Poisson or Exponential Distribution? There are $10$ questions in a two-hour test. Suppose that the time taken for Jackson to answer each equations is independent of each other and has exponential distribution with $\lambda = 0.08$ questions per min.

What is the probability that he will take more than $30$ min to answer the first $3$ questions and less than $30$ min to answer the next $3$ questions?

My initial thought would be to use exponential distribution of $\lambda =\frac2{75}$ to solve the question, by finding $$P(Y < 30)\times(1 - P(Y < 30)),$$ where Y is the time taken for Jackson to finish 3 questions.

However, I have contesting views on using Poisson distribution as well. Perhaps the community could enlighten me. Cheers!

\begin{align}
& \Pr( \,\overbrace{\text{time to answer the first three}}^{\begin{smallmatrix} \text{the sum of three exponentially} \\ \text{distributed random variables} \end{smallmatrix} }{} > 30 \text{ minutes}\, ) \\[8pt]
= {} & \Pr( \, \underbrace{ \text{number of questions answered in 30 minutes}}_\text{a Poisson-distributed random variable} {} <3 )
\end{align}
Numerically, finding the second of these two numbers is quick and easy, so my way of finding the first one would be to find the second one and rely on your understanding of why they must be equal.

For $i=1,2,\dots,10$ let $X_{i}$ denote the time needed to get an
answer on question $i$.

Then to be found is: $$P\left(X_{1}+X_{2}+X_{3}>30\wedge X_{4}+X_{5}+X_{6}<30\mid X_{1}+X_{2}+X_{3}+X_{4}+X_{5}+X_{6}\leq120\right)\tag1$$

Define $Y:=X_{1}+X_{2}+X_{3}$ and $Z:=X_{4}+X_{5}+X_{6}$.

Then $Y$ and $Z$ are iid with $P\left(Y>t\right)=P\left(Z>t\right)=e^{-\lambda}\left[1+\lambda t+\frac{1}{2}\lambda^{2}t^{2}\right]$ with $\lambda=0.08$.

Note that this expression corresponds with $P\left(N\left(t\right)\leq2\right)$
where $N\left(t\right)$ denotes a Poisson counting process with rate $\lambda$.

Likewise we have $P\left(Y+Z>t\right)=e^{-\lambda}\left[1+\lambda t+\frac{1}{2}\lambda^{2}t^{2}+\frac{1}{6}\lambda^{3}t^{3}+\frac{1}{24}\lambda^{4}t^{4}+\frac{1}{120}\lambda^{5}t^{5}\right]$.

This expression corresponds with $P\left(N\left(t\right)\leq5\right)$

We can rewrite $(1)$ as:

$$P\left(Y>30\wedge Z<30\mid Y+Z\leq120\right)=\frac{P\left(Y>30\wedge Z<30\wedge Y+Z\leq120\right)}{P\left(Y+Z\leq120\right)}\tag2$$

The denominator is straightforward and the numerator can be calculated as:

$$\begin{aligned} & \int P\left(Y>30\wedge Z<30\wedge Y+Z\leq120\mid Y=t\right)f_{Y}(t)dt\\
 & =\int_{30}^{120}P\left(Z<30\wedge t+Z\leq120\right)f_{Y}(t)dt\\
 & =\int_{30}^{90}P\left(Z<30\right)f_{Y}\left(t\right)dt+\int_{90}^{120}P\left(Z\leq120-t\right)f_{Y}\left(t\right)dt\\
 & =P\left(Z<30\right)\int_{30}^{90}f_{Y}\left(t\right)dt+\int_{90}^{120}P\left(Z\leq120-t\right)f_{Y}\left(t\right)dt\\
 & =P\left(Z<30\right)P\left(30<Y\leq90\right)+\int_{90}^{120}P\left(Z\leq120-t\right)f_{Y}\left(t\right)dt
\end{aligned}
$$
Here in the first equality it is used that $Y$ and $Z$ are independent and $f_{Y}$ denotes the PDF of $Y$ which can be found as derivative of $P\left(Y\leq t\right)$.

The working out is left to you.


Finite vector space V,W so is L(V,W) Preceding example:

  If V,W are vector spaces then $Map\left ( V,W \right )$ is an infinite dimensional vector space with operation defined by $\left ( f+g \right )\left ( v \right ):=f\left ( v \right )+g\left ( v \right )$
   and
  $ \left ( \alpha f \right )\left ( v \right ):=\alpha f\left ( v \right )$.
  The linear map $L\left ( V,W \right )$ form a subspace of Map$\left ( V,W \right )$

I believe the question below is preceded by this example:

  Prove that if V,W are finite dimensional then so is $L\left ( V,W \right )$.

Here's my attempt:

Let V,W be finite dimension n.
Hence, V,W contains a set of n linearly independent vectors but no n+1 linearly independent vectors.

Let $L:V\rightarrow W$

with
$
\left ( f+q \right )\left ( v \right ):=f\left ( v \right )+g\left ( v \right )$
and
$\left ( \alpha f \right )\left ( v \right ):=\alpha f\left ( v \right )$

Seemingly, I am unable to progress forward.

Hints are appreciated

When first learning about function structures like your $L(V,W)$, I felt it useful to think of matrices.

Remember all linear transformations $T:V\to W$ where $V$ and $W$ are finite-dimensional have matrix representations. Once you fix bases, you can find a matrix $M$ such that $T(x) = Mx$. If $T$ sends $n-$dimensional vectors to $m-$dimensional vectors then the size of $M$ is size $m\times n$. Specifically, $\dim(L(V,W)) = \dim(M_{m\times n}) = nm$.

So here's what you do. Fix bases $B_V$ and $B_W$ for $V,W$ respectively (you can even use a coordinate transformation to work in $R^k$ if you wish). Then count the number of transformations by counting the number of ways to map $B_V$ to $B_W$. Linear transformations are determined by what they do to a basis.

In case the following words make sense:

The category of finite-dimensional vector spaces over a field $k$ is symmetric monoidal closed and hence enriched over itself.


Integration separation of variable Liquid is flowing into a small tank which has a leak. Initially the tank is empty and, t minutes later,
the volume of liquid in the tank is V cm3
. The liquid is flowing into the tank at a constant rate of
80 cm3
per minute. Because of the leak, liquid is being lost from the tank at a rate which, at any
instant, is equal to kV cm3
per minute where k is a positive constant.

The equation becomes $ dv/dt=80 - kv  $ but how to separate it that dv one side and dt on other side ?

$$\frac {dv}{80-kv}=dt$$ and then carry out your work

since k is a constant you can do this

it should be :
after integration,
you will get
$t=\frac{-1}{k} \ln(80-kV)$

Let $V=0$ , $t=0$
obtain 2 equations,
combine them and you will get
$t-\frac{1}{k} \ln(80) = \frac{-1}{k} \ln(80-kV)$

convert in to exponential form,
and hence you will get the final answer as
$V=\frac{1}{k}(80-80e^{-kt}$


Integral of odd function doesn't converge? When I look up $$\int_{-1}^1 \dfrac{1}{x} dx$$

on Wolfram Alpha, it says it doesn't converge. While this is a sum of two diverging integrals, the two areas are clearly symmetric, and I'd assume the answer would be zero. How does one usually treat integrals like this?

It's undefined quite simply because you can force it to have another value. First of all, let's say by $\int_{-1}^1\frac1xdx$ you mean
$$
\lim_{s \to 0^-\,\, t \to 0^+}\left(\int_{-1}^s\frac1x dx + \int_t^1 \frac1xdx\right)
$$
Now, if we force $s = -t$, then as you say, this evaluates to $0$. However, if we say $s = -at$ for some $a > 0$, then we get something completely different (we get $\ln a$). It's this "completely different" that in the end makes the value indeterminable.

Side note: It's possible to go into subdividing the interval $[-1, 1]$ unevenly and use the definition of (Riemann) integral to see that we can get the sum to be whatever we want, and once again, this means undefined.

The integral is undefined. However consider the following formula:

$$lim_{\epsilon \to 0} \frac {x}{x^2+ \epsilon^2} = \frac {1}{x}$$

Let us try this representation of $1/x$ in your integral. Now perform the integration before taking the limit of $\epsilon$ to $0$. [This change-of-order is presumably not strictly allowed, but obviously it makes a lot of sense to do so anyway.] Note that the integrand is now divergence-free, so the integration is completely straightforward. The anti-derivative is found to be:

$$(1/2) * log (x^2 + \epsilon^2)$$

This function is symmetric, hence the cancellation that we seek indeed takes place and the integral from $-1$ to $+1$ equals zero. Of course it depends on the context of the mathematical problem at hand whether a method like this is justified.


Can every group be represented by a group of matrices? Can every group be represented by a group of matrices?

Or are there any counterexamples? Is it possible to prove this from the group axioms?

Every finite group is isomorphic to a matrix group.  This is a consequence of Cayley's theorem: every group is isomorphic to a subgroup of its symmetry group.  Since the symmetric group $S_n$ has a natural faithful permutation representation as the group of $n\times n$ 0-1 matrices with exactly one 1 in each row and column, it follows that every finite group is a matrix group.

However, there are infinite groups which are not matrix groups, for example, the symmetric group on an infinite set or the metaplectic group.

Note that every group can be represented non-faithfully by a group of matrices: just take the trivial representation.  My answer above is for the question of whether every group has a faithful matrix representation.

There are many groups which have infinite number of elements. Then of course you can't represent them with a finite bunch of stored matrices. For example the group of 2D rotations over real valued 2D vectors. But you can sometimes get arbitrarily close by using only one or a few matrices and natural numbers as exponents.

$$R_\theta = \begin{bmatrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{bmatrix}$$

Now choose $\theta = \epsilon = 2\pi\cdot 2^{-n}>0$

$${R_\theta} ^{2^n} = I$$

Given some measure of closeness, we can go on to prove we can get arbitrarily close to any element in the group if we just allow $n$ and matrix-exponents (integers) be large enough. In the example above we have in some sense a "resolution" of $n$ bits. But it's real convenient to just have to consider one "building block" $R_\theta$ and it's exponents and not try to store all possible elements separately.

( We would quickly run out of memory if we tried for any infinite group )


How do you solve for the sum of the series From the question$$\sum_{n=1}^{\infty}112*(0.065)^n$$My way of approaching is by trying to find the limit of the partial sum of a series but I can't get a general formula for the partial sum. Am I missing another easier way of approaching this?

Hint: The first term of the series is $112\times 0.065$ (not $112$). (Make sure you know why this is!) Then use the formula for an infinite geometric series.

According to the general expression for the partial sums of the geometric series:
$$\sum_{n=1}^N ar^n=ar\sum_{n=0}^{N-1} r^n=ar\frac {1-r^N}{1-r}.$$

If $|r|<1$ (as in your example) the sum will tend to
$$\frac {ar}{1-r}. $$

Now substitute $a=112$, $b=0.065$.


Example of a trigonometric series that is not fourier series? My textbook doesn't give any example of this kind of series. Could you provide some?

Trigonometric series is defined in wikipedia as :
$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$

When

$A_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \cos{nx} dx\qquad (n=0,1,2,3 \dots)$

$B_{n}=\frac{1}{\pi} \int^{2 \pi}_0 f(x) \sin{nx} dx\qquad (n=1,2,3, \dots)$

It is fourier series.

thanks.

I don't know the answer (I should have! -- see below), but in my opinion a lot of people are misinterpreting the question, so perhaps it is worth an answer to try to set this straight.

Here is an analogy: trigonometric series is to Fourier series as power series is to Taylor series.  In other words, a trigonometric series is just any series of the form

$A_{0}+\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx})$

This should be understood formally, i.e., this is a trigonometric series even if it doesn't converge anywhere.  A fourier series is the trigonometric series associated to an
$L^1$ function by taking $A_n$ and $B_n$ as above.

Just as a Taylor series need not need converge at any point except the central point, a Fourier series need not converge pointwise at any point.  Thus the Fourier series need not be a function and in particular the Fourier inversion theorem need not apply.

A theorem of Borel asserts that given any sequence $(a_n)$ of real numbers, there exists a $C^{\infty}$-function on the real line whose Taylor series at $0$ is $\sum_{n=0}^{\infty} a_n x^n$.  In particular, if the $a_n$'s grow too rapidly, the Taylor series will diverge away from zero and the function $f$ will not be analytic.

I interpret the question as asking whether the analogue of Borel's theorem is true for Fourier series: is every trigonometric series the Fourier series of some $L^1$ function (even if the trigonometric series does not converge to the function)?

P.S.: Boo to wikipedia for asserting the answer to this question without giving a reference.

Addendum: As Pierre-Yves Gaillard points out, the Fourier coefficients of any $L^1$ function $f$ are uniformly bounded (by $||f||_1$), so this answers the question as I have interpreted it.

Take a look at this wikipedia stub. A trigonometric series, which is not a Fourier series, must therefore be one that is defined by

$$ f(x) = A_{0}+\displaystyle\sum_{n=1}^{\infty}(A_{n} \cos{nx} + B_{n} \sin{nx}) $$

for which the relations

$$ A_{0}=\frac{1}{2\pi}\displaystyle\int^{2 \pi}_0 f(x)\,dx $$

$$ A_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \cos{nx} \,dx\qquad (n=1,2,3 \dots)$$

$$B_{n}=\frac{1}{\pi}\displaystyle\int^{2 \pi}_0 f(x) \sin{nx}\, dx\qquad (n=0,1,2,3, \dots)$$

do not hold. But since {$\sin(nx),\cos(nx)$} forms a complete orthonormal basis, I cannot imagine one. Except for a non-convergent one perhaps...

edit As J. Mangaldan pointed out: Lacunary trigonometric series might be the answer you search, i.e. series that cannot be analytically be continued.


Finding Launch Angle for Two Projectile Collision I'm trying to figure out the general equation for calculating the launch angle of one projectile required when trying to find a collision between that projectile and another. For the equation I've worked out so far, everything is constant/user-defined. I'm also assuming both projectiles are launched at time 0. d_T is the displacement of the target and d_P is the displacement of the projectile.

My initial vectored equations for the motion of the two particles. The user simply specifies the two velocity components for the target, and then the two projectile components are broken from the magnitude into two using sin and cos.

$$d_T= <V_{Tx}t + d_{T0x}, V_{Ty}t + d_{T0y} - \frac{1}{2}gt^2>$$

$$d_P=<V_{P}t\cos(\theta) + d_{P0x}, V_{P}t\sin(\theta) + d_{P0y} - \frac{1}{2}gt^2>$$

Setting the each component equal to the other's respective component:

$$V_{Tx}t + d_{T0x} = V_{P}t\cos(\theta) + d_{P0x}$$
$$V_{Ty}t + d_{T0y} = V_{P}t\sin(\theta) + d_{P0y}$$

Algebra:
$$t(V_{Tx}-V_{P}\cos(\theta)) = d_{P0x} - d_{T0x}$$
$$t(V_{Ty}-V_{P}\sin(\theta)) = d_{P0y} - d_{T0y}$$
$$\frac{d_{P0x}-d_{T0x}}{V_{Tx}-V_{P}\cos(\theta)}=\frac{d_{P0y} - d_{T0y}}{V_{Ty}-V_{P}\sin(\theta)}$$

Now this is where my question comes into play. Is there a way to further manipulate the expression to solve for theta? I tried playing around with it a bit to get
$$\frac{V_{Ty}-V_{P}\sin(\theta)}{V_{Tx}-V_{P}\cos(\theta)}=\frac{d_{P0y} - d_{T0y}}{d_{P0x}-d_{T0x}}$$

where I hoped I could get a sin/cos to get a tan, but no such luck cause of the pesky target velocity. Is there a possible way to algebraically/otherwise way to solve this or do I need to completely reapproach the problem?

Thanks!

Here is how I would solve this.

Let's assume that the target is launched from $(x_0, y_0)$, $x_0 \ge 0$, towards negative $x$ coordinates, at time $t = -t_0$, $t_0 \ge 0$. The muzzle velocity is $v_0$, angle $\theta_0$, and gravity $g$. Wind resistance and coriolis forces are not accounted for. The trajectory of the target projectile is
$$\begin{cases}x_T(t) = x_0 - (t + t_0) v_0 \cos\theta_0\\
y_T(t) = y_0 + t v_0 \sin\theta_0 - (t + t_0)^2 g/2\end{cases}$$

The target projectile reaches ground when $y_T(t) = 0$. (Because the trajectory is a parabola, the $t$ we are interested from is the larger one, if there are two real roots. If there are no real roots, then the target projectile is launched from below the ground level, without the projectile ever reaching ground level.)
Solving $y_T(t_{Z}) = 0$ for $t_{Z}$ yields
$$t_Z = \sin(\theta_0) \frac{v_0}{g} + \frac{\sqrt{\sin(\theta_0)^2 v_0^2 + 2 g y_0}}{g} - t_0$$
Later, when we solve for the time when the intercept is to occur, we need to compare against $t_Z$, to make sure the intercept occurs before the target projectile reaches the ground.

The intercepting projectile is launched from origin $(0,0)$ at time $t=0$ towards positive $x$ coordinates, with muzzle velocity $v_1$, and launch angle $\theta_1$. The trajectory of this intercepting projectile is
$$\begin{cases}x_I(t) = t v_1 \cos\theta_1\\
y_I(t) = t v_1 \sin\theta_1 - t^2 g/2\end{cases}$$
It is easier to do the following calculations if we use $c = \cos\theta_1$ and $\sin\theta_1 = \sqrt{1-c^2}$. The trajectory is then
$$\begin{cases}x_I(t) = t v_1 c\\
y_I(t) = t v_1 \sqrt{1-c^2} - t^2 g/2\end{cases}$$

For the intercept to occur at time $t$, the projectiles' $x$ coordinates must match: $$x_T(t) - x_I(t) = 0$$
Solving for $c$ we get
$$c(t) = \frac{x_0 - (t_0 + t)\cos\theta_0 v_0}{t v_1}$$
The launch is only possible when $-1 \le c(t) \le 1$. Solving for $t$ we get
$$\begin{align} c(t_{min}) =& 1 &\iff& t_{min} = \frac{x_0 - t_0 v_0 \cos\theta_0}{v_0 \cos\theta_0 + v_1}\\
c(t_{max}) =& -1 &\iff& t_{max} = \frac{x_0 - t_0 v_0 \cos\theta_0}{v_0 \cos\theta_0 - v_1}\end{align}$$

Using the cosine of launch angle $c(t)$ to intercept the target projectile at time $t$, the projectiles' $y$ coordinates must be equal at $t$ for them to collide:
$$y_T(t) = x_T(t), c = c(t)$$
This can be solved for $t$; it is just a quadratic equation in a funny form, with lots of terms. Move all terms to the left side, and the square root to the right side, square both sides, and move the right side back to left, and you end up with a quadratic equation in $t$. There are quite a few terms in it, though.

The two roots are of form
$$t = \frac{q \pm \sqrt{s}}{d}$$
where
$$d = 2 (g^2 t_0^2 - 2 v_T g t_0 \sin\theta_0 + v_0^2 - v_1^2)$$
$$q = v_0 (3 g t_0^2 - 2 y_0) \sin\theta_0 + 2 v_0 x_0 \cos\theta_0 - t_0(g^2 t_0^2 - 2 g y_0 + 2 v_0)$$
$$s = t_0^2 (g^2 t_0^2 v_1^2 - 4 g^2 x_0^2 - 4 g v_1^2 y_0 + 4 v_1^2 v_0^2)
    + 4 v_1^2 x_0^2 + 4 v_1^2 y_0^2 - 4 v_0^2 x_0^2
    + 4 t_0 v_0 x_0 (g^2 t_0^2 + 2 g y_0 - 2 v_1^2) \cos\theta_0
    - v_0^2 (g t_0^2 2 x_0 + 2 y_0)(g t_0^2 - 2 x_0 + 2 y_0) (\cos\theta_0)^2
    - 4 v_0^2 x_0 (g t_0^2 + 2 y_0)(\cos\theta_0)\sin\theta_0
    - 4 v_0 t_0 (g t_0^2 v_1^2 - 2 g x_0^2 - 2 v_1^2 y_0)\sin\theta_0$$

If $\max(0, t_{min}) < t < \min(t_Z, t_{max})$, then $t$ is a valid time for the collision to occur. If both $t$ are valid, you can pick one -- the earlier one is the obvious choice. (This can occur when the intercepting launcher is positioned correctly to hit the target projectile both when it is ascending, and when it is descending, I believe.)

The intercepting launcher angle is then trivially solved from $c(t)$, i.e.
$$\theta_1 = \arccos\left( \frac{x_0 - (t_0 + t)\cos\theta_0 v_0}{t v_1} \right)$$

Remember, interception launch always happens at time $t = 0$; the $t$ above is the time of the collision.

All this math is dull without example code.

Consider this simple awk script, example.awk:

#!/usr/bin/awk -f
function x_T(t)   { return x_0 - (t + t_0) * v_0 * cos_T }
function y_T(t)   { return y_0 + (t + t_0) * v_0 * sin_T - 0.5 * g * (t + t_0) * (t + t_0) }
function x_I(t,c) { return t * v_I * c }
function y_I(t,c) { return t * v_I * sqrt(1 - c*c) - 0.5 * g * t * t }

BEGIN {
    N = 200         # 200 points along the trajectory

    g = 9.81        # Standard gravity, 9.81 m/s^2

    # Target launcher
    x_0 = 1000.0    # 1000 m away
    y_0 = 30.0      # Higher elevation
    a_0 = 30.0      # 30 degree launch angle
    v_0 = 100.0     # 100 m/s muzzle velocity
    t_0 = 5.0       # Fired five seconds ago

    # Intercepting launcher
    v_I = 120.0     # 120 m/s muzzle velocity

    cos_T = cos(a_0 * 3.14159265358979323846 / 180.0)
    sin_T = sin(a_0 * 3.14159265358979323846 / 180.0)

    # In case the target never reaches ground level...
    if (sin_T*sin_T*v_0*v_0 + 2*g*y_0 < 0.0) {
        printf "Target never reaches ground level.\n" > "/dev/stderr"
        exit(0)
    }

    # Time when target projectile hits y(0) (final time)
    t_Z = sin_T * v_0 / g + sqrt(sin_T*sin_T*v_0*v_0 + 2*g*y_0) / g - t_0;
    if (t_Z < 0.0) {
        printf "Target projectile reaches destination %.3f seconds before intercept launch.\n", -t_Z > "/dev/stderr"
        exit(0)
    }

    # Minimum and maximum launch times (with valid launch angle)
    t_min = (x_0 - cos_T*t_0*v_0) / (cos_T * v_0 + v_I)
    t_max = (x_0 - cos_T*t_0*v_0) / (cos_T * v_0)
    if (t_min < 0)   t_min = 0
    if (t_max > t_Z) t_max = t_Z
    if (t_min > t_max) {
        printf "No launch window for interception.\n" > "/dev/stderr"
        exit(0)
    }

    # Coefficients for launch time solutions
    s = t_0*t_0*(g*g*t_0*t_0*v_I*v_I - 4*g*g*x_0*x_0 - 4*g*v_I*v_I*y_0+4*v_I*v_I*v_0*v_0)+4*v_I*v_I*x_0*x_0+4*v_I*v_I*y_0*y_0-4*v_0*v_0*x_0*x_0 + cos_T *4*t_0*v_0*x_0*(g*g*t_0*t_0+2*g*y_0-2*v_I*v_I) - cos_T*cos_T*v_0*v_0*(g*t_0*t_0+2*x_0+2*y_0)*(g*t_0*t_0-2*x_0+2*y_0) - cos_T*sin_T*4*v_0*v_0*x_0*(g*t_0*t_0+2*y_0) - sin_T*4*v_0*t_0*(g*t_0*t_0*v_I*v_I-2*g*x_0*x_0-2*v_I*v_I*y_0);
    d = 2*(g*g*t_0*t_0 - 2*v_0*sin_T*g*t_0 - v_I*v_I + v_0*v_0)
    c = sin_T*v_0*(3*g*t_0*t_0 - 2*y_0) + 2*cos_T*v_0*x_0 - t_0*(g*g*t_0*t_0 - 2*g*y_0 + 2*v_0*v_0)
    if (s >= 0.0 && d != 0.0) {
        t1 = (c + sqrt(s)) / d
        t2 = (c - sqrt(s)) / d
    } else {
        printf "Interception is not possible.\n" > "/dev/stderr"
        exit(0)
    }

    # Interception time
    if (t1 >= t_min && t1 <= t_max && t2 >= t_min && t2 <= t_max) {
        if (t1 < t2) t_I = t1
        else         t_I = t2
    } else
    if (t1 >= t_min && t1 <= t_max)
        t_I = t1
    else
    if (t2 >= t_min && t2 <= t_max)
        t_I = t2
    else {
        printf "Interception will fail.\n" > "/dev/stderr"
        exit(0)
    }

    # Cosine of the launch angle
    c = (x_0 - cos_T * v_0 * (t_0 + t_I)) / (t_I * v_I);
    if (c <= 0.0 || c >= 1.0) {
        printf "Incorrect launch angle.\n" > "/dev/stderr"
        exit(0)
    }

    # Launch angle. awk has no arccos()...
    a = 180.0/3.14159265358979323846 * atan2(sqrt(1-c*c), c)
    printf "Intercept in %.3f seconds using %.1f degree launch angle.\n", t_I, a > "/dev/stderr"

    # Output the trajectories using N samples.
    t = -t_0
    for (i = 0; i <= N; i++) {
        tprev = t
        t = -t_0 + (t_Z + t_0) * i / N

        # Interceptor launch?
        if (tprev < 0 && t >= 0) {
            printf "%.6f", 0
            printf " %.6f %.6f", x_T(0), y_T(0)
            printf " %.6f %.6f", 0, 0
            printf "\n"
        }

        # Intercept?
        if (tprev < t_I && t >= t_I) {
            printf "%.6f", t_I
            printf " %.6f %.6f", x_T(t_I), y_T(t_I)
            printf " %.6f %.6f", x_I(t_I, c), y_I(t_I, c)
            printf " %.6f %.6f", x_I(t_I, c), y_I(t_I, c)
            printf "\n"
        }

        # Time
        printf "%.6f", t

        # Target
        printf " %.6f %.6f", x_T(t), y_T(t)

        # Interceptor projectile?
        if (t >= 0 && t < t_I)
            printf " %.6f %.6f", x_I(t, c), y_I(t, c)

        printf "\n"
    }

    exit(0)
}

When run, it outputs $N = 200$ rows of trajectory information. Each row has format time target_x target_y interceptor_x interceptor_y (with the last two duplicated on the moment of collision). If you have both awk and gnuplot available, you can run

awk -f example.awk > out && gnuplot -p -e 'plot "out" u 2:3 t "Target" w lines lc 1, "out" u 4:5 t "Interceptor" w lines lc 3, "out" u 6:7 notitle w points pt 1 lc -1'

to compute the trajectories, saving them in file out, and plotting it on screen using gnuplot.

Using the current parameters in the file, the collision occurs at time $t = 2.931$, with $27.1$ degree launch angle:

change in the MathJax that results in every MathJax equation being doubled, one being an image and one being a text-based rendering. Usually one of them is hidden, but sometimes the doubles are positioned corner to corner, as shown here. As was already noted, a hard-refresh (Ctrl-F5) usually


Completeness of closed and open balls in a metric space Let $B_{r}(x) = \{y \in X \mid p(x,y) < r\}$ be an open ball and $\bar{B_{r}}(x) = \
[y \in X \mid p(x,y) \leq r\}$ be a closed ball.

Why is it that a closed ball is a complete metric space while an open ball may be an incomplete metric space?

I know a closed interval is complete and it makes sense why. Can we use that fact to show this for the closed ball? Can we also apply the fact that an open interval can be an incomplete metric space as well?

They seem so straightforward that I am having a hard time showing they are true :)

If $(X,d)$ is a complete metric space, and $A \subset X$ is closed, then $(A,d)$ is also a complete metric space: if $(x_n)$ is Cauchy and all points are from $A$, then in particular this is a Cauchy sequence in $X$ (as we use the same metric, and being Cauchy only depends on the distances between members of the sequence) and so has a limit $x$ in $X$. But as $A$ is closed, it must contain $x$ as well (closed sets are closed under limits). SO $d$-Cauchy sequences from $A$ converge in $(A,d)$, so it is a complete metric space.

Now show that a closed ball $\overline{B(x,r)}$ is always closed in any metric space (consider the complement and show it is open).

To see that an open ball need not be a complete metric space (in the same metric), it suffices to give a single example of this. So take $X = \mathbb{R}$, where $d$ is the usual metric $d(x,y) = |x-y|$, and show that $B(0,1)$ is not complete in this metric.

In $\mathbb Q$ with the standard metric the closed balls are of the form $[a,b]\cap\mathbb Q$ and they are NOT COMPLETE is $a<b$.
If $X\ne\varnothing$ and $d$ is the discrete metric (i.e., $d(x,y)=1$, iff $x\ne 0$),
then every open ball in $(X,d)$ is complete.


How to show these groups are isomorphic through finding an isomorphism Show, by finding a suitable isomorphism, that the two groups $(\Bbb Z,+)$ and $(\Bbb Z,∗)$, where $a∗b=a+b−1$, are isomorphic.

The first step should be to take a closer look on the group $(\mathbb{Z},\ast)$. In particular what is the neutral element?
A group homomorphism has to send the neutral element to the neutral element. That is a good starting point.
Turns out the neutral element of $(\mathbb{Z},\ast)$ is $1$.
As $a\ast 1=a+1-1=a$ and $1\ast a=1+a-1=a$.
So $\varphi(0)=1$.
The easiest function that obeys this rule, is $x\mapsto x+1$.
So now we have to show that $\varphi(x)=x+1$ is an isomorphism.
That is a bijection (easiest shown by giving an inverse, but showing injective and surjective should be easy enough too), and that it is a homomorphism.
So we have to show that $\varphi(a+b)=\varphi(a)\ast\varphi(b)$.
I leave the rest to you.


Discussing compactness Consider $\mathbb{Q}$, the set of rational numbers, with the metric  $d(p,q)=|p-q|$.

Then is
$\{q\in \mathbb{Q}\; |\; -2\leq q^2\leq 4\}$
compact?

I guess $q^2$ is effectively lying between 0 and 4. Since it is bounded and closed. So compact. But the answer key says it is not. I wonder how.

I also wonder what the importance of the given metric is.

Is it really compact? Recall that a set in a metric space is compact iff it is complete and totally bounded.

Note that the sequence $q_n=\sum_{k=1}^n\frac{1}{k!}$ belongs to that set and it is a Cauchy sequence.

What is its limit? Is it rational?

The set is not compact because it is not closed. The set has irrational limit points not contained in the set.


Probability of flopping a royal flush I have no stats training, so I am asking if I am attacking this simple statistical problem correctly.

What are the chances of flopping a royal flush in Texas hold’em?

My attempt: The first card dealt to the player must be either a $10,J,Q,K,$ or $A$, any suit.  So the probability of the first card deal would be $\frac{5}{13}$.  Then the second card dealt to the player must be the same suit, and one of the values mentioned above, meaning the probability is $\frac{4}{51}$.  It follows that the next three cards on the flop must be the three remaining cards needed to complete the royal flush, with probabilities: $(\frac{3}{50})(\frac{2}{49})(\frac{1}{48})$.  Therefore I believe the odds of flopping a royal flush is: $\frac{5!}{13\cdot 51\cdot 50\cdot 49\cdot 48}$

I would think that this probability would be independent of how many players are in the game.  The odds of the deck the be stacked in the perfect way according to how many players there are to give you the royal flush on the flop seem to be the same.

Have I made any mistakes?  Thanks in advance.

In the poker game you are describing, you are dealt two cards and there are a further three cards dealt on the "flop".  There is no opportunity to exchange cards.  So really, this is the same as asking the probability of a Royal Flush when you are dealt five cards from a deck of fifty-two.

There are ${52 \choose 5} = 2,598,960$ different equiprobable hands you can be dealt when you are dealt five cards from a standard deck of fifty-two.  There are four different hands that constitute a Royal Flush (i.e., one for each suit).  This means that the probability of a Royal Flush is:

$$\mathbb{P}(\text{Royal Flush on the flop}) = \frac{4}{{52 \choose 5}} = \frac{4}{2,598,960} = \frac{1}{649,740}.$$

You are correct.

$\dfrac {5!} {(13 \cdot 51 \cdot 50 \cdot 49 \cdot 48)} =
\dfrac {1}{13} \cdot \dfrac {5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{51 \cdot 50 \cdot 49 \cdot 48} = \dfrac {52}{13} \dfrac {5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{51 \cdot 50 \cdot 49 \cdot 48} = \dfrac {4}{C(52,5)} = \dfrac {C(4, 1)} {C(52, 5)} = 0.000001539.$

Most of the time, you'll see the probability in terms of the binomial coefficients,
$C(n, k)$.


Simultaneous equations with logarithms? Given that $a$ and $b$ are positive constants and $a>b$, solve the simultaneous equations.

$a+b=13$
$\log_6 a +\log_6 b= 2$

I have tried doing this but I can’t figure out what to do first. I am thinking that you take logs of equation $1$ but I’m not sure.

$\log_6a+\log_6b=\log_6(ab)$

$\log_6x=2$ if and only if $x=6^2=36$

Can you go on now, without peeking at the spoiler?

  \begin{cases}a+b=13\\ab=36\end{cases} has an obvious solution

Hint: Writing
$$\frac{\ln(a)}{\ln(6)}+\frac{\ln(b)}{\ln(6)}=2$$ with the first equation $$b=13-a$$ we get
$$\ln(a)+\ln(13-a)=2\ln(6)$$ Using the law of logarithm we get
$$\ln(a(13-a))=2\ln(6)$$
and $$a(13-a)=e^{2\ln(6)}$$
Can you finish?


Mean and Standard Deviation of samples consisting of all $0$'s or all $1$'s Set of numbers $\ x_1, \ldots, x_m , y_1, \ldots, y_n $ where $\ x_i=0 $ for  $i = 1,\ldots, m$ and $\ y_i=1 $ for $i = 1,\ldots, n$

Show that mean $M$ of this set is given by $\frac{n}{m+n}$ and the standard deviation $S$ by $\frac{ \sqrt{mn}} {m+n} $

I know the definitions of the mean and standard deviation and how to get them but Im really stuck at that question

The sample variance of observations denoted $x_i$ is $$S^2 = \frac{\sum_{i=1}^n (x_i - \bar x)^3}{n-1}
= \frac{\sum_{i=1}^n x_i^2\; -\; (\sum_{i=1}^n x_i)^2/n}{n-1}.$$

The middle member of the equation is the definition and the last member is
sometimes called the 'computational formula' (easily deduced from the definition). Is that the definition of $S$ you're using?

Notice that for $x_i \equiv 0,$ one has $\sum_{i-1}^m x_i = \sum_{i-1}^m x_i^2 = 0.$

Also,  for $y_i \equiv 1,$ one has $\sum_{i-1}^n y_i = \sum_{i-1}^n y_i^2 = n.$

Put it all together, and you should be off to a good start--provided I have
properly interpreted a somewhat vague question.

Calculate the zero, first and second moments
\begin{eqnarray*}
\sum 1 = ? \\
\sum z_j = ? \\
\sum z_j^2 = ? \\
\end{eqnarray*}
Then use the formulea
\begin{eqnarray*}
\mu&=&\frac{\sum z}{\sum 1 } \\
\sigma^2 &=& \frac{\sum z^2}{\sum 1 }- \frac{(\sum z)^2}{(\sum 1)^2 }.
\end{eqnarray*}

   $\sum 1 = n+m \,\,\, \sum z_j = n \,\,\,  \sum z_j^2 = n $


How to simplify ${(1+2i)}^6$? How to simplify ${(1+2i)}^6$ using De Moivre's formula?
I have found that $r=\sqrt 5$ and $\tan x=2$ but I can't find the exact value of $x$.

Hint You have that $\tan(\theta) = 2$ and you need $\cos(6\theta)$ and $\sin(6\theta).$ Drawing a triangle in the first quadrant you can see that if $\tan(\theta) = 2$ you have $\cos(\theta) = 1/\sqrt{5}$ and $\sin(\theta) = 2/\sqrt{5}.$ Now can you find $\sin(6\theta)$ and $\cos(6\theta)$?

As others have said, it might be easier to use a hybrid approach where you square or cube first and then do the rest with trig.

Using De Moivre for this computation doesn't make much sense.

Either you use it numerically and evaluate $\sqrt5^6e^{i6\arctan2}=125(0.936+i0.352)$, or you establish the analytical formulas for $\cos6x,\sin6x$. In both cases, the computation is more sophisticated than direct evaluation by the binomial formula

$$(1+i2)^6=1+i6\cdot2-15\cdot4-i20\cdot8+15\cdot16+i6\cdot32-64=117+i44$$
or, faster,
$$(1+i2)^6=(-3+4i)^3=-3^3+i3\cdot3^2\cdot4+3\cdot3\cdot4^2-i4^3=117+i44.$$


Parabola is an ellipse, but with one focal point at infinity While I was reading about conic sections, I came across the following statement:

  A parabola is an ellipse, but with one focal point at infinity.

But it is not clear to me. Can someone explain it clearly?

The equation for an ellipse with a focus at $(0,0)$ and the other at $(0,2ae)$ keeping $a(1-e)=f$ (where $f$ is distance from the vertex to the focus of the ellipse, which ends up being the focal length of the parabola) is
$$
\frac{x^2}{a^2(1-e^2)}+\frac{(y-ae)^2}{a^2}=1
$$
which is equivalent to
$$
\frac{x^2}{f(1+e)}+\frac{y^2-2aey}{a}=f(1+e)
$$
If we let $a\to\infty$ (and therefore $e=1-\frac fa\to1$), we get
$$
y=\frac{x^2}{4f}-f
$$
which is a parabola.

$\hspace{3.4cm}$

Think of a pair of cones with vertical axes and identical cone-angles, one facing up and one facing down. Place them so they're touching at one point.  I'll call this a "cone" for now.

If you slice this object with a plane perpendicular to the axis, you'll get a circle of some radius, or perhaps a single point, which you could call a circle of radius 0.

If you slice it with a slightly tilted plane, you'll get an ellipse (or a single point). Thus circules and ellipses are both "cross-sections" of a cone, or "conic sections".

If you tilt the slicing plane further, so it's nearly vertical, it'll intersect both cones, resulting in a hyperbola, or, if the plane passes through the cone-point, a pair of intersecting lines. So intersecting lines are kind of a "limit" of hyperbolas.

Go back to ellipses: place the plane to slice an ellipse from the upper half of the cone.  Tilt your slicing plane more and more, making the ellipse more and more eccentric. There's a last possible amount of tilt before you start also slicing into the bottom part of the cone. At that tilt, the intersection is no longer an ellipse, but instead a parabola.

So it's reasonable to say that a parabola is a limit of ellipses.

Of course, if you tilt a tiny bit more, you start getting hyperbolas, so it's also reasonable to say a parabola is a limit of hyperbolas.


Limit at infinity of $\frac{1}{x^{n-1}}$? We have $\lim_{x\rightarrow\infty}\frac{1}{x^n}=0$

But why is \begin{align} \lim_{x\rightarrow\infty}\frac{1}{x^{n-1}}=\lim_{x\rightarrow\infty}\frac{1}{x^n x^{-1}} =\lim_{x\rightarrow\infty}\frac{1}{\frac{x^n}{x}}=\lim_{x\rightarrow\infty}\frac{x}{x^n} \end{align} indeterminate?

Suppose I have the absolute value, is the limit still indeterminate? \begin{align} \lim_{x\rightarrow\infty}\Bigg\lvert \frac{1}{x^{n-1}}\Bigg\rvert &=\lim_{x\rightarrow\infty}\Bigg\lvert \frac{1}{x^n x^{-1}}\Bigg\rvert \\ &=\lim_{x\rightarrow\infty}\Bigg\lvert \frac{1}{\frac{x^n}{x}}\Bigg\rvert =\lim_{x\rightarrow\infty}\Bigg\lvert \frac{x}{x^n}\Bigg\rvert \end{align}

Is your question about $\displaystyle\lim_{x\to +\infty} \frac x {x^n}$ being zero, or about $\displaystyle\frac \infty \infty$ being zero? The first one is true (for $n > 1$), the other one is indeterminate. Because for instance $\displaystyle\lim_{x\to +\infty} \frac {x^{2n}} {x^n} = +\infty$ is also of the form $\displaystyle\frac \infty \infty$.

The point is that $\displaystyle\frac \infty \infty$ is just a notation for a type of limits and not an equality with the previous term.

Indeterminate simply means "we don't know". It can be any number or even infinity. For example

$$\lim_{x\to+\infty} \frac{ax+1}{x}=\lim_{x\to+\infty} a+\frac{1}{x}=a$$

$$\lim_{x\to+\infty} \frac{ax^2+1}{x}=\lim_{x\to+\infty} ax+\frac{1}{x}=+\infty$$

Note that all these limite are indeterminate limites of the forms $\frac{+\infty}{+\infty}$.


Continuous mapping from $[0,1]$ to $[0,1]^2$ As explained in this answer, it is possible to create a bijection from $[0,1]\rightarrow[0,1]^2$. However, the example provided is clearly not continuous. It seems either very complicated or impossible to create a continuous bijection between the unit interval and the unit square. Does this mapping exist? If so, what does it look like? If not, how does one prove that there does not exist a continuous bijective mapping between $[0,1]\rightarrow[0,1]^2$?

No such bijection exists: it is a standard result that any continuous bijection from a compact space (such as $[0,1]$) onto a Hausdorff space is a homeomorphism, and there's no homeomorphism from the unit interval to the unit square, since removing an interior point will make the first one disconnected, while leaving the latter connected.

Another way to see that this is not possible is to try to draw a continuous curve starting from origin that covers all the area of square without overlapping the previous path. The best you could do is to travel around the perimeter and once you approach origin after a round, no matter how close you go to origin and start the same process, there will be a few points between starting point of second round and the origin.


Is every non-square integer a primitive root modulo some odd prime? This question often comes in my mind when doing exercices in elementary number theory:

  Is every non-square integer a primitive root modulo some odd prime?

This would make many exercices much easier. Unfortunately I seem unable to discover anything interesting which may lead to an answer.

It seems likely to me that this is true. If $n\equiv2\pmod3$ then it's a primitive root modulo $3$. If $n\equiv2,3\pmod5$, it's a primitive root modulo $5$. If we would continue like this, my guess is that any non-square $n$ will satify at least one of these congruences.

This being difficult, I began considering a simplified question:

  Is every non-square integer a quadratic non-residue modulo some prime?

Or equivalently,

  If an integer is a square modulo every prime, then is it a square itself?

The second form seems easier to approach, however I still can't find anything helpful.

Let $n$ be a non-square. Write $n=a^2b$ with $b\ne 1$ square-free.
Write $b=p_1\cdot\ldots\cdot p_m$ as product of disctinct primes with $m\ge 1$.
For primes $q> n$ the factor $a^2$ can be ignored, so we have that $\left( \frac bq\right)=1$ for almost all primes $q$.
According to quadratic reciprocity law, $ \left(\frac{b}{q}\right)$ is determined by $q\bmod 8b$. Also, there is at least one residue $d\bmod 8b$ for which $q\equiv d\pmod{8b}$ implies $ \left(\frac{b}{q}\right)=-1$ (e.g. ensure $\left(\frac d{p_1}\right)=-1$ and $\left(\frac d{p_i}\right)=+1$ for all other $i$ and use the chinese remainder theorem). Especially, $d$ is relatively prime to $8b$ so that by Dirichlet there exist infinitely many primes $q$ with $q\equiv d\pmod{8b}$. For such a $q$ with $q>n$ we conclude that $n$ is not a square modulo $q$.

Your "equivalent" question seems trivial and therefore not equivalent.

"If an integer is a square modulo every prime, then is it a square itself?"

Given an integer n, there is a prime p > n. If n is a square modulo every prime, including p, then n = n (modulo p) is a square, so n is a square.

BTW. I wrote a short program that checked whether every non-square integer n <= N is the primitive root modulo some prime p > N, with N = 40 x 10^9. It found that actually every non-square n <= N is a primitive root modulo some prime p with N < p <= N + 1357, which is just the 55th prime > N. This seems to indicate that it is more promising to look for a proof than for a counterexample.


Solving matrix by hand - Gaussian-Jordan I am trying to solve the following matrix. I tried to use Gaussian-Jordan elimination, but I can figure it out. Can some one show me how to find i1, i7 and i8. I can solve this matrix with help from a math software, but I want to do it by hand.
Also I tried to solve the matrix equations by substitution, but this is very laborious.

$$
\left[ \begin {array}{ccc} r_{g}+r_{b}&-r_{b}&0
\\r_{b}&r_{{\it a1}}&r_{{\it a2}}\\
r_{g}&r_{{\it a1}}+
r_{{\it gr}}&-r_{{\it gr}}-r_{j}\end {array} \right]*
\left[ \begin {array}{ccc} i_{1}
\\i_{7}\\
i_{8}\end {array} \right]=\left[ \begin {array}{ccc} 4u
\\4u\\
u\end {array} \right]
$$

The software did not give an echelon form. I give the three equations as input and then used the solve commando. If the Gaussian-Jordan method is not possible, are the only ways to solve this matrix by usage of substitution or Cramer?

$A \cdot X=B$ can be solved (if A is invertible) as $X=A^{-1}\cdot B$

So the fastest way to solve this equation is to find $A^{-1}$and to multiply by B.

Finding an echelon form for a matrix with variable entries can be problematic, because the echelon form depends on whether this or that equals zero. To take a simpler example, one really cannot say what the reduced echelon form for the matrix $$\left[\begin{matrix}a&b\\c&d\end{matrix}\right]$$is; it could be any $2\times2$ reduced-echelon-form matrix. The software that gave you an echelon form for your matrix must be making some unwarranted assumptions.


Is mathematics one big tautology? Is mathematics one big tautology? Let me put the question in clearer terms:

Mathematics is a deductive system; it works by starting with arbitrary axioms, and deriving therefrom "new" properties through the process of deduction. As such, it would seem that we are simply creating a string of equivalences; each property can be traced back logically to the axioms. It must be that way, that's how deductive systems work!

If that be the case, then in what sense are we introducing novel or new ideas? It would seem that everything is simply equivalent to the fundamental set of axioms that we choose to start with. Is there a precise step in mathematical derivation that we can isolate as going beyond pure logic? If so, then how does this fit in with the fact that mathematics is deductive? Must we change our view of mathematics as purely deductive? And if not, then is there a way to reconcile the feeling of creativity in mathematics with the fact that it boils down to pure logic?

I'm trying to figure out the true nature of what's going on here.

Disclaimer: different people view this differently. I side with Lakatos: Logic is a tool. Proofs are a way to verify one's intuition (and in many cases to improve one's intuition) and it is a tool to check the consistency of theories in a process of refining the axioms. The fact that every proof boils down to a tautology is true but irrelevant to mathematics.

Here is an isomorphic question to the question you posed: A painting is just blobs of paint of different colour on canvas. So, are we to deduce from this fact that the art of painting is reduced to just placing paint on canvas? Technically, the answer is yes. But the painter does much more than that. In fact, it is clear that while the painter must possess quite a large amount of skill in placing paint on canvas, this skill is the least relevant (while absolutely necessary) for the creative process of painting.

So it is with mathematics. Being able to prove is essential, but is the least relevant skill for doing mathematics. In mathematics we don't deduce things from axioms. Rather we try to capture a certain idea by introducing axioms, check which theorems follow from the axioms and compare these results against the idea we are trying to capture. If the results agree we are happy. If the results disagree, we change the axioms. The ideas we try to capture transcend the deductive system. The deductive system is there to help us find consequences from the axioms, but it does not tell us how to gauge the validity of results against the idea we try to capture, nor how to adjust the axioms.

This is my personal point of view of what mathematics is (or at least what a sizable portion of it is). It is very close to what physics is. Physics is not just some theories about matter and its interactions with stuff. Rather it is trying to model reality. So does mathematics, it's just not entirely clear which reality it is trying to model.

Mathematicians seem to be persons that:

study mathematics;
pick up new ideas and formulate consistent theories that capture the essence of these ideas;
formulate and examine statements and try to prove them or find counterexamples;
publish their discoveries and results to be reviewed and used by other mathematicians;
teach mathematics to others.

I conclude that mathematics is a process of accumulation of knowledge about mathematical theories on an increasing number of topics.

About the relation between mathematics and tautologies, Kurt Gödel said (Jean van Heijenoort, 1967, From Frege to Gödel s. 601):

  "The class of provable formulas is defined to be the smallest class of
  formulas that contains the axioms and is closed under the relation
  'immediate consequence', i.e. formula $c$ of $a$ and $b$ is defined as
  an immediate consequence in terms of modus ponens or substitution."

Which in my opinion makes it clear why proved theorems doesn't have counterexamples.

Logic is the trivial part of a mathematical proof and that's the tricky part which is mathematics.


Formal Definition of the Divergence of a Vector Field I recently completed an introductory course on multivariate calculus, and I'm still trying to come to grips with the concepts taught in the vector calculus segment. Right now, I'm reviewing the concept of divergence.

I understand the lexical definition of divergence, that (in $\mathbb{R}^3$ at least) it's the volumetric density of the outward flux of a vector field. The formulaic definition that Wikipedia and Wolfram offer makes similar sense:
$$\mathrm{div}\,\mathbf{F}=\lim_{V \to 0}{\iint_S{\frac{\mathbf{F}\cdot\mathbf{n}}{V}dS}}$$

What I don't understand, however, is how
$\mathrm{div}\,\mathbf{F}=\nabla\cdot\mathbf{F}$ follows from the above statement. Something like $\mathrm{div}\,\mathbf{F}=|\nabla\mathbf{F}|$ seems to make more sense to me, because $\mathrm{div}\,\mathbf{F}=\nabla\cdot\mathbf{F}$ just adds up the partials in what are essentially three random directions ($\mathbf{i},\,\mathbf{j},\mathbf{k}$, after all, are the conventional basis vectors for $\mathbb{R}^3$), whereas $\mathrm{div}\,\mathbf{F}=|\nabla\mathbf{F}|$ isn't as... arbitrary? I suppose.

In short: how do you proceed from the formal definition of divergence as the limit of flux over volume as volume goes to zero to getting that divergence is equal to the dot product of nabla and the vector field?

Let $V$ be a cubical volume of edge length $\ell$.  As $\ell$ is small, $F$ is approximately constant over each of the faces (note we do not say it is constant over the box).

Let $r$ be the center of the box.  The divergence is then

$$\mathrm{div} F = \lim_{\ell \to 0} \frac{1}{\ell^3} [F(r + \ell \hat x/2) \ell^2 - F(r - \ell \hat x/2)\ell^2 + \text{terms like this for $y$, $z$}]$$

A simple rearrangement shows that you get exactly the definition of $\partial F/\partial x$ from the term I've written out, and similarly for the other terms.  This is all obtained by evaluating the integral for a particularly chosen geometry of the volume.

As the logic is much the same for the curl, I'd say more this is a way of defining $\nabla$ and how it works, what its expression in various coordinate systems must be.  In fact, this construction is used to derive $\nabla$ and its coordinate expression for more general manifolds.  It's not that $\nabla$ is defined out of the blue and we show that divergence is related to it.  Rather, we found divergence and curl and realized that the geometric arguments used lent themselves to relating them both to a vector differential operator.

You can use Gauss(Divergence) theorem:

$\mathrm{div}\,\mathbf{F}=\lim_{V \to 0}{\iint_S{\frac{\mathbf{F}\cdot\mathbf{n}}{V}dS}} = \lim_{V \to 0}{\iiint_V{\frac{\nabla\cdot\mathbf{F}}{V}dV}} = \nabla\cdot\mathbf{F}$

Second equality follows from Divergence theorem and third equality is simple exercise(assuming $\nabla\cdot\mathbf{F}$ is continuous).


Modified version of $\log_2(x)$ Show that the function $f(x) = \begin{cases}\log_2(x),&\text{ if $x$ is a power of $2$}\\
\text{undefined},&\text{ otherwise}\end{cases}$ is recursive but not primitive recursive.

I'm not sure how to show this. To show it is not primitive recursive, I think I need to show that it must be defined using minimization (not bounded); primitive recursive functions are the successor function, zero functions, projection functions, and any compositions of the above functions or functions $f$ such that there exist primitive recursive functions $g$ and $h$ so that $f(x, 0)  = g(x)$ and $f(x,n+1) = h(x,n,f(n))$. I know how to show that the function $f(x) = \lfloor \log_2(x+1) \rfloor$ is primitive recursive. I think I need to define a recursive predicate to determine whether $x$ is a power of $2$ (including negative powers). That's easy because that's just repeated division and addition. However, I'm not sure how to show this is not primitive recursive.

This is a rather silly situation: the only reason $f$ is not primitive recursive is that $f$ is sometimes undefined. All primitive recursive functions are defined everywhere, basically by definition.
In particular, if we modify the definition of $f$ so that $f(x)=17$ (say) whenever $x$ is not a power of $2$, we do indeed get a primitive recursive function.

Clearly, $f(x)$ can be computed recursively by repeated multiplications or divisions, as you said.  There is a computable bound to the number of multiplies or divisions, one can use $x$ for $x >= 1$ and $1/x$ for $x<1$.  The upper bound does not need to be a close one to fit the requirement, it just has to be an upper bound.
This implies that $f(x)$ is primitive recursive.


Second order ODE using method of undetermined coefﬁcients Question:
$$y''-2y'+2y=4e^x\sin x$$

Clearly the complementary function here is $(A+Bx)e^{2x}$ where $A$ and $B$ are constant. However, I am not sure for the Particular function here. Should the Particular function be $(C+Dx)x^2e^{2x}$?

Many Thanks.

Normally, using the method of undetermined coefficients you would be looking for a particular solution $y_p= (e^x)*(Acox(x)+Bsin(x))$ for the diff-equation you posted. Now when you have a product of sinusoids and an exponential as your "template" solution you have to check if $r+\mu*i$ is a solution to the characteristic equation. This comes about because the general form for the particular solution I presented is $(e^{rx})*(cos(\mu x)+sin(\mu x)) $ where this solution came about as linear combination of $e^{r+i *\mu}$. You should be able to convince yourself that this just makes the diff-eqn zero.

Thus we have a problem is if $r+\mu*i$ is a solution to the characteristic equation. Because, if $r+\mu i$ is a solution, then the particular solution I posted is a solution to homogeneous equation. So to stop this we multiply by the whole thing by $x$ or by $(Gx+H)$, which makes no difference, but I like just multiplying it by $x$. So the particular solution is $y_p = (x)*(e^x)*(Acos(x)+Bsin(x))$.

Now to explain some more things, if you have $y"+9y=x^2+cos(2x)$ as you asked. The particular solution would be $(Ax^2+Bx+C)+(Dcos(2x)+Esin(2x))$. As a result of the linear superposition principle. But let's make it more interesting, what if instead of y"+9y=... you had y'''+y"+9y'=... Now we have a problem with the polynomial part of the particular solution. we would then multiple just the polynomial part by $x$,so
$y_p$ becomes $x(Ax^2+Bx+C)+$... (the same sinusoids). Of course replacing $x(Ax^2+Bx+C)$ with $(Ax^3+Bx^2+Cx+F)$ makes no difference, but I just feel the first way is more of a generalization.

Hope this helps

First observe that $ y''_H - 2y'_h+2y_h = 0$ ($y$ homogenoeus) have solution $ y_h =e^{x}(c_1 \sin(x) + c_2 \cos(x)) $ and $y_p$ (y particular) should be linear independent of $y_h$. So that you are looking for solutions of the form:

$$ (C+Dx)\cdot e^{x} \cdot( \sin(x) + cos(x)) $$


What is probability that at least $2$ people have same birthday from group of $N$ people? Question is not that simple.

There are also leap year included.Leap year will be $366$ days and normal year will be $365$ days.

There is a statement in question that : there are exactly  $\lfloor{\dfrac{n}{4}}\rfloor$ people who born in leap year.

Now we have to find probability of at least $2$ people having same birthday from group of $N$ people .

Edit : sorry i forgot a word to add!

Any help or hints and calculation and edit in questions are appreciated. Thanks..

Let's rephrase the question.

  Find the probability that any $2$ people have their birthday in the same day of the year in a group $4n$ people, of which $n$ born were in the leap year and $3n$ people were born in the regular year.

The complement of the event "any 2 people have their birthday" is "all people have birthday on different days of the year".

Consider two mutually exclusive possibilites: $\mathcal{A}_1$ - one of these birthdays falls on "Feb. 29", and $\mathcal{A}_2$ -- none of these birthdays fall on the leap day.

The number of configurations of the second kind is
$$
   N\left(\mathcal{A}_2\right) = \prod_{k=1}^{4 n} \left( 365 - (k-1) \right)
$$
and the number of configurations of the first kind is
$$
   N\left(\mathcal{A}_1\right) = n \prod_{k=1}^{4 n-1} \left( 365 - (k-1) \right)
$$
The total configuration size is $N_T=366^n 365^{3n}$, hence the probability of having no birthday duplicates is

$$
   p_c = \frac{ N\left(\mathcal{A}_1\right) +  N\left(\mathcal{A}_2\right)}{N_T}
$$

Here is a confirmation in Mathematica:

(* computed probability expression *)
p[n_Integer] = (Product[365 - (k - 1), {k, 1, 4 n}] +
     n Product[365 - (k - 1), {k, 1, 4 n - 1}])/(366^n 365^(3 n));

(* simulation result *)
In[45]:= With[{n=7,reps=10^8},
   Sum[Boole[DuplicateFreeQ[Join[
       RandomChoice[Range[1,366],n],
       RandomChoice[Range[1,365],3n]]]],{k,1,reps}]/reps]//N

Out[45]= 0.34599694

In[53]:= N[p[7]]

Out[53]= 0.3460041650022472

(* verify that computed value belongs to the sampling
    confidence interval *)
In[55]:= With[{pSample = Out[45], pExpected = p[7]},
 IntervalMemberQ[
  Interval[pSample + {-1, 1} Sqrt[(pSample (1 - pSample))/10^8]],
  pExpected]]

Out[55]= True

Suppose we find the negation.
$p'(n)$ - is the probability when there is no two person with equal birthday.
$$p'(n) = (1-\frac{1}{365})(1 - \frac{2}{365})\cdots (1-\frac{n}{365}) = \frac{365!}{365^{n}(365-n)!}$$ (suppose we choose we choose one person from group, then choose another one with different birthday and continue this before n)

So the answer is: $$1-p'(n)$$ and it's ~$\frac{1}{2}$, when $n=23$

UPD : the probability of choosing person with leap year birthday is $s(n) = \frac{\lfloor  {\frac{N}{4}}\rfloor }{N}$ so the total probability will be :
$$(1-s(n))p_{365}(N-\lfloor \frac{N}{4}\rfloor) + s(n)p_{366}(\lfloor \frac{N}{4}\rfloor)$$ where $p_{365}(n)$ is probability for normal year


Quantum group notation I was jumping into the deep end and reading a few papers and lectures on quantum groups. My knowledge on Lie algebras is a bit thin but I was just wondering the notation used in the starting of this document:

  One should have $U_\hbar(g) = U(g)[[\hbar]]$ as a vector space [...]

My question is, is the $\hbar$ the Cartan subalgebra $\mathfrak{h}$ I have been seeing in other papers, and what does the $U(g)[[\hbar]]$ mean precisely? Usually I see the quantum group notated as $U_q(g)$, so the sudden change in notation concerned me. Thanks in advance!

In this case $\hbar$ is just a parameter. Frequently, one also denotes $U_{\hbar}(g)$ as $U_q(g)$.

$U(g)[[\hbar]]$ denotes the space of formal power series over $U(g)$, i.e. if $f\in U(g)[[\hbar]]$, then $$f=\sum_{n\in\mathbb{Z}}a_n \hbar^n$$ with $a_n\in U(g)$ for every $n$. (Sometimes people reindex so that the sum is over $\mathbb{N}$ or some other countable set.)

Usually $\hbar$ is simply the reduced Planck constant $h/(2\pi)$.

I have sadly no idea why it is featured such prominently in your context.


For two p.d. matrices $A$ and $B$, prove that $\lambda_1(AB)\leqslant \lambda_1(A) \cdot\lambda_1(B)$ If $A$ and $B$ are two nxn positive definite matrices, then show that

$$\lambda_1(AB) \leqslant \lambda_1(A) \cdot \lambda_1(B),$$

where $\lambda_1(\cdot)$ denotes the largest eigenvalue.

Let $C$ be a positive symmetric $n\times n$ matrix, and let $|\cdot|$ be the standard euclidean norm on $\Bbb R^n$. For all $X\in\Bbb R^n$,
$$|CX|\leq\lambda_1(C)|X|$$
with equality iff $X$ is eigenvector of $C$ with eigenvalue $\lambda_1(C)$, so that, again, for $C$ positive symmetric,
$$\lambda_1(C)=\max_{|X|\leq 1}|CX|=\|C\|$$ where $\|\cdot\|$ is the matrix norm subordinate to the standard euclidean norm on $\Bbb R^n$.
The answer to your question now follows, as for any matrix $D$ and any eigenvalue $\lambda$,  $|\lambda|\leq \|D\|$ so that
$$|\lambda_1(AB)|\leq\|AB\|\leq\|A\|\cdot\|B\|=\lambda_1(A)\lambda_1(B)\,.$$

Let the eigen values of $AB$ be $c_1, c_2, \ldots,c_n$ and those of $A$ and $B$ be $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ respectively.
Now, we know that, $\det(A)$ is the product of the eigenvalues of $A$, counted with their algebraic multiplicities.

Also, we have, $\det(AB) = \det(A)\det(B)$

$$
\Rightarrow c_1 c_2 \cdots c_n = (a_1 a_2 \cdots a_n)(b_1 b_2 \cdots b_n)
$$

=> $c_1$ * $c_2$ * ... * $c_n$ = ($a_1$ * $a_2$ * ... * $a_n$) * ($b_1$ * $b_2$ * ... * $b_n$ )

Using the above equality/identity(since all eigenvalues are positive, $A$ & $B$ being p.d.), we can evidently write:

$$\lambda_1(AB) \le \lambda_1(A) \lambda_1(B).$$


Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ I am trying to find a closed form for

$$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$

It seems that the answer is
$$\frac{\pi^2}{12}\left( 1-\sqrt{3}\right)+\log(2) \log \left(1+\sqrt{3} \right)$$

Mathematica is unable to give a closed form for the indefinite integral.

How can we prove this result? Please help me.

EDIT

Apart from this result, the following equalities are also known to exist:
$$\begin{align*}
\int_0^1 \frac{\log \left( 1+x^{4+\sqrt{15}}\right)}{1+x}\mathrm dx &=\frac{\pi^2}{12} \left( 2-\sqrt{15}\right)+\log \left( \frac{1+\sqrt{5}}{2}\right)\log \left(2+\sqrt{3} \right) \\ &\quad +\log(2)\log\left( \sqrt{3}+\sqrt{5}\right)
\\ \int_0^1 \frac{\log \left( 1+x^{6+\sqrt{35}}\right)}{1+x}\mathrm dx &= \frac{\pi^2}{12} \left( 3-\sqrt{35}\right)+\log \left(\frac{1+\sqrt{5}}{2} \right)\log \left(8+3\sqrt{7} \right) \\
&\quad +\log(2) \log \left( \sqrt{5}+\sqrt{7}\right)
\end{align*}$$

Please take a look here.

$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$I will evaluate the sum at the end of Jim Belk's post.

There is clearly a lot more going on here; I hope a number theorist will come and explain what is really going on.
I will be sloppy about convergence issues.

Set
$$S:= \sum_{k,n=1}^{\infty} \frac{(-1)^{k+n}}{k^2+4kn+n^2}.$$
We set $m = k+2n$, in order to diagonalize the quadratic form:
$$S = \sum_{0 < 2n < m} \frac{(-1)^{m-n}}{m^2 - 3 n^2}.$$
It turns out to be more elegant to work with
$$T : = \sum_{0 \leq 2n < m} \frac{(-1)^{m-n}}{m^2 - 3 n^2} = S + \sum_{m=1}^{\infty} \frac{(-1)^m}{m^2} = S - \frac{\pi^2}{12}.$$

Let $R$ be the ring $\ZZ[\sqrt{3}]$. Let $D \subset R$ be $\{ m+n\sqrt{3} :  0 \leq 2n < m \}$. So, simply as a matter of formal rewriting
$$T = \sum_{m+n \sqrt{3} \in D} \frac{(-1)^{m+n}}{m^2-3 n^2}.$$

We will now try to interpret each of these quantities in terms of the ring $R$. Recall that, for $m+n \sqrt{3} \in R$, the norm $N(m+n \sqrt{3})$ is $m^2 - 3 n^2$.
Define $R_{+} = \{ a \in R : N(a) > 0\}$ and $R_{-} = \{ a \in R : N(a) < 0 \}$.

For $m+n \sqrt{3} \in R$, define $\sigma_2(m+n \sqrt{3}) = (-1)^{m+n}$.

Let $\Gamma$ denote the unit group of $R$. Explicitly, $\Gamma$ is $\{ \pm 1 \} \times (2+\sqrt{3})^{\ZZ}$.

Note that multiplication by $\Gamma$ takes $R_{+}$ and $R_{-}$ to themselves. Here is the first miracle: $D$ is a fundamental domain for the action of $\Gamma$ on $R_{+}$! For example, multiplication by $2+\sqrt{3}$ maps the ray $\RR_{\geq 0} (1+0 \sqrt{3})$ bounding one side of $D$ to the ray $\RR_{\geq 0} (2+\sqrt{3})$ on the other side. Moreover, multiplication by units leaves $N( \ )$ and $\sigma_2(\ )$ unchanged.
So we can view the sum as
$$T = \sum_{a \in R_{+}/\Gamma} \frac{\sigma_2(a)}{N(a)}$$
where the sum means to pick one representative for each orbit of the $\Gamma$ action.

I'd rather sum over $R$ than $R_{+}$. Define $\chi_{\infty}$ to be $\pm 1$ on $R_{\pm}$. So we can rewrite
$$ T = \frac{1}{2} \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a) \left( 1+\chi_{\infty}(a) \right)}{|N(a)|} = \frac{1}{2}\left(  \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a)}{|N(a)|} + \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a) \chi_{\infty}(a)}{|N(a)|}  \right) = : \frac{1}{2} (U+V).$$

Now, $a$ and $b$ in $R$ are in the same $\Gamma$ orbit if and only if the ideals $(a)$ and $(b)$ are equal. So the above sums are running over all principal ideals of $R$.
Moreover, $R$ is a PID! And, finally, for a principal ideal $I = (a)$, we have $N(I) = |N(a)|$. So we can write:
$$U = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I)}{N(I)} \quad V = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I) \chi_{\infty}(I)}{N(I)} .$$
We set
$$U(s) = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I)}{N(I)^s} \quad V(s) = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I) \chi_{\infty}(I)}{N(I)^s} .$$

Note that $\sigma_2(a)$ is $1$ if and only if $1+\sqrt{3}$ divides $a$.
So we have the Euler factorization
$$U(s) = \left(-1 + 2^{-s} + 2^{-2s} + 2^{-3s} +  \cdots \right) \prod_{\pi \neq (1+\sqrt{3})} \frac{1}{1-N(\pi)^{-s}} =$$
$$\frac{-1+2^{-s}+2^{-2s}+ 2^{-3s} + \cdots}{1+2^{-s}+2^{-2s}+2^{-3s}+\cdots} \prod_{\pi}  \frac{1}{1-N(\pi)^{-s}} =(-1+2\cdot 2^{-s}) \zeta_R(s).$$
where $\pi$ runs over prime ideals of $R$.

So
$$\lim_{s \to 1^{+}} U(s) = \lim_{s \to 1^{+}} \frac{-1+2 \cdot 2^{-s}}{s-1} \lim_{s \to 1^{+}} (s-1) \zeta_R(s) = - \log 2  \lim_{s \to 1^{+}} (s-1) \zeta_R(s).$$
From the class number formula, this last limit is
$$\frac{2^2 \log (2+\sqrt{3})}{2 \cdot \sqrt{12}}.$$
So
$$U(1) = - \frac{\log 2 \log(2+\sqrt{3})}{\sqrt{3}}.$$

We now run the same trick with $V$. Again, we start with the Euler product:
$$V(s) =  \left(-1 - 2^{-s} + 2^{-2s} - 2^{-3s} + 2^{-4s} -  \cdots \right) \prod_{\pi \neq (1+\sqrt{3})} \frac{1}{1-\chi_{\infty}(\pi) N(\pi)^{-s}} =$$
$$\frac{-1 - 2^{-s} + 2^{-2s} - 2^{-3s} +2^{-4s} -  \cdots}{1-2^{-s} + 2^{-2s} - 2^{-3s} + 2^{-4s} -  \cdots} \prod_{\pi} \frac{1}{1-\chi_{\infty}(\pi) N(\pi)^{-s}} =
\left( -1 - 2^{1-s} \right)   L(s, \chi_{\infty}).$$

So
$$V = V(1) = - 2 L(1, \chi_{\infty}).$$

We now must evaluate $L(1, \chi_{\infty})$. This $L$-function is defined by a sum over the ideals of $R$; we will rewrite it in terms of $L$-functions for $\ZZ$.
Let $p \geq 5$ be prime. I claim that $p$ splits in $R$ if and only if $p \equiv \pm 1 \bmod 12$. For such a $p$, if we write $p = \pi \bar{\pi}$, I claim that $N(\pi) = N(\bar{\pi}) = p$ if $p \equiv 1 \bmod 12$ and $N(\pi) = N(\bar{\pi}) = -p$ if $p \equiv -1 \bmod 12$. Proof Sketch: The prime $p$ splits in $R$ if and only if $\left( \frac{3}{p} \right) = 1$, which is easily computed to be equivalent to $p \equiv \pm 1 \bmod 12$. Since $R$ is a PID, we can write such a $p$ as $\pi \bar{\pi}$ for $\pi$ and $\bar{\pi} \in R$. Then $N(\pi) = N(\bar{\pi}) = \pm p$. But, for $a=m+n \sqrt{3} \in R$, we have $N(a) = m^2-3 n^2 \not \equiv -1 \mod 3$. So only one of the two possibilities for $\pm p$ can occur. $\square$

So $L(1, \chi_{\infty})$ is
$$\left(1+2^{-1} \right)^{-1} \left( 1+3^{-1} \right)^{-1} \prod_{p \equiv 1 \bmod 12} \left( 1- p^{-1} \right)^{-2} \prod_{p \equiv -1 \bmod 12}  \left( 1+ p^{-1} \right)^{-2} \prod_{p \equiv \pm 5 \bmod 12} \left( 1- p^{-2} \right)^{-1}.$$

Define $\chi_4(n)$ to be $0$ if $n$ is even, $1$ if $n \equiv 1 \bmod 4$ and $-1$ is $n \equiv -1 \bmod 4$. Define $\chi_3(n)$ to be $1$, $-1$ or $0$ according to whether $n \equiv 1$, $2$ or $0 \bmod 3$. Then we have
$$L(1, \chi_{\infty}) = \prod_p \left( 1- \chi_4(p) p^{-1} \right)^{-1}  \prod_p \left( 1-\chi_3(p) p^{-1} \right)^{-1} = \left( \sum_{n=1}^{\infty} \frac{\chi_4(n)}{n} \right)  \left( \sum_{n=1}^{\infty} \frac{\chi_3(n)}{n} \right) .$$
The sum $\sum \chi_4(n) = 1-1/3+1/5-1/7 + \cdots$ is well known to be $\pi/4$. The sum $\sum \chi_3(n)/n$ is only slightly less well known; it is $\pi/(3 \sqrt{3})$.

So I get $L(1,\chi_{\infty}) = \frac{\pi}{4} \frac{\pi}{3 \sqrt{3}} = \frac{\pi^2}{12 \sqrt{3}}$ and $V = - \frac{\pi^2}{6 \sqrt{3}}$.

Putting it all together,
$$T = \frac{1}{2} \left( - \frac{\log 2 \log(2+\sqrt{3})}{\sqrt{3}}  - \frac{\pi^2}{6 \sqrt{3}} \right) = - \frac{\log 2 \log(2+\sqrt{3})}{2 \sqrt{3}} -  \frac{\pi^2}{12 \sqrt{3}} $$
$$S = \frac{\pi^2}{12}  - \frac{\log 2 \log(2+\sqrt{3})}{2 \sqrt{3}} -  \frac{\pi^2}{12 \sqrt{3}} .$$
The original integral is
$$\frac{1}{2} \left( \log(2)^2 -2 \sqrt{3} S \right) = \frac{\log(2)^2}{2} - \frac{\pi^2 \sqrt{3}}{12} +  \frac{\log 2 \log(2+\sqrt{3})}{2} + \frac{\pi^2}{12}$$
$$=\frac{\pi^2(1-\sqrt{3})}{12} + \log(2) \frac{\log(4 + 2 \sqrt{3})}{2} = \frac{\pi^2(1-\sqrt{3})}{12} + \log(2) \log(1+\sqrt{3})$$
as desired.

I have found an interesting reference related to this integral. It may be of interest to many users:

A restricted Epstein zeta function
and the evaluation of some definite integrals by Habib Muzaffar and Kenneth S. Williams


Two triangles with two equal sides and equal area will have the third size also equal? Consider two triangles $\triangle abc$ and $\triangle def$ such that $ab=de$ and $ac=df$.Also area of $\triangle abc$ is equal to area of $\triangle def$.Now draw $cm$ perpendicular to $ab$ and $fn$ perpendicular to $de$.$ab$ and $de$ are equal and area of triangles is also equal so $cm$ should be equal to $fn$.Now $\triangle amc$ and$\triangle dnf$ are congruent by right angle triangle congruence(since hypoteneous $ac$ and $df$ are equal).therefore $\angle bac$ is equal to $\angle edf$.Now in $\triangle abc$ and $\triangle def$ by SAS both $\triangle abc$ and $\triangle def$ are congruent so $bc=ef$.I don't know where i am wrong.

edit:Please read my proof and point out what's wrong

The answer is no. See the following counter-example. The two triangles have the same altitude, and equal bases (and hence equal in area) but the third sides (i.e. BC, EF) are different.

This fact can also be verified by applying the formula:- area of a triangle = 0.5 ab sin C.

Yes I know you got confused to it. but before you go further you should know that your  both triangle are acute angle and hence your altitude exist inside the triangle. but when you draw two triangle one acute and one obtuse then your altitude get outside the triangle and you are unable to prove that third side is equal.
Your explanation is possible only when both triangle would be acute angled


On the proof of a theorem in real analysis The following text is from the book Advanced Calculus by Patrick M. Fitzpatrick :

There are two things that I don't understand with the proof :

1- In order that the last equation (last line) to hold, the proof must also show that $f(x)$ is continuous but it doesn't. The sequence ${\{f_n}\}$ to be pointwise convergent to $f$ doesn't guarantee that $f$ is contentious on $I$. How to show that $f$ is contentious on $I$?

2- A part of the first claim of the two-claim of the theorem is $f$ is differentiable but it doesn't prove that. How to show that $f$ is differentiable on $I$?

1- $g$ is a uniform limit of continuous functions, therefore it is continuous.

2- $f$ is the integral of a continuous function (see point 1), so it is differentiable.


A theorem in the paper "Noncommuting Random Products" by Furstenberg I have a question concerning the proof of theorem 2.5 at page 395 of the paper Noncommuting Random Products, by H. Furstenberg, Trans. Amer. Math. Soc., 1963.

The statement is as follows:

Let $\mu$ be a probability measure on a group $G$. Let $M$ be a locally compact $G$-space, which does not admit a stationary measure for $\mu$, that is, there is no $\nu$ on $M$ such that $\mu*\nu=\nu$. Let $Z_n$ be a $\mu$-process on $M$, that is $Z_i$ are $M$ valued random variables such that the conditional distribution of $Z_{n+1}$ given $Z_{n},\dots,Z_0$ is $\mu*\delta_{Z_n}$. Let $\Delta\subset M$ be compact. Define $n(k)$ as the index $n$ for which $Z_n\in\Delta$ for the $k$-th time. Then $n(k)/k\to\infty$ with probability $1$.

The proof starts with stating that it is enough to show that for each compactly supported $\psi$ we have that with probability $1$
\begin{equation}
\frac{1}{n}\sum_0^{n-1}\psi(Z_k)\to 0
\end{equation}
This I do understand. However, Furstenberg proceeds to say that if this formula does not hold for some $\psi'$, then there exists a subsequence $n_i$ such that
\begin{equation}
\frac{1}{n_i}\sum_0^{n_i-1}\psi'(Z_k)\to \alpha\neq 0,
\end{equation}
and he seems to assume in what follows that neither $n_i$ nor $\alpha$ are random.

**
Question:
**
How is it that the second equation is the negation of the first? Is Furstenberg implicitly using here some "strong law of large numbers"?

During the proof Furstenberg picked (and fixed) a sample sequence $Z_n$. More precisely,

1). It has been proved that (2.8) $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} F(Z_k) \to 0$ holds with probability $1$ for any $F=\tau f- f$.

2). Now suppose (2.9) $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} \psi'(Z_k) \to 0$ fails for some $\psi'$. It means there is a positive probability such that the limit is $\displaystyle \frac{1}{n}\sum_{k=0}^{n-1} \psi'(Z_k) \not\to 0$.

3). Taking the intersection of these two probabilities, one concludes that there is a positive probability (and take a sample sequence $Z_n$ from this) such that (2.8) holds along this sample sequence for all $F=\tau f- f$, while (2.9) gives a nonzero limit.

For this fixed sample sequence $Z_n$, passing to subsequences of $n_k$ if necessary, one gets a positive linear functional that is nonzero. At this moment we are done with the sample sequence and consider the measure induced by the linear functional.

This is my understanding. Hope it makes sense.


Are the exercises necessary to understand the subject of a mathematical textbook? When I am reading a mathematical textbook, I tend to skip most of the exercises.
Generally I don't like exercises, particularly artificial ones.
Instead, I concentrate on understanding proofs of theorems, propositions, lemmas, etc..

Sometimes I try to prove a theorem before reading the proof.
Sometimes I try to find a different proof.
Sometimes I try to find an example or a counter-example.
Sometimes I try to generalize a theorem.
Sometimes I come up with a question and I try to answer it.

I think those are good "exercises" for me.

EDIT
What I think is a very good "excercise" is as follows:

(1) Try to prove a theorem before reading the proof.

(2) If you have no idea to prove it, take a look a bit at the proof.

(3) Continue to try to prove it.

(4) When you are stuck, take a look a bit at the proof.

(5) Repeat (3) and (4) until you come up with a proof.

EDIT
Another method I recommend rather than doing "homework type" exercises:
Try to write a "textbook" on the subject.
You don't have to write a real one.
I tried to do this on Galois theory.
Actually I posted "lecture notes" on Galois theory on an internet mathematics forum.
I believe my knowledge and skill on the subject greatly increased.

For example, I found this while I was writing "lecture notes" on Galois theory.
I could also prove that any profinite group is a Galois group.
This fact was mentioned in Neukirch's algebraic number theory.
I found later that Bourbaki had this problem as an exercise.
I don't understand its hint, though.
Later I found someone wrote a paper on this problem.
I made other small "discoveries" during the course.
I was planning to write a "lecture note" on Grothendieck's Galois theory.
This is an attractive plan, but has not yet been started.

EDIT
If you want to have exercises, why not produce them yourself?
When you are learning a subject, you naturally come up with questions.
Some of these can be good exercises. At least you have the motivation not given by others. It is not homework.
For example, I came up with the following question when I was learning algebraic geometry.
I found that this was a good problem.

Let $k$ be a field.
Let $A$ be a finitely generated commutative algebra over $k$.
Let $\mathbb{P}^n = Proj(k[X_0, ... X_n])$.
Determine $Hom_k(Spec(A), \mathbb{P}^n)$.

As I wrote, trying to find examples or counter-examples can be good exercises, too.
For example, this is a good exercise in the theory of division algebras.

EDIT
Let me show you another example of self-exercises.
I encountered the following problem when I was writing a "lecture note" on Galois theory.

Let $K$ be a field.
Let $K_{sep}$ be a separable algebraic closure of $K$.
Let $G$ be the Galois group of $K_{sep}/K$.

Let $A$ be a finite dimensional algebra over $K$.
If $A$ is isomorphic to a product of fields each of which is separable over $K$, $A$ is called a finite etale algebra.
Let $FinEt(K)$ be the category of finite etale algebra over $K$.

Let $X$ be a finite set.
Suppose $G$ acts on $X$ continuously.
$X$ is called a finite $G$-set.
Let $FinSets(G)$ be the category of finite $G$-sets.

Then $FinEt(K)$ is anti-equivalent to $FinSets(G)$.

This is a zero-dimensional version of the main theorem of Grothendieck's Galois theory.
You can find the proof elsewhere, but I recommend you to prove it yourself.
It's not difficult and it's a good exercise of Galois theory.
Hint: Reduce it to the the case that $A$ is a finite separable extension of $K$ and X is a finite transitive $G$-set.

EDIT
If you think this is too broad a question, you are free to add suitable conditions.
This is a soft question.

If your goal is to become a research mathematician, then doing exercises is important.  Of course, there will be the rare person who can skip exercises with no detriment to their development, but (and I speak from the experience of roughly twenty years of involvement in training for research mathematics) such people are genuinely rare.

The other kinds of exercises that you describe are also good, and you should do them too!

The point of doing set exercises is to practice using particular techniques, so that you can recognize how and when to use them when you are confronted with technical obstacles in your research.

In my own field, two books whose exercises I routinely recommend to my students are Hartshorne's Algebraic geometry text and Silverman's Elliptic curves text.  The exercises at the end of Cassels and Frolich are also good.

Atiyah and MacDonald also is known for its exercises.

One possible approach (not recommended for everyone, though) is to postpone doing exercises if you find them too difficult (or too time-consuming, but this is usually equivalent to too difficult), but to return to them
later when you feel that you understand the subject better.  However, if upon return, you still can't fairly easily solve standard exercises on a topic you think you know well, you probably don't know the topic as well as you think you do.

If your goal is not to become a research mathematician, then understanding probably has a different meaning and purpose, and your question will then possibly have a different answer, which I am not the right person to give.

I'm with the OP on this, I skip exercises too.

Here's the logic: A true understanding of maths is about being creative in its applications and not just the material initself.

Exercises, by definition, stifle creativity by presenting a sandbox in which to think.

It's a bit like Rocky 3, do you learn your craft by working out in a gym or by lifting logs down in the forest?

...and another thing, I reckon following examples in books leads to degenerate mathematics. They perpetuate particular styles of thinking about problems.


diagonalize a non-normal matrix , without distinct eigenvalues I wonder how to diagonalize a matrix that is non-normal, and does not have distinct eigenvalues.

Let  $\lambda_i$ be the eigenvalue, and $v_i$ be the eigenvector with that eigenvalue.

I think the process would go like this:

Determine if $\dim(\mathrm{span}(v_i)) = $ multiplicity of $\lambda_i$. If no, then it is not diagonolizable. If yes, go to 2
Is the eigenvectors linearly independent? If yes, we can diagonalize. If no, ... I don't know.

Non-normal matrices may or may not be diagonalizable.

Given $n\in \mathbb N$ and $A\in \mathcal M_{n\times n}(\mathbb C)$, it holds that
$A$ is diagonalizable (over) $\mathbb C$ if, and only if, there exists a basis $\{v_1, \ldots ,v_n\}$ of $\mathbb C^{n\times 1}$ such that $v_1, \ldots ,v_n$ are all eigenvectors of $A$.

If $A$ is diagonalizable, then to find a diagonalizing matrix $P$, you just have to find the vectors $v_1, \ldots ,v_n$ above and define $P$ by letting its columns be the eigenvectors found. This implies that $P^{-1}AP$ is a diagonal matrix.

This works whether $A$'s eigenvalues have all multiplicty $1$ or not.

Is this correct?

I'm given a matrix
I find out that it is non-normal
I find its eigenvalues. Let's say I have m of them.
I find out it does not have distinct eigenvalues
I find some set of eigenvectors to the eigenvalues, call this set S.
If and only if dim(span(the eigenvectors)) =  m , I can diagonalize A. (To put the equation in words, just to make sure "If and only if I have as many linearly independent vectors in S, as I have number of eigenvalues, the matrix is diagnoalizable").


extension to an ultrafilter Let X be a topological space. How to show a filter T on X can be extended to a unique ultrafilter on X iff T is an ultrafilter.
All I know is that every filter can be extended to an ultrafilter. I think this can be useful here. But I cannot find the link.
Any help would be aporeciated. Thanks in advance.

The key idea is this:

  If $\mathcal{F}$ is a filter on $X$, and $A \subset X$ is such that both $A \notin \mathcal{F}$ and $X\setminus A \notin \mathcal{F}$, then $\mathcal{F} \cup \{A\}$ has the finite intersection property.

The proof is simple: suppose not. Then there are finitely many sets of $\mathcal{F} \cup \{A\}$ that have empty intersection. As $\mathcal{F}$ is closed under finite intersections, this can only happen if there is some $F_1 \in \mathcal{F}$ such that $F_1 \cap A = \emptyset$. But this is equivalent to $F_1 \subseteq X\setminus A$, which implies (by the enlargement axiom for filters) that $X\setminus A \in \mathcal{F}$, quod non. This contradiction shows that $\mathcal{F} \cup \{A\}$ has the fip, so can be extended to a filter $\mathcal{F}_A$ which can be explicitly defined as $$\mathcal{F}_A := \{B \subseteq X: \exists F \in \mathcal{F}: F \cap A \subseteq B\}$$

One easily shows this is a filter on $X$ that contains $A$ and all of $\mathcal{F}$.

So if for some filter $\mathcal{F}$ we have that there is some set $A$ such that neither $A$ nor $A^c = X\setminus A$ are not in $\mathcal{F}$, we can also apply the above idea (by symmetry) to get a filter $\mathcal{F}_{A^c}$ that properly extends $\mathcal{F}$ and then $\mathcal{F}_{A^c}$ and $\mathcal{F}_{A}$ are distinct (one contains $A$, the other does not) proper extensions of $\mathcal{F}$. We can extend both of them to ultrafilters, that are thus also distinct.

We conclude by contraposition that if $\mathcal{F}$ has a unique ultrafilter extension, then for all sets $A \subseteq X$, either $A \in \mathcal{F}$ or $X\setminus A \in \mathcal{F}$, and this property characterises ultrafilters (which follows from the observation that if $\mathcal{G} \supsetneq \mathcal{F}$ for some filter, any $B \in \mathcal{G}\setminus \mathcal{F}$ will fail this criterion, as $B$ is not in $\mathcal{F}$ and $X\setminus B$ isn't either (or it would be in $\mathcal{G}$ which cannot be)).

If T is not an ultrafilter, then there is some A subset X with
A not in T and X - A not in T.  Thus T can be extended to an
ultrafilter F with A in F and to an ultrafilter G with X - A in G.


how to find subdifferential of a function $x^2+ |x-1|+|x-2|$ Given a function
$f(x) =  x^2+ |x-1|+|x-2| $
find it's subdifferential.
My approach to solving this was to divide the answer into 5 parts:

For |x-1|>1 and |x-2|>2
$f(x) = x^2+ x-1+x-2$ and $f'(x) = 2x+2$

For |x-1|<1 and |x-2|<2
$f(x) = x^2-(x-1)-(x-2)$ and $f'(x) = 2x-2$

For |x-1|>1 and |x-2|<2
$f(x) = x^2+(x-1)-(x-2)$ and $f'(x) = 2x$

For |x-1|<1 and |x-2|>2
$f(x) = x^2-(x-1)+(x-2)$ and $f'(x) = 2x$

For |x-1|=1 and |x-2|=2
$ f(x) = x^2$ and $f'(x) = 2x$

Does this look right? Is this the correct approach?

Here's an approaching using the subdifferential sum rule:
\begin{align}
\partial f(x) &= \partial f_1(x) + \partial f_2(x) + \partial f_3(x) \\
&= f_1'(x) + \partial f_2(x) + \partial f_3(x)
\end{align}
where $f_1(x) = x^2, f_2(x) = |x - 1|$, and $f_3(x) = |x - 2|$.

The function $f$ is differentiable everywhere except at $x = 1$ and $x = 2$. If $x \neq 1$ and $x \neq 2$, then we have
$$
\partial f(x) = \{f_1'(x) + f_2'(x) + f_3'(x)\}.
$$
If $x = 1$, we have
\begin{align}
\partial f(x) &= \{f_1'(x) \} + \partial f_2(x) + \{f_3'(x)\} \\
&= \{f_1'(x)\} + [-1,1] + \{f_3'(x)\} \\
&= \{f_1'(x) + f_3'(x) + g \mid -1 \leq g \leq 1 \}.
\end{align}
If $x = 2$, we have
\begin{align}
\partial f(x) &= \{f_1'(x)\} + \{f_2'(x)\} + \partial f_3(x) \\
&= \{f_1'(x)\} + \{f_2'(x)\} + [-1,1] \\
&= \{ f_1'(x) + f_2'(x) + g \mid -1 \leq g \leq 1\}.
\end{align}

First, I have to admit I am not familiar with a "subdifferential".  It look to me like you are just taking the derivative.  Second, dividing with things like "|x- 1|> 1 and |x- 2|> 2" is confusing and probably not what you want.  Instead divide the real line into three intervals: $x\le 1$, $1< x\le 2$, and $x> 2$.

For $x\le 1$ both x-1 and x-2 are negative: $x^2+ |x- 1|+ |x- 2|= x^2- (x- 1)- (x- 2)= x^2- 2x+ 3$.  The derivative is $2x- 2$.

For $1< x\le 3$ x-1 is positive but x-2 is still negative:$x^2+ |x- 1|+ |x- 2|= x^2+ (x- 1)- (x- 2)= x^2+ 10$.  The derivative is $2x$.

For $x> 2$ both x-1 and x-2 are positive: $x^2+ |x- 1|+ |x- 2|= x^2+ (x- 1)+ (x- 2)= x^2+ 2x- 3$.  The derivative is $2x+ 2$.


Name of this control strategy Assume that we have a nonlinear dynamical system on $\mathbb{R}^n$ of the form
$$
\dot{x}(t) = f(x(t),u),
$$
where $f(0)=0$, i.e. the origin is an equilibrium. My question is, is there a particular control strategy which aims to stabilize such an equilibrium point with two controls $A(t)\in\mathbb{R}^{n\times n},\;b(t)\in\mathbb{R}^n$ passing to the system
$$ \dot{x} = f(x,\tilde{u}(t,x))=\tilde{f}(A(t)x+b(t))? $$
For example, if $f(x,u) = M(x+u) := \tilde{f}(x+u)$, for a matrix $M$, then I would just set $\tilde{u}=A(t)x+b(t)-x$, i.e. I get
$$f(x,\tilde{u}) = f(x,A(t)x+b(t)-x)=\tilde{f}(A(t)x+b(t))=M(A(t)x+b(t)).$$
More precisely, is there a known control strategy based on the idea of combining the nonlinearity with a time-dependent affine transformation of the state variable? I would just like the name of this strategy, if it exists, then I will figure out where to understand and study it.
It would still be interesting to know if a similar strategy as
$$\dot{x} = f(A(t)x)+b(t)$$
is used.

No, I don't think so. Saying something does not exist, is always a big stretch. I don't know every method ever conceived, but let me explain why I think such a strategy does not exist:
The progression of the dynamical system only indirectly depends on $t$. Directly it depends only on your state $x$ and your control input $u$ ("Markov Property"). So why should I inject a time-dependency into my controller, if the system itself does not care about the time?
Now, there is one important part of your problem missing: Optimal control always looks to minimise a cost function. What is your cost function? That is a modelling assumption, you have to make! You write, you want $x$ to stay close to the origin, so maybe we put something like $J(x, u, t) = \int_{t_0}^{t_1}||x||^2$.
Now, if you drop the time dependency for your controller $u$, and assume $f$ is linear, you'd arrive at a Linear-quadratic regulator. I think that would be a reasonable controller for the problem you describe. Or at least a good point to start, to find more sophisticated control algorithms.


Integrating body angular velocity I've been reading over some very comprehensive notes on attitude representation, which were compiled by James Diebel, a Stanford student: http://www.swarthmore.edu/NatSci/mzucker1/e27/diebel2006attitude.pdf

What is of particular interest to me is equation $266$, which states that the rotation vector representation of an attitude is the integral of the body angular velocities over the time frame of interest (assuming the body and inertial frames start out coincident)

I see no proof of this anywhere in the paper, can someone help me understand how this is possible?

Edit

To clarify, this is my issue. Say I want to represent that attitude of the difference between two coordinate systems (say inertial and body) using an angle/axis vector that rotates a vector in the inertial frame to one in the body frame:
$$v_{bi}(t)$$
I have measurements of body angular rate (from, lets say a gyroscope)
$$\omega_b(t)$$
I'm curious if the following is generally true:
$$\dot{v}_{bi}(t)=\omega_b(t)$$

Equation $266$ suggests that it is, equation $265$ seems to suggest otherwise.

TL;DR: Ignore what you read in that paper; use either Eq. (2) or (3) in this post if you want to "integrate" angular velocity.

There is a solution to your problem, in the sense that it is possible to "integrate" the angular-velocity vector $\omega$ to obtain the rotation taking the inertial "world" frame onto the rotating "body" frame.  In fact, I've written a paper titled "The integration of angular velocity".  It turns out that it's pretty simple to get the frame from the angular velocity, but it's not quite as simple as just integrating $\omega$ as a vector.

First, to clear up the confusion about that paper: Diebel uses some unfortunate notation.  The equation (266) you point to says
\begin{equation}
  \mathbf{v}_{\omega'} := \int_{t_0}^{t_1} \boldsymbol{\omega}' dt
  \tag{1}
\end{equation}
There are two important things to note here.  First, $\boldsymbol{\omega}$ is primed, meaning (as you seem to understand) that this is measured with respect to the body.  That is, the basis with respect to which it is defined is different at each instant.  So this is not what we normally mean when we're talking about the angular-velocity vector.  Second, the quantity on the left is most definitely not the rotation vector required to go from the inertial frame to the body frame.  Instead, it is some kind of rotation-rate vector which, if you multiply it by some infinitesimal time, would generate that rotation that takes the body frame at this instant into the body frame at that infinitesimal time in the future.  Basically, Diebel can define any vector he wants—and he did so in Eq. (1).  But it's only any use if $t_1$ and $t_0$ are infinitesimally close.  Of course, this is all he uses it for in the following paragraph, so it's okay for him.  But in general, I suggest you ignore that equation entirely.

The reason that Eq. (1) is not useful in general is because, at different instants, $\boldsymbol{\omega}'$ is defined with respect to different frames.  So it doesn't make a whole lot of sense to add vectors from different instants together, which is just what that integral tries to do, unless the rotation is about a fixed axis.  However, it is possible to rotate that body-fixed vector into the inertial frame.  (You also probably don't want to write it as an integral equation; it's better as a standard differential equation.)  The best way to integrate angular velocity is to use quaternions.  Of course, quaternions are basically always the right way to do anything with rotations anyway.  I don't understand why Diebel insists on converting quaternions into matrices, so I'll just get away from him, and write things like any normal person would.

Let's suppose $R(t)$ is a unit quaternion function of time that takes your inertial frame onto the body frame.  For example:
\begin{equation}
  \mathbf{x}'(t) = R(t)\, \mathbf{x}\, \bar{R}(t),
\end{equation}
and similarly for $\mathbf{y}$ and $\mathbf{z}$.  Using the usual definition of angular velocity as that thing which, when crossed with a vector in the body frame, gives the rate of change of that vector.  In particular, we have:
\begin{equation}
  \frac{d \mathbf{x}'} {dt}(t) = \boldsymbol{\omega}(t) \times \mathbf{x}',
\end{equation}
and similarly for $\mathbf{y}$ and $\mathbf{z}$.  Now, we can combine these equations (and use the fact that they are generally true for any vectors, not just the basis vectors) to find a relation between the quaternion and the angular velocity:
\begin{equation}
  \boldsymbol{\omega}(t) = 2\, \dot{R}(t)\, \bar{R}(t).
\end{equation}
I'm sure that this has been done elsewhere, but my favorite derivation is in Section 3 of my paper.

There are a few ways to go from here, but the most obvious is to just use this as the differential equation you were looking for.  Just rearrange a little to find
\begin{equation}
  \dot{R}(t) = \frac{1}{2}\, \boldsymbol{\omega}(t)\, R(t).
  \tag{2}
\end{equation}
The right-hand side is a valid quaternion product, so you can just integrate the four components that result to get your frame quaternion $R(t)$.  You also need an initial condition to solve this, of course, but that's to be expected.

If, instead, you have $\boldsymbol{\omega}'$ (measured with respect to the body), you can simply rotate that back into the inertial frame as $\boldsymbol{\omega}(t) = R(t)\, \boldsymbol{\omega}'(t)\, \bar{R}(t)$, in which case you have
\begin{equation}
  \dot{R}(t) = \frac{1}{2}\, R(t)\, \boldsymbol{\omega}'(t).
  \tag{3}
\end{equation}
This is the more relevant equation if, for example, you measure angular velocity using some gyroscope embedded in the body.

Finally, I'll point out that it's also possible to integrate to find the generator of the rotation, which I'll label $\boldsymbol{r}(t)$.  This is precisely the angle/axis vector, where the axis is given by the direction of the generator and the angle is proportional to the length of the generator.  It is related to the quaternion by $R(t) = \exp[\boldsymbol{r}(t)]$.  In this case, the differential equation is a bit more complicated:
\begin{equation}
  \dot{\boldsymbol{r}} = \frac{1}{2} \boldsymbol{\omega} \times \boldsymbol{r} + \boldsymbol{\omega} \frac{\lvert\boldsymbol{r}\rvert \cot \lvert\boldsymbol{r}\rvert} {2} + \boldsymbol{r} \frac{\boldsymbol{r} \cdot \boldsymbol{\omega}} {2 \lvert\boldsymbol{r}\rvert^2} \left(1 - \lvert\boldsymbol{r}\rvert \cot \lvert\boldsymbol{r}\rvert \right).
\end{equation}
Besides looking uglier, this equation also requires some careful handling, because $\boldsymbol{r}$ typically goes through some weird values that are precisely analogous to branch cuts of the complex logarithm, but which cause serious numerical problems.  If you're careful about handling this behavior, you can make this system as good as (and occasionally even better than) the systems of Eq. (2) and (3), but it's probably not worth the effort.  These issues are all discussed at length in my paper.

The proof is by "Fundamental Theorem Of Calculus".
By definition: $\dfrac{d\vec \theta}{dt}=\vec\omega$. By Fundamental Theorem Of Calculus $\vec \theta=\int\vec \omega dt$.


Prove that there are no such positive integers $a,b,c,d$ such that $a^2 + b^2 = 3(c^2 + d^2)$ Prove that there are no positive integers a, b, c, d such that $a^2 + b^2 = 3(c^2 + d^2)$. Hint: What can you say about divisibility of a and b by 3? Look at solution with smallest possible a.

Here are some hints for you regarding this question:

Use the fact that $2$ isn't quadratic residue modulo $3$. The RHS is divisible by ... $a,b$ are of the form ...

Then assume that $(a_0,b_0,c_0,d_0)$ is the smallest solution, using the things we already stated prove there's another smaller solutions $(a_1,b_1,c_1,d_1)$, which means that ...

This method is called infinite descent, you can learn something more if you just put some little bit more effort.

The places with three dots are left for you to fill them.

Well, $A^2+B^2=1,0$ ( mod 4) so it couldn't be $A^2+B^2=4k+3$. But (4k+1)$(4k+3)=16k^2+16k+3=4(4k^2+4k)+3=4r+3$, while (4k+3)$(4k+3)=16k^2+24k+8+1=4(4k^2+6k+2)+1=4s+1$. So if a prime $p|A^2+B^2$ and p=4k+3, $p^2|A^2+B^2!$ But the exponent of number 3 is odd while it should be even. Hence there no positive integers A,B,C,D such that $A^2+B^2=3(C^2+D^2)$.


Isomorphism on labeled graphs where the vertices in the same "position" must have equivalent labels? I am looking for information about a specific type of equivalence relation on labeled graphs. Specifically, I want two labeled graphs to be considered equivalent if their underlying unlabeled graphs are isomorphic and vertices in the same "position" have equivalent labels (according to some equivalence relation defined on the vertex labels).

I hope the following example illustrates the type of equivalence relation on labeled graphs that I am looking for.

Example:
The set of possible vertex labels are the letters of the English alphabet, which may be uppercase or lowercase. Now I consider two vertex labels to be equivalent if they are either both an uppercase letter or both a lowercase letter. Now, as an example, consider the following three graphs:

Graph $G_{abC}$:

Graph $G_{deF}$:

Graph $G_{GHi}$:

Let $[a]$ be the equivalence class of lowercase letters and $[A]$ the equivalence class of uppercase letters. If I relabel each vertex with the corresponding equivalence class, then for the first two graphs I obtain the same labeled graph $G_{[a][a][A]}$:

For the third graph, on the other hand, I get a different relabeled graph $G_{[A][A][a]}$:

So although all three graphs are isomorphic, I want to say that only the first two graphs are equivalent. Specifically, $G_{abC}$ and $G_{deF}$ are equivalent because the relabeling with vertex classes yields the same relabeled graph $G_{[a][a][A]}$. The third graph $G_{GHi}$,  on the other hand, is not equivalent to the others because its relabeled graph $G_{[A][A][a]}$ is different from $G_{[a][a][A]}$.

So can anyone tell me if there is a standard name for such a kind of equivalence relation on labeled graphs? Moreover, does anyone know of any references where such an equivalence relation is discussed? In particular, I am interested in algorithms to determine if two labeled graphs are equivalent in the manner I described.

You want to think of your graph vertices as being "weighted" or "colored" (according to your equivalence of labels).  Then when you compute isomorphisms, you want to respect the coloring.  The program nauty has support for this.  The default way this is implemented is that a specific colored vertex can only be sent to a vertex that is the same color, so this should accomplish exactly what you want.

Suppose $G_1$ and $G_2$ are labeled graphs. Let the vertex set of a graph $G=(V,E)$ be $V(G)$ where each $v \in V(G)$ is assumed to have a unique label.

We say that two graphs $G_1$ and $G_2$ are equivalent if there exists an $f(G_1)=(f_v(V_1),f_e(E_1))=G_2$ such that (1) (preservation of adjacency, graph isomorphism) $(v_1,v_2) \in E(G_1) \iff (f_v(v_1),f_v(v_2))\in E(G_2)$ and (2, equivalence) $\forall v \in V(G_1)\; ,\;  f_v(v) \sim v$.

(On the rewrite $\sim$ means equivalent to.)

Also, I have abusively used $f_v(V_1) = \{f_v(v) | v \in V(G_1)\}$

Sorry, if I'm being unclear.