other even when they are not in contact. The electric force acts over the distance separating the two objects. Electric force is an action-at-a-distance force.
Field Forces Action-at-a-distance forces are sometimes referred to as field forces. The concept of a field force is used to explain this force that occurs in the absence of physical contact. A charged object creates an electric field - an alteration of the space surrounding it. Whether a charged object enters that space or not, the electric field exists. Other charges in that field would feel the alteration of the space. As another charged object enters the space and moves deeper and deeper into the field, the affect of the field becomes more and more noticeable.
Direction of Electric Field
Electric field is a vector quantity whose direction is defined as the direction that a positive test charge go when placed in the field. So, the electric field direction about a positive source charge is directed away from the positive source. And the electric field direction about a negative source charge is directed toward the negative source.
Electric Field, Work, & Potential Energy
External Forces When we discussed gravitational force, it was mentioned that the force of gravity is an internal or conservative force. When gravity does work upon an object, the object's mechanical energy is conserved. Remember, ME = PE + KE -- there was a loss of PE and a gain of KE but ME remains the same. When gravity does work upon an object to move it in the direction of the gravitational field, then the object loses PE because PE is related to the object’s height above the ground. However, energy would be required to move an object against its gravitational field. A stationary object would not naturally move against the field and gain potential energy. Energy in the form of work would have to be given to the object by an external force in order for it to gain this height and potential energy.
Under the Influence…of the Field Force
The important point is that work must be done by an external force to move an object against nature - from low potential energy to high potential energy. On the other hand, objects naturally move from high potential energy to low potential energy under the influence of the field force. It is simply natural for objects to move from high energy to low energy; but work is required to move an object from low energy to high energy.
Moving a Charge in an Electric Field
Similarly, moving a charge in an electric field against its natural direction of motion would require work. The exertion of work by an external force would add PE to the object. The natural direction of motion of an object is from high energy to low energy; but work must be done to move the object against nature. On the other hand, work would not be required to move an object from a high potential energy location to a low potential energy location.
Natural vs. Forced Movement
In Diagram C, the movement of the positive test charge from location A to location B is in the direction of the electric field – this movement is natural, like a mass falling towards earth. Work would not be required to cause such a motion and it would be accompanied by a loss of potential energy. In Diagram D, the movement of the positive test charge from location B to location A is against the electric field -- work would be required to cause this motion; it would be analogous to raising a mass within Earth's gravitational field. Since energy is given to the test charge in the form of work, the positive test charge would be gaining potential energy as the result of the motion. The low energy location for a positive test charge is a location nearest a negative source charge and the high energy location is a location furthest away from a negative source charge.
Review As we begin to discuss circuits, we will apply these principles regarding work and potential energy to the movement of charge about a circuit. Just as we reasoned here, moving a positive test charge against the electric field will require work and result in a gain in potential energy. On the other hand, a positive test charge will naturally move in the direction of the field without the need for work being done on it; this movement will result in the loss of potential energy. Before making this application to electric circuits, we need to first explore the meaning of the concept electric potential.
What is an Electric Circuit?
Electric Potential Moving a positive test charge against the direction of an electric field is like moving a mass upward within Earth's gravitational field. Both movements would be like going against nature and would require work by an external force. This work would increase the potential energy of the object. On the other hand, the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy. Potential energy is the stored energy due to the position of an object and it is related to the location of the object within a field. In this section, we will introduce the concept of electric potential and relate this concept to the potential energy of a positive test charge at various locations within an electric field.
Electric Potential Electric potential is the amount of electric potential energy that would be possessed by a charged object if placed within an electric field. The concept of potential is a location-dependent quantity - it expresses the quantity of potential energy such that it is independent of the amount of charge actually present on the object possessing the electric potential. Electric potential difference is simply the difference in electric potential between two different locations within an electric field.
Illustration of Electric Potential Difference
To illustrate the concept of electric potential difference and the nature of an electric circuit, consider the following situation. There are two metal plates parallel to each other and each being charged with an opposite type of charge . This arrangement of charged plates would create an electric field between the plates that is directed away from the positive plate and towards the negative plate. This movement of a positive test charge would occur without the need of energy in the form of work; it would occur naturally and lower the potential energy of the charge. The positive plate would be the high potential location and the negative plate would be the low potential location. There would be a difference in electric potential between the two locations.
Exchange of + and - Charges
Now suppose that the two oppositely charged plates are connected by a metal wire. What would happen? The wire serves as a pipe through which charge can flow. Over the course of time, positive charges would move from the positive plate through the charge pipe (wire) to the negative plate. positive charge would naturally move in the direction of the electric field that had been created by the arrangement of the two oppositely charged plates. As a positive charge leaves the upper plate, that plate would become less positively charged and as a positive charge reaches the negative plate, that plate would become less negatively charged. Over the course of time, the amount of positive and negative charge on the two plates would slowly diminish. Eventually, the electric field between the plates would become so small that there would be no observable movement of charge between the two plates. The plates would ultimately lose their charge and reach the same electric potential. In the absence of an electric potential difference, there will be no charge flow.
True Circuits The above illustration comes close to demonstrating the meaning of an electric circuit. However, to be a true circuit, charges must continually flow through a complete loop, returning to their original position and cycling through again. If there were a way of moving positive charge from the negative plate back up onto the positive plate, then the movement of positive charge downward through the charge pipe (i.e., the wire) would occur continuously. In such a case, a circuit or loop would be established. A common lab activity that illustrates the necessity of a complete loop utilizes a battery pack (a collection of D cells), a light bulb, and some connecting wires. The activity involves observing the affect of connecting and disconnecting a wire in a simple arrangement of the battery pack, light bulbs and wires. When all connections are made to the battery pack, the light bulb lights. In fact, the lighting of the bulb occurs immediately after the final connection is made. There is no perceivable time delay between when the last connection is made and when the light bulb is perceived to light up.
Internal & External Circuits
When the light bulb lights, charge is moving through the electrochemical cells of the battery, the wires and the light bulb filaments. It can be said that there is a current - a flow of charge within the circuit. The electric circuit demonstrated by the combination of battery, light bulb and wires consists of two distinct parts: the internal circuit and the external circuit. The part of the circuit containing electrochemical cells of the battery is the internal circuit. The part of the circuit where charge is moving outside the battery pack through the wires and the light bulb is the external circuit. In the next lesson, we will focus on the movement of charge through the external circuit. We will explore the requirements that must be met in order to have charge flowing through the external circuit.
Bell work If an electric circuit could be compared to a water circuit at a water park, then the ... ... battery would be analogous to the ____. ... positive terminal of the battery would be analogous to the ____. ... charge would be analogous to the ____. ... electric potential difference would be analogous to the ____. Choices: A. water pressure B. gallons of water flowing down slide per minute C. water D. bottom of the slide E. water pump F. top of the slide
Bell work Answer If an electric circuit could be compared to a water circuit at a water park, then the ... ... battery would be analogous to the water pump (E). ... positive terminal of the battery would be analogous to the top of the slide (F). ... charge would be analogous to the water (C). ... electric potential difference would be analogous to the water pressure (A).
Electric Current
Requirements of a Circuit
Suppose that you were given a small light bulb, an electrochemical cell and a bare copper wire and were asked to find the four different arrangements of the three items that would result in the formation of an electric circuit that would light the bulb. What four arrangements would result in the successful lighting of the bulb? And more importantly, what does each of the four arrangements have in common that would lead us into an understanding of the two requirements of an electric circuit?
Unsuccessful Attempts to Light the Bulb
The successful attempts are characterized by the production of a complete conducting loop from the positive terminal to the negative terminal, with both the battery and the light bulb being part of the loop. A complete conducting loop is made with the light bulb being part of the loop. A circuit exists and charge flows along the complete conducting path, lighting the bulb in the process. Compare the arrangement of the cell, bulb and wire on the left to the unsuccessful arrangements to the right. In attempt A, the wire does not loop back to the negative terminal of the cell. In attempt B, the wire does form a loop but not back to the negative terminal of the cell. In attempt C, there is no complete loop at all. Attempt D resembles attempt B in that there is a loop but not from the positive terminal to the negative terminal. And in attempt E, there is a loop and it does go from positive terminal to negative terminal; this is a circuit but the light bulb is not included as part of it.
Light Bulb Anatomy A light bulb is a relatively simple device consisting of a filament resting upon or somehow attached to two wires. The wires and the filament are conducting materials that allow charge to flow through them. One wire is connected to the ribbed sides of the light bulbs. The other wire is connected to the bottom base of the light bulb. The ribbed edge and the bottom base are separated by an insulating material that prevents the direct flow of charge between the bottom base and the ribbed edge. The only pathway by which charge can make it from the ribbed edge to the bottom base or vice versa is the pathway that includes the wires and the filament. Charge can either enter the ribbed edge, make the pathway through the filament and exit out the bottom base; or it can enter the bottom base, make the pathway through the filament and exit out the ribbed edge. As such, there are two possible entry points and two corresponding exit points.
Alternative Methods The successful means of lighting the bulb as shown involved placing the bottom base of the bulb on the positive terminal and connecting the ribbed edge to the negative terminal using a wire. Any charge that enters the light bulb at the bottom base exits the bulb at the location where the wire makes contact with the ribbed edge. Yet the bottom base does not have to be the part of the bulb that touches the positive terminal. The bulb will light just as easily if the ribbed edge is placed on top of the positive terminal and the bottom base is connected to the negative terminal using a wire. The final two arrangements that lead to a lit light bulb involve placing the bulb at the negative terminal of the cell, either by making contact to it with the ribbed edge or with the bottom base. A wire must then connect the other part of the bulb to the positive terminal of the cell.
Requirement of a Closed Conducting Path
There are two requirements that must be met to establish an electric circuit: There must be a closed conducting path that extends from the positive terminal to the negative terminal. An electric circuit is like a water circuit at a water park. The flow of charge through wires is similar to the flow of water through the pipes and along the slides at a water park. If a pipe gets plugged or broken such that water cannot make the complete path through the circuit, then the flow of water will soon cease. In an electric circuit, all connections must be made by conducting materials capable of carrying charge. Metallic materials are conductors and can be inserted into the circuit to successfully light the bulb. On the other hand, paper and plastic materials are typically insulators and their insertion within the circuit will hinder the flow of charge to such a degree that the current ceases and the bulb no longer lights. There must be a closed conducting loop from the positive to the negative terminal in order to establish a circuit and to have a current.
Burned Out Bulb With an understanding of this first requirement of an electric circuit, it becomes clear what is happening when an incandescent light bulb in a table lamp or floor lamp no longer works. Over time, a light bulb filament becomes weak and brittle can often break or simply become loose. When this occurs, the circuit is opened and a closed conducting loop no longer exists. Without a closed conducting loop, there can be no circuit, no charge flow and no lit bulb.
The Requirement of an Energy Supply
The second requirement of an electric circuit is that there must be an electric potential difference across the two ends of the circuit. This is most commonly established by the use of an electrochemical cell, a pack of cells (i.e., a battery) or some other energy source. It is essential that there is some source of energy capable of increasing the electric potential energy of a charge as it moves from the low energy terminal to the high energy terminal. It takes energy to move a positive test charge against the electric field. As applied to electric circuits, the movement of a positive test charge through the cell from the low energy terminal to the high energy terminal is a movement against the electric field. This movement of charge demands that work be done on it in order to lift it up to the higher energy terminal. An electrochemical cell serves the useful role of supplying the energy to do work on the charge in order to pump it or move it through the cell from the negative to the positive terminal. By doing so, the cell establishes an electric potential difference across the two ends of the electric circuit.
Household Circuits In household circuits, the energy is supplied by a local utility company. The company makes sure that the hot and the neutral plates maintain an electric potential difference of about 110 Volts to 120 Volts (in the United States). In typical lab activities, an electrochemical cell or group of cells is used to establish an electric potential difference across the two ends of the external circuit of about 1.5 Volts (a single cell) or 4.5 Volts (three cells in a pack). Analogies are often made between an electric circuit and the water circuit at a water park or a roller coaster ride at an amusement park. In all three cases, there is something that is moving through a complete loop or a circuit. And in all three cases, it is essential that the circuit include a section where energy is put into the water, the coaster car or the charge in order to move it uphill against its natural direction of motion from a low potential energy to a high potential energy. A water park ride has a water pump that pumps the water from ground level to the top of the slide. A roller coaster ride has a motor-driven chain that carries the train of coaster cars from ground level to the top of the first drop. And an electric circuit has an electrochemical cell, battery or some other energy supply that moves the charge from ground level (the negative terminal) to the positive terminal. By constantly supplying the energy to move the charge from the low energy, low potential terminal to the high energy, high potential terminal, and a continuous flow of charge can be maintained.
Establishing Potential Difference
By establishing this difference in electric potential, charge is able to flow downhill through the external circuit. Like the movement of water at a water park or a roller coaster car at an amusement park, the downhill motion is natural and occurs without the need for energy from an external source. It is the difference in potential that causes the water, the coaster car and the charge to move. This potential difference requires the input of energy from an external source. In the case of an electric circuit, one of the two requirements to establish an electric circuit is an energy source.
Conclusion In conclusion, there are two requirements that must be met in order to establish an electric circuit. The requirements are: There must be an energy supply capable of doing work on a charge to move it from a low energy location to a high energy location. This establishes an electric potential difference across the two ends of the external circuit. There must be a closed conducting loop in the external circuit that stretches from the high potential, positive terminal to the low potential, negative terminal.
CYU #1 Utilize your understanding of the requirements of an electric circuit to state whether charge would flow through the following arrangements of cells, bulbs, wires and switches. If there is no charge flow, then explain why not. a. b. c. d.
Answer CYU #1 A. Charge Flow: No
Explanation: The switch is in the open position, so there is no closed conducting loop. B. Charge Flow: No Explanation: There is no closed conducting loop from + to - terminal. C. Charge Flow: Yes Explanation: In this case, there is a closed conducting loop stretching from the + to the - terminal. Thus, there is a charge flow. However, the flow does not pass through the light bulb and so the bulb will not light. D. Charge Flow: Maybe Explanation: If two cells are pumping charge in opposite directions as indicated by the fact that there + terminals are connected to each other. If they have different voltage ratings (e.g., 1.5 V and 9 V), then there will be a charge flow. However, if their voltage ratings are identical, then there will be no charge flow.
CYU #2 In the movie Tango and Cash, Kurt Russell and Sylvester Stallone escape from a prison by jumping off the top of a tall wall through the air and onto a high-voltage power line. Before the jump, Stallone objects to the idea, telling Russell "We're going to fry." Russell responds with "You didn't take high school Physics did you. As long as you're only touching one wire and you're feet aren't touching the ground, you don't get electrocuted." Is this a correct statement?
In order for there to be a sustained flow of charge from one location to another, there must be a difference in electric potential. In this case, there would be a momentary flow of charge between the wire and the actor until they reach the same electric potential. Once at the same potential, charge flow would cease and there would be no electrocution. However, if the actor's feet touched the ground (an electric potential of 0) or another wire of a different potential, then there would be a sustained charge flow which likely would lead to electrocution.
VA _____ VB _____ VC _____ VD
Bell Work The diagram at the right shows a light bulb connected to a 12-V car battery. The + and - terminals are shown. a. As a + charge moves through the battery from D to A, it ________ (gains, loses) potential energy and ________ (gains, loses) electric potential. The point of highest energy within a battery is the ______ (+, -) terminal. b. As a + charge moves through the external circuit from A to D, it ________ (gains, loses) potential energy and ________ (gains, loses) electric potential. The point of highest energy within the external circuit is closest to the ______ (+, -) terminal. c. Use >, <, and = signs to compare the electric potential (V) at the four points of the circuit. VA _____ VB _____ VC _____ VD
Bell Work Answer a. As a positive charge moves through the battery from D to A, it gains potential energy and gains electric potential. The point of highest energy within a battery is the positive terminal. b. As a positive charge moves through the external circuit from A to D, it loses potential energy and loses electric potential. The point of highest energy within the external circuit is closest to the positive terminal. c. Va = Vb > Vc = Vd
Battery Lab Think about the battery lab that we did on Tuesday.
Why didn’t the ‘thick’ wire conduct the charge? Did any part of the thick wire work better? Why didn’t the ‘thin’ wire conduct the charge? Why did the ‘coated’ wire do the best job of conducting the charge? Why didn’t the alligator clip wires allow the charge to flow?
Power
Putting Charges to Work
Electric circuits are designed to serve a useful function. The movement of charge from terminal to terminal is of little use if the energy is not transformed into another useful form. When a circuit is equipped with a battery and a wire leading from positive to negative terminal without an electrical device (light bulb, beeper, motor, etc.) would lead to a high rate of charge flow – short circuit. charge is flowing rapidly between terminals and the rate at which energy is consumed is high. This circuit would heat the wires to a high temperature and drain the battery of its energy rather quickly. When a circuit is equipped with a light bulb, beeper, or motor, the electrical energy supplied to the charge by the battery is transformed into other forms in the electrical device. A light bulb, beeper and motor are generally referred to as a load. In a light bulb, electrical energy is transformed into useful light energy (and some non-useful thermal energy). In a beeper, electrical energy is transformed into sound energy. And in a motor, electrical energy is transformed into mechanical energy.
Energy Transformation
An electrical circuit is simply an energy transformation tool. Energy is provided to the circuit by an electrical energy source. And energy is delivered by the circuit to the load at the location of the load. The rate at which this energy transformation occurs is of great importance to those who design electrical circuits for useful functions. Previously, power has been described as the rate at which mechanical work is done. In electrical terms, power is the rate at which electrical energy is supplied to a circuit or consumed by a load. The electrical energy is supplied to the load by an energy source such as an electrochemical cell. Recall that a cell does work on a charge to move it from the low energy to the high energy terminal. The work done on the charge is equivalent to the electrical potential energy change of the charge. So, electrical power, like mechanical power, is the rate at which work is done. Like current, power is a rate quantity. Its mathematical formula is expressed on a per time basis.
Power’s Mathematical Formula
Whether the focus is the energy gained by the charge at the energy source or the energy lost by the charge at the load, electrical power refers to the rate at which the charge changes its energy. In an electrochemical cell the change is a positive change (i.e., a gain in energy) and at the load, the change is a negative change (i.e., a loss in energy). So, power is often referred to as the rate of energy change and its equation is expressed as the energy change per time. Like mechanical power, the unit of electrical power is the watt, abbreviated W. (Quite obviously, it is important that the symbol W as the unit of power not be confused with the symbol W for the quantity of work done upon a charge by the energy source.) A watt of power is equivalent to the delivery of 1 joule of energy every second. In other words: 1 watt = 1 joule / second When it is observed that a light bulb is rated at 60 watts, then there are 60 joules of energy delivered to the light bulb every second. A 120-watt light bulbs draws 120 joules of energy every second. The ratio of the energy delivered or expended by the device to time is equal to the wattage of the device.
The Kilowatt-Hour Electrical utility companies who provide energy for homes provide a monthly bill charging those homes for the electrical energy that they used. A typical bill can be very complicated with a number of line items indicating charges for various aspects of the utility service. But somewhere on the bill will be a charge for the number of kilowatt-hours of electricity that were consumed. Exactly what is a kilowatt-hour? Is it a unit of power? time? energy? or some other quantity? And when we pay for the electricity that we use, what exactly is it that we are paying for?
Kilowatt-Hour A careful inspection of the unit kilowatt-hour reveals the answers to these questions. A kilowatt is a unit of power and an hour is a unit of time. So a kilowatt • hour is a unit of Power • time. If Power = Energy / time, then Power • time = Energy. So a unit of power • time is a unit of energy. The kilowatt • hour is a unit of energy. When an electrical utility company charges a household for the electricity that they used, they are charging them for electrical energy. A utility company in the United States is responsible for assuring that the electric potential difference across the two main wires of the house is 110 to 120 volts. Maintaining this difference in potential requires energy.
Mobile Electrons It is a common misconception that the utility company provides electricity in the form of charge carriers or electrons. The fact is that the mobile electrons that are in the wires of our homes would be there whether there was a utility company or not. The electrons come with the atoms that make up the wires of our household circuits. The utility company simply provides the energy that causes the motion of the charge carriers within the household circuits. And when they charge us for a few hundred kilowatt-hours of electricity, they are providing us with an energy bill.
Calculating Power The rate at which energy is delivered to a light bulb by a circuit is related to the electric potential difference established across the ends of the circuit (i.e., the voltage rating of the energy source) and the current flowing through the circuit. The relationship between power, current and electric potential difference can be derived by combining the mathematical definitions of power, electric potential difference and current. Power is the rate at which energy is added to or removed from a circuit by a battery or a load. Current is the rate at which charge moves past a point on a circuit. Electric potential difference across the two ends of a circuit is the potential energy difference per charge between those two points. In equation form, these definitions can be stated as:
Rearranging Equations
Equation 3 above can be rearranged to show that the energy change is the product of the electric potential difference and the charge - V • Q. Substituting this expression for energy change into Equation 1 will yield the following equation: In the equation above, there is a Q in the numerator and a t in the denominator. This is simply the current; and as such, the equation can be rewritten as:
Electric Power The electrical power is simply the product of the electric potential difference and the current. To determine the power of a battery or other energy source, you simply take the electric potential difference and multiply it by the current in the circuit. To determine the power of an electrical device or a load, you simply take the electric potential difference across the device (sometimes referred to as the voltage drop) and multiply it by the current in the device.
Investigate As discussed above, the power delivered to an electrical device in a circuit is related to the current in the device and the electrical potential difference (i.e., voltage) impressed across the device. Use the Electric Power widget below to investigate the affect of varying current and voltage upon the power.
CYU #1 The purpose of every circuit is to supply the energy to operate various electrical devices. These devices are constructed to convert the energy of flowing charge into other forms of energy (e.g., light, thermal, sound, mechanical, etc.). Use complete sentences to describe the energy conversions that occur in the following devices. a. Windshield wipers on a car b. Defrosting circuit on a car
Answer #1 a. Windshield wipers on a car: Electrical energy is converted to mechanical energy. b. Defrosting circuit on a car: Electrical energy is converted to thermal energy.
CYU #2 Determine the ... ... current in a 60-watt bulb plugged into a 120-volt outlet. b. ... current in a 120-watt bulb plugged into a 120-volt outlet. c. ... power of a saw that draws 12 amps of current when plugged into a 120-volt outlet. d. ... power of a toaster that draws 6 amps of current when plugged into a 120-volt outlet. e. ... current in a 1000-watt microwave when plugged into a 120-volt outlet.
Answer #2 For each problem, use the P = V • I equation to solve for the unknown quantity. In a, b, and e, the unknown quantity is current (I); and in c and d, the unknown quantity is power (P). a. The current in a 60-Watt bulb plugged into a 120-Volt outlet is 0.5 A. I = P / V = (60 W) / (120 V) = 0.5 A b. The current in a 120-Watt bulb plugged into a 120-Volt outlet is 1.0 A. I = P / V = (120 W) / (120 V) = 1.0 A c. The power of a saw that draws 12 amps of current when plugged into a 120-Volt outlet is 1440 W. P = V • I = (120 V) • (12 A) = 1440 W d. The power of a toaster that draws 6 amps of current when plugged into a 120-Volt outlet is 720 W. P = V • I = (120 V) • (6 A) = 720 W e. The current in a 1000-Watt microwave when plugged into a 120-Volt outlet is 8.3 A. I = P / V = (1000 W) / (120 V) = 8.3 A
CYU #3 Your 60-watt light bulb is plugged into a 110-volt household outlet and left on for 3 hours. The utility company charges you \$0.11 per kiloWatt•hr. Explain how you can calculate the cost of such a mistake.
Answer #3 The energy consumed can be determined from knowledge of the power (60 Watts) and the time (3 hrs). The energy consumed is 180 Watt•hr or kW•hr. Each kW•hr costs 11 ¢ pr \$0.11. Now simply multiply the cost per ¢ by the amount of ¢ of energy consumed. The result is just a little short of 2 cents.
CYU #4 Alfredo deDarke often leaves household appliances on for no good reason. The deDarke family pays 10¢/kilowatt-hour (i.e., \$.10/kW•hr) for their electrical energy. First, you must convert watts into kilowatts. Equations Power Rating (watt) Time (hrs) Energy Used (kW*hr) Cost (cents) Cost (\$\$) P * t = E 60 watt bulb 1 0.060 kW*hr 0.6 cents \$0.006 4 120 watt bulb 2 t = E / P 100 watt bulb 10 kW*hr E = 1000/cost per hr 1000 cents \$10 P = E / t 100 60 kW*hr
Answer #4 The following relationships exist:
The Power • time = Energy Used. Units are important! The third column is expressed in kiloWatt•hour; thus it is important to convert the Watts to kiloWatts. That is, the first column divided by 1000 multiplied by the second column equals the third column. The cost in cents is equal to the energy used (in kWhr) multiplied by 10 ¢ / kWhr. The cost in dollars is 1/100-th the cost in cents. Power Rating (watt) Time (hrs) Energy Used (kW*hr) Cost (cents) Cost (\$\$) 60 watt bulb 1 0.060 kW*hr 0.6 cents \$0.006 4 0.240 kW*hr 2.4 cents \$0.024 120 watt bulb 2 100 watt bulb 100 10 kW*hr 100 cents \$1.00 1667 100 kW*hr 1000 cents \$10.00 600 watt bulb 60 kW*hr 600 cents \$6.00
Conventional Current Direction
The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. So, these negatively charged electrons move in the direction opposite the electric field. But while electrons are the charge carriers in metal wires, the charge carriers in other circuits can be positive charges, negative charges or both. In fact, the charge carriers in semiconductors, street lamps and fluorescent lamps are simultaneously both positive and negative charges traveling in opposite directions.
Direction of an Electric Current
Ben Franklin, who conducted extensive scientific studies in both static and current electricity, envisioned positive charges as the carriers of charge. As such, an early convention for the direction of an electric current was established to be in the direction that positive charges would move. The convention has stuck and is still used today. The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery. Electrons would actually move through the wires in the opposite direction. Knowing that the actual charge carriers in wires are negatively charged electrons may make this convention seem a bit odd and outdated. Nonetheless, it is the convention that is used worldwide and one that a student of physics can easily become accustomed to.
Current vs Drift Speed Current has to do with the number of coulombs of charge that pass a point in the circuit per unit of time. Because of its definition, it is often confused with the quantity drift speed. Drift speed refers to the average distance traveled by a charge carrier per unit of time. Like the speed of any object, the drift speed of an electron moving through a wire is the distance to time ratio. The path of a typical electron through a wire could be described as a rather chaotic, zigzag path characterized by collisions with fixed atoms. Each collision results in a change in direction of the electron. Yet because of collisions with atoms in the solid network of the metal conductor, there are two steps backwards for every three steps forward. With an electric potential established across the two ends of the circuit, the electron continues to migrate forward. Progress is always made towards the positive terminal. Yet the overall affect of the countless collisions and the high between-collision speeds is that the overall drift speed of an electron in a circuit is abnormally low. A typical drift speed might be 1 meter per hour. That is slow!
How Many Charges How can there by a current on the order of 1 or 2 ampere in a circuit if the drift speed is only about 1 meter per hour? There are many, many charge carriers moving at once throughout the whole length of the circuit. Current is the rate at which charge crosses a point on a circuit. A high current is the result of several coulombs of charge crossing over a cross section of a wire on a circuit. If the charge carriers are densely packed into the wire, then there does not have to be a high speed to have a high current. That is, the charge carriers do not have to travel a long distance in a second, there just has to be a lot of them passing through the cross section. Current does not have to do with how far charges move in a second but rather with how many charges pass through a cross section of wire on a circuit.
Example To illustrate how densely packed the charge carriers are, we will consider a typical wire found in household lighting circuits - a 14-gauge copper wire. In a 0.01 cm-long (very thin) cross-sectional slice of this wire, there would be as many as 3.51 x 1020 copper atoms. Each copper atom has 29 electrons; it would be unlikely that even the 11 valence electrons would be in motion as charge carriers at once. If we assume that each copper atom contributes just a single electron, then there would be as much as 56 coulombs of charge within a thin 0.01-cm length of the wire. With that much mobile charge within such a small space, a small drift speed could lead to a very large current.
Gravitational Potential Depends On…
The amount of gravitational potential energy stored in an object depended upon the amount of mass the object possessed and the amount of height to which it was raised. Gravitational potential energy depended upon object mass and object height. An object with twice the mass would have twice the potential energy and an object with twice the height would have twice the potential energy. It is common to refer to high positions as high potential energy locations. A glance at the diagram at the right reveals the fallacy of such a statement. Observe that the 1 kg mass held at a height of 2 meters has the same potential energy as a 2 kg mass held at a height of 1 meter. Potential energy depends upon more than just location; it also depends upon mass. In this sense, gravitational potential energy depends upon at least two types of quantities: 1) Mass - a property of the object experiencing the gravitational field, and 2) Height - the location within the gravitational field
Electric Potential Energy Depends On…
The direction of the electric field is in the direction that a positive test charge would be pushed. The amount of force involved in doing the work is dependent upon the amount of charge being moved (according to Coulomb's law of electric force). The greater the charge on the test charge, the greater the repulsive force and the more work that would have to be done on it to move it the same distance. If two objects of different charge - with one being twice the charge of the other - are moved the same distance into the electric field, then the object with twice the charge would require twice the force and thus twice the amount of work. This work would change the potential energy by an amount that is equal to the amount of work done. Thus, the electric potential energy is dependent upon the amount of charge on the object experiencing the field and upon the location within the field. Just like gravitational potential energy, electric potential energy is dependent upon at least two types of quantities: 1) Electric charge - a property of the object experiencing the electrical field, And 2) Distance from source - the location within the electric field
Common Misconceptions
In these first two lessons of this unit, an effort has been made to present a model of how and why electric charge flows within an electric circuit. The goal has been to help students of physics to construct an accurate mental model of the world of current electricity. This goal is often impeded by preconceived ideas regarding the nature of charge flow and the role of a battery in a circuit. In many instances, these preconceived notions are incorrect ideas and are completely inconsistent with the model presented here. Like all misconceptions in physics, they must be directly confronted in order to successfully build an accurate mental model of the physical world.
What do you Believe? True or False?
a. When an electrochemical cell no longer works, it is out of charge and must be recharged before it can be used again. b. An electrochemical cell can be a source of charge in a circuit. The charge that flows through the circuit originates in the cell. c. Charge becomes used up as it flows through a circuit. The amount of charge that exits a light bulb is less than the amount that enters the light bulb. d. Charge flows through circuits at very high speeds. This explains why the light bulb turns on immediately after the wall switch is flipped. e. The local electrical utility company supplies millions and millions of electrons to our homes everyday.
Charge Flow An electrochemical cell supplies the energy needed to move a charge from a low potential location to a high potential location. The charge that flows through a circuit originates in the wires of the circuit. The charge carriers in wires are simply the electrons possessed by the atoms that make up the wires. Charge moves abnormally slowly - on average, about 1 meter in an hour - through a circuit. Yet as soon as a switched is turned to ON, charge located everywhere within the circuit begins to move. The rate at which charge flows is the same everywhere within an electric circuit. The rate at which charge flows into a light bulb is the same as the rate at which charge flows out of a light bulb. An electrical appliance such as a light bulb transforms the electrical energy of moving charge into other forms of energy such as light energy and thermal energy. So, the amount of electrical energy possessed by a charge as it exits an appliance is less than it possessed when it entered the appliance.
Are Batteries Rechargeable?
If an electrochemical cell is not rechargeable, then why do stores sell rechargeable cells for a higher cost? What kind of rip-off is that? The fact is that electrochemical cells that are referred to as rechargeable can be bought in stores. And these batteries can be placed in small machines that are called rechargers. And the process of doing so can extend the life of the battery. So as far as the consumer is concerned, it really isn't a rip-off at all. But as far as physics is concerned, it is a major offense because batteries should never be referred to as rechargeable.
Energy, Not Charge Electric circuits are all about energy, not charge.
When a battery no longer works, it is out of energy. A battery operates by packing a collection of reactive chemicals inside. These chemicals undergo an oxidation-reduction reaction that produces energy. This energy-producing reaction is capable of pumping the charge through the battery from low energy to high energy and establishing the electric potential difference across the external circuit. When a battery no longer works, it is because the chemicals have been consumed to the point that the battery’s ability to move the charge has been severely diminished. When a battery no longer works, it is because the conversion of reactants to products have occurred to the extent that the energy- producing reaction is no longer able to do its job of pumping charge.
Re-Energizing Some batteries are said to be rechargeable because this problem of the consumption of chemical reactants can be easily fixed. Such so-called rechargeable batteries rely upon a reversible reaction. The reaction can be run in the reverse direction, turning the chemical products back into chemical reactants within the cell. Since the usual reaction which powers the circuit is an exothermic reaction (a fancy chemistry name for energy-producing), the reverse reaction is an endothermic reaction which requires energy in order to work. By placing the cell into a so-called recharger, the energy of a household electrical circuit can be used to drive the reaction in the reverse direction and transform the chemical products back into chemical reactants. This reverse process requires energy; it is the recharger which supplies the energy. With reactants replenished, the cell can now be used again to power the electric circuit. A true understanding of this process would lead one to refer to such cells as reversible or re-energizable; and the machines that are used to reverse the reaction would be properly referred to as re-energizers.
Charge Migration Electric circuits are all about energy, not charge.
The charge is simply the medium which moves the energy from location to location. The energy source does work upon the charge to supply it with energy and place it at a high electric potential. Charge at high electric potential will spontaneously begin its very slow migration towards the low potential terminal of the cell. Charge everywhere within the circuit moves together, like soldiers marching in step. As an individual charge moves through circuit elements such as light bulbs, its electrical energy is transformed into other forms of energy such as light energy and thermal energy. With many, many charges moving through the light bulb at the same time, there is a significant transformation of electrical energy to light energy to cause the light bulb filament to noticeably glow. Upon passage through a light bulb filament, an individual charge is less energized and at a lower electric potential. The charge completes its slow migration back to the low potential terminal where the electrochemical cell does work upon the charge again to move it back up to high electric potential. Once at high potential, the charge can begin its loop again through the external circuit.
Electrical Resistance
Journey of a Typical Electron
As mentioned in Lesson 1, an electrochemical cell supplies energy to move a charge from its low energy, low potential terminal to the high energy, high potential terminal. In this sense, the cell supplies the energy to establish an electric potential difference across the two ends of the external circuit. Charge will then flow through the external circuit in the same manner that water will flow from an elevated position to a low position. It is the difference in potential that causes this flow.
Charge Carriers In the wires of electric circuits, an electron is the actual charge carrier. As mentioned in Lesson 2, an electron's path through the external circuit is far from being a straight path. An electron's journey through a circuit can be described as a zigzag path that results from countless collisions with the atoms of the conducting wire. Each collision results in the alteration of the path, thus leading to a zigzag type motion. While the electric potential difference across the two ends of a circuit encourages the flow of charge, it is the collisions that discourages the flow of charge. Different types of atoms offer a different degree of hindrance to the flow of the charge carriers that pass through it.
Loss of Energy In all cases, these collisions result in a loss of energy. While most of the electrical energy possessed by a charge carrier is lost when it passes through an electrical device (often referred to as the load), even the wires of the circuit themselves act to remove energy from a charge. It is because of this energy loss in the load and in the wires themselves that the electric potential of a charge carrier is decreased as it traverses the external circuit. The electric energy supplied by the electrochemical cells becomes entirely used up in the external circuit.
More Electrical Devices = Greater Loss of Electric Potential
In an electric circuit with several electrical devices, there may be multiple losses of electric potential as the charge traverses the circuit. The total loss of electric potential of a single charge as it passes through the external circuit is equal to the gain in electric potential that it experiences in the battery. As depicted in the diagram below, a charge carrier traversing the external circuit from A to H passes through three different light bulbs. Each light bulb results in a loss of electric potential for the charge. This loss in electric potential corresponds to a loss of energy as the electrical energy is transformed into light energy and thermal energy. In addition to the changes in electric potential and electric energy that occur in the light bulbs, there is also a smaller amount of electric potential loss in the wires that connect the light bulbs. This small amount of loss in electric potential also corresponds to a small loss of energy as the electrical energy is transformed into thermal energy. The wires get hot - not as hot as the light bulb, but still measurably hot.
Voltage Drop So the journey of an electron through an external circuit involves a long and slow zigzag path that is characterized by several successive losses in electric potential. Each loss of potential is referred to as a voltage drop. Accompanying this voltage drop is a voltage boost occurring within the internal circuit - for instance, within the electrochemical cell. In the next part of Lesson 3 we will look extensively at resistance - the cause for these voltage drops.
Resistance An electron traveling through the wires and loads of the external circuit encounters resistance. Resistance is the hindrance to the flow of charge. For an electron, the journey from terminal to terminal is not a direct route. The electrons encounter resistance - a hindrance to their movement – due to their countless collisions with other atoms. While the electric potential difference established between the two terminals encourages the movement of charge, it is resistance that discourages it. The rate at which charge flows from terminal to terminal is the result of the combined affect of these two quantities.
Variables Affecting Electrical Resistance
The flow of charge through wires is often compared to the flow of water through pipes. The resistance to the flow of charge in an electric circuit is equivalent to the frictional affects between water and the pipe surfaces as well as the resistance offered by obstacles that are present in its path. It is this resistance that hinders the water flow and reduces both its flow rate and its drift speed. Like the resistance to water flow, the total amount of resistance to charge flow within a wire of an electric circuit is affected by some clearly identifiable variables.
Length & Area First, the total length of the wires will affect the amount of resistance. The longer the wire, the more resistance that there will be. There is a direct relationship between the amount of resistance encountered by charge and the length of wire it must travel across. After all, if resistance occurs as the result of collisions between charge carriers and the atoms of the wire, then there is likely to be more collisions in a longer wire. More collisions mean more resistance. Second, the cross-sectional area of the wires will affect the amount of resistance. Wider wires have a greater cross-sectional area and there will be less resistance. Water will flow through a wider pipe at a higher rate than it will flow through a narrow pipe. This can be attributed to the lower amount of resistance that is present in the wider pipe. When all other variables are the same, charge will flow at higher rates through wider wires with greater cross-sectional areas than through thinner wires. Resistance and cross-sectional area are INVERSELY PROPORTIONAL! The cross-sectional area of a circular cross-section is given by the expression Pi•r2
Material A third variable that is known to affect the resistance to charge flow is the material that a wire is made of. Some materials are better conductors than others and offer less resistance to the flow of charge. Silver is one of the best conductors but is never used in wires of household circuits due to its cost. Copper and aluminum are among the least expensive materials with suitable conducting ability to permit their use in wires of household circuits. The conducting ability of a material is often indicated by its resistivity. The resistivity of a material is dependent upon the material's electronic structure and its temperature. For most (but not all) materials, resistivity increases with increasing temperature. The table on the next slide lists resistivity values for various materials at temperatures of 20 degrees Celsius.
Resistivity (ohm•meter)
Resistivity Values Material Resistivity (ohm•meter) Silver 1.59 x 10-8 Copper 1.7 x 10-8 Gold 2.4 x 10-8 Aluminum 2.8 x 10-8 Tungsten 5.6 x 10-8 Iron 10 x 10-8 Platinum 11 x 10-8 Lead 22 x 10-8 Nichrome 150 x 10-8 Carbon 3.5 x 10-5 Polystyrene Polyethylene Glass Hard Rubber 1013
Lower Resistivity = Better Conductor
As seen in the table, there is a broad range of resistivity values for various materials. Those materials with lower resistivities offer less resistance to the flow of charge; they are better conductors. The materials shown in the last four rows of the table have such high resistivity that they would not even be considered to be conductors.
Mathematical Nature of Resistance
Resistance is a numerical quantity that can be measured and expressed mathematically. The standard metric unit for resistance is the ohm, represented by the Greek letter omega An electrical device having a resistance of 5 ohms would be represented as R = The equation representing the dependency of the resistance (R) of a wire upon the variables that affect it is: where L represents the length of the wire (in meters), A represents the cross-sectional area of the wire (in meters2), and represents the resistivity of the material (in ohm•meter). Consistent with the previous discussion, this equation shows that the resistance of a wire is directly proportional to the length of the wire and inversely proportional to the cross-sectional area of the wire. As shown by the equation, knowing the length, cross-sectional area and the material that a wire is made of (and thus, its resistivity) allows one to determine the resistance of the wire.
Investigate! Resistors are one of the more common components in electrical circuits. Most resistors have stripes or bands of colors painted on them. The colors reveal information about the resistance value. Perhaps you're doing a lab and need to know the resistance of a resistor used in the lab. Use the link below to determine the resistance value from the colored stripes.
CYU #1 Household circuits are often wired with two different widths of wires: 12-gauge and 14-gauge. The 12-gauge wire has a diameter of 1/12 inch while the 14-gauge wire has a diameter of 1/14 inch. Thus, 12-gauge wire has a wider cross section than 14-gauge wire. A 20-Amp circuit used for wall receptacles should be wired using 12-gauge wire and a 15-Amp circuit used for lighting and fan circuits should be wired using 14-gauge wire. Explain the physics behind such an electrical code.
Answer #1 A 12-gauge wire is wider than 14-gauge wire and thus has less resistance. The lesser resistance of 12-gauge wire means that it can allow charge to flow through it at a greater rate - that is, allow a larger current. Thus, 12-gauge wire is used in circuits which are protected by 20-Amp fuses and circuit breakers. On the other hand, the thinner 14-gauge wire can support less current due to its larger resistance; it is used in circuits which are protected by 15-Amp fuses and circuit breakers.
CYU #2 Determine the resistance of a 1-mile length of 12-gauge copper wire. Given: 1 mile = 1609 meters diameter = cm.
Answer #2 Answer: 7.8 ohms First find the cross-sectional area:
Use the equation: where L = 1609 m A = Pi • r2 (in meters2), = 1.7 x 10-8 ohm•meter. First find the cross-sectional area: The formula for finding the AREA of a circle is: A = Pi • r2 Pi = 3.14, r = radius If we have the diameter, how do we find the radius? We also need to convert the diameter of cm into m A = (Pi) • (diameter/2)2 -- or -- A = (Pi) • (radius)2 A = (3.14) • [( m) / 2)]2 A = x 10-6 m2 Now substitute into the above equation to determine the resistance. R = (1.7 x 10-8 ohm •m) • (1609 m) / (3.519 x 10-6 m2) R = 7.8 ohms ( ohm)
Ohm’s Law
Ohm’s Law There are certain formulas in Physics that are so powerful that they reach the state of popular knowledge. A student of Physics has written such formulas down so many times that they have memorized it without trying to. Certainly to the professionals in the field, such formulas are so central that they become engraved in their minds. In the field of Modern Physics, there is E = m • c2. In the field of Newtonian Mechanics, there is Fnet = m • a. In the field of Wave Mechanics, there is v = f • . And in the field of current electricity, there is V = I • R.
V = I * R The predominant equation which encompasses the study of electric circuits is the equation V = I • R. In words, the electric potential difference between two points on a circuit (V) is equivalent to the product of the current between those two points (I) and the total resistance of all electrical devices present between those two points (R). Through the rest of this unit, this equation will become the most common equation we see. Often referred to as the Ohm's law equation, this equation is a powerful predictor of the relationship between potential difference, current and resistance.
Ohm’s Law as a Predictor of Current
The Ohm's law equation can be rearranged and expressed as: This equation indicates the two variables that would affect the amount of current in a circuit. The current in a circuit is directly proportional to the electric potential difference across its ends and inversely proportional to the total resistance offered by the external circuit. The greater the battery voltage (i.e., electric potential difference), the greater the current. And the greater the resistance, the less the current. Charge flows at the greatest rates when the battery voltage is increased and the resistance is decreased. In fact, a twofold increase in the battery voltage would lead to a twofold increase in the current (if all other factors are kept equal). And an increase in the resistance of the load by a factor of two would cause the current to decrease by a factor of two to one-half its original value.
Illustrations The table below illustrates this relationship both qualitatively and quantitatively for several circuits with varying battery voltages and resistances. Circuit Diagram Battery Voltage Total Resistance Current 1. 1.5 V 3 0.50 Amp 2. 3.0 V 1 Amp 3. 4.5 V 1.5 Amp 4. 6 0.25 Amp 5. 0.5 Amp 6. 0.75 Amp 7. 9
Common Examples Rows 1, 2 and 3 illustrate that the doubling and the tripling of the battery voltage leads to a doubling and a tripling of the current in the circuit. Comparing rows 1 and 4 or rows 2 and 5 illustrates that the doubling of the total resistance serves to halve the current in the circuit. Because the current in a circuit is affected by the resistance, resistors are often used in the circuits of electrical appliances to affect the amount of current that is present in its various components. By increasing or decreasing the amount of resistance in a particular branch of the circuit, a manufacturer can increase or decrease the amount of current in that branch. Kitchen appliances such as electric mixers and light dimmer switches operate by altering the current at the load by increasing or decreasing the resistance of the circuit. Pushing the various buttons on an electric mixer can change the mode from mixing to beating by reducing the resistance and allowing more current to be present in the mixer. Similarly, turning a dial on a dimmer switch can increase the resistance of its built-in resistor and thus reduce the current.
The diagram below depicts a couple of circuits containing a voltage source (battery pack), a resistor (light bulb) and an ammeter (for measuring current). In which circuit does the light bulb have the greatest resistance?
Use the equation V = I • R to solve for the resistance. Diagram A:
Answer: Diagram A Use the equation V = I • R to solve for the resistance. Diagram A: R = V / I = (6 V) / (1 A) = 6 Diagram B: R = V / I = (6 V) / (2 A) = 3 The resistance is greatest in the circuit with the least current - Diagram A.
Common Lab Equipment The Ohm's law equation is often explored in physics labs using a resistor, a battery pack, an ammeter, and a voltmeter. An ammeter is a device used to measure the current at a given location. A voltmeter is a device equipped with probes that can be touched to two locations on a circuit to determine the electric potential difference across those locations. By altering the number of cells in the battery pack, the electric potential difference across the external circuit can be varied. The voltmeter can be used to determine this potential difference and the ammeter can be used to determine the current associated with this V. A battery can be added to the battery pack and the process can be repeated several times to yield a set of I-V data.
Quantities, Equations, Symbols & Units
The tendency to give attention to units is an essential trait of any good physics student. Many of the difficulties associated with solving problems may be traced back to the failure to give attention to units. As more and more electrical quantities and their respective metric units are introduced in this unit, it will become increasingly important to organize the information in your head. The table on the next slide lists several of the quantities that have been introduced thus far. The symbol, the equation and the associated metric units are also listed for each quantity. It would be wise to refer to this list often or even to make your own copy and add to it as the unit progresses. Some students find it useful to make a fifth column in which the definition of each quantity is stated.
Potential Difference (a.k.a. voltage)
Table of Information Quantity Symbol Equation(s) Standard Metric Unit Other Units Potential Difference (a.k.a. voltage) V V = PE / Q V = I • R Volt (V) J / C Current I I = Q / t I = V / R Amperes (A) Amp or C / s or V / Power P P = PE / t (more to come) Watt (W) J / s Resistance R R = • L / A R = V / I Ohm () V / A Energy E or PE PE = V • Q PE = P • t Joule (J) V • C or W • s
CYU #1 Which of the following will cause the current through an electrical circuit to decrease? Choose all that apply. a. decrease the voltage b. decrease the resistance c. increase the voltage d. increase the resistance
Answer CYU #1 Answers: A and D
The current in a circuit is directly proportional to the electric potential difference impressed across the circuit and inversely proportional to the resistance of the circuit. Reducing the current can be done by reducing the voltage (choice A) or by increasing the resistance (choice D).
CYU #2 A certain electrical circuit contains a battery with three cells, wires and a light bulb. Which of the following would cause the bulb to shine less brightly? Choose all that apply. a. increase the voltage of the battery (add another cell) b. decrease the voltage of the battery (remove a cell) c. decrease the resistance of the circuit d. increase the resistance of the circuit
Answer CYU #2 Answers: B and D
The bulb will shine less brightly if the current in it is reduced. Reducing the current can be done by reducing the electric potential difference impressed across the bulb (choice B) or by increasing the resistance of the bulb (choice D).
CYU #3 You have likely been warned to avoid contact with electrical appliances or even electrical outlets when your hands are wet. Such contact is more dangerous when your hands are wet (vs. dry) because wet hands cause ____. a. the voltage of the circuit to be higher b. the voltage of the circuit to be lower c. your resistance to be higher d. your resistance to be lower e. the current through you to be lower
Answer CYU #3 Answer: D Wet hands have less resistance and thus less hindrance to the flow of charge; the current would thus be increased. Touching an outlet with wet hands increases the risk of charge flowing through you and causing electrical shock or even electrocution.
CYU #4 If the resistance of a circuit were tripled, then the current through the circuit would be ____. a. one-third as much b. three times as much c. unchanged d. ... nonsense! There would be no way to make such a prediction.
Answer CYU #4 Answer: A Current is inversely proportional to the resistance. A threefold increase in the resistance would cause a threefold decrease in the current.
Circuit Connections
Circuit Symbols and Circuit Diagrams
Thus far, this unit has focused on the key ingredients of an electric circuit and upon the concepts of electric potential difference, current and resistance. Conceptual meaning of terms have been introduced and applied to simple circuits. Mathematical relationships between electrical quantities have been discussed and their use in solving problems has been modeled. Lesson 4 will focus on the means by which two or more electrical devices can be connected to form an electric circuit. Our discussion will progress from simple circuits to mildly complex circuits. Former principles of electric potential difference, current and resistance will be applied to these complex circuits and the same mathematical formulas will be used to analyze them.
Descriptions Electric circuits, whether simple or complex, can be described in a variety of ways. An electric circuit is commonly described with mere words. Saying something like "A light bulb is connected to a D-cell" is a sufficient amount of words to describe a simple circuit. Upon hearing (or reading) the words, a person grows accustomed to quickly picturing the circuit in their mind. But another means of describing a circuit is to simply draw it. Such drawings provide a quicker mental picture of the actual circuit. Circuit drawings like the one below have been used many times in Lessons 1 through 3. Describing Circuits with Words "A circuit contains a light bulb and a 1.5-Volt D-cell."
Schematic Diagram A final means of describing an electric circuit is by use of conventional circuit symbols to provide a schematic diagram of the circuit and its components. Some circuit symbols used in schematic diagrams are shown below.
+ and - Terminals A single cell or other power source is represented by a long and a short parallel line. A collection of cells or battery is represented by a collection of long and short parallel lines. In both cases, the long line is representative of the positive terminal of the energy source and the short line represents the negative terminal. A straight line is used to represent a connecting wire between any two components of the circuit. An electrical device that offers resistance to the flow of charge is generically referred to as a resistor and is represented by a zigzag line. An open switch is generally represented by providing a break in a straight line by lifting a portion of the line upward at a diagonal. These circuit symbols will be frequently used throughout the remainder of Lesson 4 as electric circuits are represented by schematic diagrams.
Example #1 Description with Words:
Three D-cells are placed in a battery pack to power a circuit containing three light bulbs.
Example #2 Description with Words: Three D-cells are placed in a battery pack to power a circuit containing three light bulbs.
Two Types of Connections
When there are two or more electrical devices present in a circuit with an energy source, there are a couple of basic means by which to connect them. They can be connected in series or connected in parallel. Suppose that there are three light bulbs connected together in the same circuit. If connected in series, then they are connected in such a way that an individual charge would pass through each one of the light bulbs in consecutive fashion. When in series, charge passes through every light bulb. If connected in parallel, a single charge passing through the external circuit would only pass through one of the light bulbs. The light bulbs are placed within a separate branch line, and a charge traversing the external circuit will pass through only one of the branches during its path back to the low potential terminal. The means by which the resistors are connected will have a major affect upon the overall resistance of the circuit, the total current in the circuit, and the current in each resistor.
Series Circuit So for series circuits, as more resistors are added the overall current within the circuit decreases. This decrease in current is consistent with the conclusion that the overall resistance increases.
Parallel Circuits It is clear from observing the indicator bulbs in the diagrams that the addition of more resistors causes the indicator bulb to get brighter. For parallel circuits, as the number of resistors increases, the overall current also increases. This increase in current is consistent with a decrease in overall resistance. Adding more resistors in a separate branch has the unexpected result of decreasing the overall resistance!
Example
Equivalent Resistance in Series
Charge flows together through the external circuit at a rate that is everywhere the same. The current is no greater at one location as it is at another location. The actual amount of current varies inversely with the amount of overall resistance. There is a clear relationship between the resistance of the individual resistors and the overall resistance of the collection of resistors. As far as the battery that is pumping the charge is concerned, the presence of two 6-ohm resistors in series would be equivalent to having one 12-ohm resistor in the circuit. The presence of three 6-ohm resistors in series would be equivalent to having one 18-ohm resistor in the circuit. And the presence of four 6-ohm resistors in series would be equivalent to having one 24-ohm resistor in the circuit.
Continued… This is the concept of equivalent resistance.
The equivalent resistance of a circuit is the amount of resistance that a single resistor would need in order to equal the overall affect of the collection of resistors that are present in the circuit. For series circuits, the mathematical formula for computing the equivalent resistance (Req) is: Req = R1 + R2 + R where R1, R2, and R3 are the resistance values of the individual resistors that are connected in series.
Water Analogy Throughout this unit, there has been an extensive reliance upon the analogy between charge flow and water flow. The flow of charge in wires is analogous to the flow of water in pipes. Consider the diagrams below in which the flow of water in pipes becomes divided into separate branches. At each node (branching location), the water takes two or more separate pathways. The rate at which water flows into the node (measured in gallons per minute) will be equal to the sum of the flow rates in the individual branches beyond the node. Similarly, when two or more branches feed into a node, the rate at which water flows out of the node will be equal to the sum of the flow rates in the individual branches that feed into the node.
Charge Flow The same principle of flow division applies to electric circuits. The rate at which charge flows into a node is equal to the sum of the flow rates in the individual branches beyond the node. This is illustrated in the examples shown below. In the examples a new circuit symbol is introduced - the letter A enclosed within a circle. This is the symbol for an ammeter - a device used to measure the current at a specific point. An ammeter is capable of measuring the current while offering negligible resistance to the flow of charge.
Equivalent Resistance in Parallel
The actual amount of current always varies inversely with the amount of overall resistance. To explore this relationship, let's begin with the simplest case of two resistors placed in parallel branches, each having the same resistance value of 4 ohms . Since the circuit offers two equal pathways for charge flow, only one-half the charge will choose to pass through a given branch. While each individual branch offers 4 ohms of resistance, only one-half of all the charge flowing through the circuit will encounter the 4 ohm resistance of that individual branch. Thus, as far as the battery that is pumping the charge is concerned, the presence of two 4-ohm resistors in parallel would be equivalent to having one 2-ohm resistor in the circuit.
Continued… This is the concept of equivalent resistance.
The equivalent resistance of a circuit is the amount of resistance that a single resistor would need in order to equal the overall effect of the collection of resistors that are present in the circuit. For parallel circuits, the mathematical formula for computing the equivalent resistance (Req) is: 1 / Req = 1 / R1 + 1 / R2 + 1 / R where R1, R2, and R3 are the resistance values of the individual resistors that are connected in parallel. The examples above could be considered simple cases in which all the pathways offer the same amount of resistance to an individual charge that passes through it. The simple cases above were done without the use of the equation. Yet the equation fits both the simple cases where branch resistors have the same resistance values and the more difficult cases where branch resistors have different resistance values. For instance, consider the application of the equation to the one simple and one difficult case below.
Voltage Drop in Parallel
Electric potential diagrams were introduced in Lesson 1 of this unit and subsequently used to illustrate the consecutive voltage drops occurring in series circuits. An electric potential diagram is a conceptual tool for representing the electric potential difference between several points on an electric circuit. Consider the circuit diagram below and its corresponding electric potential diagram.
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Equations
Equations with parentheses
Steps to solve an equation with parentheses:
1. Rewrite the equation without parentheses by using the distributive property.
2. Combine like terms.
3. Collect variables on one side and constants on the other.
4. Divide by the coefficient of the variable.
Solve for m:
$\fs22(m-4)=5m+7\;\Rightarrow\;\;2m-8=5m+7\;\Rightarrow\;2m-5m=7+8\;\Rightarrow\;-3m=15\;\Rightarrow\;x=\frac{15}{-3}\;\;\Rightarrow\;x=-5$
Solve for x:
$8(3x+5)-1=-68-7(2x+1)$ Solution: | 168 | 516 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 2, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-51 | latest | en | 0.657063 |
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# Stats 301 Spring 2011(1) - INTRODUCTION TO SOCIAL...
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INTRODUCTION TO SOCIAL STATISTICS SOC 301-01, Spring 2011 Tue-Th 11 am-12:15 pm LUTZ Hall Sociology Department Computer Lab Room #131 Instructor: Lauren Heberle Office: 108 Lutz Hall Office Hours: Tues-Thurs 12:30-1:30 p.m. or by appointment Office Phone: 852-4749, Fax: 852-0099 email: [email protected] Teaching Assistant: Janidean Bruner Office 113 Lutz Office phone: (502) 852-8043 Office hours: Tues & Thurs 9:30-10:30 or by appointment Important Notes : (1) Completion of general education math requirement is a prerequisite for this course. Note: Credit may not be earned for this course and MATH 109, PSYC 301 or MGMT 201. (2) This is a required course for Sociology Majors. Course Description : Be not afraid, statistics are everywhere! The goal of this course is to begin to learn how to use statistics to make sense of our social world. We will use a variety of methods to describe and analyze information about society so that we might make better decisions and critically assess decisions made by others. You will see how statistics can be used in better and not so better ways to support declarations and predictions about our social world. The course will first examine how social scientists gather numerical or quantitative data. We will discuss ways to organize, describe, and summarize numerical data so that we can see and understand patterns in the numbers. An important subject in statistics is probability theory. We will learn what it is and why it is the basis for the majority of statistical operations. Once we have built that foundation, we will cover methods or statistical tests for determining if variation exists and whether there is a demonstrable relationship between the variables in our data. Since it is important to communicate our results to others, we will learn how to talk about the results in everyday language and with accuracy. Throughout this course we will learn how to use SPSS/PASW which is a computer program that assists in analyzing data sets. Therefore the class will meet in Lutz Hall Sociology Department Computer Lab.
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```OpenStax-CNX module: m0052
1
Summary of Series and Parallel
Combination Rules
∗
Don Johnson
This work is produced by OpenStax-CNX and licensed under the
†
Abstract
Series and parallel combination rules.
Figure 1 (series and parallel combination rules) summarizes the series and parallel combination results.
These results are easy to remember and very useful. Keep in mind that for series combinations, voltage
and resistance are the key quantities, while for parallel combinations current and conductance are more
important. In series combinations, the currents through each element are the same; in parallel ones, the
voltages are the same.
∗ Version 2.5: Aug 10, 2004 11:17 am -0500
http://legacy.cnx.org/content/m0052/2.5/
OpenStax-CNX module: m0052
2
series and parallel combination rules
(a) series combination rule
Figure 1:
in =
Gn
GT
Series and parallel combination rules. (a)
(b) parallel combination rule
RT =
PN
n=1
Rn
vn =
Rn
RT
v
(b)
GT =
PN
n=1
Gn
i
Exercise 1
(Solution on p. 3.)
Contrast a series combination of resistors with a parallel one. Which variable (voltage or current)
is the same for each and which diers? What are the equivalent resistances? When resistors are
placed in series, is the equivalent resistance bigger, in between, or smaller than the component
resistances? What is this relationship for a parallel combination?
http://legacy.cnx.org/content/m0052/2.5/
OpenStax-CNX module: m0052
3
Solutions to Exercises in this Module
Solution to Exercise (p. 2)
In a series combination of resistors, the current is the same in each; in a parallel combination, the voltage
is the same. For a series combination, the equivalent resistance is the sum of the resistances, which will be
larger than any component resistor's value; for a parallel combination, the equivalent conductance is the sum
of the component conductances, which is larger than any component conductance. The equivalent resistance
is therefore smaller than any component resistance.
http://legacy.cnx.org/content/m0052/2.5/
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