text:
# What is the volume of the solid produced by revolving f(x)=sqrt(1+x^2) around the x-axis?

Nov 17, 2016

$V = \pi {\left[x + \frac{1}{3} {x}^{3}\right]}_{a}^{b}$

#### Explanation:

The volume of revolution about $O x$ is given by;

$V = {\int}_{a}^{b} \pi {y}^{2} \mathrm{dx}$

so with $f \left(x\right) = \sqrt{1 + {x}^{2}}$, then:

$V = {\int}_{a}^{b} \pi {\left(\sqrt{1 + {x}^{2}}\right)}^{2} \mathrm{dx}$
$\therefore V = \pi {\int}_{a}^{b} 1 + {x}^{2} \mathrm{dx}$
$\therefore V = \pi {\left[x + \frac{1}{3} {x}^{3}\right]}_{a}^{b}$

As you have not specified the $x$-coordinate bounds this is as far as we can proceed

format: plain


text:
# How do you find other zeros given one zero f(x)=x^3-8x^2+4x+48; 4?

Jan 27, 2017

$- 2 \mathmr{and} 6$.

#### Explanation:

$\left(x - 4\right)$ is a factor of f.

Upon division by (x - 4), the other factor of f is obtained as

${x}^{2} - 4 x - 12 = \left(x - 6\right) \left(x + 2\right)$.

So, the other roots are given by

$x - 6 = x + 2 = 0$


text:
Overfitting detecting in Binary Classification Problem

I have a binary classificaiton problem with positive label proportion is very small, 1% at most of the cases. So, recall, precision and f score for the positive label is used to evauluate my classifiers.
However, the commom overfitting detecting method uses the metric MSE, accuracy to detect the situtaion. MSE, accuracy is useless in my situation.
So, is there any methods to detecting overfitting in the small proporation positive label binary classification problem?

• Do you have enough positive label results to do cross validation? – russellpierce Dec 25 '15 at 16:03
• I have 10,000,000+ samples, so 1% mean 100,000+ positive labels, I think it enougth – bourneli Dec 26 '15 at 8:46


text:
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# Is f < g?

tagged by: Vincen

#### Is f < g?

Is f < g?

(1) f < g + 1
(2) |f|/|g| < 1

The OA is the option E.

The statement (2) is not sufficient? Why not? Experts, may you help me here?

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Vincen wrote:
Is f < g?

(1) f < g + 1
(2) |f|/|g| < 1

The OA is the option E.

The statement (2) is not sufficient? Why not? Experts, may you help me here?
Test easy cases.

Case 1: g = 1 and f = .5; (This satisfies both statements together) .5 is less than 1, so this yields a YES.
Case 2: g = -2 and f = -1.5. (Again, this satisfies both statements together) -1.5 is not less than -2, so this yields a NO.
Together the statements are not sufficient to answer the question. The answer is E

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Vincen wrote:
Is f < g?

(1) f < g + 1
(2) |f|/|g| < 1
We need to determine whether f < g.

Statement One Alone:

f < g + 1

Statement one alone is not sufficient to answer the question. For example, if f = 2 and g = 2, then f IS NOT less than g. However, if f = 1 and g = 2, then f IS less than g.

Statement Two Alone:

|f|/|g| < 1

We can multiply both sides of the inequality by |g| to get:

|f| < |g|

We can also determine that statement two alone is not sufficient to answer the question.

For example, if f = -1 and g = -2, then f IS NOT less than g. However, if f = 1 and g = 2, then f IS less than g.

Statements One and Two Together:

Using the information from statements one and two, we know that f < g + 1 and that |f| < |g|. This is still not enough information to answer the question.

For example, if f = -1 and g = -1.5, then f IS NOT less than g. However, if f = 1 and g = 2, then f IS less than g.

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text:
random-fu-0.2.7.7: Random number generation

Data.Random.List

Synopsis

# Documentation

randomElement :: [a] -> RVar a Source #

A random variable returning an arbitrary element of the given list. Every element has equal probability of being chosen. Because it is a pure RVar it has no memory - that is, it "draws with replacement."

shuffle :: [a] -> RVar [a] Source #

A random variable that returns the given list in an arbitrary shuffled order. Every ordering of the list has equal probability.

shuffleT :: [a] -> RVarT m [a] Source #

shuffleN :: Int -> [a] -> RVar [a] Source #

A random variable that shuffles a list of a known length (or a list prefix of the specified length). Useful for shuffling large lists when the length is known in advance. Avoids needing to traverse the list to discover its length. Each ordering has equal probability.

shuffleNT :: Int -> [a] -> RVarT m [a] Source #

shuffleNofM :: Int -> Int -> [a] -> RVar [a] Source #

A random variable that selects N arbitrary elements of a list of known length M.

shuffleNofMT :: Int -> Int -> [a] -> RVarT m [a] Source #


text:
# Suggested problems

## Microsoft setup bootstrapper has stopped working

Jan. 29, 2020, 6:20 p.m. by x3daudio1_7.dll is missing

## Biological Motivation

Troubleshoot [Microsoft Setup Bootstrapper has stopped Working][1] While Installing MS Office” or “Microsoft setup bootstrapper has stopped working“. So, the users just need to follow the procedure.

## Problem

A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains.

An example of an DNA string (whose alphabet contains the symbols A, C, G, and T) is ATGCTTCAGAAAGGTCTTACG.

Given: A DNA string $s$ of length at most 1000 nucleotides.

Return: Four integers corresponding to the number of times that the symbols A, C, G, and T occur in $s$.

## Sample Dataset

AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC

## Sample Output

20 12 17 21


text:
# Decision problems whose verifier is NP

We define $\mathbf P$ as the set of problems solvable in polynomial time. We define $\mathbf{NP}$ as the set of problems with a verifier $\in \mathbf P$.

Is there a name for problems whose verifiers are $\in \mathbf {NP}$ (e.g., $\mathbf{N(NP)}$)? I can't see this being a very useful complexity class, but, for example we have that $\mathbf{NP} \neq \mathbf{N(NP)} \implies \mathbf{P} \neq \mathbf{NP}$, so it might be an area of research for that reason alone.

• Why do you think that NP≠N(NP) implies P≠NP? – David Richerby May 2 '17 at 19:17
• P = NP -> NP = N(NP), so by the contrapositive. – k_g May 2 '17 at 21:33

Suppose you had a problem such that for any $x \in L$, there was a verifier $v$ such that $v$ could be checked against $x$ by a nondeterministic polynomial time algorithm. For a valid $(x, v)$ pair, there is some verifier $v'$ such that it takes polynomial time to check $((x, v), v')$ is a correct verification.
But then, you could simply combine $(v, v')$ into a single witness, and run the check in deterministic polynomial time.
Thus, the class ${\bf N(NP)}$ you describe is really just equal to ${\bf NP}$.


text:
# Grobner basis of the toric ideal $I_{A_P}$ with respect to $<_{rev}$ consists of those binomials $t_αt_β − t_{α\cap β} t_{α\cup β}$

I try to understand the proof of the Theorem. 10.1.3.(page 185) from ''Monomial Ideals'' by Herzog & Hibi.

The reduced Grobner basis of the toric ideal $I_{A_P}$ with respect to $<_{rev}$ consists of those binomials $$t_αt_β − t_{α\cap β} t_{α\cup β}$$ with neither α ⊂ β nor β ⊂ α.

Proof.

If α and β are poset ideals of $P$, then each of α∩β and α∪β is a poset ideal of P.

It is clear that the binomial $t_αt_β − t_{α∩β}t_{α∪β}$ belongs to $I_{A_P}$ and, in the case of neither α ⊂ β nor β ⊂ α, its initial monomial is $t_αt_β$.

Once we show that the set of those binomials $t_αt_β −t_{α∩β}t_{α∪β}$ with neither α ⊂ β nor β ⊂ α is a Grobner basis of $I_{A_P}$ with respect to $<_{rev}$, it follows immediately that such a set of binomials is the reduced Grobner basis of $I_{A_P}$ with respect to $<_{rev}$.

Let $G$ denote the reduced Grobner basis of $I_{A_P}$ with respect to $<_{rev}$. Let $$\begin{equation*} \prod_{i=1}^{q} t_{\alpha_i} - \prod_{i=1}^{q} t_{\beta_{i}} \end{equation*}$$ be a binomial belonging to $G$ with $\begin{equation*} \prod_{i=1}^{q} t_{\alpha_i} \end{equation*}$ its initial monomial.

What we must prove is that there are $k$ and $l$ such that one has neither $\alpha_k ⊂ α_l$ nor $α_l ⊂ α_k$.

Why is necessary and sufficiently to prove the above sentence?

I posted this question on MSE also.


text:
# Markov chain chance one state is reached before another

Hey could someone explain why this is true? I am trying to understand how to solve such a problem but I don't understand the solution.

Problem:
Given a Markov chain $\left\{X_{n}: n\in\ \mathbb{N}\right\}$ with four states 1,2,3,4 and transition matrix
$P = \begin{pmatrix} 0 & \frac{1}{2}& \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 \end{pmatrix}$
We leave from state 3. What is the chance that state 1 will be reached before state 4?

Solution:
Lets call this chance x. Lets call y the same chance but instead leaving from 2. Lets call:
$f = \begin{pmatrix} 1 \\ x \\ y \\ 0 \end{pmatrix}$
Then f satisfies P'f = f with P' the transition matrix derived from P by making state 1 and 4 absorbing states i.e.,

$\begin{pmatrix} 1 & 0& 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 1\end{pmatrix}*\begin{pmatrix} 1 \\ x \\ y \\ 0 \end{pmatrix}= \begin{pmatrix} 1 \\ x \\ y \\ 0 \end{pmatrix}$
Solving gives y = 1/2, x = 1/4.

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
Ray Vickson
Homework Helper
Dearly Missed
Hey could someone explain why this is true? I am trying to understand how to solve such a problem but I don't understand the solution.

Problem:
Given a Markov chain $\left\{X_{n}: n\in\ \mathbb{N}\right\}$ with four states 1,2,3,4 and transition matrix
$P = \begin{pmatrix} 0 & \frac{1}{2}& \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 \end{pmatrix}$
We leave from state 3. What is the chance that state 1 will be reached before state 4?

Solution:
Lets call this chance x. Lets call y the same chance but instead leaving from 2. Lets call:
$f = \begin{pmatrix} 1 \\ x \\ y \\ 0 \end{pmatrix}$
Then f satisfies P'f = f with P' the transition matrix derived from P by making state 1 and 4 absorbing states i.e.,

$\begin{pmatrix} 1 & 0& 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 1\end{pmatrix}*\begin{pmatrix} 1 \\ x \\ y \\ 0 \end{pmatrix}= \begin{pmatrix} 1 \\ x \\ y \\ 0 \end{pmatrix}$
Solving gives y = 1/2, x = 1/4.
You need to be more specific: what part do you not understand?

What you presented above is pretty standard, and is almost always part of the course material regarding Markov chains.

Also: for context: are you using a textbook and/or course notes? Do these sources not have such material in them?

If you could direct me to a place where this is explained I would be grateful. The material my teacher provided doesn't address this problem or atleast not directly, so I find it hard to understand this solution from the theory provided. I dont understand why P'f=f must be true. And I dont understand why those states are made absorbing.

Edit: I dont understand why its necessary to multiply with such a vector either.

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
If you could direct me to a place where this is explained I would be grateful. The material my teacher provided doesn't address this problem or atleast not directly, so I find it hard to understand this solution from the theory provided. I dont understand why P'f=f must be true. And I dont understand why those states are made absorbing.

Edit: I dont understand why its necessary to multiply with such a vector either.
When you don't understand something, it is a good idea to work it out from first principles. Let's do that.

Define ##f_i(n)## to be the probability that, starting from state i we reach state 1 for the first time at time n and do not reach state 4 before time n; let ##f_i = \sum_{n=1}^{\infty} f_i(n)##; this is the probability we reach state 1 before reaching state 4, starting from state i.

Note that
$$f_i(n) = P\{X_1 \neq 1,4, \: X_2 \neq 1,4,\: \ldots, X_{n-1} \neq 1,4, \: X_n = 4 |X_0=i\}$$
Conditioning on ##X_1## we have
$$f_i(n) = \sum_{j \neq 1,4} p_{ij} P\{X_2 \neq 1,4, \: \ldots, X_{n-1} \neq 1,4,\: X_n = 1 |X_1 = j, X_0 = i \} \\ = \sum_{j \neq 1,4} p_{ij} f_j(n-1).$$
Also: ##f_i(1) = p_{i1}##. Summing over n we get
$$f_i = f_i(1) + \sum_{n=2}^{\infty} f_i(n)\\ = p_{ij} + \sum_{j \neq 1,4} p_{ij} \sum_{n=2}^{\infty} f_j(n-1) = p_{i1} + \sum_{j \neq 1,4} p_{ij} f_j .$$

In other words:
$$f_2 = p_{21} + p_{22} f_2 + p_{23} f_3 \\ f_3 = p_{31} + p_{32} f_2 + p_{33} f_3.$$
Note that this is a simple 2x2 linear system that we can solve easily.

Of course, since (in this example, but not all examples) we either reach state 1 before state 4 or reach state 4 before state 1, the probability that, starting from state i we reach 4 before 1 is just ## 1 - f_i##.

Note that if we set ##f_1 = 1## and ##f_4 = 0## we can write the equations above as
$$f_1 = 1 f_1 + 0 f_2 + 0 f_3 + 0 f_4 \\ f_2 = p_{21} f_1 + p_{22}f_2 + p_{23} f_3 + p_{24}f_4\\ f_3 = p_{31} f_1 + p_{32}f_2 + p_{33}f_3 + p_{34} f_4 \\ f_4 = 0 f_1 + 0 f_2 + 0 f_3 + 1 f_4$$
This is just the set of equations ##f = \tilde{P} f## for a modified Markov chain with transition matrix
$$\tilde{P} = \pmatrix{1 &0&0&0\\p_{21}&p_{22} & p_{23} & p_{24}\\ p_{31} & p_{32} & p_{33} & p_{34} \\ 0 & 0 & 0 & 1}$$
and with
$$f = \pmatrix{1 \\f_2 \\f_3 \\0}$$

As I said: work it out in detail from first principles.

1 person
Thank you very much for taking your time and explaining this neatly and thoroughly. I'm sure this will be helpful to many others too as I haven't found an explanation of this particular problem trough Google.

Ray Vickson
Homework Helper
Dearly Missed
Thank you very much for taking your time and explaining this neatly and thoroughly. I'm sure this will be helpful to many others too as I haven't found an explanation of this particular problem trough Google.
There was a typo in my post, which I corrected in an "edit" and which was seemingly accepted, but was later not implemented. I should have written ##X_n = 1## instead of ##X_n = 4## n the definition of ##f_i(n)##.

There was a typo in my post, which I corrected in an "edit" and which was seemingly accepted, but was later not implemented. I should have written ##X_n = 1## instead of ##X_n = 4## n the definition of ##f_i(n)##.
Yes I realized that. What about the solution? Doesn't it also contain a mistake? If x is the chance leaving from state 3 shouldnt it come in the third place of the vector instead of the second?

There was a typo in my post, which I corrected in an "edit" and which was seemingly accepted, but was later not implemented. I should have written ##X_n = 1## instead of ##X_n = 4## n the definition of ##f_i(n)##.
I'm sorry but as I know sometimes replies are overlooked I'm sending another reply with the hope you will respond.

Ray Vickson


text:
## Cryptology ePrint Archive: Report 2017/1055

Cellular Automata Based S-boxes

Luca Mariot and Stjepan Picek and Alberto Leporati and Domagoj Jakobovic

Abstract: Cellular Automata (CA) represent an interesting approach to design Substitution Boxes (S-boxes) having good cryptographic properties and low implementation costs. From the cryptographic perspective, up to now there have been only ad-hoc studies about specific kinds of CA, the best known example being the $\chi$ nonlinear transformation used in Keccak. In this paper, we undertake a systematic investigation of the cryptographic properties of S-boxes defined by CA, proving some upper bounds on their nonlinearity and differential uniformity. Next, we extend some previous published results about the construction of CA-based S-boxes by means of a heuristic technique, namely Genetic Programming (GP). In particular, we propose a "reverse engineering" method based on De Bruijn graphs to determine whether a specific S-box is expressible through a single CA rule. Then, we use GP to assess if some CA-based S-box with optimal cryptographic properties can be described by a smaller CA. The results show that GP is able to find much smaller CA rules defining the same reference S-boxes up to size $7\times 7$, suggesting that our method could be used to find more efficient representations of CA-based S-boxes for hardware implementations. Finally, we classify up to affine equivalence all $3\times 3$ and $4\times 4$ CA-based S-boxes.

Category / Keywords: secret-key cryptography / ellular Automata, S-box, Cryptographic properties, Heuristics

Date: received 28 Oct 2017, last revised 17 Feb 2018

Contact author: luca mariot at disco unimib it

Available format(s): PDF | BibTeX Citation

Note: General revision fixing several errors and typos. Structure changed to better reflect the original contributions of the paper and summarize the previous work.

Short URL: ia.cr/2017/1055

[ Cryptology ePrint archive ]


text:
In mathematics, the Hilbert cube, named after David Hilbert, is a topological space that provides an instructive example of some ideas in topology. Furthermore, many interesting topological spaces can be embedded in the Hilbert cube; that is, can be viewed as subspaces of the Hilbert cube (see below).

Not to be confused with Hilbert curve.

## . . . Hilbert cube . . .

The Hilbert cube is best defined as the topological product of the intervals [0, 1/n] for n = 1, 2, 3, 4, … That is, it is a cuboid of countably infinitedimension, where the lengths of the edges in each orthogonal direction form the sequence

${displaystyle lbrace 1/nrbrace _{nin mathbb {N} }}$

.

The Hilbert cube is homeomorphic to the product of countably infinitely many copies of the unit interval [0, 1]. In other words, it is topologically indistinguishable from the unit cube of countably infinite dimension.

If a point in the Hilbert cube is specified by a sequence

${displaystyle lbrace a_{n}rbrace }$

with

${displaystyle 0leq a_{n}leq 1/n}$

, then a homeomorphism to the infinite dimensional unit cube is given by

${displaystyle h(a)_{n}=ncdot a_{n}}$

.


text:
1
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s ?

A
3 m
B
9 m
C
10 m
D
6 m
2
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is -
A
1 : 16
B
1 : 8
C
1 : 2
D
1 : 4
3
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed '$$\upsilon$$' more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then '$$\upsilon$$' is equal to :
A
$${{2{a_1}{a_2}} \over {{a_1} + {a_2}}}t$$
B
$$\sqrt {2{a_1}{a_2}} t$$
C
$$\sqrt {{a_1}{a_2}} t$$
D
$${{{a_1} + {a_2}} \over 2}t$$
4
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
The position co-ordinates of a particle moving in a 3-D coordinate system is given by
x = a cos$$\omega$$t
y = a sin$$\omega$$t and
z = a$$\omega$$t

The speed of the particle is :
A
$$\sqrt 2 \,a\omega$$
B
$$a\omega$$
C
$$\sqrt 3 \,a\omega$$
D
2a$$\omega$$
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text:
# Is F decidable, Turing-recognizable, co-Turing-recognizable

Let $F = \{\langle M \rangle : L(M)\}$, where $L(M)$ is finite.

So far I am operating under the assumption that the complement of the language is $F' = \{\langle M \rangle : L(M)\}$, where $L(M)$ is infinite.

Based on my notes from class here is what I have concluded so far:

Either $F$ is Turing recognizable or $F$ is co-Turing recognizable.

I either have to prove $F$ is Turing recognizable or $F'$ is Turing recognizable, but I am not sure how to draw a Turing machine for this. Am I on the right track? Any help is really appreciated.

Let $HP = \{\langle A,x \rangle :$ A is a TM that halts on x $\}$
and $\overline{HP} = \{\langle A,x \rangle :$ A is a TM that does not halt on x $\}$

Suppose $R$ is a recognizer for $F$.
The following reduction shows that $F$ is not Turing-Recognizable and hence not Decidable.

Loop = "On input <A, x>:
1) M = "On input <w>
10) simulate A on x
20) accept
2) if R accepts <M, w>
3)     accept
4) else
5)     reject

A loops on x $\Leftarrow\Rightarrow$ M does not accept any strings $\Leftarrow\Rightarrow L(M)$ is empty and finite
If R recognizes F then Loop recognizes $\overline{HP}$
Since $\overline{HP}$ is not recognizable, F is not Turing-Recognizable.

Suppose $R$ is a recognizer for $F'$.
The following reduction shows that $F'$ is not Turing-Recognizable and hence that $F$ is not co-Turing recognizable.

Loop = "On input (A, x)
1) M = "On input w
10) simulate A on x for |w| steps
20) if A halts on x in |w| steps
30)    reject
40) else
50)    accept

2) if R(M, x) accepts
3)     accept
4) else
5)     reject

A loops on x $\Leftarrow\Rightarrow$ M accepts all strings $\Leftarrow\Rightarrow L(M)$ is infinite
If R recognizes $F'$ then Loop recognizes $\overline{HP}$
Since $\overline{HP}$ is not recognizable, $F'$ is not Turing-Recognizable.
Hence, $F$ is not co-Turing-Recognizable.


text:
QUESTION

# How many transverse common tangents can be drawn?A. 1B. 2C. 3D. 4

Hint: The tangents intersecting two circles are known as transverse common tangents. An example is given below:

Here, AB is a transverse tangent to the two circles. Then we will apply various conditions and then come to a final conclusion.
We are given in the question that there are two circles ${{C}_{1}}$ and ${{C}_{2}}$ and then we have to find the number of transverse common tangents to the circles ${{C}_{1}}$ and ${{C}_{2}}$.

Please note that the number of tangents will depend on the relative positions of the given circles. We will get a different number of transverse common tangents in different situations. We know that, transverse common tangent is the tangent intersecting two circles. Now, we will see different types of situations based on the relative positions of the circles.
Condition 1: When one circle is lying completely inside the other circle.

In this case, there is no common tangent possible because if any line touches the inner circle then it will cut the outer circle at two points but the tangent should have only one point in common. Also, if the tangent touches the outer circle, then it would not pass through the inner circle. So, we have no common transverse tangents in this case.
Condition 2: When the two circles touch each other internally.

In this case, we have only one common tangent possible for the two circles.
Condition 3: When the two circles are intersecting each other at two points.

In this case, there are two common transverse tangents possible for the two circles.
Condition 4: When the two circles touch each other externally.

In this case, there are three common transverse tangents possible for the two circles.
Condition 5: When the two circles are separated by a distance and are not overlapping.

In this case, there are four common transverse tangents possible for the two circles.
Now in the question, we are given that the two circles, ${{C}_{1}}$ and ${{C}_{2}}$ are separated by a distance. Hence, on going through all the above discussed conditions, we can see that condition 5 matches our case. Hence, the number of common transverse tangents possible for the two circles, ${{C}_{1}}$ and ${{C}_{2}}$ is four.
Therefore, option (D) is the correct answer.

Note: We have to keep in mind that there are only five conditions possible for the position of two circles and the students can check each condition, and check which condition matches the figure given in the question in order to find the common transverse tangents.


text:
# 18.4: Atomic Orbital Basis Sets

The basis orbitals commonly used in the LCAO-MO-SCF process fall into two classes Slater-type orbitals and Cartesian Gaussian-type orbitals.

Slater-type orbitals (STO) are characterized by quantum numbers $$n$$, $$l$$, and $$m$$ and exponents (which characterize the 'size' of the basis function) $$\xi$$:

$\chi_{n,l,m}(r,\theta ,\phi )= N_{n,l,m,\xi} Y_{l,m}(\theta ,\phi )r^{n-1}e^{-\xi r},$

and $$N_{n,l,m, \xi}$$ denotes the normalization constant.

Cartesian Gaussian-type orbitals (GTO) are characterized by quantum numbers $$a$$, $$b$$, and $$c$$ which detail the angular shape and direction of the orbital and exponents $$\alpha$$ which govern the radial 'size' of the basis function akin to $$\xi$$ in STOs.

$\chi_{a,b,c}(r, \theta ,\phi )- N'_{a,b,c,\alpha}\textbf{x}^a\textbf{y}^b\textbf{z}^ce^{-\alpha r^2},$

For example, orbitals with $$a$$, $$b$$, and $$c$$ values of 1,0,0 or 0,1,0 or 0,0,1 are $$p_x$$, $$p_y$$, and $$p_z$$ orbitals; those with $$a$$, $$b$$, and $$c$$ values of 2, 0, 0 or 0, 2, 0 or 0, 0, 2 and 1, 1, 0 or 0, 1, 1 or 1, 0, 1 span the space of five d orbitals and one s orbital (the sum of the 2, 0, 0 and 0, 2, 0 and 0, 0, 2 orbitals is an s orbital because $$x^2 + y^2 + z^2 = r^2$$ is independent of $$\theta \text{ and } \phi$$).

For both types of orbitals, the coordinates $$r\, \theta, \text{ and } \phi$$ refer to the position of the electron relative to a set of axes attached to the center on which the basis orbital is located.

The Utility of Gaussians

Although STOs are preferred on fundamental grounds (e.g., as demonstrated in Appendices A and B, the hydrogen atom orbitals are of this form and the exact solution of the many-electron Schrödinger equation can be shown to be of this form (in each of its coordinates) near the nuclear centers), STOs are used primarily for atomic and linear-molecule calculations because the multi-center integrals $$\langle \chi_a \chi_b \big| g \big| \chi_e \chi_d \rangle$$ (each basis orbital can be on a separate atomic center) which arise in polyatomic-molecule calculations can not efficiently be performed when STOs are employed. In contrast, such integrals can routinely be done when GTOs are used. This fundamental advantage of GTOs has lead to the dominance of these functions in molecular quantum chemistry.

To understand why integrals over GTOs can be carried out when analogous STO based integrals are much more difficult, one must only consider the orbital products ( $$\chi_a \chi_c (r_1) \text{ and } \chi_b \chi_d (r_2)$$ ) which arise in such integrals. For orbitals of the GTO form, such products involve $$e^{-\alpha_a (\textbf{r}-\textbf{R}_a)^2}e^{-\alpha_c( \textbf{r}-\textbf{R}_c )^2}$$. By completing the square in the exponent, this product can be rewritten as follows:

$e^{-\alpha_a (\textbf{r}-\textbf{R}_a)^2} e^{-\alpha_c (\textbf{r}-\textbf{R}_c)^2} = e^{-(\alpha_a + \alpha_c )(\textbf{r}-\textbf{R'})^2}e^{-\alpha ' (\textbf{R}_a-\textbf{R}_c)^2},$

where

$\textbf{R}' = \dfrac{\alpha_a\textbf{R}_a + \alpha_c\textbf{R}_c}{\alpha_a + \alpha_c}$

and

$\alpha ' = \dfrac{\alpha_a \alpha_c}{\alpha_a + \alpha_c}.$

Thus, the product of two GTOs on different centers is equal to a single other GTO at a center $$\textbf{R}'$$ between the two original centers. As a result, even a four-center two-electron integral over GTOs can be written as, at most, a two-center two-electron integral; it turns out that this reduction in centers is enough to allow all such integrals to be carried out. A similar reduction does not arise for STOs because the product of two STOs can not be rewritten as a new STO at a new center.

To overcome the primary weakness of GTO functions, that they have incorrect behavior near the nuclear centers (i.e., their radial derivatives vanish at the nucleus whereas the derivatives of STOs are non-zero), it is common to combine two, three, or more GTOs, with combination coefficients which are fixed and not treated as LCAO-MO parameters, into new functions called contracted GTOs or CGTOs. Typically, a series of tight, medium, and loose GTOs (i.e., GTOs with large, medium, and small $$\alpha$$ values, respectively) are multiplied by so-called contraction coefficients and summed to produce a CGTO which appears to possess the proper 'cusp' (i.e., non-zero slope) at the nuclear center (although even such a combination can not because each GTO has zero slope at the nucleus).

## Basis Set Libraries

Much effort has been devoted to developing sets of STO or GTO basis orbitals for main-group elements and the lighter transition metals. This ongoing effort is aimed at providing standard basis set libraries which:

1. Yield reasonable chemical accuracy in the resultant wavefunctions and energies.
2. Are cost effective in that their use in practical calculations is feasible.
3. Are relatively transferrable in the sense that the basis for a given atom is flexible enough to be used for that atom in a variety of bonding environments (where the atom's hybridization and local polarity may vary).

## The Fundamental Core and Valence Basis

In constructing an atomic orbital basis to use in a particular calculation, one must choose from among several classes of functions. First, the size and nature of the primary core and valence basis must be specified. Within this category, the following choices are common:

1. A minimal basis in which the number of STO or CGTO orbitals is equal to the number of core and valence atomic orbitals in the atom.
2. A double-zeta (DZ) basis in which twice as many STOs or CGTOs are used as there are core and valence atomic orbitals. The use of more basis functions is motivated by a desire to provide additional variational flexibility to the LCAO-MO process. This flexibility allows the LCAO-MO process to generate molecular orbitals of variable diffuseness as the local electronegativity of the atom varies. Typically, double-zeta bases include pairs of functions with one member of each pair having a smaller exponent ($$\zeta \text{ or } \alpha$$ value) than in the minimal basis and the other member having a larger exponent.
3. A triple-zeta (TZ) basis in which three times as many STOs or CGTOs are used as the number of core and valence atomic orbitals.
4. Dunning has developed CGTO bases which range from approximately DZ to substantially beyond TZ quality (T. H. Dunning, J. Chem. Phys. 53 , 2823 (1970); T. H. Dunning and P. J. Hay in Methods of Electronic Structure Theory, H. F. Schaefer, III Ed., Plenum Press, New York (1977))). These bases involve contractions of primitive GTO bases which Huzinaga had earlier optimized (S. Huzinaga, J. Chem. Phys. 42 , 1293 (1965)) for use as uncontracted functions (i.e., for which Huzinaga varied the $$\alpha$$ values to minimize the energies of several electronic states of the corresponding atom). These Dunning bases are commonly denoted, for example, as follows for first-row atoms: (10s,6p/5s,4p), which means that 10 s-type primitive GTOs have been contracted to produce 5 separate s-type CGTOs and that 6 primitive p-type GTOs were contracted to generate 4 separate p-type CGTOs. More recent basis sets from the Dunning group are given in T. Dunning, J. Chem. Phys. 90 , 1007 (1990).
5. Even-tempered basis sets (M. W. Schmidt and K. Ruedenberg, J. Chem. Phys. 71 , 3961 (1979)) consist of GTOs in which the orbital exponents ak belonging to series of orbitals consist of geometrical progressions: $$\alpha_k \text{ = a } \beta_k$$ , where $$\alpha$$ and $$\beta$$ characterize the particular set of GTOs.
6. STO-3G bases were employed some years ago (W. J. Hehre, R. F. Stewart, and J. A. Pople, J. Chem. Phys. 51 , 2657 (1969)) but are less popular recently. These bases are constructed by least squares fitting GTOs to STOs which have been optimized for various electronic states of the atom. When three GTOs are employed to fit each STO, a STO-3G basis is formed.
7. 4-31G, 5-31G, and 6-31G bases (R. Ditchfield, W. J. Hehre, and J. A. Pople, J. Chem. Phys. 54 , 724 (1971); W. J. Hehre, R. Ditchfield, and J. A. Pople, J. Chem. Phys. 56 , 2257 (1972); P. C. Hariharan and J. A. Pople, Theoret. Chim. Acta. (Berl.) 28 , 213 (1973); R. Krishnan, J. S. Binkley, R. Seeger, and J. A. Pople, J. Chem. Phys. 72 , 650 (1980)) employ a single CGTO of contraction length 4, 5, or 6 to describe the core orbital. The valence space is described at the DZ level with the first CGTO constructed from 3 primitive GTOs and the second CGTO built from a single primitive GTO.

The values of the orbital exponents ($$\zeta s \text{ or } \alpha s$$) and the GTO-to-CGTO contraction coefficients needed to implement a particular basis of the kind described above have been tabulated in several journal articles and in computer data bases (in particular, in the data base contained in the book Handbook of Gaussian Basis Sets: A. Compendium for Ab initio Molecular Orbital Calculations , R. Poirer, R. Kari, and I. G. Csizmadia, Elsevier Science Publishing Co., Inc., New York, New York (1985)).

Several other sources of basis sets for particular atoms are listed in the Table shown below (here JCP and JACS are abbreviations for the Journal of Chemical Physics and the Journal of The American Chemical Society, respectively).

Literature Reference Basis Type Atoms
Hehre, W.J.; Stewart, R.F.; Pople, J.A. JCP 51 , 2657 (1969).
Hehre, W.J.; Ditchfield, R.; Stewart, R.F.; Pople, J.A. JCP 52 , 2769 (1970).
STO - 3G H-Ar
Binkley, J.S.; Pople, J.A.; Hehre, W.J. JACS 102 , 939 (1980). 3-21G H-Ne
Gordon, M.S.; Binkley, J.S.; Pople, J.A.; Pietro, W.J.; Hehre, W.J. JACS 104 , 2797 (1982). 3-21G Na-Ar
Dobbs, K.D.; Hehre, W.J., J. Comput. Chem. 7 , 359 (1986). 3-21G K, Ca, Ga
Dobbs, K.D.; Hehre, W.J., J. Comput. Chem. 8, 880 (1987). 3-21G Sc-Zn
Ditchfield, R.; Hehre, W.J.; Pople, J.A., JCP 54 , 724 (1971). 6-31G H
Dill, J.D.; Pople, J.A., JCP 62 , 2921 (1975). 6-31G Li, B
Binkley, J.S.; Pople, J.A., JCP 66 , 879 (1977) 6-31G Be
Hehre, W.J.; Ditchfield, R.; Pople, J.A., JCP 56 , 2257 (1972). 6-31G C-F
Francl, M.M.; Pietro, W.J.; Hehre, W.J.; Binkley, J.S.; Gordon, M.S.; DeFrees, D.J.;
Pople, J.A. JCP 77 , 3654 (1982).
6-31G Na-Ar
Dunning, T. JCP 53 , 2823 (1970). (4s/2s)
(4s/3s)
(9s5p/3s2p)
(9s5p/4s2p)
(9s5p/5s3p)
H
H
B-F
B-F
B-F
Dunning, T. JCP 55 , 716 (1971). (5s/3s)
(10s/4s)
(10s/5s)
(10s6p/5s3p)
(10s6p/5s4p)
H
Li
Be
B-Ne
B-Ne
Krishnan, R.; Binkley, J.S.; Seeger, R.; Pople, J.A. JCP 72 , 650 (1980). 6-311G H-Ne
Dunning, unpublished VDZ. (4s/2s)
(9s5p/3s2)
(12s8p/4s3p)
H
Li, Be, C-Ne
Na-Ar
Dunning, unpublished VTZ.
(5s/3s)
(6s/3s)
(12s6p/4s3p)
(17s10p/5s4p)

H
H
Li, Be, C-Ne
Mg-Ar

Dunning, unpublished VQZ. (7s/4s)
(8s/4s)
(16s7p/5s4p)
H
H
B-Ne
Dunning, T. JCP 90 , 1007 (1989). (pVDZ,pVTZ,pVQZ correlation-consistent) (4s1p/2s1p)
(5s2p1d/3s2p1d)
(6s3p1d1f/4s3p2d1f)
(9s4p1d/3s2p1d)
(10s5p2d1f/4s3p2d1f)
(12s6p3d2f1g/5s4p3d2f1g)
H
H
H
B-Ne
B-Ne
B-Ne
Huzinaga, S.; Klobukowski, M.; Tatewaki, H. Can. J. Chem. 63 , 1812 (1985). (14s/2s)
(14s9p/2s1p)
(16s9p/3s1p)
(16s11p/3s2p)
Li, Be
B-Ne
Na-Mg
Al-Ar
Huzinaga, S.; Klobukowski, M. THEOCHEM. 44 , 1 (1988).

(14s10p/2s1p)
(17s10p/3s1p)
(17s13p/3s2p)
(20s13p/4s2p)
(20s13p10d/4s2p1d)
(20s14p9d/4s3d1d)

B-Ne
Na-Mg
Al-Ar
K-Ca
Sc-Zn
Ga
McLean, A.D.; Chandler, G.S. JCP 72 , 5639 (1980). (12s8p/4s2p)
(12s8p/5s2p)
(12s8p/6s4p)
(12s9p/6s4p)
(12s9p/6s5p)
Na-Ar, P$$^-$$, S$$^-$$, Cl$$^-$$
Na-Ar, P$$^-$$, S$$^-$$, Cl$$^-$$
Na-Ar, P$$^-$$, S$$^-$$, Cl$$^-$$
Na-Ar, P$$^-$$, S$$^-$$, Cl$$^-$$
Na-Ar, P$$^-$$, S$$^-$$, Cl$$^-$$
Dunning, T.H.Jr.; Hay, P.J. Chapter 1 in 'Methods of Electronic Structure Theory',
Schaefer, H.F.III, Ed., Plenum Press, N.Y., 1977.
(11s7p/6s4p) Al-Cl
Hood, D.M.; Pitzer, R.M.; Schaefer, H.F.III, JCP 71 , 705 (1979). (14s11p6d/10s8p3d) Sc-Zn
Schmidt, M.W.; Ruedenberg, K. JCP 71 , 3951 (1979).
(regular even-tempered)
([N]s), N=3-10
([2N]s), N=3-10
([2N]s), N=3-14
([2N]s[N]p), N=3-11
([2N]s[N]p), N=3-13
([2N]s[N]p), N=4-12
([2N-6]s[N]p), N=7-15
H
He
Li, Be
B, N-Ne
C
Na, Mg
Al-Ar

## Polarization Functions

In addition to the fundamental core and valence basis described above, one usually adds a set of so-called polarization functions to the basis. Polarization functions are functions of one higher angular momentum than appears in the atom's valence orbital space (e.g, d-functions for C, N , and O and p-functions for H). These polarization functions have exponents ($$\zeta \text{ or } \alpha$$) which cause their radial sizes to be similar to the sizes of the primary valence orbitals ( i.e., the polarization p orbitals of the H atom are similar in size to the 1s orbital). Thus, they are not orbitals which provide a description of the atom's valence orbital with one higher l-value; such higher-l valence orbitals would be radially more diffuse and would therefore require the use of STOs or GTOs with smaller exponents.

The primary purpose of polarization functions is to give additional angular flexibility to the LCAO-MO process in forming the valence molecular orbitals. This is illustrated below where polarization d$$_\pi$$ orbitals are seen to contribute to formation of the bonding $$\pi$$ orbital of a carbonyl group by allowing polarization of the Carbon atom's p$$_\pi$$ orbital toward the right and of the Oxygen atom's p$$_\pi$$ orbital toward the left.

Polarization functions are essential in strained ring compounds because they provide the angular flexibility needed to direct the electron density into regions between bonded atoms.

Functions with higher l-values and with 'sizes' more in line with those of the lower-l orbitals are also used to introduce additional angular correlation into the calculation by permitting polarized orbital pairs (see Chapter 10) involving higher angular correlations to be formed. Optimal polarization functions for first and second row atoms have been tabulated (B. Roos and P. Siegbahn, Theoret. Chim. Acta (Berl.) 17 , 199 (1970); M. J. Frisch, J. A. Pople, and J. S. Binkley, J. Chem. Phys. 80, 3265 (1984)).

## Diffuse Functions

When dealing with anions or Rydberg states, one must augment the above basis sets by adding so-called diffuse basis orbitals. The conventional valence and polarization functions described above do not provide enough radial flexibility to adequately describe either of these cases. Energy-optimized diffuse functions appropriate to anions of most lighter main group elements have been tabulated in the literature (an excellent source of Gaussian basis set information is provided in Handbook of Gaussian Basis Sets , R. Poirier, R. Kari, and I. G. Csizmadia, Elsevier, Amsterdam (1985)) and in data bases. Rydberg diffuse basis sets are usually created by adding to conventional valence-pluspolarization bases sequences of primitive GTOs whose exponents are smaller than that (call it $$\alpha_{\text{diff}}$$) of the most diffuse GTO which contributes strongly to the valence CGTOs. As a 'rule of thumb', one can generate a series of such diffuse orbitals which are liniarly independent yet span considerably different regions of radial space by introducing primitive GTOs whose exponents are $$\frac{1}{3}\alpha_{\text{diff}}, \frac{1}{9}\alpha_{\text{diff}}, \frac{1}{27}\alpha_{\text{diff}},$$ etc.

Once one has specified an atomic orbital basis for each atom in the molecule, the LCAO-MO procedure can be used to determine the $$C_{\nu ,i}$$ coefficients that describe the occupied and virtual orbitals in terms of the chosen basis set. It is important to keep in mind that the basis orbitals are not themselves the true orbitals of the isolated atoms; even the proper atomic orbitals are combinations (with atomic values for the $$C_{\nu ,i}$$ coefficients) of the basis functions. For example, in a minimal-basis-level treatment of the Carbon atom, the 2s atomic orbital is formed by combining, with opposite sign to achieve the radial node, the two CGTOs (or STOs); the more diffuse s-type basis function will have a larger $$C_{i,\nu}$$ coefficient in the 2s atomic orbital. The 1s atomic orbital is formed by combining the same two CGTOs but with the same sign and with the less diffuse basis function having a larger $$C_{\nu ,i}$$ coefficient. The LCAO-MO-SCF process itself determines the magnitudes and signs of the $$C_{\nu ,i}$$.


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# Solution for Using Pythagoras Theorem Determine the Length of Ad in Terms of B and C Shown in Figure. - CBSE Class 10 - Mathematics

ConceptApplication of Pythagoras Theorem in Acute Angle and Obtuse Angle

#### Question

Using Pythagoras theorem determine the length of AD in terms of b and c shown in Figure.

#### Solution

We have,

In ΔBAC, by Pythagoras theorem

BC2 = AB2 + AC2

⇒ BC2 = c2 + b2

⇒ BC = sqrt(c^2+b^2)             ........(i)

In ΔABD and ΔCBA

∠B = ∠B                             [Common]

Then, ΔABD ~ ΔCBA           [By AA similarity]

therefore"AB"/"CB"="AD"/"CA"           [Corresponding parts of similar Δ are proportional]

rArrc/sqrt(c^2+b^2)="AD"/b

rArr"AD"="bc"/sqrt(c^2+b^2)

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [2]

Solution Using Pythagoras Theorem Determine the Length of Ad in Terms of B and C Shown in Figure. Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle.
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# Understanding recurrent SVM in volatility estimation of GARCH model

I read Chen et al. "Forecasting volatility with support vector machine-based GARCH model" (2010) where they implented a recurrent SVM procedure to estimate volatility by a GARCH based model. The model is of the form

$y_t = f(y_{t-1}) + u_t \qquad \qquad \ \ \ (1)$

$u^2_t = g(u^2_{t-1}, w_{t-1}) + w_t \qquad (2)$

At first they got estimates for $u_t$ by estimating $(1)$ by a SVM. Then, the following recurrent SVM algorithm was proposed to estimate $(2)$.

Recurrent SVM Algorithm:

Step 1: Set $i = 1$ and start with all residuals at zero: $w_t^{(1)} = 0$.

Step 2: Run an SVM procedure to get the decision function $f^{(i)}$ to the points $\{x_t, y_t\} = \{u_{t - 1}^2, u_t^2 \}$ with all inputs $x_t = \{u_{t - 1}^2, w_{t-1} \}$

Step 3: Compute the new residuals $w_t^{i+1} = u_t^2 - f^{(i)}$.

Step 4: Terminate the computaion process if the stopping criterion is satisfied; otherwise, set $i = i + 1$ and go back to Step 2.

The proposed stopping critrerion is based a Ljung-Box-Test for the residuals $w_t$. Only if the $p$-values of the test in five consecutive periods are higher than 0.1 the process is stopped.

As real world example the log-returns of the New York Stock Exchange (NYSE) composite stock index for the period from January 8, 2004 to December 31, 2007 was used. The last 60 observations where used as test sample. Hence, the estimation was done with the first 940 observations. In their study, the process converged after 121 interations. (Question:) However, my implementation in R does not converge. I think I have a misunderstanding of the concept. Because I think I implemented it exactly as stated. My R code is the following

rm(list = ls())

library(quantmod)
library(e1071)

#Get NYSE data and convert to log returns
id     <- "^NYA"
data   <- getSymbols(id, source = "yahoo", auto.assign = FALSE,
from = "2004-01-08", to = "2007-12-31")
series <- data[,6]  #Get adjusted closing prices
series <- na.omit(diff(log(series)))*100  #Compute log returns

#Lagged data for analysis
x      <- na.omit(cbind(series, lag(series)))

#Set parameters as in paper
svm_eps   <- 0.05
svm_cost  <- 0.005
sigma     <- 0.02
svm_gamma <- 1/(2*sigma^2)

#SVM to get u_t
svm     <- svm(x = x[,-1], y = x[,1], scale = FALSE,
type = "eps-regression", kernel = "radial",
gamma = svm_gamma, cost = svm_cost, epsilon = svm_eps)

u    <- svm$residuals #Extract u_t n <- 60 #Size of test set u_tr <- u[1:(nrow(u) - n)] #Subset to training set u_tr <- na.omit(cbind(u_tr, lag(u_tr)))^2 #Final training set #Recurrent SVM for vola estimation i <- 1 p_count <- 0 while(p_count < 5){ print(i) #Print number of loops #Estimate SVM for u^2 svmr <- svm(x = u_tr[,-1], y = u_tr[,1], scale = FALSE, type = "eps-regression", kernel = "radial", gamma = svm_gamma, cost = svm_cost, epsilon = svm_eps) #Test autocorrelation of residuals to lag 1 test <- Box.test(svmr$residuals, lag = 1, type = "Ljung-Box")
p_val   <- test$p.value p_count <- ifelse(p_val > 0.1, p_count + 1, 0) #Extract residuals for next estimation step w <- svmr$residuals
w        <- c(0, w[-length(w)])  #lag 1

u_tr <- cbind(u_tr[,1:2], w)

i <- i + 1
}

• I tried to highlight your question as it was a bit lost in the long text. So essentially, would you like us to go through the code and look for a mistake? Or how could we help you? – Richard Hardy May 18 '16 at 18:22
• I want to reproduce the results in the paper. So, going through the code and checking for mistakes would be helpful. Moreover, maybe I didn't understand the idea of recurrent SVM at all. Hence, I am not sure whether my implementation makes any sense in the context of recurrent SVM. I am unsure about my understanding because I read it several times but I don't see where I am wrong with my implementation. – random_guy May 18 '16 at 18:54


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# Ellipse equation

1. Mar 29, 2014

### alpha25

Hi, does exist an easy way to change the center of circle or a ellipse in polar coordinates?

thanks!

2. Mar 29, 2014

### HallsofIvy

Staff Emeritus
Yes, and it was given in any text book dealing with the conic sections that I have ever seen:
If the equation of an ellipse centered at (0, 0) is
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$$
then the same ellipse, centered at (a, b) has equation
$$\frac{(x- a)^2}{a^2}+ \frac{(y-b)^2}{b^2}= 1$$

3. Mar 29, 2014

### dextercioby

Now you have to pass from the cartesian eqns that Halls wrote to polar coordinates and you're done.

4. Mar 29, 2014

### alpha25

Yes thanks...but I need it in polar coordinates

5. Mar 29, 2014

### arildno

Insert polar representations for "x" and "y", multiply out parentheses and simplify and redefine variables/constants.
In particular, remember simplifying trig identities, such as, for example:
$$2\sin^{2}\theta=1-\cos(2\theta)$$

6. Mar 29, 2014

### arildno

Now, HallsofIvy made out a special case, with the center with the same values as the lengths of the semi-axes.

You shouldn't make that restriction here (call one of the (a,b)-pairs (c,d)-for example).

To give you the first step on your way, multiplying up and out, we get (with (c,d) centre coordinates):
$$b^{2}r^{2} \cos^{2}\theta+a^{2}r^{2} \sin^{2}\theta-2b^{2}cr \cos\theta-2a^{2} dr\sin\theta=a^{2}b^{2}-c^{2}-d^{2}$$
There would be various ways to simplify this expression further, and redefing independent constants.

One very compact way of doing so would be to transform your equation into the following form:
$$Ar^{2}\cos\gamma+Br\sin\phi=C$$
where the angle "phi" is a phase-shifted version of "gamma"/2 with a fourth constant D to be determined along with A, B and C (gamma being twice the value of "theta")

Last edited: Mar 29, 2014


text:
# Find the co-ordinates of the orthocentre of triangle ABC whose vertices are A(1,2), B(2,3) and C(4,3)

Slope of AB=$\frac{3-2}{2-1}=1$

Slope of BC=$\frac{3-3}{4-2}=0$

Slope of CA=$\frac{3-2}{4-1}=\frac{1}{3}$

Let AD, BE and CF be the altitudes in $△ABC$

Then,

Slope of BE=

Slope of CF=

We now have the vertices and slopes of AD, BE and CF. Let (x, y) be the orthocentre. So, by the equation $y-{y}_{1}=m\left(x-{x}_{1}\right)$, we have:-

Solving these equations, we find x=1, y=6

Thus, the required orthocentre is (1, 6).

• -1
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## elementary reaction

$K = \frac{k_{forward}}{k_{reverse}}$

Tiffany Cao 1D
Posts: 94
Joined: Fri Sep 29, 2017 7:03 am
Been upvoted: 1 time

### elementary reaction

Can we assume the rate law for individual elementary reactions? For an example, if a slow elementary reaction is 2I+H2==>2HI, is the rate law=k[I]^2[H2}?

Janine Chan 2K
Posts: 71
Joined: Fri Sep 29, 2017 7:04 am

### Re: elementary reaction

We can assume the rate law using the coefficients in an elementary reaction only when proposing a reaction mechanism.


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# Search for the Production of Single Sleptons through R-parity Violation in $p \bar{p}$ Collisions at $\sqrt{s}$=1.8 TeV

Abstract : We report the first search for supersymmetric particles via s-channel production and decay of smuons or muon sneutrinos at hadronic colliders. The data for the two-muon and two-jets final states were collected by the DO experiment, and correspond to an integrated luminosity of 94 +/- 5 pb^{-1}. Assuming that R-parity is violated via the single coupling lambda^{prime}_{211}, the number of candidate events is in agreement with expectation from the standard model. Exclusion contours are given in the (m_0,m_{1/2}) and (m_{tilde{chi}},m_{tilde{nu}}) planes for lambda^{prime}_{211}=0.09, 0.08 and 0.07.
Document type :
Journal articles

http://hal.in2p3.fr/in2p3-00012325
Contributor : Magali Damoiseaux <>
Submitted on : Friday, January 10, 2003 - 11:11:14 AM
Last modification on : Monday, March 29, 2021 - 3:22:09 PM

### Citation

V.M. Abazov, B. Abbott, Abdelmalek Abdesselam, M. Abolins, V. Abramov, et al.. Search for the Production of Single Sleptons through R-parity Violation in $p \bar{p}$ Collisions at $\sqrt{s}$=1.8 TeV. Physical Review Letters, American Physical Society, 2002, 89, pp.261801. ⟨in2p3-00012325⟩

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# DE: Models of Motion

1. Sep 29, 2008

### rocomath

A rocket is fired vertically and ascends with constant acceleration a=100m/s^2 for 1min. At that point, the rocket motor shuts off and the rocket continues upward under the influence of gravity.

a) Find the maximum altitude acquired by the rocket

b) The total time elapsed from the take-off until the rocket returns to the earth

Ignore air resistance.

a) The maximum height can be found through setting velocity equal to zero, and the integral of acceleration is velocity.

$$\int_0^Ta(t)dt=\int_0^{60}a(t)dt+\int_{60}^Ta(t)dt=0$$

$$\int_0^Ta(t)dt=\int_0^{60}100dt-\int_{60}^T9.8dt=0$$

T=672.244898s

Is (a) good so far?

Last edited: Sep 29, 2008
2. Sep 29, 2008

### gabbagabbahey

(a) looks good so far.

3. Sep 29, 2008

### rocomath

yay! Thanks for the confirmation.

4. Oct 6, 2008

### rocomath

Ok, so I have the time when my rocket is at it's highest. I know the position function equation, but I can't use it since my acceleration isn't constant.

Can I modify it such that $$x_1(t)$$ has g=100, and $$x_2(t)$$ has g=9.8, and the maximum height is just $$x_1+x_2$$???

5. Oct 6, 2008

### gabbagabbahey

Well $a(t)=\dot{v}(t)$ so why not integrate a(t') from t'=0 to t'=t to find the speed of the particle at a time t the same way you did to find v(T). Treat two cases, t<60min and t>60. this should give you the piecewise function for v(t). Then, repeat the process to find x(t) and finally plug in to the equation to find x(T).

Last edited: Oct 6, 2008
6. Oct 6, 2008

### rocomath

Oh ok, so just keep on integrating! Let me try it.

Thanks :)


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# why silicon carbide sic has a higher melting in slovakia

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SILICON CARBIDE Silicon carbide (SiC) exists only as a solid and, as it has no liquid phase, the material is rigid at all practical operating temperatures. This means that silicon carbide elements can be installed horizontally or vertically, without any additional supports, which simplifies the design of the equipment in which they are fitted.

Silicon carbide (SiC) is an extremely hard substa…

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text:
## Lecture 3 (05.10.2015)

To visualize knots/links, we will use certain diagrams instead of three-dimensional pictures. To produce a (two-dimensional) diagram from a (three-dimensional) link, we choose a (two-dimensional) plane in space and project the link onto the plane such that

• over each point of the diagram, there lie at most two points of the geometrical link. In particular, at each point in the diagram, at most two strands cross — these points are called the crossings of the diagram,
• at each crossing, it is indicated which strand passes over/under the other.

Using the export function of knotplot, you can check how links (curves in three-dimensional space) are converted to two-dimensional diagrams.

Here are some examples of diagrams:

And here are some examples of “not diagrams” (not to be confused with knot diagrams 😉 ) because these violate the rules we imposed above:

Clearly each geometrical link $L$ has many different diagrams, depending onto which plane we project. The diagrams of a topological link $\cal L$ are defined to be all the diagrams of all its representatives $L\in\cal L$ — that is, we can take any representative geometrical link, deform it by ambient isotopy as we like, and then project onto any plane.

Given a diagram $D$, we can not reconstruct the geometrical link $L$ it was projected from, because the relative height information in the direction perpendicular to the plane of projection is lost. However, since over/under crossings are indicated in $D$, we can unambiguously recover the type of $L$ from the diagram, i.e. the equivalence class $[L]$. That is, we can reconstruct a unique topological link from a diagram. Consequently, it makes sense to talk about equivalence (ambient isotopy) of diagrams: Two diagrams are called equivalent  (ambient isotopic) if and only if they arise as projections from links that are equivalent (ambient isotopic). We will discuss this equivalence of diagrams further in the next lecture.

For now, let us introduce a basic quantity of a diagram: Its crossing number.

Definition 1.10:

• The crossing number $|D|$ of a diagram $D$ is the number of all its crossings.
• The crossing number $C(L)$ of a geometrical link $L$ is defined as $$C(L):=\min\{|D|\;:\;D \text{ is a diagram of some geometrical link } L’ \text{ with } L’\cong L\}.$$ Sometimes the crossing number is also called the order of $L$.

There exist many diagrams that are equivalent but have different crossing number. In fact, given a diagram $D$, one can always find equivalent diagrams $D’$ such that $|D’|$ is larger (as much larger as we want) than $|D|$. To understand that mechanism, consider the following picture:

Here the dashed circle indicates that not the whole diagram of the link is shown, just the small part that is contained in the circle. In the depicted move, the part outside the circle remains the same. But the transformation in the circle clearly has the properties that a) it increases the crossing number by 1, and b) it leads to a diagram that is equivalent to the previous one.

Hence links do not have a “maximal crossing number”. But in view of the minimum taken in Definition 1.10. ii), crossing number is a well-defined quantity. It should be clear that crossing number is a link invariant. But to get used to the concepts, we give a formal proof of this fact.

Lemma 1.11: Crossing number is a link invariant.

Proof:  Let $L_1,L_2$ be two geometrical links that are equivalent, $L_1\cong L_2$. We have to show that

$$c(L_1)=\{|D|\,:\,D \text{ is a diagram of some link } L’ \text{ with } L’\cong L_1\}$$

coincides with $$c(L_2)=\{|D|\,:\,D \text{ is a diagram of some link } L” \text{ with } L”\cong L_2\}$$

But a link $L’$ that is equivalent to $L_1$ is also equivalent to $L_2$, in fact $L’\cong L_1\cong L_2$ $\Rightarrow L’\cong L_2$. Similarly, a link $L”$ that is equivalent to $L_2$ is also equivalent to $L_1$. Hence the two sets of diagrams over which the minimum is taken to compute $c(L_1)$ and $c(L_2)$ are the same. In particular, $c(L_1)=c(L_2)$. This completes the proof.

Some remarks:

• The proof shows that crossing number is in fact a property of topological links. That is always the case with link invariants (or with general equivalence classes, for that matter). Suppose you have a function $F$ from the family of all geometrical links to some set $S$, and you want to define a “corresponding” function $\cal F$ from the set of all topological links to $S$. The obvious way to define $\cal F$ in terms of $F$ is as follows: Take a topological link, i.e. an equivalence class $[L]$, and define ${\cal F}([L]) := F(L)$. Does this make sense? Remember that on the left hand side, $[L]$ is a set of many (equivalent) geometrical links, whereas on the right hand side, only a single geometrical link, namely $L$, appears. So we have to show that if we pick some other geometrical link $L’\in [L]$, then our definition is consistent, i.e. it gives the same answer $F(L’)=F(L)$. Since $[L]$ consists precisely of all links that are equivalent to $L$, the definition of $\cal F$ will be consistent if and only if $F$ is a link invariant. That is, link invariants are nothing but well-defined functions from the family of all topological links to some set $S$.
• After the somewhat boring number of components, crossing number is another and much more interesting link invariant.

For example, the unknot has crossing number 0, because it has the diagram

which has zero crossings. If a knot has crossing number zero, it means that is is equivalent to the unknot.

Are there knots $K$ with crossing number $c(K)=1$? It was demonstrated in the lecture that this is not the case – any diagram with precisely one crossing is equivalent to a diagram with zero crossings, i.e. to the unknot. (One considers the picture of one crossing with four open ends and checks that all possibilities to connect the four ends result in either a diagram with exactly one crossing, which is however seen to be equivalent to a diagram with zero crossings, or a (lin) diagram with more than one crossing).

The same question for $c=2$ is left as an exercise:

Exercise 1.12: (Problem 1) Show that there exists no topological knot with crossing number two.

The trefoil has crossing number at most three: In its standard diagram

there are three crossings. Does there exist a diagram of the trefoil that has less than three crossings? According to what we saw above for crossing number 1 and 2, that would mean the crossing number of the trefoil would be zero, i.e. it would be equivalent to the unknot. Later we will find out that this is not the case, i.e. the crossing number of the trefoil is really three.

This discussion shows that although the crossing number is a link invariant, it is often very hard to determine for a a given link.

Sometimes it will be useful to give a diagram an orientation. In the case of a knot (= 1-link) this means that we give the curve a direction, indicated by an arrow on the diagram, which it keeps throughout the whole diagram. Hence there are two possible choices of orientation for each knot.

• In the case of an $n$-link, we give a direction to each of its $n$ components – in this case, there exist $2^n$ different orientations.
• For oriented links $L_1,L_2$, we have a natural notion of equivalence by requiring the existence of an ambient isotopy matching $L_1$ and $L_2$, including their orientations.
• In a oriented link diagram, we can distinguish between two different kinds of crossings, which gives rise to the definition of two numbers (writhe and linking number):

Definition 1.13:

• A crossing $c$ in an oriented link diagram is called positive (or: a crossing of positive type), if the (oriented) overstrand has to be rotated counter-clockwise to match the (oriented) understrand. Otherwise the crossing is called negative, or a crossing of negative type. The sign of a crossing is defined as $$\varepsilon(c):=\begin{cases} +1 &\text{if } c \text{ is positive} \\ -1& \text{if } c \text{ is negative}. \end{cases}$$
• The linking number of two components $K_1,K_2$ of an oriented link diagram is defined as $$\ell(K_1,K_2):=\frac{1}{2}\sum_c\varepsilon(c),$$ where the sum runs over all crossings $c$ in which $K_1$ passes over $K_2$, or vice versa. That is, crossings of $K_1$ with itself and crossings of $K_2$ with itself are not included in the sum.
• The writhe (or twist number) of an oriented link diagram $L$ is defined as $$w(L):=\sum_c\varepsilon(c),$$ where the sum runs over all crossings of $L$.

There exist different ways for remembering how to determine the sign of a crossing. The one with the rule how the overstrand has to be rotated to match the understrand was used as the definition of the sign here, but we give two more (equivalent) characterisations:

First, one can take the point of view that the crossing in question is looked at from a direction such that the overstrand passes from the left to right (i.e., rotate the whole diagram until this is the case). If in this configuration, the understrand passes from bottom to top, the sign is positive, and negative otherwise.

Second, one can approach the crossing along the overstrand (“on the bridge”), in the direction given by its orientation. If the understrand passes below from the right to left, the sign is positive, and otherwise it is negative.

Note that writhe is not a link (or knot-) invariant. For example, below you see two equivalent diagrams (both corresponding to the knot type “unknot”)

The left diagram has writhe zero (no crossings), but the right diagram has writhe $-1$. Since both are equivalent, writhe cannot be an invariant of oriented knots.

Linking number, on the other hand, is an invariant (of oriented 2-links), as we will show next time. Regarding the definition of linking number, here is another question/exercise: Show that linking number is always an integer. This requires a proof because of the factor $\frac{1}{2}$ in the definition. If you need hints on how to do this, you can consider the book by Livingston, Exercise 2.6, page 37.


text:
# Proving loop invariant for a possibly nonterminating while loop

I started studying through Aho, Ullman - Foundations of Computer Science as a free time exercise. In the second chapter about loop invariants and inductive proofs, there is a starred exercise.

int sum = 0;
scanf("%d", &x);
while(x >= 0) {
sum = sum + x;
scanf("%d", &x);
}
printf("%d", sum);

Read a number into x, accumulate it into sum variable if x is nonnegative, and move on with the loop until user enters a negative number. Pretty simple stuff. Intuitively, we can readily see the printf output will be a nonnegative number. This is the thing we want to prove.

My loop invariant choice is:

Before every check of loop condition, value of sum is nonnegative.

Of course, in line with the aim of the chapter, we also need to prove this invariant by induction. I can't come up with a rigorous formulation for an inductive proof myself.

Here is a bogus attempt:

$$S(x): If \hspace{0.4em} x\geq0, sum \geq 0$$

Proof is by induction on the value of variable x.

1. $$S(0)$$ base case holds because adding 0 can't change the value and sum already starts with 0.
2. Assume $$S(n)$$ is true. How to use the inductive hypothesis?

I can't formulate this. I am more used to utilizing the number of iteration of the loop (i) to base my inductive proof on and say "such and such is true before the $$i^{th}$$ iteration. But in this example, there is no easy way to talk about the number of iteration in terms of program variables sum and x.

There's no need to perform an explicit induction here. Notice how your proposed invariant doesn't refer to the number of loop iterations. You're on the right track. Let $$I \equiv \text{sum}\geq 0$$ be our candidate loop invariant.

To show that $$I$$ is indeed a loop invariant it suffices to show that

• $$I$$ holds when control reaches the loop for the first time and
• $$I$$ is re-established by the loop body, assuming $$I$$ held before and the loop test (here $$x\geq 0$$) was true as well.

This already guarantees that $$I$$ will also hold upon loop termination. (Actually, even $$I\land \neg(x\geq 0)$$ since the loop test cannot be true anymore, but you don't need that in this example.)

The assignment statement $$\text{sum} = 0$$ establishes $$I$$ because $$0\geq 0$$. To do that more formally we'd use Hoare logic.

Working backwards from $$I$$ through the loop body we notice that the reading into $$x$$ doesn't affect $$\text{sum}$$, so we still have $$I$$. Pushing further backwards through the assignment statement, we arrive at $$K\equiv \text{sum}+x\geq 0$$ (by substituting the right-hand-side of the assignment for the occurrences of the left-hand-side in $$I$$). We need to prove that $$I\land x\geq 0$$ implies $$K$$, but that's trivial.

• I understand this and much appreciated but, theme of the chapter being proving loop invariants by induction, I was expecting there would be an inductive proof too. In a sense, this is for the case of being exercise-complete, while reading the book. Would be nice if you can also share an inductive proof (or showing that an inductive proof is not possible). Feb 3 at 9:01
• The essence of the proof technique goes back to Floyd, who devised it for flow charts/transition diagrams. The soundness proof of his (any every later) assertional reasoning method involves one proof by induction, typically on the number of steps in the execution of a program. When applying such a method, there's no more need for an induction.
– Kai
Feb 3 at 22:53


text:
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### fundamental theorem of calculus properties


text:
## Afrika Statistika

### A simple note on some empirical stochastic process as a tool in uniform L-statistics weak laws

Gane Samb Lô

#### Abstract

In this paper, we are concerned with the stochastic process $$\beta_n (q_t, t) = \beta_n(t) = \frac{1}{\sqrt{n}}\sum_{j=1}^n \{G_{t,n}(Y(t)) - G_t(Y_j(t))\} q_t(Y_j(t)), \tag{A}$$ where for $n \geq 1$ and $T > 0$, the sequences $\{Y_1(t), Y_2(t), \cdots, Y_n(t), t \in [0, T]\}$ are independent observations of some real stochastic process $Y(t), t \in [0,T]$, for each $t \in [0,T], G_t$ is the distribution function of $Y(t)$ and $G_{t,n}$ is the empirical distribution function based on $Y_1(t), Y_2(t), \cdots, Y_n(t)$ and finally $q_t$ is a bounded real function defined on $\mathbb{R}$. This process appears when investigating some time-dependent L-Statistics which are expressed as a function of some functional empirical process and the process (A). Since the functional empirical process is widely investigated in the literature, the process reveals itself as an important key for L-Statistics laws. In this paper, we state an extended study of this process, give complete calculations of the first moments, the covariance function and find conditions for asymptotic tightness.

#### Article information

Source
Afr. Stat., Volume 5, Number 1 (2010), 245-251.

Dates
First available in Project Euclid: 1 January 2014


text:
My Math Forum (http://mymathforum.com/math-forums.php)
-   Number Theory (http://mymathforum.com/number-theory/)
-   -   Sum of number's number (http://mymathforum.com/number-theory/346671-sum-numbers-number.html)

 Vitty June 28th, 2019 12:53 PM

Sum of number's number

Hey, I have a number theory problem: determine the sum of the digits of a natural number. I have been thinking about it for a lot time during this days, but I can't still find a solution. Can someone give me an idea or advise me a book/internet site with something useful?

Thank a lot.

 mathman June 28th, 2019 01:48 PM

Add them up? What are you looking for?

 DarnItJimImAnEngineer June 28th, 2019 03:33 PM

I can find it modulo 9 no problem. :p

 Greens June 28th, 2019 05:52 PM

This seems pretty brute force, wouldn't surprise me at all if there was something more elegant.

Let each digit $d_i \in \{0,1,2 \dots , 9\}$ be such that

$\displaystyle d_{1}d_{2}d_{3} \dots d_{N} = a \in \mathbb{N}$

(This is concatenation, but also $\displaystyle a= \sum_{i=1}^{N} d_{i}10^{(N-i)}$ )

Then we know $d_1 = floor \left(10^{-(N-1)}a \right)$

Now, we know $a-10^{(N-1)}d_{1} = d_{2}d_{3} \dots d_{N}$

thus $d_{2} = floor \left(10^{-(N-2)} \left(a-10^{(N-1)}d_{1} \right) \right)$

and $d_{3} = floor \left(10^{-(N-3)} \left(a-10^{(N-1)}d_{1} - 10^{(N-2)}d_{2} \right) \right)$

Continuing this strategy leaves

$\displaystyle d_{n} = floor \left( 10^{-(N-n)} \left( a - \sum_{i=1}^{n-1} d_{i}10^{(N-i)} \right) \right)$

So, the sum of digits should be

$\displaystyle \sum_{i=1}^{N}d_{i} = floor \left(10^{-(N-1)}a \right) + \sum_{n=2}^{N} floor \left( 10^{-(N-n)} \left( a - \sum_{i=1}^{n-1} d_{i}10^{(N-i)} \right) \right)$

For $N \geq 2$

 Vitty June 28th, 2019 11:34 PM

-Let $\displaystyle d_i \in \{0,1,2 \dots , 9\}$ and $\displaystyle \displaystyle d_{1}d_{2}d_{3} \dots d_{N} = a \in \mathbb{N}$

-Now I know that each digit is generated by: $\displaystyle d_i = \sum_{i=1}^{\left \lfloor log_{10}(A)+1 \right \rfloor}\frac{A\,\, mod\,\, 10^{i}-A\,\, mod\,\, 10^{i-1}}{10^{i-1}}$

-I expand the sum and I obtain: $\displaystyle \frac{A\, mod\,\, 10-A\,\, mod\,\, 1}{1}+\frac{A\, mod\,\, 10^{2}-A\,\, mod\,\, 10}{10}+\cdots +\frac{A\, mod\,\, 10^{i}-A\,\, mod\,\, 10^{i-1}}{10^{i-1}}$

-I have to simplify the denominator so: $\displaystyle \frac{1\cdot (A\, mod\,\, 10-A\,\, mod\,\, 1)}{1}+\frac{10\cdot (\frac{A}{10}\, mod\,\, 10-\frac{A}{10}\,\, mod\,\, 1)}{10}+\cdots +\frac{10^{i-1}\cdot (\frac{A}{10^{i-1}}\, mod\,\, 10-\frac{A}{10^{i-1}}\,\, mod\,\, 1)}{10^{i-1}}$

-I split the term: $\displaystyle A\, \, mod\, \, 10+\frac{A}{10}\, \, mod\, \, 10+\cdots+\frac{A}{10^{i-1}}\, \, mod\, \, 10-(A\, \, mod\, \, 1+\frac{A}{10}\, \, mod\; 1+\cdots +\frac{A}{10^{i-1}}\, \, mod\, \, 1)$

-Now I would have picked up $\displaystyle mod\,\,10$ and $\displaystyle mod\,\,1$: $\displaystyle \left (\sum_{i=1}^{\left \lfloor log_{10}(A)+1 \right \rfloor}\frac{A}{10^{i-1}}\right)\, \, mod\, \, 10-\left (\sum_{i=1}^{\left \lfloor log_{10}(A)+1 \right \rfloor}\frac{A}{10^{i-1}}\right)\, \, mod\, \, 1$; but I can't; some ideas to move forward?

Quote:
 Originally Posted by Greens (Post 611193) $\displaystyle \sum_{i=1}^{N}d_{i} = floor \left(10^{-(N-1)}a \right) + \sum_{n=2}^{N} floor \left( 10^{-(N-n)} \left( a - \sum_{i=1}^{n-1} d_{i}10^{(N-i)} \right) \right)$
Can we move from this sum to a function without $\displaystyle \sum$?

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# Diagonalise matrix iff eigenvectors independent (True/False)

Is it true that we can diagonalise a matrix iff eigenvectors are independent? I think so, but I'm not sure.

Clarification: lets say I, for some matrix A, find eigenvalues a, b, and c. The eigenvectors associated with those eigenvalues I will call a, b, c. We know that a ≠ b ≠ c <=> a, b, c are independent.

Now, my question is if this statement is true

a, b, c are independent <=> we can diagonalise a matrix A.

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Could you clarify what you mean by "eigenvectors are independent"? There are often infinitely many eigenvectors! – Thomas Belulovich Feb 26 '14 at 17:02
I agree with Thomas: the question should be clarified. Having said that, keep in mind that if $u_1,\dots ,u_N$ are eigenvectors associated with distinct eigenvalues, then they are independent; but not all matrices can be diagonalized. – Etienne Feb 26 '14 at 17:08
@ThomasBelulovich: Why only "often"? Do you have finite fields in mind? – Roland Feb 26 '14 at 17:15
@Roland Yes :). – Thomas Belulovich Feb 26 '14 at 17:16

The answer of sami is notable. The main point is the number of such independent eigenvectors, so i think the best is to say

The matrix $A$ ($n×n$) is diagonalizable if and only if there exist $n$ linearly independent eigenvectors.

The result in this form is true

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No, this is not true: for example the matrix $$\left(\begin{matrix}1&1\\0&1\end{matrix}\right)$$ has only one eigenvalue $1$ and the eigenspace is $\operatorname{span}\left\{e_1=(1,0)^T\right\}$ hence $(e_1)$ is linearly independent but $A$ isn't diagonalizable.

However, if the eigenvectors form a basis of the linear space then the matrix is diagonalizable and the result is true.

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It's all about that basis. I should have highlighted that too! – imranfat Feb 26 '14 at 17:22

It is true we can diagonalize a matrix IFF we have an eigenvector-basis of this matrix. Using a basis consisting of eigenvectors implies that the eigenvectors are independent Sidenote: Realize that 2 independent eigenvectors may not always have to come from distinct eigenvalues. Meaning it is possible that one eigenvalue can produce two independent eigenvectors. But disctinct eigenvalues do produce independent eigenvectors.

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## Chemistry: A Molecular Approach (3rd Edition)

$$198 km = 1.98\times10^7 cm$$
To convert from km to cm, we can first convert kilometers to meters. 1 km = 1000m 198 km = 198000 m Now we can easily convert meters to centimeters. 1 m = 100 cm 198000 m = 19800000 cm = $1.98\times10^7 cm$