# Question on when to use polar coordinates to prove existence of limit/ does the method always work?

Show that the following limit exists or does not exist (general example)

$$\lim \limits_{(x,y) \to (0,0)} \dfrac{e^{-x^2-y^2}-1}{x^2+y^2}$$

i) Direct substitution of $$x=0$$ , $$y=0$$ leads to indeterminate form of $$\frac{0}{0}$$

ii) Taking the limit along $$x$$ , $$y$$ axes and $$y=x$$ all result with the value $$0$$

iii) Convert to polar:

$$\lim \limits_{r \to 0^+} \dfrac{e^{-r^2}-1}{r^2}->\frac{0}{0}$$

$$L'Hopital's$$ $$rule$$

$$\lim \limits_{r \to 0^+} \dfrac{-2re^{-r^2}}{2r}=-1$$

So the limit exists and its value is -1

My questions:

1. After converting the limit expression to polar, why is $$\lim \limits_{r \to 0^+}$$ instead of $$\lim \limits_{r \to 0}$$ ? Both have the same computation
1. From the example above, how would I know if the limit $$DNE$$ when taking the limit after converting to polar? Would taking the limit of the polar converted expression $$DNE$$ or not give a finite number to know that the original limit $$DNE$$? This is of course if I chose to convert to polar without knowing that a different path gave a different limit.

2. When would it be appropriate to covert to polar to show the existence of a limit when not told that it existed or not in the first place? Does converting to polar always work?

$$\lim \limits_{(x,y) \to (0,0)} \dfrac{{xy^4}}{x^2+y^8}$$

• this limit $$DNE$$ as it has different limits along different paths namely $$y=0$$ and $$x = y^4$$, respectively 0 $$≠$$ $$\frac{1}{2}$$

Polar conversion: (this limit DNE, but polar conversion results in 0, a finite number)- to check

$$\lim \limits_{r \to 0^+} \dfrac{{rcosθ*r^4sin^4θ}}{r^2cos^2θ+r^8sin^8θ}$$

$$\lim \limits_{r \to 0^+} \dfrac{r^5({cosθ*sin^4θ})}{r^2(cos^2θ+r^6sin^8θ)}$$

$$\lim \limits_{r \to 0^+} \dfrac{r^3({cosθ*sin^4θ})}{cos^2θ+r^6sin^8θ}$$

$$\frac{0}{(cos^2(θ))}=0$$

The limit is 0

• Does converting to polar only work when (x,y) is approaching (0,0) or also work for say (x,y) is approaching a point like (-1,7)? Jun 23, 2020 at 11:11
• For the first question, I think it is just emphasising that $r$ cannot be negative so if you approach to $0$ you can only approach from the right Jun 23, 2020 at 11:16
• So does it mean that in polar coordinates, since it goes in counterclock-wise direction, it is same as $r->0^{+}$? Jun 23, 2020 at 11:18

For question 1, we take the limit as $$r \to 0^{+}$$ because in polar coordinates, $$r$$ represents the distance from the origin to point $$(x, y)$$ which is always non-negative.

For questions 2 and 3, keep in mind that we have

$$\lim_{(x, y) \to (0, 0)} \frac{e^{-x^2-y^2} - 1}{x^2 + y^2} = c$$

for some finite number $$c$$ if and only if

$$\lim_{r \to 0^{+}} \frac{e^{-r^2} - 1}{r^2} = c$$

In other words, the first limit is DNE if and only if the second one is DNE. Thus, if you manage to find some finite result $$c$$ for the second one, then you have also solved the first one. Sometimes, it is easier to evaluate limits in polar coordinates that in Cartesian coordinates so we take advantage of this when this applies.

An important note

Taking the limit along x , y axes and y=x all result with the value 0

It is important to note that in order for limit of a sequence to exist in a metric space like $$\mathbb{R}^2$$, all of its sub-sequences must also converge to that limit. That means that no matter how you walk your way to the limit, you must always arrive at the limit.

Hence, taking the limit along the $$x$$-axis, $$y$$-axis and the line $$y = x$$ is just one way to warn yourself early when the limit actually does not exist when these limits give different values.

But, if these limits all agree, this is not sufficient to say that the limit does converge to some finite number $$c$$ because there can be some distorted path to approach $$(0, 0)$$ for which a different limit can be computed.

However, the polar form takes into consideration all possible ways to walk to the origin because no matter how you approach $$(0, 0)$$, the distance from your point to $$(0, 0)$$ always converges to $$0$$, hence we have $$r \to 0^{+}$$.

• Also if the original limit had (x,y) approaching some other point like (-2,9), would converting to polar not work then since r is not approaching the origin? Jun 23, 2020 at 11:23
• For this specific limit in this question, yes, you cannot take advantage of the nice polar form to evaluate your limit easily if $(x, y)$ does not approach the origin. Jun 23, 2020 at 11:30
• Wait I put another example in my question now, and the limit does not exist by taking limit of different paths, but when converting to polar, the limit is 0, a finite number Jun 23, 2020 at 11:30
• Unfortunately, for that supposed counter-example, you can't take advantage of that nice polar form. Usually, the hint is the presence of $x^2 + y^2$. This is because $r$ is equal to the distance of $(x, y)$ to the origin, i.e. $r = \sqrt{x^2 + y^2}$ which is equivalent to $r^2 = x^2 + y^2$, the classic Pythagorean theorem. :) Jun 23, 2020 at 11:33
• Actually, there is a problem with the evaluation of your limit. The end result is $0/\cos^{2}\theta$ which is not always zero for all angles $\theta$. For instance, if you approach along the angle $\theta = \pi/2$, the expression evaluates to the indeterminate form $0/0$. However, in the first question, you had that nice $x^2 + y^2$ in it that you eliminated the $\theta$'s entirely when converted to polar form, so we were able to evaluate it easily to $-1$ without problems. :) Jun 23, 2020 at 11:44

If you restrict the polar argument to the range $$[0,2\pi)$$, the Cartesian-to-polar transformation is a bijection. Hence whatever limit computation you perform in polar coordinates gives exactly the same conclusion as when computed in Cartesian.

Polar coordinates are used for convenience when a polar symmetry (like in your example) or a significant simplification is apparent.


# Determine the off - diagonal elements of covariance matrix, given the diagonal elements

I have some covariance matrix $$A = \begin{bmatrix}121 & c\\c & 81\end{bmatrix}$$

The problem is to determine the possible values of $$c$$.

Now I know that the elements of this matrix are given by the usual definition of the covariance,

$$\frac{1}{N-1} \sum_{i=1}^N (X_i - \bar{x})(Y_i - \bar{y})$$

and so e.g.

$$\frac{1}{N-1} \sum_{i=1}^N (X_i - \bar{x})^2 = 121$$

$$\frac{1}{N-1} \sum_{i=1}^N (Y_i - \bar{y})^2 = 81$$

But I can't see how to go from here to determining $$c$$?

• is it always 2x2 matrix? Apr 16 at 13:58

You might find it instructive to start with a basic idea: the variance of any random variable cannot be negative. (This is clear, since the variance is the expectation of the square of something and squares cannot be negative.)

Any $$2\times 2$$ covariance matrix $$\mathbb A$$ explicitly presents the variances and covariances of a pair of random variables $$(X,Y),$$ but it also tells you how to find the variance of any linear combination of those variables. This is because whenever $$a$$ and $$b$$ are numbers,

$$\operatorname{Var}(aX+bY) = a^2\operatorname{Var}(X) + b^2\operatorname{Var}(Y) + 2ab\operatorname{Cov}(X,Y) = \pmatrix{a&b}\mathbb A\pmatrix{a\\b}.$$

Applying this to your problem we may compute

\begin{aligned} 0 \le \operatorname{Var}(aX+bY) &= \pmatrix{a&b}\pmatrix{121&c\\c&81}\pmatrix{a\\b}\\ &= 121 a^2 + 81 b^2 + 2c^2 ab\\ &=(11a)^2+(9b)^2+\frac{2c}{(11)(9)}(11a)(9b)\\ &= \alpha^2 + \beta^2 + \frac{2c}{(11)(9)} \alpha\beta. \end{aligned}

The last few steps in which $$\alpha=11a$$ and $$\beta=9b$$ were introduced weren't necessary, but they help to simplify the algebra. In particular, what we need to do next (in order to find bounds for $$c$$) is complete the square: this is the process emulating the derivation of the quadratic formula to which everyone is introduced in grade school. Writing

$$C = \frac{c}{(11)(9)},\tag{*}$$

we find

$$\alpha^2 + \beta^2 + \frac{2c^2}{(11)(9)} \alpha\beta = \alpha^2 + 2C\alpha\beta + \beta^2 = (\alpha+C\beta)^2+(1-C^2)\beta^2.$$

Because $$(\alpha+C\beta)^2$$ and $$\beta^2$$ are both squares, they are not negative. Therefore if $$1-C^2$$ also is non-negative, the entire right side is not negative and can be a valid variance. Conversely, if $$1-C^2$$ is negative, you could set $$\alpha=-c\beta$$ to obtain the value $$(1-C^2)\beta^2\lt 0$$ on the right hand side, which is invalid.

You therefore deduce (from these perfectly elementary algebraic considerations) that

If $$A$$ is a valid covariance matrix, then $$1-C^2$$ cannot be negative.

Equivalently, $$|C|\le 1,$$ which by $$(*)$$ means $$-(11)(9) \le c \le (11)(9).$$

There remains the question whether any such $$c$$ does correspond to an actual variance matrix. One way to show this is true is to find a random variable $$(X,Y)$$ with $$\mathbb A$$ as its covariance matrix. Here is one way (out of many).

I take it as given that you can construct independent random variables $$A$$ and $$B$$ having unit variances: that is, $$\operatorname{Var}(A)=\operatorname{Var}(B) = 1.$$ (For example, let $$(A,B)$$ take on the four values $$(\pm 1, \pm 1)$$ with equal probabilities of $$1/4$$ each.)

The independence implies $$\operatorname{Cov}(A,B)=0.$$ Given a number $$c$$ in the range $$-(11)(9)$$ to $$(11)(9),$$ define random variables

$$X = \sqrt{11^2-c^2/9^2}A + (c/9)B,\quad Y = 9B$$

(which is possible because $$11^2 - c^2/9^2\ge 0$$) and compute that the covariance matrix of $$(X,Y)$$ is precisely $$\mathbb A.$$

Finally, if you carry out the same analysis for any symmetric matrix $$\mathbb A = \pmatrix{a & b \\ b & d},$$ you will conclude three things:

1. $$a \ge 0.$$

2. $$d \ge 0.$$

3. $$ad - b^2 \ge 0.$$

These conditions characterize symmetric, positive semi-definite matrices. Any $$2\times 2$$ matrix satisfying these conditions indeed is a variance matrix. (Emulate the preceding construction.)

• It might be worth mentioning that $C$ here is the correlation and, as shown, is always between $-1$ and $+1$ Apr 17 at 12:00

An intuitive method to determine this answer quickly is to just remember that covariance matrices may be interpreted in the form

$$$$A = \begin{pmatrix} \sigma_1^2 & \rho_{12}\sigma_1\sigma_2 &\rho_{13}\sigma_1\sigma_3 & \cdots & \rho_{1n}\sigma_1 \sigma_n \\ & \sigma_2^2 & \rho_{23}\sigma_2\sigma_3 & \cdots & \rho_{2n}\sigma_2 \sigma_n \\ & & \sigma_3^2 & \cdots & \rho_{3n}\sigma_3 \sigma_n \\ & & & \ddots & \vdots \\ & & & & \sigma_n^2 \end{pmatrix}$$$$

where $$\rho_{ab} \in [-1,1]$$ is a Pearson Correlation Coefficient. In your case you have

\begin{align} \sigma_1^2 = 121 ,~~~ \sigma_2^2 = 81 ~\Longrightarrow ~ |c| \leq \sqrt{121\cdot 81} = 99 \end{align}

i.e. $$c \in [-99, 99]$$.

• +1 This is a great answer for readers familiar with this representation of covariance matrices. I don't think it would be remiss, though, to point out that when $n\gt 2$ the correlation coefficients are subject to additional restrictions: it doesn't suffice for them all just to lie between $-1$ and $1.$ The case $n=3$ is discussed at stats.stackexchange.com/questions/72790.
– whuber
Apr 17 at 18:05

$$A$$ is posdef so by Sylvesters criterion $$det(A) = 121 \cdot 81 - c^2 \geq 0$$. Thus, any $$c \in [-99, 99]$$ will produce a valid covariance matrix.

• Covariance matrices can be positive semi-definite right? Does the semi here change anything? Apr 16 at 15:23
• semi is the case when c is exactly 99 and det = 0. Apr 16 at 16:16
• @Hunaphu $c=+99$ and $c=-99$ both lead to zero determinant. If $-99 < c < +99$ then you would have a strictly positive definite matrix Apr 17 at 12:03
There are three main possibilities of note. One is that the variable are uncorrelated, in which case the off-diagonal entries are easy to calculate as 0. Another possibility is that you don't really have two different variables. $$y$$ is simply a scalar multiple of $$x$$ (i.e. perfect correlation). If $$y= c x$$, then $$\sigma_{xy} =\sigma_{x}\sigma_{xy}=99$$. We get a third possibility in noting that the above assumes $$c>0$$. For $$c<0$$, we get $$\sigma_{xy} =-99$$.
Geometrically, the covariance between two vectors is the product of their lengths times the cosine of the angle between them. Since the cosine varies from $$-1$$ to $$1$$, the covariance ranges from the product of their lengths to the negative of the product.
Another approach is to consider $$z_1 = \frac{x-\mu_{x}}{\sigma_{x}}$$ and $$z_2 = \frac{y-\mu_y}{\sigma_{y}}$$. $$\sigma_{xy} = \sigma_{(\sigma_x z_1)(\sigma_y z_2)}=\sigma_x \sigma_y \sigma_{z_1z_2}=99\sigma_{z_1z_2}$$ and $$\sigma_{z_1z_2}$$ is simply the correlation between $$x$$ and $$y$$, which ranges from $$-1$$ to $$1$$.


# Do all symmetric $n\times n$ invertible matrices have a square root matrix?

My question relates to the conditions under which the spectral decomposition of a nonnegative definite symmetric matrix can be performed. That is if $A$ is a real $n\times n$ symmetric matrix with eigenvalues $\lambda_{1},...,\lambda_{n}$, $X=(x_{1},...,x_{n})$ where $x_{1},...,x_{n}$ are a set of orthonormal eigenvectors that correspond to these eigenvalues (i.e. $X$ is an orthogonal matrix), and $\Lambda=\text{diag}(\lambda_{1},...,\lambda_{n})$ then

$A=X\Lambda X'$

is the spectral decomposition of $A$. If we then let $A^{1/2}=X\Lambda^{1/2}X'$, where $\Lambda^{1/2}$ is a square root matrix of $\Lambda$ - i.e. $\Lambda^{1/2}\Lambda^{1/2}=\Lambda$, then $A^{1/2}A^{1/2}=A$. Thus $A^{1/2}$ is a square root matrix of $A$.

So if a real nonnegative definite symmetric $n\times n$ matrix $A$ has $n$ eigenvalues then the matrix has a spectral decomposition and thus a square root matrix too. My question is do all symmetric $n\times n$ invertible matrices have $n$ eigenvalues, and thus a square root matrix? Furthermore it is not clear to me whether the eigenvalues have to be distinct or not.

• All real symmetric matrices are diagonalisable, does that help?? – Vishesh Jul 26 '13 at 16:37
• The eigenvalues can be repeated. If the geometric multiplicity of the eigenvalues is the same as the algebraic multiplicity then the matrix can be diagonalized. Not all symmetric matrices have distinct eigenvalues, take the identity. – Wintermute Jul 26 '13 at 16:46
• Oh yes, my bad, I totally forgot that. Thanks – Vishesh Jul 26 '13 at 16:50
• @dandar: Your question suggests you're asking about matrices with real entries, in which case perhaps $\Lambda$ is also required to be real? If that's correct, it's instructive to look at the $1 \times 1$ case. – Andrew D. Hwang Jul 26 '13 at 16:52
• Yes the OP should clarify whether the matrix is real, and (if so) whether the square-root is supposed to be real. On the other hand, all complex matrices have complex square-roots; no symmetry is required for that. – GEdgar Jul 26 '13 at 16:56

## 4 Answers

All symmetric matrices are diagonalizable, therefore they have $n$ eigenvalues (which don't have to be distinct, by the way), all of which are real. The spectral theorem says:

We can decompose any symmetric matrix $A\in S^n$ using symmetric eigendecomposition: $$A = \sum_{i=1}^n\lambda_iq_iq_i^T = Q\Lambda Q^T, \qquad \Lambda=diag(\lambda_i,\dots,\lambda_n)$$ where the matrix $Q = [q_1,\dots,q_n]$ is orthogonal (with $Q^TQ=I_n$), and contains the eigenvectors of $A$, while the diagonal matrix $\Lambda$ contains the eigenvalues of A.

The matrix "power rule": $$A^k = Q\Lambda^k Q^{-1}$$ can be used (with $k<0$ being allowed for invertible matrices, which means there should be no $\lambda_i=0$). Note that if there are negative eigenvalues, $\Lambda^{\frac{1}{2}}$ will become complex.

Note that in the complex case, the transpose operations should be replaced with the Hermitian operation (the conjugate transpose).

• Thank-you for the reply, and yes I now see that all symmetric matrices are diagonalizable. This is because if the matrix is $n\times n$ then we can always construct a set of $n$ orthonormal eigenvectors - i.e. regardless of whether the eigenvalues are repeated or not. Thus we can always compute the spectral decomposition and hence the square root matrix will exist. – dandar Jul 26 '13 at 17:12

What about the matrix $(-1){}{}{}{}{}{}{}{}{}{}{}$?

• Thank-you for your response. You have found a flaw in my inital summary of the use of the spectral decomposition to find the square root matrix of $A$. I should have stated that $A$ needs to be nonnegative definite. This ensures all the eigenvalues of $A$ are greater than or equal to $0$ which precludes your counter-example. – dandar Jul 26 '13 at 17:31

For your first question: that you can diagonalize real symmetric matrices is the so-called spectral theorem.

For the second: your argument is general; you did not use that the eigenvalues were distinct.

• Thanks for the reply. Yes you are right the theory for the spectral theorem does not care about whether or not the $n$ eigenvalues of $A$ are distinct or not. – dandar Jul 26 '13 at 17:13

Matrix square root can be defined in many ways. If you just want $X$ such that $X^2 = A$, you approach is good.

However, the principal square root is defined only for the matrices with no strictly negative eigenvalues and zero being at most nonderogatory eigenvalue (which is unimportant here, since symmetric matrices are diagonalizable).

The importance of the principal square root lies in the fact that it is a unique square root with the spectrum in the open right half-plane. If we extended this to the matrices with the real negative eigenvalues, we would either lose uniqueness, or the "open right half-plane" would have to be replaced by something less nice. Of course, there are reasons to ask for this. Read more in Higham's "Functions of Matrices".

Since you ask about symmetric matrices, your eigenvalues are real, so you can only define principal square root if your matrix has nonnegative eigenvalues, which means it is positive semidefinite. That also means that your square (or any other) root will also be positive semidefinite, which can be easily seen from the eigenvalue (spectral) decomposition.

• Thank-you for your reply. I had not realised the definition of a square root matrix has many forms. – dandar Jul 26 '13 at 17:19
• You can compare it to the square root in the set of the nonnegative real numbers. Principal square root is, for example, $\sqrt{4}=2$, but nothing prevents us to also consider $-2$ as a square root of $4$. We even do so, for example, when solving quadratic equations (the $\pm$ sign before $\sqrt{b^2-4ac}$). Unlike the real numbers, which can be considered matrices with a single element, general matrices have much more elements ($n^2$ of them), so you get a much wider variety of candidates for a square root (basically, all combinations of choices for the square root of the eigenvalues). – Vedran Šego Jul 26 '13 at 17:41


# Tricopter physics help

Hello!

I have been studying how exactly a tricopter works and I came across a problem that is not quite clear to me...

I understand that the tail rotor has to be slightly tilted for a specific angle alpha in order for the horizontal component Fx to counteract the unbalanced torques of the 3 propellers. However this Fx component of force F now causes that the sum of all forces is not zero - therefore the tricopter will constantly try to drift in the direction of Fx when trying to hoover!?
(see the attached sketch)

Am I missing something here or is my conclusion correct?
How is this problem handled? Do you constantly have to correct for this drift by rolling or is this effect negligible?

Thank you for any input! =)

Views: 2612

### Replies to This Discussion

The angle is adjusted by a servo to keep the heading constant.

I can help you out. But first I want you to draw a full tricopter diagram so that way I can better explain it to you by referencing your own figure.

So draw a tricopter from a top down. Include all the the prop rotations, the resulting moments, and resulting forces like you have above. Label each motor to make things easier. If you do that I can do my best to explain what is going on.

Conventional helicopters have this same problem too as a result of the tail rotor, and often hover at a very slight roll angle.

I had not considered the tri copter in this same way before, but at first glance, I believe you are right.

Thank you for all the responses! =)

@Johnatan Hair

Somehow I forgot that the helicopter has a very similar situation going on with the tail rotor! Thank you for reminding me on that! =)

@WuStangDan
I would be very grateful if you could do that and explain the problem from your perspective. I draw a figure by hand as you asked - I hope it is clear enough. I chose the tricopter where the front two rotors rotate in opposite directions, so that the unbalanced torque and therefore angle alpha can be smaller.

In addition I added two pages of my calculations. On first page there are the three conditions for balancing torques and on the second page I wrote the sums of forces. Clearly the force Fx remains in the end.

Thank you!

Okay perfect.

So yes your calculations are correct. If a tricopter was flying exactly like you have drawn, where the servo motor is at the exact angle $\alpha$ to balance out the the moment in the z axis, the triopter would have a single force in the x axis.

The reason why I wanted you to do the full calculations is make sure that you understood that you haven't actually missed something in your calculations. Becuase the real answer to your question is somewhat of a let down, I didn't want you to go back to your calculations because you didn't believe me.

So tricopters don't "drift" in the x direction when flying in real life. So that means there is something missing from your drawing. You have propellers, motors that can rotate the propellers, and most of the standard parts of a multirotor. But you don't have a negative feedback controller that all multirotors have. Now the IMU in a multirotor cannot detect when it is "drifting" at a constant velocity. So if you had your multirotor sitting in the trunk of your car while you drove down the highway, the gyros would report the same values as if they were flat on the ground. But a constant force on a body does not move it at a constant velocity. Newtons second law states that the tricopter would begin accelerating in the x direction, not just slowly "drift" like I'm assuming you thought it would based on the wording you used in your question. IMU's can detect acceleration so therefore the negative feed back controller would detect it, and change the speed between the two front motors to balance that acceleration.

This will make the tricopter no longer perfectly flat, but rather at a slight angle, making it so that F1 and F2 now both have x components.

First of all thank you very much for the detailed answer!

Reading your explanation confirmed my own thinking when I was doing the calculations above:
"Tricopter cannot hoover still when oriented horizontally. It has to be slightly tilted around y axis to counteract the F3x force"

Maybe "drifting" was a poorly chosen word - I understand that the tricopter would accelerate in direction of F3x until in equillibrium with air drag.

Thanks - now I really understand how a tricopter works. =)

Hi. I was hoping you still check this site or whatever but you would greatly help me if you could tell me about the sources of your knowledge and where you got to know such detailed info along with the diagrams and stuff? i desperately need them and wherever i look, its wayyy to complex for my understanding. Its imperative that i manage to find the completer working of the tricopter-equations,torque balancing and all. Thanks! :D

Vidur said:

Hi. I was hoping you still check this site or whatever but you would greatly help me if you could tell me about the sources of your knowledge and where you got to know such detailed info along with the diagrams and stuff? i desperately need them and wherever i look, its wayyy to complex for my understanding. Its imperative that i manage to find the completer working of the tricopter-equations,torque balancing and all. Thanks! :D

Hi Vidur! I was notified per email of your reply, however, I am afraid I don't have any specific sources I could recommend because I derived the equations above myself.

I started by researching the tricopters frame geometry and kinematics, then I found a suitable (simplified) relationship between propellers angular velocity and its thrust & torque. After that you just need to set the balance of forces and torques in all three directions (x,y & z) and use algebra to get to the solution you seek.

Note that my deriavation is far from complete since it is limited to a steady-state solution (tricopter hovering still). I only used it to clear up some confusion I had about the hovering state of a tricopter. If you wanted to derive a complete dynamic model for a tricopter you would also need to include acceleration terms, as well take air drag into account. Note that at higher velocities, incoming airflow could also significantly affect the thrust on the rotor, which you would somehow have to take into account and when flying at low altitutudes ground effect might also play a role.

As you probably see, there are many physical phenomenon that affect tricopter's flight, which is probably the reason most derivations of dynamics equations become so complex. You first need to consider what is actually the goal you are trying to achieve with your model (equations) and then evaluate which physical effects you will have to include and which you could neglect.

oh i see. Well that definitely seems like an uphill task. Could you provide me whatever links that you used in order to get any sort of insight into this? I would be very glad if i could get some sort of starting point to go about my research!
Primoz K said:

Vidur said:

Hi. I was hoping you still check this site or whatever but you would greatly help me if you could tell me about the sources of your knowledge and where you got to know such detailed info along with the diagrams and stuff? i desperately need them and wherever i look, its wayyy to complex for my understanding. Its imperative that i manage to find the completer working of the tricopter-equations,torque balancing and all. Thanks! :D

Hi Vidur! I was notified per email of your reply, however, I am afraid I don't have any specific sources I could recommend because I derived the equations above myself.

I started by researching the tricopters frame geometry and kinematics, then I found a suitable (simplified) relationship between propellers angular velocity and its thrust & torque. After that you just need to set the balance of forces and torques in all three directions (x,y & z) and use algebra to get to the solution you seek.

Note that my deriavation is far from complete since it is limited to a steady-state solution (tricopter hovering still). I only used it to clear up some confusion I had about the hovering state of a tricopter. If you wanted to derive a complete dynamic model for a tricopter you would also need to include acceleration terms, as well take air drag into account. Note that at higher velocities, incoming airflow could also significantly affect the thrust on the rotor, which you would somehow have to take into account and when flying at low altitutudes ground effect might also play a role.

As you probably see, there are many physical phenomenon that affect tricopter's flight, which is probably the reason most derivations of dynamics equations become so complex. You first need to consider what is actually the goal you are trying to achieve with your model (equations) and then evaluate which physical effects you will have to include and which you could neglect.

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# Number of elements of $3n$ binary tuples, where the ordinates add up to $2n$.

I have the following problem.

Take $$\Omega_n=\{(a_1, a_2 , \cdots , a_{3n})| a_i= 0\rm{\,or}\, 1\}$$. Define $$A_n=\{\omega \in \Omega_n|\exists k, \sum _{i=1}^{3k} a_i = 2k\}$$ and $$S_m=\{(a_1, a_2 , \cdots , a_{3m}) \in A_m|\inf \{ k,| \sum _{i=1}^{3k} a_i = 2k\}=m\}$$ we are interested in finding $$|S_m|$$ the cardinality of $$S_m$$

There is a straightforward recursive formula$$S_n={3n \choose 2n}-\left (S_1 {3(n-1) \choose 2(n-1)}+S_2{3(n-2) \choose 2(n-2)}+ \cdots S_{n-1} {3\cdot(1) \choose 2\cdot(1)} \right )= {3n\choose 2n}-\left (\sum _{i=1}^{n-1}S_i {3(n-i)\choose 2(n-i)} \right )$$ I was not sure how to compute that so I used OEIS to see if it has some nice formula. After finding by hand the first values $$3,6,21,90,429$$ it suggested me the formula $$\frac{2}{3n-1}{3n\choose 2n}$$. I managed to prove it by induction.

I would be interested in some combinatorial proof of that, a bijection would be highly appreciated.

• What is "inf" ? Do you mean the smallest $k$ for which the summation holds? – Doc Nov 28 '13 at 6:20
• @Doc out of habit infimum, yes exactly – clark Nov 28 '13 at 6:24
• Not at all there yet, but it seems interesting that the answer also takes the form $\frac{1}{n} {3n-1\choose n}$. – Doc Nov 28 '13 at 6:37
• oops .... that should say $\frac{3}{3n-1} {3n-1\choose n}$. – Doc Nov 28 '13 at 7:36
• @Doc: Or $\frac3n\binom{3n-2}{n-1}$, though that isn’t obviously any more useful. – Brian M. Scott Nov 29 '13 at 19:42

Code heads as 1 and tails as 0, and the problem can be phrased in the following way: Flip a fair coin until you have exactly twice as many heads as tails, and then stop. The value of $|S_n|$ is the number of such sequences of coin flips that have length exactly $3n$.

For this rephrased version, I asked the same question as the OP a couple of years ago on this site, under "Combinatorial proof of $\binom{3n}{n} \frac{2}{3n-1}$ as the answer to a coin-flipping problem." It took me a few weeks before I found an answer to my own question, so I wouldn't call this an easy combinatorial proof to come up with. (In fact, I generalized my argument, recast it in terms of counting certain lattice paths, and got the argument published in the Electronic Journal of Combinatorics as "Enumerating Lattice Paths Touching or Crossing the Diagonal at a Given Number of Lattice Points.")

Anyway, I'll reproduce my original combinatorial argument here. It uses the equivalent $|S_n| = \frac{3}{3n-1} \binom{3n-1}{n}$ as the formula to be proved. It also uses the following result (see, for example, Section 7.5 of Concrete Mathematics):

Raney's Lemma: Let $a_1, ... a_m$ be a sequence of integers such that $\sum a_i = 1$. There is a unique index $j$ such that the partial sums of the sequence $a_j, a_{j+1}, ... a_{j+m-1}$ (cyclic indices) are positive.

On to the proof.

Intro.

Consider the sequences with $2n$ occurrences of $-1$ (i.e., $2n$ heads) and $n$ occurrences of $+2$ (i.e., $n$ tails). We want to show that the number of these sequences with all partial sums nonzero is $\binom{3n-1}{n} \frac{3}{3n-1}$. The complete sum and empty sum are clearly $0$, so "partial sum" excludes those two cases. The sequences we want to count can be split into three groups: (1) all partial sums positive, (2) all partial sums negative, (3) some partial sums positive and some negative.

Group 1: The number of these sequences with all partial sums positive is $\binom{3n-1}{n} \frac{1}{3n-1}$.

This is the part that uses Raney's lemma. If all partial sums are positive, the last element in the sequence must be $-1$. Thus we want to count the number of sequences with $2n-1$ occurrences of $-1$ and $n$ occurrences of $+2$ that add to $+1$ and have all partial sums positive. Ignoring the partial sums restriction, there are $\binom{3n-1}{n}$ such sequences. If we partition these $\binom{3n-1}{n}$ sequences into equivalence classes based on cyclic shifts, Raney's lemma says that exactly one sequence in each equivalence class has all partial sums positive. Because there are $3n-1$ elements in each sequence there are $3n-1$ sequences with the same set of cyclic shifts, and so there are $3n-1$ sequences in each equivalence class. Thus the number of sequences in Group 1 is $\binom{3n-1}{n} \frac{1}{3n-1}$.

Group 2: The number of these sequences with all partial sums negative is also $\binom{3n-1}{n} \frac{1}{3n-1}$.

To see this, just reverse the sequences counted in Part 1.

Group 3: The number of these sequences with some positive partial sums and some negative partial sums is, yet again, $\binom{3n-1}{n} \frac{1}{3n-1}$.

This one is a little trickier. First, because of the $-1$'s, it is not possible to switch from positive partial sums to negative partial sums. Thus any sequence counted here must have exactly one switch: from negative partial sums to positive partial sums. The switch must occur at some point where the partial sum is $-1$ and the next element is $+2$. Thus we have some sequence $(a_1, \ldots, a_m, +2, a_{m+2}, \ldots, a_n)$ where the sums $a_1, a_1 + a_2, \ldots, a_1 + \cdots + a_m$ are all negative. Consider the sequence $(+2, a_m, \ldots, a_2, a_1, a_{m+2}, \ldots, a_n)$. Since $+2 + a_m + \cdots + a_1 = 1$, and $a_k + a_{k-1} + \cdots + a_1 < 0$ for all $k$, $1 \leq k \leq m$, it must be the case that $+2 + a_m + \cdots + a_{k+1} > 1$ for all $k$, $1 \leq k \leq m-1$. So the sequence $(+2, a_m, \ldots, a_2, a_1, a_{m+2}, \ldots, a_n)$ is in Group 1. To see that this mapping is a bijection, note that any sequence in Group 1 must start with $+2$ and have a first time that a partial sum is equal to $+1$. Thus this transformation is reversible.

Summing up.

Putting Groups 1, 2, and 3 together we see that the total number of sequences we want to count is $\binom{3n-1}{n} \frac{3}{3n-1}$.

• This is great! I didnt know about Ranney's Lemma, it was hidden all along this problem, it was nicely exploited! – clark Dec 5 '13 at 0:05


# Prove by Induction.

1. Jan 29, 2008

### PFStudent

1. The problem statement, all variables and given/known data
1.1.2

Write,

$${\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100}}$$

as a fraction in lowest terms.

3. The attempt at a solution

Rewriting the problem as a summation,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{n\cdot(n+1)}}$$

Then considering the first few terms,

$${\frac{1}{1\cdot2}} = \frac{1}{2}$$

$${\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} = \frac{2}{3}$$

$${\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} = \frac{3}{4}$$

This leads to the conjecture that,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{n}{n+1}}$$

How would I prove the above by induction?

-PFStudent

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

### rock.freak667

Assume the statement is true for n=N, then prove true for n=N+1

EDIT:

$$\sum_{i=1} ^{n} \frac{1}{i(i+1)}=\frac{n}{n+1}$$

3. Jan 29, 2008

### jdavel

the last term in your sum (where i = n) is going to be 1/n(n+1). if you take the sum one further (to i = n+1) what will the last term be now?

4. Jan 29, 2008

### PFStudent

Hey,

Thanks for the reply rock.freak667 and jdavel.

Assume,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{n}{n+1}}$$

is true for k. That is,

$${{\sum_{i = 1}^{k}}{\frac{1}{i(i+1)}}} = {\frac{k}{k+1}}$$

Then to prove it is true for k+1, consider the following,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {{\sum_{i = 1}^{k}}{\frac{1}{i(i+1)}}} + {\frac{1}{k+1((k+1)+1)}}$$

Which reduces to,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{k}{k+1}} + {\frac{1}{{k^2}+3k+1)}}$$

However, how do I know the above is true?

In other words how do I know if the proof by induction actually proved it?

Thanks,

-PFStudent

5. Jan 29, 2008

### rock.freak667

Well you basically want to get

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{(k+1)}{(k+1)+1}}$$

basically the sum is the same but instead of "n" put k+1

6. Jan 29, 2008

### jdavel

you're missing a pair of parenetheses in your 3rd equation that's leading to an error in the last term of your 4th equation. what should that last term be?

7. Jan 29, 2008

### PFStudent

Hey,

Consider the following,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{n}{n+1}}$$

Assume it is true for k,

$${{\sum_{i = 1}^{k}}{\frac{1}{i(i+1)}}} = {\frac{k}{k+1}}$$

Now, if it is true for k+1 the result expected is,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{k+1}{(k+1)+1}}$$

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{k+1}{k+2}}$$

Then to prove that, consider the following,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {{\sum_{i = 1}^{k}}{\frac{1}{i(i+1)}}} + {\frac{1}{(k+1)((k+1)+1)}}$$

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\left({\frac{k}{k+1}}\right)} + {\frac{1}{(k+1)(k+2)}}$$

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {{\frac{k(k+2)}{(k+1)(k+2)}}} + {\frac{1}{(k+1)(k+2)}}$$

Which reduces to,

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{(k+1)(k+1)}{(k+1)(k+2)}}$$

$${{\sum_{i = 1}^{k+1}}{\frac{1}{i(i+1)}}} = {\frac{k+1}{k+2}}$$

Proof.

Thanks,

-PFStudent

Last edited: Jan 29, 2008
8. Jan 30, 2008

### HallsofIvy

Staff Emeritus
Were you required to use induction?

Since
$$\frac{1}{n(n+1)}= \frac{1}{n}- \frac{1}{n+1}$$
that is a "telescoping" series and the sum is immediate.

9. Jan 30, 2008

### PFStudent

Hey,

Could I have proved,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{n}{n+1}}$$

by showing that it is a telescoping series? If so, how?

Thanks,

-PFStudent

10. Jan 30, 2008

### HallsofIvy

Staff Emeritus
$$\sum_{i=1}^{100} \frac{1}{i(i+1}= \frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100}$$
$$= (\frac{1}{1}- \frac{1}{2})+ (\frac{1}{2}-\frac{1}{3})+ (\frac{1}{3}-\frac{1}{4})+ \cdot\cdot\cdot+ (\frac{1}{99}- \frac{1}{100})$$
The last fraction in each pair cancels the first fraction in the next pair (except of course in the last pair). Every fraction except the first and last cancel so the sum is
$$\frac{1}{1}- \frac{1}{100}= \frac{100-1}{100}= \frac{99}{100}$$

More generally,
$$\sum_{i= 1}^n \frac{1}{i(i+1)}= \frac{1}{1}- \frac{1}{n+1}= \frac{n+1-1}{n+1}= \frac{n}{n+1}$$

11. Feb 4, 2008

### PFStudent

Hey,

Thanks HallsofIvy for showing how it could have been proved by showing it was a telescoping sum.

However, how would you know that the individual sums can be rewritten as the sum or difference of two fractions. In other words, how did you figure out the following,

$$\sum_{i=1}^{100} \frac{1}{i(i+1}= \frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100} = \left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right)+ \left(\frac{1}{3}-\frac{1}{4}\right) + \cdot\cdot\cdot + \left(\frac{1}{99}- \frac{1}{100}\right)$$

That is, how did you figure out that,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{A_{1}}{1}}-{\frac{B_{1}}{2}}}\right)} + {\left({{\frac{A_{2}}{2}}-{\frac{B_{2}}{3}}}\right)} +. . . + {\left({{\frac{A_{n}}{n}}-{\frac{B_{n}}{n+1}}}\right)}$$

Additionally, is finding the above the same as using the technique of partial fractions?

Also, what is the general way of finding $$A$$ and $$C$$ for the following (where: $$B, D, E, and{{.}}F$$; are all given),

$${\frac{E}{F}} = {{{\frac{A}{B}}\pm{\frac{C}{D}}}}$$

Where it can be shown that,

$${E} = {AD \pm BC}$$

$${F} = {BD}$$

Which is the same as,

$${\frac{E}{F}} = {\frac{AD \pm BC}{BD}}$$

Thanks,

-PFStudent

Last edited: Feb 4, 2008
12. Feb 6, 2008

### PFStudent

Hey,

I was looking over this and realized that the following,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{A_{1}}{1}}-{\frac{B_{1}}{2}}}\right)} + {\left({{\frac{A_{2}}{2}}-{\frac{B_{2}}{3}}}\right)} +. . . + {\left({{\frac{A_{n}}{n}}-{\frac{B_{n}}{n+1}}}\right)}$$

Was better written as,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{C_{1}}{1}}-{\frac{C_{2}}{2}}}\right)} + {\left({{\frac{C_{2}}{2}}-{\frac{C_{3}}{3}}}\right)} + {\cdot} {\cdot} {\cdot} + {\left({{\frac{C_{n-1}}{n-1}}-{\frac{C_{n}}{n}}}\right)} + {\left({{\frac{C_{n}}{n}}-{\frac{C_{n+1}}{n+1}}}\right)}$$

But, I'm still having some trouble figuring how would one decipher what the constants "C" must be.

Thanks,

-PFStudent

13. Feb 6, 2008

### rock.freak667

$$\sum_{n=1} ^{N} \frac{1}{n(n+1)} \equiv \sum_{n=1} ^{N} (\frac{1}{n}-\frac{1}{n+1})$$

Use partial fractions on 1/n(n+1)

$$\sum_{n=1} ^{N} \frac{1}{n}-\frac{1}{n+1}$$

then input n=1,2,3,...,N-2,N-1,N. then add them all up.

14. Feb 6, 2008

### PFStudent

Hey,

I see that however, how would you have figured out what the constants for the partial fractions of,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}}$$

would be?

In other words, how would you have solved the following,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{C_{1}}{1}}-{\frac{C_{2}}{2}}}\right)} + {\left({{\frac{C_{2}}{2}}-{\frac{C_{3}}{3}}}\right)} + {\cdot} {\cdot} {\cdot} + {\left({{\frac{C_{n-1}}{n-1}}-{\frac{C_{n}}{n}}}\right)} + {\left({{\frac{C_{n}}{n}}-{\frac{C_{n+1}}{n+1}}}\right)}$$

for: $$C_{1}, C_{2}, C_{3},..., C_{n-1}, C_{n}$$?

That is where I am stuck.

Thanks,

-PFStudent

15. Feb 6, 2008

### rock.freak667

I don't think you need to do all of that.

16. Feb 11, 2008

### PFStudent

Hey,

Thanks for the reply, rock.freak667. Your right I did not need to prove that it is a telescoping sum.

However, I still would like to know how would one have realized that,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}}$$

is a telescoping sum?

Further after the above realization one would have had to solve for the constants: $$C_{1}, C_{2}, C_{3},..., C_{n-1}, C_{n}$$; for the following,

$${{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{C_{1}}{1}}-{\frac{C_{2}}{2}}}\right)} + {\left({{\frac{C_{2}}{2}}-{\frac{C_{3}}{3}}}\right)} + {\cdot} {\cdot} {\cdot} + {\left({{\frac{C_{n-1}}{n-1}}-{\frac{C_{n}}{n}}}\right)} + {\left({{\frac{C_{n}}{n}}-{\frac{C_{n+1}}{n+1}}}\right)}$$

And that is, what I wanted to know. How do you do that?

Thanks,

-PFStudent

Last edited: Feb 11, 2008


Showing $\mathbb{Z}[i]/(1+2i) \oplus\mathbb{Z}[i]/(6-i)\cong\mathbb{Z}[i]/(8+11i)$

I am attempting to solve Ch 14 Problem 7.7 from Artin's algebra book.

Let $R=\mathbb{Z}[i]$ and let $V$ be the R-module generated by elements $v_1$ and $v_2$ with relations $(1+i)v_1+(2-i)v_2=0$ and $3v_1+5iv_2=0$. Write this module as a direct sum of cyclic modules.

Attempt

I have obtained $V\cong R^2/ \begin{bmatrix} 1+i & 3 \\ 2-i & 5i \end{bmatrix} R^2 \cong R/[8+11i]R=\mathbb{Z}[i]/(8+11i)$.

Now, I see that $(1+2i)(6-i)=8+11i.$ Now, I would like to show that $\mathbb{Z}[i]/(1+2i) \oplus\mathbb{Z}[i]/(6-i)\cong\mathbb{Z}[i]/(8+11i)$, so I can have $V$ as a direct sum of cyclic modules as needed, but how can I show this?

I have already shown that $(1+2i,6-i)=(1)=\mathbb{Z}[i]$ and thus $(1+2i)+(6-i)=\mathbb{Z}[i]$. Intuition would suggest that $(1+2i)\oplus(6-i)=\mathbb{Z}[i]$, although I think this is false since $(i-6)(1+2i)+(1+2i)(6-i)=0$.

I must confess that I am very new to module theory so please be patient with me. I don't even how it would be possible to have $\mathbb{Z}[i]/(1+2i) \oplus\mathbb{Z}[i]/(6-i)\cong\mathbb{Z}[i]/(8+11i)$ since $\mathbb{Z}[i]/(1+2i)$and $\mathbb{Z}[i]/(6-i)$ aren't even submodules of the same set.

• $1+2i$ and $6-i$ are coprime elements of the Euclidean domain $\Bbb{Z}[i]$. Their product is $8+11i$ then also their "least common multiple". So this is just another instance of the Chinese Remainder Theorem. Compare with isomorphisms of $\Bbb{Z}$-modules $$\Bbb{Z}_2\oplus\Bbb{Z}_3\simeq\Bbb{Z}_6, \quad\Bbb{Z}_3\oplus\Bbb{Z}_5\simeq\Bbb{Z}_{15}$$ et cetera. Mar 29 '18 at 18:54
• In other words the mapping $$a+bi+(8+11i)\mapsto (a+bi+(1+2i),a+bi+(6-i))$$ is an isomorphism. Mar 29 '18 at 18:55
• Thank you. If I can show that that mapping is an isomorphism, this will show $\mathbb{Z}[i]/(8+11i) \cong\mathbb{Z}[i]/(1+2i) \times \mathbb{Z}[i]/(6-i)$. However, I am not trying to show $\times$, but $\oplus$. Mar 29 '18 at 19:00
• Aren't those the same thing? At least up to isomorphism (assuming you use one for inner direct sum and the other for outer). Mar 29 '18 at 19:24
• For an answer following the method described in the problem statement: compute the Smith normal form of the relations matrix. Some relevant posts: 1, 2, 3. Mar 30 '18 at 4:36

In addition to everything said in the comments, I just wanted to remark one more thing that might be helpful concerning your last paragraph about the fact that $\mathbb{Z}[i]/(1+2i)$ and $\mathbb{Z}[i]/(6-i)$ are not submodules of some common module.
As was stated in the comments, the product of rings obtained from the chinese remainder theorem is in particular a product of modules, i.e. a direct sum, so this works perfectly fine. The point I wanted to add: If you want, you might still view both factors as submodules of $\mathbb{Z}[i]/(8+11i)$ by the following observation:
Consider a direct product of rings (commutative with $1$), $R=R_1\times R_2$. Then you might view any $R$-module $M$ as a direct sum of submodules $M_1,M_2\subset M$, where $M_i$ is a $R_i$-module (so in particular again a $R$-module) for $i=1,2$, namely set $M_1:=(1,0)\cdot M\subset M$ and $M_2:=(0,1)\cdot M\subset M$, then $$M=M_1\oplus M_2$$ as $R$-module. (By the way, the inverse procedure works as well, if $\tilde M_1$ and $\tilde M_2$ are $R_1$- resp. $R_2$-modules, then they in particular are both $R$-modules, via $(r_1,r_2)\cdot m_1:=r_1m_1$ for $m_1\in \tilde M_1$ and $(r_1,r_2)\cdot m_2:=r_2m_2$ for $m_2\in\tilde M_2$ and so one may build the $R$-module $\tilde M=\tilde M_1\oplus\tilde M_2$. This is also described in this question/answer.)
Applied to your situation: As you already proved $$R:=\mathbb{Z}[i]/(8+11i)\simeq \underset{=:R_1}{\underbrace{\mathbb{Z}[i]/(1+2i)}}\times\underset{=:R_2}{\underbrace{\mathbb{Z}[i]/(6-i)}},$$ we would now like to find elements $e_1,e_2\in R$ such that $e_1$ corresponds to $(1,0)\in R_1\times R_2$ and $e_2$ corresponds to $(0,1)$. To find those, we use the Euclidean algorithm to get $$1=\underset{=:e_2}{\underbrace{(1-3i)(1+2i)}}+\underset{=:e_1}{\underbrace{(-1)(6-i)}}.$$ So we might write $R$ as the direct sum of $R$-submodules $$R=e_1R\oplus e_2R=(7-i)R\oplus (i-6)R.$$ (Naturally, as a $R$-module this is of course isomorphic to the former version $R=R_1\oplus R_2$, as $e_i\cdot(\bullet)\colon R_i\rightarrow e_iR$ is an isomorphism of $R$-modules, $i=1,2$ - I just thought this slightly different point of view might possibly help to get more comfortable with the situation...)


# How can I prove that $2^{n+2}\mid(2n+3)!$?

I'm not sure where to proceed or how to go about proving this assertion holds for all natural numbers n: $$2^{n+2} \mid(2n + 3)!$$

The base case is $n=1$, where $2^{1+2}\mid(2\cdot 1+3)!$ which simplifies to $8 \mid 120$, and 8 does indeed divide 120.

Again, we shall assume the statement is true for $n = k$. $$2^{k+2}\mid(2k+3)!$$ Then we shall prove that the statement must be true for n = k + 1: $$2^{k+3}\mid (2(k+1)+3)!=(2k+5)!$$
Sorry for the poor formatting, this is the first time I've posted here. I'm not sure if I've started off on the right path, or where to go next. Any suggestions would be lovely!

There are $n+1$ even numbers in $1,\cdots,2n+3$ and one of them is $4$ (given $n\ge1$).

• So because the divisor has 2 to the power of whatever, it will always divide whatever the quotient is due to it having a 4 in the factorial? Is this the correct way to interpret your response? – knames Jul 3 '14 at 0:09
• @knames You use the phrases "the divisor" and "whatever" and "it" and "the quotient" but I don't know what you're referring to with these phrases. Please be clear and specific. – blue Jul 3 '14 at 0:10
• the divisor being 2^(n+2) and the quotient being (2n+3)!. The left part before | will always divide the right part after | because the right part has a 4 in the factorial? – knames Jul 3 '14 at 0:12
• @knames The word "quotient" refers to what you get after you've divided, it doesn't refer to the thing you're dividing. Just because something is divisible by $4$ does not mean it is divisible by $2^{n+2}$; that's where the total of $n+1$ even numbers comes in. Try thinking some more about my hint. – blue Jul 3 '14 at 0:14
• This is as explicit as I can make it: $$(2n+3)!=1\cdot2\cdot3\cdots(2n+3)=(\underbrace{2\cdot\color{Red}4\cdots2n+2}_{n+1~\rm numbers})(1\cdot3\cdots2n+3)$$ $$=2^{n+1}(1\cdot\color{Red}2\cdot3\cdots n+1)(1\cdot3\cdots 2n+3)=2^{n+2}(3\cdots n+1)(1\cdot3\cdots 2n+3).$$ Honestly, I thought my hint was pretty clear and straightforward, and this level of explication was totally unnecessary. – blue Jul 3 '14 at 0:41

$(2k+5)!=(2k+5)(2k+4)(2k+3)!=(2k+5)(2k+4)2^{k+2}\cdot c$ where the last equality is due to $2^{k+2}\mid (2k+3)!$ which you assume.

Hence, $(2k+5)!=(2k+5)(k+2)(2)(2^{k+2})$. Finish it :)

We have $(2k+5)!=(2k+3)!(2k+4)(2k+5)$. Since by the induction hypothesis, $2^{k+2}$ divides $(2k+3)!$, and since $2$ divides $2k+4$, we conclude that $2^{k+3}$ divides $(2k+5)!$.

Remark: (This refers to an earlier version of the post.) In writing up a proof, never try to travel from what you want to show to something "known."

• I'm new to number-theory, could you elaborate on what you meant by "In writing up a proof, never try to travel from what you want to show to something "known."" – knames Jul 2 '14 at 23:56
• @knames Sometimes you start from the fact that you're trying to prove and manipulate things until you've arrived at an already established fact. This is how it goes in any kind of math, number theory or no. You might even try burning both ends of the stick and meeting somewhere in the middle. At the end, though, when you actually go to write down your proof, you need to start at something known and end up at the desired claim, in that order. – blue Jul 3 '14 at 0:00
• In high school, people pick up the very bad habit of writing down what they want to be true, and then manipulating until they get something true, like $x=x$. This is logically wrong, unless one shows that every step that got us from what we want to $x=x$ is reversible. If one showed reversibility (which one often cannot show), it would be OK. But people who go from the desired to something known often lose control over the logic of the argument, and end up writing a confused circular non-proof. I am not saying you shouldn't fool around informally with the "desired." – André Nicolas Jul 3 '14 at 0:01
• But ultimately the written up argument must go "the right way" so that its validity is clear. – André Nicolas Jul 3 '14 at 0:02
• @blue, @ André Nicolas Thank you, this makes sense to me now! – knames Jul 3 '14 at 0:03

Start with $2k+2∣(2k+3)!$. This means there exists a $b \in\mathbb{N}$ such that $(2k+3)! = b2^{k+2}$.

\begin{align*} (2k+3)! = b2^{k+2} & \Rightarrow (2k+3)!(2k+4)(2k+5) = b2^{k+2}(2k+4)(2k+5)& \\ & \Leftrightarrow (2k+5)! = b2^{k+2}2(k+2)(2k+5) \\ & \Leftrightarrow (2k+5)! = (b(k+2)(2k+5))2^{k+3} \\ & \Rightarrow 2^{k+3} | (2k+5)! \\ & \Leftrightarrow 2^{(k+1)+2} | (2(k+1)+3)! \end{align*}

• Just a quick question, what is the significance of including the last line? Why break down the exponent and everything to the right of the "|"? – knames Jul 3 '14 at 1:25
• It makes the induction pedantically clear. This expression is a straight substitution of $(k+1)$ for $k$ in the equation we're trying to prove. – NovaDenizen Jul 3 '14 at 1:42

## Theory

The number of factors of the prime number, p, that divide into n! is $\sum_{i=1}^{\infty} \left \lfloor \dfrac{n}{p^i} \right \rfloor$

As formidable as this equation may look, it's not really that bad.

First of all, once $\left \lfloor \dfrac{n}{p^i} \right \rfloor$ becomes $0$, all of the terms after that are also $0$. So it's really a finite sum.

Second, it can be shown that $$\left \lfloor \dfrac{n}{p^{i+1}} \right \rfloor = \left \lfloor \dfrac{\left \lfloor \dfrac{n}{p^i} \right \rfloor}{p} \right \rfloor$$

So, knowing the value of $\left \lfloor \dfrac{n}{p^i} \right \rfloor$ makes it easier to compute $\left \lfloor \dfrac{n}{p^{i+1}} \right \rfloor$.

## Application

We make the following computations.

• $\left \lfloor \dfrac{2n+3}{2} \right \rfloor = n+1$
• $\left \lfloor \dfrac{n+1}{2} \right \rfloor \ge 1 \quad (\forall n \ge 1)$

It follows that the highest power of 2 that divides into $(2n+3)!$ is greater than or equal to $(n+1) + 1 = n+2$. In other words, $2^{n+2} | (2n+3)!$


# I Uniform circular motion -- How can radial acceleration have a calculated value?

1. Jul 10, 2017

### Gurasees

In uniform circular motion, direction of particle is changing at every moment but its speed remains the same. If the magnitude of velocity or speed remains the same, change in magnitude of velocity is zero. Then how come radial acceleration can have a calculated value since acceleration = change in magnitude of velocity/ change in time. Yes, the object has an acceleration due to change in direction but how can we possibly obtain a value?

2. Jul 10, 2017

### scottdave

This video may shed some insight for you. I was helpful for me, explaining derivatives of vectors.

3. Jul 10, 2017

### A.T.

Where did you get that defintion of acceleration from?

4. Jul 10, 2017

### Staff: Mentor

Acceleration is defined as the change in the velocity divided by the change in time, of: $\vec a = \frac{dV}{dt}$
Since velocity is changing, this requires that there be an acceleration. Note that acceleration is not defined as the change in the magnitude of the velocity, but simply the change in the velocity.

You can find a simple calculus derivation here: https://en.wikipedia.org/wiki/Centripetal_force#Calculus_derivation

5. Jul 10, 2017

### scottdave

Yes. Note that acceleration is a vector quantity and is often in a different direction than velocity. In the case of uniform circular motion, it is always at 90° to the velocity direction (towards center of the circle). I think the video does a nice job of explaining this.

6. Jul 11, 2017

### Gurasees

Actually yeah definition that i wrote for acceleration is wrong. Here acceleration is associated with change in direction. But what i am asking is how can we obtain a numerical value of acceleration if there is no change in numerical value of velocity?

7. Jul 11, 2017

### A.T.

Velocity is a vector which does change.

8. Jul 11, 2017

A vector is a "numerical" value. Just not a scalar. Speed is a Scalar, Velocity is a vector. Since the Force (vector) and the Velocity of the particle are always at 90 degrees - no work is done, but there is action/reaction.

F=ma where F and a are vectors is critical to proper analysis,

MANY- heck if not all, cases involving normal vectors are counter-intuitive, and are worth special attention. If you master the vector math behind the precession of a gyroscope, for example - you will know more physics than 99.9% of the population.

It may seem like a simple concept - but complete comprehension is very valuable.

9. Jul 11, 2017

### Staff: Mentor

The numerical value of the velocity's vector components is continuously changing. This is true in both Cartesian and polar/spherical coordinates.

10. Jul 11, 2017

### vanhees71

Instead of unclear words formulae can only help the understanding ;-)).

Take the example of a particle running around on a circle of radius $R$ around the origin in the $xy$ plane with constant angular velocity $\omega$. The position vector is given by
$$\vec{x}(t)=R \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix}.$$
The velocity is
$$\vec{v}=\dot{\vec{x}} = R \omega \begin{pmatrix} -\sin(\omega t) \\ \cos (\omega t) \\ 0 \end{pmatrix}$$
and the acceleration
$$\vec{a}=\dot{\vec{v}}=R \omega^2 \begin{pmatrix} -\cos(\omega t) \\ -\sin(\omega t) \\ 0 \end{pmatrix}=-\omega^2 \vec{x},$$
i.e., the acceleration is radially towards the center with the magnitude $a=|\vec{a}|=\omega^2 R$. To keep the particle on the circle you need the corresponding force, called the centripetal force, $\vec{F}=m \vec{a}=-m \omega^2 \vec{x}$.

Last edited: Jul 15, 2017
11. Jul 15, 2017

### Staff: Mentor

To clarify: both velocity and acceleration are vectors, so this should be written as $$\vec a = \frac {d \vec v}{dt}$$ or, component by component: $$a_x = \frac{dv_x}{dt} \\ a_y = \frac{dv_y}{dt} \\ a_z = \frac{dv_z}{dt}$$ or, in the matrix notation that vanhees71 used: $$\vec a = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} = \frac {d}{dt} \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix}$$

12. Jul 29, 2017

### Arjan82

Without formula's: acceleration in one direction is independent of the acceleration in any other direction (as long as the directions are normal to each other). Thus if you look at your object, it's velocity in horizontal direction does change all the time, and therefore it has an acceleration in horizontal direction. This is also independently true for the vertical direction.

13. Aug 1, 2017

### CWatters

Why should it be a problem to calculate a numerical value of acceleration in such a case?

Lets say you have a ball going North at 3m/s and later it's found to be going South at 3m/s. The numerical value of the velocity (aka speed) hasn't changed, only the direction has changed. To calculate the acceleration you first need to calculate the change in velocity which in this case is 6m/s South.


# Does the series $\sum 2^n \sin(\frac{\pi}{3^n})$ converge?

Check if $$\sum_{n = 1}^{\infty}2^n \sin\left(\frac{\pi}{3^n}\right)$$ converges.

I tried to solve this by using the ratio test - I have ended up with the following limit to evaluate: $$\lim_{n \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} \right)$$ And now - I am stuck and don't know how to proceed with this limit. Any hints?

• Use the fact that $\sin(x) \leq x$ – TheOscillator Jan 21 '18 at 22:00
• L'Hospital's rule will help you with the limit you pose, but you can show convergence directly as Olivier outlines. – Malcolm Jan 21 '18 at 22:01
• @Malcolm This is a sequence, not a function. A sequence is not differentiable. – Aemilius Jan 21 '18 at 22:01
• If $\lim_{x \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^x} \right)}{\sin \left(\frac{\pi}{3^x} \right)} \right)$ exists (which it does) then $\lim_{n \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} \right)$ also exists and equals the same value. – Malcolm Jan 21 '18 at 22:05

Hint. One has $$\left|\sin\left(\frac{\pi}{3^n}\right)\right|\le\frac{\pi}{3^n},\quad n=1,2,\cdots.$$ Can you take it from here?

• Your observation implies that this limit in question - say, L: $$0 \le L \le 2/3$$ and so this series does converge? – Aemilius Jan 21 '18 at 22:04
• You then have $$\left|\sum_{n = 1}^{\infty}2^n \sin \frac{\pi}{3^n} \right| \le \sum_{n = 1}^{\infty}\left|2^n \sin \frac{\pi}{3^n} \right| \le \pi \sum_{n = 1}^{\infty}\frac{2^n}{3^n}<\infty$$ and the given series is absolutely convergent then it is convergent. – Olivier Oloa Jan 21 '18 at 22:07

Note that

$$2^n \sin\left(\frac{\pi}{3^n}\right) \sim \pi\frac{2^n}{3^n}$$

then use comparison test with $$\sum \frac{2^n}{3^n}$$

If you want to stick with your approach using the ratio test, you can (although Olivier's answer is more direct). Caveat: I'll detail every step of the derivation, which is not actually necessary for a proof.

We will only rely on elementary arguments, specifically the fact that $\sin'(0)=\cos 0 = 1$ — which is equivalent to $$\lim_{x\to 0} \frac{\sin x}{x} = 1 \,.\tag{1}$$ From there, you can write $$\lim_{n \to \infty} \frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} = \lim_{n \to \infty} 2\cdot \frac{1}{3}\cdot \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot\frac{\frac{\pi}{3^n}}{\sin \left(\frac{\pi}{3^n} \right)} = \frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot\left(\frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} \tag{2}$$ and, by (1), we get \begin{align*} \lim_{n \to \infty} \frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} &= \frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot \lim_{n \to \infty} \left(\frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} =\frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot \left(\lim_{n \to \infty} \frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} \\ &= \frac{2}{3}\cdot 1\cdot 1^{-1} = \boxed{\frac{2}{3}} \end{align*} and you can conclude with the ratio test.

Same idea as Olivier express in a different way, you know that $$\sin\left(x\right)\underset{(0)}{=}x+o\left(x\right)$$ Hence

$$2^n\sin\left(\frac{\pi}{3^n}\right) \underset{(+\infty)}{\sim}\pi \left(\frac{2}{3}\right)^n$$

What can you say about $\displaystyle \sum_{n \geq 0}\left(\frac{2}{3}\right)^n$ ?


Why doesn't the quadratic equation contain $2|a|$ in the denominator?

When deriving the quadratic equation as shown in the Wikipedia article about the quadratic equation (current revision) the main proof contains the step: $$\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}$$ the square root is taken from both sides, so why is $$\sqrt{4a^2} = 2a$$ in the denominator and not $$\sqrt{4a^2} = 2\left |a \right |$$ Could somebody explain this to me? Thank you very much

migrated from mathoverflow.netJun 25 '18 at 17:26

This question came from our site for professional mathematicians.

One could take the square root as $2|a|$ instead, which would lead to:

$$x+\frac {b}{2a} = \pm{\frac {\sqrt{b^{2}-4ac}}{2|a|}} \quad\iff\quad x = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{2|a|}} \tag{1}$$

However, given that $\,|a|\,$ is either $\,a\,$ or $\,-a\,$ it follows that $\,\pm|a|=\pm a\,$, so the formula simplifies to:

$$x = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{2|a|}} = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{\color{red}{2a}}} = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \tag{2}$$

$(1)\,$ and $\,(2)\,$ are entirely equivalent, but $\,(2)\,$ is more convenient to use.

When taking the square root we put a $\pm$ on the right hand side to account for the two roots, so it is unnecessary to strip off the sign of $a$, as we will put it back anyways.

The two square roots of $a^2$ are $a$ and $-a$, sometimes written together as $\pm a$.

For real numbers $\pm a$ is equivalent to $\pm |a|$ but this is not true for complex numbers. So putting the absolute value operation in would make the proof less general.

We could write the proof as

$$\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}$$

$$\pm\left(x+{\frac {b}{2a}}\right)={\frac {\pm\sqrt{b^{2}-4ac}}{\pm2a}}$$

$$x+{\frac {b}{2a}}={\frac {\pm\sqrt{b^{2}-4ac}}{2a}}$$

But generally it is considered sufficient to put in just a single $\pm$ from the start rather than putting in one for each square root and then removing the redundant ones.

If you put $x =\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ into $ax^2+bx+c$, since $x^2 =\dfrac{b^2\mp2b\sqrt{b^2-4ac}+(b^2-4ac)}{4a^2} =\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2}$ you get

$\begin{array}\\ ax^2+bx+c &a\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2} +b\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}+c\\ &=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a} +\dfrac{-2b^2\pm 2b\sqrt{b^2-4ac}}{4a}+c\\ &=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}-2b^2\pm 2b\sqrt{b^2-4ac}+4ac}{4a}\\ &=0\\ \end{array}$

If you use $|a|$, it won't work since you can't combine the terms.

• The OP is only putting the absolute value when taking the square root, not putting it on $-b/2a$. This is correct, but unnecessary, see dxiv's answer. – Fan Zheng Jun 25 '18 at 18:29

my preference for remembering and using the quadratic formula (and electrical engineers seem to do that often) is to remember the root quadratic equations as:

$$x^2 \ + \ b\,x \ + \ c \ = \ 0$$

which has solution:

$$x \ = \ \begin{cases} -\tfrac{b}{2} \pm \sqrt{\left(\tfrac{b}{2}\right)^2 - c} \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 > c \\ \\ -\tfrac{b}{2} \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 = c \\ \\ -\tfrac{b}{2} \pm i \sqrt{c - \left(\tfrac{b}{2}\right)^2} \qquad & \text{for }\left(\tfrac{b}{2} \right)^2 < c \\ \end{cases}$$

normalizing out the "$a$" does not make the quadratic equation less general. the only degrees of freedom are $b$ and $c$, so that means normally (except for a double root), there are two independent solutions.

Alternatively, noting $a\ne 0$: \begin{align}\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}-4ac}{4a^{2}}} \iff \\ 4a^2\left(x+{\frac {b}{2a}}\right)^{2}&=b^{2}-4ac \iff \\ \left(2ax+b\right)^{2}&=b^{2}-4ac \iff \\ 2ax+b&=\pm \sqrt{b^2-4ac} \iff \\ x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.\end{align}


What is the distribution of this random series?

Let $\xi_n$ be iid and uniformly distributed on the three numbers $\{-1,0,1\}$. Set $$X = \sum_{n=1}^\infty \frac{\xi_n}{2^n}.$$ It is clear that the sum converges (surely) and the limit has $-1 \le X \le 1$..

What is the distribution of $X$?

Does it have a name? Can we find an explicit formula? What else can we say about it (for instance, is it absolutely continuous)?

We can immediately see that $X$ is symmetric (i.e. $X \overset{d}{=} -X$). Also, if $\xi$ is uniformly distributed on $\{-1,0,1\}$ and independent of $X$, we have $X \overset{d}{=} \frac{1}{2}(X+\xi)$. It follows that for the cdf $F(x) = \mathbb{P}(X \le x)$, we have $$F(x) = \frac{1}{3}(F(2x+1) + F(2x) + F(2x-1)). \quad (*)$$

The cdf of $\sum_{n=1}^{12} \frac{\xi_n}{2^n}$ looks like this:

It looks something like $\frac{1}{2}(1+\sin(\frac{\pi}{2}x))$ but that doesn't quite work (it doesn't satisfy (*)).

-
I think the equation (*) together with the boundary conditions $F(x\leq -1)=0$ and $F(x\geq 1)=1$ determines the cdf $F$ uniquely. –  Fabian Apr 10 '12 at 20:20
Remotely related is the Fabius random variable (see wiki). –  Sasha Apr 10 '12 at 20:25
It can be shown easily that $F(0)=1/2$, $F(1/2)=5/6$, $F(-x)=1-F(x)$. –  Fabian Apr 10 '12 at 20:32
$F(1/4)=2/3$, $F(1/3)=8/11$, $F(2/3)=10/11$, $F(3/4)=17/18$, ... The values of $F$ for rational arguments can be obtained recursively (but I did not yet figure out an explicit formula). –  Fabian Apr 10 '12 at 20:48
For $x=-1+2^{-k}$ you can get that $F(x)=\frac{1}{2(3^k)}$. The first $k$ have to be negative, and then half of those end up on the left. Thus, for small $\epsilon$, $F(-1+\epsilon)$ is roughly $\frac{1}{2}\epsilon^{\log_2 3}$. –  Thomas Andrews Apr 10 '12 at 20:59

This is more of a comment, than an answer, yet it's too big, and graphics can't be used in comments.

It is not hard to work out cumulants of $X$: $$\kappa_X(r) = \sum_{n=1}^\infty \frac{\kappa_\xi(r)}{2^{n r}} = \frac{\kappa_\xi(r)}{2^r-1} = \frac{3^r-1}{2^r-1} \cdot \frac{B_r}{r} \cdot [ r \geqslant 2]$$ Obviously, due to symmetry, odd cumulants and moments vanish. This implies the following low order moments: $$m_2 = \mathbb{E}(X^2) = \frac{2}{9}, \quad m_4 = \frac{14}{135}, \quad m_6 = \frac{106}{1701}, \quad \ldots$$

The distribution itself appears to not be absolutely continuous, based on simulations:

Following Fabian's footsteps, it is easy to code computation of CDF at rational points:

ClearAll[cdf];
cdf[x_?ExactNumberQ] /; x >= 1 := 1;
cdf[0] = 1/2;
cdf[x_?Negative] := 1 - cdf[-x];
cdf[x_Rational /; EvenQ[Denominator[x]]] /; -1 < x < 1 :=
cdf[x] = (cdf[2 x] + cdf[2 x - 1] + cdf[2 x + 1])/3;
cdf[x_Rational] := (* set up linear equations and solve them *)
Block[{f, den = Denominator[x], ru1, ru2, vals, sol, ru3},
ru1 = {f[z_] :> Divide[f[2 z] + f[2 z + 1] + f[2 z - 1], 3]};
ru2 = {f[z_ /; z <= -1] :> 0, f[z_ /; z >= 1] :> 1,
f[z_?Negative] :> 1 - f[-z]};
ru3 = f[r_Rational /; Denominator[r] < den] :> cdf[r];
vals = Table[f[k/den], {k, den - 1}] /. ru3;
sol = Solve[(((vals /. ru1) //. ru2) /. ru3) == vals,
Cases[vals, _f]];
Function[{arg, res}, Set[cdf[arg], res]] @@@
Cases[vals, f[a_] :> {a, f[a] /. First[sol]}];
cdf[x]
]

So, for example,

In[101]:= cdf[1/5]

Out[101]= 31/49

The agreement with simulations is excellent:

-


# Compact notation for one-hot indicator vectors?

Many machine learning approaches use one-hot vectors to represent categorical data. This is sometimes called using indicator features, indicator vectors, regular categorical encoding, dummy coding, or one-hot encoding (among other names).

I'm searching for a compact way to denote a one-hot vector within a model.

Say we have a categorical variable with $$m$$ categories. First, apply some arbitrary sorting to the categories. A one-hot vector $$v$$ is then a binary vector of length $$m$$ where only a single entry can be one, all others must be zero. We set the $$i^\text{th}$$ entry to 1, and all others to 0, to indicate that this $$v$$ represents the categorical variable taking on the $$i^\text{th}$$ possible value.

One clunky attempt based on misguided set notation;

$$v \in \{0, 1\}^m \qquad\qquad \sum_{i=1}^m v_i = 1$$

I've also seen math-oriented people refer to a one-hot vector using the notation

$$\mathbf{e}_i$$

But I don't understand where this notation comes from or what it is called.

Can anyone help me out? Is there a paper that does a good job of this?

Thank you,

There are several ways to note dummy variables (or one-hot encoded), one of them is the indicator function :

$$\mathbb{1}_A(x) := \begin{cases} 1 &\text{if } x \in A, \\ 0 &\text{if } x \notin A. \end{cases}$$

For $$e_i$$ it is a vector of the standard base, where $$e_i$$ denotes the vector with a $$1$$ in the $$i$$ ith coordinate and $$0$$'s elsewhere. For example, in $$\mathbb{R}^5$$, $$e_3 = (0, 0, 1, 0, 0)$$

• This is a good answer! Thanks @Fisher. Mar 19 '20 at 0:17

Found some relevant threads to your question. Hope this helps.

"Dummy variable" versus "indicator variable" for nominal/categorical data

What is "one-hot" encoding called in scientific literature?


Difference between minus one and plus one induction?

I recently started a Combinatorics class, in which my teacher (grad student) has instructed us to Prove by induction that $$1^2+2^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$$ this is trivial in the fact that it has been solved many times before, however my professor has insisted I solve it by using $P(n-1)$ as opposed to $P(n+1)$, which I've done below.

Basis

$$\frac{1(1+1)(2*1+1)}{6} = 1$$

Inductive Step $(n-1)$

$$1^2+2^2+\ldots + (n-1)^2 = \frac{(n-1)(n)(2(n-1)+1)}{6}$$ Which Simplifies to $$\frac{(n-1)(n)(2n-1)}{6} \rightarrow \frac{2n^3-3n^2+n}{6}$$ Add $6\frac{n^2}{6}$ to both sides and we've proven by induction.

My question is do there exists any mathematical proofs for which solving by Induction with $n+1$ and $n-1$ are not interchangeable and should I petition my professor to be able to use them interchangeably. I am aware that solving using $n-1$ and $n+1$ is identical, at least for every scenario I've come across (we're working with positive integers so I'm not expecting any variance from that), however given the overwhelming amount of resources, I can't for the life of me figure out why I am being instructed to use a method opposite what seems to be the norm for any other reason besides my teacher's personal preference.

• Are you asking about the difference between $P(n-1)\Rightarrow P(n)$ and $P(n)\Rightarrow P(n+1)$ in the inductive step? Those are identical. I am not quite sure what you mean by induction with $n+1$ vs $n-1$. Jan 28 '18 at 18:20
• I know that they are the same as far as my example goes, I was just wondering if there was any reason why my teacher has instructed us using $n-1$ for anything other than his own personal preference, given that I couldn't find any resources online for "Solving induction step with n minus one" or similar. Such that is what I'm being taught actually a mirror of a theorem? I'll update my question for clarity.
– jfh
Jan 28 '18 at 18:24
• Perhaps you can put side-by-side your approach and your instructor's approach. Sometimes one direction is preferred over another due to simplicity or clarity. Jan 28 '18 at 18:28
• Well, to be fussy neither are either "with n+1" or "with n-1". One is "with n implying n+1" and the other is "with n-1 implying n". Which if you replace $m$ with $n-1$ become "with n implying n+1" and the other "with m implying m+1" which are obviously the same. Jan 28 '18 at 19:44
• You can do induction with n implying n-1 to prove for example if P(237) is true and P(n) implies P(n-1) for natural number then P(n) is true for all natural numbers less than or equal to 237. That's kind of ... useless. ... Unless you are proving something integers. For example we can prove $b^{n+m} = b^nb^m$ for all integers by induction by "inducing down" if we need to. Jan 28 '18 at 19:47

As Michael Burr noted in the comments, the two conventions are identical; it's just a change of name for the variable. You could equally assume $P(k+12)$ holds and prove $P(k+13)$ from that.

The advantage of using $P(n-1) \implies P(n)$ is that your target formula is already expressed in terms of $n$, so you don't have to rewrite the target in terms of $n+1$ to figure out what you're looking for; the advantage of using $P(n) \implies P(n+1)$ is that the inductive hypothesis is already expressed in terms of $n$.

One other advantage of using $P(n-1) \implies P(n)$ is that it transfers better to "strong induction," where you can assume $P(k)$ for all $k < n$ to prove $P(n)$. Here there is definitely less rewriting going on if you use $P(n)$ as opposed to $P(n+1)$ as your target.

I've seen both used widely. I personally usually prefer $P(n-1) \implies P(n)$.

• Thank you, I know my question sounds like its teetering on the border between naivety and caution, but I think your answer does well to say that noticing the distinction between the terms of the right hand side can help identify which method to use.
– jfh
Jan 28 '18 at 18:38

I would argue that, if anything, there are reasons to prefer $P(n) \Rightarrow P(n+1)$.

Natural numbers can be defined in many ways, but the usual inductive definition is the following:

1. $0$ is a natural number;
2. If $n$ is a natural number, then $s(n)$ is a natural number.

Here $s(n)$ denotes the successor of $n$.

These two rules define a set $\mathbb N$ together with an induction principle (which allows us to prove properties of all elements of $\mathbb N$ and is in fact the usual mathematical induction) and a recursion principle (which allows us to construct new objects from the elements of $\mathbb N$).

Then $\mathbb N$ can be endowed with the usual operations satisfying all the well-known properties. In particular, it is customary to write the successor $s(n)$ of $n$ as the sum $n+1$, although it is the sum between two natural numbers that is actually defined by recursion using the successor.

There are of course many inductive structures other than the set of natural numbers. For example, binary trees are defined by:

1. $v$ is a binary tree (a single vertex, which is also the root);
2. If $t_1$ and $t_2$ are binary trees, then $t_1 \bullet t_2$ is a binary tree (the graph formed by taking $t_1$ and $t_2$, adding a new vertex as a root and joining the roots of $t_1$ and $t_2$ to the new root).

How does the induction principle look like for binary trees? If you want to prove that $P(t)$ holds for any binary tree $t$, you have to prove:

1. (Basis) $P(v)$ holds;
2. (Inductive step) If $P(t_1)$ and $P(t_2)$ hold, then $P(t_1 \bullet t_2)$ holds.

In this case there is no equivalent to the predecessor of a natural number.

I think the only reason s/he did it this way is s/he wanted to have the final simplification end in the form $\frac {2n^3 + 3n^2 + n}6$

Had he done $P(n) \implies P(n+1)$ it would have involve a lot of factoring to get in the form $\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$.

Try it:

$1 + 2 + .... + n^2 = \frac {2n^3 + 3n^2 + n}6$

So $1 + 2 + ...... + n^2 + (n+1)^2 = \frac {2n^3 + 3n^2 + n}6 + n^2 + 2n + 1$

$= \frac {2n^3 + 9n^2 + 13n + 6}6$

... and we have to somehow get that to .... $=\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$

... which isn't impossible ...

$\frac {2n^3 + 9n^2 + 13n + 6}6= \frac {2n^3 + 6n^2 + 6n + 2 + 3n^2 + 7n + 4}6$

$=\frac {2(n+1)^3 + 3n^2 + 6n + 3 + n + 1}6$

$= \frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$

... but.... why schlep? That factoring and working backwords isn't the point.

It's easier to follow it and to go to the conclusion if you you work toward simplifying.

=====

I suppose an easier compromise would be to do:

$\frac {2(n+1)^2 + 3(n+1)^2 + (n+ 1)}{6} = \frac {2n^3 + 6n^2 + 6n + 2 + 3n^2 + 6n + 3 + n+1}6 = \frac {2n^3 + 9n^3 + 13n + 6}6$

first.

Then:

$1+2 + ..... + n^2 + (n+1)^2 =$

$\frac {2n^3 + 3n^2 + n}6 + (n+1)^2 = \frac {2n^3 + 3n^2 + n}6 +n^2 + 2n + 1=$

$\frac {2n^3 + 3n^2 + n + 6n^2 + 12n + 6}6 = \frac {2n^3 + 9n^3 + 13n + 6}6$

$= \frac {2(n+1)^2 + 3(n+1)^2 + (n+ 1)}{6}$

• Whoa! I just read your professors arithmetic. Whoa. That is is clean and pretty with the $(n^3 - bn^2 + n)+2bn^2 = (n^3 + bn^2 +n)$. This use of conjugates just.... I don't know.. it's elegant. And gives a potential geometric insight as to why this works. It doesn't put one method of induction over another but for this particular proof.... it sure is pretty. Jan 28 '18 at 20:08


# A poker hand contains five cards. Find the probability that a poker hand can be....

a) Four of a kind (Contains four cards of equal face value)

So for this one, we want four cards that have the same face value, different suit. And the last card can be any remaining card.

There are 13 ranks, (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K). We have $$\binom{13}{1}$$ ways to pick 1 rank out of 13. For each of these ranks, we want to pick 4 cards that are all the same rank (and it is forcefully implied that these 4 cards will differ in suits). So $$\binom{13}{1}\binom{4}{4}$$, and for each of these ways, we have $$\binom{48}{1}$$ way to pick 1 card out of the remaining deck of 48 cards, because we picked 4 cards already. Thus, $$\frac{\binom{13}{1}\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}$$ is the total number of ways.

b) Full House (Three cards of equal face value, and two others of equal face value). So i.e: 3, 3, 3, 2, 2 would be a full house where the three 3's and two 2's are distinguishable (different suit).

So there are 13 ranks again, $$\binom{13}{1}$$ ways to pick 1 rank out of 13 total. For each of these ways, we want to pick three cards of the same rank. So $$\binom{13}{1}\binom{4}{3}$$. Now we want two more cards that are of equal face value that differ from the other three cards picked earlier, so there are 12 ranks left, and $$\binom{12}{1}$$ ways to pick 1 rank out of 12 remaining. For each of these ways, we have $$\binom{4}{2}$$ ways to pick 2 cards out of the 4 suits belonging to the same rank, thus $$\binom{12}{1}\binom{4}{2}$$. The total number of ways is: $$\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}}{\binom{52}{5}}$$

c) Three of a kind. (Three cards of equal face value, and two cards with face values that differ from each other and the other three).

13 ranks, we want to pick 3 cards of the same rank. $$\binom{13}{1}\binom{4}{3}$$. Now we want two cards that have different face values from the three picked, and each other. Now here is where I'm a bit confused, the ranks MUST be different, but the suits can be the same. So proceeding, $$\binom{12}{1} \binom{4}{1}$$ to pick 1 card of a different face value, but could be same suit, and then $$\binom{11}{1}\binom{4}{1}$$ to pick another card with a different face value, but could be a different suit. Multiply these altogether and divide by the denominator, and we have our total ways.

Is my work correct?

• Very nearly. Your numerator is the number of total ways to achieve these combinations. Dividing by $52 \choose 5$ gives you the probability of achieving those hands. Feb 14 '19 at 19:36
• That's what I said.....
– Stuy
Feb 14 '19 at 19:39
• Then you'll want to clarify the last line of sections (a) and (b), which both refer to the quotient as "[t]he total number of ways" rather than the probability. Feb 14 '19 at 20:00

You correctly calculated the probabilities of four of a kind and a full house, not the total number of ways. The number of ways these hands can be obtained is given by the numerators of your probabilities.

What is the probability of three of a kind?

As you know, there are $$\binom{52}{5}$$ possible five-card hands that can be drawn from a standard deck.

However, your count of the favorable cases is not quite right. There are $$\binom{13}{1}$$ ways to choose the rank from which three cards are drawn and $$\binom{4}{3}$$ ways to select cards of that rank, as you found. The remaining two cards must come from different ranks. There are $$\binom{12}{2}$$ ways to select two ranks from which one card is drawn and $$\binom{4}{1}$$ ways to choose one card from each of the two ranks. Hence, the number of favorable cases is $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2$$ Therefore, the probability of three of a kind is $$\frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{2}\dbinom{4}{1}^2}{\dbinom{52}{5}}$$

You count each hand twice since the order in which the singletons are selected does not matter. For instance, if the hand is $$\color{red}{5\heartsuit}, \color{red}{5\diamondsuit}, 5\spadesuit, 7\clubsuit, \color{red}{10\diamondsuit}$$, you count it once when you select $$7\clubsuit$$ as the card you are drawing from the remaining $$12$$ ranks in the deck and $$\color{red}{10\diamondsuit}$$ as the card you are drawing from the remaining $$11$$ ranks in the deck and once when you select $$\color{red}{10\diamondsuit}$$ as the card you are drawing from the remaining $$12$$ ranks in the deck and $$7\clubsuit$$ as the card you are drawing from the remaining $$11$$ ranks in the deck. However, both selections result in the same hand. Therefore, you need to divide your answer by $$2$$. Notice that $$\frac{1}{2}\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{11}{1}\binom{4}{1} = \binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2$$
• What matters is which cards are selected. We need to choose the rank from which three cards are drawn, three cards of that rank, the two ranks from which a single card is drawn, and one card from each of those ranks. In your attempt, you distinguished between the first single card you draw and the second one you draw by first selecting a card from one of the remaining $12$ ranks and then selecting a card from one of the remaining $11$ ranks. However, you still get the same hand if you select the cards in the opposite order, which means you have counted every hand twice. Feb 17 '19 at 0:07


1. ## Formula for 1+(1+2)+(1+2+3)+(1+2+3+4)+...

Can someone show me the formula for calculating 1+(1+2)+(1+2+3)+(1+2+3+4)+...

2. Originally Posted by OReilly
Can someone show me the formula for calculating 1+(1+2)+(1+2+3)+(1+2+3+4)+...
Yes, I presume you have a finite sum.
---
The key in the the fact that,
$\displaystyle 1+2+...+x=\frac{x(x+1)}{2}$
---
Therefore, that entire sum is,
$\displaystyle \frac{1(2)}{2}+\frac{2(3)}{2}+...+\frac{n(n+1)}{2}$
Thus,
$\displaystyle \sum_{k=1}^{n}\frac{k(k+1)}{2}=\frac{1}{2}\sum_{k= 1}^n k^2+k$
Thus, subdivide the summation,
$\displaystyle \frac{1}{2}\sum_{k=1}^n k^2+\frac{1}{2}\sum_{k=1}^n k$
Use the fact above, and the sum of squares formula,

$\displaystyle \frac{1}{2}\cdot \frac{n(n+1)}{2}+\frac{1}{2}\cdot \frac{n(n+1)(2n+1)}{6}$
Simplify,
$\displaystyle \frac{n(n+1)}{4}+\frac{n(n+1)(2n+1)}{12}$
$\displaystyle \frac{3n(n+1)+n(n+1)(2n+1)}{12}$
Thus, (factor),
$\displaystyle \frac{n(n+1)(3+2n+1)}{12}$
Simplify again,
$\displaystyle \frac{n(n+1)(2n+4)}{12}$
Simplify again,
$\displaystyle \frac{n(n+1)(n+2)}{6}$
~~~
Just for refernce, the sum of squares formula is,
$\displaystyle \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$
----
I think it is cool that,
$\displaystyle \left( \begin{array}{c}n+2\\3 \end{array} \right)=\frac{(n+2)(n+1)n}{3!}$

That means you can calculate the sum by taking 2 more than number of terms and finding the number of combinations of forming three.

3. Thanks!

I didn't know sum of squares formula.

4. Hello,OReilly!

TPHacker is absolutely correct . . . Nice job, T.P. !

If we are desperate, we could find the formula "from scratch".

Can someone show me the formula for calculating:
. . . $\displaystyle 1+(1+2)+(1+2+3)+(1+2+3+4)+\hdots$

Crank out a list of the first few sums:

$\displaystyle \begin{array}{ccccccccc}S_1\:=\:1 \\ S_2\:=\:1 + 3\:=\:4\\S_3\:=\:1 + 3 + 6\:=\:10\\ S_4\:=\:1 + 3+6+10 \:=\:20\\ S_5\:=\:1+3+6+10+15\:=\:35 \\S_6\:=\:1+3+6+10+15+21\:=\:56\\ S_7\:=\:1+3+6+10+15+21+28 \;=\;84\\ \vdots\end{array}$

We have a function $\displaystyle f(n)$. .For $\displaystyle n = 1,2,3,\hdots$

. . the function has consecutive values: .$\displaystyle 1\quad4\quad10\quad20\quad56\quad84\;\hdots$

Take differences of consecutive terms: . . $\displaystyle 3\quad6\quad\;10\;\quad15\quad21\;\hdots$

Take differences again: . . . . . . . . . . . . . .$\displaystyle 3\quad\;4\quad\;\;5\quad\;\;6\;\hdots$

Take differences again: . . . . . . . . . . . . . . . $\displaystyle 1\quad\;1\quad\;\:1\;\hdots$

We have a series of constants at the third differences.
. . Hence, $\displaystyle f(n)$ is a third-degree polynomial ... a cubic.

Hence, the function is of the form: .$\displaystyle f(n)\:=\:an^3 + bn^2 + cn + d$
. . and we must determine $\displaystyle a,b,c,d.$

Use the first four values from our list.

$\displaystyle S_1\,=\,1:\;\;a\cdot1^3 + b\cdot1^2 + c\cdot1 + d\:=$ $\displaystyle \:1\quad\;\;\;\Rightarrow\quad\;\; a+b+c+d\;=\;1$
$\displaystyle S_2\,=\,4:\;\;a\cdot2^3 + b\cdot2^2 + c\cdot2 + d \:= \:$ $\displaystyle 4\;\;\;\quad\Rightarrow\quad\; 8a + 4b +2c + d \;= \;4$
$\displaystyle S_3\,=\,10:\;a\cdot3^3 + b\cdot3^2 + c\cdot3 + d \:=$ $\displaystyle \:10\quad\Rightarrow\quad 27a + 9b + 3c + d \;= \;10$
$\displaystyle S_4\,=\,20:\;a\cdot4^3 + b\cdot4^2 + c\cdot4 + d\:=$ $\displaystyle \:20\quad\Rightarrow\quad 64a + 16b + 4c + d \;= \;20$

Now we must solve this system of equations . . . but it's easy!

. . $\displaystyle \begin{array}{cccc}a+b+c+d\:=\:1 \\ 8a + 4b + 2c + d \:=\:4 \\ 27a + 9b + 3c + d \:= \:10\\ 64a + 16b + 4c + d \:=\:20\end{array}\;\begin{array}{cccc}(1)\\(2)\\( 3)\\(4)\end{array}$

Subtract (1) from (2): .$\displaystyle 7a + 2b + c \:= \:3\quad\;\;\;(5)$
Subtract (2) from (3): .$\displaystyle 19a + 5b + c \:= \:6\quad\;(6)$
Subtract (3) from (4): .$\displaystyle 376a + 7b + c \:= \:10\;\;(7)$

Subtract (5) from (6): .$\displaystyle 12a + 2b\:=\:3\;\;(8)$
Subtract (6) from (7): .$\displaystyle 18a + 2b\:=\:4\;\;(9)$

Subtract (8) from (9): .$\displaystyle 6a\,=\,1\quad\Rightarrow\quad \boxed{a = \frac{1}{6}}$

Substitute into (8): .$\displaystyle 12\left(\frac{1}{6}\right) + 2b\:=\:3\quad\Rightarrow\quad \boxed{b = \frac{1}{2}}$

Substitute into (5): .$\displaystyle 7\left(\frac{1}{6}\right) + 3\left(\frac{1}{2}\right) + c\:=\:3\quad\Rightarrow\quad \boxed{c = \frac{1}{3}}$

Substitute into (1): .$\displaystyle \frac{1}{6} + \frac{1}{2} + \frac{1}{3} + d\:=\:1\quad\Rightarrow\quad \boxed{d = 0}$

Therefore: .$\displaystyle f(n)\:=\:\frac{1}{6}n^3 + \frac{1}{2}n^2 + \frac{1}{3}n\quad\Rightarrow\quad \boxed{f(n)\;=\;\frac{n(n+1)(n+2)}{6}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This "scratch"method can be used
. . when we suspect we have a polynomial function.
It is long and tedious, but it works.

Of course, it is more efficient to learn a few sums-of-powers formula:

. . $\displaystyle \sum_{k=1}^n k \;= \;\frac{n(n+1)}{2}$

. . $\displaystyle \sum_{k=1}^n n^2\;=\;\frac{n(n+1)(2n+1)}{6}$

. . $\displaystyle \sum_{k=1}^n k^3\;= \;\frac{n^2(n+1)^2}{4}$

Note that the third formula is the square of the first formula,
. . making it easier to memorize.

. . This means: .$\displaystyle (1 + 2 + 3 + 4 + 5)^2\:=\:1^3+2^3+3^3+4^3+5^3$

It looks like a bad joke or a terrible blunder, doesn't it?

5. Originally Posted by Soroban
We have a series of constants at the third differences.
. . Hence, $\displaystyle f(n)$ is a third-degree polynomial ... a cubic.
Tell me something, is this a famous method? This is the second time I seen it used on this forum. When I was younger I developed many theorems concerning differences of a sequence. One of them you just used. (A sequence is a polynomial of degree n if and only if it takes n steps to reach a constanct sequence).
---
I have a elegant prove with this rule that,
$\displaystyle a^{(m)}_k=\sum_{k=1}^nk^m$ is always a polynomial sequence.

Consider an infinite sequence,
$\displaystyle 0^m,1^m+0^m,2^m+1^m+0^m,...$
Its diffrence sequence is,
$\displaystyle 0^m,1^m,2^m,3^m,...$
A polynomial sequence of degree $\displaystyle m$ therefore $\displaystyle m$ subtractions are required. In total we used $\displaystyle m+1$ subtractions on our original sequence. Thus there exists a polynomial sequence of degree $\displaystyle m+1$. Furthermore, the first coefficient is $\displaystyle \frac{1}{(m+1)!}$. Based on the fact the constant sequence follows the factorial.

6. Originally Posted by ThePerfectHacker
Tell me something, is this a famous method?
Yes, I seem to recall that it is associated with Newton's name, but that
be a manifestation of false memory syndrome, but it is in Acton's
"Numerical Methods that Work", which I swear by

RonL

7. Hello, TPHacker!

Tell me something, is this a famous method?

I'm sure it is . . .

I ran across many years ago in some book. .Since then, I've seen more efficient methods,
but I like that very primitive method. .(But I don't use it unless I am forced to.)

In case anyone is interested . . .
I was shown this method in graduate school . . . quite an eye-opener.

To find, for example, $\displaystyle \sum^n_{k=1} k^4$, we are expected to know the three "preceding" formulas:
. . $\displaystyle \sum^n_{k=1} k \:=\:\frac{n(n+1)}{2}\qquad\sum^n_{k=1} k^2 \:=\:\frac{n(n+1)(2n+1)}{6}\qquad\sum^n_{k=1}$$\displaystyle k^3\:=\:\frac{n^2(n+1)^2}{4}$

Consider the next-higher power and form: .$\displaystyle k^5 - (k-1)^5$

We have: /$\displaystyle k^5 - (k-1)^5\:=\;5k^4 - 10k^3 + 10k^2 - 5k + 1$

Now let $\displaystyle k \,= \,n,\,n\!-\!1,\,n\!-\!2,\,...\,,\,3,\,2,\,1$ and "stack" the equations.

. . . . $\displaystyle n^5 - (n-1)^5\;=\quad\;\;5n^4\quad -\quad\;\; 10n^3\quad +$ . .$\displaystyle 10n^2\quad\; -\quad\; 5n\quad\; + 1$
$\displaystyle (n-1)^5-(n-2)^5 \;=$ $\displaystyle \:5(n-1)^4 - 10(n-1)^3 + 10(n-1)^2 - 5(n-1) + 1$
$\displaystyle (n-2)^5-(n-3)^5 \;=$ $\displaystyle \:5(n-2)^4 - 10(n-2)^3 + 10(n-2)^2 - 5(n-2) + 1$
. . . . . . $\displaystyle \vdots$ . . . . . . . . . . .$\displaystyle \vdots$ . . . . . . . .$\displaystyle \vdots$ . . . . . . . $\displaystyle \vdots$ . . . . . . . $\displaystyle \vdots$ . . . $\displaystyle \vdots$
. . $\displaystyle 3^5\quad -\quad 2^5 \qquad= \quad\;\;5(3^4)\;\;\; -\;\;\; 10(3^3)$ . $\displaystyle +\;\;\; 10(3^2)\quad -\quad 5(3)\quad + 1$
. . $\displaystyle 2^5\quad -\quad 1^5\qquad = \quad\;\;5(2^4)\;\;\; -\;\;\; 10(2^3)$ . $\displaystyle +\;\;\; 10(2^2)\quad -\quad 5(2)\quad + 1$
. . $\displaystyle 1^5\quad -\quad 0^5 \qquad = \quad\;\;5(1^4)\;\;\; -\;\;\; 10(1^3)$ . $\displaystyle + \;\;\;10(1^2)\quad -\quad 5(1)\quad + 1$

Add the stack (most of the left side cancels out):

. . $\displaystyle n^5\;=\;5\left(\sum k^4\right) - 10\left(\sum k^3\right) +$ $\displaystyle 10\left(\sum k^2\right) - 5\left(\sum k\right) + \left(\sum 1\right)$

We have: .$\displaystyle n^5 \;= \;5\left(\sum k^4\right)-10\cdot\frac{n^2(n+1)^2}{4} +$ $\displaystyle 10\cdot\frac{n(n+1)(2n+1)}{6} - 5\cdot\frac{n(n+1)}{2} + n$

Then, after an enormous amount of algebra, we get:

. . . . $\displaystyle \boxed{\sum^n_{k=1} k^4\;=\;\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}}$


# Pointwise converging subsequence on countable set

Let $A \subseteq \mathbb R$ be countable and let $f_n: A\to \mathbb R$ be a sequence of functions such that there exists $M \ge 0$ with $\,\lvert\,f_n(x)\rvert\le M$ for all $n$. I am trying to show that there exists a subsequence $f_{n_k}$ of $f_n$ that converges pointwise.

Here is what I have so far: If $A = \{a_1, a_2, \dots \}$ then $f_n(a_1)$ is a bounded sequence hence by Bolzano Weierstrass theorem contains a convergent subsequence $f_{n_{k_1}}$. By the same argument $f_{n_{k_1}}(a_2)$ contains a convergent subsequence $f_{n_{k_2}}$.

Next I want to define $$f_{n_k} (x) = \lim_{j \to \infty} f_{n_{k_j}}(x),$$ the pointwise limit. Then $f_{n_k}(a_j)$ converges for every $a_j \in A$ (it's clear by how it was defined).

Am I done now or am I missing something? Is there anything left to show?

Let $A=\{a_n:n\in\mathbb N\}$. Using Bolzano-Weierstrass for the bounded sequence $\{f_n(a_1)\}_{n\in\mathbb N}$ we can find a convergent subsequence which we denote as $\{f_{1,n}(a_1)\}_{n\in\mathbb N}$. Next, as $\{f_{1,n}(a_2)\}_{n\in\mathbb N}$ is bounded, it also contains a convergent subsequence which we denote as $\{f_{2,n}(a_2)\}_{n\in\mathbb N}$.
In this way we construct recursively the following convergent sequences: \begin{align} f_{1,1}(a_1),&f_{1,2}(a_1),\ldots f_{1,n}(a_1),\ldots,\\ f_{2,1}(a_2),&f_{2,2}(a_2),\ldots f_{2,n}(a_2,\ldots,\\ \vdots&\\ f_{n,1}(a_n),&f_{n,2}(a_n),\ldots f_{n,n}(a_n),\ldots,\\ \vdots& \end{align} with $\{f_{k,n}\}_{n\in\mathbb N}$ a subsequence of all the sequences $\{f_{j,n}\}_{n\in\mathbb N}$, for $j<k$, and say that $\lim_{n\to\infty}f_{j,n}(a_j)=f(a_j)$.
The sequence $\{f_{n,n}\}_{n\in\mathbb N}$ is finally a subsequence of all the above, and hence $\{f_{n,n}(a_j)\}_{n\in\mathbb N}$ converges for all $j\in\mathbb N$, and in particularly $$\lim_{n\to\infty}f_{n,n}(a_j)=f(a_j),$$ for all $j\in\mathbb N$.
• Nice answer. But I don't understand why $f_{k,n}$ is subsequence of $f_{j,n}$ please can you illustrate? – MathLover Nov 17 '18 at 11:46


Problems

 Compute the magnitude and phase of $(1-j) \bigg(\frac{1}{2} + j \frac{\sqrt{3}}{2}\bigg)$ Compute the magnitude and phase of $e^{j\pi/2} (1+j) (1+3j)$ Compute the magnitude and phase of $j e^{j \pi/3}$ Compute the magnitude and phase of $e^{j \pi/4} + e^{j 3 \pi/4}$ Compute the magnitude and phase of $(1+3j)^2$ Compute the magnitude and phase of $(1-3j)/(1+3j)^2$ Compute the magnitude and phase of $e^{j \pi/5} \times e^{j 2 \pi/5} \times e^{j 3 \pi/5} \ldots e^{j 9 \pi/5}$ Compute the magnitude and phase of $e^{j \pi/5} \times e^{j 2 \pi/5} \times e^{j 3 \pi/5} \ldots e^{j 9 \pi/5} \times e^{j 10 \pi/5}$ Let $z_1 = 1, z_2 = -\frac{1}{2} + j \frac{\sqrt{3}}{2}, z_3 = -\frac{1}{2} - j \frac{\sqrt{3}}{2}$ [a)] What are $z_1^3, z_2^3$ and $z_3^3$? [b)] Show that $z_3 = z_2^2$ [c)] Show that $z_1 + z_2 + z_3 = 0$ Can you now see why $z_1,z_2,z_3$ can be called the cube roots of unity. They are usually expressed as $1, \omega, \omega^2$. Part c shows that the sum of the cube roots of unity is zero. In one of the homework problems, we will show that this true for $n$th roots of unity for any $n$. Let $z_1 = 2 e^{j\pi/4}$ and $z_2 = 8 e^{j\pi/3}$. Find and express your answer in Cartesian and polar form [a)] $2z_1-z_2$ [b)] $\frac{1}{z_1}$ [c)] $\frac{z_1}{z_2^2}$ [d)] $\sqrt[3]{z_2}$ Prove that $\int e^{ax} \ \cos(bx) \ dx = \frac{e^{ax}}{a^2+b^2} \left(a \cos(bx) + b \sin(bx) \right)$ You can use integration tricks you learned in your calculus class to solve this problem. That is not the point of the exercise. Try using Euler's formula and then using integration of exponentials to see if you can solve the problem.

 OWN 1.25d Is the even part of $x(t) = \cos \left( 4 \pi t\right) \ u(t)$ periodic? If it is, what is the time period? OWN 1.25e Is the even part of $x(t) = \sin \left( 4 \pi t\right) \ u(t)$ periodic? If it is, what is the time period? OWN 1.26b Is the discrete-time signal $x[n] = \cos \left(\frac{n}{8}\right)$ periodic? If it is, what is the time period? OWN 1.26c Is the discrete-time signal $x[n] = \cos \left(\frac{\pi n^2}{8}\right)$ periodic? If it is, what is the time period?

 OWN 1.27a Is the system $y(t) = x(t-2)+x(2-t)$ memoryless, linear, stable, time-invariant and causal? OWN 1.27c Is the system $y(t) = \int_{-\infty}^{2t} x(\tau) \ d \tau$ memoryless, linear, stable, time-invariant and causal? OWN 1.27d Is the system $y(t) = \begin{cases} 0, \ \ t < 0 \\ x(t)+x(t-2), t \geq 0 \end{cases}$ memoryless, linear, stable, time-invariant and causal? OWN 1.27g Is the system $y(t) = \frac{d}{dt}x(t)$ memoryless, linear, stable, time-invariant and causal? Consider the system defined by $y(t) = x(t) \ u(t)$. Is this linear, time invariant, stable, invertible?

 OWN 2.4 Compute the convolution of (1) \begin{align} x[n] = \left\{ \begin{array}{ll} 1, & 3 \leq n \leq 8 \\ 0, & \hbox{otherwise.} \end{array} \right. \ \hbox{and} \ h[n] = \left\{ \begin{array}{ll} 1, & 4 \leq n \leq 15 \\ 0, & \hbox{otherwise.} \end{array} \right. \end{align} OWN 2.6 Compute the convolution of $x[n] = \left(\frac{1}{3} \right)^{-n} \ u[-n-1]$ and $h[n] = u[-n-1]$

Bandpassfromoldexam
DTFS
Phasefunction
Two systems
Problem5fromoldexam - There is a small mistake in the solutions to the last part. The correct answer is 450Hz,500Hz,550 Hz

Review done on Sunday, December 10, 2017 before the final exam

 Compute Discrete-time Fourier transform of $x[n] = \sin \left( \frac{\pi n/5}{\pi n}\right) \ \cos \left(\frac{5 \pi n}{2} \right)$
page revision: 43, last edited: 21 Aug 2021 10:45