text: Which of the following has two stereoisomers? So let's say that our paramagnetic sample is in here. Which of the following species is not paramagnetic? If oxygen is present, it would displace the dumb-bell sphere. Which kind of isomerism is exhibited by octahedral $[Co(NH_3)_4Br_2]Cl$ ? Given that the ionic product of $Ni(OH)_2$ is $2 \times 10^{-15}$. The following types of substances are paramagnetic: 1. Join the 2 Crores+ Student community now! Which of the following is paramagnetic ? The correct explanation comes from Molecular Orbital theory. I have this picture of this balance drawn down here. answered Jan 7, 2019 by ramesh (82.7k points) selected Jan 9, 2019 by Vikash Kumar . Check Answer and Solution fo Answer: b Explanation: Nitrogen has diamagnetic property. The original atom is also paramagnetic. a) Oxygen b) Nitrogen c) Nitrogen dioxide d) Nitric oxide. Other components that have diamagnetic property are carbon, boron, etc. And we can figure out if a sample is paramagnetic or not by using this special balance that I have. And so something that's paramagnetic is pulled into an external magnetic field. Give reason Mn2+ is more paramagnetic than fe2+ brainly.in/question/6370745. It shows that all the electrons in oxygen are paired, so oxygen should be diamagnetic. It is paramagnetic and high spin complex O b. Which one of the following … 1.2k VIEWS. Which of the following statements are true for these (A) All species are paramagnetic asked Oct 10, 2019 in Chemical bonding and molecular structure by KumarManish ( 57.6k points) chemical bonding Which of the following species is paramagnetic ? Most elements and some compounds are paramagnetic. Explanation: CN-, CO and NO + are isoelectronic with 14 electrons each and there is no unpaired electrons in the MO configuration of these species. So these are diamagnetic. Which among the following gases have diamagnetic property (ability to be repelled by magnetic fields)? BHU 1997: Which of the following is paramagnetic ? Which of the following is paramagnetic? 1.1k VIEWS. It includes mainly metals like iron, copper iron, etc. When exposed to an external magnetic field, internal induced magnetic fields form in these materials that are ordered in the same direction as the … On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be: An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. Which of the following is paramagnetic B2 C2 N2 F2 To determine the magnetic nature of molecules first we have to draw MO energy level diagrams of all the m Answer to: Which of the following atoms or ions is paramagnetic? 300+ SHARES. It's like our paramagnetic sample has gained weight. Which of the following paramagnetic in nature 10.9k LIKES. So, this is paramagnetic. Can you explain this answer? Yet oxygen is paramagnetic. But when the strong magnetic field is removed, the net magnetic alignment is lost as the dipoles relax back to their normal random motion. Reaction between acetone and methyl magnesium chloride followed by hydrolysis will give : Identify the correct statements from the following: Which of the following set of molecules will have zero dipole moment ? 3:14 7.3k LIKES. No valence electrons are paired here. Correct option (a) O-2. Paramagnetic Oxygen Analyzers Questions & Answers. When a material is placed within a magnetic field, the magnetic forces of the material's electrons will be affected. The paramagnetic materials include certain chemical elements which has a magnetic permeability higher than 1 and therefore attracted by magnetic forces. 5. $\ce{CH_3 C \equiv CCH_3 ->[H2/Pt] A ->[D_2 /Pt] B}$, In Wolff‐Kishner reduction, the carbonyl group of aldehydes and ketones is converted into. Identify compound X in the following sequence of reactions: Identify a molecule which does not exist. The hypothetical complex chloropentaamminecobalt (III) chloride can be represented as. Moreover, enthusiasts are also catered with the detailed breakdown of the atomic, optical and chemical behaviour of the metals. Can you explain this answer? 25 ml of this solution neutralises 20 ml of 0.1N $\ce{H_2SO_4}$.The % of $\ce{Na_2CO_3}$ in this sample is. 1. Paramagnetism: Paramagnetic is basically a type of magnetism in which substances are getting attracted by an extrinsic magnetic field. Knowing that the chemistry of lanthanoids(Ln) is dominated by its + 3 oxidation state, which of the following statements is incorrect? To keep reading this solution for FREE, Download our App. Given that the ionic product of $Ni(OH)_2$ is $2 \times 10^{-15}$. 1.1k SHARES. Paramagnetic substances are those substances that gets weakly magnetized in the presence of an external magnetic field. 9.5k VIEWS. However, materials can react quite differently to … (A) [Co(NH3)6]3 + (B) [Ni(CO)4] (C) [Ni(NH3)4]2+ (D) [Ni(CN)4]2-. Diamagnetic, Paramagnetic, and Ferromagnetic Materials. (a) NO 2 (b) N 2 O (c) N 2 O 3 (d) N 2 O 5. the p block elements; aiims; neet; Share It On Facebook Twitter Email. 9.5k SHARES. Which of the following complexes are paramagnetic in nature ? | EduRev Chemistry Question is disucussed on EduRev Study Group by 122 Chemistry Students. Paramagnetic Metals + properties give you a broad overview of these metals from multiple angels. Since there are no unpaired electro Asked on January 17, 2020 by Mathi Vaibhav. Learn more about paramagnetic. Which of the following compound has all the four types $(1^{\circ}, 2^{\circ},3^{\circ} \; \text{and} \; 4^{\circ})$ of carbon atoms? The correct option is (a) NO 2. The normality of 26% (wt/vol) solution of ammonia (density = 0.855 ) is approximately : 7. Select one: O a. Which of the following spec... chemistry. The atomic radiusis: Find out the solubility of $Ni(OH)_2$ in 0.1 M NaOH. The hypothetical complex chloropentaamminecobalt (III) chloride can be represented as. 2:10 46.3k LIKES. Use The Molecular Orbital Diagram Shown To Determine Which Of … It is diamagnetic and high spin complex c. It is diamagnetic and low spin complex d. It is paramagnetic and low spin complex 6. This effect is known as Faraday's Law of Magnetic Induction. 1.2k SHARES. The normality of 26% (wt/vol) solution of ammonia (density = 0.855 ) is approximately : 1.25 g of a sample of $\ce{Na_2CO_3}$ and $\ce{Na_2SO_4}$ is dissolved in 250 ml solution. . In other terms we can say that these substances tend to get weakly attracted to a permanent magnet. How to solve: Which one of the following species is paramagnetic? We now turn to a molecular orbital description of the bonding in $$\ce{O2}$$. According to quantum mechanics, such a system cannot have a total spin equal to zero; therefore, each atom or molecule has a net magnetic moment which arises from the electron spin angular momentum. Dec 27,2020 - Which of the following ions is paramagnetic:a)La3+ (Z = 57)b)Lu3+ (Z = 71)c)Sm3+ (Z = 62)d)Yb2+ (Z = 70)Correct answer is option 'C'. Paramagnetic, lanthanide shift reagents were first introduced in the 1960s 115,116 with the use of chiral lanthanide complexes to determine enantiomeric excess following soon after. > The Lewis structure of "O"_2 gives a misleading impression. O-2 is paramagnetic due to the presence of one unpaired electron. "O"_2 is paramagnetic because it has two unpaired electrons. When a paramagnetic material is placed in a strong magnetic field, it becomes a magnet, and as long as the strong magnetic field is present, it will attract and repel other magnets in the usual way. 1 Answer +1 vote . Classification of Elements and Periodicity in Properties, The decreasing order of the ionization potential of the following elements is. Paramagnetism, kind of magnetism characteristic of materials weakly attracted by a strong magnet, named and extensively investigated by the British scientist Michael Faraday beginning in 1845. Magnetism: The most common magnetic states of elements in the ground state are diamagnetic and paramagnetic. Which of the following compounds show optical isomerism? The atomic orbitals of the O atoms overlap to form the σ and π orbitals of the O2 molecule as per the molecular orbital theory. It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. It's attracted to an external magnetic field. Paramagnetic Metals Properties. In the presence of an external magnetic field, these substances tend to move from a region of a weak to a strong magnetic field. Which of the following species is paramagnetic ? 300+ VIEWS. Paramagnetic properties are due to the realignment of the electron paths caused by the external magnetic field. Which of the following arrangements does not represent the correct order of the property stated against it ? The magnetism of an element is known from its electron configuration. This capability allows paramagnetic atoms to be attracted to magnetic fields. Among the following the paramagnetic compound is a)Na2O2,b)O3,c)N2O,d)kO2. Which of the following complexes exists as pair of enantiomers? How to solve: Which of the following atoms or ions is diamagnetic? Correct increasing order for the wavelength of absorption in the visible region for the complexes of $Co^{3+}$ is: When $Ag^+$ reacts with excess of sodium-thiosulphate then he obtained species having charge and geometry respectively : The hybridization of central metal ion in $K_2[Ni(CN)_4]\, and\, K_2[NiCl_4]$ are respectively, In Wolff‐Kishner reduction, the carbonyl group of aldehydes and ketones is converted into. 000+ LIKES. The paramagnetic nature of O 2 is due to unpaired electrons. Which of the following statements about Fe(CO)5 is correct? Diatomic oxygen, $$O_2$$ is a good example of paramagnetism (described via molecular orbital theory). Identify compound X in the following sequence of reactions: Identify a molecule which does not exist. Molecular Oxygen is Paramagnetic. Best answer. In a particular isomer of $\left[Co\left(NH_{3}\right)_{4}Cl_{2}\right]^{0},$ the $Cl-Co-Cl$ angle is $90^\circ$, the isomer is known as, Compounds, ${{[PtC{{l}_{2}}N{{H}_{3}})}_{4}}]B{{r}_{2}}$ and $[PtB{{r}_{2}}{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$ , shows the following type of isomerism. In a paramagnetic material, the individual atoms possess a dipole moment, which when placed in a magnetic field, interact with one another, and get spontaneou… All atoms and molecules which have an odd number of electrons. On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be: An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. Reaction between acetone and methyl magnesium chloride followed by hydrolysis will give : Identify the correct statements from the following: Which of the following set of molecules will have zero dipole moment ? Classification of Elements and Periodicity in Properties, Which of the following complex is optically inactive, The compound $\ce{[Pt(NH3)2Cl2]}$ can exhibit. The atomic radiusis: Find out the solubility of $Ni(OH)_2$ in 0.1 M NaOH. Paramagnets do not retain any magnetization in the absence of an externally applied magnetic field, because thermal motion randomizes the spin orientations. Paramagnetism refers to a property of certain materials that are weakly attracted to magnetic fields. The #4s# subshell contains #1# electron (in one #4s# orbital) and the #3d# subshell contains #5# electrons, one in each #3d# orbital. 'S say that these substances tend to get which of the following is paramagnetic attracted to a permanent magnet a magnetic.... Weakly attracted to magnetic fields therefore attracted by magnetic fields ) to be repelled magnetic... 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## Two remarks on graded nilpotent classes

Last update: 15 April 2014

## Notes and References

This is an html version of the paper Two remarks on graded nilpotent classes by A. V. Zelevinskii.

1.Let $\mathrm{\Gamma }=⨁{V}_{i}$ $\text{(}i\in ℤ\text{)}$ be a finite-dimensional graded vector space over an algebraically closed field and $E\left(V\right)$ a set of linear operators $N:V\to V$ such that $N\left({V}_{i}\right)\subset {V}_{i+1}$ for $i\in ℤ\text{.}$ Graded nilpotent classes of type $V$ are defined as the orbits of the natural action of $\text{Aut} V=\mathrm{\Pi }GL\left({V}_{i}\right)$ on the set $E\left(V\right)\text{.}$ Their properties are analogous to those of ordinary nilpotent classes, that is, orbits of $GL\left(V\right)$on the set $N\left(V\right)$ of all nilpotent operators $V\to V\text{.}$ The results in this note corroborate this analogy.

It was proved independently in [ADe1980] and [Zel1981] that there are only finitely many graded nilpotent classes of a given type, and their degeneracies were described. Generally speaking, the closures of classes are singular varieties. Their geometry is studied in [ADK1981]. In [Zel1981] the author raised the question of computing the Deligne-Goreski-Macpherson cohomology sheaves ${ℋ}^{i}\left(X\right)$ in these varieties (see [KLu1980], [Lus1981]), and made a conjecture about the connection between these sheaves and the representations of the group $GL\left(n\right)$ over a $p\text{-adic}$ field. The non-graded analogue of this problem is considered in [Lus1981], where the ${ℋ}^{i}\left(X\right)$ are calculated for the closures of ordinary nilpotent classes and their connection with the representations of $GL\left(n\right)$ over a finite field is indicated.

Proceeding as in [Lus1981], we construct in §2 a compactification of the graded nilpotent classes, which is a Schubert variety. As in [Lus1981], this enables us to compute the ${ℋ}^{i}\left(X\right)$ for the closures of graded nilpotent classes in terms of the Kazhdan-Lusztig polynomials [KLu1979]; in contrast to [Lus1981], the resulting polynomials are connected with the ordinary symmetric group.

In §3 we construct a parametrization of an arbitrary graded nilpotent class that enables us to compute very easily the number of its elements over a finite field. The non-graded analogue of this construction enables us to give a simple geometrical proof of Steinberg's well-known result that the number of nilpotent operators in an $n\text{-dimensional}$ space over the field ${F}_{q}$ is equal to ${q}^{n\left(n-1\right)}\text{.}$

2. We put $n=\text{dim} V$ and ${n}_{i}=\text{dim} {V}_{i}$ and assume, without loss of generality, that ${n}_{i}\ne 0$ only for $1\le i\le k\text{.}$ For every operator $N\in E\left(V\right)$ and all pairs $i,j$ with $i\le j,$ we denote by ${N}_{ij}$ the compositum ${V}_{i}\stackrel{N}{\to }{V}_{i+1}\stackrel{N}{\to }\dots \stackrel{N}{\to }{V}_{j}$ (in particular, ${N}_{ii}$ is the identity operator ${V}_{i}\to {V}_{i}\text{)}$ and we put ${r}_{ij}={r}_{ij}\left(N\right)=rk{N}_{ij}\text{.}$ We denote by $M\left(V\right)$ the set of matrices $M=\left({m}_{ij}\right),$ $1\le i,j\le k,$ having non-negative integral entries such that ${m}_{ij}=0$ for $i>j+1,$ and with the row- and column-sums ${n}_{1},\dots ,{n}_{k}\text{.}$

Proposition 1 (see [ADe1980], [Zel1981]). Graded nilpotent classes of type $V$ are parametrized by the set $M\left(V\right)\text{:}$ to a matrix $M$ there corresponds the class ${X}_{M}=\left\{N\in E\left(V\right):{r}_{ij}\left(N\right)=\sum _{\epsilon \le i\le j\le t}{m}_{st} \text{for} i\le j\right\}\text{.}$ The numbers ${m}_{ij}$ are recovered from the ${r}_{ij}$ by the formulae ${m}_{ij}={r}_{ij}-{r}_{i-1,j}-{r}_{i,j+1}+{r}_{i-1,j+1}$ $\text{(}i\le j\text{)},$ and ${m}_{i,i-1}={r}_{i-1,i}\text{.}$ The closure of the class of the operator $N$ is the set of those $N\prime$ for which ${r}_{ij}\left(N\prime \right)\le {r}_{ij}\left(N\right)$ for all $i\le j\text{.}$

We denote by $F\left(V\right)$ the variety of flags $0={U}^{0}\subset {U}^{1}\subset \dots \subset {U}^{k}=V$ such that $\text{dim} {U}^{i}/{U}^{i-1}={n}_{i}$ for $1\le i\le k\text{.}$ We put ${V}^{i}={V}_{1}\oplus \dots \oplus {V}_{i},$ so that the flag $F$ consisting of the spaces ${V}^{l}$ lies in $F\left(V\right)\text{.}$ Let $P$ be the stabilizer of $F$ in $GL\left(V\right)\text{.}$ The $P\text{-orbits}$ on $F\left(V\right)$ are called Schubert cells, and their closures Schubert varieties, of type $V\text{.}$ An important example for our purposes is the Schubert variety $G\left(V\right)=\left\{\left({U}^{l}\right)\in F\left(V\right):{U}^{i}\supset {V}^{i-1} \text{for} 1\le i\le k\right\}\text{.}$ It is known that the Schubert cells contained in $G\left(V\right)$ are parametrized by the set $M\left(V\right):$ to the matrix $M$ there corresponds the cell ${O}_{M}=\left\{\left({U}^{i}\right)\in F\left(V\right):\text{dim}\left({U}^{i}\cap {V}^{j}\right) \left({U}^{i}\cap {V}^{j-1}+{U}^{i-1}\cap {V}^{j}\right)={m}_{ij} \text{for} 1\le i,j\le k\right\}\text{.}$

We define a map $E\left(V\right)\to F\left(V\right)$ that associates with $N\in E\left(V\right)$ the flag $\left({U}^{i}\right),$ where ${U}^{i}=\left\{\left({v}_{1},\dots ,{v}_{k}\right)\in {V}_{1}\oplus \dots \oplus {V}_{k}:{v}_{j+1}=N\left({v}_{j}\right) \text{for} j\ge i\right\}\text{.}$

Theorem 1. This map is an isomorphism between $E\left(V\right)$ and an open subvariety of $G\left(V\right)$ for which ${X}_{M}={O}_{M}\cap E\left(V\right)$ for all $M\in M\left(V\right)\text{.}$

We split the set $\left\{1,2,\dots ,n\right\}$ in running order into blocks ${B}_{1},\dots ,{B}_{k},$ where $\text{Card} {B}_{i}={n}_{i}\text{.}$ For every $M=\left({m}_{ij}\right)\in M\left(V\right)$ we put ${S}^{M}=\left\{w\in {S}_{n}:\text{Card}\left(w\left({B}_{i}\right)\cap {B}_{j}\right)={m}_{ij} \text{for} 1\le i,j\le k\right\}\text{.}$ It is known that ${S}_{M}$ has a unique element $w\left(M\right)$ of maximal length.

Corollary 1 (see [Lus1981]). The sheaves ${ℋ}^{i}\left({\stackrel{‾}{X}}_{M}\right)$ are $0$ for odd $i,$ and for $x\in {X}_{M\prime }\subset {\stackrel{‾}{X}}_{M}$ we have $\sum _{i\ge 0}{q}^{i} \text{dim} {ℋ}^{2i}{\left({\stackrel{‾}{X}}_{M}\right)}_{x}={P}_{w\left(M\prime \right),w\left(M\right)}\left(q\right),$ where the right-hand side is the Kazhdan-Lusztig polynomial ([KLu1979], [KLu1980]).

3. Let $M=\left({m}_{ij}\right)\in M\left(V\right),$ $N\in {X}_{M},$ and ${r}_{ij}={r}_{ij}\left(N\right)\text{.}$ For all $i,j$ with $i\le j,$ let ${I}_{ij}=\text{Im} {N}_{ij},$ so that $\text{dim} {I}_{ij}={r}_{ij}\text{.}$ For $1\le j\le k$ we denote by ${F}_{j}$ the flag $\text{(}{I}_{1j}\subset {I}_{2j}\subset \dots \subset {I}_{jj}={V}_{j}\text{)}$ in the space ${V}_{j}\text{.}$ It is clear that the map $N↦\left({F}_{1},\dots ,{F}_{k}\right)$ of ${X}_{M}$ into the product of flag varieties of this type is a fibration. Let us describe a typical fibre, that is, the set of operators $N$ with fixed ${I}_{ij}\text{.}$ We choose a complement ${C}_{ij}$ to ${I}_{i-1,j}$ in ${I}_{ij}\text{.}$ The operator $N$ is determined by its restrictions to all the ${C}_{ij},$ which can be chosen independently of one another, as the sum of an arbitrary operator ${C}_{ij}\to {I}_{i-1,j+1}$ and an arbitrary surjective operator ${C}_{ij}\to {C}_{i,j+1}\text{.}$ This leads to the following result.

Proposition 2. The number of elements of ${X}_{M}$ over ${F}_{q}$ is ${q}^{d}\prod _{i}{\mathrm{\Phi }}_{q}\left({n}_{i}\right)/\prod _{i\le j}{\mathrm{\Phi }}_{q}\left({m}_{ij}\right)$ where ${\mathrm{\Phi }}_{q}\left(r\right)=\left(q-1\right)\left({q}^{2}-1\right)\dots \left({q}^{r}-1\right),$ and $d=∑i

We now consider the non-graded analogue of our construction. The nilpotent classes in $N\left(V\right)$ are parametrized by the collections $\lambda =\left({m}_{i}\right)$ of non-negative integers with $\sum \left(i+1\right){m}_{i}=n\text{:}$ the class ${X}_{\lambda }$ consists of operators with ${m}_{i}$ Jordan blocks of order $\left(i+1\right)\text{.}$ (The numbers ${m}_{ij}$ above have a similar meaning.) For $N\in {X}_{\lambda }$ and $0\le i\le n$ we put ${I}_{i}=\text{Im} {N}^{i}$ and ${r}_{i}=\text{dim} {I}_{i}\text{.}$ By assigning to $N$ the flag $\left({I}_{i}\right)$ we see that ${X}_{\lambda }$ is a fibration over the variety of flags of this type. As above, this readily yields the well-known formula for the number of elements of ${X}_{\lambda }$ over ${F}_{q}$ (see [SSt1970]).

In conclusion, we show how to break up $N\left(V\right)$ into pieces from which we can build an affine space of dimension $n\left(n-1\right)\text{.}$ (The assumption that such a decomposition exists is made in [SSt1970].) We consider an $\left(n-1\right)\text{-dimensional}$ vector space $U$ with a preferred complete flag $0={U}^{0}\subset {U}^{1}\subset \dots \subset {U}^{n-1}=U\text{.}$ To any nilpotent class ${X}_{\lambda }$ we assign the variety ${Y}_{\lambda }=\left\{A\in \text{Hom}\left(U,V\right):\text{dim} A\left({U}^{{r}_{i}-1}\right)={r}_{i+1} \text{for} 0\le i\le n-1\right\}\text{.}$ It is easy to see that the ${Y}_{\lambda }$ are pairwise disjoint and that their union is $\text{Hom}\left(U,V\right)\text{.}$ By assigning to any $A\in {Y}_{\lambda }$ the flag $\left(A\left({U}^{{r}_{i}-1}\right)\right)$ we see that ${Y}_{\lambda }$ is a fibration over the same variety as ${X}_{\lambda }$ and it is easy to see that their fibres are isomorphic. From this it is clear that ${X}_{\lambda }$ breaks into pieces from which we can build ${Y}_{\lambda }\text{.}$ By taking these decompositions for all ${X}_{\lambda },$ we obtain the required decomposition of $N\left(V\right)\text{.}$

## References

[ADe1980] S. Abeasis and A. Del Fra, Degenerations for the representations of an equi-oriented quiver of type ${A}_{m}$, Boll. Un. Mat. Ital. Suppl. 1980, no. 2, 157-171. MR84e:16019

[ADK1981] S. Abeasis, A. Del Fra, and H. Kraft, The geomerty of representations of ${A}_{m}$, Math. Ann. 256 (1981), 401-418. MR83h:14038

[Zel1981] A. V. Zelevinskii, A $p\text{-adic}$ analogue of the Kazhdan-Lusztig conjecture, Funktsional. Anal, i Prilozhen. 15:2 (1981), 9-21. MR84g:22039

[KLu1979] D. Kazhdan and G. Lusztig, Representations of Coxeter groups and Hecke algebras, Invent. Math. 53 (1979), 165-184. MR81j:20066

[KLu1980] D. Kazhdan and G. Lusztig, Schubert Varieties and Poincare Duality, Proc. Symp. Pure Math. 36 (1980), 185-203. MR84g:14054

[Lus1981] G. Lusztig, Green polynomials and singularities of nilpotent classes, Adv. in Math. 42 (1981). 169-178. MR83c:20059

[SSt1970] T.A. Springer and R. Steinberg, Conjugacy classes, Lecture Notes in Math. 131 (1970), 167-266. MR42:3091
Translation: Seminar po algebraicheskim gruppam, Mir, Moscow 1973, 162-262.

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Example

$\frac{d}{dx}\left(\sqrt{x}\ln\left(x\right)\right)$
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Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=\sqrt{x}$ and $g=\ln\left(x\right)$

$\sqrt{x}\cdot\frac{d}{dx}\left(\ln\left(x\right)\right)+\ln\left(x\right)\frac{d}{dx}\left(\sqrt{x}\right)$
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The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\sqrt{x}\cdot\frac{d}{dx}\left(\ln\left(x\right)\right)+\frac{1}{2}x^{-\frac{1}{2}}\ln\left(x\right)$
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The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\sqrt{x}\cdot\frac{1}{x}\cdot\frac{d}{dx}\left(x\right)+\frac{1}{2}x^{-\frac{1}{2}}\ln\left(x\right)$
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The derivative of the linear function is equal to $1$

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Multiplying the fraction and term

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Simplifying the fraction by $x$

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Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$

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# How do you find the slope of the line containing the indicated points: ( -4, 8) and ( -3, -6)?

Jun 4, 2018

$- 14$

#### Explanation:

The slope of a straight line is given by:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

where:

• $m$ is the slope of the line

• $\left({x}_{1} , {y}_{1}\right)$ are the first coordinates

• $\left({x}_{2} , {y}_{2}\right)$ are the second coordinates

So, we get:

$m = \frac{- 6 - 8}{- 3 - \left(- 4\right)}$

$= - \frac{14}{1}$

$= - 14$


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# VeGa – Updated 11/25

I’ve got another batch of algorithms coming out soon, so I thought I’d get some more work done on my new book first:

Enjoy!

# Vectorized Deep Learning

I’ve put together a brief summary of my most recent work in vectorized deep learning, and uploaded an updated zip file of my algorithms.

Enjoy!

The ZIP file is available on my ResearchGate Homepage.

# VeGa – Updated

Here’s an updated draft of my new book, VeGa.

Enjoy!

# Visualizing Diophantine Equations

I was thinking about one-off states of systems, since these are arguably bad for predictions –

Just imagine assuming that a new state is similar to some known one-off state, because they are for example nearest neighbors in the space of the dataset. It could very well be the case that the dataset is simply not consistent around the known one-off state, and that’s why it was clustered alone.

This in turn led me to think about the possibility of one-off states of physical systems, which ultimately led me to the following idea:

Take a discrete integer-labeled subset of Euclidean space, and draw a point at a given index $(i,j)$ only if the following equation can be satisfied:

$B^i + G^j = A$,

where $B$ and $G$ are luminosity values from $[0,255]$, and $A$ is some integer. This will produce patterns defined by a Diophantine equation, and it turns out the solutions look a lot like the outputs generated by cellular automata. The code works by using a vectorized Monte Carlo method to find solutions to the equations.

Below are the first 50 solutions, beginning with $A = 1$, up to $A = 50$, though there are only 49 images, since there was no solution for one of the values of $A$.

Because $A$ is the sum of the two terms, $A$ operates as a cap on total luminosity, which is interesting to me, because luminosity is, physically, energy. So perhaps these types of equations could be used to describe the behavior or structure of physical systems as a function of increasing energy –

This was actually part of my thought process, since electron orbitals are wildly different shapes as a function of energy.

I thought this was interesting, since the allure of automata is that they produce patterns that don’t have obvious closed-form equations. These patterns however don’t appear to be the product of a closed-form equation, but they are, and it’s a simple one at that. Moreover, they get really aesthetically interesting for larger values of $A$, allowing for a wider variety of colors and luminosities.

Here are some examples for $A \approx 2500$:

You can of course express a three-dimensional analog, and this was the original idea, but I really like the way the two-dimensional version looks:

$R^i + B^j + G^k = A$.

The code for both the two-dimensional and three-dimensional version is attached.

# A Note on Random Variables

Imagine you’ve got a vector of integers, $v = [1 2 3 4 5]$. You could use this vector as the source of the elements for a random variable, that draws upon this vector of numbers. So for example, the sequence $[3 2 5 4]$ could be generated by tolling such a random variable 4 times. More formally, $x$ is a random variable over the set $\{1, 2, 3, 4, 5\}$.

But now imagine instead we permute the original vector $v$, generating the sequence,

$[3 4 2 1 5]$.

If we keep doing this, we can generate up to $5! = 120$ sequences of length $5$, without repetition.

This suggests a question:

Is there a difference between doing exactly that, and tolling the random variable $x$, $120$ times?

The answer is, yes, and it’s easily measurable.

For the string generated by permutation, the average over every sequence of length five is exactly $3$, if you begin at the first entry of the resultant string. This will not be the case for the string generated by $120$ tolls of the random variable $x$. Said otherwise, the string generated by permutation has a cycle length of five, even though it never repeats, in that each of the five elements appear over every cycle of five, though in different orders. For a longer cycle length over a larger set of numbers, you might not be able to notice any pattern, but it’s easy to test for –

Take the average over the entire string, and then iterate through cycle lengths, taking the average over each resultant cycle, until every cycle average equals the overall average.

So in our example above, we would iterate through cycle lengths, beginning at $1$, and up. Once we arrive at a cycle length of $5$, taking the average over the resultant cycles would produce an answer of exactly three for every cycle, which is equal to the overall average, over the entire string.

For a much larger set of numbers, you can have an enormous number of permutations, until you necessarily have repetition. For example, if $v$ contains just $10$ numbers, then there are $3,628,800$ permutations possible. As a practical matter, it would likely be difficult to distinguish between the observations generated by a true random variable, and permuting a fixed set of numbers, once the set of numbers gets sufficiently large, despite it being very easy to test for the difference between the two, even by hand.

A lot of people have considered the issue of superficial randomness, and I’m certainly not the first person to do so, though what I’m really interested in is –

Whether or not we should find random variables in nature, if physics computable.

My answer is, I’m not sure.

It’s obvious that basic physics is computable, and this has some consequences that are unavoidable, as a matter of pure math.

For example, despite what people say about physical entropy increasing over time, a computable process of nature simply cannot increase beyond a bounded rate of Kolmogorov complexity over time. Specifically, if $F$ is the computable rule of physics that describes a system, and $X$ is the initial state of the system, then $F(X,t)$ should tell you the state of the system at time $t$.

This can be specified with $K(X) + K(F) + K(t)$ bits –

I.e., the sum over the complexities of the initial conditions, the rule of the system, and the amount of time elapsed. $K(X)$ and $K(F)$ are constants, and $K(t) < \log(t)$, which implies that the system’s complexity is bounded by $\log(t) + C$.

This is simply the way it is, and there’s no debating this, and it implies that the complexity of a computable system is bounded over time.

There is some room to disagree about what constitutes a random variable, but my personal view is that a random variable should produce Kolmogorov-random observations. Assuming otherwise implies that there are computable patterns in the resultant data.

You could allow for this, but is it really random, if a machine can compress the output? In the extreme case, allowing for non-Kolmogorov-random, random variables, would allow for obvious symmetries in observations generated by a random variable, and that just doesn’t satisfy intuition.

For that reason, I’m going to assume that observing a random variable for a sufficiently long time will generate a Kolmogorov-random string.

So in this view, where could physically real, Kolmogorov-random objects come from, if the rules of physics are computable?

I think one answer is exhaustive processes, that generate all possible outcomes over a range. This a computable way of generating random outcomes:

Simply generate every possible outcome in a range, and some of them will be Kolmogorov-random.

This is a strange consequence of computing, in that it’s easy to generate all strings of a given length, many of which will be Kolmogorov-random, but generating a particular Kolmogorov-random string cannot be done, without an input that is at least as long as the output. Physically, this means that you can’t produce a Kolmogorov-random object using a computable rule of physics, unless you start with an object of equivalent or larger size.

I can’t point to anything that actually generates all objects in a possible range of, e.g., energies, and so I am instead pointing out that even with computable physics, you can still generate Kolmogorov-random objects, though you can’t generate exactly one without starting with a larger object.

This allows for two possibilities:

1. You have a process that generates all possible objects in a given range of sizes, many of which will be Kolmogorov-random;

2. You have a process that starts with a massive object, and winnows it down to a smaller Kolmogorov-random object.

Over enough time, you could at least conceive of an object that is Kolmogorov-random, generated as part of some massive process, that gets separated from its source, and appears as a single, inexplicable object, but I can’t think of any examples, and I’m instead simply working through the underlying mathematics of what’s possible with computable physics.

I also spend a lot of time thinking about the complexity of life, and what you find there is very similar to what you find in art –

Highly complex components, and a symmetric overall structure.

For example, the structures of a bird, butterfly, or human face, are all highly symmetrical, but extremely complex in their individual components. To make the analogy to computer theory, it’s similar to starting with a Kolmogorov-random string, and performing basic operations on that string, such as reading it backwards, or perhaps subdividing it, and repeating some of its components. This would generate an overall string that could be significantly compressible, that nonetheless consists of individual components that are in isolation, Kolmogorov-random.

So even with all of this said, the arguments above point only to the possibility of Kolmogorov-random objects, which are distinct from systems that generate truly random observations.


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# How to pass a variable from BeforeTest to Test annotation in TestNG?

A TestNG class can have various TestNG methods such as @BeforeTest,@AfterTest, @BeforeSuite, @BeforeClass, @BeforeMethod, @test, etc. Thereare various scenarios where we need to carry forward a few variables from thesemethods to the main @Test method. Since none of these methods support return type, the best way to pass a variable is using class /instance variable rather thanlocal variable.

The scope of class/instance variable is within the entire class. As a result, whatevervalue is set up at @BeforeTest or @BeforeMethod can be utilized at @Test method.

In this article, we will see how to pass a variable from @BeforeTest to @Test annotation in TestNG.

### Approach/Algorithm to solve this problem

• Step 1 − Create a TestNG class NewTestngClass and declare a variable as employeeName.

• Step 2 − Write the following code inside @BeforeTest

public void name() {
employeeName = "Ashish";
System.out.println("employeeName is: "+employeeName);
}
• Step 3 − Write a @Test method in the class NewTestngClass and use the variable employeeName as described in the code.

• Step 4 − Now create the testNG.xml as given below to run the TestNG classes.

• Step 5 − Finally, run the testNG.xml or directly testNG class in IDE or compile and run it using command line.

## Example

The following code for common TestNG class, NewTestngClass

### src/ NewTestngClass.java

import org.testng.annotations.*;
public class NewTestngClass {
String employeeName;
// test case 1
@Test()
public void testCase1() throws InterruptedException {
System.out.println("in test case 1 of NewTestngClass");
String employeeFullName = employeeName + " Anand";
System.out.println("employeeFullName is: "+employeeFullName);
}
@BeforeTest
public void name() {
employeeName = "Ashish";
System.out.println("employeeName is: "+employeeName);
}
}

### testng.xml

This is a configuration file that is used to organize and run the TestNG test cases. It is very handy when limited tests are needed to execute rather than the full suite.

<?xml version = "1.0" encoding = "UTF-8"?>
<!DOCTYPE suite SYSTEM "http://testng.org/testng-1.0.dtd" >

<suite name = "Suite1">
<test name = "test1">
<classes>
<class name = "NewTestngClass"/>
</classes>
</test>
</suite>

### Output

employeeName is: Ashish
in test case 1 of NewTestngClass
employeeFullName is: Ashish Anand
===============================================
Suite1
Total tests run: 1, Passes: 1, Failures: 0, Skips: 0
===============================================


text:
# Teddy - Node generation in arc shape between two reference nodes

Hallo zusammen!

I am trying to generate a node sequence between two reference points NR1 and NR2. By using the commands NR1 and NR2 the created nodes are placed in a straight virtual line connecting both reference points. However I need those nodes to be placed outside this straight line, forming an arc. In other words, I need to transform the virtual straight line between NR1 and NR2 into a virtual arc line, with a pre-defined radius.

Could somebody help me with that? Is that possible at all?

Hau

I’m guessing you are referring to sofimsha code.
In that case (I think) you have to do it manually by calculating the coordinates yourself.
Perhaps you can use the command NODE with spherical coordinates.
There is an example in your installation folder:
C:\Program Files\SOFiSTiK\2020\SOFiSTiK 2020\sofimsha.dat\english\sofimsha9_sphere.dat

You could also switch to sofimshc and use SLNB which gives you an arc right away.