问:
Column rank of a matrix
Let $a\in (0,1)$. Let $N,T,J$ be positive integers with $N>J, N>T$, $J>T$. Consider the following matrix
$$
B\equiv \begin{pmatrix}
I_N & A_1\\
I_N & A_2\\
... & ...\\
I_N & A_T\\
\end{pmatrix}
$$
where $I_N$ is the $N\times N$ identity matrix and, for $t=1,...,T$,  $A_t$ is an $N\times J$ matrix such that:

Each row of $A_t$ contains one and only one element equal to $a$, while the other elements are $0$. (This implies that the sum of all the elements of $A_t$ is $N*a$).

My questions:

(Q1) Show that the columns of $B$ are not linearly independent.

(Q2) Determine what is the column-rank of $B$.

(Q3) My suspect is that the column-rank of $B$ is $N+J-1$. Consider the system of equations
$$
Y=(B'B)X
$$
where $Y$ and $X$ are $(N+J)\times 1$ vectors.
Given that $B$ is not full column-rank, this system will have infinitely many solutions (assuming that it is consistent).
I want to show that we can freely set anyone element of $X$ equal to a known value (for example, zero) and, in turn, the system of equations will have a unique solution with respect the other $N+J-1$ unrestricted elements of $X$.

My thoughts:

[Q1] I think that Q1 can be answered as follows (your advise would neverthless be appreciated).

As suggested here, $B$ is full column rank if and only if
$$
C\equiv \begin{pmatrix}
A_2-A_1\\
A_3-A_1\\
\cdots\\
A_T-A_1\\
\end{pmatrix}
$$
is full column rank.

Consider any $t\in \{2,...,T\}$.
Given the structure of each $A_1$ and $A_t$, we have that each row of $A_t-A_1$ is:

either $0_{1\times J}$;

or, one of its elements is $a$, one of its elements is $-a$, all the other elements are $0$.

Therefore, $C$ is not full column rank if and only if there exists scalars $p_1,...,p_J$ not all equal to zero solving a system of at most $N(T-1)$ equations, where each equation looks like
$$
p_k-p_j=0 \hspace{1cm} \text{ for some $k,j\in \{1,...,J\}$ with $k\neq j$}
$$

This system admits as solution $p_1=p_2=...=p_J\neq 0$. Hence, $C$ is not full column rank. Therefore, $B$ is not full column rank.

[Q2] I don't know how to answer this. But, from doing some simulations, I believe that the column-rank should be $N+J-1$.

[Q3] I don't know how to answer this. It is true that, by correctly setting one of the elements of $X$ equal to zero, the system will have a unique solution with respect to the other $N+J-1$ elements of $X$. However,  in general, we are not completely free in deciding which element of $X$ can be normalised. Nevertheless, I believe that in my specific example we are free. How can I show it?

答:
Q1:

If you sum the first $N$ columns of $B$ you get the vector of all ones.  If you sum the last $J$ columns you get the vector of all $a$'s.  These two vectors are proportional, hence the columns of $B$ are linearly dependent.

Q2:

The rank is not uniquely determined by the conditions you have on $B$.  For example, if $T=1$ then the rank of $B$ is $N$, the row rank.  For $T>1$ it's easy to make examples with e.g. rank $N+J-1$ as suggested.

Title: Column rank of a matrix


问:
Prove the solution to a linear pde is 0 on an open unit ball
Given linear PDE $$x \cdot \nabla u +u=0$$ on a unit open ball, $u$ is $C^1$ on the ball and $C^0$ on the boundary, prove $u$ is $0$ on the unit ball.

I solved that $u$ is constant on the boundary, $u(0)=0$. I am thinking about using proof by contradiction and applying for the mean value theorem.

答:
This PDE can be solved using the method of characteristics. We first solve the PDE, and then deduce the claimed result.

Solution of PDE

The characteristics $X$ are the solution to the parameterised ODE
$$ \partial_tX(x_0,t) = X(x_0,t), \\ X(x_0,0) = x_0$$
where $x_0$ ranges over the points in $\partial B_1(0)$ i.e. $|x_0|=1$. (we remark that $\partial B_1(0)$ is non-characteristic, i.e. none of the characteristics are tangent to it.)
The solution is
$$X(x_0,t) = x_0 e^t.$$

Lets say that on $\partial B_1(0)$, $u=g$ for some continuous function $g:\partial B_1(0)\to\mathbb R$. Setting $U(x_0,t):=u(X(x_0,t))$, we note that
$$ \partial_t U = ((\nabla u)\circ X)\cdot\partial_t X = X\cdot u(X) = -U$$
which gives
$$U(x_0,t) = g(x_0)e^{-t}.$$
$X$ has the inverse $(x_0(x),t(x)) =(\frac x{|x|} , \log |x|)$. so we can use this to write the solution as
$$u(x) = U(x_0(x),t(x)) =g\left(\frac x{|x|}\right) \frac1{|x|}$$
We can easily verify this factor of $\frac1{|x|}$ is correct: say $g\equiv 1$, i.e. $u(x)=\frac1{|x|}$ then
$$ \nabla u(x) = \frac{-x}{|x|^3} \implies x\cdot\nabla u(x) = \frac{-1}{|x|} = -u(x).$$

Deducing that $u$ vanishes

Once we have the above form, it is easy to conclude. If the boundary data $g(x)=g_0\neq 0$ at any point $x$, then along the line joining $0$ to $x$, the solution takes the form $\frac{g_0}{|x|}$, which explodes as $x\to 0$. Since we assumed the solution to be continuous, this means $g$ must be identically zero.

However, from the form of $u$, this means that $u\equiv 0$.

ContentLicense: CC BY-SA 4.0


问:
Not sure of my answers on a specific probability problem
Sixty percent of the consumers' population is reputed to prefer a particular brand A of toothpaste. If a group of consumers are selected at random and interviewed, what is the probability that exactly 5people must be selected until interviewing the first consumer who prefers brand A? What is the expected number of consumers to be selected until the first one who prefers brand A is interviewed?

$Answers$

a. $P(5) = (0.6)(0.4)^4$ =$0.0514$

b. $E(x) = (5)(0.6) = 3$

Did I answer it correctly?

答:
Seems to be a geometric random variable with
$$P(X = k) = (.6)(.4)^{k-1},$$ for $k = 1,2,3,\dots.$
Then one can show that $E(X) = 1/.6 = 1.6667.$

The analytic demonstration that the infinite series
for $E(X)$ sums as claimed is not trivial, but a method is
shown in most elementary probability books and in
the link. (Commonly, the method involves differentiation--either with or without use of
moment generating functions.)

Informally, for the current example it is more
than sufficient to use R to sum the first thousand
terms, as shown below.

```<code>k = 1:1000
sum(k*.6*.4^(k-1))
[1] 1.666667
</code>```

However, just from this, it seems your answer (b) is not correct.

Note: As you can see in the link, there are two
parameterizations of a geometric distribution. The one implemented in R as dgeom is not the one you're
using here.

The difference is whether the random variable
$X$ counts the number of trials until the first success (as here) or the number of trials before the first success.

Body:
Seems to be a geometric random variable with
$$P(X = k) = (.6)(.4)^{k-1},$$ for $k = 1,2,3,\dots.$
Then one can show that $E(X) = 1/.6 = 1.6667.$

The analytic demonstration that the infinite series
for $E(X)$ sums as claimed is not trivial, but a method is
shown in most elementary probability books and in
the link. (Commonly, the method involves differentiation--either with or without use of
moment generating functions.)

Informally, for the current example it is more
than sufficient to use R to sum the first thousand
terms, as shown below.

k = 1:1000
sum(k*.6*.4^(k-1))
[1] 1.666667

However, just from this, it seems your answer (b) is not correct.

Note: As you can see in the link, there are two
parameterizations of a geometric distribution. The one implemented in R as dgeom is not the one you're
using here.

The difference is whether the random variable
$X$ counts the number of trials until the first success (as here) or the number of trials before the first success.


问:
Strange triangle perimeter problem - Russian Olympiad 2010
This problem came from Crux Mathematicorum - Canadian Mathematical Society.

The perimeter of triangle $\triangle ABC$ is equal to $4$. Points $X$ and $Y$ are marked on the rays $AB$ and $AC$ in such a way that $AX = AY = 1$. The segments $BC$ and $XY$ intersect at point $M$ in their interior. Prove that the perimeter of one of the triangles $\triangle ABM$ or $\triangle AMC$ is equal to $2$.

I created $MN$ parallel to $AB$, $N$ is on $AC$. Then I calculated the two pairs of similar triangles introduced by this parallel segment. I also used Stewart to calculate $AM$ It didn't go anywhere.

答:
What strange conditions have been given? We are told the lengths AX = AY = 1 and
the perimeter of ABC is 4, and effectively nothing else. The conclusion, which is an
either-or statement, is equally puzzling.
Let us reflect the point A over both X and Y to two points U and V so that AU = AV =
2. This seems slightly better, because AU = AV = 2 now, and the “two” in the perimeter
is now present. But what do we do? Recalling that s = 2 in the triangle, we find that U and
V is the tangency points of the excircle, call it a. Set IA the excenter, tangent to BC at TA. See Figure 2.7E.

Looking back, we have now encoded the AX = AY = 1 condition as follows: X and
Y are the midpoints of the tangents to the A-excircle. We need to show that one of ABM
or ACM has a perimeter equal to the length of the tangent.
Now the question is: how do we use this?
Let us look carefully again at the diagram. It would seem to suggest that in this case,
ABM is the one with perimeter two (and not ACM). What would have to be true in
order to obtain the relation AB + BM + MA = AU? Trying to bring the lengths closer to the triangle in question, we write AU = AB + BU = AB + BT. So we would need
BM + MA = BT , or MA = MT .
So it would appear that the points X, M, Y have the property that their distance to A
equals the length of their tangents to the A-excircle. This motivates the last addition to our
diagram: construct a circle of radius zero at A, say ω0. Then X and Y lie on the radical axis
of ω0 and Ta; hence so does M! Now we have MA = MT, as required.
Now how does the either-or condition come in? Now it is clear: it reflects whether T
lies on BM or CM. (It must lie in at least one, because we are told that M lies inside the
segment BC, and the tangency points of the A-excircle to BC always lie in this segment
as well.) This completes the solution, which we present concisely below.
Let IA be the center of the A-excircle, tangent to BC at T ,
and to the extensions of AB and AC at U and V. We see that AU = AV = s = 2. Then XY
is the radical axis of the A-excircle and the circle of radius zero at A. Therefore AM = MT.
Assume without loss of generality that T lies on MC, as opposed to MB. Then $ AB +
BM + MA = AB + BM + MT = AB + BT = AB + BU = AU = 2 $ as desired.


问:
Can we find Nash Equilibrium payoffs in degenerate games
In a 2 person, constant sum game, all NE strategies have identical payoff, the 'value' of the game. Typically this is calculated by first calculating one NE strategy.

In many algorithms for enumerating NE of a bimatrix game, we assume that the game is nondegenerate. For example, the simplest algorithm I know of, support enumeration, requires this hypothesis.

What is the simplest algorithm to calculate the payoff that discards this hypothesis?

答:
Yes we can using a modification to the Lemke-Howson algorithm. Normally when running the Lemke-Howson algorithm in tableau form on a non degenerate game, in each iteration there is a unique variable that wins the minimum-ratio test. In the degenerate game we  will  be  faced  with  the  problem  of  breaking  ties and if we have a bad tie-breaking rule the program can end up running forever. The approach used, instead of the basic minimum-ratio test, is called the lexicographic minimum ratio test. In the article Equilibrium points of bimatrix games by Lemke and Howson they describe this under lexicographic method. There is also a description of this in chapter three of the book Algorithmic Game Theory by Nisan, Roughgarden, Tardos, and Vaziran.

The simplest algorithm for any game though would be given a tie during the  solving procedure. To break it, choose randomly and if we later come back to the same case, then break the tie in a different way.


问:
Comparing coefficients of polynomials
If we have two polynomials, say for example, $a_0+a_1x+a_2x^2 = 1+2x+3x^2$ for all $x$, why must it follow that $a_0 = 1,a_2 = 2,$ and $a_3 = 3$? Can we generalize this for any polynomials and infinite sum polynomials as well?</p>

答:
This follows from the fact that if a polynomial is identically vanishing, then it mist be the null polynomial, all of whose coefficients vanish.

Now transpose all terms to one side to have a polynomial which vanishes identically. Setting the coefficients of this polynomial each to $0$ gives you your result.

ContentLicense: CC BY-SA 3.0

Title: Comparing coefficients of polynomials


问:
Recurrence Relations with single roots
I have the following recurrence: $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n$
with initial values $a_1 = 1, a_2 = 4, a_3 = 9$

I have found the characteristic equation to be $x^3-3x^2+3x-1$ and the root to be 1.

My text book is not helpful in how I should go about solving this when I have a single root and don't have the $a_0$ value given.

Any tips on how I could move forward to solve this?

答:
If the characteristic equation has roots $x_1,x_2,\ldots,x_k$, where ($x_i \neq x_j, \, \forall i \neq j$) the root $x_j$ has multiplicity $m_j$, then the solution is of the form
$$a_n = \sum_{j=1}^k P_j(n)x_j^n$$
where $P_{j}(n)$ is a polynomial in $n$ of degree $m_j-1$. Hence, in your case,
$$a_n = (c_0 + c_1 n + c_2 n^2) 1^n, \text{i.e., } a_n = c_0 + c_1 n + c_2 n^2$$ With the initial conditions, we get that
$$a_n = n^2$$</p>

Title: Recurrence Relations with single roots


问:
What would be the graph of $P(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)\cdots$ (with infinitely-many factors)?
Suppose A polynomial $P(x)$, such that,

$$P(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)\cdots$$
having infinite roots. Where $x_1$, $x_2$, $x_3$, $x_4$, $\ldots$ are roots of the polynomial.

My question is what would be the graph of this function? Will it be just the $x$-axis or there will some other curve?

答:
When you have something like this you have to deal with infinite series and radius of convergence and the result is very sensitive to the way in which you define your objects analytically.

I think a good example for this comes from number theory and the theory of partitions. In there the following equation (due to Euler) is proven as a formal expansion:

$\prod_{k = 1}^{\infty}(1 - x^k) = 1 + \displaystyle\sum_{j = 1}^{\infty}(-1)^j(x^{j(3j-1)/2} + x^{j(3j+1)/2})$.

This is a formality between series, without reference to actual values of x, and is obtained byt multiplying the product and collecting powers of $x$ (the proof of it is a different story, and it's called pentagonal number theorem).

Nevertheless you can study both sides and study radius of convergence. For example, with the right hand side, if $|x| \le r < 1$ then we have

$\displaystyle\sum_{j = 1}^{\infty}\left|(-1)^j(x^{j(3j-1)/2} + x^{j(3j+1)/2})\right| \le 2\displaystyle\sum_{j = 1}^{\infty}r^{j(3j-1)/2} \le 2(1 + r  + r^2 + ...) = \dfrac{2}{1 - r}$,

which proves this functions converges absolutely and uniformly in compacts of $(-1, 1)$. This means there exists a continuous function to which this series converges and so it makes sense to graph it, but it might not be the graph of a polynomial and it only makes sense in the region of convergence.

Analogously you can check what happens with the product and obtain a similar convergence phenomenon (by expanding the products into polynomials you can basically repeat the argument above). You can see the first products here:

If instead you do it with the polynomials you get (I did it for k = 4):

These are starting to look similar and it indeed seems some function is at the limit indeed.

Notice how all these graphs go through (0, 1) since in both the product and the series version, in the terms that form the sequence these graphs all go thorugh there. On the other hand, in the product version all go through $(\pm, 0)$ since these are zeros of all finite products, meanwhile in the series version it doesnt have those zeroes yet, it just approaches to it in the limit.

Finally, notice how outside $[-1, 1]$ the graphs do not seem to go nowhere, and that makes sense since  in there the behaviour at $\pm\infty$ dominates and the degree of the polynomial changes the behaviour.


问:
Closed/open components confusion
I'm studying point-set topology from the John B. Conway book and in the chapter about connectedness there is a result that states that the closure of a connected set is connected and each component is closed, and then right away the next result states that the components of an open set in $\mathbb R^q$ are all open, both make intuitive sense at some level but they don't make sense together, what am I missing? Or is it just the fact that connected sets are both open and closed?

答:
There’s no conflict when you realize that they’re talking about two different spaces.

For a simple example consider the open set $U=(0,1)\cup(2,3)$ in $\Bbb R$. Let $\tau$ be the usual topology on $\Bbb R$, and let $\tau_U$ be the relative (subspace) topology on $U$. The components of $U$ are $(0,1)$ and $(2,3)$. These components are open both in $\tau_U$ and in $\tau$. They are also closed in $\tau_U$, but they are not closed in $\tau$. In both $\tau_U$ and $\tau$, however, their closures are connected: they are their own closures in $\tau_U$, and their closures in $\tau$ are $[0,1]$ and $[2,3]$.

To sum up, a component of a set is always closed in that set, but it need not be closed in the ambient space. A component of an open set in $\Bbb R^n$ is always both open and closed in that set, and it is always open in $\Bbb R^n$, but it is closed in $\Bbb R^n$ if and only if it is all of $\Bbb R^n$.

Title: Closed/open components confusion


问:
Definite integral of a product of normal pdf and cdf
Denote the pdf of the standard normal distribution as $\phi(x)$ and cdf as $\Phi(x)$. Does anyone know how to calculate $\int_{-\infty}^y \phi(x)\Phi(\frac{x−b}{a})dx$?

Notice that this question is similar to an existing one,

https://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution

the only difference being that I'm computing the integral over $(-\infty, y)$ for some real $y$, rather than over the entire real line.

Thank you!

答:
What follows isn't exactly the answer to your question but should be what you need after you make the necessary change of variables.

\begin{align}
&\int_{l}^{h} \textrm{d} x\, \frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right) \Phi(x)\nonumber\\
=& \tfrac{1}{2}(\Phi(z_h) - \Phi(z_l)) \\
& - \tfrac{1}{2}\left[\frac{\mu}{z_h} < 0\right] \\
& + \tfrac{1}{2}\left[\frac{\mu}{z_l} < 0\right] \\
& - T\left(z_h, \frac{h}{z_h}\right) \\
& + T\left(z_l, \frac{l}{z_l}\right) \\
& - T\left(\frac{\mu}{\rho}, \frac{\mu \sigma + z_h \rho^2}{\mu}\right) \\
& + T\left(\frac{\mu}{\rho}, \frac{\mu \sigma + z_l \rho^2}{\mu}\right)
\end{align}
where
\begin{align}
\rho &=\sqrt{1+\sigma^2} \\
z_l &= \frac{l - \mu}{\sigma} \\
z_h &= \frac{h - \mu}{\sigma} \\
\end{align}
and where $\phi,\Phi$ are the pdf and cdf of the standard Normal distribution (i.e. Normal with mean $0$ and std dev $1$) and $T$ denotes the Owen's T function.

Here's an implementation of this integral using TensorFlow Probability; it should be easily adaptable to any numerical backend which offers Owen's T function.

```<code>import tensorflow as tf
import tensorflow_probability as tfp
from tensorflow_probability.python.internal import dtype_util
tfd = tfp.distributions

def gaussian_expected_standard_gaussian_cdf(m, s, lo, hi):
  """Computes:
     E[ Normal(0,1).cdf(X) | lo<X<hi, X~Normal(m,s)] * (
       Normal(m,s).cdf(hi) - Normal(m,s).cdf(lo))
  """
  m, s, lo, hi, dtype = _prepare_args(m, s, lo, hi)
  d = tfd.Normal(dtype(0), 1)
  # We derive our implementation by beginning with the result 10,010.3 (pg 403) of "A Table of Normal Integrals" by DB Owen.
  #
  # rho = sqrt(1+b^2)
  #
  # int phi(x) Phi(a + bx) dx                   # Result #10,010.3, pg 403.
  # =   T(x, a / (x rho)) + T(a / rho, x rho / a)
  #   - T(x, (a+bx)/x)
  #   - T(a / rho, (a b + x rho**2)/a)
  #   + Phi(x) Phi(a / rho)
  #   + C
  # =   ([Phi(x) + Phi(a/rho)]/2                # Property 2.7, pg 414. See also
  #      - Phi(x) Phi(a/rho)                    # wikipedia page on Owen's T.
  #      - [a/x < 0]/2)
  #   - T(x, (a+bx)/x)
  #   - T(a / rho, (a b + x rho**2)/a)
  #   + Phi(x) Phi(a / rho)
  #   + C
  # =  Phi(x)/2 - [a/x < 0]/2
  #   - T(x, (a+bx)/x)
  #   - T(a / rho, (a b + x rho**2) / a)
  #   + C + Phi(a/rho)/2
  #
  # Evaluate the indefinite integral from u to v:
  #
  #  + (Phi(v) - Phi(u))/2
  #  - [a/v < 0]/2
  #  + [a/u < 0]/2
  #  - T(v, (a+bv)/v)
  #  + T(u, (a+bu)/u)
  #  - T(a / rho, (a b + v rho**2) / a)
  #  + T(a / rho, (a b + u rho**2) / a)
  #
  # Make substitutions a=m, b=s and change of variables
  # v=(hi-m)/s=:zhi  and u=(lo-m)/s:=zlo:
  #
  #  + (Phi(zhi) - Phi(zlo)) / 2
  #  - [m / zhi < 0]/2
  #  + [m / zlo < 0]/2
  #  - T(zhi, hi / zhi)
  #  + T(zlo, lo / zlo)
  #  - T(m / rho, (m s + zhi rho**2) / m)
  #  + T(m / rho, (m s + zlo rho**2) / m)
  rho2 = s**2 + 1.
  m_over_rho = m * tf.math.rsqrt(rho2)
  zhi = (hi - m) / s
  zlo = (lo - m) / s
  return (
    0.5 * (d.cdf(zhi) - d.cdf(zlo))
    - tf.where(tf.equal(m, 0) | tf.equal(m < 0, zhi < 0),
               dtype(0.), dtype(0.5))
    + tf.where(tf.equal(m, 0) | tf.equal(m < 0, zlo < 0),
               dtype(0.), dtype(0.5))
    - _safe_owens_t(zhi, hi, zhi)
    + _safe_owens_t(zlo, lo, zlo)
    - _safe_owens_t(m_over_rho, m * s + zhi * rho2, m)
    + _safe_owens_t(m_over_rho, m * s + zlo * rho2, m)
  )

# We need to be able to call Owen's T with some potentially "odd"
# inputs; the following function ensures no NaNs and preserves gradients
# using the "double where trick."
def _safe_owens_t(h, numer, denom):
  # Property 2.4 (pg 414) states that:
  #   T(h, inf) = (1 - Phi(|h|))/2
  # We use this property to skip the Owen's T thus avoiding nans.
  # We'll interpret 0/x=0 and inf/inf=inf; use what may seem like convoluted
  # logic to preserve gradients.
  is_numer_zero = tf.equal(numer, 0)
  is_denom_nonzero = tf.not_equal(denom, 0)
  has_finite = tf.math.is_finite(denom) | tf.math.is_finite(numer)
  safe_denom = tf.where(is_denom_nonzero & has_finite, denom, 1.)
  dtype = dtype_util.as_numpy_dtype(h.dtype)
  return tf.where(
      is_numer_zero | is_denom_nonzero,
      tfp.math.owens_t(h, numer / safe_denom),
      0.5 * tfd.Normal(dtype(0), 1).survival_function(tf.math.abs(h)))

def _prepare_args(m, s, lo, hi=None):
  dtype = dtype_util.common_dtype([m, s], dtype_hint=tf.float32)
  m = tf.convert_to_tensor(m, dtype, name='m')
  s = tf.convert_to_tensor(s, dtype, name='s')
  lo = tf.cast(lo, dtype, name='lo')
  dtype = dtype_util.as_numpy_dtype(dtype)
  if hi is None:
    return m, s, lo, dtype
  hi = tf.cast(hi, dtype, name='hi')
  return m, s, lo, hi, dtype
</code>```

You can test the code like:

```<code>m,s = -0.25,1.43
lo,hi = -np.inf,1.
d = tfd.Normal(m,s)
x = d.sample(int(100e6))
x = tf.boolean_mask(x,(x>lo)&(x<hi))
[
  tf.reduce_mean(tfd.Normal(0,1).cdf(x)) * (d.cdf(hi) - d.cdf(lo)),
  gaussian_expected_standard_gaussian_cdf(m,s,lo,hi)
]
</code>```

Body:
What follows isn't exactly the answer to your question but should be what you need after you make the necessary change of variables.

\begin{align}
&\int_{l}^{h} \textrm{d} x\, \frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right) \Phi(x)\nonumber\\
=& \tfrac{1}{2}(\Phi(z_h) - \Phi(z_l)) \\
& - \tfrac{1}{2}\left[\frac{\mu}{z_h} < 0\right] \\
& + \tfrac{1}{2}\left[\frac{\mu}{z_l} < 0\right] \\
& - T\left(z_h, \frac{h}{z_h}\right) \\
& + T\left(z_l, \frac{l}{z_l}\right) \\
& - T\left(\frac{\mu}{\rho}, \frac{\mu \sigma + z_h \rho^2}{\mu}\right) \\
& + T\left(\frac{\mu}{\rho}, \frac{\mu \sigma + z_l \rho^2}{\mu}\right)
\end{align}
where
\begin{align}
\rho &=\sqrt{1+\sigma^2} \\
z_l &= \frac{l - \mu}{\sigma} \\
z_h &= \frac{h - \mu}{\sigma} \\
\end{align}
and where $\phi,\Phi$ are the pdf and cdf of the standard Normal distribution (i.e. Normal with mean $0$ and std dev $1$) and $T$ denotes the Owen's T function.

Here's an implementation of this integral using TensorFlow Probability; it should be easily adaptable to any numerical backend which offers Owen's T function.

import tensorflow as tf
import tensorflow_probability as tfp
from tensorflow_probability.python.internal import dtype_util
tfd = tfp.distributions

def gaussian_expected_standard_gaussian_cdf(m, s, lo, hi):
  """Computes:
     E[ Normal(0,1).cdf(X) | lo<X<hi, X~Normal(m,s)] * (
       Normal(m,s).cdf(hi) - Normal(m,s).cdf(lo))
  """
  m, s, lo, hi, dtype = _prepare_args(m, s, lo, hi)
  d = tfd.Normal(dtype(0), 1)
  # We derive our implementation by beginning with the result 10,010.3 (pg 403) of "A Table of Normal Integrals" by DB Owen.
  #
  # rho = sqrt(1+b^2)
  #
  # int phi(x) Phi(a + bx) dx                   # Result #10,010.3, pg 403.
  # =   T(x, a / (x rho)) + T(a / rho, x rho / a)
  #   - T(x, (a+bx)/x)
  #   - T(a / rho, (a b + x rho**2)/a)
  #   + Phi(x) Phi(a / rho)
  #   + C
  # =   ([Phi(x) + Phi(a/rho)]/2                # Property 2.7, pg 414. See also
  #      - Phi(x) Phi(a/rho)                    # wikipedia page on Owen's T.
  #      - [a/x < 0]/2)
  #   - T(x, (a+bx)/x)
  #   - T(a / rho, (a b + x rho**2)/a)
  #   + Phi(x) Phi(a / rho)
  #   + C
  # =  Phi(x)/2 - [a/x < 0]/2
  #   - T(x, (a+bx)/x)
  #   - T(a / rho, (a b + x rho**2) / a)
  #   + C + Phi(a/rho)/2
  #
  # Evaluate the indefinite integral from u to v:
  #
  #  + (Phi(v) - Phi(u))/2
  #  - [a/v < 0]/2
  #  + [a/u < 0]/2
  #  - T(v, (a+bv)/v)
  #  + T(u, (a+bu)/u)
  #  - T(a / rho, (a b + v rho**2) / a)
  #  + T(a / rho, (a b + u rho**2) / a)
  #
  # Make substitutions a=m, b=s and change of variables
  # v=(hi-m)/s=:zhi  and u=(lo-m)/s:=zlo:
  #
  #  + (Phi(zhi) - Phi(zlo)) / 2
  #  - [m / zhi < 0]/2
  #  + [m / zlo < 0]/2
  #  - T(zhi, hi / zhi)
  #  + T(zlo, lo / zlo)
  #  - T(m / rho, (m s + zhi rho**2) / m)
  #  + T(m / rho, (m s + zlo rho**2) / m)
  rho2 = s**2 + 1.
  m_over_rho = m * tf.math.rsqrt(rho2)
  zhi = (hi - m) / s
  zlo = (lo - m) / s
  return (
    0.5 * (d.cdf(zhi) - d.cdf(zlo))
    - tf.where(tf.equal(m, 0) | tf.equal(m < 0, zhi < 0),
               dtype(0.), dtype(0.5))
    + tf.where(tf.equal(m, 0) | tf.equal(m < 0, zlo < 0),
               dtype(0.), dtype(0.5))
    - _safe_owens_t(zhi, hi, zhi)
    + _safe_owens_t(zlo, lo, zlo)
    - _safe_owens_t(m_over_rho, m * s + zhi * rho2, m)
    + _safe_owens_t(m_over_rho, m * s + zlo * rho2, m)
  )

# We need to be able to call Owen's T with some potentially "odd"
# inputs; the following function ensures no NaNs and preserves gradients
# using the "double where trick."
def _safe_owens_t(h, numer, denom):
  # Property 2.4 (pg 414) states that:
  #   T(h, inf) = (1 - Phi(|h|))/2
  # We use this property to skip the Owen's T thus avoiding nans.
  # We'll interpret 0/x=0 and inf/inf=inf; use what may seem like convoluted
  # logic to preserve gradients.
  is_numer_zero = tf.equal(numer, 0)
  is_denom_nonzero = tf.not_equal(denom, 0)
  has_finite = tf.math.is_finite(denom) | tf.math.is_finite(numer)
  safe_denom = tf.where(is_denom_nonzero & has_finite, denom, 1.)
  dtype = dtype_util.as_numpy_dtype(h.dtype)
  return tf.where(
      is_numer_zero | is_denom_nonzero,
      tfp.math.owens_t(h, numer / safe_denom),
      0.5 * tfd.Normal(dtype(0), 1).survival_function(tf.math.abs(h)))

def _prepare_args(m, s, lo, hi=None):
  dtype = dtype_util.common_dtype([m, s], dtype_hint=tf.float32)
  m = tf.convert_to_tensor(m, dtype, name='m')
  s = tf.convert_to_tensor(s, dtype, name='s')
  lo = tf.cast(lo, dtype, name='lo')
  dtype = dtype_util.as_numpy_dtype(dtype)
  if hi is None:
    return m, s, lo, dtype
  hi = tf.cast(hi, dtype, name='hi')
  return m, s, lo, hi, dtype

You can test the code like:

m,s = -0.25,1.43
lo,hi = -np.inf,1.
d = tfd.Normal(m,s)
x = d.sample(int(100e6))
x = tf.boolean_mask(x,(x>lo)&(x<hi))
[
  tf.reduce_mean(tfd.Normal(0,1).cdf(x)) * (d.cdf(hi) - d.cdf(lo)),
  gaussian_expected_standard_gaussian_cdf(m,s,lo,hi)
]


问:
Show that a group of order $2^2 \cdot 5 \cdot 23^r$ is solvable but not simple.
Let $G$ be a group of order $2^2 \cdot 5 \cdot 23^r$ where $r \gt 0$

a) Show that $G$ is not simple.

b) Show that $G$ is solvable.

My attempt:

a) Assume that:

$S$ is the set of sylow-$23$-subgroups and $n_{23}$ is the number of conjugates of $S$

By Sylow's 3rd theorem, we have

$n_{23} | 2^2 \cdot 5$ and $n_{23}\equiv 1 \pmod {23}$

Which implies that $n_{23}=1,\ 2^a,\ 5,\ 2^b \cdot 5$ where $a,b\leq 2$, and $n_{23}=1+23k$

Since $23>2^a,\ 5,\ 2^b.5$, if we assume that $n_{23}= 2^a,\ 5, $ or $2^b \cdot 5$, that would lead us to a contradiction. Hence, $n_{23}=1$, and then $S$ is normal.

As a result, $G$ is not simple.

b) $S$ is a $23$-group $\implies S$ is nilpotent $\implies S$ is solvable.

Since $S$ is normal, $G/S$ is a group. We have

$o(G/S)=23$ $\implies G/S$ is cyclic $\implies G/S$ is abelian $\implies G/S$ is solvable $\implies G$ is solvable.

Edit:

$S$ is a $23$-group $\implies S$ is nilpotent $\implies S$ is solvable.

Since $S$ is normal, $G/S$ is a group. We have

$o(G/S)=2^2 \cdot 5$

Suppose that $G/S=H$.

We have $o(H)=2^2 \cdot 5$

Suppose that

$R$ is the set of Sylow-${5}$-subgroups and $n_{5}$ is the number of conjugates of $R$

From the 3rd Sylow's theorem, we have

$n_5\equiv 1 \pmod 5$ and $n_5 | 2^2$

which implies that

$n_5=1+5k$ and $n_5=1,\ 2,$ or $2^2$.

Since $5>\gt 2,\ 2^2$, the only possible value of $n_5$ is $1$. So we can conclude that $R$ is normal in $H$.

$R$ is a $5$-group $\implies R$ is nilpotent $\implies R$ is solvable.

Since $R$ is normal, $H/R$ is a group. We have

$o(H/R)=4$ $\implies H/R$ is abelian $\implies H/R$ is solvable $\implies H=G/S$ is solvable  $\implies G$ is solvable.

答: The Sylow $5$-subgroup of $G/S$ must be normal because $n_5(G/S)=1, 2, \text{or } 4$.


问:
Number of functions
Let $F$ denote the set of all functions from $\{1,2,3\}$ to $\{1,2,3,4,5\}$

a) Find and simplify the number of functions $f$ in $F$ so that $f(1)=4$.

b) Find and simplify the number of one-to-one functions $f$ in $F$ so that $f(1)$ is greater than or equal to $4$.

c) Find and simplify the number of functions $f$ in $F$ so that $f(1)$ does not equal $f(2)$.

For (a) I have no idea how to write it out, and for (c) can't I make $f(1)=1$ and $f(2)=2,3,4,5$ that way they don't equal each other?

答:
This whole problem is basically just an exercise in using the multiplication principle, also sometimes called the Chinese menu principle:

Suppose that you are making a sequence of $k$ choices. There are $n_1$ different ways to make the first choice, $n_2$ different ways to make the second choice, and so on, through $n_k$ different ways to make the $k$-th choice. Then there are altogether $n_1n_2\dots n_k$ different ways to make the whole sequence of choices.

(a) Once you set $f(1)=4$, you must assign values to $f(2)$ and $f(3)$. In each case this means picking one of the $5$ numbers $1,2,3,4$, and $5$, so there are $5\cdot5=25$ ways to make the two choices. Once you’ve done that, you’ve completely defined the function $f$, so there are $25$ functions $f:\{1,2,3\}\to\{1,2,3,4,5\}$ such that $f(1)=4$.

(b) The same basic approach to solving the problem and to explaining the solution can be used here as well.

(c) No, you can’t simply make $f(1)=1$ and choose $f(2)\in\{2,3,4,5\}$: if you count only those functions, you’re missing all of the functions that have $f(1)=3$ and $f(2)=5$, for instance. Once again look at it in terms of building $f$ one value at a time. You can let $f(1)$ be any of the five numbers $1,2,3,4$, and $5$. No matter which value you choose, you can let $f(2)$ be any of the four remaining numbers. Finally, you can let $f(3)$ be any of the five numbers. This sequence of three choices can be made in $5\cdot4\cdot5=100$ ways.

Title: Number of functions


问:
If $A$ is a matrix with rank($A$)=1,then the minimal polynomial of $A$ is of degree $2$.

If  $n \ge 2$ and if  $A$ is a matrix with rank($A$)=1,then the minimal polynomial of $A$ is of degree $2$.

So, we see that $rank (A) =1$.So, nullity of $A \ge 1$.Now we try to find the range space of $A$.If $y \in R(A)$ then $A(x)=y$.

We see that null space  of $A$ is being spanned by $n-1$ linearly independent vectors. We can extend this to a basis of $\mathbb{R}^n$.Let $x_1,...,x_n$ be the basis where $x_1,..,x_{n-1} \in ker(A)$.We see that $x_n$ must not be in ker($A$) otherwise $A$ is the zero matrix so that $A=0$.

Let $x_n$ be the vector which spans the range space Of $A$ such that $A(\sum(c_ix_i))=x_n$ where $1 \le i \le n$.So from here we can conclude that $0,(c_n)^{-1}$ are the only eigenvalues values. Hence the minimal polynomial becomes $x(x-(c_n)^{-1})$ .

答: Let $x_1,...,x_n$ a basis such that $(x_1,...,x_{n-1})$ is a basis of $ker A$ and $x_n$ a basis of $Im A$. We have $A(x_n)=cx_n$, $c\neq 0$ since $A\neq 0$. Write $P=X(X-c)$, $P(A)(e_i)=(A-cI)(A(x_i)=0$ if $i<n$. $P(A)(x_n)=A(A-cI)(x_n)=0$. Implies that $P(A)=0$.


答:
Correct. $(.4)^4$ because we want the first four to be people who don't prefer brand A, and then we want the 5th person, $.6$ chance, to prefer brand A.

Your probability of success is $60\%$. This can be written as $\frac{6 \text{ successes}}{10 \text{ trials}}$. To find the expected value, you flip it. You need $\frac{10 \text{ trials}}{6 \text{ successes}}=5/3 \text { trial per success}$.

Title: Not sure of my answers on a specific probability problem

Body:
Sixty percent of the consumers' population is reputed to prefer a particular brand A of toothpaste. If a group of consumers are selected at random and interviewed, what is the probability that exactly 5people must be selected until interviewing the first consumer who prefers brand A? What is the expected number of consumers to be selected until the first one who prefers brand A is interviewed?

$Answers$

a. $P(5) = (0.6)(0.4)^4$ =$0.0514$

b. $E(x) = (5)(0.6) = 3$

Did I answer it correctly?


问:
Asymptotic or Exact formula for Harmonic Number
I know Faulhaber's formula for positive integers. However, is there an asymptotic or exact formula for generalized Harmonic number. For example,
how can I calculate
\begin{align*}
\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}.
\end{align*}
I am looking forward to an exact formula or asymptotic formula.
Any help or references will be appreciated.</p>

答:
There is an exact formula, that I just created.
It is pretty similar to the Euler-Mclaurin formula, except it gives you the exact residual terms.

For all complex $k$, except -1:
$$\sum_{j=1}^{n}j^{k}=\frac{n^{k+1}}{k+1}+\frac{n^{k}}{2}+\zeta(-k)+\frac{\pi\,n^{k+2}}{k+1}\int_{0}^{\pi/2}\left(\sec{v}\,\text{csch}{(n\pi\tan{v})}\right)^2\left(1-\frac{\cos{(k+1)\,v}}{(\cos{v})^{k+1}}\right)\,dv$$

Please refer to my RG paper for the detailed proof:

Analytic continuation for various functions


问:
Choose $4$ students out $15$ but Andy can only be selected if Jordan is chosen as treasurer. What is the probability of this event happening?
A committee of four people will be selected among a group of 15 students . The committee will have the following positions: president, secretary, treasurer and marshal. Andy, one of the students , will accept a position in a committee as long as Jordan is selected as the treasurer. What is the probability that Andy serves in a committee? Your answer must be rounded to the nearest hundredth.

In my answer I first found the total ways to select 4 out of 15 students. Next since Jordan takes one position there are three positions left and if one of the three is given to Andy, only two are left. Thus, this equals to:
(12C2)(3!)/(15C4)

答:
It is simpler to just compute the Pr of Jordan as treasurer and Andy in another post in the committee

Thus P($\mathbf J$  in committee)$\cdot$ P($\mathbf J$ gets treasurer post)$\cdot$ P($\mathbf A$ in commitee)

$ = \dfrac4{15}\cdot\dfrac 1 4 \times \dfrac3{14} = \dfrac 1 {70}$

Who and where the rest in the committee are doesn't matter !


问:
What is the smallest solution to $x = 2 \operatorname{mod} (6y+1)$, $x = 3 \operatorname{mod} (12y+1)$, $x= 3 \operatorname{mod} (18y+1)$?
I'd like to solve a system of modular equations, but it's a somewhat unusual system. Specifically, what is the smallest natural number $x$ for which $x = 2\pmod{6y+1}$, and $x=3\pmod{12y+1}$, and $1 \pmod{18y+1}$ for some natural number $y$?

If $y=1$, then WolframAlpha tells me that $x=1654$.  But is it possible that a bigger choice of $y$ could yield a smaller value of $x$?  This came up in the context of Godel coding, by the way.

答:
I don't have reputation to comment, so trying to answer here.

To make all 3 modulo co-prime to each other. we take smallest y>0. naturally if we take y=1 then we have modulo as {7,13,19} and they are prime numbers. so, choice of y seems good.

Now the smallest number should be within ${7*13*19=1729}$, so 1654 is the smallest answer you get.

I don't know what could be the technique which gives  higher value of y and smaller value of x.


问:
Find the probability mass function of $V=\min(X,Y)$
Let $X$ and $Y$ be independent $\mathrm {Geom}(p)$ random variables.

Find the probability mass function of $V=\min(X,Y).$

I don't understand what I am supposed to find the probability for. I know that $X\sim \mathrm{Geom}(p)$ for $p\in[0,1]$ if $\mathbb{P}(X=k)=(1-p)^{k-1}p$. But how would I use this to find a p.m.f?

答:
$$P(V=k)=P(X=k, Y \geq k)+P(Y=k, X \geq k)$$ $$=2P(X=k, Y \geq k)-P(X=k,Y=k)$$ $$= \sum\limits_{i=k}^{\infty}(1-p)^{k-1}p (1-p)^{i-1}p-(1-p)^{k-1}p (1-p)^{k-1}p.$$

Use the formula for a geometric sum to finish the calculation.


问:
Question on Summation Formula for Fermat Quotients
According to Wolfram (https://mathworld.wolfram.com/FermatQuotient.html),

$\displaystyle \frac{2^{p-1}-1}{p}=\frac{1}{2}\sum_{n=1}^{p-1}\frac{\left(-1\right)^{n-1}}{n}$

I am struggling to understand how the sum is derived because the RHS converges to $\frac{1}{2}\ln2$ as $p\rightarrow\infty$. However, Fermat quotients diverge. I would be grateful for any help on this.

答:
The meaning of $\frac1a \pmod m$

The idea behind the notation $\frac 1a$ modulo $m$ comes from ring theory (Please use the Wikipedia pages for groups and rings to get in touch with these concepts). Indeed, the integers $0,...,m-1$ form a group under operation $a + b = c$ where $c$ is the remainder when $a+b$ (the usual sum) is divided by $m$, and under the multiplication operator $a \cdot b =  c$ where $c$ is the remainder when $a\cdot b$ (the usual product) is divided by $m$.

Now, $1$ is called the multiplicative identity , because $a \cdot 1 =a$ for any $a$. In correspondence with group theoretic reasoning, we call an element $a \in \{0,...,m-1\}$ invertible if there is a $b$ such that $a \cdot b = 1$. That is, if there is an integer $b \in \{0,1,...,m-1\}$ such that $a \cdot b$ leaves a remainder of $1$ when divided by $m$.

It can be proven that the inverse, if it exists, is unique. The notation of the inverse of $a$ , if it exists, is either $\frac 1a$ or $a^{-1}$.

Now, there is an easy result for when $a$ is invertible in $\{0,...,m-1\}$ (we call this as invertibility modulo $m$).

$a$ is invertible if and only if $\gcd(a,m) = 1$.

So $\frac 1a$ is a valid element of $\{0,1,...,m-1\}$ whenever $a$ is coprime to $m$.

$\textbf{A SMALL CAVEAT}$ : The quantity $\frac{2^{p-1}-1}{p}$ is an actual division by $p$. You can see that $\gcd(p,p) = p \neq 1$ so $\frac 1p$ doesn't make sense modulo $p$. That is an actual division by $p$. I will show this in my example below. It is an abuse of notation of $\frac{\cdot}{\cdot}$ to represent both actual division and modulo division, and we should be careful of this in general. You can ask if $\frac{2^{p-1}-1}{p}$ will always be an integer : this is true by Fermat's little theorem.

Examples

Let $m=5$ and $a=2$. Then $\gcd(2,5) = 1$ so $\frac 12$ is a well defined element modulo $5$. Can we find it? By experimenting, $2 \cdot 3 = 6$ leaves remainder $1$ upon division by $5$, so $3 = \frac 12$ modulo $5$. In particular, when we are dealing with the ring of integers modulo $5$, and we see $\frac 12$ somewhere, it actually stands for $3$.

Let $m =15$ and $a = 8$. Then $\gcd(8,15) = 1$ and you can see that $\frac 18 = 2$.

Let $m = 16$ and $a=10$. Then $\gcd(10,16) \neq 1$ so $\frac 1{10}$ cannot be defined or talked about modulo $16$.

Let $m$ be a prime, and $a \neq 0$. Then $\gcd(a,m) = 1$ so $\frac 1a$ always exists, although one can find it only by experimentation or algorithms like reversing the Euclidean algorithm.

On the identity in the question

Thus, what does the identity mean finally? Well, the RHS is not a real number, but rather an integer modulo $p$. The only way to illustrate this is to actually compute both the sides for a certain prime $p$. We will compute each term separately. Let's go with $p=5$.

LHS : $\frac{2^{p-1}-1}{p} = \frac{2^4-1}{5} = 3$, an actual division, remember the caveat.

RHS : We have $\frac 12 = 3$. Then for each of $n=1$ to $4$ we compute $\frac{(-1)^{n-1}}{n}$. For $n=1$, $\frac {(-1)^0}{1} = \frac 11 = 1$. Then for $n=2$ we have $\frac{(-1)^1}{2} = -\frac{1}{2}= -3 = 2$. For $n=3$ we have $\frac{(-1)^2}{3} = \frac 13 = 2$. For $n=4$ we have $\frac{(-1)^3}{4} = \frac {-1}4 = -4 = 1$ (since $4\times 4 = 16 = 1$). So the sum finally is $3(1+2+2+1) = 3 \cdot 6 = 18 =3$.

So both RHS and LHS are equal to $3$ modulo $5$. That is how you would read that identity.

Historical significance

So what's the fuss about this whole identity? Well, it comes from Fermat's last theorem. Recall the Fermat's last theorem : for every $n \geq 3$ there are no positive integers $x,y,z$ with $x^n+y^n=z^n$.

It is easy to prove that Fermat's last theorem is true if and only if it is true for $n$ taking just prime values, $n>2$. So we can study each prime. While studying each prime, the Fermat quotient comes up, in a way that I can't explain but is illustrative.

Wieferich proved in 1909 that if the FLT was contradicted for $n=p$, then it is necessary that $\frac{2^{p-1}-1}p = 0 \pmod{p}$. That is, we must have that the Fermat quotient for $p$ with base $2$ must be equal to $0$. This generated interest in the identity given because it naturally provided a different way of computing the Fermat quotient.

Only two primes (such primes with Fermat quotient with base $2$ equal to $0$ are called Wieferich primes) are known! They are $1093$ and $3511$.

The result does not stay true for only base $2$ : it was later extended to all prime bases up to $89$. That is : if there is a counterexample of FLT for $n=p$ then the Fermat quotient of $p$ with base $k$ must be zero for all primes $k$ from $2$ to $89$! If only we could show that no such prime existed, we'd be done, but this has been elusive.

Proof of the identity

The proof of the identity is not easy, because it involves some manipulations that you will not be able to get at first glance. So I will not include it here. Instead, I will ask you to visit the post here, which goes through the proof of a very similar identity. You can attempt to use the proof there to prove this.

The main idea is to get yourself comfortable with arithmetic modulo $p$. This is necessary before you can understand the above identity.


答: Hint: $P(V=k) = P(\min(X, Y)=k)$ is equal to $$P(X=k\cap Y>k)+P(X>k\cap Y=k)+P(X=k\cap Y=k)$$

Title: Find the probability mass function of $V=\min(X,Y)$

Body:
Let $X$ and $Y$ be independent $\mathrm {Geom}(p)$ random variables.

Find the probability mass function of $V=\min(X,Y).$

I don't understand what I am supposed to find the probability for. I know that $X\sim \mathrm{Geom}(p)$ for $p\in[0,1]$ if $\mathbb{P}(X=k)=(1-p)^{k-1}p$. But how would I use this to find a p.m.f?


问:
How to show this map is a chart map?

For a smooth $n$-manifold $M$ and smooth, real-valued functions $y_1, \cdots , y_n$ defined on a neighborhood $U$ of some $p \in M$, such that $\{dy^1|_p, \cdots , dy^n|_p\}$ is a basis of $T^∗_pM$, show that $\phi = (y_1,\cdots, y_n): U \to \mathbb{R}^n$ is a chart map on a sub-neighborhood of $U$ containing $p$.

So I tried taking the differential of each $y^i$ and writing it as a linear combination of $\{dy^1|_p, \cdots , dy^n|_p\}$ with a change of basis. Then I fixed a dual basis to $\{dy^1|_p, \cdots , dy^n|_p\}$ (basis in the double dual, which is identified with the tangent space). I wanted to show that the kernel of each $dy^i_p$ is trivial, which would show that $d\phi_p$ is invertible. Then I use the implicit function theorem ( a hint I was given).

My issue is that, I don't know how to show that $dy^i_p$ have a trivial kernel. I'm afraid that I get lost in all the notation, and with the reason for why I need to perform a change of basis. If anyone could clarify this problem for me, that would be great.

答:
If $\{dy^1|_p,\ldots, dy^n|_p\}$ is a basis for $T_p^*M$, then this means that the differential of $\phi = (y^1,\ldots, y^n)\colon U \to \Bbb R^n$ at $p$, i.e., the linear map $d\phi_p\colon T_pM \to \Bbb R^n$ is an isomorphism. (It is not true in general that the kernel of each $dy^i|_p$ is trivial, this is impossible by the rank-nullity theorem.)

This is because $d\phi_p = (dy^1|_p,\ldots, dy^n|_p)$. Namely, if we write $(e_1,\ldots, e_n)$ for the dual basis of $(dy^1|_p,\ldots,dy^n|_p)$, any $v \in T_pM$ can be written as $v = \sum_{i=1}^n dy^i|_p(v) e_i$, but $v \in \ker d\phi_p$ means that $dy^i|_p(v) = 0$ for all $i$, so that $v=0$. Then $\ker d\phi_p = \{0\}$ says that $d\phi_p$ is injective, but $\dim T_pM = \dim \Bbb R^n$, so $d\phi_p$ is an isomorphism, as claimed.

Now the inverse function theorem kicks in and we conclude that there is an open neighborhood $U'$ of $p$, contained in $U$, for which $\phi|_{U'} \colon U' \to \phi[U']\subseteq \Bbb R^n$ is a diffeomorphism onto the open set $\phi[U']$.

Title: How to show this map is a chart map?


问:
calculating surface area of an arbitrary portion of a hypersphere
This is coming from a statistics background.  I have several distributions defined on the hypersphere (really, the positive orthant of the hypersphere).  I'm trying to compare these candidate distributions against the original data I used to fit the distribution.

For this, I'm intending to use something like Hellinger distance.  I can calculate the probability of falling into a particular area numerically--using the posterior predictive and empirical, I just compute the number that fell into a said area and divide by total sample size.  But for Hellinger distance, I still need the area that I'm integrating across--that means the surface area of an arbitrary subset of the hypersphere.

This $n$-dimensional geometry is somewhat outside my current understanding.  Is there any established method for calculating surface area of an arbitrary portion of the hypersphere?

答:
All very well understood:

$$ S_n = \frac{2^{(n+1)/2} \pi^{(n-1)/2)}}{(n-2)!!} $$

for $n$ odd;

$$S_n = \frac{2 \pi^{n/2}}{(n/2 -1)!}$$

for $n$ even.

And the area of the positive orthant is a trivial fraction of this.

(Computer that orthant area as a proportion for $n = 2,3,4...$ and see if you see the rather obvious rule.)


答:
The probability that the minimum of X, Y is k is the probability that they're both k, X is k and Y is larger, or Y is k and X is larger. Fortunately, the latter two are the same because the distributions are the same.

$$\begin{split}P(V=k)&=P(X=k,Y>k)+P(Y=k,X>k)+P(X=k,Y=k)\\
&=2P(X=k,Y>k)+P(X=k,Y=k)\\
&=2\left((1-p)^{k-1}p\left[1-\sum_{i=0}^{k-1}(1-p)^ip\right]\right)+\left((1-p)^{k-1}p\right)^2\end{split}$$

Use $\sum_{k=0}^n r^k=\frac{1-r^{n+1}}{1-r}$ to get $\sum_{i=0}^{k-1}(1-p)^i=\frac{1-(1-p)^{k}}{p}$

$$\begin{split}2\left((1-p)^{k-1}p\left[1-\left(1-(1-p)^{k}\right)\right]\right)+\left((1-p)^{k-1}p\right)^2\\
=2(1-p)^{k-1}p(1-p)^{k}+\left((1-p)^{k-1}p\right)^2\\
=2(1-p)^{2k-1}p+(1-p)^{2k-2}p^2\\
=(1-p)^{2k-2}p(2-p)\end{split}$$


问:
Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$
Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$

答:
I thought it might be instructive to present an approach that relies on straightforward contour integration.  To that end, we now proceed.

Let $I$ be the integral given by

$$I=\int_0^\infty \frac{\log(x)}{x^3-1}\,dx$$

Now, moving to the complex plane, we analyze the contour integral $J$ given by

$$J=\oint_C \frac{\log^2(z)}{z^3-1}\,dz$$

where $C$ is the classical keyhole contour with a semi-circular deformation at $z=1+i0^-$.  That is, $J$ is given by

$$\begin{align}
J&=\int_0^\infty \frac{\log^2(x)}{x^3-1}\,dx-\text{PV}\int_0^\infty\frac{(\log(x)+i2\pi)^2}{x^3-1}\,dx+i\frac{4\pi^3}3\\\\
&=-i4\pi I+4\pi^2\text{PV}\int_0^\infty \frac1{x^3-1}\,dx+i\frac{4\pi^3}3\\\\
&=-i4\pi I-\frac{4\pi^3}{3\sqrt3}+i\frac{4\pi^3}3\tag1
\end{align}$$

Rearranging $(1)$ reveals that the integral of interest can by written in terms of $J$ as

$$I=i\frac1{4\pi}J+i \frac{\pi^2}{3\sqrt 3}+\frac{\pi^2}{3}\tag2$$

Now, applying the residue theorem we see that $\frac{iJ}{4\pi}$ is given by

$$\begin{align}
\frac{iJ}{4\pi}&=-\frac12 \left(\text{Res}\left(\frac{\log^2(z)}{z^3-1}, z=e^{i2\pi/3}\right)+\text{Res}\left(\frac{\log^2(z)}{z^3-1}, z=e^{i4\pi/3}\right)\right)\\\\
&=-\frac{4\pi^2}{27}\left(\frac54+i\frac{3\sqrt{3}}{4}\right)\tag3
\end{align}$$

Using $(3)$ in $(2)$ yields the coveted result

$$I=\frac{4\pi^2}{27}$$

as expected!


问:
In how many ways can a person buy $10$ biscuits if he decides to take at least one biscuit of each variety?
In a bakery four types of biscuits are available. In how many ways a person can buy $10$ biscuits if he decides to take at east one biscuit of each variety?

Let the four types of biscuits be A, B, C, and D.

Let us say he bought $x_1$ of A,  $x_2$ of B, $x_3$ of C, and $x_4$ of D.

Then , $^{x2 +x3 +x4+x1 }C_{x_1} + ^{x2+x3+x4 }C_{x_2} \ldots + x_4C_{x_4}$.

Now, solution in my book tells me that this way of solving is wrong. I don’t understand where am I wrong. I have arranged the way to buy the quantities also.

答:
Hints:

First see Stars and Bars.

The equation $x_1 + x_2 + x_3 + x_4 = 10$, where each variable is a positive integer bijects to:

$y_1 + y_2 + y_3 + y_4 = 6$, where each variable is a non-negative integer.

The bijects verb refers to the fact that it is easy to set up a bijection between the set of $(x_1,x_2,x_3,x_4)$ solutions and the set of $(y_1, y_2, y_3, y_4)$ solutions.


问:
Existence of solution of equation continuous $f(x)=0$
$f:[0,+\infty)\rightarrow \mathbb{R}$ is continuous function. s.t $f(0)=1$, $f(1)=0$
then existence of minimal $x>0$ s.t $f(x)=0$ is guaranteed?

If it has counterexample, for $C_1$ function $f$, is guaranteed the existence of minimal $x>0$  which satisfy above condition?

Please give me a proof or a counterexample for these.

答:
Since $f$ is continuous, the set $A = f^{-1}[\{0\}]$ is closed. It is bounded below since $A \subset [0, \infty)$ and thus, $c := \inf A$ exists and is an element of $A$.

Since $f(0) \neq 0$, $c \neq 0$ and thus, $c > 0$ is the minimal element you were looking for.


问:
Closed Form of $a_n = \int_0^1 \ln(1+x^n) dx$
I want to know the closed form of :
$$a_n = \int_0^1 \ln(1+x^n)dx, \quad \forall n \in \mathbb{N}$$

I found :

$$0<a_n<\frac{1}{n+1}, \quad \lim_{n\to\infty} a_n =0$$

I started from
\begin{align}
&\ln (1+x)=\sum_{k=0}^\infty \frac{(-1)^k x^{k+1}}{k+1} \\
& \Rightarrow \ln(1+x^n)=\sum_{k=0}^\infty \frac{(-1)^k x^{nk+n}}{k+1} \\
& a_n = \int_0^1 \ln(1+x^n)dx = \sum_{k=0}^\infty \frac{(-1)^k}{k+1}\int_0^1 x^{n+nk}dx \\
&=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{n+nk+1} \\
&=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{(n+1)(k+1)-k}
\end{align}

But I stucked here. Is there any closed (or approximated) from exist?

These are some results for litte $n$ :
\begin{align}
& a_1 = 2\ln 2 - 1 \\
& a_2 = \ln2 - 2 + \frac{\pi}{2} \\
& a_3 = 2\ln 2 - 3 + \frac{\pi \sqrt{3}}{3} \\
\end{align}

答:
Integrate by parts
$$a_n=\int_0^1 \ln({x^n+1})dx= -n+\ln2+ \int_0^1 \frac n{x^n+1}dx$$
With the roots
$x_k= e^{i\frac{(2k-1)\pi}n} $ for $x^n+1=0$,  integrate
\begin{align}
\int_0^1 \frac{n}{x^n+1}dx & = -\sum_{k=1}^n\int_0^1 \frac{x_k}{x-x_k} dx
=  -\sum_{k=1}^nx_k\ln(1-x_k^{-1})\\
 &= -\sum_{k=1}^n x_k \left[i \frac{n-2k+1}{2n}+ \ln \left(2 \sin\frac{(2k-1)\pi}{2n}\right) \right]
\end{align}
Per $\sum_{k=1}^nx^k=0$ and the symmetry of $x^k$ to arrive at the close-form
$$\int_0^1 \frac n{x^n+1}dx =\sum_{k=1}^{[\frac n2]} ( \pi-\theta_k )\sin\theta_k -\cos\theta_k \ln\sin^2\frac{\theta_k}2,\>\>\>
\theta_k= \frac{2k-1}{n} \pi$$
Listed below are the results for the first few $n’s$

\begin{align}
& a_2 = - 2 + \ln 2 + \frac\pi2\\
 & a_3 = -3 +2\ln2 + \frac\pi{\sqrt3}\\
 & a_4 = -4 +\ln 2+ \frac\pi{\sqrt2}+\sqrt2\ln(1+\sqrt2)\\
 & a_5 = -5 +2\ln2 +\frac\pi{\sqrt{10}}\sqrt{5+\sqrt5}+\sqrt5\ln\frac{1+\sqrt5}2\\
 & a_6 =-6+\ln2 + \pi +\sqrt3\ln(2+\sqrt3) \\
 & a_7 =-7+\ln2 + \frac\pi2\csc\frac\pi7
+\frac{\ln\csc^2\frac\pi{14}}{\sec\frac\pi7}
+\frac{\ln\csc^2\frac{3\pi}{14}}{\sec\frac{3\pi}7}
+\frac{\ln\csc^2\frac{5\pi}{14}}{\sec\frac{5\pi}7}
\end{align}


答:
I don't understand what I am supposed to find the probability for. I know that $X\sim \mathrm{Geom}(p)$ for $p\in[0,1]$ if $\mathbb{P}(X=k)=(1-p)^{k-1}p$. But how would I use this to find a p.m.f?

Such a random variable is the count of trials in a sequence of iid Bernoulli trials until a success occurs.   Such as, for example, the count of flips of a biased coin until the first head occurs.

If both $X,Y$ as identical and independent geometric random variables, then the minimum is the count of trials in two parallel sequences of iid Bernoulli trials until the first success occurs.   Such as, for example, the count of flips of two biased coins until at least one head occurs.

What kind of random variable is $\min(X,Y)$, and what parameter(s) will it have?

 $\min(X,Y)\sim\mathcal{Geo}_1( 1-(1-p)^2)$  $$\mathsf P(\min(X,Y){=}k)=(1-(1-p)^2)(1-p)^{2k-2}\mathbf 1_{k\in\Bbb N^+}$$


答:
Start with:

$$\begin{align}P(V\geq k) &=P(X\geq k \land Y\geq k)\\&=P(X\geq k)\cdot P(Y\geq k)\end{align}$$

and

$$P(V=k)=P(V\geq k)-P(V\geq k+1)$$

So you just need to show $$P(X\geq k)=P(Y\geq k)=(1-p)^{k-1}.$$

Then you get:

$$\begin{align}P(V=k)&=(1-p)^{2k-2}-(1-p)^{2k}\\
&=(1-p)^{2k-2}p(2-p)\end{align}$$

So $V \sim\mathrm{Geom}(2p-p^2).$

You can see that last relationship to $\mathrm{Geom}(2p-p^2)$ by remembering the most common model of $X.$

If you have a coin with probability $p$ For heads, and $X$ is the number of tosses before you get your first heads, then $X\sim\mathrm{Geom}(p).$

Then $V$ can be seen as the number of tosses of a pair of such coins, until at least one of them comes up heads. But this is the same as one coin with the probability of a heads being $1-(1-p)^2=2p-p^2.$

More generally, let $V=\min(X_1,\dots,X_n),$ where the $X_i$ are independent and $X_i\sim \mathrm{Geom}(p_i).$

Then $V\sim \mathrm{Geom}\left(1-(1-p_1)(1-p_2)\cdots(1-p_n)\right).$


问:
Can $\int_{2}^{\infty} [\zeta(x)-1] dx $ be evaluated?
I wonder whether an exact evaluation of $$\int_{2}^{\infty} [\zeta(x)-1] dx $$ can be obtained. Its sum analogue is already known: $$\sum_{n=2}^{\infty} [\zeta(n)-1] = 1. $$
Presumably, finding the indefinite integral is pretty difficult or even impossible, but I suspect the definite version could be found. So far, I have only found results on the mean values of the Riemann zeta function, but nothing on the integral mentioned above.

The two expressions are related via the Euler-Maclaurin summation formula, but thus far I have not been able to use it to calculate the integral.

Body:
This is not an answer
\begin{align*}\begin{split}\int_2^{\infty}[\zeta(x)-1]\,dx &= \int_2^{\infty} \sum_{n=2}^{\infty} \frac{1}{n^x}\,dx=\sum_{n=2}^{\infty} \frac{1}{n^2\ln(n)}\\ &\sim \int_2^{\infty} \frac{1}{x^2\ln(x)}\,dx+\frac{0+\frac{1}{4\ln(2)}}{2}-\sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!}f^{(2k-1)}(2)\\ &= -\mathrm{Ei}(-\ln(2))+\frac{1}{\ln(2)}\bigg[\frac{1}{8}+\frac{1}{48}-\frac{1}{960}\bigg]\\ &+\frac{1}{\ln^2(2)}\bigg[\frac{1}{96}-\frac{13}{11520}\bigg]-\frac{1}{\ln^3(2)}\bigg[\frac{1}{1280}-\frac{29}{96768}\bigg]+\cdots\\ &\approx 0.60544739 \end{split}\end{align*}


问:
Loan and Annuity problem - finding X
I am new in finance math, and I am working on this problem. I was wondering if someone could help me solve it.

A loan of $10,000$ is to be repaid with annual payments, at the end of
each year, for the next $21$ years. For the first $7$ years the payments
are $X$ per year, the second $7$ years they are $2X$ per year, and for the
third $7$ years they are $3X$ per year.

a) Find $X$ if $i^{(2)} =10% $

b) Find $X$ if $i^{(2)} = 8%$ for the first 10 years and $i^{(2)} = 12%$ afterwards.

My attempt was

a) My timeline looked like $7$ Xs starting from $t=1$ to $7$, then $2X$ from $8$ to $14$ and then $3X$ from $15$ to $21$.

$$PV = 10000 $$

$$PV = X a_{\overline{7} \rceil.05} + (1.05)^{-7} *2X a_{\overline{14} \rceil.05} + (1.05)^{-14} * 3X a_{\overline{21} \rceil.05} $$

$$10000 = \frac{X[1-(1.05)^{-7}]}{\frac{0.05}{1.05}} + (1.05)^{-7}\frac{2X[1-(1.05)^{-14}]}{\frac{0.05}{1.05}} + (1.05)^{-14}\frac{3X[1-(1.05)^{-21}]}{\frac{0.05}{1.05}}  $$

When I calculate it I got $243.70$

I wasn't sure if $(1.05)^{-7}$ for example was needed since I am starting at year 8 and bring it to time 0 , where the present value of 2X is 7 payments. Same for $(1.05)^{-14}$

I appreciate your help

答:
The clearest way to proceed is to first calculate the effective periodic rate--in this case, the rate is annual.  Since you are given a nominal rate of interest $i^{(2)}$, we must convert this to the effective annual rate $i$ via the formula $$\left(1 + \frac{i^{(m)}}{m}\right)^m = 1 + i.$$
In the first question, the interest rate is constant throughout the repayment term, which is simpler; we have $$i = \left(1 + \frac{0.10}{2}\right)^2 - 1 = 0.1025.$$  Then our equation of value is $$\require{enclose} 10000 = X a_{\enclose{actuarial}{7}i} + 2X v^7 a_{\enclose{actuarial}{7}i} + 3X v^{14} a_{\enclose{actuarial}{7}i}. \tag{1}$$  This can also be written
$$10000 = X a_{\enclose{actuarial}{21}i} + X v^7 a_{\enclose{actuarial}{14}i} + X v^{14} a_{\enclose{actuarial}{7}i}. \tag{2}$$
Your equation is not correct because it is computing the present value of cash flow in which the first $7$ payments are $X$, the next $7$ payments are $2X$, the next $7$ payments are $5X$, and then another $14$ payments are $3X$.  This is because the term of the second deferred annuity is $14$ years, so you are paying at times $t = 8, 9, \ldots 21$ the amount $2X$, and then starting from $t = 15$, you are paying another $3X$ from the third annuity.

To avoid this confusion, it is helpful to write out the cash flow:
$$PV = (Xv + Xv^2 + \cdots + Xv^7) + (2Xv^8 + 2Xv^9 + \cdots + 2Xv^{14}) + (3Xv^{15} + 3Xv^{16} + \cdots 3Xv^{21}).$$  And now if you factor things out, we get $$PV = X(v + v^2 + \cdots + v^7) + 2Xv^7(v + v^2 + \cdots + v^7) + 3Xv^{14}(v + v^2 + \cdots + v^7).$$  And we can factor this even further to get $$PV = X(1 + 2v^7 + 3v^{14})a_{\enclose{actuarial}{7}i}. \tag{3}$$  We can also factor Equation $(1)$ in the same way.

So where does Equation $(2)$ come from?  Well, we could have written the cash flow instead as
$$PV = X(v + v^2 + \cdots + v^{21}) + X(v^8 + v^9 + \cdots + v^{21}) + X(v^{15} + v^{16} + \cdots + v^{21}).$$  You can see that the first annuity of $X$ has a $21$-year term, and the second has a $14$-year term deferred by $7$ years, so that these contributions of $X$ will contribute the proper amount to the final $7$ years of the overall repayment.

How do we do the second part?  In this case, we must calculate two effective rates of interest.  One approach is to calculate the effective annual rate for the first $10$-year period, call that $i = (1 + 0.08/2)^2 - 1 = 0.0816$, and then the effective annual rate for the remainder of the term, call that $j = (1 + 0.12/2)^2 - 1 = 0.1236$.  Then the equation of value becomes
$$10000 = Xa_{\enclose{actuarial}{7}i} + 2Xv_i^7 (a_{\enclose{actuarial}{3}i} + v_i^3 a_{\enclose{actuarial}{4}j}) + 3Xv_i^{10} v_j^4 a_{\enclose{actuarial}{7}j}, \tag{4}$$  where $v_i = 1/(1+i)$ and $v_j = 1/(1+j)$ are the effective annual present value discount factors for each interest period.  Note we must split up the second $7$-year payment period into two deferred annuities due to the change in interest rate after time $t = 10$.  Also note that the total present value discount factor of the last $7$ payments is neither $v_i^{14}$ nor $v_j^{14}$ but rather $v_i^{10} v_j^4$ because for the first $10$ years of deferral, we are under rate $i$, and for the next $4$ years of deferral, we are under rate $j$.

This leads us to ask whether there is an analogue of Equation $(2)$ for this question.  The answer is yes, and as usual, to understand how to get it, we write out the cash flow:
$$\begin{align}
PV &= X(v_i + v_i^2 + \cdots v_i^7) + 2X(v_i^8 + v_i^9 + v_i^{10} + v_i^{10}v_j + v_i^{10}v_j^2 + v_i^{10}v_j^3 + v_i^{10}v_j^4) \\
&\quad + 3X(v_i^{10}v_j^5 + \cdots v_i^{10}v_j^{11}) \\
&= X(v_i + v_i^2 + \cdots v_i^{10} + v_i^{10}v_j + \cdots + v_i^{10}v_j^{11}) \\
&\quad + X(v_i^8 + v_i^9 + v_i^{10} + v_i^{10}v_j + \cdots v_i^{10}v_j^{11}) \\
&\quad + X(v_i^{10}v_j^5 + \cdots v_i^{10}v_j^{11}). \end{align} \tag{5}$$
This gives the equation of value
$$\begin{align}10000 &= X(a_{\enclose{actuarial}{10}i} + v_i^{10} a_{\enclose{actuarial}{11}j}) + X(v_i^7 a_{\enclose{actuarial}{3}i} + v_i^{10} a_{\enclose{actuarial}{11}j}) + X v_i^{10} v_j^4 a_{\enclose{actuarial}{7}j}.\end{align}\tag{6} $$

Title: Loan and Annuity problem - finding X


问:
Finding the Second Derivative of an integral within an integral
Looking at the following equations, I think that g'' is some form of the equation within f(x). However, it's a bit too much to wrap my head around and I am not exactly sure how to get there.

答:
Hint:

You will use the Fundamental Theorem of Calculus twice, which states that $$\frac{d}{dx}\int_a^xf(t)dt=f(x).$$

So to get you started, can you say what $g(y)$? Notice:

$$g'(y)=\frac{d}{dy}g(y)=\frac{d}{dy}\int_3^yf(x)dx$$


问:
Covariance between Bernoulli and Binomial Random Variable
I am given that $X$ is a Bernoulli random variable. I am also given vector $Y$ where $Y_1 = X$ and other $Y_i$ are also in Bernoulli Distribution.

Then, I am given $Z = Y_1 + ... + Y_n$ and am asked to calculate covariance between $X$ and $Z$.

So it seems like they are not independent. $E[X] = p$ and $E[Z] = np$ so $Cov(X,Z) = E[XZ] - np^2$
but when I tried to expand this out, I eventually ended up getting $0$ so does that mean the random variables are actually independent from each other or am I making some math error there? How would I go about getting $E[XZ]$? Help would be appreciated. Thanks.

答:
If I understand correctly, you are given vector $(Y_1, Y_2, ..., Y_n)$ but $Y_1=X$ so this is just $(X, Y_2, ..., Y_n)$ with all of them following an independent bernoulli distribution with parameter p. And you have $Z=X+Y_2+...+Y_n$. Thus covariance is linear in each of its arguments and we have

$$\begin{split}Cov(X,Z)&=Cov(X, X+Y_2+...+Y_n)\\
&=Cov(X,X)+Cov(X,Y_2)+...+Cov(X,Y_n)\\
&=Var(X)\\
&=p(1-p)\end{split}$$

You could also have done it your way, but your calculations are not right. We know $Cov(X,Z)=E(XZ)-\underbrace{E(X)}_{np}\underbrace{E(Z)}_{np*n}$.  Now $E(XZ)=E(X^2+XY_2+...+XY_n)$. We need to find $E(X^2)$. This is $E(X^2)=Var(X)+[E(X)]^2=p(1-p)+n^2p^2$. So $E(XZ)=p(1-p)+n^2p^2+(n-1)*n^2p^2$. Thus

$$\begin{split}Cov(X, Z)&=p(1-p)+n^2p^2+(n-1)*n^2p^2-n^3p^2\\
&=p(1-p)\end{split}$$


问:
compute $ \iiint_Kxyz\ dxdydz$
The question is:
$$ \iiint_Kxyz\ dxdydz\quad k:=\{(x,y,z):x^2+y^2+z^2\leq1, \  \ x^2+y^2\leq z^2\leq 3(x^2+y^2), \ x,y,z\geq 0\} $$

Here how i have tried to solve this:
$$\iint_{x^2+y^2\leq1}\int_{\sqrt{x^2+y^2}}^{\sqrt{3(x^2+y^2)}}xyz \ dz\ dxdy=\frac{1}{2}\iint xy\left(3(x^2+y^2)-(x^2+y^2)\right)=...=0$$
But the answer that i got is zero which is obviously wrong what is wrong with my solution?
any suggestion would be great, Thanks

答:
Basically we are trying to find the volume of the sphere $x^2+y^2+z^2 \leq 1$ between two cones $z^2 = x^2 + y^2$ and $z^2 = 3(x^2+y^2)$.

So the bounds are much easier to set up in spherical coordinates.

$x = \rho \cos \theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi$

$0 \leq \rho \leq 1, 0 \leq \theta \leq \frac{\pi}{2}$ are obvious as we are in first octant and radius of the sphere is $1$.

Now to find the bounds of $\phi$, we observe that

$x^2 + y^2 \leq z^2 \leq 3(x^2+y^2)$

plugging in $x,y,z$, $ \frac{1}{\sqrt3} \leq \tan \phi \leq 1 \implies \frac{\pi}{6} \leq \phi \leq \frac{\pi}{4}$

$xyz = \rho^3 \cos\theta \sin \theta \sin^2\phi \cos\phi$

So the integral becomes,

$\displaystyle \int_{0}^{\pi/2} \int_{\pi/6}^{\pi/4} \int_0^1 \rho^5 \cos\theta \sin \theta \sin^3\phi \cos\phi \ d\rho \  d\phi \ d\theta$

I did not do the integral by hand but WolframAlpha shows the result as $\displaystyle \frac{1}{256}$.

Title: compute $ \iiint_Kxyz\ dxdydz$


问:
Counterexample of Borsuk-Ulam theorem (Discontinuous Functions)
I'm looking to build a function $f:S^2 \to \mathbb R^2$ such that $f(x)\neq f(−x)$ for all $x\in S^2$.

By Borsuk-Ulam Theorem, this function must be discontinuous. I was trying to build a not too complicated function, but I always encountered a problem.

I appreciate any help.

答:
$S^2=\{(\cos u,\sin u \cos v,\sin u \sin v):u\in [0,2\pi)\land |v|\le \pi /2\}.$

Consider the equator $E=\{(\cos u, \sin u,0):u\in [0,2\pi)\}.$

For $u\in [0,2\pi)$ let $g(\cos u, \sin u,0)=u.$

If $\sin u \sin v >0$  let $g(\cos u,\sin u \cos v,\sin u \sin v)=3\pi.$

If $\sin u \sin v <0$  let $g(\cos u,\sin u \cos v,\sin u \sin v)=-3\pi.$

Let $f(x)=(g(x),0)$ for all $x\in S^2.$

$f$ maps $E$ bijectively to $[0,2\pi)\times \{0\}.$ And $x\in E\iff x\ne -x\in E.$

If $x\in S^2$ \ $E$ then $\{f(x),f(-x)\}=\{(3\pi,0),(-3\pi,0)\}$ so $f(x)\ne f(-x).$


问:
Counting outcomes of drawing two balls at a time
Question: A bag contains 4 black balls and 4 white balls. Suppose one draws two balls at a time, until the bag is empty. Count the number of outcomes where balls are pairwise the same color?

I am given the solution:

I don't get the multiplication of the two terms. Can someone help me understand the reasoning?

答:
Me neither, it looks like it should be 6 to me. B B W W where B stands for 2 black balls, W for 2 white balls

${4\choose 2}=6$

B B W W, B W B W, B W W B, W B B W, W B W B, W W B B


问:
A series for $\frac{355}{113}-\pi$

Series for sums $\pi+\dfrac{p_n}{q_n}$

Lehmer's interesting series relating $\pi$ to its early convergents $3$, $\dfrac{22}{7}$ and $\dfrac{355}{113}$ may be written as follows.

$$\begin{align}
\sum_{n=1}^\infty \dfrac{n2^n}{\dbinom{2 n}{n}}&=\pi+3\tag{A.1}\\
\dfrac{2}{7}\sum_{n=1}^\infty \dfrac{n^22^n}{\dbinom{2 n}{n}}&=\pi+\dfrac{22}{7}\tag{A.2}\\
\dfrac{2}{35}\sum_{n=1}^\infty \dfrac{n^32^n}{\dbinom{2 n}{n}}&=\pi+\dfrac{22}{7}\tag{A.3}\\
\dfrac{1}{113}\sum_{n=1}^\infty \dfrac{n^42^n}{\dbinom{2 n}{n}}&=\pi+\frac{355}{113}\tag{A.4}\\
\end{align}$$

Series for differences $\pi-\dfrac{p_n}{q_n}$ or $\dfrac{p_n}{q_n}-\pi$

Series that prove the sign of the error when approximating $\pi$ by its convergents include:

$$\begin{align}
\pi-3&=4·24\sum_{k=0}^\infty \frac{(4k+1)!}{(4k+6)!}(k+1)\tag{B.1}\\
\frac{22}{7}-\pi&=4^2·240\sum_{k=0}^\infty \frac{(4k+3)!}{(4k+11)!}(k+1)(k+2)\tag{B.2}\\
\frac{22}{7}-\pi&=4^3·285120\sum_{k=0}^\infty \frac{(4k+1)!}{(4k+14)!}(k+1)(k+2)(k+3)\tag{B.3}\\
\end{align}$$

A series to prove $\frac{22}{7}-\pi>0$

Series and integrals for inequalities and approximations to $\pi$

Changing the initial value for summations leads to results as in A.1, A.2 and A.3.

$$\begin{align}
\pi+3&=4·24\sum_{k=-1}^\infty \frac{(4k+1)!}{(4k+6)!}(k+1)\tag{C.1}\\
\frac{22}{7}+\pi&=-4^2·240\sum_{k=-3}^\infty \frac{(4k+3)!}{(4k+11)!}(k+1)(k+2)\tag{C.2}\\
\frac{22}{7}+\pi&=-4^3·285120\sum_{k=-3}^\infty \frac{(4k+1)!}{(4k+14)!}(k+1)(k+2)(k+3)\tag{C.3}\\
\end{align}$$

Let us assume there is a series for the fourth convergent

$$\sum_{k=0}^\infty \frac{q}{\Pi_{i}(4k+d_i)}=\frac{355}{113}-\pi\tag{B.4}$$

for some positive rational $q$ and positive integers $d_i$.

Question:
  Is there a technique to obtain the difference series (B.1, B.2, B.3) from their sum counterparts (A.1, A.2, A.3) that allows to determine B.4 from A.4?

答:
Since the second series for $\frac{22}{7}-\pi$ comes from

$$\frac{1}{28}\int_0^1 \frac{x(1-x)^8(2+7x+2x^2)}{1+x^2} dx = \frac{22}{7}-\pi$$

the following integral could be a starting point.

$$\frac{1}{99440} \int_0^1 \frac{x^{11}(1-x)^{12}( 858416(1+x^2)+6215x)}{1+x^2}dx = \frac{355}{113}-\pi$$

Maybe smaller coefficients would help such as in

$$\frac{1}{3164}\int_0^1 \frac{x^9(1-x)^8(791x+50(1+x^2))}{1+x^2} dx = \frac{355}{113} -\pi,$$

but the series obtained applying the technique in https://math.stackexchange.com/a/1657416/134791 is

$$\frac{4561920}{113} \sum_{k=0}^\infty \frac{ (81 k^2 + 648 k + 1145) (4 k + 16)}{ (4 k + 10)_{13} } = \frac{355}{113} - \pi$$

not as simple as the ones in the question.

Title: A series for $\frac{355}{113}-\pi$


问:
Find $∠ADC$ given $A,D$ on circle with diameter $BC$ and $∠BAO = 50°$
If $BC$ is a diameter of the circle and $∠BAO = 50°$. Then find the value of $∠ADC$.

This stumped me a little, I think there is a rule that mentions something about the center of the circle and it's relation to angles on the circumfrence, but I'm not sure, could someone help me point out the laws that would help to solve this problem? (Do not give me the solution, since I want to approach this myself)

Thanks in advance! :)

答:
See $ \angle  OAB = \angle OBA  = 50^\circ$ (why?)

$\angle AOC = 100^\circ  $ (why?)

Now see angle subtended in the centre is double the angle subtended at any other point on the circle. Join AC. Now $ \angle AOC = 2\angle ADC$

 $\angle ADC= 50^\circ $


问:
A bag contains 4 black balls and 4 white balls. Suppose one draws two balls at a time, until the bag is empty...
 Question: A bag contains 4 black balls and 4 white balls. Suppose one draws two balls at a time, until the bag is empty. What is the probability that each drawn pair is of the same color?

I have been provided a solution, but I find it difficult to understand. If someone could decipher it for me it would be greatly appreciated.

Background information: order is not important so we use the formula:
$$n!\over (n-r)!r! $$ which is also written in the notation: $$
        \begin{pmatrix}
        n \\
        r  \\
        \end{pmatrix}
$$
where n is the total number of elements, and r is the elements we draw from it
For example, if we wanted to know how many combinations of 2 there were in 5 elements, we'd use n=5 and r=2.

Given solution

1) The number of such doubly unordered outcomes S is:
$$ {1\over4!} * \begin{pmatrix}
        8 \\
        2  \\
        \end{pmatrix} * \begin{pmatrix}
        6 \\
        2  \\
        \end{pmatrix} * \begin{pmatrix}
        4 \\
        2  \\
        \end{pmatrix} *
     \begin{pmatrix}
        2 \\
        2  \\
        \end{pmatrix}
 = {1 \over4!}*{8!\over2!6!}
*{6!\over2!4!}
 * {4!\over2!2!}
 * {2!\over2!0!} = {1\over4!}{8!\over(2!)^4} = 105$$

2) The number of such outcomes with pairwise the same color is
$${1\over2!}
\begin{pmatrix}
        4 \\
        2  \\
        \end{pmatrix}
    \begin{pmatrix}
        2 \\
        2  \\
        \end{pmatrix}
* {1\over2!}
\begin{pmatrix}
        4 \\
        2  \\
        \end{pmatrix}
    \begin{pmatrix}
        2 \\
        2  \\
        \end{pmatrix} = 3 * 3 = 9$$

Thus, the probability each pair is of the same color is 9/105 = 3/35

My main confusion stems from the initial $1\over4!$ where does that come from? Similarly, the ${1\over2!}$ in step 2. Other than that, I grasp step 1, in step 2 I'm unsure of why we have to multiply the equation by itself.

答:
I don't get the 9 count of pairwise balls of same color. To me 9 seems wrong. I can count 6 pairs and not 9:

B B W W, B W B W, B W W B, W B B W, W B W B, W W B B

where B stands for 2 black balls, W for 2 white balls. So to me the response from @angryavian is not correct.


问:
Calculating time of flight of object (Mechanics - horizontal elevated launch)
I need help with the following question:

A smooth spherical object is projected horizontally from a point of vertical height H = 25.78 metres above horizontal ground with a launch speed of u = 24.23m/s . Calculate the time of flight of the object, in seconds, from its point of launch till it hits the ground. Assume that air resistance is negligible and take g = 9.81.

Appreciate any comments and feedback on this. Thanks.

答: An object is fired at 98m/s, 30degrees abouv the horizontal. If the acceleration due to gravity is 9.8m/s^2 downward find the range of projectile and the velocity on impact.</p>


答:
Hints

Mark the congruent line segments in the figure (a circle has a constant radius) to observe that:

 $$\measuredangle ABC = \measuredangle BAO$$

Then, try to use the fact that the measures of the angles in the same segment of a circle are equal (the measures of two angles at the circumference subtended by the same arc are equal):

$$\measuredangle ABC = \measuredangle ADC$$

Title: Find $∠ADC$ given $A,D$ on circle with diameter $BC$ and $∠BAO = 50°$

Body:
If $BC$ is a diameter of the circle and $∠BAO = 50°$. Then find the value of $∠ADC$.

This stumped me a little, I think there is a rule that mentions something about the center of the circle and it's relation to angles on the circumfrence, but I'm not sure, could someone help me point out the laws that would help to solve this problem? (Do not give me the solution, since I want to approach this myself)

Thanks in advance! :)


答:
Since $AO=OB$, $\angle A =\angle B=50^\circ $. Thus $\angle O =100^\circ $. Since $\angle ADC= \frac1 2 \angle AOC$,

 $\angle ADC= 50^\circ $.


问:
Open cover of ℚ that cannot be reduced to a finite subcover
I am attempting to prove $ℚ$ does not have the Heine-Borel property. I thought to prove that $ℚ$ is not compact, such that every open cover of $ℚ$ does not contain a finite subcover.

How would I go about this? Would it be the same as finding an open cover of $ℚ ∩ [0, 1]$ that cannot be reduced to a finite subcover?

I'm new to these concepts, so I do not fully understand everything. Any help will be greatly appreciated. Thanks!

答:
That would be one way to do it. Specifically, you can let $\alpha$ be an irrational in $\left(\frac12,1\right)$, and for $n\in\Bbb Z^+$ let $\alpha_n=\alpha-\frac1{n+1}$. Let $U_0=[0,\alpha_1)\cap\Bbb Q$, and for $n\in\Bbb Z^+$ let $U_n=(\alpha_1,\alpha_{n+1})\cap\Bbb Q$. Finally, let $V=(\alpha,1]$, and let

$$\mathscr{U}=\{V\}\cup\{U_n:n\ge 0\}\,.$$

Now show that $\mathscr{U}$ is an open cover of $[0,1]\cap\Bbb Q$ that has no proper subcover (and hence no finite subcover), even theough $[0,1]\cap\Bbb Q$ is a closed, bounded subset of $\Bbb Q$.


问:
Let $n$ be a positive integer. Determine the number of solutions to $x_1 + \cdots + x_k \leq n$ with nonnegative integer solutions.
Let $n$ be a positive integer. Determine the number of solutions to $x_1 + \cdots + x_k \leq n$ with nonnegative integer solutions. Determine the number of solutions with positive integer solutions.

I understand why every positive solution of $x_1 + · · · + x_k = m$ corresponds to an ordered partition of $m$. However, I do not understand why the number of positive solutions of $x_1 + \cdots + x_k \leq n$ is

$$\sum_{m=k}^{n}\binom{m-1}{k-1}.$$

Similarly, I do not understand how using multisets we obtain $\binom{n+k}{k}$ for the nonnegative integer solutions. Can someone explain this?

答:
Consider some positive solution of the inequality $$x_1+\cdots+x_k\le n\tag1.$$ It is necessarily a solution of an equation $$x_1+\cdots+x_k=m\tag2$$ for some $m:\ k\le m\le n$. Therefore the number of solutions of the inequality (1) is the sum of the numbers of solutions of the equations (2) for all $m:\ k\le m\le n$:
$$\sum_{m=k}^n\binom{m-1}{k-1}=\binom{n}k.
$$
The last expression can be obtained also in a direct though somewhat tricky way.

The same formula work for the number of non-negative solutions as well. Just replace $n(m)$ with $n(m)+k$.


答:
What about the open cover

$$\left\{\;A_n:=\Bbb Q\,\cap\,(n,n+1)\;\;,\;\;B_n=\Bbb Q\,\cap\,\left(n-0.01,\,n+0.01\right)\;\right\}\;,\;\;n\in\Bbb Z\;\;?$$

Can you see why not only you can't have a finite subcover of the  above that'll cover all of $\;\Bbb Q\;$ , but in fact you cannot even take one single open set $\;A_n\,,\,\,B_n\;$ without leaving out at least one rational number?

Title: Open cover of ℚ that cannot be reduced to a finite subcover

Body:
I am attempting to prove $ℚ$ does not have the Heine-Borel property. I thought to prove that $ℚ$ is not compact, such that every open cover of $ℚ$ does not contain a finite subcover.

How would I go about this? Would it be the same as finding an open cover of $ℚ ∩ [0, 1]$ that cannot be reduced to a finite subcover?

I'm new to these concepts, so I do not fully understand everything. Any help will be greatly appreciated. Thanks!


问:
Model theory/stability theory
I was flipping through Baldwin's Stability Theory book and saw an example that has me confused...

The example is a 1st-order theory $T$ of refining equivalence relations $E_i(x, y), i< ω$, where

$E_j$ refines $E_i$ when $i < j$

each $E_{i+1}$ relation splits each $E_i$-equivalence class into two $E_{i+1}$-equivalence classes.

Here's what I'm picturing: $E_0$ splits the model into two classes, call them $a_0/E_0$ and $a_1/E_0$. Then $E_1$ splits $a_0/E_0$ into two pieces $a_{00}/E_1$ and $a_{01}/E_1$, and $E_1$ splits the class $a_1/E_0$ into two pieces $a_{10}/E_1$ and $a_{11}/E_1$, ad infinitum.

So the classes are arranged like the tree ${}^{<ω}2$ of finitely long $0-1$ sequences (the tree ordering is sequence extension), where the $n$th level of the tree has the $E_n$ classes as nodes.

My confusion:

Each infinite branch of the tree (there are $2^\omega$ of them) should correspond to an incomplete type of an element, right? And those types are pairwise contradictory. So, if $M$ is a countable model of $T$, there are $2^ω$ types over $M$. If all that is right, then $T$ is $\omega$-unstable. $T$ is supposed to be (some kind of) stable however.

But can't the type space still blow up over uncountable sets of parameters for cardinals of countable cofinality? Is it that such a T is stable at some cardinals, but doesn't “stay stable”?

Sorry if this is boring or obtuse!
-M

答:
This theory is stable in all cardinals that are at least $\mathfrak c$. To see this, notice that a type is determined entirely by which class of each $E_n$ it belongs in (of which there are $\mathfrak c$ many possible combinations in total) and which elements it is (not) equal to.

Such a theory is called superstable.

There are theories which are stable, but not superstable, and do explode precisely for parameters of size $\kappa$ such that $\kappa^\omega>\kappa$. For example, the theory of (countably) infinitely many independent equivalence relations with infinitely many classes. With $\kappa$ parameters you can have each relation have $\kappa$ many classes, yielding $\kappa^\omega$ types.

(Edit: the theory I mentioned is of infinitely many independent relations with infinitely many classes; the theory of infinitely many independent equivalence relations with two classes is, I believe, bi-interpretable with the one you described in your question.)

Title: Model theory/stability theory


问:
Definite integration - how to find area under a curve?
so I was doing my homework and go to the "complex unfamiliar" questions which arent required to be completed but I still want to understand and get an answer.

The question refers to a valuation of a torn banknote as in Australia if greater than 80% of the note is still there full face value is given, between 80 and 20% the percentage value of the note is given and below 20% no value is given.

I can do the translation of area to percent to get an answer, I just don't know how to get the area in the first place, and that's where I am stuck.

this is the graph that I generated from the points below, I measured in 0.5cm increments and then measured the height which is the y, the polynomial regression is there solely for visual stimulus and comunication

These are the dimension (points/ measurements) of the banknote all of it is in cm

I know for a fact my calculations are incorrect as 67.2cm^2 is 100% unreasonable so I would like some of you guys to have a go, I took an assumption maybe its not possible to automate with desmos, either way, I think I may have to do it with working out so if that is possible that would be great. good luck to you all if you attempt this!

答:
I do not know what is going on. I checked the parameters of the polynomial regression and they are correct.

Now, what I have is
$$\int_0^{8.5}f(x)\,dx=26.8718$$

Using an interpolating function (needing to extrapolate it between $x=0$ and $x=0.5$), I obtain
$$\int_0^{8.5}f(x)\,dx=\frac{6403}{240}=26.6792$$


问:
Not every open set in $\mathbb{R}^n$, for $n\geq2$, is a disjoint union of countable open balls.
I know that every open set in $\mathbb{R}$ can be decomposed as the union of countably many disjoint open intervals and I'm trying to show that this property no longer holds in higher dimensions. For this I'm trying to show that there is no countable, pairwise disjoint open balls whose union is the
open square $(0, 1)^2 ⊂ \mathbb{R}^2$. I know that it is because of the corners, but I don't know how to write this in mathematical terms. Thanks for any help.

答: Any open subset $O$ of $\Bbb R^n$ is a finite or countable disjoint union of open connected sets  and that’s all that can be said, really. This is true in any separable locally connected space. In $\Bbb R$ those open connected sets can only be intervals or segments. In $\Bbb R^n, n>1$ there can be more variation (open rectangles minus finitely many points, interiors of closed curves in all sorts of shapes etc) but you do know the connected open sets are path-connected too.


答:
You can immediately see that $\mathbb Q$ is not compact, since it is not bounded. So for instance the cover $\{(-n,n)\}_{n\in\mathbb N}$ does not admit a finite subcover.

When you consider $\mathbb Q\cap[0,1]$, you can use that it is not closed (since a compact subset of $\mathbb R$ is closed). Or you can construct an open cover that does not admit an open subcover, like $$\Big\{\Big(-1,\tfrac1{\sqrt2}-\tfrac1n\Big)\cup\Big(\tfrac1{\sqrt2},2\Big)\Big\}_{n\in\mathbb N}.$$


问:
$g'(x) \ge f'(x)$ implies $h(x) = \frac{f(x)}{g(x)}$ is decreasing?
Conjecture:

If $f(x)$ and $g(x)$ are continuous and differentiable in $(a,b)$ and $g'(x) \ge f'(x)$ for all $x \in(a,b)$ then $h(x) = \frac{f(x)}{g(x)}$ is decreasing in $(a,b)$ if $h(x)$ is continuous in $(a,b)$.

I came up with this myself when I was studying the nature of functions however I don't know if it's true. (It seems very obvious intuitively)

$h(x) = \frac{f(x)}{g(x)}$.
To prove $h'(x) < 0$ for $ x \in(a,b)$

$h'(x ) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$

To prove: $g(x)f'(x) < f(x)g'(x)$

Here I don't know what to continue with.

Some examples:

If $g(x) = k$(Constant function) and $k$ is not $0$

$\frac{\sin x}{x}$ for $x \in (0 , \infty)$ or $x \in (-\infty,0)$

I found many other examples but no counterexamples.

Any counterexample in case this is not true will help.

答: Some assumptins are missing. A counter-example is $f(x)=x-1$ and $g(x)=2x$ in which case $h$ is increasing.


问:
Decay estimate of $L^2$ integral.
Let $C_{R} = [\frac{-R}{2} , \frac{R}{2}]^n$, the cube centered at origin in $\mathbb{R}^n$ of length $R > 0$
and $\|x\|_{\infty} = \max_{i = 1}^n \{x_i \}$, $x \in \mathbb{R}^n$. Suppose $\phi$ is a complex valued function on $\mathbb{R}^n$ and has the decay estimate: for some $\epsilon > 0$,

$|\phi(x)| \leq C_1 (1 + \|x\|_{\infty})^{-n - \epsilon}$, for all $x \in \mathbb{R}^n$.

Then prove that $\| \phi ( 1 - \chi_{C_R} )\|_{L^2(\mathbb{R}^n)}^2 \leq C_2 R^{-n}$.

P.S: When sup norm $\|.\|_{\infty}$ is replaced by $l^2$ norm and cube $C_{R}$ replaced by ball, then I can estimate the above integral by a well known formula of integration of radial function.

答:
Since $n^{-1/2}\lVert x\rVert_2\leq\lVert x\rVert_\infty$, you also have the bound
\begin{align*}
\lvert\phi(x)\rvert&\leq C_1(1+\lVert x\rVert_\infty)^{-n-\epsilon}\\
&\leq C_1(1+n^{-1/2}\lVert x\rVert_2)^{-n-\epsilon}\\
&\leq \underbrace{n^{(n+\epsilon)/2}C_1}_{C'_1}(1+\lVert x\rVert_2)^{-n-\epsilon}.
\end{align*}
Also, $C_R$ contains the $2$-ball of radius $R/2$, so
$$
\lVert\phi(1-\chi_{C_R})\rVert_{L^2}^2\leq
\lVert\phi(1-\chi_{B_{R/2}})\rVert_{L^2}^2
$$
and you can use your 2-norm result.

Title: Decay estimate of $L^2$ integral.


问:
Limit $a_n = \frac{n!}{\sqrt{(2n+1)!}}, \quad n\in\mathbb{N}$
How do I show that the result is zero?

$\lim(a_n)= \frac{n!}{\sqrt{(2n+1)!}}, \quad n\in\mathbb{N}$

$\frac{(n+1)!}{\sqrt{(2(n+1)+1)!}}/
\frac{n!}{\sqrt{(2n+1)!}}
$

$\frac{(n+1)}{\sqrt{(2n+3)!}}*
{\sqrt{(2n+1)!}}
$

I will use the ratio test, but in the calculations I will get to the result 1/2 which is quite wrong, see the previous procedure.

答:
$$a_n = \frac{n!}{\sqrt{(2n+1)!}}\implies \log(a_n)=\log(n!)-\frac 12 \log((2n+1)!)$$

Use Stirling approximation twice and continue with Taylor series
$$\log(a_n)=-n \log (2)+\frac{1}{4} \log \left(\frac{\pi }{4 n}\right)-\frac{3}{16
   n}+O\left(\frac{1}{n^2}\right)$$
$$a_n= e^{\log(a_n)}\sim 2^{-n}\,\frac{\sqrt[4]{\pi } }{\sqrt{2}\sqrt[4]{n }}\exp\left(-\frac{3}{16 n} \right)$$


问:
What conditions ensure that a functor has a left / right inverse.
A functor $F : C \to D$ has a left inverse $G : D \to C$ if $G \circ F : C \to C$ is naturally isomorphic to $\mathrm{id}_C$ and ditto for a right inverse. Are there nice criteria for when a functor admits a left inverse or a right inverse?

I know that a functor is an equivalence of categories (i.e. admits both a left and a right inverse) if and only if it is fully faithful and essentially surjective.

Is it true that a functor admits a left inverse if and only if it is faithful and a right inverse if and only if it is full and essentially surjective?

It is clear that if $G \circ F \simeq \mathrm{id}_{\mathrm{C}}$ then $G$ must be full and essentially surjective and $F$ must be faithful.

However, I can't show that these conditions are sufficient.

答:
Usually what you ask for is that a fully faithful functor $F$ has a left adjoint $L$ (and in that case, the counit $LF \to 1$ of the adjunction is an isomorphism), or a right adjoint $R$, and in that case the unit $1\to RF$ of the adjunction is an isomorphism.

Point is, you need $F$ to be fully faithful both times!


问:
Was it necessary to have $g(x)\neq4$ and $x\neq4$ in the assumptions of this problem?

Suppose $f'(4)=3 , g'(4)=7,g(4)=4,g(x)=4,g(x)\neq4$ for $x\neq4$ find:

$1)$$\lim_{x\to4}\frac{f(x)-f(4)}{g(x)-4}\quad\quad\quad\quad\quad\quad\quad\quad2)\lim_{x\to4}\frac{f(g(x))-f(4)}{x-4}$

I know how to solve this problem ( we can do it by rewriting first one as $\lim_{x\to4}\dfrac{{\frac{f(x)-f(4)}{x-4}} }{\frac{g(x)-g(4)}{x-4}}$ and second one as $\lim_{x\to4}\dfrac{f(g(x))-f(g(4))}{g(x)-g(4)}\times\dfrac{g(x)-g(4)}{x-4}$ and using defenition of derivative.)

My question : Was it necessary to say $g(x)\neq4$ and finding the values for $x\neq4$? I also don't understand how $g(x)\neq4$ if we have $g(x)=4$ in the assumptions.

答:
For the first limit to exist, it's necessary that there exists a small interval $(4-\delta,4+\delta)$ in which $g(x)\ne4$ except for $g(4)=4$ itself. (If there is no such interval, then the quotient is undefined at a sequence of points tending to $x=4$, which means the limit cannot be defined.)

Rather than bother saying the bit about the interval $(4-\delta,4+\delta)$, I believe the writer chose a simpler hypothesis, namely that $g(x)\ne4$ for $x\ne4$. This is stronger than is really required (as discussed above), but simpler to say and to deal with. But the real point is to just ensure that the quotient itself is defined enough to make the limit make sense in the first place.


问:
Good calculus exercises/problems?
I can't enroll in a university this year, so I'm studying calculus at home, but the only exercises about calculus that I find are the easy ones. Do you know a great page where I can find not only calculus exercises, but problems as well?

I want to find about:

hard limits,

derivatives and integrals,

and some problems

答:
Barring any more information, I'd suggest you acquire Michael Spivak's Calculus, and begin working your way through the material. There are plenty of exercises and examples, but you'll also deepen your conceptual understanding at the same time, which will help ensure the material "sticks with you" over the long-run, which I presume is a goal of yours, as you seem committed to enrolling at a university, perhaps even to pursuing math, and/or math-related fields.

Spivak will prepare you well.

Title: Good calculus exercises/problems?


问:
Proof of discontinuity at a point
I am trying to prove that the function $f(x) = \frac{1}{x}$ for $x \neq 0$ and $0$ if $x = 0$ is continuous everywhere except $x = 0$. Here is my attempt.

We first prove continuity. Let $x_0 > 0$ and $\epsilon > 0$. Set $\delta = \min\left(\frac{x_0}{2}, \frac{\epsilon x_0^2}{2}\right)$. Then $|x - x_0| < \delta$ implies
\begin{align*}
|f(x) - f(x_0)| & = \left \lvert \frac{1}{x} - \frac{1}{x_0} \right \rvert \\
& = \frac{|x-x_0|}{xx_0} \\
& < \frac{\frac{x_0^2}{2} \epsilon}{\frac{x_0}{2}} \\
& = \epsilon.
\end{align*}

I'm stuck on the $x < 0$ case. In the above work, the step $|xx_0| = xx_0$ breaks down in that case unless $x < 0$, but $\delta$ can't depend on $x$.

To prove discontinuity at $x = 0$, I need to find $\epsilon > 0$. so that $\forall \delta > 0$, there exists $y$ such that $|y| < \delta$ and $|f(x) - f(y)| \geq \epsilon$. Here is my attempt.

Let $\epsilon = 0$, $\delta > 0$, and pick $0 > y > \delta$. Then $|y| = -y < \delta$. Also:
$$f(x) - f(y)| = \left \lvert f(0) - \frac{1}{y} \right \rvert = \frac{1}{y} < \delta.$$

This fails because $\delta$ has to be arbitrary, unless I can begin with the case where $\delta < 1$ and then take the case where $\delta \geq 1$.

答:
Your 3rd display line's denominator should be $x_0^2/2,$ not $x_0/2.$ (Typo?) This step should be justified by stating that if $x_0>0$ and $|x-x_0|<\delta$ then $x>x_0-\delta\ge x_0-x_0/2=x_0/2>0,$ so $(x_0x)^{-1}\le (x_0\cdot x_0/2)^{-1}.$

You can do the cases $x_0>0$ and $x_0<0$ at once. Let $\delta=\min (|x_0|/2,\epsilon x_0^2/2).$  Now if $x_0\ne 0$ and $|x-x_0|<|x_0|/2$ then $|x|>|x_0|/2>0$. Hence for any $x_0\ne 0$  we assert that $$|x-x_0|<\delta \implies |x-x_0|<|x_0|/2 \implies |x_0x|=|x_0|\cdot |x|> |x_0|\cdot |x_0|/2=x_0^2/2>0.$$

So if $x_0\ne 0$ and $|x-x_0|<\delta$ then  $$|f(x)-f(x_0)|=|x_0-x|/|x_0x|\le|x_0-x|/ (x_0^2/2)<(\epsilon x_0^2/2)/(x_0^2/2)=\epsilon.$$

Title: Proof of discontinuity at a point


问:
find absolute maximum and minimum with explanation
if $f(x,y,z)=x+y-z$ and constraint is sphere $g(x,y,z)=x^2+y^2+z^2=81$. I have to find the absolute maximum and minimum. And I have to explain how both an absolute maximum and minimum must be attained.

my solution: I obtained critical points $(-3\sqrt{3} ,-3\sqrt{3},3\sqrt{3}) and (3\sqrt{3} ,3\sqrt{3},-3\sqrt{3})$ by solving the system $\nabla f= \lambda \nabla g$ and constraint equation.

Then I used the equation $L(l;x,y,z) =x+y-z-l(x^2+y^2+z^2-81)$ and found the matrix $\begin{bmatrix} 0 & -2x & -2y & -2z\\ -2x & -2l & 0 & 0\\ -2y & 0 & -2l & 0\\ -2z & 0 & 0 & -2l \end{bmatrix}$.

I plugged in the critical points, found the sequence of determinants and found that $(-3\sqrt{3} ,-3\sqrt{3},3\sqrt{3})$ is local minimum and $(3\sqrt{3} ,3\sqrt{3},-3\sqrt{3})$ is local maximum.

How do I say that they are absolute maximum and minimum, as well as that they both must be attained?

答:
The problem can also be interpreted geometrically.  We have the sphere  $ \ x^2 \ + \ y^2 \ + \ z^2 \ = \ 81  \ $ of radius $ \ 9 \ $ centered on the origin.  The plane $ \ x \ + \ y \ - \ z \ = \ 0 \ $ passes through the origin and all planes $ \ x \ + \ y \ - \ z \ = \ d \ $ are parallel to it.

It is a reasonably well-known result (and not difficult to prove) that the perpendicular distance between parallel planes  $ \ ax \ + \ by \ + \ cz \ = \ d_1 \ $ and  $ \ ax \ + \ by \ + \ cz \ = \ d_2 \ $  is  $ \ D \ = \ \frac{| \ d_2 \ - \ d_1 \ |}{\sqrt{a^2 \ + \ b^2 \ + \ c^2}} \ $ .  The planes parallel to $ \ x \ + \ y \ - \ z \ = \ 0 \ $ are displaced along a line with direction vector $ \ \langle \ 1 \ , \ 1 \ , \ -1 \ \rangle \ $ .

We will want the tangent planes to the sphere which are the result of "parallel displacement" along the radii lying on this line.  Those tangent planes then lie at a distance $ \ D \ = \ 9 \ $ from the plane  $ \ x \ + \ y \ - \ z \ = \ 0 \ $ , which tells us that
$$  \ 9 \ = \ \frac{| \ d \ - \ 0 \ |}{\sqrt{1^2 \ + \ 1^2 \ + \ (-1)^2}} \ \ \Rightarrow \ \ |d|  \ = \ 9 \sqrt{3} \ \ . $$

(The line is a diameter of the sphere, so we would expect the symmetry of the sphere to produce two such planes.)

These planes are thus  $ \ x \ + \ y \ - \ z \ = \ 9 \sqrt{3} \ $ and  $ \ x \ + \ y \ - \ z \ = \ -9 \sqrt{3} \ \ , $  giving us the maximum and minimum values of $  \ f (x,y,z) \ $ , subject to the "spherical" constraint.

Examples of this sort were likely involved in the reasoning that led to the "Lagrange method".


问:
Find process correlation function
I have following task:

There are two normal centered random processes $X(t)$ and $Y(t)$ given. They have correlation functions $K_{x}(t_{1}, t_{2}), K_{y}(t_{1}, t_{2})$ and cross-correlation function $R_{xy}(t_{1}, t_{2})$. Get correlation function of process $Z(t) = X(t)Y(t)$.

What did I do:

$\bar Z = M[XY]$

$K_{Z} = M[(X(t_{1})Y(t_{1}) - M[X(t_{1})Y(t_{1})])\cdot (X(t_{2})Y(t_{2})-M[X(t_{2})Y(t_2)])] = M[X(t_1)X(t_2)Y(t_1)Y(t_2)]-M[X(t_1)Y(t_1)M[X(t_2)Y(t_2)]]-M[X(t_2)Y(t_2)M[X(t_1)Y(t_1)]]+M[M[X(t_1)Y(t_1)]M[X(t_2)Y(t_2)]] = M[X(t_1)X(t_2)Y(t_1)Y(t_2)] + M[X_1Y_1]M[X_2Y_2]$

But the answer is:

$K_x(t_1, t_2)\cdot K_y(t_1, t_2) + R_{xy}(t_1, t_2)R_{xy}(t_2, t_1) = M[X(t_1)X(t_2)]\cdot M[Y(t_1)Y(t_2)] + M[X(t_1)Y(t_2)]\cdot M[X(t_2)Y(t_1)]$

Honestly I don't know how to reduce my ideas to correct answer. Any ideas?

答:
To calculate the correlation function of $ Z $, the following formula for 4th order moment of centered Gaussian random variables is useful.
If $ (X_1,X_2,X_3,X_4) $ is a centered Gaussian 4d-random variable, then
\begin{align*}
 \mathsf{M}[X_1X_2X_3X_4]&=\mathsf{M}[X_1X_2]\mathsf{M}[X_3X_4]\\
 &\quad +\mathsf{M}[X_1X_3]\mathsf{M}[X_2X_4]+\mathsf{M}[X_1X_4]\mathsf{M}[X_2X_3].\tag{1}
\end{align*}
(cf.  E. Parzen, Stochastic Processes, Holden-Day, 1962. p.92, Ch3, Th4C.  )
Using (1),
\begin{align*}
 &\mathsf{M}[X(t_1)X(t_2)Y(t_1)Y(t_2)] =\mathsf{M}[X(t_1)X(t_2)]\mathsf{M}[Y(t_1)Y(t_2)]\\
 &\quad +\mathsf{M}[X(t_1)Y(t_1)]\mathsf{M}[X(t_2)Y(t_2)] +\mathsf{M}[X(t_1)Y(t_2)]\mathsf{M}[X(t_2)Y(t_1)].
\end{align*}
Hence
\begin{align*}
K_Z(t_1,t_2)&= \mathsf{M}[X(t_1)Y(t_1)X(t_2)Y(t_2)]-\mathsf{M}[X(t_1)Y(t_1)]\mathsf{M}[X(t_2)Y(t_2)]\\
&=\mathsf{M}[X(t_1)X(t_2)]\mathsf{M}[Y(t_1)Y(t_2)] + \mathsf{M}[X(t_1)Y(t_2)]\mathsf{M}[X(t_2)Y(t_1)]\\
&=K_X(t_1,t_2) K_Y(t_1,t_2)+R_{XY}(t_1,t_2)R_{XY}(t_2,t_1).
\end{align*}

Title: Find process correlation function


问:
Closure of the set $\{1 \} $ in $ \tau_{1} \cap \tau_{2} $
Let $\tau_{1} $ and $ \tau_{2} $ be two topologies on $X = \Bbb R$ defined as:

$ \tau_{1}= \{ U \in X \mid 0 \not\in U \text{ or } U = \Bbb R\} $ and
$ \tau_{2} = \{ U \in X \mid 0 \in U \text{ or } U= \varnothing\}$.

Now define another topology  $\tau_{3}: = \tau_{1} \cap \tau_{2} $.

What is the closure of the set $\{1\}$ in $ \tau_{3} $?

My attempt

For an open set $ U  \in \tau_{1} $, $ 0\not\in U $ or $ U= \Bbb R$, while for an open set
$ U^{'}  \in \tau_{2}$, we have $ 0 \in U $ or $ U= \varnothing $.
Thus $ \Bbb R $ belongs to both topologies but no other non trivial open set belongs to both of them.

Is $ \tau_{3} = \{ \Bbb R, \varnothing \} $?

Is the closure of
$ \{1\} $ the whole $\Bbb R$?

答:
Indeed $\tau_1 \cap \tau_2 = \{\varnothing, \Bbb R\}$: the right to left inclusion is obvious, and for the left to right inclusion (contrapositive): if $\varnothing \neq O \subsetneq \Bbb R$, then if $0 \in O$, then $O \notin \tau_1$, and if $0 \notin O$, then $O \notin \tau_2$, and in either case $O \notin \tau_1 \cap \tau_2$.

And in this indiscrete topology the closure of any non-empty subset is $\Bbb R$ as its the only closed superset of a non-empty set.

So your solution is correct. Note that the closure of $\{1\}$ in $\tau_1$ is $\{0,1\}$ and in $\tau_2$ it's $\{1\}$. So these are unrelated to the closure of $\{1\}$ in the intersection topology.


答:
Shaum's Outline Calculus Problems

http://www.amazon.com/Schaums-000-Solved-Problems-Calculus/dp/0071635343

Title: Good calculus exercises/problems?

Body:
I can't enroll in a university this year, so I'm studying calculus at home, but the only exercises about calculus that I find are the easy ones. Do you know a great page where I can find not only calculus exercises, but problems as well?

I want to find about:

hard limits,

derivatives and integrals,

and some problems