# Is this formula for $\frac{e^2-3}{e^2+1}$ known? How to prove it?

I found an interesting infinite sequence recently in the form of a 'two storey continued fraction' with natural number entries:

$$\frac{e^2-3}{e^2+1}=\cfrac{2-\cfrac{3-\cfrac{4-\cdots}{4+\cdots}}{3+\cfrac{4-\cdots}{4+\cdots}}}{2+\cfrac{3-\cfrac{4-\cdots}{4+\cdots}}{3+\cfrac{4-\cdots}{4+\cdots}}}$$

The numerical computation was done 'backwards', starting from some $x_n=1$ we compute:

$$x_{n-1}=\frac{a_n-x_n}{a_n+x_n}$$

And so on, until we get to $x_0$. The sequence converges for $n \to \infty$ if $a_n>1$ (or so it seems).

For constant $a_n$ we seem to have quadratic irrationals, for example:

$$\frac{\sqrt{17}-3}{2}=\cfrac{2-\cfrac{2-\cfrac{2-\cdots}{2+\cdots}}{2+\cfrac{2-\cdots}{2+\cdots}}}{2+\cfrac{2-\cfrac{2-\cdots}{2+\cdots}}{2+\cfrac{2-\cdots}{2+\cdots}}}$$

For $a_n=2^n$ we seems to have:

$$\frac{1}{2}=\cfrac{2-\cfrac{4-\cfrac{8-\cdots}{8+\cdots}}{4+\cfrac{8-\cdots}{8+\cdots}}}{2+\cfrac{4-\cfrac{8-\cdots}{8+\cdots}}{4+\cfrac{8-\cdots}{8+\cdots}}}$$

I found no other closed forms so far, and I don't know how to prove the formulas above. How can we prove them? What is known about such continued fractions?

There is another curious thing. If we try to expand some number in this kind of fraction, we can do it the following way:

$$x_0=x$$

$$a_0=\left[\frac{1}{x_0} \right]$$

$$x_1=\frac{1-a_0x_0}{1+a_0x_0}$$

$$a_1=\left[\frac{1}{x_1} \right]$$

However, this kind of expansion will not give us the above sequences. We will get faster growing entries. Moreover, the fraction will be finite for any rational number. For example, in the list notation:

$$\frac{3}{29}=[9,28]$$

You can easily check this expansion for any rational number.

As for the constant above we get:

$$\frac{e^2-3}{e^2+1}=[1,3,31,74,315,750,14286,\dots]$$

Not the same as $[1,2,3,4,5,6,7,\dots]$ above!

We have similar sequences growing exponentially for any irrational number I checked.

$$e-2=[1,6,121,284,1260,3404,25678,\dots]$$

$$\pi-3=[7,224,471,2195,10493,46032,119223,\dots]$$

By the way, if we try CF convergents, we get almost the same expansion, but finite:

$$\frac{355}{113}-3=[7,225]$$

$$\frac{4272943}{1360120}-3=[7,224,471,2195,18596,227459,\dots]$$

So, the convergents of this sequence are not the same as for the simple continued fraction, but similar.

Comparing the expansion by the method above and the closed forms at the top of the post, we can see that, unlike for simple continued fractions, this expansion is not unique. Can we explain why?

Here is the Mathematica code to compute the limit of the first fraction:

Nm = 50;
Cf = Table[j, {j, 1, Nm}];
b0 = (Cf[[Nm]] - 1)/(Cf[[Nm]] + 1);
Do[b1 = N[(Cf[[Nm - j]] - b0)/(Cf[[Nm - j]] + b0), 7500];
b0 = b1, {j, 1, Nm - 2}]
N[b0/Cf[[1]], 50]

And here is the code to obtain the expansion in the usual way:

x = (E^2 - 3)/(E^2 + 1);
x0 = x;
Nm = 27;
Cf = Table[1, {j, 1, Nm}];
Do[If[x0 != 0, a = Floor[1/x0];
x1 = N[(1 - x0 a)/(x0 a + 1), 19500];
Print[j, " ", a, " ", N[x1, 16]];
Cf[[j]] = a;
x0 = x1], {j, 1, Nm}]
b0 = (1 - 1/Cf[[Nm]])/(1 + 1/Cf[[Nm]]);
Do[b1 = N[(1 - b0/Cf[[Nm - j]])/(1 + b0/Cf[[Nm - j]]), 7500];
b0 = b1, {j, 1, Nm - 2}]
N[x - b0/Cf[[1]], 20]

Update

I have derived the forward recurrence relations for numerator and denominator:

$$p_{n+1}=(a_n-1)p_n+2a_{n-1}p_{n-1}$$ $$q_{n+1}=(a_n-1)q_n+2a_{n-1}q_{n-1}$$

They have the same form as for generalized continued fractions (a special case). Now I understand why the expansions are not unique.

• I am always thoroughly impressed with the length and detail in your posts. On the other hand, you might want to key in a little more on what your main question is here. – Brevan Ellefsen Feb 7 '17 at 22:25
• @BrevanEllefsen, if these are known results, I would like a reference. If not, I would like a hint about a proof. And the most important question - why is the expansion not unique, even though the properties seem to be similar to simple continued fractions. Thanks, by the way. I'm glad I'm not the only one who likes my posts :) – Yuriy S Feb 7 '17 at 22:26
• @Nemo, I'm not sure I understand what you did here, but I figured out the same thing after deriving the forward recurrence relations for numerator and denominator: $$p_{n+1}=(a_n-1)p_n+2a_{n-1}p_{n-1}$$ $$q_{n+1}=(a_n-1)q_n+2a_{n-1}q_{n-1}$$ Obviously, we have a special case of a generalized continued fraction. Which is why it is not unique – Yuriy S Feb 8 '17 at 9:18

For the first one, you could write $$f(n) = \frac{n-f(n+1)}{n+f(n+1)}$$ Then you suggest $$f(2) = \frac{e^2-3}{e^2+1}$$ But this then gives \begin{align} f(1) = \frac{2}{e^2-1}\\ f(3) = \frac{4}{e^2-1} \\ f(4) = \frac{3e^2-15}{e^2+3}\\ f(5) = \frac{-2e^2+18}{e^2-3} \end{align} I don't know if there is a recurrence relation that solves this, but you have a few more closed forms...
The second one we have $$g(2) = \frac{2 - g(2)}{2+g(2)}$$ so we can solve the quadratic $x^2+3x-2$ to get $(\sqrt{17}-3)/2$.
For the third one, we have $$h(n) = \frac{n-h(2n)}{n+h(2n)}$$ using the trial version of $h(2)=1/2$, we get \begin{align} h(4)=\frac{2}{3}\\ h(8)=\frac{4}{5}\\ h(16)=\frac{8}{9} \end{align} then it is likely that $$h(n)=\frac{n}{n+2}$$ as this satisfies the recurrence and that $h(2)=1/2$.
• Thank you, when I look back it's clear that quadratic radicals or rational numbers are easily proven (provided the limit exists), just as you show here. For irrational numbers see the comments and the update, linking this to generalized continued fractions, and the proofs for $e$ related numbers can be found elsewhere – Yuriy S May 31 '17 at 11:29


Deriving the derivative of $\tan^{2}x$ by quotient rule using (sinx/cosx) identity I am getting a different value (2secx(tanx+tan^2(x))

Deriving the derivative of $$\tan^{2}x$$ by quotient rule using $$\frac{\sin x}{\cos x}$$ identity, i am getting a different value $$2\sec x[\tan x+\tan^{2}x]$$ than by directly getting chain rule is $$2\sec^{2}x\tan x$$

Whats wrong?

Here is a correct derivation of the derivative of $$\tan^2 x$$ using the quotient rule:
$$\dfrac d {dx} \tan^2 x = \dfrac d {dx} \dfrac {\sin^2 x}{\cos^2x }=\dfrac{2\sin x \cos x \cos^2x + \sin^2 x 2 \cos x \sin x}{\cos^4x}$$
$$=2\dfrac{\sin x(\cos^2x+\sin^2x)}{\cos^3x}=2\dfrac{\sin x}{\cos^3 x}=2\tan x \sec^2x$$
• Yes, the chain rule yields $2 \tan x \sec^2x$ too – J. W. Tanner Oct 11 '19 at 6:58
• $\cos^2 x+\sin^2 x = 1$ is the Pythagorean identity – J. W. Tanner Oct 11 '19 at 7:08


# Is there a way to randomly distribute points within a circle on the surface of a sphere?

I'm attempting to set up a situation where on a 3D sphere, I choose a random point and construct a circle around this point with some radius. I then want to randomly distribute points within this circle. Is there a straightforward way to do this?

I have no trouble finding random points on the surface of the sphere; however I cannot seem to find a way to distribute points randomly on a closed region of the sphere.

Many thanks for any help in advance.

• Two related questions. – J. M.'s ennui Nov 18 '19 at 5:17
• You could brute force it by finding random points on the sphere, then throwing away any that aren't inside the desired region. – Foo Bar Nov 18 '19 at 12:59
• @Foo, yes, one can certainly use the rejection method for this. The tricky part is in figuring out how not to throw away too many points. – J. M.'s ennui Nov 18 '19 at 14:37
• Depends on your underlying distribution. I assume you want uniform random over the sphere's area. For contrast, consider uniform random points in a circle and then project the circle onto the sphere. That's still random but with a different distribution function. – Carl Witthoft Nov 19 '19 at 19:39

You can intersect the sphere and a cylinder, and then use RandomPoint. For example, here is a random point on the sphere:

sphere = Sphere[];
SeedRandom[1]
pt = RandomPoint[sphere]

{0.707037, 0.595614, 0.381239}

Then, you create a cylinder in he direction of the random point with a radius:

r = .7;
cylinder = Cylinder[{{0,0,0},pt}, r];

Now, intersect the cylinder and the sphere and use RandomPoint:

reg = RegionIntersection[cylinder, sphere];
pts = RandomPoint[reg, 1000];

Visualization:

Graphics3D[{Sphere[], Red, Point @ pts}]

• Can you verify that this produces uniform random per unit sphere area (as opposed to some projection of the intersecting circle)? I"m not familiar with how this Mathematica function actually selects points. – Carl Witthoft Nov 19 '19 at 19:41
• @CarlWitthoft That's what the documentation says: "RandomPoint will generate points uniformly in the region reg." In this case the region is the spherical cap. – Carl Woll Nov 19 '19 at 20:36

Using Christian Blatter's results from this math.SE answer, here is how to randomly sample a spherical cap:

randomCapPoint[{r_, r2_}, dir_?VectorQ] := With[{h = RandomReal[{Sqrt[1 - (r2/r)^2], 1}]},
RotationTransform[{{0, 0, 1}, Normalize[dir]}][r
Append[Sqrt[1 - h^2] Normalize[RandomVariate[NormalDistribution[], 2]], h]]]

For example,

BlockRandom[SeedRandom[1337]; (* for reproducibility *)
With[{r = 1, r2 = 2/5, d = {1.3, -2.4, 2}, n = 5000},
Graphics3D[{{Opacity[1/2], Sphere[{0, 0, 0}, r]},
{Blue, Arrow[Tube[{{0, 0, 0}, Normalize[d]}]]},
{Directive[AbsolutePointSize[2], Orange],
Point[Table[randomCapPoint[{r, r2}, d], {n}]]}}]]]

• Same question as at Carl Woll's answer: how do you verify the spatial distribution is "uniform" with respect to sphere area? – Carl Witthoft Nov 19 '19 at 19:43
• @Carl, using the results in Christian's answer, you could take a histogram of the longitude and colatitude of the sample points, and verify that they have the required distribution. – J. M.'s ennui Nov 19 '19 at 20:51

We can also take RandomPoints in boolean region obtained by the RegionIntersection of a Sphere and a Ball with radius r centered at a random point on the sphere:

SeedRandom[1]
r = .7;

ctr = RandomPoint[Sphere[]];

pts = RandomPoint[RegionIntersection[Ball[ctr, r], Sphere[]], 1000];

Graphics3D[{Red, Point@pts, White, Opacity[.5], Sphere[]}]

You can use Archimedes' result (https://en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder) that the area between two lines of latitude is the same as the corresponding area of the enclosing cylinder.

This gives a fairly simple result for sampling the top of the sphere, above height Z

pt[Z_] := Module[{z, θ, r},
z = RandomVariate[UniformDistribution[{Z, 1}]];
θ = RandomVariate[UniformDistribution[{0, 2 π}]];
r = Sqrt[1 - z^2];
{r Cos[θ], r Sin[θ], z}]

• This is basically what my method does, except I replaced {r Cos[θ], r Sin[θ]} with r Normalize[RandomVariate[NormalDistribution[], 2]]. – J. M.'s ennui Nov 19 '19 at 20:49


# Trying to answer this counting question

In a ballroom dance class, participants are divided into couples for each drill session. One partner leads and the other follows for three minutes, and then the couple switches roles for the next three minutes.

(a) Only four people show up on time. How many ways are there to pair them up?

My answer here is $4C2 = \binom42 = \tfrac{4!}{2!2!}$

(b) If instead six people show up on time, how many ways are there to pair them up?

My answer here is $6C2 = \binom62 = \tfrac{6!}{4!2!}$

(c) Assume all m people in the class arrive on time. (There are an even number of people in the class.) How many ways are there to pair them up?

My answer here is $mC2 = \binom m2= \tfrac{m!}{(m-2)!2!}$

(d) Consider this time assuming that we specify which member of each couple leads first. How many ways are there to pair-and- specify the dancers

My answer here is $mP2 = \tfrac{m!}{(m-2)!}$

## UPDATE:

Part (a) the answer is $\tfrac{1}{2}(4C2 * 2C2) = 3$

Part (b) the answer is $\tfrac{1}{2}(6C2 * 4C2 * 2C2) = 45$

• Yup, they are all correct. Just a typo in the second answer, you meant 6C2, I think. – learner Dec 8 '15 at 4:15
• Yes, I corrected it, Thanks! – Coheen Dec 8 '15 at 4:16
• No, they are not correct. See theophile's answer below. – fleablood Dec 8 '15 at 4:28
• @lucidgold What is the conflict with the answers? As far as I can see, all three are saying the same thing. – Théophile Dec 8 '15 at 18:17

For part (b), the number of ways to pick a first pair, then a second pair, then a third pair, is $\binom 62 \binom 42 \binom 22$, or in your notation $6C2 \times 4C2 \times 2C2$. But that not only pairs up all the dancers but distinguishes $\{AB, CD, EF\}$ from $\{AB, EF, CD\}$, $\{CD, AB, EF\}$, and several other permutations of the pairs. In fact you have counted the same three pairs $3!$ times, which is the number of permutations of three things. So the correct answer is not to divide by $2$ but rather to divide by $3! = 6$:

$$\frac{\binom 62 \binom 42 \binom 22}{6}.$$

Part (d) seems easier to me than part (c). First you choose who will be leaders in the first part of the lesson. There are $\binom{m}{m/2}$ ways to do this. Then you line up these $m/2$ leaders in a line and assign each of the remaining $m/2$ dancers to one of the leaders. That is, each ordering of the remaining $m/2$ dancers produces a unique pairing of followers with leaders. There are $(m/2)!$ such orderings, so the total number of possible pairings is ...

Once you have part (d), I would go back to part (c). Clearly there are more pairings counted in part (d) than in part (c). How many more?

For each way you can pair up dancers in part (c), within each pair there are two ways to choose who will lead during the first part of the lesson. Since there are $m/2$ pairs, there are therefore $2^{m/2}$ ways to choose which $m/2$ dancers will lead at first. But the choice of pairs for part (c), followed by choosing leaders for the first three minutes, can give us every choice of ordered pairs that exists in part (d). Therefore the number of choices in part (c) must be ...

No, these are not quite correct—at least not the way I understand the question. When there are four people, for example, there are only three ways to pair them up: $\{AB,CD\}, \{AC,BD\}$, and $\{AD,BC\}$. The reason this is different from your answer is that you have counted how many ways there are to choose two people from a set of four. This is ${4 \choose 2} = 6$, namely, $AB, AC, AD, BC, BD, CD$. But the number of ways to form one pair isn't the same as the number of ways to split the group into pairs.

One way to do the count for part (a) is to pick one pair (as you have done), then to form a second pair from the remaining two people. The number of ways to do this is $${4 \choose 2}{2 \choose 2} = 6 \cdot 1 = 6.$$ But in doing so we have accounted for each pairing twice; e.g., we have counted both $\{AB,CD\}$ and $\{CD,AB\}$, when these are in fact the same pairing. Therefore we should divide by $2$ to get $\frac62 = 3$ pairings.

• This is a great answer. it shows the problem with part a and leaves the rest to OP. There are similar problems with the rest, but if OP understands this answer, s/he should be able to deal with the rest. If not, that is another question. This is not an easy question. – Ross Millikan Dec 8 '15 at 4:28
• So what happened if the number of people is huge like 300? Is there is any formula I can use to solve this problem? So what I understand is I should use combinations and divide the answer by 2? is this correct? – Coheen Dec 8 '15 at 4:30
• @Coheen: you asked four different questions. All your answers were not correct. If the number of people increases, yes there is a formula you can use. The answer for the first three differs greatly from the answer to the fourth. For ones like the first three, you pick one pair, then another, then another, and so on. Then you need to consider that you could pick the pairs in any order. Your formulas just pick the first pair. – Ross Millikan Dec 8 '15 at 4:36
• For part (b), would the following be correct: (6C2* 4C2* 2C2 )/2 = 45 – Coheen Dec 8 '15 at 4:56
• @Coheen The reason for dividing by two was because there were two pairs. When there are three, you should divide by 3!. This is described in more detail in David K's answer. – Théophile Dec 8 '15 at 18:16

If I am not mistaken this is applications of the Multinomial Theorem. A lot of good examples are given in Sheldon Ross's A First Course in Probability.

Also note with part $(d)$ that uniqueness matters. So if we had 4 people show up and we chose who was going to lead who we would have a permutation for all of the members dancing. For example, let the 4 people be named: John, Sally, Jordan, Steve. If John lead Steve for the first 3 minutes it would not be the same as Steve leading John. So essentially we would have the same combination below, but without the denominator in the Multinomial, since order matters.


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Eigenvalues and Eigenvectors of a Matrix Description Calculate the eigenvalues and corresponding eigenvectors of a matrix. Eigenvalue and Eigenvector for a 3x3 Matrix Added Mar 16, 2015 by Algebra_Refresher in Mathematics Use this tool to easily calculate the eigenvalues and eigenvectors of 3x3 matrices. Display decimals, number of significant digits: … Relevance. FINDING EIGENVALUES AND EIGENVECTORS EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . The ideal is to express a given vector as a linear combination of eigenvectors. asked Jan 25 '15 at 23:57. user3435407 user3435407. How to find eigenvalues quick and easy – Linear algebra explained . Eigen vector, Eigen value 3x3 Matrix Calculator. In these examples, the eigenvalues of matrices will turn out to be real values. The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic polynomial (i.e., the polynomial whose roots are the eigenvalues of a matrix). Any help is greatly appreciated. I'm writing an algorithm with a lot of steps (PCA), and two of them are finding eigenvalues and eigenvectors of a given matrix. then the characteristic equation is . In order to do this, I need the eigenvectors but I am kind of lost how to compute them without using a huge library. Rewrite the unknown vector X as a linear combination of known vectors. Check the determinant of the matrix. It will find the eigenvalues of that matrix, and also outputs the corresponding eigenvectors.. For background on these concepts, see 7.Eigenvalues and Eigenvectors by Marco Taboga, PhD. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. If the resulting V has the same size as A, the matrix A has a full set of linearly independent eigenvectors that satisfy A*V = V*D. FINDING EIGENVALUES • To do this, we ﬁnd the values of λ which satisfy the characteristic equation of the matrix A, namely those values of λ for which det(A −λI) = 0, stream That example demonstrates a very important concept in engineering and science - eigenvalues and eigenvectors- which is used widely in many applications, including calculus, search engines, population studies, aeronautics … EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . which satisfy the characteristic equation of the. Any help is greatly appreciated. Since the zero-vector is a solution, the system is consistent. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. Eigenvalues and Eigenvectors of a Matrix Description Calculate the eigenvalues and corresponding eigenvectors of a matrix. 3xx3 matrices and their eigenvalues and eigenvectors. The eigenvectors for D 0 (which means Px D 0x/ ﬁll up the nullspace. Finding of eigenvalues and eigenvectors. 2 Answers. To find all of a matrix's eigenvectors, you need solve this equation once for each individual eigenvalue. We compute a power of a matrix if its eigenvalues and eigenvectors are given. Favorite Answer. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. In general, for any matrix, the eigenvectors are NOT always orthogonal. <> The result is a 3x1 (column) vector. FINDING EIGENVALUES • To do this, we ﬁnd the values of … The eigenvalues are immediately found, and finding eigenvectors for these matrices then becomes much easier. Notice how we multiply a matrix by a vector and get the same result as when we multiply a scalar (just a number) by that vector.. How do we find these eigen things?. How do you find the eigenvectors of a matrix? Find 2 linearly independent Eigenvectors for the Eigenvalue 0 c.) The e-value 0 has both geometric and algebraic multiplicity 2. If $\theta \neq 0, \pi$, then the eigenvectors corresponding to the eigenvalue $\cos \theta +i\sin \theta$ are In this section we will introduce the concept of eigenvalues and eigenvectors of a matrix. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. Visit http://ilectureonline.com for more math and science lectures!In this video I will find eigenvector=? Solve the system. The eigenvector v of an operator A is a vector such that: A = To do this, we find the values of ? What is the trace of a matrix? We start by finding the eigenvalue: we know this equation must be true:. This is a linear system for which the matrix coefficient is . If you love it, our example of the solution to eigenvalues and eigenvectors of 3×3 matrix will help you get a better understanding of it. To find eigenvectors, take $M$ a square matrix of size $n$ and $\lambda_i$ its eigenvalues. matrices eigenvalues-eigenvectors. The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. Ʋ�ψ�o��|�ߛ�z?cI���4��^?��R9���(/k����k The projection keeps the column space and destroys the nullspace: Relevance. 1 decade ago. Set the characteristic determinant equal to zero and solve the quadratic. Find the eigenvalues and eigenvectors. Yes, finding the eigenvectors should be straightforward. Remark. The result is a 3x1 (column) vector. Notice, however, that you have x=1 as a double root. The matrix A has an eigenvalue 2. If the determinant is 0, the matrix has no inverse. This pages describes in detail how to diagonalize a 3x3 matrix througe an example. Find the eigenvalues and bases for each eigenspace. For example, say you need to solve the following equation: First, you can rewrite this equation as the following: I represents the identity matrix, with 1s along its diagonal and 0s otherwise: Remember that the solution to […] Free Matrix Eigenvectors calculator - calculate matrix eigenvectors step-by-step This website uses cookies to ensure you get the best experience. ��~�?.����(x�$ׄ��;�oE|Ik�����$P���?�Iha��֦�BB')���q�����d�z��I;E���k��y� �@���9P}����T���3�T׸�2q�w8�{�T�*�N�mk�ǟJBZ�em���58j��k������~���-lQ9i�[$aT$A�_�1#sv;q吺��zz{5��iB�nq��()���6�au�޼ ���)��F�ܐQXk�jhi8[=���n�B�F��$.�CFZН.�PҷD����GօKZ����v��v��ʀ~��|rq�ٷ����3B�f��ٲ��l More: Diagonal matrix Jordan decomposition Matrix exponential. EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix. By using this website, you agree to our Cookie Policy. Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. View all posts by KJ Runia, 4. The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. If$ \mathbf{I} $is the identity matrix of$ \mathbf{A} $and$ \lambda $is the unknown eigenvalue (represent the unknown eigenvalues), then the characteristic equation is \begin{equation*} \det(\mathbf{A}-\lambda \mathbf{I})=0. Find the characteristic polynomial of a matrix – What is the fastest way to find eigenvalues? On the previous page, Eigenvalues and eigenvectors - physical meaning and geometric interpretation appletwe saw the example of an elastic membrane being stretched, and how this was represented by a matrix multiplication, and in special cases equivalently by a scalar multiplication. There may be two independent eigenvectors corresponding to that. If you need a refresher, check out my other lesson on how to find the determinant of a 2×2.Suppose we are given a square matrix A where, Eigenvalue is the factor by which a eigenvector is scaled. $$\tag{1}$$ , which is a polynomial equation in the variable$\lambda$. I tried to find the inverse of the eigenvectors, but it brought a wrong matrix. Without having to make extensive calculations explain why 0 is an eigenvalue of A b.) I implemented an algorithm that computes three eigenvalues given a 3x3 Matrix. In quantum physics, if you’re given an operator in matrix form, you can find its eigenvectors and eigenvalues. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. How do you find the eigenvectors of a 3x3 matrix? so clearly from the top row of the equations we get. Enter a matrix. Do you know how to solve it? [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. Calculate eigenvalues and eigenvectors. Some of my solutions do not match answers in my differential equations text (Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons). This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. SOLUTION: • In such problems, we ﬁrst ﬁnd the eigenvalues of the matrix. Syntax: eigen(x) Parameters: x: Matrix Example 1: filter_none. by Marco Taboga, PhD. Av = λIv. For Example, if x is a vector that is not zero, then it is an eigenvector of a square matrix … When I try to solve for the eigenvectors I end up with a 3x3 matrix containing all 1's and I get stumped there. Eigenvalue is the factor by which a eigenvector is scaled. Favorite Answer. Eigenvalues and eigenvectors calculator. Finding of eigenvalues and eigenvectors. Matrix A: Find. Note that if we took the second row we would get . Eigenvalues and eigenvectors calculator. /�7P=š� EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix. Eigenvectors for: Now we must solve the following equation: First let’s reduce the matrix: This reduces to the equation: There are two kinds of students: those who love math and those who hate it. Syntax: eigen(x) Parameters: x: Matrix … Source(s): eigenvectors 3x3 matric: https://tinyurl.im/fNPuM. All that's left is to find the two eigenvectors. The code for this originally is … Please check my work in finding an eigenbasis (eigenvectors) for the following problem. The process for finding the eigenvalues and eigenvectors of a 3xx3 matrix is similar to that for the 2xx2` case. λ 1 =-1, λ 2 =-2. https://www.khanacademy.org/.../v/linear-algebra-eigenvalues-of-a-3x3-matrix The determinant of matrix M can be represented symbolically as det(M). The Formula of the Determinant of 3×3 Matrix. If the determinant is 0, then your work is finished, because the matrix has no inverse. I do not wish to write the whole code for it because I know it is a long job, so I searched for some adhoc code for that but just found 1 or 2 libraries and at first I prefer not to include libraries and I don't want to move to matlab. In summary, when$\theta=0, \pi$, the eigenvalues are$1, -1$, respectively, and every nonzero vector of$\R^2$is an eigenvector. I'm having a problem finding the eigenvectors of a 3x3 matrix with given eigenvalues. Calculate the eigenvalues and the corresponding eigenvectors of the matrix. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. Display decimals, number of significant digits: Clean. In other words, the eigenvalues and eigenvectors are in$\mathbb{R}^n$. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. �������lMOK���� ��� n��h vx{Vb�HL����%f;bz\5� This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. Find the eigenvalues and bases for each eigenspace. Eigenvalue$ \boldsymbol{\lambda = 6} \$, 4.3. FINDING EIGENVALUES • To do this, we ﬁnd the values of λ which satisfy the characteristic equation of the matrix A, namely those values of λ for which det(A −λI) = 0, where I is the 3×3 identity matrix. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. I implemented an algorithm that computes three eigenvalues given a 3x3 Matrix. →Below is a calculator to determine matrices for given Eigensystems. Thanks! Answer Save. The determinant of a triangular matrix is easy to find - it is simply the product of the diagonal elements. I am trying to find the eigenvectors for the following 3x3 matrix and are having trouble with it. 27.7k 1 1 gold badge 25 25 silver badges 52 52 bronze badges. You need to calculate the determinant of the matrix as an initial step. The values of λ that satisfy the equation are the generalized eigenvalues. Av = λv. Illustrate the process of finding eigenvalues and corresponding eigenvectors of a 3x3 matrix. Notice, however, that you have x=1 as a double root. To find the inverse of a 3x3 matrix, first calculate the determinant of the matrix. Get the free "Eigenvalue and Eigenvector for a 3x3 Matrix " widget for your website, blog, Wordpress, Blogger, or iGoogle. The only eigenvalues of a projection matrix are 0 and 1. I have to find 4 things for the Matrix A which is a 3x3 matrix with all values equal to 1 A= 1 1 1 1 1 1 1 1 1 a.) In this page, we will basically discuss how to find the solutions. How to find the eigenspace of a 3x3 matrix - Suppose A is this 2x2 matrix: [1 2] [0 3]. which satisfy the characteristic equation of the.

## how to find eigenvectors of a 3x3 matrix

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# simple cryptography examples

In the examples above, statement (A) is plaintext, while statement (B) is a reverse cipher text. Thus, cryptography is defined as the art of manipulating or scrambling plaintext into ciphertext. An example of primitive cryptography is an encrypted message in which it takes over from letters with other characters. I will discuss a simple method of enciphering and deciphering a message using matrix transformations and modular arithmetic, and show how elementary row operations can sometimes be used to break an opponent's code. Cryptography is the technique of protecting information by transforming it into a secure format. Introduction. Simple substitution cipher is the most commonly used cipher and includes an algorithm of substituting every plain text character for every cipher text character. Select primes p=11, q=3. Cryptography is the science of keeping information secret and safe by transforming it into form that unintended recipients cannot understand. Solved Examples 1) A very simple example of RSA encryption This is an extremely simple example using numbers you can work out on a pocket calculator (those of you over the age of 35 45 can probably even do it by hand). Final Example: RSA From Scratch This is the part that everyone has been waiting for: an example of RSA from the ground up. Hence the modulus is $$n = p \times q = 143$$. 2. n = pq â¦ In this process, alphabets are jumbled in comparison with Caesar cipher algorithm. I am first going to give an academic example, and then a real world example. A simple example of an encryption algorithm would be changing all Ns to a 3, or all Zs to a 1. Classic Encryption - The Caesar Cipher. Background: Many of the ideas we use to keep secrets in the digital age are far older than the Internet. Cryptography, or cryptology (from Ancient Greek: ÎºÏÏÏÏÏÏ, romanized: kryptós "hidden, secret"; and Î³ÏÎ¬ÏÎµÎ¹Î½ graphein, "to write", or -Î»Î¿Î³Î¯Î±-logia, "study", respectively), is the practice and study of techniques for secure communication in the presence of third parties called adversaries. Cryptography originated approximately 4000 years ago in Egypt. Keys for a simple substitution cipher usually consists of 26 letters. The word cryptography comes from the word: KRYPTOS and GRAPHEIN. The process of encoding a plain text message in some secret way is called Encryption. Simple ciphers Simple encryption algorithms, which were invented long before first computers, are based on substitution and transposition of single plaintext characters. The following are common examples of encryption. Encryption is the conversion of information into an cryptographic encoding that can't be read without a key.Encrypted data looks meaningless and is extremely difficult for unauthorized parties to decrypt without the correct key. 1. The RSA encryption system is the earliest implementation of public key cryptography. Quick Background. This project for my Linear Algebra class is about cryptography. Meanwhile, the operations performed in modern encryption algorithms are usually similar but they affect single bits and bytes. The routine may perform several passes and changes, called permutations, on the plaintext. An example key is â Calculation of Modulus And Totient Lets choose two primes: $$p=11$$ and $$q=13$$. Example. The purpose of this note is to give an example of the method using numbers so small that the computations can easily be carried through by mental arithmetic or with a simple calculator. Key cryptography of the ideas we use to keep secrets in the digital age far. 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# probability or rolling 1 1 2 with six dice

What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?

Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.

There are $6^6=46656$ possible rolls. I believe that there are 6896 'winning' rolls among these as I wrote some code to check every roll. This makes the probability around 0.1478

I'm not sure how to get this without the help of a computer though.

• What have you tried? Also, since you have six dice, is unclear whether, for instance, $(1,1,2,3,1,2)$ is valid. Is it valid? – Arthur Jun 18 '17 at 20:29
• yes, 1,1,2,x,x,x where x could be any value – mjmdavis Jun 18 '17 at 20:38
• Common practice is to say at least two 1's and one 2 in your case and exactly " ... " otherwise. – infinitylord Jun 18 '17 at 20:44
• @infinitylord like that? – mjmdavis Jun 18 '17 at 20:48
• If you want positive responses, I recommend stating the question as: "What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die" and then (and this part is really huge for this site) show your attempt at the problem. – infinitylord Jun 18 '17 at 21:01

Approach via inclusion-exclusion and De Morgan's laws.

Let $A$ be the event that you roll at least two $1$'s. Let $B$ be the event that you roll at least one $2$.

We are attempting to calculate then $Pr(A\cap B)$

$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c) = 1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$

We calculate each term on the right now that they are in simpler form.

$Pr(A^c)$ refers to the probability of strictly fewer than two $1$'s occurring, i.e. at most one $1$ occurring. Either exactly one $1$ occurs or no $1$'s occur. In the case of exactly one $1$, pick its location and then pick each remaining digit for a total of $6\cdot 5^5$ possibilities. In the case of no $1$'s, pick each digit for a total of $5^6$ possibilities. Taking the ratio of this compared to the $6^6$ equally likely dice rolls, we calculate $Pr(A^c)=\frac{6\cdot 5^5+5^6}{6^6}$.

$Pr(B^c)$ refers to the probability of strictly fewer than one $2$ occurring, i.e. no $2$'s. This occurs with probability $Pr(B^c)=\frac{5^6}{6^6}$

$Pr(A^c\cap B^c)$ refers now to the probability of at most one $1$ and no $2$'s, which similarly to before we break into cases for either exactly one $1$ or no $1$'s. For exactly one $1$, first pick the location, then pick each remaining digit for $6\cdot 4^5$ possibilities and for no $1$'s pick each digit for $4^6$ possibilities for a probability of $Pr(A^c\cap B^c)=\frac{6\cdot 4^5+4^6}{6^6}$

This gives a final probability of:

$$Pr(A\cap B) = 1-\frac{6\cdot 5^5+5^6}{6^6}-\frac{5^6}{6^6}+\frac{6\cdot 4^5+4^6}{6^6}$$

$$=1-\frac{39760}{46656} = \frac{431}{2916}\approx 0.147805$$

You can consider the possibility to find probability directly without inclusion-exclusion principle. This is a longer way, but there are those who like it. Consider all possibilities to roll at least two ones and at least one two with six 6-sided die. We can have exactly two $1$ and one $2$. Denote this event by $A_{21}$. Or we can have exactly two $1$ and two $2$: event $A_{22}$. There are $10$ disjoint events total: $A_{21}$, $A_{22}$, $A_{23}$, $A_{24}$, $A_{31}$, $A_{32}$, $A_{33}$, $A_{41}$, $A_{42}$, $A_{51}$.

Find and then add their probabilities. The probability of $A_{21}$ is $$\mathbb P(A_{21})=\frac{\binom{6}{2}\binom{6-2}{1}4^{6-2-1}}{6^6}=\frac{15\cdot 4\cdot 4^3}{6^6}.$$ By first multiplier $\binom{6}{2}=15$ we choose the rolls which result in $1$, next multiplier $\binom{6-2}{1}=4$ we choose the roll among the rest which results in $2$, and the last multiplier $4^3$ is number of possible outcomes in the rest $6-2-1=3$ rolls.

Similar, $$\mathbb P(A_{22})=\frac{\binom{6}{2}\binom{4}{2}4^{2}}{6^6}=\frac{15\cdot 6\cdot 16}{6^6},$$ $$\mathbb P(A_{23})=\frac{15\cdot 4\cdot 4}{6^6},\quad \mathbb P(A_{24})=\frac{15}{6^6},\quad \mathbb P(A_{31})=\frac{20\cdot 3\cdot 16}{6^6},$$ $$\mathbb P(A_{32})=\frac{20\cdot 3\cdot 4}{6^6}, \quad \mathbb P(A_{33})=\frac{20}{6^6}, \quad \mathbb P(A_{41})=\frac{15\cdot 2\cdot 4}{6^6},$$ $$\mathbb P(A_{42})=\frac{15}{6^6}, \quad \mathbb P(A_{51})=\frac{6}{6^6}.$$ The total probability of rolling at least two ones and at least one two with six 6-sided die is equal to the sum of probabilities above: $$\frac{6896}{6^6}=0,147805213.$$

As I understood your question, you are asking about probability that between $6$ values will be at least two $1$'s and one $2$ (don't looking at other values and order).

Well, so the probability of $(1,1,2,x,x,x)$ (excluding order and remarking that there may be other $1$'s and $2$'s between $x$'s) is exactly the number of sets of six digits from $1$ to $6$ that start with $1,1,2$ divided by total number of sets of six digits from $1$ to $6$. There are as many (disordered) sets of six digits from $1$ to $6$ that start with $1,1,2$ as just (disordered) sets of three digits from $1$ to $6$.

Thus the answer is $P = \frac{(3^3)/3!}{(6^6)/6!} = \frac{5}{72}$.

• Hasek I'm not sure if this result is correct. – mjmdavis Jun 18 '17 at 23:07
• The line "There are as many (disordered) sets of six digits from 1 to 6 that start with 1,1,2 as just (disordered) sets of three digits from 1 to 6." is incorrect. After all, what does it mean for a "disordered set to start with something"? Further, your calculations don't appear to make much sense. What would $3^3/3!$, your supposed numerator, represent? Notice that $3!$ is even but $3^3$ is odd and so $3^3/3!$ is not an integer, so it could not possibly represent the answer to a counting problem. Similarly for the denominator. – JMoravitz Jun 18 '17 at 23:28
• Yes, I'm definitely wrong, sorry. – Hasek Jun 19 '17 at 6:44