text:
An alternative to the breeder’s and Lande’s equations

An alternative to the breeder’s and Lande’s equations

format: plain


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# Math Help - Quadratics - sequences and squares

1. ## Quadratics - sequences and squares

Okay, I've got this question and I've spent about an hour trying to figure it out and it's doing my head in:

Write a formula for the nth term of this quadratic sequence:
7, 15, 27, 43, 63
nth term = ??

Write this expression in completed square form:
x² - 10x +2

Hence solve
x² - 10x +2 = -5

Ok, any help would be appreciated as I've tried many ways of doing it....
Thanks =D

2. Hello, Soriku Strife!

Write a formula for the $n^{th}$ term of this quadratic sequence:
. . 7, 15, 27, 43, 63, . . .
Here's one approach to this problem (bear with me, it's quite long).

We write a general quadratic function: . $f(n) \;=\;an^2 + bn + c$
. . and we must determine $a,\,b,\,c.$

We have three "unknowns", so we need three equations.
We'll use the first three terms of the sequence.

The first term is 7. .That is: . $f(1) = 7$
. . $a\cdot1^2 + b\cdot1 + c \:=\:7\quad\Rightarrow\quad a + b + c \:=\:7\;\;{\color{blue}[1]}$

The second term is 15. .That is: . $f(2) = 15$
. . $a\cdot2^2 + b\cdot2 + c\:=\:15\quad\Rightarrow\quad 4a + 2b + c \:=\:15\;\;{\color{blue}[2]}$

The third term is 27. .That is: . $f(3) = 27$
. . $a\cdot3^2 + b\cdot3 + c \:=\:27\quad\Rightarrow\quad 9a + 3b + c \:=\:27\;\;{\color{blue}[3]}$

$\begin{array}{ccccc}\text{Subtract {\color{blue}[1]} from {\color{blue}[2]}:} & 3a + b & = & 8 & {\color{blue}[4]} \\
\text{Subtract {\color{blue}[2]} from {\color{blue}[3]}:} & 5a + b & = & 12 & {\color{blue}[5]}\end{array}$

Subtract [4] from [5]: . $2a \:=\:4\quad\Rightarrow\quad\boxed{ a \:=\:2}$

Substitute into [4]: . $3(2) + b \:=\:8\quad\Rightarrow\quad\boxed{ b \:=\:2}$

Substitute into [1]: . $2 + 2 + c \:=\:7\quad\Rightarrow\quad\boxed{ c \:=\:3}$

Therefore, the function is: . $f(n) \;=\;2n^2 + 2n + 3$

3. Thanks, made more sense than the way I did it

4. Hello again, Soriku Strife!

Write this expression in completed square form: . $x^2 - 10x + 2$

We have: . $x^2 - 10x + 2$

To complete the square: take one-half of the x-coefficient and square it.

. . $x^2 - 10x \,{\color{blue}+\,25} + 2 \,{\color{blue}-\,25}\;=\;(x^2-10x + 25) - 23 \;=\;(x-5)^2-23$

Hence, solve: . $x^2 - 10x +2 \:= \:-5$

We have the equation: . $(x-5)^2 - 23 \:=\:-5$

Add 23 to both sides: . $(x - 5)^2 \:=\:18$

Take square roots: . $x - 5 \;=\;\pm\sqrt{18}\quad\Rightarrow\quad x - 5\;=\;\pm3\sqrt{2}$

Add 5 to both sides: . $\boxed{x \;=\;5 \pm3\sqrt{2}}$


text:
# Summation

A summation is a concise expression describing the addition of an arbitrarily large set of numbers. A capital sigma is denotes summation.

$$\sum_{i=1}^{n} f(i) = f(1) + f(2) + ... + f(n)$$

Here, $$m$$ is the lower bound and $$n$$ is the upper bound. The summation starts at $$m$$ and ends at $$n$$. $$i$$ is the index of summation.

## Examples

Simple closed forms exist for many algebraic functions.

$$\sum_{i=1}^{n} c = nc$$ $$\sum_{i=1}^{n} 1 = n$$ $$\sum_{i=1}^{n} i = \frac{n(n + 1)}{2}$$ $$\sum_{i=1}^{3} i^2 = 1^2 + 2^2 + 3^2 = 14$$

## Product

A similar notation is used to convey iterative multiplication.

$$\prod_{i=a}^{b} f(i)$$


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# Do the octonions contain infinitely many copies of the quaternions?

Note: by "infinitely many", I'm confident I always mean $$\beth_1$$ many herein.

We can easily show the quaternions contain infinitely many copies of $$\Bbb C$$ because, given any unit vector $$\in\Bbb R^3$$ of components $$b,\,c,\,d$$, $$\Bbb R[h]$$ is isomorphic to $$\Bbb C$$ with $$h:=bi+cj+dk$$. Sure, these aren't "independent" copies of $$\Bbb C$$ in the same way $$\Bbb R[i],\,\Bbb R[j],\,\Bbb R[k]$$ are. But it's still of interest because, for example, a family of tensor products over matrices using different copies of $$\Bbb C$$ provide an easy definition of a determinant, even though general matrices of quaternions prohibit this. For example, if $$A_1,\,\cdots,\,A_n$$ are matrices and $$O$$ denotes a contextually appropriate zero matrix, the block matrix $$\left(\begin{array}{cccc} A_{1} & O & \cdots & O\\ O & A_{2} & \cdots & O\\ \vdots & \vdots & \ddots & O\\ O & O & \cdots & A_{n} \end{array}\right)$$can be said to have determinant $$\prod_l\det A_l$$. The order is important, but one in particular is natural.

I may be overlooking some details of the above benefit to arbitrarily many copies of $$\Bbb C$$ in $$\Bbb H$$, wherein a very rich commuting family of matrices is constructed, but my question isn't about that. I'm wondering how we'd prove there are infinitely many copies of $$\Bbb H$$ in $$\Bbb O$$. (Again, block matrices provide a benefit, in this case inheriting the associativity of the $$A_l$$.) I suspect a proof exists that admits the following sketch:

• In $$\Bbb O$$, create some $$h_1,\,h_2$$ each analogous to $$i\in\Bbb C$$, generating an associative algebra and satisfying $$h_1h_2=-h_2h_1$$;
• Note that any such pair of square roots of $$-1$$ can model the quaternions viz. $$i=h_1,\,j=h_2,\,k=h_1h_2$$;
• Show the above can be done in infinitely many different ways.

Of course there are infinitely many ways to choose the pair $$(h_1,\,h_2)$$, but $$\Bbb R[h_1,\,h_2]$$ won't always be a different set for such pairs. That's why I suspect the proof requires a few clever i-dotting t-crosses.

• In fact, we can identify the space of copies of $\Bbb H$ in $\Bbb O$ with the compact Riemannian symmetric space of type $G$, namely, $G_2 / SO(4)$; in particular this space has dimension $8$. – Travis Feb 2 at 5:29

Yes. One can show the following 2 things:

1. Any octonion $$x$$ not in $$\mathbb R$$ generates a subalgebra $$A$$ isomorphic to $$\mathbb C$$.

2. For any octonion $$y$$, the subalgebra generated $$B$$ by $$x$$ and $$y$$ is associative.

This should be covered in treatments of composition algebra, e.g. it is probably in Springer-Veldkamp.

Hence if we take $$y$$ outside of $$A$$ in 1, by Frobenius' classification of $$\mathbb R$$-division algebras, $$B$$ must be isomorphic to $$\mathbb H$$ (a priori it may not be obvious $$B$$ is division, but you can get this a posteriori from Frobenius' theorem as adjoining inverses to $$B$$, which necessarily lie in $$\mathbb O$$, still gives you an associative algebra), and thus the 4-dimensional space spanned by $$1, x, y, xy$$. Since no finite collection of $$\mathbb R^4$$ subspaces of $$\mathbb R^8$$ cover $$\mathbb R^8$$, you get infinitely many copies of $$\mathbb H$$.

To say that you get (at least) the cardinality of the continuum copies of $$\mathbb H$$, it suffices to show

1. For any nonreal octonion $$z$$, there is a subalgebra $$B \simeq \mathbb H$$ as above not containing $$z$$.

To see this, take $$x$$ as above not in $$\mathbb R z$$. Then $$x$$ and $$z$$ generate a space $$B' \simeq \mathbb H$$. Simply take $$y \not \in B'$$. Then the algebra $$B$$ generated by $$x$$ and $$y$$ cannot contain $$B'$$ since $$B \ne B'$$ but $$\dim B = \dim B'$$.

• For what it's worth, property (2) is called alternativity. – Travis Feb 2 at 5:30
• I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry – Vincent Mar 19 at 22:55


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Dataplot Vol 1 Vol 2

# COCHRAN TEST

Name:
COCHRAN TEST
Type:
Analysis Command
Purpose:
Perform a Cochran test that c treatments have identical effects.
Description:
The Cochran test is a non-parametric test for analyzing randomized complete block designs where the response variable is a binary variable (i.e., there are only two possible outcomes, which are coded as 0 and 1).

The Cochran test assumes that there are c experimental treatments (c >= 2). The observations are arranged in r blocks, that is

Treatment
Block 1 2 ... c
1 X11 X12 ... X1c
2 X21 X22 ... X2c
3 X31 X32 ... X3c
... ... ... ... ...
r Xr1 Xr2 ... Xrc

The Friedman test is the usual non-parametric test for this kind of design. The Cochran test is applied for the special case of a binary response variable (i.e., it can have only one of two possible outcomes).

Then the Cochran test is

 H0: The treatments are equally effective. Ha: There is a difference in effectiveness among treatments. Test Statistic: The Cochran test statistic is $$T = c(c-1) \frac{\sum_{i=1}^{c}{(C_j - \frac{N}{c})^2}} {\sum_{i=1}^{r}{R_i(c - R_i)}}$$ with c, Ci r, Ri and N denoting the number of treatments, the column total for the ith treatment, the number of blocks, the row total for the ith block, and the grand total, respectively. Significance Level: $$\alpha$$ Critical Region: T > $$\chi_{1-\alpha,c-1}^{2}$$ where $$\chi^{2}$$ is the chi-square percent point function. Note that this is based on a large sample approximation. In particular, it assumes that r is "large". Conclusion: Reject the null hypothesis if the test statistic is in the critical region.

Syntax:
COCHRAN TEST <y> <block> <treat>
<SUBSET/EXCEPT/FOR qualification>
where <y> is the response variable;
<block> is a variable that identifies the block;
<treat> is a variable that identifies the treatment;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
Examples:
COCHRAN TEST Y BLOCK TREATMENT
COCHRAN TEST Y X1 X2
COCHRAN TEST Y BLOCK TREATMENT SUBSET BLOCK > 2
Note:
In Dataplot, the variables should be given as:

Y BLOCK TREAT

X11 1 1
X12 1 2
... 1 ...
X1c 1 c
X21 2 1
X22 2 2
... 2 ...
X2c 2 c
... ... ...
Xr1 r 1
Xr2 r 2
... r ...
Xrc b c

If your data are in a format similar to that given in the DESCRIPTION section (i.e., you have colums Y1 to Yc, each with r rows), you can convert it to the format required by Dataplot with the commands:

LET NTREAT = 5
LET NBLOCK = SIZE Y1
LET NTOTAL = K*NBLOCK
LET BLOCK = SEQUENCE 1 1 NBLOCK FOR I = 1 1 NTOT
LET Y2 TREAT= STACK Y1 Y2 Y3 Y4 Y5
COCHRAN TEST Y2 BLOCK TREAT
Note:
The Cochran test is based on the following assumptions:

1. The blocks were randomly selected from the population of all possible blocks.

2. The outcomes of the treatments can be coded as binary responses (i.e., a "0" or "1") in a way that is common to all treatments within each block.
Note:
The case where there are exactly two treatments is equivalent to the McNemar test. The McNemar test is equivalent to a two-tailed sign test.
Note:
If the Cochran test rejects the null hypothesis of equally effective treatments, pairwise multiple comparisons can be made by applying the Cochran test on the two treatments of interest. For example, to test treatments 3 and 5, you can do something like the following

COCHRAN TEST Y BLOCK TREATMENT SUBSET TREATMENT = 3 5
Default:
None
Synonyms:
None
Related Commands:
 FRIEDMAN TEST = Perform a Friedman test. KRUSKAL WALLIS TEST = Perform a Kruskal Wallis test. ANOVA = Perform an analysis of variance. SIGN TEST = Perform a sign test. MEDIAN POLISH = Carries out a robust ANOVA. T TEST = Carries out a t test. RANK SUM TEST = Perform a rank sum test. SIGNED RANK TEST = Perform a signed rank test.
Reference:
Conover (1999), "Practical Nonparametric Statistics", Third Edition, Wiley, pp. 250-256.
Applications:
Analysis of Binary Two-Way Randomized Block Designs
Implementation Date:
2005/12
Program:

. Following example from p. 253 of Conover
1 1 1
1 1 1
0 1 0
1 1 0
0 0 0
1 1 1
1 1 1
1 1 0
0 0 1
0 1 0
1 1 1
1 1 1
END OF DATA
LET N1 = SIZE Y1
LET NTOT = 3*N1
LET BLOCK = SEQUENCE 1 1 N1 FOR I = 1 1 NTOT
LET Y TREAT = STACK Y1 Y2 Y3
COCHRAN Y BLOCK TREAT

This example generates the following output

Cochran Test for Two-Way Randomized Complete Block Designs
(Dichotomous Data)

Response Variable: Y
First Group-ID Variable: BLOCK
Second Group-ID Variable: TREAT

H0: Treatments Have Identical Effects
Ha: Treatments Do Not Have Identical Effects

Summary Statistics:
Total Number of Observations:                36
Number of Blocks:                            12
Number of Treatments:                        3

Test:
Cochran Test Statistic:                        2.800000
CDF of Test Statistic:                        0.7534030
P-Value:                                      0.2465970

Percent Points of the Chi-Square Reference Distribution
-----------------------------------
Percent Point               Value
-----------------------------------
0.0    =          0.000
50.0    =          1.386
75.0    =          2.773
90.0    =          4.605
95.0    =          5.991
97.5    =          7.378
99.0    =          9.210
99.9    =         13.816

Conclusions (Upper 1-Tailed Test)
----------------------------------------------
Alpha    CDF   Critical Value     Conclusion
----------------------------------------------
10%    90%            4.605      Accept H0
5%    95%            5.991      Accept H0
2.5%  97.5%            7.378      Accept H0
1%    99%            9.210      Accept H0

NIST is an agency of the U.S. Commerce Department.

Date created: 12/5/2005
Last updated: 10/13/2015


text:
## Thomas' Calculus 13th Edition

The definition of average rate of change is $\frac{R(θ2) - R(θ1)}{θ2-θ1}$. R(θ) = $\sqrt 4θ +1$ Interval = [0, 2] $\frac{R(2) - R(0)}{2-0}$ R(2) = $\sqrt 4.2 +1$ = $\sqrt 9$ = 3 R(0) = $\sqrt 4.0 +1$ = $\sqrt 1$ = 1 $\frac{3 - 1}{2-0}$ = 1


text:
# The 59Fe (n,γ) 60Fe Cross Section from the Surrogate Ratio Method and Its Effect on the 60Fe Nucleosynthesis

@article{Yan2021The5,
title={The 59Fe (n,$\gamma$) 60Fe Cross Section from the Surrogate Ratio Method and Its Effect on the 60Fe Nucleosynthesis},
author={S. Q. 胜权 Yan 颜 and X. Y. 鑫悦 Li 李 and K. Nishio and Maria Lugaro and Zhiyuan 志远 Li 李 and H. Makii and Marco Pignatari and Y. B. 友宝 Wang 王 and Riccardo Orlandi and Kentaro Hirose and Kazuaki Tsukada and P. Mohr and Gongshun 恭顺 LI 李 and J. S. 建松 Wang 王 and B. S. 丙水 Gao 高 and Y. L. 银录 Han 韩 and B. 冰 Guo 郭 and Y. J. 云居 Li 李 and Y. P. 阳平 Shen 谌 and T. K. Sato and Y. Ito and Fumi Suzaki and Jun 军 Su 苏 and Y. Y. 彦云 Yang 杨 and J. S. 建松 Wang 王 and Jun 军 Ma 马 and Pengfei 鹏飞 Ma 马 and Z. 真 Bai 白 and S. W. 世伟 Xu 许 and Jie 捷 Ren 任 and Q. W. 启文 Fan 樊 and S. 晟 Zeng 曾 and Zhicong 志聪 HAN 韩 and W. K. 威克 Nan 南 and W. K. 威克 Nan 南 and C. 晨 Chen 陈 and G. 钢 Lian 连 and Q. 强 Hu 胡 and F. F. 芳芳 Duan 段 and S. Y. 树亚 Jin 金 and X. D. 晓东 Tang 唐 and W. P. 卫平 Liu 柳},
journal={The Astrophysical Journal},
year={2021},
volume={919}
}
The long-lived 60Fe (with a half-life of 2.62 Myr) is a crucial diagnostic of active nucleosynthesis in the Milky Way galaxy and in supernovae near the solar system. The neutron-capture reaction 59Fe(n,γ)60Fe on 59Fe (half-life = 44.5 days) is the key reaction for the production of 60Fe in massive stars. This reaction cross section has been previously constrained by the Coulomb dissociation experiment, which offered partial constraint on the E1 γ-ray strength function but a negligible… Expand

#### References

SHOWING 1-10 OF 68 REFERENCES
The 95zr(n, gamma)96zr cross section from the surrogate ratio method and its effect on the s-process nucleosynthesis
The 95Zr(n,gamma)96Zr reaction cross section is crucial in the modelling of s-process nucleosynthesis in asymptotic giant branch stars because it controls the operation of the branching point at theExpand
Measurement of the 60Fe(n, gamma)61Fe Cross Section at Stellar Temperatures.
A first measurement of the stellar ( 60)Fe(n, gamma)(61)Fe cross section, the predominant destruction mechanism of (60)Fe, has been performed by activation at the Karlsruhe Van de Graaff accelerator, showing a significant reduction in uncertainty with respect to existing theoretical discrepancies. Expand
60Fe in core-collapse supernovae and prospects for X-ray and gamma-ray detection in supernova remnants
We investigate 60Fe in massive stars and core-collapse supernovae focussing on uncertainties that influence its production in 15, 20, and 25 M$\odot$ stars at solar metallicity. We find that theExpand
First Experimental Constraint on the Fe-59(n, gamma)Fe-60 Reaction Cross Section at Astrophysical Energies via the Coulomb Dissociation of Fe-60
The radionuclide Fe-60 has been of great interest to the nuclear astrophysics community for over a decade. An initial discrepancy between the observed and modeled Galactic Fe-60/Al-26 ratio motivatedExpand
Settling the half-life of 60Fe: fundamental for a versatile astrophysical chronometer.
An independent measurement with a new method that determines the 60Fe content of a material relative to 55Fe with accelerator mass spectrometry substantially improves the reliability of this chronometer as an important chronometer for astrophysical applications in the million-year time range. Expand
The Nucleosynthesis of 26Al and 60Fe in Solar Metallicity Stars Extending in Mass from 11 to 120 M☉: The Hydrostatic and Explosive Contributions
• Physics
• 2006
We present the 26Al and 60Fe yields produced by a generation of solar metallicity stars ranging in mass between 11 and 120 M☉. We discuss the production sites of these γ-ray emitters and quantify theExpand
Determination of the 233Pa(n, γ) capture cross section up to neutron energies of 1 MeV using the transfer reaction 232Th(3He, p)234Pa∗
Abstract The γ -ray emission probability distribution of 234 Pa ∗ has been measured between 5.2 and 6.2 MeV excitation energy using the transfer reaction 232 Th( 3 He, p) 234 Pa ∗ at an incident 3 HeExpand
Measurement of the 54,57Fe(n,γ) Cross Section in the Resolved Resonance Region at CERN n_TOF
Abstract 54 Fe and 57 Fe are stable iron isotopes, which play an important role in the nucleosynthesis of the slow neutron capture process ( s process). In addition, these nuclei are present in manyExpand
Detection of γ-ray lines from interstellar 60Fe by the high resolution spectrometer SPI
It is believed that core-collapse supernovae (CCSN), occurring at a rate ∼once per century, have seeded the inter- stellar medium with long-lived radioisotopes such as 60 Fe (half-life 1.5 Myr),Expand
Estimation of M1 scissors mode strength for deformed nuclei in the medium- to heavy-mass region by statistical Hauser-Feshbach model calculations
• Physics
• 2017
Radiative neutron capture is an important nuclear reaction whose accurate description is needed for many applications ranging from nuclear technology to nuclear astrophysics. The description of suchExpand


text:
Number of Random Fractional Reals whose Total Exceeds 1

Let real numbers be selected at random following a continuous uniform distribution from the interval $\closedint 0 1$ until their total sum is greater than $1$.
The expectation of the number of selections is Euler's number $e$.


text:
Redirected from page "Figures/Perrinet06fens"

Clear message

## abstract

The machinery behind Ocular Following Response (OFR) is confronted to ambiguities which are efficiently resolved in the primate visual system. We study here a model of center-surround motion integration in a probabilistic framework and try to identify its different dynamical components by using contrast gain responses to the particular bipartite stimuli. Motion integration may be modeled as an ideal observer in a probabilistic framework by using bayesian modeling [Weiss et al., 2002] or statistical inference which proved to be successfully applied to the ocular following response [Perrinet et al., 2005]. Experiments on primates’ oculo-motor recordings concentrated on bipartite stimuli optimized to study the dynamics of information integration for different levels of noise which provide evidence for an orientation selective suppressive effect of the surround on the contrast gain control of local stimuli [Barthélemy et al., 2006]. We extend here our previous model to integrate different spatial cues: the information propagates to give a command response for ocular response that we could compare with the human behavioral response. We present results which show that the hypothesis of independence of local measures succesfully accounts for the monotonic integration of the spatial motion signal but that another mechanism should be added to account for suppressive saturation. Adding this, we observed similar dynamics for the contrast gain control mechanisms observed in the behavioral data and in neuro-physiological through in-vivo cortical recording by optical imaging (see accompanying poster number 21 by Reynaud et al. [2006]).

 Figure 1: Effects of size of a disk grating on the monkey's OFR response. We present here the gain of the oculo-motor response to a central grating (temporal frequency 10 Hz) as a function of its diameter for the macaque monkey (open circles) and the model (continuous line). Fits were performed as a function of the diameter d thanks to equations derived in the text. (Left) At low frequencies (0.12~cpd) and contrasts, the gain increases monotonically with the diameter. The curves are well fitted by the simpler equation with only an excitatory pooling. (Right) However in more general conditions (here 0.7~cpd), the initial gain decreases after a given diameter suggesting a suppressive effect. This corresponds to a surround inhibition which is well captured by the equation, the inhibition being more pronounced when contrast is higher and contrary to intuition, the diameter $\omega$ extracted from the fits remains constant across curves. This provided a general explanation of the ratio-Of-Guassian model in a probabilistic framework.

## reference

• Laurent Perrinet, Jens Kremkow, Frédéric Barthélemy, Guillaume S. Masson, Frédéric Chavane. Input-output transformation in the visuo-oculomotor loop: modeling the ocular following response to center-surround stimulation in a probabilistic framework. In FENS, 2006 abstract.

see accompanying poster no 21 :

• Alexandre Reynaud, Frédéric Barthélemy, Ivo Vanzetta, Guillaume S. Masson, Frédéric Chavane. Input-output transformation in the visuo-oculomotor loop: comparison of real-time optical imaging recording in V1 to the ocular following response to center-surround stimulation. In FENS, 2006 abstract.

All material (c) L. Perrinet. Please check the copyright notice.

welcome:
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# Directional derivative of a function

Could anyone help me with this?

Which directional derivative exists at the Point $(0,0)$ of function $$f(x,y)=\sqrt{|xy|}$$

Well i made this: Assume $(v,w)$ a vector with lenght $v^2+w^2=1$ and with formula $$\lim_{t\to0} \frac{f(t(v,w)) - f(0,0)}{t} = \lim_{t\to0} \frac{f(tv,tw)-f(0,0)}{t} \\ = \lim_{t\to0} \frac{\sqrt{|tvtw|}}{t} = \lim_{t\to0} \frac{t\sqrt{|vw|}}{t} = \sqrt{|vw|}$$ so Limit of this function at the Point $(0,0)$ exists it means all directional derivatives exist

I'm not sure with this though. Can anyone please revise it if i ever made a mistake!

thanks

• Note that $\sqrt{|tvtw|} = |t|\sqrt{|vw|}$. – Arthur Apr 18 '16 at 17:55
• yeah it is sorry – suugii02 Apr 18 '16 at 17:56


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• Adsorption effect of NO$_2$ on ZnO (100 nm) nanowires, leading towards reduced reverse leakage current and voltage enhancement

• # Fulltext

https://www.ias.ac.in/article/fulltext/boms/043/0267

• # Keywords

Piezoelectric potential; VING; nanogenerator; adsorption of NO$_2$.

• # Abstract

Here, we report the adsorption effect of NO$_2$ on ZnO (100 nm) nanowires. We have studied the effect of adsorbed NO$_2$ molecules on ZnO nanowire-based energy harvester for an exposure time of 1, 2, 3, 4, 5 and 6 h in a sealedchamber at 50 ppm which yielded piezoelectric voltage of 543.6 mV, 834.6 mV, 1.071 V, 1.78 V, 1.969 V and 2.835 V, respectively. We have thoroughly investigated the behaviour of ZnO nanowires in the presence of NO$_2$ and observed a maximum output piezoelectric voltage of 2.835 V with a power density 158.2 mW cm$^{-2}$. This is the first time that ZnO based piezoelectric energy harvester is being used for the voltage enhancement in the presence of NO$_2$. We have used vertically integrated nanowire generator (VING) structure. X-ray diffraction pattern revealed the growth orientation ofZnO nanowires were along the $c$-axis from the substrate. ZnO nanowires were grown on indium tin oxide-coated polyethylene terephthalate substrates via a hydrothermal route. Surface morphology has been examined by scanning electron microscopy images and diameter of ZnO nanowires was found to be around 100 nm. Piezoelectric voltage has been generated by the VING by applying minute external force of $\sim$50 nN. Periodic output voltage peaks were being measured by picoscope 5204.

• # Author Affiliations

1. Microelectronic and Nanotechnology–Shamsuddin Research Centre (MiNT-SRC), Faculty of Electrical and Electronic Engineering, Universiti Tun Hussein Onn Malaysia (UTHM), 86400 Parit Raja, Batu Pahat, Johor, Malaysia
2. Department of Physics, Nanotechnology Research Centre, Faculty of Science and Mathematics, Universiti Pendidikan Sultan Idris, 35900 Tanjung Malim, Perak, Malaysia
3. School of Sciences, University of Management and Technology, Lahore, Pakistan
4. Department of Physics, Goverment Degree College for Women, Haveli Lakha, Pakistan

• # Bulletin of Materials Science

Volume 44, 2021
All articles
Continuous Article Publishing mode

• # Dr Shanti Swarup Bhatnagar for Science and Technology

Posted on October 12, 2020

Prof. Subi Jacob George — Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur, Bengaluru
Chemical Sciences 2020


text:
# probability – question about the plan of the region when we transform pairs of random variables

Leave $$X$$ Y $$Y$$ They will be two independent random variables.

Leave $$Z = g (X, Y)$$ A function of two random variables.
To find the CDF of $$Z$$ we have

begin {align} P {Z leq z } = & P {g (X, Y) leq z } \ = & iint_ {g (x, y) leq z} ^ {} f_X (x) f_Y (y) dxdy. end {align}

In the books, Inequality. $$g (x, y) ≤ z$$ defines a region in the $$(x, y)$$ airplane.
How do we find this region? For example $$Z = X + Y$$.

Another question is: Why do we move in capital letters? $$(g (X, Y))$$ a lowercase $$(g (x, y))$$?

Also for example the probability that a random variable $$X$$ It is less then or equal to another random variable. $$Y$$ one is
begin {align} P {X leq Y } = & int_ {y = – infty} ^ {+ infty} int_ {x = – infty} ^ {y} f_X (x) f_Y (y) dxdy. end {align}
Because it is not
begin {align} P {X leq Y } = & int_ {y = – infty} ^ {+ infty} int_ {x = – infty} ^ {Y} f_X (x) f_Y (y) dxdy. end {align}
I mean
$$int_ {x = – infty} ^ {Y}$$
do not

$$int_ {x = – infty} ^ {y}.$$


text:
# How to/Can you get a PhD position when you need more experience first?

My friend loves mathematics and wants to continue research as a mathematics PhD student, but this doesn't seem possible!

She did a prestigious but inadequate bachelors program (almost no math) and because of this she is having a very hard time getting good grades for her master courses. Thus she feels she has inadequate knowledge for a PhD position; which is probably right.

Her thesis adviser told her that she cannot continue a career in research mathematics: no university would consider her given the situation now nor if she took an extra year or two to study more.

• Question: Is this true? Does it really stop here?
• I'm interested in similar experiences, opinions and suggestions (from experienced professionals).
• I'm particularly not looking for a rosy picture!

In my opinion she is very talented, motivated, original and quick! She would flourish given more time to take specific foundational bachelor and deepening masters courses. Until now she performed extremely well (near perfect score) and this is confirmed by for example two scholarships she won (BA and MA).

PS. I tried to keep it terse but please ask for further details if needed.

Thanks a lot!

Update: Thank you all very much for the responses and useful comments.

It turns out that there is much more "staff rotation" in applied PhD programs and it seems plausible that she can find a nice position there; from which to continue her studies.

Thanks again.

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She needs to consult someone who is realistic but more supportive than her advisor. It is not true that taking an extra year or two (or three) to study will reduce her chances. In fact, if she can afford the cost, I would encourage her to take as much time as she needs before applying to and entering a Ph.D. program to develop her mathematical knowledge and skills as much as possible before starting the Ph.D. program. To me the time matters far less than her strength when she does enter the Ph.D. program. – Deane Yang Feb 23 '11 at 23:46
You left out an important detail, which is geography: where is this taking place? How far would she be willing to move to do a PhD (including another continent)? etc. All these are huge factors. – Thierry Zell Feb 24 '11 at 1:05
I understood that she was in almost no math BA program. By indication that she ahas a hard time in getting good grades in MA program I suppose the MA program is closer to math. But what is really her MA program about, this is not clear to me from the description, and it seems relevant. – Zoran Skoda Feb 24 '11 at 14:49
I would suggest a PhD degree in applied math and/or biosimulation. The following career history will be familiar to many. Undergraduate major in history, minor in linguistics: did not pursue further. Worked as a political operative for a losing campaign. Grad school in economics: dropped out. Grad school in applied math: dropped major. Grad school in physics: received PhD, found tenure-track job following two post-docs, but ultimately was denied tenure eleven years post-PhD. That student's name: Edward Witten. – John Sidles Feb 25 '11 at 16:36
Ha :-) Great comments everyone... thanks! – wires Mar 15 '11 at 1:48

To supplement Kevin Lin's point about the Berkeley pre-PhD program, Smith College (in the U.S.) runs an analogous post-baccalaureate program that has been successful in placing students in various graduate programs. It seems a good fit for the situation and aspirations of the person described:

"This program is open to all women with a serious interest in pursuing a higher degree in the mathematical sciences."

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If I may: Our program was given an AMS award for Mathematics Programs That Make a Difference: ams.org/programs/diversity/citation2011 :-) – Joseph O'Rourke Aug 28 '11 at 0:19

I'm not sure if we do it anymore, but Berkeley has (used to have?) a "pre-PhD" program. See page 25 of this. Probably some other schools have similar programs as well.

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Berkeley is the only school I've heard about with such a program, so it may be unique. It would be nice to introduce more, but I guess one problem is how to fund them. – Peter Shor Feb 24 '11 at 18:14
Peter- Berkeley may be the only such school to have a separate program, but at Oregon, the default first year of graduate school is roughly comparable to the pre-Ph.D. program at Berkeley, in that one is taking essentially advanced undergrad courses (see this webpage: math.uoregon.edu/graduate/courses.php; the 500-level courses are also cross-listed as senior-level undergrad courses). I assume things are in practice similar at other comparable schools (but of course, I don't know them as well). – Ben Webster Feb 24 '11 at 20:56

Her thesis advisor's opinion of her chances may weigh heavily in any PhD application packets, since the advisor would normally be expected to write a recommendation letter. Usually, if you are advising a very strong Master's candidate, you would not be dismissive of her chances, especially if she has offered to study for an extra two years to make up any gaps in her knowledge. I'm sorry that this isn't rosier.

But perhaps there is more to understand... There seems to be a contradiction in your question when you say she is having trouble getting good grades, but has at the same time won two scholarships and has near perfect scores.

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Thanks a lot for you comments; the bachelor she took was not in mathematics, but a University College''. There were little mathematics courses; Groups, Algorithms, linear optimization and I think that was it. She was admitted to a pure maths masters with single semester preparation where she took at least number theory and topology. All-in-all way less that what a regular bachelor student does; she does pass the masters courses, but obviously has a very hard time getting along. I hope this clears it up a bit. Again, thanks a lot; highly appreciated. – wires Feb 23 '11 at 23:40

I don't know whether this is true today, but 30 years ago I had the impression that there were Math Ph. D. programs in the United States that would accept anybody with a pulse, just so they would have warm bodies to put in front of their Calculus classes. Naturally, we're not talking about Princeton here, but there were (and are) a lot of Ph. D. programs in the US, and they couldn't all be real choozy. Where there's a will, there's a way.

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Teaching Calculus classes - they needed warm bodies to serve as Teaching Assistants. – Gerry Myerson Feb 23 '11 at 23:53
@unknown: Gerry is probably talking about the students having TA positions to teach the undergrad calculus classes. – jd.r Feb 23 '11 at 23:55
@unknowngoogle: Math departments use grad students to teach calculus, not take it. – Jim Conant Feb 23 '11 at 23:59
@Ben, 30 years ago a foreign student asked me to write letters for her. I told her I would write that she was the 16th best student in a class of 23, and that I didn't think such a letter would help her much. She said that was perfect - she knew she wasn't cut out for graduate work, but her country required her to apply to grad school as a condition of paying her college bills, so she wanted a letter that would get her rejected. So I wrote the letters as requested - and a couple of places accepted her! I'll grant you that "pulse" and "warm body" were overstatements, but not by a whole lot.... – Gerry Myerson Feb 24 '11 at 11:31
Gerry- I'm not disputing your experience. (Though you definitely should have said "a pulse and a BA.") But I still think it makes your answer less helpful and less convincing. Someone with a master's degree in math, even if they are a bit underprepared, can get into a much better graduate program than that, and that's a point worth emphasizing. – Ben Webster Feb 24 '11 at 19:10

Here is a wild and crazy idea: answers to similar MathOverflow questions suggested going for a Master's degree en route to a Ph.D. Perhaps she (after finishing her current Masters) can enter another Master's program at the university in which she wants to do Ph.D. research. She can at least ask at that university what her chances are for trying such. (It may be possible to do cross discipline; for example a Masters in statistics or computer science might be acceptable before trying for a Ph.D. in mathematics.)

Gerhard "Thinking Outside The Normal Framework" Paseman, 2011.02.23

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Does she really need to become PhD Student in Mathematics? Why doesn't she try with something different?

I mean, a mathematician is, in my experience, not just a person who studies abstract mathematics. There are plenty of other occupations that involve the same skills and duties as pure mathematics. There is a lot of mathematics needed into other similar subjects, like computer science. Moreover, the interdisciplinarity of computer science should help her in getting such a position. Finally, original thinking and ability to formalize and being a hardworker are skills really needed and rewarded in this field!

To say my experience, I graduated in Mathematics (in the field of arithmetic geometry) but now I'm a PhD Student in computer science... and despite my lack of knowledge in specific topics, like programming, I feel I'm really helpful here!

Moreover, in some areas of Mathematics the research borders are quite far, so starting "late" could imply facing with difficulties in the future... or at least, I feel that this is what the most mathematicians think about this, and probably is the reason why she's having hard times in finding a PhD position in Maths...

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Getting top grades in GRE will probably help her a long way to get into a PhD program. Trying past and practice GRE exams will probably give her a good idea whether she is ready to face a graduate program in mathematics.

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-1: Sorry. But I really think the GRE is a terrible indicator of whether or not someone should be working towards a PHD in Mathematics for the next 4-5 years. There are an enormous number of things that will help you much more than studying for the GRE. – Eric Naslund Feb 24 '11 at 0:32
An exceptional score in the subject GRE might help, but even then I'm not sure and it would be hard to get. The general GRE would not help at all. – Thierry Zell Feb 24 '11 at 0:59
I'm ambiguous about this advice, but I'm not sure I vote against it -- I think a lot of top programs essentially ignore GRE scores as they count as a low-level indicator compared to someone who's already written a research paper, or has a really wide basis of knowledge in the field, but outside of this handful of places a stellar GRE score might grab someone's attention. I also slightly disagree with Thierry in that I think a stellar score on the general GRE with an unusual background would interest me more than a high score on the subject test (which is pretty mundane as a mathematics test). – Kevin McGerty Feb 24 '11 at 3:34
In my experience with graduate admissions (Harvard, University of Washington) the math subject GRE score is important---certainly much more so than other comments above suggest. For example, the Berkeley math grad application page says "Experience has shown that the score on the Mathematics Subject GRE is a partial indicator of preparation for Berkeley's PhD program. A score below the 80th percentile suggests inadequate preparation and must be balanced by other evidence if a favorable admission decision is to be reached." See math.berkeley.edu/graduate_phd.html – William Stein Feb 24 '11 at 19:20
I'm going to second the impression that this probably deoends on the program. I was once asked if I wanted to read some graduate applications at a top program. Once we sat down with these big files full of scores and statements and what-not, I asked what to do and was told "just read the letters" – Ramsey Feb 24 '11 at 20:28


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### Home > GB8I > Chapter 5 Unit 6 > Lesson INT1: 5.1.1 > Problem5-10

5-10.

Plot $ΔABC$ on graph paper with points $A(1,1)$, $B(-1,-1)$, and $C(4,-1)$.

1. Use the function ($x→-x$, $y→y$) to transform $ΔABC$. Label this triangle $ΔA′B′C′$. What kind of transformation does this produce? Name the coordinates of the image.

Review the Math Notes box in Lesson 3.1.4.

This transformation produces a reflection over the $y$-axis.
Name the coordinates of each image point.

2. Now translate $ΔABC$ upward and right a total distance of five units, parallel to the line $y = \frac { 4 } { 3 }x + 5$ to create $ΔA″B″C″$. Name the coordinates of the image.

This translation is to be parallel to the line, meaning you should translate every point up $4$ and to the right $3$.

Use the eTool below to solve the problem.
Click the link at right for the full version of the eTool: 5-10 HW eTool


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# Atmospheric and Astrophysical Neutrinos above 1 TeV Interacting in IceCube

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### Abstract

The IceCube Neutrino Observatory was designed primarily to search for high-energy (TeV--PeV) neutrinos produced in distant astrophysical objects. A search for $$\gtrsim 100$$~TeV neutrinos interacting inside the instrumented volume has recently provided evidence for an isotropic flux of such neutrinos. At lower energies, IceCube collects large numbers of neutrinos from the weak decays of mesons in cosmic-ray air showers. Here we present the results of a search for neutrino interactions inside IceCube's instrumented volume between 1~TeV and 1~PeV in 641 days of data taken from 2010--2012, lowering the energy threshold for neutrinos from the southern sky below 10 TeV for the first time, far below the threshold of the previous high-energy analysis. Astrophysical neutrinos remain the dominant component in the southern sky down to 10 TeV. From these data we derive new constraints on the diffuse astrophysical neutrino spectrum, $$\Phi_{\nu} = 2.06^{+0.4}_{-0.3} \times 10^{-18} \left({E_{\nu}}/{10^5 \,\, \rm{GeV}} \right)^{-2.46 \pm 0.12} {\rm {GeV^{-1} \, cm^{-2} \, sr^{-1} \, s^{-1}} }$$, as well as the strongest upper limit yet on the flux of neutrinos from charmed-meson decay in the atmosphere, 1.52 times the benchmark theoretical prediction used in previous IceCube results at 90\% confidence.

### Most cited references2

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### Evidence for High-Energy Extraterrestrial Neutrinos at the IceCube Detector

(2013)
We report on results of an all-sky search for high-energy neutrino events interacting within the IceCube neutrino detector conducted between May 2010 and May 2012. The search follows up on the previous detection of two PeV neutrino events, with improved sensitivity and extended energy coverage down to approximately 30 TeV. Twenty-six additional events were observed, substantially more than expected from atmospheric backgrounds. Combined, both searches reject a purely atmospheric origin for the twenty-eight events at the $$4\sigma$$ level. These twenty-eight events, which include the highest energy neutrinos ever observed, have flavors, directions, and energies inconsistent with those expected from the atmospheric muon and neutrino backgrounds. These properties are, however, consistent with generic predictions for an additional component of extraterrestrial origin.
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### BL Lac Objects in the Synchrotron Proton Blazar Model

(2002)
We calculate the spectral energy distribution (SED) of electromagnetic radiation and the spectrum of high energy neutrinos from BL Lac objects in the context of the Synchrotron Proton Blazar Model. In this model, the high energy hump of the SED is due to accelerated protons, while most of the low energy hump is due to synchrotron radiation by co-accelerated electrons. To accelerate protons to sufficiently high energies to produce the high energy hump, rather high magnetic fields are required. Assuming reasonable emission region volumes and Doppler factors, we then find that in low-frequency peaked BL Lacs (LBLs), which have higher luminosities than high-frequency peaked BL Lacs (HBLs), there is a significant contribution to the high frequency hump of the SED from pion photoproduction and subsequent cascading, including synchrotron radiation by muons. In contrast, in HBLs we find that the high frequency hump of the SED is dominated by proton synchrotron radiation. We are able to model the SED of typical LBLs and HBLs, and to model the famous 1997 flare of Markarian 501. We also calculate the expected neutrino output of typical BL Lac objects, and estimate the diffuse neutrino intensity due to all BL Lacs. Because pion photoproduction is inefficient in HBLs, as protons lose energy predominantly by synchrotron radiation, the contribution of LBLs dominates the diffuse neutrino intensity. We suggest that nearby LBLs may well be observable with future high-sensitivity TeV gamma-ray telescopes.
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### Author and article information

###### Journal
2014-10-07
2015-01-13
1410.1749
10.1103/PhysRevD.91.022001

Phys. Rev. D 91, 022001 (2015)
18 pages, 12 figures
astro-ph.HE

High energy astrophysical phenomena


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# Let $L$ be the splitting field of $f(x) = x^3 + 10x + 15$ over $\mathbb{Q}$, and $a_1$ (real), $a_2, a_3$ (complex conjugates) the roots of $f$

For $H$ = {$a_1$, $a_2$, $a_3$}, let $S_H$ be the symmetric group on $H$.

Identifying $Gal(L/\mathbb{Q})$ with a subgroup of $S_H$, how do we know if the transposition ($a_2$ $a_3$) the same as complex conjugation?

• Just because knowing $a_1$ is real then the two complex roots are complex conjugates – marwalix Jul 31 '16 at 17:03
• For example, if $a_2 = 1+i$ and $a_3 = a-i$, then the transposition $(a_2 a_2)$ will map $1+i$ to $1-i$ and map $1-i$ to $1+i$. – 3x89g2 Jul 31 '16 at 17:18

First you should think about what you mean by 'it is the same as complex conjugation', so that the proof becomes evident. If $\sigma \in H$ is complex conjugation this means that $\sigma=\tau|_{\Omega}$, where $\tau: \mathbb{C} \rightarrow \mathbb{C}$ is complex conjugation, and $\Omega \subset \mathbb{C}$ is the splitting field of $f$. Now, a field homomorphism $\varphi: \Omega \rightarrow \mathbb{C}$ is completely determined by the images of $a_1$, $a_2$ and $a_3$. $\tau|_{\Omega}$ clearly is such a homomorphism, and even more clearly $(a_2 a_3)$ is such a homomorphism (by assumption). $\tau(a_1)=a_1=(a_2 a_3) (a_1)$, $\tau(a_2)=\overline{a_2}=a_3=(a_2 a_3) (a_2)$ and $\tau(a_3)=a_2=(a_2 a_3) (a_ 3)$, so both homomorphisms coincide on $\{a_1, a_2, a_3 \}$, so they must be the same.


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BEAST 2 Help Me Choose

# BEAST 2 Help Me Choose Standard template -- clock model panel

## Standard template – clock model panel

Here, we will consider the strict clock, uncorrelated relaxed clock (Drummond et al, 2006, Douglas et al, 2021) and random local clock (Drummond & Suchard, 2010).

## Initial exploration

For an initial run when no experience with the type of data is available, running with the strict clock and simple substitution model and tree prior is a good way to find out whether there is signal in the data.

A relaxed clock analysis can then be added if there are a good number of mutations in the data and a reasonable number of taxa. If the distribution of the coefficient of variation is hugging zero, this is a signal that there is little rate variation among branches, and a strict clock is supported by the data. If this distribution is not touching zero, this is a signal there is rate variation and a relaxed clock is supported by the data. If you are unsure, you can run a path sampling or nested sampling analysis to perform Bayesian hypothesis testing.

## Which relaxed clock model to use

There are three implementations of relaxed clock models in BEAST 2

• the classic uncorrelated relaxed clock (Drummond et al, 2006) that works well when rates and branch lengths are not very correlated, but is not efficient when they are.
• a version that introduces new operators (Zhang & Drummond, 2020) that works well when branch lengths and rates are highly correlated, but is less efficient if they are not.
• the optimised relaxed clock(ORC) version that adapt operators (Douglas et al, 2021) and works well whether the rates and branch lengths are correlated or not and due to new operators is more efficient than the other two.

Unless a specific feature of the first two implementation is of interest, the ORC should be used for best performance.

## Further exploring the relaxed clock

If a relaxed clock is supported, there is a choice of different clock rate distributions. The default for ORC is a log-normal distribution, but exponential and gamma distributions are options as well. Make sure to make the mean of the distribution 1, since the rates are relative to the mean clock rate in ORC. For example, for the exponential set the mean to 1 and do not estimate this parameter. For a gamma, set mode to OneParameter and estimate the alpha parameter (beta is ignored in this parameterisation) and specify a prior on alpha.

If no substantial differences are observed in the posterior distribution (trees appear to be similar and parameter estimates appear the same in Tracer) choose one of the rate distributions and report the analysis is robust for the choice of distribution.

However, if substantial differences are observed, a path sampling or nested sampling analysis may be required for selecting the most appropriate model.

## Further explorations

A random local clock allows the clock model to behave like a strict clock (if there are no rate changes) and a relaxed clock (if there are many rate changes). The number of rate changes is sampled during MCMC, and is drive by the data and prior. From this perspective, it is superior to the strict clock and relaxed clock models. However, it may suffer from convergence issues.

## References

Douglas J, Zhang R, Bouckaert R. Adaptive dating and fast proposals: Revisiting the phylogenetic relaxed clock model. PLoS computational biology. 2021 Feb 2;17(2):e1008322. doi:10.1371/journal.pcbi.1008322.

Drummond AJ, Ho SY, Phillips MJ, Rambaut A. Relaxed phylogenetics and dating with confidence. PLoS biology. 2006 May;4(5):e88. doi:10.1371/journal.pbio.0040088

Drummond AJ, Suchard MA. Bayesian random local clocks, or one rate to rule them all. BMC biology. 2010 Dec;8(1):1-2. doi:10.1186/s12862-020-01609-4.

Zhang R, Drummond A. Improving the performance of Bayesian phylogenetic inference under relaxed clock models. BMC evolutionary biology. 2020 Dec;20(1):1-28.

## Bayesian evolutionary analysis by sampling trees

Disclaimer: The above is the opinion of the author RB. If you do not agree, or spot a mistake, contact the author, or discuss this in the issues area or raise a new issue. A link will be added from this page to make sure others can find it.

Served through Jekyll, customised theme based on the twentyfourteen wordpress theme.


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# Latex girls

Latex clothing should correctly be called rubber clothing, because latex is used to describe various substances, including the latex of the rubber tree (Hevea brasiliensis). After pre-treatment by smoking/rolling, it becomes rubber. Only through vulcanization, i.e. the introduction of sulfur and heating, is the material stabilized. This end product is then called rubber. The same term is also used for synthetic production from crude oil, since the end product is chemically identical. In general, however, one still speaks of latex clothing to make a clear distinction from car tires etc. Meanwhile, some manufacturers also use recycled latex sheets, the basic material for this is, for example, the recycling of old tires. Furthermore there are also small manufacturers who cast their latex sheets, like structural latex, also by hand. They use liquid latex, dye it and fill it into a mould up to the desired material thickness. This liquid latex usually has a low ammonia content.


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# An immediately inspiration

Algebra Level pending

Define a complex number $$z$$ which satisfied below expression:

$z=\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(\frac{1}{\sqrt2} +\frac{1}{\sqrt2}(...)^2)^2)^2$

and the coefficient of the imagine part of $$z$$ is positive.

Find argument of complex number $$z$$ in degrees.

Clarification: You may assume that the argument is between $$0^\circ$$ to $$360^\circ$$.

×


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# Welcome to Linear Algebra!

Here is a really typical application.

import pylab A_image = pylab.mean(pylab.imread(DATA + 'Jenks.png'), 2) @interact def svd_image(i=(5,(1..100))): u,s,v = pylab.linalg.svd(A_image) A = sum(s[j]*pylab.outer(u[0:,j],v[j,0:]) for j in range(i)) g = graphics_array([[matrix_plot(A),matrix_plot(A_image)]]) show(g,axes=False, figsize=[8,10]) pretty_print(html('
Compressed using %s eigenvalues
'%i))

## Click to the left again to hide and once more to show the dynamic interactive window

### Section One.I.1

So what kinds of solutions can happen, anyway?  As in Section One.I.2, we write things in the convenient matrix/vector format.

var('x1,x2') @interact def _(M = matrix(RR,[[1,5],[-2,-7]]), V = matrix(RR,[[7,-5]]),range=(5,(1,100)),show_sol=False): a, b, c, d = M[0,0],M[0,1],M[1,0],M[1,1] e, f = V[0,0],V[0,1] E1 = a*x1+b*x2==e E2 = c*x1+d*x2==f eq1 = solve(E1,x2)[0].rhs() eq2 = solve(E2,x2)[0].rhs() pretty_print(html("Solving:")) pretty_print(html('Eq 1: '+latex(E1))) pretty_print(html('Eq 2: '+latex(E2))) P = plot(eq1,(x1,-range,range))+plot(eq2,(x1,-range,range),color='red',figsize=5) show(P) if show_sol: try: pretty_print(html(solve([E1,E2],[x1,x2])[0])) except: pretty_print(html('No solution?'))

## Click to the left again to hide and once more to show the dynamic interactive window

var('x1,x2,x3') @interact def _(M = matrix(RR,[[1,5,2],[-2,-7,3],[0,1,1]]), V = matrix(RR,[[7,-5,0]]),range=(5,(1,100)),show_sol=False): a1, a2, a3, b1, b2, b3, c1, c2, c3 = M[0,0],M[0,1],M[0,2],M[1,0],M[1,1],M[1,2],M[2,0],M[2,1],M[2,2] e, f, g = V[0,0],V[0,1],V[0,2] E1 = a1*x1+a2*x2+a3*x3==e E2 = b1*x1+b2*x2+b3*x3==f E3 = c1*x1+c2*x2+c3*x3==g eq1 = solve(E1,x3)[0].rhs() eq2 = solve(E2,x3)[0].rhs() eq3 = solve(E3,x3)[0].rhs() pretty_print(html('Eq 1: '+latex(E1))) pretty_print(html('Eq 2: '+latex(E2))) pretty_print(html('Eq 3: '+latex(E3))) P = plot3d(eq1,(x1,-range,range),(x2,-range,range))+plot3d(eq2,(x1,-range,range),(x2,-range,range),color='red')+plot3d(eq3,(x1,-range,range),(x2,-range,range),color='green') #show(P,viewer='canvas3d') show(P) if show_sol: try: pretty_print(html(solve([E1,E2,E3],[x1,x2,x3])[0])) except: pretty_print(html('No solution?'))

## Click to the left again to hide and once more to show the dynamic interactive window

### Section One.I.2

We can solve these systems, including parameters, using Sage.  Here is the syntax; you define your variables, and then solve using one list for the equations and one for the variables you want to solve for.

var('x1,x2,x3') solve([6*x1+5*x2+3*x3==2,x1-4*x3==8],[x1,x2,x3])
 [[x1 == 4*r1 + 8, x2 == -27/5*r1 - 46/5, x3 == r1]] [[x1 == 4*r1 + 8, x2 == -27/5*r1 - 46/5, x3 == r1]]

The "r1" syntax means "any real number" - that is the parameter. We use "==" to indicate equality because Python (the language underlying Sage) uses it for something else (variable assignment).

Note our example from class (correctly) has $x1$ as being fixed at $9/5$.

solve([2*x1+x2-x3==5,3*x1-x2+x3==4],[x1,x2,x3])
 [[x1 == (9/5), x2 == r2 + 7/5, x3 == r2]] [[x1 == (9/5), x2 == r2 + 7/5, x3 == r2]]

If we want to assume a certain variable is a parameter, we can indicate that by leaving it out of the list of variables to be solved for.

solve([2*x1+x2-x3==5,3*x1-x2+x3==4],[x1,x2])
 [[x1 == (9/5), x2 == x3 + 7/5]] [[x1 == (9/5), x2 == x3 + 7/5]]

### Section One.I.3

It's not as easy to demonstrate the "General=Particular+Homogeneous" principle using a computer system, because it will not automatically check for equality of solution sets (this could be an arbitrarily hard problem).

However, there is a way to represent at least some of this using Sage.

M = matrix([[2,1,-1],[3,-1,1]]) V = vector([5,4]) M\V
 (9/5, 7/5, 0) (9/5, 7/5, 0)

That was a particular solution; the next cell gives us the vector from the homogeneous solution.

M.right_kernel()
 Free module of degree 3 and rank 1 over Integer Ring Echelon basis matrix: [0 1 1] Free module of degree 3 and rank 1 over Integer Ring Echelon basis matrix: [0 1 1]

As you probably are already suspecting, we don't yet have the theoretical chops to deal with all of this yet, but at least you can see that it's possible to compute these things programmatically.

### Section One.II.1

Sage makes it easy to define vectors and do things with them.

V = vector([1,2])
V+V
 (2, 4) (2, 4)
-3*V
 (-3, -6) (-3, -6)

Be sure to use the right syntax - brackets inside parentheses.

Here are a couple plots from class.

plot(vector([1,2]))+plot(vector([-2,1]))+plot(vector([-1,3]),color='green')
plot(vector([-2,1]))+plot(vector([-6,3]),color='red')

However, for the purposes of this text, it might be more interesting to plot some of the solutions that we put in vector notation in the text.  This is tricky, because Sage (unlike Hefferon) treats vectors as pretty much always being based at the origin.  So instead we can just use ordered pairs or points to plot some vectors.

parametric_plot((1-2*x,2+x),(x,-3,3))+parametric_plot((-2*x,x),(x,-3,3),color='red',linestyle='--')

I've put the homogeneous solution in red dashes.

We can do the same thing in three dimensions.

parametric_plot3d((9/5,7/5+x,x),(x,-3,3))+parametric_plot3d((0,x,x),(x,-3,3),color='red')+point((0,0,0),color='green',size=20)

That was the "line as vectors" interpretation of the general and homogeneous solutions.

var('y,z') parametric_plot3d((3-2*y-z,y,z),(y,-3,3),(z,-3,3))+parametric_plot3d((-2*y-z,y,z),(y,-3,3),(z,-3,3),color='red')+point((0,0,0),color='green',size=20)

### Section One.II.2

Vectors have lengths.  And you can multiply them in this weird way.

V = vector([1,2]) W = vector([2,-3])

Sage calls this the "norm" for good reasons ("normalization", for instance).

norm(V)
 sqrt(5) sqrt(5)
norm(W)
 sqrt(13) sqrt(13)

Dot product is the only good product we can define on vectors, so:

V*W
 -4 -4

Sage also allows the dot product in this way, but we haven't done much programming yet so the notation may seem odd.

V.dot_product(W)
 -4 -4

Here's an example similar to one from the book.  Are these vectors orthogonal?

V = vector([1,3,2]) W = vector([5,-2,0])
plot(V)+plot(W)

Gosh, they could be... is the angle $\pi/2$?

N(arccos((V*W)/(norm(V)*norm(W))))
 1.62044588932890 1.62044588932890
N(pi/2)
 1.57079632679490 1.57079632679490

Looks like it's a bit bigger.  But you might not be able to tell that from the picture, eh?

### Section One.III.1

Sage has the row-reduced echelon form built in, but you have to do a little more work to get it.  Also, one tends to do it with just the matrix of coefficients, not the augmented one, for reasons which will become clear when we revisit these systems later.

M = matrix([[2,2],[4,3]])
 [2 2] [4 3] [2 2] [4 3]

This syntax gets you an echelon form, but not the reduced one.

M.echelon_form()
 [2 0] [0 1] [2 0] [0 1]

For full echelon reduced form, you use ".rref()" for "row-reduced echelon form".

M.rref()
 [1 0] [0 1] [1 0] [0 1]

We can confirm that all the examples in question are equivalent.

M1 = matrix([[2,2],[0,-1]]) M2 = matrix([[1,1],[0,-1]]) M3 = matrix([[2,0],[0,-1]])
M1.rref(); M2.rref(); M3.rref()
 [1 0] [0 1] [1 0] [0 1] [1 0] [0 1] [1 0] [0 1] [1 0] [0 1] [1 0] [0 1]

### Topic: Accuracy of Computations

Suppose we try to solve some system as in the topic in the book.

var('x,y') @interact def _(number_zeros=(8,[1..20])): pretty_print(html("Solving $x+2y=3$ and $1."+"0"*number_zeros+"1x+2y=3."+"0"*number_zeros+"1$")) pretty_print(html(solve([x+2*y==3, (1+.1^number_zeros)*x+2*y== (3+.1^number_zeros)],[x,y])))

## Click to the left again to hide and once more to show the dynamic interactive window

Notice that this goes from fine at 13 zeros to weird at 14 zeros to a free variable at 15 zeros!  What the heck?

solve([x+2*y==3, (1+.1^14)*x+2*y== (3+.1^14)],[x,y])
 [[x == (100079991719344/97904339725445), y == (193633027456991/195808679450890)]] [[x == (100079991719344/97904339725445), y == (193633027456991/195808679450890)]]

What follows is essentially the example on page 67 of what can happen if you try to do row reduction.

M = matrix([[.00000000000000000001,1,1],[1,-1,0]])
 [1.00000000000000e-20 1.00000000000000 1.00000000000000] [ 1.00000000000000 -1.00000000000000 0.000000000000000] [1.00000000000000e-20 1.00000000000000 1.00000000000000] [ 1.00000000000000 -1.00000000000000 0.000000000000000]
 [1.00000000000000e-20 1.00000000000000 1.00000000000000] [ 0.000000000000000 -1.00000000000000e20 -1.00000000000000e20] [1.00000000000000e-20 1.00000000000000 1.00000000000000] [ 0.000000000000000 -1.00000000000000e20 -1.00000000000000e20]

Oops, looks like $y=1$, eh?  But if $x=0$ it won't solve the original system, where clearly $x=y$.

Here is how we might do the examples in the book, which have much less accuracy of the computing assumed:

M = matrix(RealField(10),[[0.001,1,1],[1,-1,0]])

I created a matrix with only ten bits of accuracy for each number, which corresponds to maybe three digits of accuracy.  Not enough, apparently:

 [ 0.0010 1.0 1.0] [0.00098 -1000. -1000.] [ 0.0010 1.0 1.0] [0.00098 -1000. -1000.]


text:
# How to do this question ?

### Related Question

South America has a total land area of 17997182 square kilometers. If we round it off this value to the nearest one lakh, we get

South America has a total land area of 17997182 square kilometers. If we round it off this value to the nearest one lakh, we get

Dear Shreya,

There are quick tricks to solve this question, but I would explain here using fundamentals of rounding-off.

Purpose of rounding off a number to x is to be able to write the number in terms of x without using decimal.

For rounding a number to nearest x (which is lakh in this case), we need to think of a number line with unit x.

For this case number line would look like this,

----------|----------|----------|----------|----------|
178        179        180        181
(178, 179 ... are in lakhs)

Now lets find the place for 179.97182 lakh on this line (remember we want to get rid of decimals).

----------|----------|----------|----------|----------|
178        179       * 180       181

Place for 17.997182 lakh is approximately shown as * on the line.

Here you could see that this number is closer to 180 lakh, so rounded-off number will be 180 lakhs (18000000).

regards,

ExamHelp

You just have to write in a simpler way Like 180000 or 180 thousand

goog

### Respond to this discussion

You need to login to write a comment


text:
# Synopsis: Neutron Test for Newton’s Gravity

Experiments with neutrons search for violations of gravity’s inverse square law at subnanometer distances.

Newton’s law of universal gravitation predicts that the gravitational force between two objects is proportional to the objects’ masses and inversely proportional to the square of the distance between them. The law, which applies to weakly interacting objects traveling at speeds much slower than that of light, has survived test after test. However, some quantum gravity theories anticipate that the law might break down at small distances. Now, through experiments with a pulsed neutron beam, Christopher Haddock of Nagoya University, Japan, and colleagues have checked Newton’s law on subnanometer scales. So far, the team has found no deviations from Newtonian predictions.

The team fired pulses of neutrons at a chamber filled with either helium or xenon gas and monitored both the travel time of the neutrons through the gas and the neutrons’ scattering angles. From these measurements, they reconstructed the scattering process with the aid of simulations. They found that the scattering-angle distribution fit the predictions—based only on known laws of physics—for neutrons bouncing off gas nuclei. This result indicates that, within the sensitivity of the experiment, no unexplained force—be it modified gravity or another type of interaction—acts on length scales below 0.1 nm. However, the researchers were only able to determine an extremely large upper limit on the strength of such a force: $1{0}^{24}$ times that of gravity. Still, this is the strictest limit set by any experiment on these spatial scales. The team is currently upgrading the setup to reduce sources of noise and envisions achieving order-of-magnitude sensitivity improvements in the near future.

This research is published in Physical Review D.

–Katherine Wright

Katherine Wright is a Contributing Editor for Physics.

More Features »

### Announcements

More Announcements »

Gravitation

Nuclear Physics

Cosmology

## Related Articles

Gravitation

### Synopsis: Mirror, Mirror—Which Coating is the Quietest of Them All

Gravitational-wave detectors may benefit from an alternative coating material that is less noisy at low temperatures than currently used materials. Read More »

Gravitation

### Synopsis: On-Demand Solutions for Black Hole Mergers

An analytical model may provide a faster and more accurate way to analyze the gravitational-wave signals from black hole mergers. Read More »

Gravitation

### Synopsis: Persistence of Gravitational-Wave Memory

Researchers predict the existence of three new long-lived signatures of gravitational waves, as part of a unified mathematical framework for identifying such effects.   Read More »


text:
# C:\Users\kccha\virtual\1.7.10\minecraft\sounds\mob\chicken\step2.ogg

#### Chef_Beardardee

-
What is step2.ogg? I go to that file and there is no ogg file. Where would I go about to get the .ogg file to run the modpack. (This happens with all ModdedNetwork Mod Packs on the launcher).

-


text:
# How do you write .000265 in scientific notation?

It is $2.65 \cdot {10}^{- 4}$

Sep 11, 2015

$.000265 = 2.65 \times {10}^{- 4}$
$\textcolor{w h i t e}{\text{xx}}$a single non-zero digit, followed by a decimal point followed by as sequence of digits
$\textcolor{w h i t e}{\text{xx}}$10 to an exponential power which will cause the decimal point to shift an appropriate number of places (left with a negative exponent, right with a positive) to form the original number.


text:
# Lightning-powered vehicles - are they possible? Are they realistically usable for everyday transport?

Note - I have tagged this as because, apart from feasibility, I'm hoping for some actual numbers that will tell me the capabilities of such vehicles.

Background

On a planet far away there are constant thunderstorms over the plains of Groth.

Until the advent of science and technology, the locals kept well away from the area, despite what would have provided excellent trading routes. This was because the chances of you and your pack-animals surviving were minimal.

Now however planet-wide electrical technology has reached the level of early 20th century Earth. Gasoline-powered vehicles have not been invented.

Someone has the idea that the Grothlanders would be the ideal people to test out the newly invented electrical motors because they live all around the perimeter of the Plains. There is a continual although not exactly continuous supply of electricity from lightning.

How the vehicles work

They are 4-wheeler carts steered from the front and powered by an electric motor driving the rear axle.

They have an aerial sticking out of the top to catch the frequent lightning strikes. The driver and passengers are protected by a Faraday-type cage to avoid being killed. They have padded helmets to muffle the constant deafening noise.

Question

Given a lightning strike every few minutes, would such a vehicle be feasible? It has to travel up to 100 miles across the Plains of Groth on any particular journey.

In particular, is it feasible to catch, store and use lightning in this way and would the weight of all the necessary equipment to make this work be excessive?

Assuming that the idea is feasible at all: Given the energy of a typical Earth lightning bolt, how often on average would a minimal-sized vehicle have to be struck in order to keep going pretty much continuously at, say, 30mph (approx 50kph)?

Assumptions

• A minimal-sized vehicle has condensers to store the electricity, an electric motor, is capable of driving hundreds of miles whilst being struck by lightning, carries a human-sized driver and one passenger, and a Faraday cage to protect them. The plains are more-or-less flat with firm but wet going. Some suspension is necessary but there is no rough terrain. Ball-bearings are available.

• There are future plans for some kind of railway network. But but this is still years away. For now, all they have are self-contained 4-wheeler motor carts. Most of these are home-built apart from the electrical motors that they have to purchase.

• The lightning storms are equivalent to those on Earth but go on day and night through some climate anomaly not disclosed here. If you venture onto the Plains you can expect to be struck pretty regularly. Exactly how often is part of my question.

• The hard-science tag was added while I was writing my question. Let me know if you want harder numbers than I gave you.
– JBH
Sep 26, 2020 at 23:29
• You need a flux capacitor. Sep 27, 2020 at 0:01
• Did I say question? I meant answer.
– JBH
Sep 27, 2020 at 0:56
• It seems more practical to have fuel stations scattered about that collect the lightning. It seems like too much energy to power a car safely, especially with energy dissipation. A static station could deal with that, a home made vehicle might not.
– Muz
Sep 28, 2020 at 7:13
• Is it necessary to use an electric motor for propulsion? The problem with lightning is its power is hard to capture. But you could use it to heat up water and create steam (water is available from the rain), and you could run a steam engine. Or heat up one side of a Stirling engine (and cool the other side with the rain).
– Nyos
Sep 29, 2020 at 17:58

So the maths works
A lightning bolt contains ~5 gigajoules of energy, or as Wikipedia helpfully tells us 38 gallons of gasoline. So if you could get a lightning bolt to hit you, and could store all the energy at 100% efficiency, that's enough to drive a 2019 Acura 1000 miles. (Exactly 1000 - it's rated as 3.8 gal / 100 miles).

Even if you lose 90% of the energy, that's still 100 miles per strike. Even if your go-cart things guzzle fuel at 7.6 gal / 100 miles, that's still miles per strike.

But the lightning doesn't

The trouble is getting the lightning to strike you. Because of this, the exact design wont work.

Lightning travels in steps of about 60m - looking for the path of least resistance greedily like water flowing down hill. (This is why lightning can strike a tree a few hundred meters away from a tall building.)

You attract all lightning to you over distance d based on this equation (from XKCD's "what if").

$$d = \sqrt{ - h ( h - 120)}$$

If your car has a 3m rod sticking up - it will cast a lightning shadow 18.7m wide - so anything going to hit the surrounding 19m will come for your lightning rod.

10m? 33m radius. Once you hit 60m, the shadow is at its peak. A lightweight 4-wheeler cart with a 60m tower is going to fall over over on the slightest slope.

So what will?

However; Some 60m towers, spaced about 100m apart with lightning rods powering a nearby electric train network. That's starting to become possible.

4 towers holding the corners of a large 100x100m mesh of thick wire will capture all lightning coming for the region under the mesh.

Also possible is lightning mixed with some technology to convert the power to liquid or gas fuel. Eg electrolysis to create a hydrogen fuel, and pumps to compress it. You lose a bit of power in the conversions but JBHs hypothetical maths can be a reality with a big factory cranking out compressed hydrogen.

• +1! I like the idea of a stationary lightning rod that's powering electric rail. Especially if you expand the idea to inductive power where there's no rails, just power lines in the roads and cars that tap that power. Clever!
– JBH
Sep 27, 2020 at 6:16
• Not sure about this, but what about balloons for lightning rods? Is it absolutely implausible to have a say 50m cable with a ballon at the top that can take and conduct lightning strikes? Would you need handwavium for the cable or the material of the ballon? Sep 27, 2020 at 18:16
• @L.Dutch I'm referencing existing mathematics. Is it really best that's duplicated rather than referenced? If you want redundant calculations say so and I'll do it in the morning.
– Ash
Sep 27, 2020 at 18:25
• The mix of US customary and SI units makes my head spin… ! Sep 28, 2020 at 8:27
• I think the problem might be in that lightning follows the path of least resistance. If you storing the energy or doing work with the energy there is going to be resistance. If it's too difficult for the lightning to go through your apparatus, it will find another way to the ground. Sep 29, 2020 at 15:43

I don't think it would work in practice...

While lightning has more than enough energy, you'd have problems harnessing it. Problem is enormous power (high voltage and current), ie. lots of energy which is delivered in very very short period of time.

And to transmit/store that high power, you need big conductors (to carry high current), with extremely low resistance (think ideally: superconductors). Otherwise, you'd have big power losses in small space, which amounts to extreme heating, which amounts to metal becoming vapor in very small fraction of a seconds, ie. explosion (not chunky enough conductors would be blowing like overloaded fuses)

What might work for mobile units is using lightning indirectly, by being isolated enough so lightning never strikes vehicle, and using other effects to harness its energy, perhaps like:

• induction which is usually responsible for frying your electronics when lightning strikes. eg. any moving current (like lightning) creates changing EM field, and any voltage is induced in any conductor in changing EM field. Longer the conductor, higher the voltage. That's how modern wireless phone charging works (although phone wireless charging uses much lower voltages and MUCH more rapid changing)

• making use of step voltages - ground potential is NOT the same on different radiuses from center point of lightning strike. Bigger the distance (and more perpendicular to the point of origin of lightning strike), higher the voltage difference.

Those two indirect methods generate MUCH less power than direct lightning strike, so they bring needed electrical devices into realm of possibility for car-like vehicle. However, unless the ground close to you gets hit by lightning VERY often, energy harvested would probably be too low to use such car (unless you park it for days in such stormy area to charge before using it).

It won't work, but it's cool to think about

The good news: Oh, yeah, you could operate a car on lightning strikes. Lightning can deliver 10 GW of electricity. Now, that article downplays the power of lightning by telling you that when you extrapolate the power to KWh, it isn't much (and, in terms of running a whole city, it ain't). But that's still 10 GW to work with (you don't divide the power into the length of the lightning strike or the claim would be 50 MW).

So, 10 GW. That's a honking lot of power. From this source we learn that to drive 400 miles an "average" car (and that's one wild guess. Cars are all over the map, so you should treat this as a ball park figure) will burn 560 KWh. And... 10 GW / 60 / 60 kinda = 2,775 KWh or almost 5X the energy needed to run the car for 400 miles. So, you can easily run for 800 miles (10 hours of driving at 80 MPH) with the A/C on, every light you can hook up, and your stereo running loud enough to be heard in Iceland.

EDIT: Curiously, no matter how poorly I did the math, it appears I was basically right, anyway. From this source we learn that "A single bolt of lightning carries a relatively large amount of energy (approximately 5 gigajoules or about the energy stored in 38 gallons of gasoline)." Average passenger car gets 30 mpg so one bolt is worth 1,140 miles - well beyond a single day's supply.

But that's not the problem

The problem is how to attract the lightning. Electricity follows the path of least resistance to the lowest electrical potential — which will almost never be the car (and it's not just the high-insulation vulcanized rubber tires). Even if you paved the roads with silver (great conductor) and used gold instead of rubber for the tires (great conductor), the best you're going to get is an average electrical potential equal to the ground within about a 3 meter radius of you. If you're cresting a hill, you have a great chance of being struck. If you're at the lowest point around you... not so much.

But hope is not lost for good science fiction!

Because what you could do is operate your car on negative voltage compared to "ground." And in a car, that's "chassis ground" or the voltage of the chassis metal. (Said another way, to keep passengers safe, the chassis of the car is used as the 0V reference.) Normally a car operates at +6VDC, +12VDC, or +24VDC (depending on what kind of vehicle it is).

Yours operates at -480VDC. And that honker can haul the mail! The benefit? You can put the -480VDC charging terminal on the top of the vehicle where it will (science fiction) draw all the lightning bolts you can handle!

And when your car is charged, a +24VDC plate covers the connector so that the lightning is pushed away from it. And we'll ignore what happens when two plates with a potential difference of 504VDC do. It's called a capacitor. Just ignore it. Science fiction. :-)

But is it safe?

It's your world. Declare it to be safe. Those folks grew up on it so lightning management would be a fundamental technology. The operator's cabin would be electrically isolated (very well insulated) from the rest of the car. There'd be some way of discharging electrical "overburden" (me, I want an arc lamp that rivals The Luxor Hotel).

And keep in mind, you'd need to only be struck once a day at the most to keep this car rolling. You can't run a city on lightning — but a car? Not hard at all.

Big ol' trucks might need to be hit multiple times a day.

A motorcycle might need to be hit once in your lifetime. Your lifetime, not the motorcycle's. Your people probably abandoned motorcycles a long time ago. The 2 Km search radius for the rider was too much work for the Search & Rescue team.

• You are confusing GW and GWh. 10GW/60/60 is 2,775 kW/h which is the wrong unit and doesn't mean anything. You need to find the number of joules and divide that by an hour. Sep 27, 2020 at 10:24
• "Lightning can deliver 10 GW of electricity." Surely you mean 1.21 giggawatts. Surely. :-) Sep 27, 2020 at 10:52
• 3 meters? That seems way too low when the car has a giant metal rod sticking into the air. It might still be too low to expect to get hit in an Earth thunderstorm, but increasing the frequency of strikes (quite) a bit might make it more realistic. Sep 27, 2020 at 11:23
• 10GW * 5/1000s = 50MJ. 50MJ/110.25kWh = 0.126. That means, you need about 8 lightning strikes to cover 40miles. Sep 27, 2020 at 11:56
• The most powerful laser in the world is 2 petawatts, which is about three hundred times the total global electricity generating capacity. But we can’t power everything using it (and we can power it), because it only runs at that power for a trillionth of a second. Lightning is the same — very high power for a very short time — and so its power is meaningless in this context. You need to be talking about the energy. Sep 27, 2020 at 20:09

Yes, it is possible to make such a machine. A lightning strike would bring about one billion (1,000,000,000) joules of energy. Aviation fuel contains 43 million joules per kilogram. Hence we have the equivalent of 23 kg of fuel from the single strike, 6 gallons. BMW i3 Giga can drive 846 miles, 1361 km on that! Even a small aircraft like Cessna 172 only requires about 10 gallons for a whole hour. So under any bearable efficiency of the lightning collector, a vehicle getting a strike per minute could not only ride, but even fly, even if heavy and inefficient. Or, probably, a strike per half an hour should be enough.

Some sources estimate the energy of the single strike to be about 20 times less (and there are different types of lightnings). This would still work even for the mentioned Cessna aircraft if we count on a strike per minute.

It is however technically challenging to extract all power from the lightning. Some kind of advanced supercapacitor is required that could take lots of energy very quickly and be charged to a very high voltage. Then no less advanced voltage converter needs to convert the energy into something that a normal electric engine could use. So probably not with the steampunk technology and not with 100 % efficiency. But if really a strike per minute, it seems that there is a huge efficiency margin for building something that just moves over ground.

• Where do you take this number from? (This)[energy-cast.com/41-lightning.html] only mentions 50MJ. Factor of 20 less. Sep 27, 2020 at 16:09
• All numbers are provided with the sources that are referenced in this answer, links under the numbers. Other sources may give different numbers, there are multiple types of lightnings after all. However even if we have just 1 kg of fuel rather than 23, this is still more than enough to run a car for a one minute. Sep 28, 2020 at 11:09

This is likely to be far more 'practical' if the lightning strike were to generate steam, and not be stored as a pile of electrons.

• This is a very short answer. You should expand on it with more details and possibly sources and diagrams. Sep 27, 2020 at 17:21

Let's approach it from the other direction.

## Let's audit the tech available at the time.

Electric cars were abundant, readily available, and just worked. They were the vehicle of choice for society ladies who wished personal mobility. Gasoline engines required a "driver" who was as much a mechanic to keep the fidgety engines working - on Downton Abbey this was Tom Branson's original job.

Without any application for engine starting, lead-acid batteries become a poor choice, done only on the cheapest vehicles. So other tech, notably Edison's nickel-iron batteries, are a huge winner. (Nickel-iron batteries are long-lived and almost indestructible, but have high internal resistance, so do poorly with "engine cranking". Not an issue for electric cars, but it may be an issue here.)

However, I'm burying the lede.

The 600 pound gorilla of the early 20th century was electric railroads. This technology had exploded onto the scene, with streetcars replacing horse cars, but much more importantly, interurban electric railroads.

These are high-speed (for the time) trains that ran at 70-100 miles per hour. They ran as small as single cars (a bit longer than a streetcar), or fairly long trains. The electric power ran normally at 600 volts, but as high as 4000 volts DC or 12,000 volts AC.

This is dragging us to a conclusion: The right way to do this is ferry the automobiles on flatcars on the interurban railway.

## Catching the lightning

The trolley wire will be a magnet for lightning strikes. You build this system with a huge amount of capacitance at megavolt levels - so you make the rail network as vast as possible, without circuit breaks.

Now, here is where we need to know the polarity of the lightning. If your lightning is consistently the same polarity, that will play nicely with a DC system. Otherwise some of your strikes will work against you, and those tend to be the more powerful strikes. Remember, we don't get diodes in this half of the century.

It won't work to have one track positive and the other track negative polarity. The lightning either wouldn't care or would be drawn to the more opposite.

AC is right out. First, half your strikes will be the wrong polarity (because of timing), and the net energy would be nil. Second, a vast system with a lot of capacitance will have a lot of problems both due to the capacitance and phase drift due to the vast distance.

Substations, in as many places as possible, will have some sort of unknown ultra-capacitor technology of some kind, including possibly a very tall stack of those lovely, indestructible Edison batteries.

Another option is to have a transmission line well above (vertically) the trolley wire. The transmission line would catch the lightning. This transmission would be the highest voltage DC that can be mustered given the tech of the age. It would have DC-DC conversions at substations. This was not easy in 1915, but not impossible.

• "AC is right out. First, half your strikes will be the wrong polarity (because of timing)," Three-phase AC might fix this problem. Sep 28, 2020 at 5:38
• @nick012000 The problem is the lightning is probably going to go for the phase with the greatest voltage difference (highest attraction), so then it'll always strike the wrong phase. Same problem with having one track be + and the other -. Sep 28, 2020 at 7:03

I would say no. Safety is probably the core issue.

Air around lightning is heated to 50,000 degrees F. Air is a terrible conductor but unless the devices used to store AND ground the lightning are superconductors, there's bound to be tremendous heat around it. Since it's possible to expect one every few minutes, catching two in a row within seconds might be too hazardous.

Shock waves from the thunder are enough to cause property damage and bruises. It's similar to that of an explosion or supersonic aircraft 'boom'.

Let's assume the heat and shock waves don't kill you. We're looking at 10 GW of electricity. Living within 50m of a 765 kV power line carries risk of cancer. Lightning is over 1000 times stronger than that, and at a much closer proximity. You'll likely get a lot of near fatal radiation over one drive.

• Downvote for citing "emfacademy"'s nonsense about power lines causing cancer. This has been repeatedly debunked. May 11, 2021 at 22:14
• @jdunlop The article you linked clearly states "Although the studies we have included do not show an association with MF, our results are broadly consistent with previous pooled analyses of MF and childhood leukaemia in that the elevated risk we found was limited to < 50 m of a 200 + kV lines, a distance at which MF are more likely to be elevated." There's a table: nature.com/articles/s41416-018-0097-7/tables/4
– Muz
May 17, 2021 at 3:06
• The formula for an electric field is also k * Q / r², meaning that as r (radius) becomes slightly higher, the risk drops drastically. A field at 1m is 2500 times stronger than a field at 50m, and 10,000 times stronger than a field at 100m. And with 1000x the power, that field has another 1000x multiplier.
– Muz
May 17, 2021 at 3:09


text:
# How to prevent Mathematica from complete evaluation of special functions?

Recently I tried to check a result of integration which should be a linear combination of the Riemann $\zeta\;$-function at various arguments, the coefficients of the combination being of primary interest.

Unfortunately Mathematica have expanded all occurrences of $\zeta\;$-functions with even integer arguments.

Is there a way to prevent it from doing that?

A sample code:

 Integrate[ Log[1-x]^5/x^5, {x, 0, 1}]

This is even simpler (thanks to @CarlWoll):

ClearSystemCache[];
Block[{Zeta = Inactive[Zeta]},
Integrate[Log[1 - x]^5/x^5, {x, 0, 1}]]

Using the Gayley-Villegas trick:

InternalInheritedBlock[{Zeta},
Unprotect[Zeta];
Zeta[a_Integer /; a > 1] /; ! TrueQ[$in] := Block[{$in = True},
Inactive[Zeta][a]
];
Protect[Zeta];
Integrate[Log[1 - x]^5/x^5, {x, 0, 1}]
]

Note: The integral is actually computed in terms of limits of an expression involving PolyLog[n, 1-x] (for n = 2,3,4,5); we could do something similar for PolyLog as was done for Zeta above. However PolyLog actually evaluates to Zeta first, which is why the above works in this case. I mention it in case PolyLog does not evaluate to Zeta in a future version. And it's possible the integral will not be computed in terms of PolyLog in the future, too, I suppose. In any case, I thought some explanation of why the code happens to work in this case would be helpful. (It will certainly help me a year from now when someone asks why this doesn't work, and I won't remember a thing about it.)

• Simpler is ClearSystemCache[]; Block[{Zeta = Inactive[Zeta]}, Integrate[Log[1 - x]^5/x^5, {x, 0, 1}]]. I tried using this Block earlier and it didn't work. When I saw your answer, I realized that the cache was interfering. – Carl Woll Apr 14 '18 at 15:50
• @CarlWoll Ditto for me. But I didn't realize the cache was causing the problem, even though I mentioned it to the OP. (Ha-ha). Thanks. – Michael E2 Apr 14 '18 at 15:55
• After my comment, I read the comments to the OP, and saw that! lol – Carl Woll Apr 14 '18 at 15:59
• @MichaelE2 This is a nice suggestion, however I would like to recognise when it's superior over a simple pattern matching approach. +1. – Artes Apr 14 '18 at 16:35
• @Artes This fiddling with the internals slows down Integrate, even worse if we inactivate PolyLog in the same way. Makes me wonder how safe it is. In fact, I got a wrong answer with one thing I tried (not shown). On the other hand, pattern matching can be made much harder if there are coefficients in the integrand that obfuscate or destroy the pattern. But done your way, it is perfectly safe. (+1 to your answer a while ago.) – Michael E2 Apr 14 '18 at 17:49

Mathematica is a term rewriting system and there are various ways to suppress automatic evaluation of expressions. The system doesn't evaluate Zeta for odd integer arguments (see e.g. Zeta[Range[2, 20]]). Zeta[n] yields expressions involving n-th powers of Pi, thus one of possible ways to achieve the goal would be e.g.

Integrate[Log[1 - x]^5/x^5, {x, 0, 1}] /.
Times[x_, Pi^n_Integer] :> x Pi^n Inactivate[Zeta[n]]/Zeta[n] //

Another way is to use HoldForm[Zeta[n]] instead of Inactivate[Zeta[n]] however the latter is more universal and handy. You can use Activate (also with appropriate patterns) to evaluate the expression, e.g.

Activate[%]

-((5 Pi^2)/6) - (11 Pi^4)/18 - 30 (Zeta[3] + Zeta[5])

Nevertheless using such a replacement should be appropriately restricted to avoid possible ambiguities with expressions involving symbolic results in terms of powers of Pi.

• For safety, you can restrict the transformation to even powers of $\pi$: Times[x_, π^n_?EvenQ] – J. M. will be back soon Apr 13 '18 at 18:11
• @J.M.needshelp. Right, however the OP might explain if expressions like e.g. Integrate[BesselK[0, x]^2, {x, 0, Infinity}] are subject to transform or not. – Artes Apr 13 '18 at 18:23
• Thanks for the answer. However we are just lucky that in this special case we can recognize our function by a power of $\pi$. It may be however a more general function with a "simple" value at a special point. Is there a general way to prevent Mathematica from evaluation of some listed functions? – drer Apr 13 '18 at 19:57
• @drer I've mentioned about pattern option in Inactivate which could be exploited. With an example at hand this could be worked out, nonetheless there is no obvious general way. – Artes Apr 13 '18 at 20:21
• @drer Re "in this special case we can recognize our function by a power of π": It's not clear that it is because Zeta is evaluating that we have powers of π in the answer. Executing Block[{Zeta = Inactive[Zeta]}, Integrate[Log[1 - x]^5/x^5, {x, 0, 1}]] suggests that the powers of π are computed without using Zeta`. If so, one cannot blame the evaluation of special functions for returning a correct answer in a certain form. – Michael E2 Apr 14 '18 at 1:41


text: dc.creator Vasudev, R. en_US dc.creator Zare, R. N. en_US dc.date.accessioned 2007-08-31T14:10:54Z dc.date.available 2007-08-31T14:10:54Z dc.date.issued 1982 en_US dc.identifier 1982-FB-9 en_US dc.identifier.uri http://hdl.handle.net/1811/29135 dc.description.abstract The variation with the rotational quantum number N of the widths of lines in the $(0, 9^{0}, 0) - (0, 0^{1}, 0)$ band of the $HCO \tilde{A}^{2}A""-\tilde{X}^{2}A'$ transition is recorded using a single-mode dye Laser. The HCO radical is produced in a radio frequency discharge is acetaldehyde $(CH_{3}CHO)$ and Ar/He, and the $\tilde{A}-\tilde{X}$ transition in the 570-620 nm region is detected by laser-induced changes in the discharge impedance. The variation of the observed line widths with N is interpreted in terms of Coriolis interaction and K-type resonance of the $(0, 9^{0}, 0) \Sigma (K'=0)$ level with nearby strongly predissociated $\Pi (K'=1)$ and $\Delta (K'=2)$ levels, respectively, of the upper state. en_US dc.format.extent 55230 bytes dc.format.mimetype image/jpeg dc.language.iso English en_US dc.publisher Ohio State University en_US dc.title LASER OPTOGALVANIC STUDY OF HCO $\tilde{A}$ STATE PREDISSOCIATION en_US dc.type article en_US


text:
# Cosmology - Exoplanet Properties

First direct image of an extrasolar planet, in 2004, was of a planet of mass 3-10 Mjupiter orbiting around a brown dwarf (2M1207) with a mass of 25 Mjupiter. Techniques like Radial velocity, Transit, Gravitational microlensing, Imaging, Astrometry, etc., has been used for detection of exoplanets. The number of detections have been increasing every year.

Until around 2010, the Radial velocity method was used extensively, but now most of the detections are done by the Transit method. There was a spike in the number of detections in 2014, which was when Kepler Space Telescope (KST) started giving the results.

A mass-period distribution shows that the Radial velocity method is more biased towards detection of massive planets with a larger period, whereas using the Transit method, planets with lower period are only detected as shown in the following image (Courtesy: NASA Exoplanet Archive).

There is a colossal increase in the number of detection of smaller mass planets since the advent of KST. This is evident from the figure given below. The planets detected by KST are divided into two groups: hot massive planets called “Hot Jupiters” and lower mass planets called “Hot Super Earths” (as they are more massive than Earth).

When we plot the number of extrasolar planets detected versus distance to them, we find that most of these planets are within 2kpc, which is well within our galaxy. Maybe the planets are not so uncommon in the universe, since our detection is limited to only certain type of planets in a very small part of the universe.

Planets are formed from circumstellar disc or proto planetary disc. If planets are formed as a by-product during star formation, maybe the number planets in the universe exceeds the number of stars in universe!!

## Habitable Zones

A Habitable Zone can be defined as the zone around the star where water can exist in its liquid form. Consider a planet at distance $a_p$ from the star as shown in the following figure. A simple method for calculating the temperature of the planet is described as follows.

$$\left ( \frac{L_\ast}{4\pi a^2_p} \right )\pi R^2_p(1 - A) = 4\pi R^2_p \sigma T^4_p$$

and

$$\frac{L_\ast}{4\pi R^2_\ast} = \sigma T^4_\ast$$

$$\therefore T_p = (1 - A)T_\ast \sqrt{\frac{R_\ast}{2a_p}}$$

In our case substituting

• Lsun = 3.83 x 1026

• ap = 1.5 ∗ 1011 and

• A = 0.3

Will give $T_{Earth} = 255K$. The actual calculation is very involved which includes cloud physics. The Habitable zone in our solar system lies between 0.9 AU and 1.7 AU.

The Sun’s luminosity was found to be increasing with time because of the decreasing gas pressure. It was 30% less bright when it started burning hydrogen. This would result is shift of habitable zone away from the Sun. Since Earth is near the inner edge of the Habitable zone, maybe one day it will move out of the zone!

### Continuously Habitable Zone

In short it is termed as CHZ can be defined as the region in which liquid water can exist over the entire Main Sequence lifetime of a star. The KST has detected many extrasolar planets which do lie in the habitable zone.

A bio-signature is any substance − such as an element, isotope, molecule, or phenomenon that provides scientific evidence of past or present life. An example is detection of both O2 and CO2 on a planet, which is usually not possible through geological processes alone. This detection is done by analyzing the absorption spectra.

### Points to Remember

• Techniques like Radial velocity, Transit, Gravitational microlensing, Imaging, Astrometry, etc., have been used for detection of exoplanets.

• Radial velocity method is more biased towards detection of massive planets with larger period.

• Hot massive planets are called “Hot Jupiter’s” and lower mass planets are called “Hot Super Earths”.

• The number of planets in the universe exceeds the number of stars in the universe.

• A Habitable zone can be defined as the zone around the star where water can exist in its liquid form.