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Peskin & Schroeder eq. 9.26 and functional methods

I have been reading chapter 9 in Peskin & Schroeder's QFT book and has been stuck in transition from equation 9.26 to 9.27. Equation 9.26 reads: $$\frac{1}{V^2} \Sigma_{m,l} \exp{[-i(k_m.x_1+k_l .x_2)]}(\prod_{k_{n}>0}\int{d\ Re\phi_{n}\ d\ Im\ \phi_{n}}) * (Re\ \phi_{m}+iIm\ \phi_{m})(Re \ \phi_{l}+iIm\ \phi_{l}) * \exp[\frac{-i}{V}\Sigma_{k^{0}_{n}>0}(m^2-k^2_{n})[(Re\ \phi_{n})^2+(Im\ \phi_{n})^2]] \tag{9.26}$$

and with $$k_l=+k_m$$, equation vanishes while with $$k_l = -k_m$$ the equation becomes,

$$\frac{1}{V^2}\Sigma_{m} \exp[-ik_m\ .(x_1-x_2)](\prod_{k^{0}_{n>0}}\frac{-i\pi V}{m^2-k^{2}_n})\ \frac{-iV}{m^2-k^2_{m}-i\epsilon}$$

I can work out the term in parenthesis using Gaussian integrals but I can't seem to wrap my head around how we get the term

$$\frac{-iV}{m^2-k^2_{m}-i\epsilon}$$

Can anybody point me in the right direction please? Apparently $$(Re\ \phi_{m})^2$$ adds up to the $$(Im\ \phi_m)^2$$ when $$k_l = - k_m$$ and we get this contribution. I don't understand how.

• Try to see the discussion of replacement $t\rightarrow (t-i\epsilon)$ after (9.23). It seem that the gaussian integral has only imaginary exp, therefore we have to do analytic continuation. May be it will help. – Artem Alexandrov Mar 15 '19 at 18:58
• I mean that $\epsilon$-prescription does not come directly in (9.26)$\rightarrow$(9.27). P&S state that in all formulas after (9.21) one should replace $k^2-m^2\rightarrow k^2-m^2+i\epsilon$ to make integrals convergent – Artem Alexandrov Mar 15 '19 at 19:01
• I understand that but what I fail to grasp is how the two parts of $\phi$ add up and become this term. I tried to follow along from the definition of fourier expansion of $\phi$ in 9.21 but I am probably too naive to see it... – Blitz Mar 17 '19 at 2:34

Consider the case when $$k_l=-k_m$$. Peskin and Shroeder equation 9.26 would then read
$$\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\int d\mathrm{Re}\phi_n\ d\mathrm{Im}\phi_n \right)\times (\mathrm{Re}\phi_m\ \mathrm{Re}\phi_{-m} +i\ \mathrm{Re}\phi_m\ \mathrm{Im}\phi_{-m}+i\ \mathrm{Im}\phi_m\ \mathrm{Re}\phi_{-m}-\mathrm{Im}\phi_m\ \mathrm{Im}\phi_{-m}) \times \exp\left[-\frac{i}{V}\sum_{k_n^0>0}(m^2-k_n^2)[(\mathrm{Re}\phi_n)^2+(\mathrm{Im}\phi_n)^2] \right] .$$
In the first term, using $$\mathrm{Re}\phi_{-m}=\mathrm{Re}\phi_m^* = \mathrm{Re}\phi_m$$, we get
\begin{align} &\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\int d\mathrm{Re}\phi_n\ d\mathrm{Im}\phi_n \right)\times (\mathrm{Re}\phi_m)^2 \times e^{-\frac{i}{V}\sum_{k_n^0>0}(m^2-k_n^2)[(\mathrm{Re}\phi_n)^2+(\mathrm{Im}\phi_n)^2]}\\ &=\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0,n\neq m}\int d\mathrm{Re}\phi_n\ e^{-\frac{i}{V}(m^2-k_n^2)(\mathrm{Re}\phi_n)^2} \right) \times\left(\int_{k_m^0>0} d\mathrm{Re}\phi_m\ (\mathrm{Re}\phi_m)^2\ e^{-\frac{i}{V}(m^2-k_m^2)(\mathrm{Re}\phi_m)^2}\right) \left( \prod_{k_n^0>0}\int d\mathrm{Im}\phi_n\ e^{-\frac{i}{V}(m^2-k_n^2)(\mathrm{Im}\phi_n)^2} \right). \end{align}
Using the identities $$\int d\xi\ e^{-b\xi^2} = \sqrt{\frac{\pi}{b}}$$ and $$\int d\xi\ \xi^2e^{-b\xi^2} = \frac{1}{2b}\sqrt{\frac{\pi}{b}}$$, the equation simplifies as \begin{align} &\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0,n\neq m}\sqrt{\frac{-i\pi V}{m^2-k_n^2}} \right) \left(\frac{1}{2}\frac{-iV}{m^2-k_m^2-i\epsilon}\sqrt{\frac{-i\pi V}{m^2-k_m^2}}\right) \left( \prod_{k_n^0>0}\sqrt{\frac{-i\pi V}{m^2-k_n^2}} \right)\\ &=\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\frac{-i\pi V}{m^2-k_n^2} \right)\frac{1}{2}\frac{-iV}{m^2-k_m^2-i\epsilon} . \end{align} The second and the third terms are of the form $$\int d\xi\ \xi\ e^{-b\xi^2} =0$$. Finally for the fourth term we use $$\mathrm{Im}\phi_{-m} = \mathrm{Im}\phi_m^* = -\mathrm{Im}\phi_m$$ and then it gives the same contribution as the first term which cancels the factor of $$1/2$$.

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### Single-Channel (Mono) Versus Channel-Separated Audio

It is important to distinguish between single-channel (mono) and channel-separated audio.

• In mono audio, all speakers are recorded on a single channel.

• In channel-separated audio, each speaker is isolated to a distinct channel.

Channel-separated audio makes it possible to transcribe each channel independently and maintain a perfect correspondence between the person speaking and the words spoken. For analytic purposes, it is important to have each speaker on a separate channel.

For example, channel-separated audio not only decouples overtalk from overall accuracy, but it also allows for an objective measurement of the overtalk in calls. However, in single-channel (mono) audio, the greater the overtalk, the lower overall accuracy will be.

Voci employs a process called diarization of mono audio to separate speakers into separate channels. The effectiveness of diarization is decreased when source audio includes hold music, voice recordings, or more than two speakers. Overtalk may also negatively impact diarization accuracy. However, for typical agent and caller situations with only two speakers, diarization is very effective for separating speakers to their own channels for enhanced analytics.

### Note

Recording channel-separated source audio instead of mono will typically generate a 10% accuracy increase. Voci highly recommends using channel-separated audio for transcription.


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Ask Your Question

# Revision history [back]

Please give us a special case. So far, try EllipticCurve? to see how ot initialize an elliptic curve in sage. There are many examples. $E[p]$ can be "computed" more or less starting for intsance from the corresponding division_polynomial. For instance:

p = 3

for a in [3..22]:
print a

E = EllipticCurve( QQ, [-1, a] )
P = E.division_polynomial(3)
L.<a> = NumberField(P)
Cl_L = L.class_group()

if p.divides( Cl_L.order() ):

print "E =", E
print "RANK ::", E.rank()
print "P =", P
print "L =", L
print "Cl(L) ~", L.class_group()

break

The above rudimentary search finds:

E = Elliptic Curve defined by y^2 = x^3 - x + 19 over Rational Field
RANK :: 1
P = 3*x^4 - 6*x^2 + 228*x - 1
L = Number Field in a with defining polynomial 3*x^4 - 6*x^2 + 228*x - 1
Cl(L) ~ Class group of order 15 with structure C15 of Number Field in a with defining polynomial 3*x^4 - 6*x^2 + 228*x - 1


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# Multivariate Linear Regression and the Octave

Posted by Jerry on October 31, 2016

## Multivariate Linear Regression

### Gradient Descent in Practice (Feature Scaling)

Make sure features are on a similar scale.

Get every Feature into approximately a -1 ≤ xi ≤ 1 range 将数据集（训练集）进行缩放，原因是直接使用原始的数据集进行梯度下降，迭代次数会非常多而且相当复杂。

Mean normalization Replace xi with xi - ui (average value) / si (range) to make features have approximately zero mean (Do not apply to x0 = 1)

### Gradient Descent in Practice (Learning Rate)

1. “Debugging”: How to make sure gradient descent is working correctly

example automatic convergence test: Declare convergence if J(θ) decrease by less than 10 -3 in one iteration Gradient descent not working : use smaller α

α 很足够小，那么每次迭代后的梯度都会下降，但是收敛的速度很慢 α 太大了，不能够每次迭代梯度都能下降，可能造成不收敛（may not converge or slow converge)

1. How to choose learning rate α

### Features and polynomial Regression(多项式回归)

That is, how to fit a polynomial, like a quadratic function, or a cubic function, to your data

## Computing Parameters Analytically

### Normal Equation(标准方程法)

Normal Equation: Method to solve for θ Analytically. θ=(XTX)-1XTy

m training examples, n features the disadvantages and advantages between Gradient Descent and Normal Equation Gradient Descent Normal Equation • Need to choose α • No need to choose α
• Needs many iterations. • Don’t need to iterate. • Works well even • Need to compute (XTX)-1 时间复杂度 O(n3) when n is large. • Slow if n is very large (N>10000)

### Normal Equation Noninvertibility

Normal Equation : θ=(XTX)-1XTy what if XTX is non-invertible?(singular/degenerate)

1. Redundant features(linearly dependent). E.g. x1=size in feet 2 x2 = size in m2
2. Too many features(e.g.m≤n) · Delete some features, or use regularization
prinv(x'*x)*x'*y
pinv and inv
pinv函数可以伪逆(Return the pseudoinverse of X.)

## Octave Tutorial

format long/short
v = 1:0.1:2
v = 1:6
ones(1,3)
C = 2*ones(2,3)
v = rand(3,3) 随机随机矩阵
w = randn(1,1000)高斯随机矩阵
eye(4)单位矩阵
help rand
size(A)返回矩阵大小，size(A,1)返回矩阵的行数
who/whos
v = princeY(1:10)
load princeY.dat
save hello.mat v
save hello.txt v -ascii

A = [1 2;3 4;5 6]
A(3,2)
A(2,:)
A = [A,[100;101;102]]
A(:)

A*C
A.*C
abs(v)
log(v)
exp(v)
A'
max(a)
[val,ind]=max(a)
max(A)
magic(Number)

sum
prod
floor
ceil
max(A,[],1)
max(max(A))
sum(sum(A.*eye(9)))
flipud(eye(9))

## Word List

Feature Scaling: 特征缩放 Mean normalization：均值归一化 iteration： 迭代 converge：使收敛 Noninvertibility：不可逆性 round off your answer to two decimal ：保留两位小数 unvectorized/verctorized : 未向量化的/向量化的


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# Thread: Some problems with series

1. Originally Posted by Hweengee
How do I determine the convergence/divergence of the following:

$\displaystyle \sum _{n=1}^{\infty }\left({\sqrt[n] 3}-1 \right)$

I can't use limit comparison test with $\displaystyle {\sqrt[n] 3}$ because it gives me 0. Root and ratio tests don't seem appropriate and I'm struggling to find a divergent series which is smaller than this to compare it with. Would it be possible to compare it with $\displaystyle \sum _{n=1}^{\infty }\left(\ln\left(n \right)-1 \right)$?

As for

$\displaystyle \sum _{n=3}^{\infty }\frac{1}{\left( lnln\left(n \right) \right)^{\ln \left( n \right) }}$

I'm more or less completely stuck.

Thanks for the help in advance.
I see that your second series was never actually solved. Well I would suggest Cauchy's Condensation test. Since the sequence in the series is positive and decreasing it only converges iff $\displaystyle \sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2^n))^{\ln(2 ^n)}}=\sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2)n)^{ \ln(2)n}}\sim\sum_{n=3}^{\infty}\frac{2^n}{\ln(n)^ n}$. Now it is pretty clear this converges.

2. would you please explain furthur that why it converge. as a glance i see that it is divergant.

3. sorry with te RATIO TEST i found it convergant

4. Originally Posted by Mathstud28
I see that your second series was never actually solved. Well I would suggest Cauchy's Condensation test. Since the sequence in the series is positive and decreasing it only converges iff $\displaystyle \sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2^n))^{\ln(2 ^n)}}=\sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2)n)^{ \ln(2)n}}\sim\sum_{n=3}^{\infty}\frac{2^n}{\ln(n)^ n}$. Now it is pretty clear this converges.
It is possible to compare with a simple series: we have $\displaystyle (\ln\ln n)^{\ln n}=e^{\ln n\ln\ln\ln n}=n^{\ln\ln\ln n}\geq n^2$ as soon as $\displaystyle n\geq e^{e^{e^2}}\simeq 5.8\cdot 10^{702}$ (!!), hence $\displaystyle \frac{1}{(\ln\ln n)^{\ln n}}\leq \frac{1}{n^2}$ for these pretty large $\displaystyle n$. Hence it converges.

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# Limits

• May 3rd 2011, 10:56 AM
prasum
Limits
lim x tending to infinity ((2-x)^40*(4+x)^5)/(2-x)^45
• May 3rd 2011, 11:06 AM
Plato
Quote:

Originally Posted by prasum
lim x tending to infinity ((2-x)^40*(4+x)^5)/(2-x)^45

HINT: $\frac{{\left( {2 - x} \right)^{40} \left( {4 + x} \right)^5 }}{{\left( {2 - x} \right)^{45} }} = \left( {\frac{{4 + x}}{{2 - x}}} \right)^5$
• May 3rd 2011, 11:09 AM
TheChaz
Quote:

Originally Posted by prasum
lim x tending to infinity ((2-x)^40*(4+x)^5)/(2-x)^45

Do you mean

$\lim_{x \to \infty} \frac{(2-x)^{40}(4+x)^5}{(2-x)^{45}}$

? There's a language for that.
• May 4th 2011, 05:36 PM
prasum
i am sorry in the numerator it is (2+x)^45
• May 4th 2011, 07:42 PM
HallsofIvy
Much the same. IF you were to multiply everything out, the numerator would be x^{45} (NOT -x^45) plus terms involving lower powers of x and the denominator would be -x^{45} plus terms involving lower powers of x.

General rule: If a rational function has polynomials of the same degree in numerator and denominator, the limit,as x goes infinity, is the ratio of those leading coefficients. If the numerator has lower degree than the denominator, the imit, as x goes to infinity, is 0. If the numerator has higher degree than the denominator, the limit, as x goes to infinity, does not exist.


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# Question which has me stumped

1. Mar 8, 2008

### rushil01

Hey guys, im currently studying Engieering in Australia and there is one question which was given to us which i have no idea on how to complete and i was wondering if you guys could give me a hand with it;

What is the equilibrium seperation of HCl? ITs measured electric dipole moment is 3.60 x 10^-30C.m and the percentage of covalent bonding in HCl is 82.5%

ANy help would be greatly appreciated. Thanx

2. Mar 9, 2008

### lightarrow

$$r =\ \mu_D/q\cdot P\ =\ 6.25\cdot10^{18}\ \mu_D/P\ =\ 6.25\cdot10^{18}\cdot3.6\cdot10^{-30}/0.825\ =\ 2.73\cdot10^{-11}\ m\ =\ 27.3\ pm$$

3. Mar 16, 2008

### mate1000

What are those formulas you came up with and where did 6.25*10^18 come from.
Thanks

4. Mar 16, 2008

### rushil01

i got my lecturer to do it and he used the formula p=qd. I'll post the full working out when i have a little more time


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Announcements
Expression of concern regarding paper by Park et al, published on 25 June 2015: “Epidemiological investigation of MERS-CoV spread in a single hospital in South Korea, May to June 2015”, Euro Surveill. 2015;20(25):pii=21169. It has been brought to our attention that some of the authors may not have been informed about the content of the above paper. There is a lack of clarity regarding rights to use the data. The editorial team are investigating what action needs to be taken.

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 Home Eurosurveillance Edition  2010: Volume 15/ Issue 37 Article 2 Printer friendly version
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Eurosurveillance, Volume 15, Issue 37, 16 September 2010
Rapid communications
Quantifying benefits and risks of vaccinating Australian children aged six months to four years with trivalent inactivated seasonal influenza vaccine in 2010
1. Victorian Infectious Diseases Reference Laboratory, Melbourne, Australia
2. Communicable Disease Control Directorate, Department of Health Western Australia, Perth, Australia

Citation style for this article: Kelly H, Carcione D, Dowse G, Effler P. Quantifying benefits and risks of vaccinating Australian children aged six months to four years with trivalent inactivated seasonal influenza vaccine in 2010. Euro Surveill. 2010;15(37):pii=19661. Available online: http://www.eurosurveillance.org/ViewArticle.aspx?ArticleId=19661
Date of submission: 12 August 2010

Australian and New Zealand health authorities identified seasonal trivalent inactivated influenza vaccines manufactured by CSL Biotherapies as the probable cause of increased febrile convulsions in children under five within 24 hours of vaccination and recommended against their use in this age group. We quantified the benefit-risk profile of the CSL vaccines using the number needed to vaccinate and suggest they might have caused two to three hospital admissions due to febrile convulsions for every hospital admission due to influenza prevented.

Introduction

The recognition of an unexpectedly high number of febrile convulsions (not defined precisely initially) in children aged less than five years within 24 hours of receipt of trivalent inactivated influenza vaccine (TIV) in Australia in April 2010, led to the initial suspension of the childhood vaccination campaign for seasonal influenza in our country [1]. This was followed by more general discussion about the safety of influenza vaccines in children [2].

In 2010, trivalent vaccines from three manufacturers, CSL Biotherapies, Solvay Pharmaceuticals and Sanofi-Pasteur, had been licensed for use in Australian children. All vaccines contained the strains recommended by the World Health Organization (WHO) [3]. The influenza A(H1N1) virus vaccine component of these vaccines was the 2009 pandemic strain.

The increase in febrile convulsions was first recognised in the state of Western Australia [4]. This was possible because at the beginning of 2008 Western Australia had implemented a vaccine programme aimed at immunising children aged from six months to four years against influenza [5]. This programme acknowledged (i) the importance of children in the spread of influenza, (ii) the high hospitalisation rate due to laboratory-confirmed influenza in this age group and (iii) three childhood deaths associated with influenza in the state in 2007 [5]. Influenza vaccine coverage of 20%–30% had been achieved in this age group for 2008 and 2009 and also in 2010, prior to the suspension of the programme.

Given this background and following concern about the possible risk of febrile convulsions associated with influenza vaccination, the childhood influenza vaccination programme was suspended by the health authorities in Western Australia on 22 April 2010. For the whole of Australia, the precise number of febrile convulsions associated with the administration of TIV was not clear at the time. However, as a precautionary measure, the Australian Department of Health and Ageing recommended against using any formulation of 2010 TIV in young children on 23 April 2010 [1].

At the same time, the Ministry of Health in New Zealand recommended specifically against the use of Fluvax or Fluvax Junior (both manufactured by CSL Biotherapies) in children from six months to four years, but recommended the continued use of other influenza vaccines licensed for children in this age group, specifically Influvac (Solvay Pharmaceuticals) and Vaxigrip (Sanofi-Pasteur) [6], two vaccines that were also licensed in Australia. Recommendations for other age groups remained unchanged. Three months later, on 30 July 2010, the Australian Department of Health and Ageing published revised recommendations indicating it was safe to use either Influvac or Vaxigrip in all children aged from six months to four years but recommended against the use of Fluvax or Fluvax junior in children aged under five years [7].

A timeline of recommendations from health authorities in Australia and New Zealand is shown in Table 1.

Table 1. Timeline of decisions communicated by health authorities in Australia and New Zealand following recognition of febrile convulsions in children under five years occurring within 24 hours of receipt of trivalent inactivated vaccine, 2010

Decisions regarding the suspension of an immunisation programme are important because they may have consequences for individuals and a long-term impact on vaccination programmes overall. It is therefore important that these decisions are based on objective assessments of the benefit and risk of all administered vaccines. We present an approach for quantifying the benefit-risk profile for vaccines from two manufacturers, CSL Biotherapies and Solvay Pharmaceuticals, in 2010. We briefly compare with data collected in Western Australia on the frequency of febrile convulsions following administration of vaccines from Sanofi-Pasteur and CSL Biotherapies in 2008 and 2009.

Methods

We compared the benefit-risk profile of vaccination using data for the two vaccines (CSL Biotherapies and Solvay Pharmaceuticals) that had been administered in sufficient numbers in 2010 for meaningful comparisons to be made. To compare benefit and risk, we used hospitalisation data from 2009, obtained from the Department of Health in Western Australia, and adverse event data from 2010, sourced from a detailed investigation of adverse events in Western Australia (unpublished data).

We assessed benefit by estimating the number of children that would have required vaccination to prevent one hospital admission due to laboratory-confirmed influenza. We chose hospitalisation due to influenza of any type or subtype during the 2009 influenza A(H1N1) pandemic because viral testing for children hospitalised with febrile illnesses was intensive and relatively few children hospitalised due to influenza in 2009 would have remained undiagnosed.

We assessed risk using hospital admissions for febrile convulsions, which were defined by a systematic review of cases with reference to the criteria of the Brighton Collaboration for febrile convulsions following immunisation [8]. Receipt of at least one dose of influenza vaccine in 2010 was verified from the Australian Childhood Immunisation Register [9] and direct questioning of parents or guardians.

Quantifying benefit
To quantify benefit, we calculated the number of children that would need to be vaccinated to prevent one hospital admission due to influenza, that is, the number needed to vaccinate (NNV). The NNV is calculated as the reciprocal of the absolute risk reduction. The absolute risk reduction is calculated as the product of the absolute risk in the unexposed (in this case the unvaccinated) and the relative risk reduction. Vaccine effectiveness (VE) is the standard expression of the relative risk reduction for vaccine preventable diseases [10]. Hence, the NNV to prevent one hospitalisation = 1/(hospitalisation rate in the unvaccinated x VE). We therefore needed to estimate two parameters: the hospitalisation rate in the unvaccinated and the VE.

To estimate the hospitalisation rate in the unvaccinated, we ascertained the vaccine status of children admitted to hospital for influenza in 2009, thus allowing us to calculate the number of these children who were not vaccinated. We then estimated vaccine coverage for influenza vaccine in 2009 for children aged between six months and four years from vaccine usage reported by immunisation providers state-wide. This allowed us to estimate the size of the unvaccinated population in this age group = (1- the proportion vaccinated) x (the population in that age group). The number of unvaccinated children admitted to hospital divided by the estimated number of unvaccinated children in the population gave the hospitalisation rate in the unvaccinated.

The VE for prevention of influenza infection in children aged between 2 and 16 years has been estimated in a systematic review as 59% (95% confidence interval (CI): 41 to 71) [11]. This estimate is supported by findings from a study of children aged 0 to 4 years in Western Australia in 2008, in which the VE was 68% (95% CI: 26 to 86) [5]. The study found no significant difference in VE by age group (less than two years compared with two to four years) for children with an influenza-like illness proven to be caused by influenza [5]. We assumed the VE for TIV in 2010, the year for which we calculated the benefit-risk profile, would have been similar to previous estimates of VE when vaccine and circulating strains matched. This VE estimate should be appropriate for the TIV for 2010, as vaccine manufacturing processes were the same in 2010 as for previous years. We therefore used a VE of 60% for all children aged six months to four years in this analysis.

Quantifying risk
To quantify risk, we calculated the number of hospitalisations that could be attributed to adverse events following vaccination of the NNV to prevent one hospital admission. We used vaccine-specific rates of febrile convulsions determined during the detailed investigation into suspected adverse outcomes in 2010 in Western Australia (unpublished). We compared these with vaccine-specific rates of febrile convulsions in 2008 and 2009, determined by passive surveillance in both years but augmented by active questioning of parents/guardians of children admitted to hospital in 2009. We quantified the risk of admission to hospital for a febrile convulsion following receipt of TIV as the product of the NNV, the absolute risk of a febrile convulsion and the risk of hospital admission following a febrile convulsion.

Results

Quantifying benefit: the NNV to prevent one hospitalisation due to influenza

The number of children living in Western Australia aged between six months and four years in 2009 was approximately 130,000 of whom 30.3% were estimated to have received at least one dose of TIV in that year. The number of unvaccinated children was therefore 90,610.

In that year, 432 cases of influenza were notified in children in this age range, of whom 383 (89%) were infected with 2009 pandemic influenza A(H1N1) and 49 (11%) were infected with seasonal influenza strains or influenza A not subtyped. One hundred and twenty children were hospitalised for any type of influenza, of whom 74 were unvaccinated. Vaccine status was unknown for a further 13 children. We allocated the children with unknown vaccine status to one of the two known groups in the same proportion as those whose status was known, giving an estimated total of 82 unvaccinated children admitted to hospital. The hospitalisation rate in unvaccinated children aged six months to four years was therefore 90 per 100,000.

To prevent one hospitalisation due to any strain of circulating influenza in 2009 would have required the vaccination of 1,852 children (1/[90/1000,00 x 0.6]),  the NNV.

Quantifying risk: hospital admission for febrile convulsions following receipt of TIV
Prior to cessation of the immunisation programme in April 2010, more than 10,000 doses of Fluvax or Fluvax Junior and more than 3,300 doses of Influvac had been administered to children aged six months to four years in Western Australia in 2010 [4]. Detailed follow up investigation of adverse events following immunisation identified 56 children with febrile convulsions that occurred within 24 hours of receipt of TIV between 8 March and 25 April 2010, of whom 19 (34%) required overnight hospital admission. All 55 cases with information on vaccine formulation had received Fluvax or Fluvax Junior. The estimated risk of a febrile convulsion following receipt of Fluvax or Fluvax Junior was 0.39%, compared to 0% for Influvac (unpublished data).

Applying these estimates of risk to the NNV to prevent one hospitalisation due to influenza indicates that vaccination with Fluvax or Fluvax Junior would have been likely to have caused seven febrile convulsions (1,852 x 0.0039), with two to three of these children requiring hospitalisation (34% of seven children). Based on the data available, Influvac would have caused no febrile convulsions and no hospital admissions.

The upper limit of the 95% CI for Influvac estimated a febrile convulsion risk of 0.08% (unpublished data). If the real risk were this high, Influvac may have caused one to two febrile convulsions, and zero or one hospital admission, in preventing one hospitalisation due to laboratory-confirmed influenza. The corresponding risk of febrile convulsions for the Fluvax products (that is, estimating risk as the upper limit of the 95% CI= 0.51%, unpublished data) would have seen nine febrile convulsions attributed to these vaccines, causing three hospital admissions for febrile convulsions for each hospital admission for influenza that was prevented.

Using routine passive surveillance in 2008 and 2009, plus direct enquiry to parents or guardians of children admitted to hospital with laboratory-confirmed influenza in 2009, we estimated the risk of a febrile convulsion following receipt of any seasonal TIV to have been 0.003% in 2008 and zero in 2009. In those two years, these vaccines would have caused no hospital admissions due to febrile convulsions for each hospital admission due to influenza that was prevented (Table 2).

Table 2. The number needed to vaccinate to prevent one hospital admission for laboratory-confirmed influenza, all subtypes, and the risk of hospital admission for febrile convulsion following receipt of trivalent inactivated influenza vaccinea by year and vaccine manufacturer

Discussion and conclusion

We have demonstrated a method that can be used to quantify the benefit and risk of vaccinating children aged six months to four years against influenza. In 2009, with relatively high levels of hospital admission, it would have been necessary to have vaccinated about 1,850 West Australian children with a vaccine that was 60% effective to have prevented one hospital admission due to laboratory-confirmed influenza.

If the hospitalisation rate for influenza in 2010 was the same as that in 2009, we estimated that vaccination with Fluvax or Fluvax Junior in 2010 may have caused two to three hospital admissions due to febrile convulsions for every hospital admission due to influenza prevented. Although the influenza season is not yet over in Western Australia, current data indicate that influenza virus circulation has been considerably lower in 2010 than in 2009 [12] and fewer children will have been hospitalised for influenza in 2010 than were hospitalised in 2009. This implies that the NNV to prevent one hospital admission in 2010 would be higher than in 2009.

We did not see this same risk profile with another TIV licensed for this age group in 2010, or with the vaccines manufactured by CSL Biotherapies in 2008 or 2009, despite similarly high vaccine coverage in these two years in Western Australia. Our results therefore indicate there is no excess risk over benefit with a childhood influenza vaccination programme in general. The problem identified in 2010 was related to a vaccine produced by a single manufacturer. Investigations published to date have failed to find a cause for the problem [13].

A comprehensive approach to this analysis would involve assessment of other outcomes due to laboratory-confirmed influenza and a detailed sensitivity analysis. Moreover, we have chosen to quantify the NNV using all children aged six months to four years. The benefit-risk profile would be improved if only children who were at increased risk of hospitalisation following influenza infection were targeted for vaccination. However, we have shown that a good past benefit-risk profile for a vaccine may not guarantee a favourable profile in future years. This highlights once again the importance of continued and comprehensive safety monitoring of influenza vaccines post-marketing.

Acknowledgements
We acknowledge the contributions of many staff in the Department of Health in Western Australia to the investigation of adverse events in children following influenza vaccination in 2010.

References

1. Australian Government. Department of Health and Ageing. Seasonal Flu Vaccine and young children. 23 April 2010. Media Release. Available from: http://www.health.gov.au/internet/main/publishing.nsf/Content/mr-yr10-dept-dept230410.htm
2. Collignon P, Doshi P, Jefferson T. Child influenza vaccination. Ramifications of adverse events in children in Australia. BMJ. 2010;340:c2994.
3. World Health Organization (WHO). Recommended composition of influenza virus vaccines for use in the 2010 southern hemisphere influenza season. WHO; Revised 8 Oct 2009. Available from: http://www.who.int/csr/disease/influenza/200909_Recommendation.pdf
4. Government of Western Australia. Department of Health. Ministerial Review into the Public Health Response into the Adverse Events to the Seasonal Influenza Vaccine. July 2010. Final Report to the Minister for Health. [Accessed 16 Aug 2010]. Available from: http://www.health.wa.gov.au/publications/documents/Stokes_Report.pdf
5. Kelly H, Jacoby P, Dixon G, Moore HC, Carcione D, Williams S, et al. Vaccine effectiveness against laboratory-confirmed influenza in healthy young children: a case control study. Pediatric Inf Dis . Forthcoming 2010.
6. New Zealand Ministry of Health. Letter to General Practitioners, Practice Nurses, Practice Managers and Health Professionals. 26 April 2010. [Accessed 14 Jul 2010]. Available from: http://www.moh.govt.nz/moh.nsf/pagesmh/9164/$File/gp-fluvax-fax-apr2010.doc 7. Department of Health and Ageing. Seasonal flu vaccination for young children can be resumed. Updated advice from the Chief Medical Officer. [Accessed 9 Sep 2010]. Available from: http://www.health.gov.au/internet/main/publishing.nsf/Content/mr-yr10-dept-dept300710.htm 8. Bonhoeffer J, Menkes J, Gold MS, de Souza-Brito G, Fisher MC, Halsey N, et al. Brighton Collaboration Seizure Working Group. Generalized convulsive seizure as an adverse event following immunization: case definition and guidelines for data collection, analysis, and presentation. Vaccine 2004; 22(5-6):557-62. 9. Australian Government. Medicare Australia. Australian Childhood Immunisation Register. Available at http://www.medicareaustralia.gov.au/public/services/acir/index.jsp 10. Kelly H, Attia J, Andrews R, Heller RF. The number needed to vaccinate (NNV) and population extensions of the NNV: comparison of influenza and pneumococcal vaccine programmes for people aged 65 years and over. Vaccine 2004;22(17-18):2192-8. 11. Smith S, Demicheli V, Di Pietrantonj C, Harnden AR, Jefferson T, Matheson NJ, et al. Vaccines for preventing influenza in healthy children. Cochrane Database Syst Rev. 2006;(1):CD004879. 12. Australian Government. Department of Health and Ageing. Australian Influenza Surveillance Report. No. 35, 2010. Reporting period 28 August-3 September 2010. Available from: http://www.healthemergency.gov.au/internet/healthemergency/publishing.nsf/Content/ozflu2010-jul-sep-pdf-cnt.htm/$File/ozflu-no35-2010.pdf
13. Australian Government. Department of Health and Ageing. Therapeutic Goods Administration. Investigation into febrile reactions in young children following 2010 seasonal trivalent influenza vaccination. 2 Jul 2010. Available from: http://www.tga.gov.au/alerts/medicines/fluvaccine-report100702.htm

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# zbMATH — the first resource for mathematics

## da Silva Graca Arroja Neves, André

Compute Distance To:
 Author ID: aroja-neves.andre Published as: Neves, A.; Neves, Andre; Neves, André Homepage: http://math.uchicago.edu/~aneves/ External Links: MGP · Wikidata · GND
 Documents Indexed: 37 Publications since 2004
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#### Co-Authors

 8 single-authored 15 Marques, Fernando Codá 3 Bray, Hubert L. 3 Brendle, Simon 3 Tian, Gang 2 Ammann, Bernd Eberhard 2 Hanke, Bernhard 1 Agol, Ian 1 Akutagawa, Kazuo 1 Eichmair, Michael 1 Irie, Kei 1 Lamm, Tobias 1 Lee, Dan Archibald 1 Liokumovich, Yevgeny 1 Lotay, Jason Dean 1 Metzger, Jan
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#### Journals

 5 Annals of Mathematics. Second Series 3 Inventiones Mathematicae 3 Oberwolfach Reports 2 Journal of Differential Geometry 1 Communications in Mathematical Physics 1 Communications on Pure and Applied Mathematics 1 Jahresbericht der Deutschen Mathematiker-Vereinigung (DMV) 1 Duke Mathematical Journal 1 Journal für die Reine und Angewandte Mathematik 1 Transactions of the American Mathematical Society 1 Journal of the American Mathematical Society 1 Geometric and Functional Analysis. GAFA 1 Notices of the American Mathematical Society 1 Communications in Analysis and Geometry 1 Mathematical Research Letters 1 European Mathematical Society Newsletter 1 Geometry & Topology 1 Bulletin of the Brazilian Mathematical Society. New Series 1 Cambridge Journal of Mathematics
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#### Fields

 35 Differential geometry (53-XX) 9 Calculus of variations and optimal control; optimization (49-XX) 7 Manifolds and cell complexes (57-XX) 4 General mathematics (00-XX) 4 Partial differential equations (35-XX) 4 Global analysis, analysis on manifolds (58-XX) 2 $K$-theory (19-XX) 2 Measure and integration (28-XX) 1 Relativity and gravitational theory (83-XX)


text:
Plotting a user function?
02-09-2014, 01:09 AM (This post was last modified: 02-09-2014 01:16 AM by veeblefester.)
Post: #1
 veeblefester Junior Member Posts: 43 Joined: Dec 2013
Plotting a user function?
I have a user function defined and working.

Forgive my ignorance, I cannot figure out how to plot a USER function.

I can enter and 'check' and plot a function in one of the V0 thru V9 built in functions in Advanced Graphing Symbolic View.

How do I plot a USER function from a PROGRAM?

Thanks in advance for any help.

Happy coding and may you see ' i No errors in the program ' with every compilation.
02-09-2014, 11:01 AM
Post: #2
 Giancarlo Member Posts: 243 Joined: Dec 2013
RE: Plotting a user function?
Hello,
You could:
- assign your equation to V1,V2,.. with a V1:= your_eq_goes_here;
- invoke the advanced graphic application through the STARTAPP("name_of_the_app") command
- uncheck all the other V# with a loop with the command UNCHECK(4); UNCHECK(5),....
- invoke the plot screen via STARTVIEW(1);

I was not very happy when I didn't see a simple calculation program in the prime manual in the programming chapter but now that i am able to do it thanks to this forum, i can appreciate the very long DICE program which shows how to bend the builtin applications for my needs.

Hope it helps,

Giancarlo
02-10-2014, 06:51 AM (This post was last modified: 02-10-2014 06:58 AM by veeblefester.)
Post: #3
 veeblefester Junior Member Posts: 43 Joined: Dec 2013
RE: Plotting a user function?
Thanks for the response.

The solution, I realized, is that the proper variable in my equation must be named X.

As in f(X).

Happy coding and may you see ' i No errors in the program ' with every compilation.
02-28-2014, 02:21 AM
Post: #4
 jte Member Posts: 69 Joined: Feb 2014
RE: Plotting a user function?
(02-09-2014 11:01 AM)Giancarlo Wrote:
- uncheck all the other V# with a loop with the command UNCHECK(4);

Just a comment: you can pass lists to UNCHECK. For example, UNCHECK(MAKELIST(X,X,0,9)) will uncheck all 10 user formulas.
02-28-2014, 07:00 AM
Post: #5
 Giancarlo Member Posts: 243 Joined: Dec 2013
RE: Plotting a user function?
This shinks the code even more,

Thanks for the trick

Giancarlo
10-01-2015, 07:17 PM
Post: #6
 StephenG1CMZ Senior Member Posts: 870 Joined: May 2015
RE: Plotting a user function?
(02-28-2014 02:21 AM)jte Wrote:
(02-09-2014 11:01 AM)Giancarlo Wrote:
- uncheck all the other V# with a loop with the command UNCHECK(4);

Just a comment: you can pass lists to UNCHECK. For example, UNCHECK(MAKELIST(X,X,0,9)) will uncheck all 10 user formulas.

Interestingly, on the Android emulator UNCHECK seems able to handle lists for Function, but not for Solve. The embedded help for UNCHECK only describes the UNCHECK(n) syntax.

Code:
 EXPORT TRYS() BEGIN  SolveTRY.E1:=QUOTE(X^2+X=0);  SolveTRY.E2:="X^2+X=0";  SolveTRY.UNCHECK(1); //WORKS  Function.UNCHECK({1,2});//LIST WORKS ON FUNCTION  //SolveTRY.UNCHECK(MAKELIST(X,X,1,9)); //DOESNT WORK ON SOLVE END;

(The problem relates to the use of a list, the MAKELIST is not significant).

Stephen Lewkowicz (G1CMZ)
ANDROID HP Prime App broken offline on some mobiles
10-02-2015, 05:26 AM
Post: #7
 cyrille de brébisson Senior Member Posts: 1,009 Joined: Dec 2013
RE: Plotting a user function?
Hello,

> The embedded help for UNCHECK only describes the UNCHECK(n) syntax.

The list use of most function is NOT described in the function help itself (as a mater of fact, I do not know where it is described)...

The reason for this is that, deep down, the functions do NOT implement list processing.

List processing is implemented at the level evaluating the function, not in the function evaluation.

So, if the system sees a + and see that arg 1is a list and arg 2 is NOT a list, then it will call + n times, with each element of list 1 and with a copy of arg 2 and construct the resulting list. The + function is NOT aware that it is doing list processing.

This is great, because any new function created WILL implement list processing automatically/by itself, without the function creator having to do anything.

But it means that it is often an overlooked function when doing documentation.

Some functions specifically flag the evaluator as: Not list processing friendly. In this case, list processing does get disabled. As a general rule, this is the case for all the functions that have special processing for lists.

Cyrille

Although I work for the HP calculator group, the views and opinions I post here are my own. I do not speak for HP.
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# Is it unusual for the MEAN to outperform ARIMA?

I recently applied a range of forecasting methods (MEAN, RWF, ETS, ARIMA and MLPs) and found that MEAN did surprisingly well. (MEAN: where all future predictions are predicted as been equal to the arithmetic mean of the observed values.) MEAN even outperformed ARIMA on the three series I used.

What I want to know is if this is unusual? Does this mean the times series I'm using are strange? Or does this indicate that I've set something up wrong?

• en.wikipedia.org/wiki/Martingale_%28probability_theory%29 – user541686 Nov 21 '14 at 16:27
• @Mehrdad one could definitely craft a nice answer around Martingales. – shadowtalker Nov 21 '14 at 19:18
• It's quite common for simple methods to perform well, especially out-of-sample (which is where it matters). This effect becomes stronger on short series. If there's not much serial correlation or trend, then we'd expect the mean to do very well even with relatively long series. – Glen_b Nov 25 '14 at 9:07

I'm a practitioner, both producer and user of forecasting and NOT a trained statistician. Below I share some of my thoughts on why your mean forecast turned out better than ARIMA by referring to research article that rely on empirical evidence. One book that time and time again I go back to refer is the Principles of Forecasting book by Armstrong and its website which I would recommend as an excellent read for any forecaster, provides great insight on usage and guiding principles of extrapolation methods.

To answer you first question - What I want to know is if this is unusual?

There is a chapter called Extrapolation for Time-Series and Cross-Sectional Data which also available free in the same website. The following is the quote from the chapter

"For example, in the real-time M2-competition, which examined 29 monthly series, Box-Jenkins proved to be one of the least-accurate methods and its overall median error was 17% greater than that for a naive forecast"

There lies an empirical evidence on why your mean forecasts was better than ARIMA models.

There is also been study after study in empirical competitions and the third M3 competition that show Box - Jenkins ARIMA approach fails to produce accurate forecast and lacks evidence that it performs better for univariate trend extrapolation.

There is also another paper and an ongoing study by Greene and Armstrong entitled "Simple Forecasting: Avoid Tears Before Bedtime" in the same website. The authors of the paper summarize as follows:

In total we identified 29 papers incorporating 94 formal comparisons of the accuracy of forecasts from complex methods with those from simple—but not in all cases sophisticatedly simple—methods. Eighty-three percent of the comparisons found that forecasts from simple methods were more accurate than, or similarly accurate to, those from complex methods. On average, the errors of forecasts from complex methods were about 32 percent greater than the errors of forecasts from simple methods in the 21 studies that provide comparisons of errors

To answer your third question: does this indicate that I've set something up wrong? No, I would aconsider ARIMA as complex method and Mean forecast as simple methods. There is ample evidence that simple methods like Mean forecast outperform complex methods like ARIMA.

To answer your second question: Does this mean the times series I'm using are strange?

Below are what I considered to be experts in real world forecasting:

• Makridakis (Pioneered Empirical competition on Forecasting called M, M2 and M3, and paved way for evidence based methods in forecasting)
• Armstrong (Provides valuable insights in the form of books/articles on Forecasting Practice)
• Gardner (Invented Damped Trend exponential smoothing another simple method which works surprisingly well vs. ARIMA)

All of the above researchers advocate, simplicity (methods like your mean forecast) vs. Complex methods like ARIMA. So you should feel comfortable that your forecasts are good and always favor simplicity over complexity based on empirical evidence. These researchers have all contributed immensely to the field of applied forecasting.

In addition to Stephan's good list of simple forecasting method. there is also another method called Theta forecasting method which is a very simple method (basically Simple Exponential smoothing with a drift that equal 1/2 the slope of linear regression) I would add this to your toolbox. Forecast package in R implements this method.

• I really like the perspective you provide on forecasting and the evidence you adduce to support it, but elements of this answer are objectionable because they read too much like a rant against "statisticians" or perhaps against formal statistical training--and are flat out wrong. For instance, Makridakis' advanced degrees are in (guess what?) statistics, that's what he teaches, and that's what he does. – whuber Nov 21 '14 at 23:02
• Got it, I'll remove it. I was trying to make a point that major advances on evidence based methods came from non statisticians. But I see your point that it might come across like that – forecaster Nov 21 '14 at 23:04
• Done, Also, Makridakis PHD was in Management Information Systems according to this published interview – forecaster Nov 21 '14 at 23:18
• +1. However, you write "There lies an empirical evidence on why your mean forecasts was better than ARIMA models." - no, this only tells us that the mean was better (in this specific instance), not why. I'd turn the argument around and put the burden of proof on ARIMA and other models. I have never understood quite why the data-generating process should care about past errors against my model, which is what MA terms essentially model. My personal suspicion is that ARIMA is so popular because you can actually prove stuff, like unit roots and stationarity - not because it forecasts well. – Stephan Kolassa Nov 25 '14 at 19:19
• One more reason why arima could be popular be is that it is able to incorporate regressors/causal variables while exponential smoothing is not able to or never will be able to incorporate regressors even with the state space formulation a.k.a ets. As far as univariate forecasting is concerned, simple adhoc methods work best vs. ARIMA. – forecaster Nov 25 '14 at 19:46

This is not at all surprising. In forecasting, you very often find that extremely simple methods, like

• the overall mean
• the naive random walk (i.e., the last observation used as a forecast)
• a seasonal random walk (i.e., the observation from one year back)
• Single Exponential Smoothing

outperform more complex methods. That is why you should always test your methods against these very simple benchmarks.

A quote from George Athanosopoulos and Rob Hyndman (who are experts in the field):

Some forecasting methods are very simple and surprisingly effective.

Note how they explicitly say they will be using some very simple methods as benchmarks.

In fact, their entire free open online textbook on forecasting is very much recommended.

EDIT: One of the better-accepted forecast error measures, the Mean Absolute Scaled Error (MASE) by Hyndman & Koehler (see also here) measures how much a given forecast improves on the (in-sample) naive random walk forecast: if MASE < 1, your forecast is better than the in-sample random walk. You'd expect this to be an easily beaten bound, right?

Not so: sometimes, even the best out of multiple standard forecasting methods like ARIMA or ETS will only yield a MASE of 1.38, i.e., be worse (out-of-sample) than the (in-sample) random walk forecast. This is sufficiently disconcerting to generate questions here. (That question is not a duplicate of this one, since the MASE compares out-of-sample accuracy to in-sample accuracy of a naive method, but it is also enlightening for the present question.)

• Thanks for accepting, but maybe you want to wait for a day - if a question has accepted answers, fewer people will even read it, let alone comment or answer. And other people may have different takes on this. Feel free to un-accept ;-) – Stephan Kolassa Nov 21 '14 at 14:05
• Its very honest of you :) I'll give it a day. Thanks. – Andy T Nov 21 '14 at 14:06
• "complex" is a close relative of "overfitted." – shadowtalker Nov 21 '14 at 19:18
• +1 nice answer. If forecasting was an evidence based field like medicine, ARIMA method would be history. – forecaster Nov 21 '14 at 22:47
• simple minded ARIMA without validating the Gaussian Assumptions is already history for most of us BUT apparently not for all ! – IrishStat Nov 21 '14 at 22:59


text:
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main-content

Thermodynamics is a subject that all engineering students have to face and that most of them treat with great respect. This makes it all the more important to offer a good and easy-to-understand approach to the laws of energy conversion. This is what this textbook is intended to do: It covers the basics of classical technical thermodynamics as they are typically taught at universities: The first and second law of thermodynamics as well as equations of state are explained for idealized and real fluids which are subject to a phase change. Thermodynamic mixtures, e.g. humid air, are treated as well as chemical reactions. Components and thermodynamic cycle that convert energy are presented. The book attaches great importance to drawings and illustrations, which should make it easier to comprehend complex matter. Technical applications and apparatus are presented and explained. Numerous exercises and examples conclude the book and contribute to a better understanding of the theory.

### Chapter 1. Introduction

Abstract
Thermodynamics is one of the most difficult and probably even one of the most challenging disciplines in mechanical engineering. However, thermodynamics is necessary to understand the principles of energy conversion, e.g. when designing thermal machines such as gas turbines, internal combustion engines or even fuel cells in modern applications.
Achim Schmidt

### Chapter 2. Energy and Work

Abstract
It is well known from physics lessons that energy is conserved. The principle of energy conservation
Achim Schmidt

### Chapter 3. System and State

Abstract
In this chapter it is clarified what a thermodynamic system is and how its state can be described. Any system is separated from an environment by a system boundary as it is done in other technical disciplines, e.g. technical mechanics, as well. First, the permeability of the system boundary is categorised. However, after classifying the system and the system boundary, it is the next step to identify the internal state. This leads to so-called state values, e.g. pressure and temperature, that fix the state of a system. Our everyday experience shows, that the state can be varied by external impacts across the boundary.
Achim Schmidt

### Chapter 4. Thermodynamic Equilibrium

Abstract
For the further understanding of thermodynamics the principle of thermodynamic equilibrium is essential. Systems in thermodynamic equilibrium do not change their state without any impact from outside, i.e. they are perfectly in rest.
Achim Schmidt

### Chapter 5. Equations of State

Abstract
Chapter 3 has shown, how to quantify the internal state of a system. Basically, a distinction has been made between thermal and caloric state values. Since not all state values can be measured directly, equations to calculate them are required. The Paddle Wheel experiment for instance, see Sect. 2.4.1, has shown, that the internal energy describes the energetic state of a fluid.
Achim Schmidt

### Chapter 6. Thermal Equation of State

Abstract
In this chapter, the correlation between the thermal state variables pressure p, specific volume v and temperature T is investigated. As already discussed in the previous chapter, thermal state values can easily be measured. Furthermore, according to Gibbs’ phase law for ideal gases two intensive state values, i.e. pressure and temperature, fix the state of a thermodynamic system unequivocally. Thus, a mathematical function to calculate the third thermal state value v must exist:
$$v=f(p, T)$$
Achim Schmidt

### Chapter 7. Changes of State

Abstract
So far thermodynamic systems have been discussed and state values to quantify their internal state have been categorised. In this chapter the focus will be on changes of state, i.e. bringing a system from an initial state to a new state.
Achim Schmidt

### Chapter 8. Thermodynamic Processes

Abstract
So far it has been clarified what a thermodynamic system is and how its state can be determined. Within conventional thermodynamics the principle of thermodynamic equilibrium forms the basis for thermodynamic calculations. However, due to external impacts a system can be shifted from one equilibrium state to another.
Achim Schmidt

### Chapter 9. Process Values HeatHeat and WorkWork

Abstract
Over the previous chapters the internal state of a system has been described. A thermodynamic system can change its state and several possible changes of state have been discussed. For the further understanding it is essential to be familiar with the terms reversibility/irreversibility as well as with the idea of a quasi-static change of state. These approaches have been clarified in the previous Chaps. 7 and 8: By external impacts a system’s thermodynamic equilibrium can be disturbed and the system moves into a new balanced state.
Achim Schmidt

### Chapter 10. State Value Versus Process Value

Abstract
The previous chapters have shown, that it is essential to distinguish between state and process values. State values specify the state of a thermodynamic system. Furthermore, they can be visualised in a pv-diagram for instance. The state of an ideal gas is fixed by two independent state values, so that each pv-pair in a pv-diagram defines the state of a system unambiguously.
Achim Schmidt

### Chapter 11. First Law of Thermodynamics

Abstract
In this chapter the causality between process variables and the change of state variables of a thermodynamic system is derived. The first section introduces the principle of equivalence between work and heat—an essential prerequisite in order to formulate the energy conservation principle known as first law of thermodynamics. Finally, the first law of thermodynamics is applied for closed respectively open systems as well as for thermodynamic cycles. Anyhow, thermodynamic cycles play an important role in the following chapters.
Achim Schmidt

### Chapter 12. Caloric Equations of State

Abstract
In the previous chapters the law of energy conservation has been thoroughly discussed. By this, it was possible to evaluate thermodynamic systems energetically. A distinction has been made between closed and open systems. However, next to thermal state values, such as pressure p or temperature T, that can be measured easily, a new category of state values has been introduced: These state values, i.e. specific internal energy u and specific enthalpy h for instance, can not be determined by a sensor and thus need to be calculated.
Achim Schmidt

### Chapter 13. Meaning and Handling of Entropy

Abstract
In the previous Chap. 12 the caloric equations of state have been introduced and derived thermodynamically. A new state value, the specific entropy s has been introduced. However, at that point it was not clear how to utilise entropy in order to evaluate thermodynamic systems. This chapter focuses on clarifying why entropy is beneficial. First, a comparison is made with the first law of thermodynamics, so that entropy balancing is comprehensible: In contrast to energy, entropy is not a conservation value, since entropy can be generated within a system. Nevertheless, entropy can be balanced. A distinction is made between closed/open systems and thermodynamic cycles. Furthermore, a new state diagram, i.e. the Ts-diagram, is derived. Such a state diagram visualises the process values specific heat q and specific dissipation $$\psi$$. Obviously, together with the pv-diagram, that illustrates the other process value, i.e. the specific work, it is an important diagram in thermodynamics. Finally, two new changes of state, namely the isentropic and polytropic change of state are treated in this chapter.
Achim Schmidt

### Chapter 14. Transient Processes

Abstract
Irreversibility plays an important role in evaluating thermodynamic processes. In case friction occurs, a system can not be operated reversibly. This has been treated in Chap. : A wire pendulum for instance does not reach its initial, starting position after a number of oscillations. The amplitude decreases by time until the pendulum finally stops in its rest position. This is due to dissipation, i.e. friction at the mounting point as well as due to interactions between environment and pendulum. Consequently, kinetic energy is dissipated and transferred to the environment from where it can not be gained back into the system. Consequently, the state value entropy in such a case rises due to dissipation. It has been shown, that the rate of generation of entropy is a quantitive measure for the degree of irreversibility.
Achim Schmidt

### Chapter 15. Second Law of Thermodynamics

Abstract
Though the balance of entropy has been denoted as Second Law of Thermodynamics in the previous chapters, its classical formulation comes along with thermodynamic cycles. Thus, the focus now is on these cycles. However, there are two different types of thermodynamic cycles: Clockwise cycles on the one hand convert heat into mechanical energy and are named thermal engines. Counterclockwise cycles on the other hand are fridges and heat pumps, that lift thermal energy from a lower to a higher temperature level. However, at that stage cycles are represented in a so-called black-box notation, i.e. the fluxes at its border are balanced without focussing on the detailed physical processes that run inside the machine.
Achim Schmidt

### Chapter 16. ExergyExergy

Abstract
So far the law of energy conservation, i.e. the first law of thermodynamics, has been discussed, that quantified the changeability of energy. However, this changeability is limited, so that thermal energy can not be converted completely into mechanical energy in a steady state process. The second law of thermodynamics can be applied to evaluate the constraints of energy conversion. In Chap. 15 a clockwise Carnot machine was introduced: A machine that operates between two thermal reservoirs in order to convert thermal energy into maximum mechanical work. Its efficiency in best case is given by the minimum and maximum temperature the machine is working in-between. This principle is essential to understand the thermodynamic idea of exergy as maximum working capability of any form of energy. The significance of the exergy is presented in this chapter.
Achim Schmidt

### Chapter 17. Components and Thermodynamic Cycles

Abstract
Although many technical components, that are required to run energy conversion processes, have already been discussed previously, the focus in this chapter is on thermal turbo-machines as well as on heat exchangers. However, in this chapter the technical components are treated in steady state operation. In order to quantify the efficiency of turbine and compressor the so-called isentropic efficiency is defined. In addition, relevant thermodynamic cycles are introduced and discussed. Cyclic processes have been discussed in the previous chapters as well, but mostly in a black-box notation. A distinction has been made between clockwise cycles, i.e. thermal engines, and counterclockwise cycles, i.e. cooling machines respectively heat pumps. This chapter concludes the first part of this book.
Achim Schmidt

### Chapter 18. Single-Component Fluids

Abstract
This chapter deals with single-component fluids, i.e. pure fluids not being mixed with other fluids. In Part I idealised single-component fluids have already been introduced: Ideal gases, that follow the well-known thermal and caloric equations of state, and incompressible liquids, whose specific volume remains constant. For incompressible liquids the caloric equations of state can be adapted easily. The behaviour of real fluids, however, deviates from the idealised models. In particular, real fluids can change their state of aggregation.
Achim Schmidt

### Chapter 19. Mixture of Gases

Abstract
So far single component fluids, i.e. fluids of one chemical composition, have been treated. Part I has focused on ideal gases and incompressible liquids. In Chap. 18 fluids underlying a phase change have been investigated, i.e. a fluid can occur in solid, liquid and gaseous state—however, its chemical composition remains constant even when a phase change takes place.
Achim Schmidt

### Chapter 20. Humid Air

Abstract
In the previous chapter it has been clarified how to handle mixtures of ideal gases. Definitions for characterising the mixture’s composition have been introduced as well as the irreversibility of the mixing of different ideal gases. However, in this chapter the focus is on a technical relevant mixture of fluids: humid air.
Achim Schmidt

### Chapter 21. Steady State Flow Processes

Abstract
Steady state flow processes have already been discussed when introducing the first law of thermodynamics for open systems, see Sect. 11.3. Figure 21.1 shows an example for a simple open system with a single inlet and a single outlet. Characteristic for open systems in steady state is, that the mass inside the system remains constant with respect to time, so that the mass flux into the system needs to be balanced by the mass flux out of the system, i.e.
\begin{aligned} \boxed {\dot{m}_{1}=\dot{m}_{2}}. \end{aligned}
Achim Schmidt

### Chapter 22. Thermodynamic Cycles with Phase Change

Abstract
Though thermodynamic cycles have already been introduced in Sect.  17.2, this chapter covers cycles, in which the working fluid is subject to phase changes. If a system after several changes of state finally reaches the initial state, it is called a thermodynamic cycle. Thus, all state values reach their initial value, i.e.
Achim Schmidt

### Chapter 23. Combustion Processes

Abstract
In Part I ideal gases and incompressible liquids have been introduced and the basic thermodynamic correlations have been derived. Part II has shown, that real fluids can be subject to a change of aggregate state and that many thermodynamic cycles are based on phase change, e.g. a steam power plant. Furthermore, Part II covered mixture of fluids, e.g. humid air or mixtures of ideal gases. However, within these mixture each component was stable and not part of a chemical reaction, i.e. decomposition of the present atoms and molecules. In Part III the focus now is on chemical reacting systems: First, the stoichiometry of a chemical reaction is investigated, i.e. the principle of mass conservation is applied to reactants and products of a chemical reaction. In doing so it is possible to predict the composition of the products. This is important for instance, when the composition of an exhaust gas of a combustion process needs to fulfil technical thresholds. Second, a chemical reacting system is investigated in terms of an energy balance. In this chapter the heating-value approach is followed with focus on technical combustions, i.e. combustions based on fossil fuels. In Chap. 24 another energetic approach based on absolute enthalpy respectively entropy is introduced. Major advantage of this method is, that the irreversibility of chemical reactions can be quantified.
Achim Schmidt

### Chapter 24. Chemical Reactions

Abstract
This chapter examines chemical reactions more generally than the previous chapters. In Chap. 23 the focus has been on the combustion of fossil fuels and the lower/upper heating value has been introduced to handle a chemical decomposition energetically. However, this method used to be rather impractical, since the oxidisation has to be split into several parts and a distinction has been required whether condensation occurs. Now, a more straightforward method is preferred by introducing the so-called absolute specific enthalpy/entropy. In doing so, the specific enthalpy does not only imply a caloric effect, as it has been done in part I and II, but it includes the specific chemical bonded energy as well.
Achim Schmidt


text:
# What stops charges from jumping across a capacitor?

I've learnt in class that charges store up on capacitor plates (the electrons). But why don't the electrons just jump across capacitor, what stops them? At first I thought perhaps it might just be the air molecules, as they are insulating and have no charge carriers, the electrons wouldn't have enough energy to barge their way through.

How about if I had a capacitor in a vacuum though? Would the electrons just be able to leap across the capacitor because there is nothing there to stop them? If this were the case however, the capacitor could theoretically be any separation and the electrons would simply float across, which seems absurd, so there must be something stopping the electrons from leaving?

• What makes you think that a vacuum is a conductor? A vacuum is a better insulter than air. Generally, it is the ion content of the medium that determines conductivity. (excluding known elemental conductors, like copper.) For example, pure water is a fairly good insulter. Impurities and things like salt, which breaks into its component ions, are generally why we think of water as conductive. -See: google.com/… and en.wikipedia.org/wiki/… – CoilKid Mar 13 '15 at 23:57

This is quite an interesting question.

First - air is a (poor) conductor. See this earlier answer for some details on how well (or poorly) air conducts (especially when the relative humidity increases).

Next - vacuum as an insulator. You are right that once electrons are "in space", a vacuum doesn't provide much impediment. This is why cathode ray tubes (old TVs, oscilloscopes) and other vacuum tubes could work. But you might know that these typically contain a hot cathode: a filament that is heated, and that then gives off ("evaporates") electrons. The hotter the filament, the more electrons. This is a good way to regulate current in vacuum tubes (to this day, it's how X ray tubes work!).

The analogy of "boiling off" the electrons is quite good. Just as water, when heated, will evaporate more vigorously, so electrons will escape material more readily when they are hot. And just as some liquids have lower boiling points than others, different materials have different "work functions" - the amount of energy that an electron needs to be given before it can escape.

This was demonstrated, of course, in the early 1900's when it was shown that shining light of a certain wavelength or shorter could excite electrons and cause them to escape from metals (the photoelectric effect) but that longer wavelength light, regardless of intensity, could not. This was actually later understood to be caused by the particle nature of light and one of the first piece of experimental evidence of quantum mechanics (for which Einstein got a Nobel prize).

Back to the electrons and their work function. I wrote an answer about this a while ago: that was actually more detailed, looking at the velocity distribution of electrons that do escape. Taking a step back from that, electrons in a metal behave a bit like a gas: they move around randomly, bump into each other, and their velocities follow the Boltzmann distribution. This means there are always a few highly energetic ones around - more when the metal is hot. There is a natural energy barrier keeping the electrons bound to the metal, and electrons with energy greater than that can and will escape. This is described in this wiki article.

Perhaps you have already understood that a capacitor in vacuum can also lose charge when exposed to light: so if you want your vacuum capacitor to hold charge for a long time, you have to keep it cold and dark.

The same question could be asked about electrons in atoms. Why on Earth do they remain bound to the nucleus instead of floating around freely? The answer to this question and to yours is that the nucleus (in the case of atoms) and the metal (in your case) is offering a quite cozy place for the electrons to be and leaving it is quite costly in term of energy.

From an energetic point of view, electrons have to move across a huge barrier to go to the other side and in normal capacitors the applied potential difference is simply not enough to provide this missing energy.

Sometimes however, the applied voltage can reach what is called the breakdown voltage which is a voltage that is strong enough to help the electrons climb the energetic barrier and cross the gap: this generates a spark that is not often wanted. This voltage breakdown can be due to either a too high voltage to begin with or a change in the external conditions (humidity, damaging of the electrodes etc..) that will lower the barrier to cross.

Lastly, you are right to say that the vacuum is conducting in some sense (at least in the sense that an electron in vacuum with a certain speed does not need a constant supply of energy to continue moving) and the presence of some atoms that do not favour charge transport in air can worsen the crossing of the gap. Experiments have shown that the breakdown voltage was lowered as the quality of the vacuum was increased.

• I thought the electrons were roughly bound to nucleus because of electrostatic force (nucleus being positive)? But in the case of capacitor, don't the electrons gather all on one plate, and instead of having an attractive electrostatic force, you have repulsive because you have a buildup of negative charge? And why does such an energetic barrier exist that prevents the electrons from jumping across? thanks – Joshua Lin Mar 14 '15 at 0:12
• the build up of the potential difference created between two capacitor plates (because of the charge accumulation you talk about) is only as much as what you put in with your DC or AC generator. If we call this potential difference $V$, then if $eV < \Delta E$, where $\Delta E = E_{vacuum} - E_{metal}$ is the height of the energetic barrier to cross the vacuum gap, then your electrons have no reason (and no physical mean) to go to the other side. This is both because of the electrostatic interaction and some chemical bonding that the electrons do not want to leave the metal. – gatsu Mar 14 '15 at 9:39


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# Tag Info

2

How about dividing the problem? You can first train a classification model that predicts the type of function (linear or exponential). Then you can use your seperately trained nn depending on the classification output. P.S. I'm not sure why you would use a neural network for this problem. Fitting a linear/exponential function seems to be a relatively simple ...

2

Transformer models have limited sequence length at inference time because of positional embeddings. But there are workarounds. Self-attention in transformer does not distinguish the order of keys/values, it works as if the sequence is a bag of words. So to expose the sequence order to the model, one typically adds an extra "positional embedding" ...

2

What you refer to as logic AI is a subset of what is called symbolic AI, as you manipulate symbols, according to certain rules (which could be rules of logic). These rules are either authored by a human being, or they can be learned from examples. There are algorithms to derive decision trees (eg ID3) or other rule sets from training data. But the important ...

1

I would say that in general situation more estimators are better. RandomForest fits a lot of estimators - decision trees that take a subset of data (obtained sampling with replacement) and subset of features (by default sqrt(n_features) in sklearn). Each of these estimators is noisy and prone to overfitting, producing a complicated decision surface. But when ...

1

Is this a valid implementation of second-order regression? No, but it is not far off. To perform a full second-order regression, you will need all terms for $x_{i,j}x_{i,k}$ where the first index is the example and the second index the feature. This includes every combination of two input variables. In your element-wise squaring you only produce terms $x_{i,... 1 Intuitively, I feel like if there are 30 foods, each with 2 states, then that is 60 states, no$2^{30}$. Let's try it with 3 pellets. If you are right there would be$2 \times 3 = 6$states, if the authors are right there would be$2^3 = 8$states. Using * for a pellet, and - for a space, we have the following states: * * * * * - * - * * - - - * * - * - - -... 1 For deep learning models, embedding vectors have become the standard way of encoding text features almost immediately after their introduction. The reason for this is that neural networks work with data encoded with continuous values ranging from 0 to 1 (or sometimes from -1 to 1). Bag of Words and TF-IDF can be modified to produce values in this range, but ... 1 Here is a quick idea: first calculate the count of how many times each word occurs in these documents (I don't know whether to lowercase them or not, do interface and Interface mean different things?), and sort them in the descending order of occurrence. Most frequent words can be called "keywords" of your configurations (such as vlan), or maybe ... 1 I have found the answer to this doubt I had. Here, 0.007709330413490534 = 1 / S, q = input, Z = 3. Basically, this is the formula to quantize the input value. If you pull out 1/S then it becomes clear. Here, is an article related to this topic. 1 As DKDK said, Indeed one could fit both linear and exponential function and see which one has smaller residual, without using any complex AI. But OTOH this could be a great toy-problem for learning about neural networks. You could have a network with these parts: A network with a final sigmoid activation, which predicts whether the function is linear or not.... 1 I will give you a few scenarios where matrix factorisation stills works pretty well. Topic Modelling : Given a matrix of Document as Rows and Terms/Words as column you can use Non Negative Matrix factorisation to identify Topics. Number of Topics is defined by user or can be treated as hyperparameter. Image Ref : https://towardsdatascience.com/nmf-a-visual-... 1 This is the case as the loss doesn't have to monotonically decrease when it's updated in the negative direction. For example: Let$L(\theta) = \theta^2 $and$\theta_0= 3$Let the subscript n in$\theta_n$denote the iteration number. Then$\nabla_{\theta}L(\theta_0) = 2*\theta = 2*3 = 6$For the loss to decrease in this case$\epsilon < 1\$ needs to hold ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# Bibliotek

Musik » Black Eyed Peas »

## Just Can't Get Enough

66 spelade låtar | Gå till låtsida

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Just Can't Get Enough 3:41 27 jul 2012, 14:28
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Just Can't Get Enough
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Just Can't Get Enough 3:41 25 mar 2011, 14:45
Just Can't Get Enough 3:41 25 mar 2011, 13:29
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Just Can't Get Enough 3:41 24 mar 2011, 19:53
Just Can't Get Enough 3:41 24 mar 2011, 19:26
Just Can't Get Enough 3:41 24 mar 2011, 17:57
Just Can't Get Enough 3:41 24 mar 2011, 17:53
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Just Can't Get Enough 3:41 21 mar 2011, 19:43
Just Can't Get Enough 3:41 21 mar 2011, 19:39
Just Can't Get Enough 3:41 21 mar 2011, 19:35


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# Two force problems

1. Sep 18, 2005

### kevinlikesphysics

3. [SFHS99 4.P.22.] An 1680 kg car is moving to the right at a constant velocity of 2.94 m/s.

(a) What is the net force on the car?
wrong check mark N

(b) What would be the net force on the car if it were moving to the left?
in Newtons = ??

4. [SFHS99 4.P.23.] A freight train has a mass of 5.5 107 kg. If the locomotive can exert a constant pull of 15 105 N, how long would it take to increase the speed of the train from rest to 85 km/h?
in seconds =??

thanks alot

2. Sep 18, 2005

### bayan

The net force will be zero if there is no acceleration! Newtons first law of motion.

same would apply to part b if it moves at constant velocity towards left.

For second question.

we know the mass and we know the force it can apply.
$$F=ma$$

$$\frac{F}{m}=a$$

$$\frac{15105}{5.5*10^7}=a$$

$$a=2.75*10^-^4$$

$$v=u+at$$

velocity is 85KM/h which is 23.61 m/s

$$23.61=0+2.75*10^-^4t$$

$$t=\frac{23.61}{2.75*10^-^4}$$

$$t=85972 s$$

23 hours 52 mins 52 seconds!

if the force applied by engine is 15105N.

if it is $$15*10^5$$ then the answer is 865.7 seconds

3. Sep 18, 2005

### kevinlikesphysics

thanks alot they are correct, your explanations helped alot

Last edited: Sep 18, 2005


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# Three events A, B and C are defined in the sample space S. The events A and B are mutually exclusive and A and C are independent. What is P(A|C)?

## $P \left(A\right) = 0.2 , P \left(B\right) = 0.4$ and $P \left(A \cup C\right) = 0.7$

Jan 28, 2017

$P \left(A | C\right) = 0.2$

#### Explanation:

We use the definition of Conditional Probability;

$P \left(X | Y\right) = \frac{P \left(X \cap Y\right)}{P \left(Y\right)}$

Se we have

$P \left(A | C\right) = \frac{P \left(A \cap C\right)}{P \left(C\right)}$

We are also told that $A$ and $C$ are independent events

$\therefore P \left(A \cap C\right) = P \left(A\right) \cdot P \left(C\right)$

So:

$P \left(A | C\right) = \frac{P \left(A\right) \cdot P \left(C\right)}{P \left(C\right)}$
$\text{ } = P \left(A\right)$
$\text{ } = 0.2$


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## DMPG '16 B6 - Counting Money

View as PDF

Points: 7 (partial)
Time limit: 1.0s
Memory limit: 64M

Author:
Problem types
Allowed languages
Ada, Assembly, Awk, Brain****, C, C#, C++, COBOL, CommonLisp, D, Dart, F#, Forth, Fortran, Go, Groovy, Haskell, Intercal, Java, JS, Kotlin, Lisp, Lua, Nim, ObjC, OCaml, Octave, Pascal, Perl, PHP, Pike, Prolog, Python, Racket, Ruby, Rust, Scala, Scheme, Sed, Swift, TCL, Text, Turing, VB, Zig

Bob is now so rich that he is no longer able to keep track of his own wealth. Instead, he now uses a computer to do it for him. Most computers use binary, or base to count. However, Bob's computer needs to handle negative numbers as well, so instead, it uses base . Base is very similar to base , except the digits represent increasing powers of instead of . This means that the digits (from right to left) represent , , , , and so on, instead of , , , . However, there is still a problem: his computer can't convert between base and base yet. Can you help Bob write a program to do this?

There are two operations that will be needed:

• A x Change , a number in base , into base .
• B y Change , a number in base , into base .
Base -2 Conversion Base 10

#### Input Specification

The first line of input will contain , the number of operations to be performed.

The next lines will each contain an operation in the format described above.

For at least of the marks, you will only need to convert from base to base .

For at least another of the marks, .

#### Output Specification

For each operation, print out the result of the operation on a single line.

#### Sample Input

2
A 11011
B 6

#### Sample Output

7
11010


text:
Metafor

ULiege - Aerospace & Mechanical Engineering

Sidebar

doc:user:elements:volumes:hyper_materials

Hyperelastic materials

NeoHookeanHyperPk2Material

Description

Neo-Hookean hyperelastic law, using a PK2 tensor.

The potential per unit volume is computed based on the average compressibility over the element, ($\theta$):

$$U^{vol}=\dfrac{k_0}{2} \left[\ln\right(\theta\left)\right]^2$$

The deviatoric potential is computed based on a Cauchy tensor with a unit determinant:

$$U^{dev}=\dfrac{g_0}{2} \left[\text{tr}\right(\hat{\mathbf{C}}\left)-3\right]$$

Parameters

Name Metafor Code
Density MASS_DENSITY
Initial bulk modulus ($k_0$) HYPER_K0
Initial shear modulus ($g_0$) HYPER_G0

LogarihtmicHyperPk2Material

Description

Logarithmic hyperelastic law, using a PK2 tensor.

The potential per unit volume is computed based on the average compressibility of the element, ($q$):

$$U^{vol}=\dfrac{k_0}{2} \left[\ln\right(\theta\left)\right]^2$$

The deviatoric potential is computed based on a Cauchy tensor with a unit determinant:

$$U^{dev}= \dfrac{g_0}{4} \ln \left(\hat{\mathbf{C}}\right):\ln \left(\hat{\mathbf{C}}\right)$$

Parameters

Name Metafor Code
Density MASS_DENSITY
Initial bulk modulus ($k_0$) HYPER_K0
Initial shear modulus ($g_0$) HYPER_G0

EvpIsoHLogarithmicHyperPk2Material

Description

Logarithmic hyperelastic law, using a PK2 tensor.

The potential per unit volume is computed based on the average compressibility of the element, ($\theta$):

$$U^{vol}=\dfrac{k_0}{2} \left[\ln\right(\theta\left)\right]^2$$

The deviatoric potential is computed based on a Cauchy tensor with a unit determinant:

$$U^{dev}= \dfrac{g_0}{4} \ln \left(\hat{\mathbf{C}}^{el}\right):\ln \left(\hat{\mathbf{C}}^{el}\right)$$

Parameters

Name Metafor Code Dependency
Density MASS_DENSITY -
Initial bulk modulus ($k_0$) HYPER_K0 -
Initial shear modulus ($g_0$) HYPER_G0 -
Number of the material law which defines the yield stress $\sigma_{yield}$ YIELD_NUM -

FunctionBasedHyperPk2Material

Description

Hyperelastic law, using a PK2 tensor. Its function applied on the strain spectral decomposition is a user law.

The potential per unit volume is computed based on the average compressibility of the element, ($\theta$):

$$U^{vol}=\dfrac{k_0}{2} \left[\ln\right(\theta\left)\right]^2$$

The deviatoric potential is computed based on a hyperelastic user function defined in Viscoelastic laws.

Parameters

Name Metafor Code Dependency
Density MASS_DENSITY -
Initial bulk modulus ($k_0$) HYPER_K0 -
Number of the hyperelastic law HYPER_FUNCTION_NO -

VeIsoHyperPk2Material

Description

Viscoelastic hyperelastic law, using a PK2 tensor. The law includes a main branch (spring and dashpot in parallel) and one or several Maxwell branches (spring and dashpot in series).

Each branch has its behavior corresponding to a viscoelastic law, supplied by the user.

The potential per unit volume is computed based on the average compressibility of the element, ($\theta$):

$$U^{vol}=\dfrac{k_0}{2} \left[\ln\right(\theta\left)\right]^2$$

The deviatoric potential is computed based on the viscoelastic laws :

$$U^{dev}= U^{dev}_{\text{main,elastic}}\left(\hat{C}\right) + \sum_{Maxwell} U^{dev}_{\text{Maxwell,elastic}}\left(\hat{C}^{\text{el}}\right)$$

The dissipation potential is written as:

$$\Delta t \phi^{dev}= \Delta t \phi^{dev}_{\text{main,viscous}}\left( \exp{\frac{\ln{\Delta\hat{C}}}{\Delta t}} \right) + \sum_{Maxwell} \Delta t \phi^{dev}_{\text{Maxwell,viscous}}\left(\exp{\frac{\ln{\Delta C^{\text{vis}}}}{\Delta t}} \right)$$

where $$\Delta\hat{C} = {\hat{F}^n}^{-T} \hat{C}^{n+1} {\hat{F}^n}^{-1}$$

$$\Delta C^{\text{vis}} = {{F^{\text{vis}}}^n}^{-T} {C^{\text{vis}}}^{n+1} {{F^{\text{vis}}}^n}^{-1}$$

The potentials $U^{dev}_{\text{main,elastic}},~~U^{dev}_{\text{Maxwell,elastic}},~~\phi^{dev}_{\text{main,viscous}},~~\phi^{dev}_{\text{Maxwell,viscous}}$ are hyperelastic functions defined in Viscoelastic laws.

Parameters

Name Metafor Code Dependency
Density MASS_DENSITY -
Initial bulk modulus ($k_0$) HYPER_K0 -
Number of the main viscoelastic law MAIN_FUNCTION_NO -
Number of the first Maxwell viscoelastic law MAXWELL_FUNCTION_NO1 -
Number of the second Maxwell viscoelastic law (optional) MAXWELL_FUNCTION_NO2 -
Number of the third Maxwell viscoelastic law (optional) MAXWELL_FUNCTION_NOI -


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# Step-by-step Solution

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asin
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asinh
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## Step-by-step explanation

Problem to solve:

$\int\left(\right)dx\sin\left(\right)^2$

Learn how to solve trigonometric integrals problems step by step online.

$\int0.9167dx$

Learn how to solve trigonometric integrals problems step by step online. Solve the trigonometric integral int(sin(2))dx. Calculating the sine of 2 degrees. The integral of a constant is equal to the constant times the integral's variable. As the integral that we are solving is an indefinite integral, when we finish we must add the constant of integration.

$\frac{11}{12}x+C_0$

### Problem Analysis

$\int\left(\right)dx\sin\left(\right)^2$

### Main topic:

Trigonometric integrals

~ 1.37 seconds


text:
# zbMATH — the first resource for mathematics

Geodesic equivalence via integrability. (English) Zbl 1017.37029
Given two symplectic manifolds $$(M,\omega)$$ and $$(M', \omega')$$ with Hamiltonian functions $$H$$ and $$H'$$, let $$\Phi: M\to M'$$ be a diffeomorphism mapping an isoenergy surface $$Q$$ in $$M$$ to an isoenergy surface $$Q'$$ in $$M'$$. The diffeomorphism is said to be orbital if its restriction to $$Q$$ takes the orbits of the Hamiltonian system $$(M,\omega, H)$$ to the orbits of $$(M',\omega',H')$$. In the paper under review, the authors shows how to construct a one-paramter family of integrals from a given orbital diffeomorphism between two Hamiltonian systems.
As a significant application of this result, whenever two geodesically equivalent Riemannian metrics $$g$$ and $$g'$$ are given on the same manifold $$M$$, a set of $$n=\text{dim} M$$ integrals for the geodesic flow is obtained. Such integrals Poisson commute as a consequence of a classical result proved by Levi-Civita in 1896. Let $$G$$ be the endomorphism of the tangent bundle defined by raising one index of $$g'$$ by means of $$g$$. If the eigenvalues of $$G$$ are all distinct at least at one point and $$M$$ is connected, then they are all distinct at almost every point, and the integrals are proved to be functionally independent. Therefore, the geodesics flows of the two metrics are completely integrable in the sense of Arnol’d-Liouville.
As a corollary of this result, the authors find out a topological condition preventing a closed connected real-analytic manifold from admitting a pair of geodesically equivalent metrics such that the eigenvalues of $$G$$ are all distinct at least at one point.

##### MSC:
 37J35 Completely integrable finite-dimensional Hamiltonian systems, integration methods, integrability tests 53C22 Geodesics in global differential geometry 70H06 Completely integrable systems and methods of integration for problems in Hamiltonian and Lagrangian mechanics 53A20 Projective differential geometry
Full Text:


text:
2013-14 Nilpotent matrix

Let $$A, B$$ are $$N \times N$$ complex matrices satisfying $$rank(AB – BA) = 1$$. Prove that $$(AB – BA)^2 = 0$$.

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text:
# Why does the man at the front desk tell saul to visit the reference desk? a) he says that the reference

###### Question:

Why does the man at the front desk tell saul to visit the reference desk?
a) he says that the reference desk is the best place to take a nap.
b) he thinks that saul would enjoy talking to the person at the reference desk.
c) he wants saul to relay a message to the person at the reference desk
d) he believes that the person at the reference desk will best be able to saul find what he needs.

### Level2.2 Identy functions: graphs AEBLook at this graph:AYIs this relation a function?yesno

Level 2.2 Identy functions: graphs AEB Look at this graph: AY Is this relation a function? yes no...

### Now factor the expression. 1/2 a + 6

Now factor the expression. 1/2 a + 6...

### The school is 3.2 blocks from hector's house. how many blocks does hector walk to go to and from school

The school is 3.2 blocks from hector's house. how many blocks does hector walk to go to and from school...

### True/False Indicate whether the sentence or statement is true or false. Historians and archaeologists

True/False Indicate whether the sentence or statement is true or false. Historians and archaeologists did not know hieroglyphics existed until the Rosetta Stone was discovered. Please select the best answer from the choices provided T F...

### What kind of reaction occurs when potassium is placed in water? a. a single-displacement reaction c. a decomposition reaction

What kind of reaction occurs when potassium is placed in water? a. a single-displacement reaction c. a decomposition reaction b. a double-displacement reaction d. electrolysis...

### Lgbtq study-ishAre pansexual asexual omnisexual valid ?

Lgbtq study-ish Are pansexual asexual omnisexual valid ?...

### What's does the author want the reader to infer about the nature of the war from this selection?

What's does the author want the reader to infer about the nature of the war from this selection?...

### Acartoon lasts 18 minutes. how many seconds is the cartoon?

Acartoon lasts 18 minutes. how many seconds is the cartoon?...

### Is this correct? just double checking. let me know asap

Is this correct? just double checking. let me know asap $Is this correct? just double checking. let me know asap$...

### In round-robin scheduling, a) the time quantum should be larger than the context switch time. b) the time quantum should be as

In round-robin scheduling, a) the time quantum should be larger than the context switch time. b) the time quantum should be as smaller than the the context switch time. c) the time quantum should be equal to 200 ms divided by the number of precess waiting. d) the time quantum should be as small as p...

### Earth's most recent ice age began approximately how long ago?

Earth's most recent ice age began approximately how long ago?...

### In this lab you will use a cart and track to explore various aspects of motion. you will

In this lab you will use a cart and track to explore various aspects of motion. you will measure and record the time it takes the cart to travel certain distances, and then complete some calculations. in the space below, write a scientific question that you will answer by doing this experiment....

### Nistu is cooking rice in a pot. When she leaves the pot on the stove for a while, the temperature of the rice increases.

Nistu is cooking rice in a pot. When she leaves the pot on the stove for a while, the temperature of the rice increases. What happens to the molecules of the rice when the temperature of the rice increases?...

### Use rhombus IJKL and the given information to find each value. M angle 3 = If m angle 1 = 62 degree, find m angle 3. x = If m

Use rhombus IJKL and the given information to find each value. M angle 3 = If m angle 1 = 62 degree, find m angle 3. x = If m angle 4 = (3x - 1) degree and m angle = (2x + 30) degree, find the value of x'....

### Fuel economy estimates for automobiles built in a certain year predicted a mean of 27.8 mpg & a

Fuel economy estimates for automobiles built in a certain year predicted a mean of 27.8 mpg & a standard deviation of 5.2 mpg for highway driving. Assume that a normal distribution can be applied. Within what range are 95% of the automobiles?...

### A motorboat can go 8 miles downstream on a river in 20 minutes. it takes 30 minutes for the boat to go upstream the same

a motorboat can go 8 miles downstream on a river in 20 minutes. it takes 30 minutes for the boat to go upstream the same 8 miles. find the speed of the current....

### What problems may global warming cause?

What problems may global warming cause?...

### Can someone by answering these i’m having trouble

Can someone by answering these i’m having trouble...
Which of the equations below represents a line parallel to the x-axis $Which of the equations below represents a line parallel to the x-axis$...