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# An arithmetic sequence is a sequence in which each term

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An arithmetic sequence is a sequence in which each term [#permalink]  25 Jan 2008, 12:53
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p, r, s, t, u

An arithmetic sequence is a sequence in which each term after the first term is equal to the sum of the preceding term and a constant. If the list of numbers shown above is an arithmetic sequence, which of the following must also be an arithmetic sequence?

I. 2p, 2r, 2s, 2t, 2u
II. p-3, r-3, s-3, t-3, u-3
III. p^2, r^2, s^2, t^2, u^2

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Nov 2012, 07:44, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.
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Re: arithmetic sequence [#permalink]  25 Jan 2008, 13:00
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blog wrote:
An arithmetic sequence is a sequence in which each term after the first term is equal to the sum of the preceding term and a constant. If the list of numbers shown above is an arithmetic sequence, which of the following must also be an arithmetic sequence?

1.) 2p, 2r, 2s, 2t, 2u
2.) p-3, r-3, s-3, t-3, u-3
3.) p square, r square, s square, t square, u square.

(A)1 only
(B)2 only
(C)3 only
(D)1 and 2
(E)2 and 3

D.

Picking numbers is great for this problem:

For example: if the arithmetic sequence is 5,7,9,11,....
1) 10,14,18,22.... is still an arithmetic sequence
2) 2,4,6,8....... is still an arithmetic sequence
3) is not an arithmetic sequence
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Re: arithmetic sequence [#permalink]  25 Jan 2008, 13:09

a(n) = a(n-1) + d

1) b(n) = 2*a(n) = 2*a(n-1)+2*d = b(n-1)+2*d -> arithmetic (d_new = 2*d)
2) b(n) = a(n) - c = a(n-1)+d-c = a(n-1)-c+d = b(n-1) + d -> arithmetic
3) b(n) = a(n)^2 = a(n-1)^2 + 2*a(n-1)*d + d^2 = b(n-1) + 2*a(n-1)*d + d^2 <> b(n-1) + const -> not arithmetic
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Re: OG12 - PS - Q228 [#permalink]  21 Jul 2010, 21:42
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Hi masland,

The first sentence defines an arithmetic sequence. For example, {5, 10, 15, 20, 25} is an arithmetic sequence.

When you have a roman numeral question, start with either a) the roman numeral that is easiest to evaluate or else b) the roman numeral that appears most frequently among the answer choices.

Let's start with II because it shows up the most (three times). Using our example above ({5, 10, 15...}), we can see that {2, 7, 12...} will also be an arithmetic sequence....eliminate A and C (because they don't contain II).

Let's look at I because it is easier than III. If {5, 10, 15...} is an arithmetic sequence, then clearly {10, 20, 30...} is also an arithmetic sequence....eliminate B and E (because they don't contain I).

The correct answer must be D!

(And there is no need to evaluate III--which is fortunate since I wasn't sure what you meant by "p2, r2" although I guess you mean squares).
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Re: OG12 - PS - Q228 [#permalink]  09 Aug 2010, 07:49
From the definition you can clearly see that II is an arithmetic sequence. III is clearly not.

Option I is a trap! If you expand it to $$p+p, r+r, s+s, t+t, u+u$$ it appears it doesn't increase by a constant.

If you rewrite the original sequence with an increase of $$n$$ you get, $$p, p + n, p + 2n, p+3n, p+4n$$.

Now multiply it by 2, and get $$2p, 2p + 2n, 2p + 4n, 2p + 6n, 2p + 8n$$. Now you can see that I increases by $$2n$$!
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Re: An arithmetic sequence is a sequence in which each term [#permalink]  01 Nov 2012, 00:52
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1) multiples all the terms by 2
2) subtracts all the terms by 3..

They gotto be in arithmetic sequence

whereas 3 squares .. anythign when squared will increase the value by a huge value. . and arithmetic sequence may not be maintained
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Re: An arithmetic sequence is a sequence in which each term [#permalink]  27 Jan 2013, 03:31
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I followed a mixed approach. Started by picking numbers and realized soon that option 3 would not give me the results
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Re: An arithmetic sequence is a sequence in which each term [#permalink]  23 Mar 2014, 17:05
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Re: An arithmetic sequence is a sequence in which each term [#permalink]  09 May 2014, 14:26
Hi,

Can someone please clarify a nagging issue:

For a sequence to be an arithmetic sequence, does the different between the units have to be constant? To elaborate, are all three examples below considered arithmetic sequences?

-[2,4,6,8] = difference of 2
-[3,9,81...] = all the units are squared but the differences are not constant
-[2, 5, 9, 14]= the difference is 2+1 so (5= 2 + 3), (9=3+4), (14 = 9 + 5)
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Re: An arithmetic sequence is a sequence in which each term [#permalink]  10 May 2014, 04:12
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russ9 wrote:
Hi,

Can someone please clarify a nagging issue:

For a sequence to be an arithmetic sequence, does the different between the units have to be constant? To elaborate, are all three examples below considered arithmetic sequences?

-[2,4,6,8] = difference of 2
-[3,9,81...] = all the units are squared but the differences are not constant
-[2, 5, 9, 14]= the difference is 2+1 so (5= 2 + 3), (9=3+4), (14 = 9 + 5)

Arithmetic Progression is a special type of sequence in which the difference between successive terms is constant.

{2, 4, 6, 8} is an arithmetic progression (the difference = 2).
{3, 9, 81} is neither arithmetic not geometric progression.
{2, 5, 9, 14} is neither arithmetic not geometric progression.

Theory on sequences problems: sequences-progressions-101891.html

All DS sequences problems to practice: search.php?search_id=tag&tag_id=111
All PS sequences problems to practice: search.php?search_id=tag&tag_id=112

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Re: An arithmetic sequence is a sequence in which each term [#permalink]  10 May 2014, 04:59
blog wrote:
p, r, s, t, u

An arithmetic sequence is a sequence in which each term after the first term is equal to the sum of the preceding term and a constant. If the list of numbers shown above is an arithmetic sequence, which of the following must also be an arithmetic sequence?

I. 2p, 2r, 2s, 2t, 2u
II. p-3, r-3, s-3, t-3, u-3
III. p^2, r^2, s^2, t^2, u^2

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III

Best thing should be plug in:

p = 1, q = 2, r = 3, s = 4, t = 5

I. 2, 4, 6, 8, 10 (Arithmetic Progression)
II. -2, -1, 0, 1, 2 (Arithmetic Progression)
III. 1, 4, 9, 16, 25 (Not AP)

Hence I and II
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Arithmetic Sequence question= Need Help! [#permalink]  09 Mar 2015, 14:58
p,r,s,t,u

An arithmetic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant. if the list of letters shown above is an arithmetic sequence, which of the following must also also be an arithmetic sequence?

I. 2p,2r,2s,2t,2u
II. p-3,r-3,s-3,t-3,u-3
III. p^2,r^2,s^2,t^2,u^2

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
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Re: An arithmetic sequence is a sequence in which each term [#permalink]  09 Mar 2015, 15:00
Expert's post
mawus wrote:
p,r,s,t,u

An arithmetic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant. if the list of letters shown above is an arithmetic sequence, which of the following must also also be an arithmetic sequence?

I. 2p,2r,2s,2t,2u
II. p-3,r-3,s-3,t-3,u-3
III. p^2,r^2,s^2,t^2,u^2

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III

Merging topics.

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Re: An arithmetic sequence is a sequence in which each term [#permalink]  10 Mar 2015, 00:35
mawus wrote:
p,r,s,t,u

An arithmetic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant. if the list of letters shown above is an arithmetic sequence, which of the following must also also be an arithmetic sequence?

I. 2p,2r,2s,2t,2u
II. p-3,r-3,s-3,t-3,u-3
III. p^2,r^2,s^2,t^2,u^2

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III

hi mawus,

as we know an arithmetic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant...
this means the difference between each consecutive number is constant....
lets look at the three choices...
I. 2p,2r,2s,2t,2u....
since the difference is constant,say x and each number has been multiplied by a constant 2, the difference too will remain 2x... so it will be an arithmetic sequence...

II. p-3,r-3,s-3,t-3,u-3
since the difference is constant,say again x and each number has been subtracted by a constant 3 , the difference too will remain x-3... so it will be an arithmetic sequence...

III. p^2,r^2,s^2,t^2,u^2
since the difference is constant in initial sequence ,say x and now, each number has been multiplied by itself. Basically it means that each term is being multiplied by a different number, which is equal to itself... the difference now will change for each two consecutive number ... so it will not be an arithmetic sequence...

only l and ll.. ans D....
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Re: An arithmetic sequence is a sequence in which each term [#permalink]  16 Aug 2015, 16:09
The list of letters is bugging me, specifically because there is a gap between p and r (q). Is this relevant or just a distraction in the question? A simple explanation would be very helpful! Thank you!
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Re: An arithmetic sequence is a sequence in which each term [#permalink]  16 Aug 2015, 16:16
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jmaz88 wrote:
The list of letters is bugging me, specifically because there is a gap between p and r (q). Is this relevant or just a distraction in the question? A simple explanation would be very helpful! Thank you!

Valid question, but it is an official question. It is what it is and you can not question the language or the OA for the question.

I dont believe knowing the alphabets in a particular sequence is relevant for this question. GMAT can even give us strange symbols to stand for these variables. You just have to understand what is an arithmetic progression and after that it is all a matter solving the question with whatever variables are given to you.

If GMAT wants to give you a,z,y,b,c and say that they are in arithmetic progression (ie difference between 2 consecutive terms is constant and is the same value), then you have to stick to this pattern of variables.

Hope this helps.
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Re: An arithmetic sequence is a sequence in which each term   [#permalink] 16 Aug 2015, 16:16
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22.6.4: iii. Problems

Q1

This problem will be concerned with carrying out an SCF calculation for the HeH$$^+$$ molecule in the $$^1\sum_g^+ (1\sigma^2)$$ ground state. The one- and two-electron integrals (in atomic units) needed to carry out this SCF calculation at R = 1.4 a.u. using Slater type orbitals with orbital exponents of 1.6875 and 1.0 for the He and H, respectively are:
\begin{align} & S_{11} = 1.0, & & S_{22} = 1.0, & & S_{12} = 0.5784 & \\ & h_{11} = -2.6442, & & h_{22} = -1.7201, & & h_{12} = -1.5113, & \\ & g_{1111} = 1.0547, & & g_{1121} = 0.4744, & & g_{1212} = 0.5664, & \\ & g_{2211} = 0.2469, & & g_{2221} = 0.3504, & & g_{2222} = 0.6250, & \end{align}
where 1 refers to $$1s_{He} \text{ and 2 to } 1s_H.$$ Note that the two-electrons integrals are given in Dirac notation. Parts a. - d. should be done by hand. Any subsequent parts can make use of the QMIC software provided.
a. Using $$\phi_1 \approx 1s_{He}$$ for the initial guess of the occupied molecular orbital, form a 2x2 Fock matrix. Use the equation derived above in question 1 for $$F_{\mu\nu}$$.
b. SOlve the Fock matrix eigenvalue equations given above to obtain the orbital energies and an improved occupied molecular orbital. In so doing, note that $$\langle \phi_1 |\phi_1 \rangle = 1 = C_1^T SC_1$$ gives the needed normalization condition for the expansion coefficients of the $$\phi_1$$ in the atomic orbital basis.
c. Determine the total SCF energy using the result of exercise 3 above at this step of the iterative procedure. When will this energy agree with that obtained by using the alternative expression for E(SCF) given in exercise 2?
d. Obtain the new molecular orbital, $$\phi_1$$, from the solution of the matrix eigenvalue problem (part b).
e. A new Fock matrix and related total energy can be obtained with this improved choice of molecular orbital, $$\phi 1$$. This process can be continued until a convergence criterion has been satisfied. Typical convergence criteria include: no significant change in the molecular orbitals or the total energy (or both) from one iteration to the next. Perform this iterative procedure for the HeH$$^+$$ system until the difference in total energy between two successive iterations is less than $$10^{-5}$$ a.u.
f. Show, by comparing the difference between the SCF total energy at one iteration and the converged SCF total energy, that the convergence of the above SCF approach is primarily linear (or first order).
g. Is the SCF total energy calculated at each iteration of the above SCF procedure (via exercise 3) an upper bound to the exact ground-state total energy?
h. Using the converged self-consistent set of molecular orbitals, $$\phi 1 \text{ and } \phi 2$$, calculate the one- and two-electron integrals in the molecular orbital basis. Using the equations for E(SCF) in exercises 2 and 3 calculate the converged values of the orbital energies making use of these integrals in the mo basis.
i. Does this SCF wavefunction give rise (at R $$\rightarrow \infty$$) to proper dissociation products?

Q2

This problem will continue to address the same HeH$$^+$$ molecular system as above, extending the analysis to include "correlation effects." We will use the one- and two-electron integrals (same geometry) in the converged (to 10$$^{-5}$$ au) SCF molecular orbital basis which we would have obtained after 7 iterations above. The converged mos you would have obtained in problem 1 are:
$\phi_1 = \begin{bmatrix} & -0.89997792 \\ &-0.15843012 \end{bmatrix} \phi_2 = \begin{bmatrix} & -0.83233180 \\ & 1.21558030 \end{bmatrix}$

a. Carry out a two configuration CI calculation using the $$1\sigma^2 \text{ and } 2\sigma^2$$ configurations first by obtaining an expression for the CI matrix elements $$H_{ij} (i,j = 1\sigma^2, 2\sigma^2)$$ in terms of one- and two-electron integrals, and secondly by showing that the resultant CI matrix is (ignoring the nuclear repulsion term):
$\begin{bmatrix} -4.2720 & 0.1261 \\ 0.1261 & -2.0149 \end{bmatrix}$
b. Obtain the two CI energies and eigenvectors for the matrix found in a part a.
c. Show that the lowest energy CI wavefunction is equivalent to the following two-determinant (single configuration) wavefunction:
$\dfrac{1}{2} \left[ \big| \left( a^{\dfrac{1}{2}}\phi_1 + b^{\dfrac{1}{2}}\phi_2 \right)\alpha \left( a^{\dfrac{1}{2}}\phi_1 - b^{\dfrac{1}{2}\phi_2} \right)\beta \big| + \big| \left( a^{\dfrac{1}{2}}\phi_1 - b^{\dfrac{1}{2}}\phi_2 \right)\alpha \left( a^{\dfrac{1}{2}}\phi_1 + b^{\dfrac{1}{2}}\phi_2 \right)\beta \big| \right]$
involving the polarized orbitals: $$a^{\dfrac{1}{2}}\phi_1 \pm b^{\dfrac{1}{2}}\phi_2$$, where a = 0.9984 and b = 0.0556.
d. Expanding the CI list to 3 configurations by adding the $$1\sigma 2\sigma$$ to the original $$1\sigma^2$$ and $$2\sigma^2$$ configurations of part a above. First, express the proper singlet spin-coupled $$1\sigma 2\sigma$$ configuration as a combination of Slater determinants and then computer all elements of this 3x3 matrix.
e. Obtain all eigenenergies and corresponding normalized eigenvectors for this CI problem.
f. Determine the excitation energies and transition moments for HeH$$^+$$ using the full CI result of part e above. The nonvanishing matrix elements of the dipole operator r(x,y,z) in the atomic basis are:
$\langle 1_{SH} | z | 1_{SHe} \rangle = 0.2854 \text{ and } \langle 1_{SH} |z| 1_{SH} \rangle = 1.4$

First determine the matrix elements of r in the SCF orbital basis then determine the excitation energies and transition moments from the ground state to the two excited singlet states of HeH$$^+$$.
g. Now turning to perturbation theory, carry out a RSPT calculation of the first-order wavefunction $$|1\sigma^2\rangle^{(1)}$$ for the case in which the zeroth-order wavefunction is taken to be the $$1\sigma^2$$ Slater determinant. Show that the first-order wavefunction is given by:
$|1\sigma^2 \rangle^{(1)} = -0.0442|2\sigma^2 \rangle .$
h. Why does the $$|1\sigma 2\sigma \rangle$$ configuration not enter into the first-order wavefuncion?
i. Normalize the resultant wavefunction that contains zeroth- plus first-order parts and compare it to the wavefunction obtained in the two-configuration CI study of part b.
j. Show that the second-order RSPT correlation energy, $$E^{(2)}\text{, of HeH}^+$$ is -0.0056 a.u. How does this compare with the correlation energy obtained from the two- configuration CI study of part b?

Q3

Using the QMIC programs, calculate the SCF energy of $$HeH^+$$ using the same geometry as in problem 1 and the STO3G basis set provided in the QMIC basis set library. How does this energy compare to that found in problem 1? Run the calculation again with the 3- 21G basis basis provided. How does this energy compare to the STO3G and the energy found using STOs in problem 1?

Q4

Generate SCF potential energy surfaces for $$HeH^+ \text{ and } H_2$$ using the QMIC software provided. Use the 3-21G basis set and generate points for geometries of R = 1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.5, and 10.0. Plot the energies vs. geometry for each system. Which system dissociates properly?

Q5

Generate CI potential energy surfaces for the 4 states of $$H_2$$ resulting from a CAS calculation with 2 electrons in the lowest 2 SCF orbitals $$(1\sigma_g \text{ and } 1\sigma_u)$$. Use the same geometries and basis set as in problem 4. Plot the energies vs. geometry for each system. Properly label and characterize each of the states (e.g., repulsive, dissociate properly, etc.).

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text:
# All Maths formulas for class 9 : CBSE Class 9 Maths Formulas : Get Chapter-wise Formulas

#### Bywisdomira

(Last Updated On: November 25, 2022)

Maths Formulas for Class 9: Class 9 is one of the most important phase in a student’s life. Here lies the foundation of board exams. Board exams are a challenge in themselves. Mathematics is a difficult and intimidating subject because of its mathematical formulas. It is a bit difficult but not too difficult yet many students could not achieve it.

Due to the mathematical formula, they avoid this subject and focus on other subjects. This reluctance causes them to lose marks in the overall scenario, although may perform quite well in other subjects.

Mathematics often intimidates students because it comes with the complexity of learning the formulas. A glimpse of the formulas can raise the anxiety of the students and this is the reason why many of them are hesitant to study the subject.

Best Free Online Education Sites 2023

But we are here with one such solution locker of formulas that to help calm the anxiety of the students and have an easy approach to learn the formulas of class 9 maths, this article is a summary list of all the important formulas and how to remember them. Provides some useful tips and tricks for

It would be great if students had the formulas for class 9 polynomials and other such topics in Ashan language and in one place. This will enable them to solve problems and thus improve their grades. In this article, candidates can get the formulas for all the chapters of Maths in Class 9. Read chapter-wise important formulas for class 9.

## Maths Formulas for Class 9

Mathematical formulas are not just meant to be learned blindly. You have to focus on understanding, applying and analyzing all the formulas of mathematics. All Maths Formulas make it easy for you to solve maths questions. You can learn such formulas logically.

Before getting into the list of formulas, let us examine the major chapters in class 9 maths that require formulas.

## List of Important Class 9 Math Formulas

Let us look at some important chapter-wise lists of formulas for all Polynomial Identities for Class 9 and All other Identities of Class 9 in Mathematics.

Know 9th CBSE Exam Concepts

## Class 9 Maths Topic Wise All Formulas

Any number that can be written as p / q where p and q are integers and q ≠ 0 are rational numbers. Irrational numbers cannot be written in the form p / q.

1. There is a unique real number that can be represented on a number line.
2. If r is such a rational number and s is an irrational number, then (r + s), (r – s), (r × s) and (r/ s) are irrational.
3. For positive real numbers, the corresponding identities hold together:
1. ab−−√ab = a−−√×ba×b
2. ab−−√ab = abab
3. (a−−√+b)×(a−−√b)=ab(a+b)×(a−b)=a−b
4. (a+b)×(ab)=a2b(a+b)×(a−b)=a2−b
5. (a−−√+b)2=a2+2ab−−√+b(a+b)2=a2+2ab+b
4. If you want to rationalize the denominator of 1 ⁄ √ (a + b), we need to multiply it by √ (a – b)/√ (a – b), where a and b are both integers.
5. Let a be a real number (greater than 0) and p and q be rational numbers.
1. ap x b= (ab)p+q
2. (ap)q = apq
3. ap / a= (a)p-q
4. ap / bp = (ab)p

### Polynomial Formula Class 9

The polynomial p(x) for a variable ‘x’ has an algebraic expression:

p(x) = anxn + an-1xn-1 + ….. + a2x2 + a1x + a0 ; where a0, a1, a2, …. an are constants where an ≠ 0

1. Any real number; Let ‘a’ be considered the zero of the polynomial ‘P(x)’ if P(a) = 0. In this case, a is said to be the of the equation P(x) = 0.
2. Each one-variable linear polynomial will have a unique zero, a real number that is the zero of the zero polynomial and a non-zero constant polynomial with no zeros.
3. Remainder Theorem: If the power of p(x) is greater than or equal to 1 and dividing p(x) by the linear polynomial x – a will give the remainder in the form p(a).
4. Factoring Theorem: x – a will be a factor of the polynomial p(x) whenever p(a) = 0. The opposite is also true every time.

### Class 9 Maths Formulae for Coordinate Geometry

Whenever you have to trace an object on a plane, you need to divide the plane into two perpendicular lines, making it a Cartesian plane.

• The horizontal line is called the x-axis and the vertical line is called the y-axis.
• The coordinates of a point are in the first quadrant as (+, +), in the second quadrant as (-, +), in the third quadrant as (-, -) and in the fourth quadrant as (+, -) ; where + and – denote positive and negative real numbers respectively.
• The coordinates of the origin are (0, 0) and thus it moves up to move in positive and negative numbers.

Practice 9th CBSE Exam Questions

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## Class 9 Algebraic Identities

Once the students get hold of all the Algebraic Identities of Class 9, they will be able to solve all the algebra related problems in their exam. Given below are the algebraic identities for class 9 which are considered to be very important maths formulas for class 9:

1. (a + b)2 = a2 + 2ab + b2
2. (a – b)2 = a2 – 2ab + b2
3. (a + b) (a – b) = a2 -b2
4. (x + a) (x + b) = x2 + (a + b) x + ab
5. (x + a) (x – b) = x2 + (a – b) x – ab
6. (x – a) (x + b) = x2 + (b – a) x – ab
7. (x – a) (x – b) = x2 – (a + b) x + ab
8. (a + b)3 = a3 + b3 + 3ab (a + b)
9. (a – b)3 = a3 – b3 – 3ab (a – b)
10. (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2xz
11. (x + y – z)2 = x2 + y2 + z2 + 2xy – 2yz – 2xz
12. (x – y + z)2 = x2 + y2 + z2 – 2xy – 2yz + 2xz
13. (x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz
14. x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz -xz)
15. x+ y2 =  1212 [(x + y)2 + (x – y)2]
16. (x + a) (x + b) (x + c) = x+ (a + b + c)x2 + (ab + bc + ca)x + abc
17. x3 + y3 = (x + y) (x– xy + y2)
18. x3 – y3 = (x – y) (x+ xy + y2)
19. x2 + y2 + z2 – xy – yz – zx = 1212 [(x – y)2 + (y – z)2 + (z – x)2]

### Maths formulas for class 9 for Triangles

A triangle is a closed geometric figure made up of three sides and three angles.

1. Two figures are congruent if they have the same shape and size.
2. If two triangles ABC and DEF are congruent under the correspondence that A ↔ D, B ↔ E and C ↔ F, then symbolically, they can be expressed as ∆ ABC ≅ ∆ DEF.

### Right Angled Triangle: Pythagoras Theorem

Let ∆ ABC be a right angled triangle in which AB is perpendi ∆ ABC ≅ ∆ DEF.cular, BC is base and AC is hypotenuse; Then the Pythagorean theorem would be expressed as:

(Hypotenuse)2 = (Perpendicular)2 + (Base)2
i.e. (AC)2 = (AB)2 + (BC)2

### Class 9 Maths Formulas for Areas of Parallelograms & Triangles

A parallelogram is a type of quadrilateral that contains parallel opposite sides.

1. Area of parallelogram = Base × Height
2. Area of Triangle = 1212 × Base × Height

### Class 9 Maths Formulae for Circle

A circle is a closed geometric figure. All points on the boundary of a circle are equidistant from a fixed point inside the circle (called the center).

1. Area of a circle (of radius r) = π × r2
2. The diameter of the circle, d = 2 × r
3. Circumference of the circle = 2 × π × r
4. Sector angle of the circle, θ = (180 × l ) / (π × r )
5. Area of the sector = (θ/2) × r2; where θ is the angle between the two radii
6. Area of the circular ring = π × (R2 – r2); where R – radius of the outer circle and r – radius of the inner circle

### Class 9 Maths Heron’s Formula

Heron’s formula is used to calculate the area of a triangle whose three sides are known. Let the lengths of three sides be a, b and c.

• Step 1 – Calculate the semi-perimeter, s=a+b+c2s=a+b+c2
• Step 2 – Area of the triangle = s(sa)(sb)(sc)−−−−−−−−−−−−−−−−−√s(s−a)(s−b)(s−c)

### Class 9 Maths Formulae for Surface Areas & Volumes

Here, LSA stands for Lateral/Curved Surface Area and TSA stands for Total Surface Area.

### Class 9 Maths Formulas for Statistics

Some facts or figures which can be collected or converted into some useful purpose are known as data. These figures can be represented graphically to increase readability for people.

Three measures of formulas for the interpretation of unclassified data:

### Class 9 Maths Formula for Probability

Probability is the probability that any event will occur. The probability of any event can only range from 0 to 1, with 0 being no probability and 1 being the probability of that event occurring.

Probability=NumberoffavourableoutcomesTotalNumberofoutcomesProbability=NumberoffavourableoutcomesTotalNumberofoutcomes

These are some of the important maths formulas for class 9 which will help you in making your preparation journey easier. Take Embibe’s Class 9 Maths mock tests for free. Refer to sources whenever required. Make the best use of all available resources. It will be easy for you to get high marks in maths.

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### FAQ

Q.1: Where can I practice for more Class 9 Maths questions?
Answer: You can practice on Embibe for class 9 maths questions. NRACETJOB provides you topic wise questions which are available free of cost.

Question 2: Is NCERT Maths enough for class 9?
Answer: Yes, for class 9. NCERT maths book is enough. Just make sure you understand all the concepts and solve all the questions diligently. Note that regular exercise is a must.

Q.3: How can I learn these math formulas?
Answer: Mathematics is a matter of reasoning. Therefore it should be interpreted in the same way. You can learn these formulas by understanding them logically. Then, you can try to solve the questions by applying these formulas.

Q.4: Are the maths formulas for class 9 based on NCERT?

Answer: We have compiled these formulas for class 9 maths so that students can understand them. These formulas are based on NCERT, ICSE and all other related boards.

Q.5: How can you learn Maths NCERT All Polynomial Class 9 Formulas?

Answer: You can try to remember everything you are trying to learn in story form. Sequencing will help you memorize formulas in a specific order. Also, be sure to understand the etymologies of formulas rather than cramming them. In this way you will be able to remember all the formulas of class 9 maths for a long time.


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Home How to find the origin of some invasive html?
Reply: 1

# How to find the origin of some invasive html?

Arepo
1#
Arepo Published in 2018-01-12 13:19:30Z
 I've been working on the site https://founderspledge.com/ and just noticed that if you go to the main page, and open the Chrome console to view the source, there's an element that I certainly didn't put there:   It also doesn't show up when I run the site's code in localhost, or when I load the page source separately using Chrome's 'View Source' option. I Googled the div's class list, and the top few hits were from anti-ad blocker software (eg https://github.com/sitexw/FuckAdBlock), which doesn't seem malign, but doesn't explain why it would be on this site. So 1) How can I get rid of this element? 2) How might it have got there? 3) Does it imply there's a hole in the site's security?
Marko
2#
Marko Reply to 2018-01-12 14:16:00Z
 UPDATE: Facebook have acknowledged this as a bug: https://developers.facebook.com/bugs/336662116810282/?hc_location=ufi I'm so glad to find this issue - have been tearing my hair out to work out the same exact thing! I found it's coming from fbevents.js (https://connect.facebook.net/en_US/fbevents.js) version 2.8.7 - you can clearly see the code that adds the div in. It seems like this is something that has happened very recently? But likewise, I couldn't find anything other than references to anti adblocking scripts which had me concerned!
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News: Did you know ? you can use LaTex inside Postings on fractalforums.com!

## The All New FractalForums is now in Public Beta Testing! Visit FractalForums.org and check it out!

 Pages: 1 [2] 3 4   Go Down
 Author Topic: The "operator" i is more complex than that!  (Read 12823 times) Description: numbers and things as roots of unity and complex factors 0 Members and 1 Guest are viewing this topic.
jehovajah
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May a trochoid in the void bring you peace

 « Reply #15 on: February 01, 2011, 03:17:03 PM »

Finally i can ask the sensible question: what is the square root of a magnitude that is directed in the direction π!

On the face of it there should be no difference to the square root of a magnitude whichever direction it is measured in, but there is and the reason is because the magnitude is rotating!

So the question is : what is the square root of a magnitude that is rotating between 0 and π? we have to take the orientation of the resultant into consideration, because all our magnitudes are in free rotation.

Thus the answer is :half the rotation and the square root of the magnitude!

You can generalise it to any root.
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jehovajah
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 « Reply #16 on: February 02, 2011, 06:50:49 AM »

I have found the most wonderful site for any researcher!

Here i have found Riemanns collected works in german! And of course understandings of the foundation of mathematics untrammeled by ignorance.

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jehovajah
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 « Reply #17 on: February 10, 2011, 09:55:06 AM »

Watched a programme on Fermats last theorem, the modern proof. Well Fermat would certainly not have proved it that way!
I suspect Fermat's proof would not satisfy todays rigorous standards. Fermat knew about Diophantine numbers, and i suspect he conjectured his proof based on the relation $(cos\theta)^n+(sin\theta)^n \neq R^n$ the radius of a circle
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jehovajah
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 « Reply #18 on: July 20, 2011, 05:25:43 PM »

Complex vector algebra is how i refer to this topic now. i.j,k are orthogonally oriented vectors. The properties of a complex vector space reflect all the properties of so called complex numbers through the product rules of complex vectors. The myth of number is what has caused the difficulty in accepting directed vectors, and the √-1 is a red herring. People hate negative numbers and that hatred passes on to anything to do with them. But everybody thinks they know what number is and yet they cannot explain what it is easily, so how are they going to do with negative numbers or roots of negative numbers.

Numbers have a place but not at the foundation of mathematics. Euclid placed "numbers"about 2/3rd of the way through his elements, because they actually derive from geometry.

Trig ratios are fundamental, and vectors are representations of trig ratios.
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jehovajah
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 « Reply #19 on: September 03, 2011, 09:35:37 PM »

i have done a lot of work on generalising the notion of sign without realising that i was in fact generalising it to unit vectors!

Thus i can entirely replace sign by the notion of a unit vector. However to reference the orientation of a unit vector i used the convention of radials emanating from a spherical centre with the radian measure specifying the direction from a pole and a perpendicular rod along the great circles.

Now here is the trick, by prefixing or postfixing a radial to a magnitude we create a vector.

The radial clearly records rotational orientation , thus the notion of a vector is linked to rotation, clock arithmetics and moduli. The notion of reflection in a centre of rotation is therefore to be distinguished from the notion of radial orientation .

We do not need to worry about addition subtraction and general aggregation structures, the familiar rules apply, but with consistent significance. We do not get the notion of the √-1 we instead get the notion of the √L1  if L is the left pointing unit vector.

We only get i if we understand that √ is notation for the geometric mean, and in that case we expect not only√1 but also  the U vector giving U√1.

We then  need to define U•U=r•R•r•R =L that is some aggregation operation produces L from r•R aggregated against r•R. "r" is some action on vectors. (confused? i know i am! And that is the point. The definitions of i are in confusion!! instead of using the notion of "multiplication" we in fact just rotate. The action r is in fact "rotate to vector radial + π/2", and we do not use i as a number but as an instruction)

These unit vectors when employed as gnomons for aggregation have to produce a consistent result. The gnomon  may be a curved gnomon and it acts on a vector.

When looked at in this way the complex decisions and aggregation structures that are being chosen are revealed, but they are all understandable and not mysterious or imaginary.

This also seems to be done in isolation but in fact it is done in analogy with a fractal scale of other relations. This is the genius of Hamilton's exposition of complex algebra.

Thus i replace sign by a vector and the sign rules by a set of operators or actions that give the required results . Unit Vectors with a set of actions replace the complex number notion.
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jehovajah
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 « Reply #20 on: January 01, 2012, 11:19:04 AM »

I have recently posted some links to material that makes intelligible the actual proceses involved in apprehending√-1, and places Hamilton and Grassmann in Context.

http://jehovajah.wordpress.com/jehovajah/blog/the-ideas-that-led-to-quaternions

Although the material contains mathematical notation, the sense is well explained without needing to understand the notation. By this you may come to realise the struggle that took place to find an intelligible notation for the field of "sets" , and  how expediency has led to your mathematical education being a melange of half baked ideas.

In any case enjoy and deepen your knowledge of where fractals "come" from in space.
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jehovajah
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 « Reply #21 on: May 13, 2012, 10:15:24 AM »

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 « Reply #22 on: May 13, 2012, 05:03:34 PM »

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jehovajah
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 « Reply #23 on: June 02, 2012, 03:14:11 AM »

...

You know me too well!
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jehovajah
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 « Reply #24 on: March 08, 2013, 11:53:11 AM »

I was in conversation with a research scientist just the other day and he was bemoaning $sqrt( -1)$
And I thought
" Haven't I dealt with this subject in one of my posts?"

Well I have revisited it here thinking hmm I have forgotten so much!
The confusion created by this particular piece of conventional notation is profound. It turns the world of mathematics upside down and sets it into a flat spin!
I refuse to engage with this notation nowadays. Instead I view it as a procedural call in a computational language. The procedure or algorithm it specifically calls depends on the argument of the call ie the context.

Now, when I say I refuse to engage with this notation I do not mean I do not note and value it's significance. It is an invitation to engage in what Hamilton called the mathesis of the imaginaries. This mathesis is hidden away in algebra, as a subject boundary and particularly in abstract algebras. If we take Algebra to be a blind persons geometry then the symbol becomes a braille mark alerting the reader to this singular fact. The plane is not the real world! One cannot model a 3d world in a 2d plane! We have to go off piste!( pun also intended).

In reality $i$ is a label as is $\pi$ and they both alert to the rotational dynamic in Spaciometry. The procedural call is to some process algebra of rotation.
$sqrt( -1)$ is also a functional symbol that has to hang about in process algebras awaiting it's sequential call. In these cases it highlights a meta algebra, a meta mathematics which more prosaically I identify with any sufficiently high level computer programming language that has a mathematical and statistical sub routine library.
This is fair, and in fact invigorating. It means that we can dispense with mathematics as a singular discipline and place the content securely within the contexts of computational science and philosophy.

The only recognised person, after Newton to attempt such a transformation is Hermann Grassmann in his Ausdehnungslehre of 1844. The version his brother Robert masterminded in 1862 is far removed from Hermanns original conception. That he was able to draw attention to the similarities and developments of his original ideas is perhaps what illustrates the applicability of his original notions.

You see, by my meanderings, where this topic leads and always must lead away from mathematics so called to a broader process algebra, even beyond group and ring theoretical analyses.

That we can watch a 3d animation is a consequence of the actual meaning of $sqrt( -1)$!
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jehovajah
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 « Reply #25 on: May 23, 2013, 09:55:17 AM »

The nature of i is discussed in this thread
http://www.fractalforums.com/new-theories-and-research/geometric-algebra-geometric-calculus/msg61550/#msg61550
in which the symbolic complexity is demonstrated, as it represents the geometrical form and trans-form-ation.

I think i have already made the statement that i and √-1 shhould now be clearly distinguished. One is algebraic and a constant proportion, the other is a complex label that acts as a mnemonic during actual arithmetical calculations.
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Roquen
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 « Reply #26 on: July 11, 2013, 02:24:02 PM »

I always advise completely ignoring 'i' except when (in)formally demonstrating the product.  Once that's done, you never need it again.  It's nothing more than a comma or period past that point...or a typographic element to denote which of the two linearly independent 'parts' one's talking about.  i = sqrt(-1) is at best very misleading and at worst completely wrong (depending on how you interpret the expression).
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jehovajah
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 « Reply #27 on: July 16, 2013, 01:41:04 AM »

Euclids Stoikeioon book 5 helps to explain where the confusion derives from..

The Greeks and the Indians had no concept of negative magnitudes. There was also no concept of multiplication. They also had two concepts of duality.
http://jehovajah.wordpress.com/jehovajah/blog/2013/07/14/isos-and-analogos-different-notions-of-duality

Because of these precursory facts, one has to understand Brahmagupta and Bombelli exactly as they wrote and thought. Quantity was factorised into multiples, and they were aggregated or disintegrated.

Brahmagupta did not introduce negative magnitudes, rather he demonstrated how Shunya was at the back of everything. If we subtract what we have we can easily see Shunya. And what we have to subtract is multiple if what we have  is multiple. In explaining this he utilised analogous duality, straight from the Eudoxian concept of analogos and logos.

The question of the rule that a rectangle formed from a pair of magnitudes that are going to be subtracted should be aggregated does not make sense in an isos duality. The issue never arises practically. It is only in an analogos duality that this rule makes sense. This is because of the single rule that magnitudes must be of the same kind to be compared, aggregated or factorised.

Logos analogos constraints , applied rigorously mean that algebraic rules are the foundation of Arithmetic. The creation of natural numerals, and then fractional numerals and finally negative numerals were on the basis of the logos analogos duality. The isos duality was restricted to direct geometrical comparisons. In this case negative quantities are geometrically rotations of magnitudes.

On the page they look no different to any other geometrical construction. However analogos duality requires 4 magnitudes, and a negative magnitude is drawn within a gnomonic form, where the parts to be subtracted actually have relative positions!

This formal presentation of a magnitude with a subtracted magnitude, forces the requirement of an additional magnitude. This rule is expressed as – * – = + in Bombelli's notational rhetoric. But Brahmagupta had explained this in his treatment of analogos duality.

These rules actually arise in the cubic setting, and it is only in that case where 3 pairs of magnitudes are compared and each magnitude has a part subtracted, that the full rule appears. However, to reduce it to a rule of signs requires the creation of negative magnitudes, as opposed to parts to be subtracted in the process.

In isos duality the $sqrt(-1)$ is no different to the $sqrt 1$. All quadrature is done in the circle, using chords crossing the diameter perpendicularly. Only by using rotation orientation as a distinguisher can any distinction be made , but it does not actually uniquely define a value. On the other hand analogous duality does enable a unique representation in notation..

The rise of coordinates and complex combinations as vectors has meant that the imaginary quantities can be distinguished in coordinate notation. The myth created around so called imaginary numbers can be dispensed with and more accessible rules and notation can be used.

We do not, as a rule seek to confuse our colleagues if we make a mathematical discovery in these days. However, in the past it was a recognised way of determining primacy to send out or publish a confusing or coded message.
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jehovajah
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 « Reply #28 on: October 23, 2013, 12:44:04 PM »

Believing that I was some mathematical entity, Clifford went on to use it in the further development of his algebras.
http://rsta.royalsocietypublishing.org/content/356/1740/1123.full.pdf

In this regard, I has become a behavioural marker, guaranteeing a predictable mathematical behaviour.
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jehovajah
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 « Reply #29 on: April 30, 2014, 01:28:30 AM »

I have placed this here to avoid jamming any ones computer! It is very relevant to the work of Grassman as it occurs historically after he has written and published the Ausdehnungslehre 1844. However it shows little trace of Grassmanns concepts, and consequently it is quite hard to follow . After Riemann's death Mathematicians turned to Grassmanns work to make sense of Riemanns !

The work of Gauss similarly in 1827 the year Justus published a paper on Trigonometry in space, contains some concepts of Euler regarding parameter switching and surface analysis by tangent planes. The linear combination of the map from the parameter plane to the surface, and then from the surface to the unit sphere for the normal to the surface shows ideas that are similar to Hermans but not expressed in terms of vectors, or dot products etc. Thus Hermanns concepts are in keeping with his times, but add that extra insight that makes the difference for the general mathematician.

Gauss may have had a good grasp of what he meant but without the presentation of Hermann Grassmann it was very difficult to follow his argument or disquisition


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# Time, Speed & Distance Questions for RRB NTPC Set-3 PDF

0
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### Time, Speed & Distance Questions for RRB NTPC Set-3 PDF

Download RRB NTPC  Time, Speed & Distance Questions Set-3 PDF. Top 10 RRB NTPC questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: With an average speed of 40 km/ hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. The total journey is

a) 35 km

b) 60 km

c) 70 km

d) 50 km

Question 2: A person can row a distance of one km upstream in ten minutes and downstream in four minutes. What is the speed of the stream ?

a) 4.5 km/h

b) 4 km/h

c) 9 km/h

d) 5.6 km/h

Question 3: A train 150 metres long crosses a milestone in 15 seconds and crosses another train of the same length travelling in the opposite direction in 12 seconds. The speed of the second train in km/hr is

a) 52

b) 56

c) 54

d) 58

Question 4: A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. The distance (in metres) between them after 6 minutes is

a) 190

b) 200

c) 100

d) 150

Question 5: Pipe A alone can fill a tank in 8 hours. Pipe B alone can fill it in 6 hours. If both the pipes are opened and after 2 hours pipe A is closed, then the other pipe will fill the tank in

a) 6 hours

b) 3.5 hours

c) 4 hours

d) 2.5 hours

Question 6: If I walk at 5 km/hour, I miss a train by 7 minutes. If, however, I walk at 6 km/hour, I reach the station 5 minutes before the departure of the train. The distance (in km between my house and the station is

a) 6

b) 5

c) 4

d) 3

Question 7: Walking 6/7 th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is

a) 1 hour

b) 1 hour 12 minutes

c) 1 hour 15 minutes

d) 1 hour 20 minutes

Question 8: A does half as much work as B in three fourth of the time. If together they take 18 days to complete a work, how much time shall B take to do it alone ?

a) 30 days

b) 35 days

c) 40 days

d) 45 days

Question 9: A student goes to school at the rate of 2.5 km/h and reaches 6 minutes late. If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is

a) 5

b) 4

c) 3

d) 1

Question 10: A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water (in litres) will fall into the sea in a minute?

a) 4,00,000

b) 40,00,000

c) 40,000

d) 4,000

Let the required distance be $x$ km

Acc to ques :

=> $\frac{x}{35} – \frac{x}{40} = \frac{15}{60}$

=> $\frac{8x – 7x}{280} = \frac{1}{4}$

=> $x = \frac{280}{4} = 70$ km

Let the speed of the person be $x$ km/h and speed of stream be $y$ km/h

=> Speed when going upstream = $(x – y)$ km/h

and speed while going downstream = $(x + y)$ km/h

=> $\frac{1}{x – y} = \frac{10}{60}$

=> $x – y = 6$

Also, $\frac{1}{x + y} = \frac{4}{60}$

=> $x + y = 15$

Subtracting the two equations, we get :

$2y = 9$

=> Speed of stream = $y = 4.5$ km/h

1st train crosses milestone in 15 seconds.

=> Speed of 1st train = $\frac{150}{15}$ = 10 m/s

Length of 2nd train = 150 m

Let speed of 2nd train = $x$ m/s

Acc to ques :

=> $\frac{150 + 150}{x + 10} = 12$

=> $x + 10 = 25$

=> $x = 25 – 10 = 15$ m/s

=> Speed of 2nd train = $15 * \frac{18}{5}$ km/h

= 54 km/h

When two bodies move in the same direction their relative speed is the difference of their speeds.
The relative speed between the police and the thief is 1 km per hour.
1km per hour = 1 km in 60 minutes.
Therefore 100 m in 6 minutes.This is the distance between them after 6 minutes.
Hence Option C is the correct answer.

Let us consider the capacity of the tank to be 48 litres.
Pipe A fills it in 8 hours at 6 litres per hour.
Pipe B fills it in 6 hours at 8 litres per hour.
Total of 28 litres would be filled in 2 hours with both pipes open.
Afetr 2 hours Pipe A is closed.Hence Pipe B has to fill the remaining tank.
Quantity of water needed to fill the tank is =48-28 = 20 litres.
Pipe B fills at the rate of 8 litres per hour.
Hence it takes 2.5 hours for Pipe B to fill remaining 20 litres of the tank.
Hence Option D is the correct answer.

When speed is 5 km/hour time is t + (7/60) hours
When speed is 6 km/hour time is t – (5/60) hours
Distance is same.Hence the product of speed and time must be same.
5 (t + (7/60)) = 6(t – (5/60)
5t + (7/12) = 6t – (1/2)
t=(7/12)+(1/2)
t = 13/12 hours
Distance = 5 ( t + (7/60) )
=5t + (7/12)
=5 (13/12) + (7/12) = 72/12
=6 kms.
Hence Option A is the correct answer.

Speed = $s$
Time =$t mins$
Time taken when the man walks at $\frac{6}{7}$ th of his speed $= t+12$
Since distance is same the product of speed and time must be same.
$\frac{6\times (t+12)}{7} = t$
t= 72 mins = 1 hour and 12 minutes
Hence Option B is the correct answer.

Three fourth of the work A is equal to half of what B had done.
Work done by A $=\frac{1}{2} \times \frac{3}{4}$ Work done by B $= \frac{2}{3}$ Work done by B
Work is completed in 18 days when both A and B work together.
$18 \times (A+B)$ $=1$
Replacing A with $\frac{2}{3}B$
$18 \times (\frac{2}{3}B+B) =1$
$18 \times (\frac{5}{3}B) =1$
$B=30$
Hence Option A is the correct answer.

When speed is 2.5 km/hour time = t + (6/60) hours
When speed is 3 km/hour time is t – (10/60) hours
Distance is same.Hence the product of speed and time must be same.
2.5 (t + (6/60)) = 3(t – (10/60)
2.5t + (15/60) = 3t – (30/60)
0.5t=(15/60)+(30/60)
0.5t = 45/60 hours
t = 3/2 hours
Distance = 3 ( t – (10/60) )
=3t – (30/60)
=3 (3/2) – (30/60) = (9/2)-(1/2) = (8/2)=4 kms
Hence Option B is the correct answer.

Speed $= 2km/hr = \frac{100}{3} m/min$
Volume of water flowing into sea = Depth of river $\times$ Width of River $\times$ Rate of flow
$=3 \times 40 \times \frac{100}{3} = 4000 m^3$
$1 m^3 = 1000$ litres
Volume of water flowing into sea $= 40,00,000$ litres


text:
# CutePDF command line or equivilant

Is it possible to run a command that will automatically print to file using CutePDF?

I have tried this and still no luck: print C:\test.doc /D:CPW2

If not any other suggestion on how to automate a print to pdf / convert to pdf through a command line silently (freeware)

-
What kind of files are they? The program used to display, edit, and print the file will determine what terminal command should be used, if the command exists at all. –  daxlerod Jul 3 '13 at 12:34
what error message do you get? –  golimar Jul 3 '13 at 12:50
Do you need a specific output path or could it be the same path as the input path is? For example print C:\test.doc would create a new PDF file under C: –  nixda Jul 3 '13 at 12:59
The path doesn't really matter, would prefer it to be the same as the input but that can be changed accordingly, the error im getting is: "Unable to initialize device CPW2" Although in printers this is the correct port. Im not sure whether its something to do with cutepdf itself –  Ash King Jul 3 '13 at 13:06
@AshKing Hello, I would be pleased if you could leave some feedback to know if I have to improve my answer. –  nixda Aug 17 '13 at 9:09

### Simple solution

If the output path doesn't matter, you could try PDFCreator with the following commandline:

C:\Program Files (x86)\PDFCreator>PDFCreator.exe /NOSTART /PF"C:\test.doc"

1. Per GUI you have to enable Use Auto-save and set a auto-save path once.
From now on, this path will be used every time you execute the command above
2. You have to set PDFCreator as your default printer

Tip:
Per GUI you can save your settings in a profile (.INI file) which can then be selected when you execute your command line. This way, you don't have to use Auto-save as default for every normal print (e.g. from Word or Excel). Read here for more about PDFCreators commandline usage

PDFCreator.exe /NoStart /OptionsFile"C:\myprofile.ini" /PF"C:\test.doc"

The advantage of the second solution is, that you don't have to set PDFCreator as your default printer and you have full control over the output file path and name

1. Install PDFCreator together with its COM module

2. Copy & paste this code to a textfile and save it as Convert2PDF.vbs

Set PDFCreator = Wscript.CreateObject("PDFCreator.clsPDFCreator", "PDFCreator_")
With PDFCreator
.cStart "/NoProcessingAtStartup"
.cOption("UseAutosave") = 1
.cOption("UseAutosaveDirectory") = 1
.cOption("AutosaveFormat") = 0
.cOption("AutosaveStartStandardProgram") = 0
DefaultPrinter = .cDefaultprinter
.cDefaultprinter = "PDFCreator"
.cClearcache
.cPrinterStop = false
.cOption("AutosaveDirectory") = WScript.Arguments(1)
.cOption("AutosaveFilename") = WScript.Arguments(2)
.cPrintfile cStr(WScript.Arguments(0))
c = 0
Do While (ReadyState = 0) and (c < 120)
c = c + 1
Wscript.Sleep 250
Loop
.cDefaultprinter = DefaultPrinter
.cClearcache
WScript.Sleep 200
.cClose
End With

Convert2PDF.vbs "C:\input.doc" "C:\outputfolder" "outputfilename"
Personally I use a slightly different version where the input and output folder+filename stays the same. I created a shortcut in my shell:sendto folder to easily convert files per right-click


text:
The Prague seminar on function spaces

## forthcoming program

June 13, 2018

Abdulhamit Küçükaslan (Mathematical Institute, Czech Academy of Sciences, Prague):

Boundedness of Hilbert transform in local Morrey-Lorentz spaces
Abstract:

In this talk, we investigate the boundedness of the Hilbert transform $H$ in the local Morrey-Lorentz spaces $M_{p,q;\lambda}^{loc}$
$1\le q \le \i$, $\frac{q}{q+\lambda} \leq p \leq \frac{q}{\lambda}$. We prove that the operator $H$ is bounded in $M_{p,q;\lambda}^{loc}$ under the condition $1< q <\i$, $\frac{q}{q+\lambda} < p < \frac{q}{\lambda}$.

In the limiting case $p =\frac{q}{q+\lambda}, 1<q<\i$, we prove that the operator $H$ is bounded from the space $M_{p,q;\lambda}^{loc}$ to the weak local Morrey-Lorentz space $WM_{p,q;\lambda}^{loc}$. Also we show that for the limiting case  $p =\frac{q}{\lambda}, 0<q\leq \i$, the modified Hilbert transform $\widetilde{H}$ is bounded from the space $M_{p,q;\lambda}^{loc}$ to the bounded mean oscillation space $BMO$.

back to the main page


text:
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# Rearrangement of Carbocation for JEE

## Introduction to Carbocation

Last updated date: 24th Mar 2023
Total views: 77.4k
Views today: 1.54k

A carbocation is a cation with a positively charged carbon atom. Some examples are methenium CH3+, methanium CH5+, and vinyl C2H3+ cations. Carbocations containing more than one positively charged carbon can be found very rarely. For example, ethylene dication C2H42+. According to the IUPAC, a carbocation is defined as any cation that contains an even number of electrons with a significant portion of the positive charge residing on a carbon atom. The carbon atom of a carbocation is sp2 hybridised. The three hybridised orbitals of the carbon atom are used for single bonding to three substituents, and the remaining p-orbital is empty.

Carbocations are important reactive intermediates in many organic synthesis reactions. Carbocation rearrangements are generally reactions in organic Chemistry where the movement of a carbocation from an unstable state to a more stable state occurs through the molecular rearrangement in the structure. After the carbocation rearrangement, the new molecule is the structural isomer of the initial molecule.

## Classification of Carbocations

1. Primary Carbocation: Here, the carbon having the positive charge is attached to one carbon of an alkyl or aryl group.

2. Secondary Carbocation: In this carbon, a positive charge is attached to two other carbons.

3. Tertiary Carbocation: In this case, the carbon having the positive charge is attached to three carbon atoms.

In methyl carbocation, the carbon bearing a positive charge is attached to only three hydrogen atoms.

Type of Carbocation

## Generation of Carbocation

The different methods available to generate stable or unstable carbocation are:

1. Ionisation: When direct ionisation of an alkyl or aryl compound occurs, a leaving group attached to a carbon atom leaves with its pair of electrons, forming a carbocation. Stable carbocations can be produced in super acids. For example, the addition of alkyl fluorides to SbF5 forms a stable carbonation.

$\mathrm{RF}+\mathrm{SbF}_{5} \rightarrow \mathrm{R}^{+}+\mathrm{SbF}_{6}^{-}$

2. Ionisation After an Initial Reaction: Occasionally, the functional groups in an organic compound may undergo an initial reaction first. During this reaction, the functional group converts to a good leaving group, and the intermediate formed then ionises into a carbocation after the removal of the leaving group. Protonation of organic alcohol generates an oxonium ion which ionises to a carbocation by losing a water molecule.

$\mathrm{R}-\mathrm{OH} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{R}-\stackrel{+}{\mathrm{OH}_{2}} \rightarrow \mathrm{R}^{+}+\mathrm{H}_{2} \mathrm{O}$

3. Attack of π-system on Electrophiles: When an electrophile attacks one of the atoms of a π-system in an organic molecule, the adjacent atom acquires a positive charge. In such cases, if the positive atom is carbon, then a carbocation is formed. When the π-system of an alkene or alkyne gets attacked with an electrophile like proton, bromonium ion, etc., a positive charge develops on the adjacent carbon and leads to the formation of a carbocation intermediate. For example,

Attack of π-system in Alkene on an Electrophile.

## Stability of Carbocation

The stability of a carbocation is decided by three main factors, inductive effect, hyperconjugation, or resonance effect. Carbocations are the most stable when the charge is on a tertiary carbon and the least stable on a primary carbon. The order of stability is 3o carbocation > 2occarbocation > 1o carbocation > Methyl carbocation.

Order of Stability of Carbocation

1. Inductive Effect: If the groups attached to positively charged carbon are electron releasing, they will delocalise the positive charge on carbon and will decrease the intensity of the positive charge on carbon. This stabilises the carbocation. The more alkyl groups are attached, the more stable the carbocation will be as alkyl groups release electrons.

2. Hyperconjugation: If in any carbocation, the sigma bond in conjugation with vacant p orbital on carbon has a positive charge, then it participates in delocalisation. Thus, they spread the charge all over the alkyl groups. This effect is known as hyperconjugation or no bond resonance. The structure of the ethyl carbocation has the following series of resonance forms, according to the theory of hyperconjugation.

Resonating Structure of Ethyl Carbocation

(Note: in the above image, numbers in subscript don’t indicate the number of atoms but used to distinguish between hydrogen atom.)

3. Resonance: Conjugation with multiple bonds or lone pairs of electrons increases the stability of a carbocation. For example, triphenylmethyl cation has higher stability than diphenylmethyl cation. In the case of triphenylmethyl cation, delocalisation of positive charge is more as it has more resonating structures.

Resonance Effect in Carbocation

## Rearrangement of Carbocation

Carbocation rearrangements are the movement of a carbocation from an unstable state to a more stable state through structural rearrangement to form a structural isomer of the initial molecule. Rearrangements of carbocation are of two types: hydride shift and alkyl shift. When rearranged, the molecules can also undergo unimolecular substitution (SN1) or unimolecular elimination (E1) reactions.

### Hydride Shift

In hydride shift, hydrogen moves over from one carbon to a neighbouring, which is less substituted carbon. Here, the hydrogen migrates towards the positively charged carbon with its electron pair, leaving a stable carbocation behind. Generally, hydride shift occurs in the reaction of an alcohol with HBr, HCl, and HI. Examples of carbocation rearrangement are given below.

1-2 Hydride Shift

Below is a carbocation rearrangement example, which is a reaction between an alcohol and hydrogen chloride.

• Protonation and Loss of Water Molecule: In this SN1 reaction, -OH which leaves the organic alcohol forms a carbocation on carbon (C-3) after receiving a proton from the nucleophile to produce an alkyloxonium ion.

• Hydride Shift: The hydrogen atom attached to the carbon atom adjacent to the original carbon (C-2) undergoes a hydride shift in this case.

• Attack of Cl Atom: The Cl atom can now attack the carbocation and form a more stable structure because of hyperconjugation. The carbocation here is the rearranged most stable carbocation, which is tertiary carbon.

Hydride Shift Rearrangement of Carbocation

### Alkyl Shift

Alkyl Shift reactions are similar to hydride shifts. Here, instead of the proton, the alkyl group shifts with the nucleophile. The group that shifts takes its electron pair with it to form a bond with the nearby carbocation. Finally, the shifted alkyl group and the carbocation with positive charge switch positions on the molecule.

Alkyl Shift Rearrangement of Carbocation

## Conclusion

A carbocation, which was earlier known as carbonium, is an ion with a positively charged carbon atom. They are reactive intermediates, an electron deficient species with a sextet of electrons. We can obtain stable carbocations in super acids. The order of stability is tertiary carbocations > secondary carbocations > primary carbocations > methyl carbocation.

The stability of the carbocation can be ensured by factors like inductive effect (+I), hyperconjugation, and resonance, which can decrease the positive charge on carbon. Carbocations generally undergo two types of rearrangements: hydride shift and alkyl shift. The rearranged carbocations can also undergo reactions like unimolecular substitution (SN1) or unimolecular elimination (E1) and form simple or complex mixtures of products.

Competitive Exams after 12th Science

## FAQs on Rearrangement of Carbocation for JEE

1. Discuss the reactivity of carbocations.

In general, carbocations undergo three types of reactions, which are listed below:

Nucleophile Capture: Carbocations will react with even mild nucleophiles like water to form a new bond. The carbocation may combine with a negative ion or species possessing an electron pair (nucleophiles such as OH-, Cl-, Br-, water ) and they are very fast reactions.

Elimination to form a pi Bond: Carbons near the carbocation can lose a proton to form a double or triple bond with the carbocation.

Rearrangement: Carbocations undergo rearrangement via 1,2-hydride or 1,2-alkyl shifts and generate a more stable carbocation.

2. What is rearrangement in carbocation?

The migration of a carbocation from an unstable state to a more stable state is known as a carbocation rearrangement and is a common phenomenon in organic Chemistry. Carbocation rearrangements can be of two different types: hydride shifts and alkyl shifts. Hydride Shift indicates that a hydrogen moves from one carbon to a nearby carbocation that has less substitution. Alkyl shift refers to when an alkyl group moves and takes its electron pair along with it to create a bond with a nearby carbocation.


text:
# How sub-exponential time does $\text{3SAT}$ have to be to make $\text{NP} \neq\text{EXP}$? What else would imply $\text{NP} \neq\text{EXP}$?

The exponential-time hypothesis posits that if $$\mathsf{3SAT}$$ has NO subexponential time algorithm (i.e. one in $$\mathcal O(2^{o(n)})$$), then $$\mathsf{P}\neq \mathsf{NP}$$. However, I am interested in the case that it does have such an algorithm.

From posts such as this, it seems that simply finding such an algorithm will not prove $$\mathsf{NP} \neq\mathsf{EXP}$$, so I am wondering just how subexponential the algorithm needs to be in order to guarantee this.

More generally, is there some sort of lists of conjectures/hypotheses that would imply $$\mathsf{NP} \neq\mathsf{EXP}$$?

• In order to prove that NP$\neq$EXP you would need to find a problem that is in EXP and not in NP, so investigating the complexity of SAT is not a correct approach because it is known to be in both. – Laakeri Apr 27 at 5:04
• @Laakeri For example, if we know that 3SAT can be solved in $O(n^k)$ time for some k, then clearly $\text{NP}\neq \text{EXP}$. So I'm wondering if there is some intermediate level (between $2^{\log n}$ and $2^n$, say $O(2^{\sqrt n})$ s.t. knowing that 3SAT can be solved in this time implies $\text{NP}\neq \text{EXP}$ – D.R Apr 27 at 5:44
• I see, this is by time hierarchy theorem, right? I guess for it to apply you need to have a complexity class that is closed under polynomial-time reductions. – Laakeri Apr 27 at 7:00
• The first sentence in this post is not an accurate statement of the ETH. – D.W. Apr 27 at 7:32
• If there is a $2^{n^{o(1)}}$ algorithm, then the time hierarchy theorem would separate NP from EXP. – Yuval Filmus Apr 27 at 7:35


text:
# The original paper by Kutta

The Runge-Kutta method is a milestone of computational mathematics. I can find the papepr by Runge, but not the one by Kutta.

By wikipedia, the title is Beitrag zur näherungweisen Integration totaler Differentialgleichungen (1901)

Can anyone find it?


text:
## Cover Letters for Publications in LaTeX

I admit it. Although I compose most of my journal publications in LaTeX, I use Microsoft Word for writing the cover letters. The university's letterhead is provided in Word, and setting up and compiling a LaTeX document for a single text page is not worth the trouble.

However, a cover letter I was writing today was an exception. The corresponding paper is a major revision of a submission; about half of the text is new or updated material. Describing these changes in the cover letter proved challenging, because I kept referring to sections, tables, figures, and equations that I was moving around. Using the symbolic labels I had defined in the document would really help, allowing me to write things like:

Table \ref{tab:id} has been expanded to include the six largest systems,
while the values for all 32 systems now appear in the new Figure
\ref{fig:unused}.

Using the labels of one document in another proved quite easy. I first filtered with a shell command only the references I was interested in (the newlfm letter style I used was choking on the table of contents commands).

egrep "newlabel|bibcite" paper.aux >labels.aux

I then included the file with those labels in the cover letter.
\input{labels.aux}

I expected to spend a lot of time composing our letterhead in LaTeX, but I adopted a much simpler solution. I converted the Word document to PDF, cropped the PDF page to the size of the letterhead, and included the PDF file as a graphic.

\newsavebox{\Caueb}
\sbox{\Caueb}{\parbox[t]{6 in}{%


text:
# Graph theory exporting problems

This post is a wiki. Anyone with karma >750 is welcome to improve it.

I type in the following SAGE code and get the graph on the screen that I would expect to get.

g = Graph({0,[1,2,2]}); g

g.plot(graph_border = True)

When I try to export this to a file with the following code I get an "AssertionError" message.

g = Graph({0,[1,2,2]}); g

g.plot(graph_border = True).save('graph.pdf')

How can I fix this so that I can output this graph as a pdf file. I love how easy I can make nice looking graphs in SAGE, but it is useless if I can't export them to use in Latex.

edit retag close merge delete

Sort by » oldest newest most voted

This post is a wiki. Anyone with karma >750 is welcome to improve it.

Thanks for the reply John. I tried outputting it as a png file and I don't get the error message anymore. But, when I evaluate the two lines of code I get the graph without a link to the file. Where do I find the file?

It is strange I was watching Will Stein's graph theory tutorial and typing exactly as he did and it didn't work for me on my computer. When he typed the code I suggested in the original question he got a link to a pdf file, which he could click on and then have access to the file.

more

This post is a wiki. Anyone with karma >750 is welcome to improve it.

Save it as a png file instead of a PDF file. What should work is

g.plot(graph_border = True).save('graph.png')

Then your LaTeX file should look like

\documentclass{article}
\usepackage{graphicx}
\begin{document}

Hello.

\includegraphics{graph.png}

Goodbye.

\end{document}

For this particular graph, I get a nicer picture doing g.plot(graph_border = True) than doing g.plot(graph_border = True).save('graph.png'), so I would actually do the first command to view the picture, and then save that picture manually into graph.png.

more


text:
# Distinct eigenvalues for integral operator?

Is there some sufficient condition on the kernel $K$ of a (say, finite rank, to simplify) integral operator $$\mathcal K:f(x)\in L^2(\mathbb R)\mapsto \int K(x,y)f(y)dy$$ so that it has all its non-zero eigenvalues $\textbf{distinct}$ (i.e. with multiplicity one) ?

-
I can refer you to this paper. –  Mhenni Benghorbal Aug 25 '12 at 12:34
A finite-rank operator on an infinite-dimensional Banach space has infinite-dimensional kernel. Are you talking about the nonzero eigenvalues? –  Qiaochu Yuan Aug 25 '12 at 23:17

The OP added to his question that "the multiplicity of eigenvalues is strictly one". The following answer was given before that addition which does not assure the multiplicity of the eigenvalues to be one.

Note that, if $A$ is a compact operator, then the spectrum of $A$, excluding $0$, consists of a countable family of isolated points with finite multiplicity $|\lambda_1| \geq |\lambda_2| \geq \dots$ and either the spectrum of $A$ is finite or $\lim_{n\rightarrow \infty} \lambda_n = 0$.

Now, you can use the fact that "integral operators with kernels of finite rank are compact operators".

-
This shows that the nonzero eigenvalues have finite multiplicity. I think the OP wants to know when they have multiplicity $1$. –  Qiaochu Yuan Aug 25 '12 at 23:18
I don't follow. What do you mean by "distinct eigenvalues" if it isn't "the eigenvalues have multiplicity $1$"? –  Qiaochu Yuan Aug 26 '12 at 0:40
That isn't what I asked. What do you mean by "distinct eigenvalues"? –  Qiaochu Yuan Aug 26 '12 at 3:31
Okay, so what would it mean to not have distinct eigenvalues? Does this just mean "has at least $2$ eigenvalues?" I don't think this is how anyone uses the word "distinct." –  Qiaochu Yuan Aug 26 '12 at 4:01
Sorry but I fail to see how this is, even remotely, an answer to the question asked. –  Did Sep 5 '12 at 20:43


text:
# Question on the closed string propagator calculated in Blumenhagen et. al. Basic Concepts in String Theory

You may find Basic Concepts in String Theory by Blumenhagen et. al. here. In page 37, they display the propagators of a closed string.

I also did calculation to verify their results, but my results show that the propagators should contain Heaviside step functions. For example,

$\langle X_L^\mu(\bar z)X_L^\nu(\bar w)\rangle=[\frac{\alpha'}{4}\eta^{\mu\nu}\ln \bar z-\frac{\alpha'}{2}\eta^{\mu\nu}\ln(\bar z-\bar w)]\theta(\tau-\tau')+[\frac{\alpha'}{4}\eta^{\mu\nu}\ln \bar w-\frac{\alpha'}{2}\eta^{\mu\nu}\ln(\bar w-\bar z)]\theta(\tau'-\tau),$

$\langle X_R^\mu(z) X_L^\nu(\bar w)\rangle=-\frac{\alpha'}{4}\eta^{\mu\nu}\ln\bar w\,\theta(\tau'-\tau)-\frac{\alpha'}{4}\eta^{\mu\nu}\ln z\,\theta(\tau-\tau')$.

Here, $\tau$ is the time associated with $z$ and $\tau'$ associated with $w$. The reason that there are Heaviside functions is simply that the time ordered product by definition contains Heaviside functions. Please look at my results carefully, and you will find out that if we assume $\tau>\tau'$, my results agree with Blumenhagen et. al..

So what's going wrong with my calculation?

• Can you describe the procedures to calculate above propagators, or give some references about the calculations about these propagators? Thanks! – phys_student Aug 7 '17 at 4:51
• I found on page 47 of this lecture notes the propagators are derived, which maybe helpful for you. stringworld.ru/files/Arutyunov_G._Lectures_on_string_theory.pdf – phys_student Aug 7 '17 at 5:40
• @phys_student Thanks! It's been a long time ago. I just worked out details very carefully, using very conventional treatment. Nothing fancy. But I will look into the lecture notes. – Drake Marquis Aug 7 '17 at 8:07

Your Result is right, Blumenhagen Lüst Theisen does only look the case where $\tau>\tau'$.


text:
# Automated Crop Production For the $21^{St}$ Century

• Lu, F.M. (Dept. of Agricultural Machinery Engineering College of Agriculture, Taiwan National University, Taiwan)
• Published : 2000.06.01

#### Abstract

After ten years of implementing the agricultural automation program in Taiwan, some positive effects and satisfactory results have been recognized by both the agricultural industry and local administrative bureaux. The automation of agriculture is a response to sophisticated demands for production and quality in countries with high labor costs. The development of sensor systems, control systems, precision agriculture systems, and engineering for plant culture systems will determine the degree of automation used for crop production in the 21st century. The engineering system will capitalize upon expertise from physiologists, pathologists, systems analysts, agronomists, horticulturists, computer programmers, economists, crop producers and managers in order to efficiently implement automated crop production.


text:
# Are there any dangerous commands in LaTeX?

I encountered this question: What is the most dangerous command in SQL?. And a similar one came to my mind,

Are there any dangerous commands in LaTeX?

By dangerous we may mean any or all of these scenarios:

1. When a particular command is used inadvertently by someone in a file, the command may harm one's system or files.
2. A command may be put in a file with malicious intent, then the file is given to someone, and when that file is compiled in the remote system, a harm will occur.
3. A command in a file will cause the compilation process of the file to consume prohibitively high resources.
4. A command will cause LaTeX or TeX to malfunction and produce unintended results.

When you answer, please include relevant examples, with explanations of how exactly it will cause harm.

• \renewcommand – egreg Feb 13 at 10:28
• Can you define 'dangerous' here? The obvious one is \write18 followed by other \write operations ... – Joseph Wright Feb 13 at 10:30
• \dagger is a dangerous thing – David Carlisle Feb 13 at 10:43
• @DavidCarlisle surely \ddagger is more dangerous?! However I'd agree that the things that are generally dangerous for computers is the ability to write arbitrary things to arbitrary files... – rbrignall Feb 13 at 10:55
• For 1 and 2 you might want to look at some questions under the security tag (tex.stackexchange.com/q/10418/134574 and tex.stackexchange.com/q/358642/134574 for example), and the answer basically is: if you compile \write18{rm -rf /*} with --shell-escape (and Linux) you'll get what you asked for. 3 depends: you can make TeX loop forever with \def~{~}~, but it won't consume that much resources. With Lua you may get better (bad) results. 4 can be made into a list: the limit is your creativity and TeX knowledge :-) – Phelype Oleinik Feb 13 at 15:05

### Category 1

\newwrite\out
\immediate\openout\out=\jobname
\immediate\write\out{Fool!}
\bye

This will overwrite the TeX file.

TeX will obey your instructions even if they're foolish. However, it will refuse to write file above the current working directory (under the standard settings of the most common distributions).

Obviously, if the current directory is / and you run TeX with superuser privileges, you can overwrite anything on your system; but this is true of every scripting language interpreter that's able to write out files, so not a specific thing of TeX.

### Category 2

If you run TeX with -shell-escape enabled and sufficient privileges, you can execute any program, including rm -f /. So be careful when enabling -shell-escape. Some packages, notably minted require -shell-escape. Examine the input file you receive from other people before processing it.

### Category 3

TeX engines different from LuaTeX only allocate a maximum amount of memory.

It's easy to make TeX loop forever without wasting resources with

\def\fool{\fool}\fool

On the other hand, complicated graphics with LuaTeX, that has dynamic allocation of memory, can lead to consuming vast amount of resources.

### Category 4

Redefine a primitive in a wrong way and you'll be welcomed by puzzling error messages such as

! A <box> was supposed to be here.
-
l.9 \end{document}

which is produced by

\documentclass{article}

\renewcommand{\box}[1]{-#1-}

\begin{document}

\box{abc}

\end{document}

The most dangerous command is \renewcommand (or \def). Another instance is the following

\documentclass{article}
\renewcommand{\fi}{whatever}
\begin{document}
\end{document}

that makes LaTeX stop with the very puzzling error message

! LaTeX Error: Missing \begin{document}.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
...

l.3 \begin{document}

• In cat.4, what happens if I compile the code, then delete the line \renewcommand and recompile the code without that line? – manooooh Feb 14 at 1:58
• @manooooh Other errors – egreg Feb 14 at 8:48
• @Quuxplusone I added “scripting program” – egreg Feb 14 at 17:37
• @Quuxplusone Sorry, that's outside my expertise. On non-LuaTeX engines you can't change the memory allocation during the run: it is fixed at launch time. On the other hand, with -shell-escape you can fork as many processes as you want. – egreg Feb 14 at 17:55
• About category 4, redefining a primitive: One of my favorites is: \documentclass{article}\begin{document}\def\else{Ha! Ha!}\end{document} You get: ! LaTeX Error: \begin{document} ended by \end{document}. – Ulrich Diez Feb 14 at 18:56

If you have a LaTeX-allergy, then all LaTeX is dangerous. ;-)

To be more serious:

You might be interested in these documents:

A platform independent computer virus
by Keith Allen McMillan
URL: http://vxheaven.0l.wtf/lib/vkm00.html
URL: ftp://ftp.cerias.purdue.edu/pub/doc/viruses/KeithMcMillan-PlatformIndependantVirus.ps

Summary:

In this master-thesis the author elaborates on his research regarding the question whether a platform-independent virus can be implemented in LaTeX.
The author defines viruses as follows: A computer virus is a fragment of a computer program whereof the user of the program is not aware. When the program is executed, the viral fragment takes control and performs the actions it is programmed to perform. Purposes of viruses:

1. Propagation on the infected computer-system by copying themselves or causing other programs to copy them.
2. Protecting themselves from detection.

The author distinguishes viruses from rabbits—the latter are not fragments but they are entire programs—and worms. Worms are self-contained self-replicating computer programs whose spreading is not focused on the files of the infected computer-system but whose spreading is focused on a network-environment, i.e., making copies of themselves on as many machines as possible.

Towards the end of the master-thesis, the author outlines how to program a virus in LaTeX which can without modification run and spread on a variety of computer platforms as long as these platforms support LaTeX and GNU emacs. In order to prevent unleashing such a virus, the actual code used in his research is not presented. The LaTeX/GNU emacs virus outlined does not have a payload and does only infect LaTeX-files with filename-extension .tex in the current directory where the \documentstyle-directive contains a comment %DoNotInfectMe by placing itself behind that directive.

The author explains that

• GNU emacs is used for providing the virus with a file containing a list of target-files/.tex-files in the current directory that can be infected.
• LaTeX is used for processing those macros that do the work of infection.
• The author outlines how to structure the LaTeX code executed by the virus and the LaTeX macros that constitute the routines of the virus.

Don’t take LaTeX files from strangers
by Steven Checkoway, Hovav Shacham, and Eric Rescorla

Quote from the abstract:

TeX, LaTeX, and BibTeX files are a common method of collaboration for computer science professionals. It is widely assumed by users that LaTeX files are safe; that is, that no significant harm can come of running LaTeX on an arbitrary computer. Unfortunately, this is not the case: In this article we describe how to exploit LaTeX to build a virus that spreads between documents on the MiKTeX distribution on Windows XP as well as how to use malicious documents to steal data from web-based LaTeX previewer services.

Are Text-Only Data Formats Safe? Or, Use This LaTeX Class File to Pwn Your Computer
by Steven Checkoway, Hovav Shacham, and Eric Rescorla
URL: https://hovav.net/ucsd/dist/texhack.pdf

Quote from the abstract:

We show that malicious TeX, BIBTeX, and METAPOST files can lead to arbitrary code execution, viral infection, denial of service, and data exfiltration, through thefile I/O capabilities exposed by TeX’s Turing-complete macro language. This calls into doubt the conventional wisdom view that text-only data formats that do not access the network are likely safe. We build a TeX virus that spreads between documents on the MiKTeX distribution on Windows XP; we demonstrate data exfiltration attacks on web-based LaTeX previewer services.

Hacking with LaTeX
by Sebastian Neef
URL: https://0day.work/hacking-with-latex/

The author elaborates on the capability of (La)TeX to create/write and read text files via \input/\read/\write and on \write18. \input/\read/\write could be used for reading/copying/overwriting sensitive files. \write18 can be used for doing harm by exexuting programs.

LaTeX Malicious PDF Generation
URL: http://blog.9bplus.com/latex-malicious-pdf-generation/

The author outlines that LaTeX is used for creating pdf-files and that there are LaTeX packages like the package movie15 which can be used for integrating .swf-files (Shockwave Flash Files) into .pdf-files. .swf-files in turn can be malicious.
(Remark by me, Ulrich Diez: The "danger" lies in the fact that like many other pdf-creation-tools LaTeX can be used for integrating objects/items into .pdf-files. Integrated objects/items in turn do not necessarily need to be created by means of LaTeX, and can be malicious.)

Summa summarum:

With many of the things explained in these elaborates, the danger does neither come per se from using LaTeX, nor has something to do with producing LaTeX-code with inherent malignancy, but the danger comes from applying LaTeX/LaTeX-code to resources (.swf-files, scripts that get executed by the pdf-viewer, etc) that are not trustworthy/that are compromised.

Someone might (ab)use (La)TeX's capability of writing external files for producing huge garbage-files again and again until the SD-card/the SSD is damaged due to the amount of writing-cycles.

Besides this you can use (La)TeX for writing things in cryptic ways.
When writing my answer to the question "Macro for mass hyper-reference?" I took the opportunity to show some of them.
I leave judging whether this is dangerous to others.
In any case you can write in cryptic ways (where one doesn't immediately see what the code does) directives for creating garbage-files or carrying out \write18-calls.

\outer is a good candidate for generating unintended results.

Making a command \outer after having added it to a hook can cause trouble when it comes to appending to that hook:

The minimal example

\documentclass{article}

\AtEndDocument{\one}
\outer
\def\one{one}
\AtEndDocument{\two}
\def\two{two}

\begin{document}
\end{document}

delivers

Runaway text?
! Forbidden control sequence found while scanning text of \toks@.
<inserted text>
}
l.6 \AtEndDocument{\two}

\outer-tokens inside \if..\else..\fi also trigger error-messages:

The minimal example

\documentclass{article}

\outer\def\macro{}

\begin{document}
\iffalse\macro\fi
\end{document}

delivers

! Incomplete \iffalse; all text was ignored after line 6.
<inserted text>
\fi
l.6 \iffalse\macro
\fi
?

By the way:

All this does not answer the question whether there is a (single) command in LaTeX that is dangerous.

So I don't understand why this answer of mine gets so many "likes" although it does not really answer the question. ;->

(But using LaTeX's macro-capabilities, you can create macros that can be used as single commands which trigger the execution of a lot of code/of a lot of directives/instructions which have an effect which can probably be considered dangerous.)

• Question: Are those links infected? – manooooh Feb 14 at 2:00
• Can you make a small summary of each of the internet links? – AndréC Feb 14 at 7:00
• For those who are allergic to LaTeX, I recomment NiTriLe – Tristan Feb 14 at 14:33
• @Tristan All the readers of this forum already are in contact with LaTeX. Can NiTriLe alleviate unwanted sequelae/aftereffects? ;-) – Ulrich Diez Feb 14 at 18:32
• @manooooh I can't tell you. ;-) (By the way I nowadays tend to use virtual machines that are disposed of in "data-nirvana" after the session.) – Ulrich Diez Feb 14 at 18:40

In my opinion the most dangerous command is \globaldefs1. Don't try! ;-)

\documentclass{article}
%\globaldefs1
\begin{document}
\section{Test}
\end{document}

It changes the way definitions work in LaTeX fundamentally. Definitions become global by default. That is, it undermines the way LaTeX works: groups are no longer efficient. As a consequence, everything goes berserk. Of course, if you really know what you are doing, you can use it. Ironically, you may only use it locally.

Of course, as every cat knows, \catcode can also be used to do maximal damage. It can be used in sneaky ways, which must be the reason why it is called cat code 😸.

• Since we should not try it, could you elaborate? – Sigur Feb 13 at 23:48
• \globaldefs-1 doesn't break as bad, but it's equally fun when it does :-) – Phelype Oleinik Feb 13 at 23:51
• @sigur I changes the way definitions work in LaTeX fundamentally. Definitions become global by default. That is, it undermines the way LaTeX works: groups are no longer efficient. As a consequence, everything goes berserk. – user194703 Feb 13 at 23:54
• @PhelypeOleinik Yes, that’s another good example, both of which I consider more dangerous than \renewcommand. The latter does generically only isolated damage. – user194703 Feb 13 at 23:58
• Of course, as every cat knows, \catcode can also be used to do maximal damage. It is also sneaky, which is why it is called 🐱 code😸 – user194703 Feb 14 at 0:01

In old times on DOS file systems, a really bad command to use in LaTeX was

\include{chapter.tex}

That looks innocuous, but the spurious extension causes it to open chapter.tex.aux for writing and truncate it, and the old DOS file systems abbreviated the name to be replaced with an empty file as chapter.tex.

At some point of time emTeX, then the prevalent TeX system, refused to open a file named in that manner, but not before many documents had been lost by the unwary.

• Though not a comprehensive answer, but a very interesting one. – Masroor Feb 15 at 1:55