答:
Let $I$ be the midpoint of $AD$.

Then $|IE| = \frac{1}{2} |CD| = \frac{1}{2} |AB| = |IF|$.

So $IEF$ is an isosceles triangle.

It follows that $ \angle DHG = \angle IEG = \angle IFH = \angle AGH$.

Title: Midpoints of diagonals

Body:
Let $ABCD$ be convex quadrilateral such that $AB=CD$. And $E\neq F$ where $E, F$ is midpoint of $AC, BD$ respectively. Then prove that angle between$(AB, EF)$ and $(CD, EF)$ are equal.

I'll prove $\vec{AB}\cdot\vec{EF}=\vec{CD}\cdot\vec{EF}$. Using $\vec{EF}=\frac{1}{4}(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})$, it's sufficient to prove that $$(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})\cdot\vec{AB}=(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})\cdot\vec{CD}\iff (\vec{CB}+\vec{AD})\cdot\vec{AB}=(\vec{CB}+\vec{AD})\cdot\vec{CD}\iff BC\cdot CDcos\angle C+AD\cdot CDcos\angle D=BC\cdot ABcos\angle B+BA\cdot ADcos\angle A\iff BCcos\angle C+ADcos\angle D=CBcos\angle B+ADcos\angle A$$. I can't continue after this. Can anyone help me?


问:
Order of a dilation and a reflection of a function
$Parent function$: $f(x)$ = $\sqrt{x}$

$A$: the function is reflected in the y-axis and then translated 2 units to the right.

$B$: the function is translated 2 units to the right, then reflected in the y-axis.

for $A$, my working outs:

```<code>    f(-x) - reflected in the y-axis
</code>```

Now, I have a doubt.  After reflected in the y-axis, will the transformation be included in the reflection?

i.e should it be: $f(-(x-2))$ or $f(-x+(-2))$

My understanding suggests that it should be $f(-x-2)$.

Similarly for $B$,

```<code>    f(x-2) - two units to the right
</code>```

Now, should it be

$f(-(x-2))$ or $f(-x-2)$

I think it should be $f(-x+2)$

I know, if it was reflected in the x-axis, and translated two units upwards, then it would have been:

$-f(x) + 2$

And, Reversing the operation: $-[f(x) + 2]$ = $-f(x) - 2$

That is why I have doubts regarding the order of reflection in the y-axis.

I would appreciate if you could provide an answer to my doubts.

答:
The instruction for A clearly says "reflect, then translate". So you reflect first:
$$f_1(x) = f(-x)$$
Then you translate the result:
$$f_2(x) = f_1(x -2)$$
Similarly for B. Do the operations in the order that the problem says to do them.

ContentLicense: CC BY-SA 4.0


问:
Trig substitution for a triple integral
This problem involves calculating the triple integral of the following fraction, first with respect to $p$:

$$
\int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^{2}  \frac{p^2\sin(\phi)}{\sqrt{p^2 + 3}} dp d\phi d\theta
$$

This involves trig-substitution (I believe), and I am just hoping for an explanation of how to do it.

答:
Update:  I don't use a trig-substitution because I don't see which one could work. To integrate $p^{2}/\sqrt{p^{2}+3}$ I applied a technique derived from this more general one.

If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that

$$\int \frac{P(x)}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x=Q(x)\sqrt{ax^{2}+bx+c}+\int \frac{C}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x.$$

(Described in Cálculo Integral em $\mathbb{R}$ by M. Olga Baptista.)

EDIT: simplified the exposition. The integral is separable, as already noted by J. M. in a comment. Assuming the limits of integration are as follows, we have

$$\begin{eqnarray*}
I &:&=\int_{\theta =0}^{2\pi }\int_{\phi =0}^{\pi
}\int_{p=0}^{2}\frac{p^{2}\sin (\phi )}{\sqrt{p^{2}+3}}\;\mathrm{d}p\;%
\mathrm{d}\phi \;\mathrm{d}\theta \\&=&\left( \int_{0}^{2\pi }\mathrm{d}\theta \right) \left(
\int_{0}^{\pi }\sin(\phi)\;\mathrm{d}\phi\;\right)\left(\int_{0}^{2}\frac{p^{2}}{\sqrt{%
p^{2}+3}}\;\mathrm{d}p\right) \\
&=&2\pi \left(
\int_{0}^{\pi }\sin(\phi)\;\mathrm{d}\phi\;\right)\left(\int_{0}^{2}\frac{p^{2}}{\sqrt{%
p^{2}+3}}\;\mathrm{d}p\right).
\end{eqnarray*}$$

The antiderivative of $\sin (\phi )$ is $-\cos \left( \phi \right) $; hence

$$I =4\pi \int_{0}^{2}\frac{p^{2}}{\sqrt{p^{2}+3}}\;\mathrm{d}p.$$

For the evaluation of this last integral we are going to apply the following property:

If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that

$$\int \frac{P(x)}{\sqrt{x^{2}+c}}\;\mathrm{d}x=Q(x)\sqrt{x^{2}+c}+\int \frac{C}{\sqrt{%
x^{2}+c}}\;\mathrm{d}x.$$

In the present case, we have $P(x)=x^{2}$ and $Q(x)$ is of the form $Q(x)=Ax+B$

$$\int \frac{x^{2}}{\sqrt{x^{2}+c}}\;\mathrm{d}x=\left( Ax+B\right) \sqrt{%
x^{2}+c}+\int \frac{C}{\sqrt{x^{2}+c}}\;\mathrm{d}x.$$

Differentiating both sides, we get

$$\begin{eqnarray*}
\frac{x^{2}}{\sqrt{x^{2}+c}} &=&A\sqrt{x^{2}+c}+\frac{x\left( Ax+B\right) }{%
\sqrt{x^{2}+c}}+\frac{C}{\sqrt{x^{2}+c}} \\
&=&\frac{1}{\sqrt{x^{2}+c}}\left[ A\left( x^{2}+c\right) +x\left(
Ax+B\right) +C\right],
\end{eqnarray*}$$

and so

$$x^{2} =A\left( x^{2}+c\right) +x\left( Ax+B\right) +C=2Ax^{2}+Bx+Ac+C.$$

Comparing coefficients, we get $A=1/2,\;B=0,\;C=-c/2$. Hence

$$\begin{eqnarray*}
\int \frac{x^{2}}{\sqrt{x^{2}+c}}\;\mathrm{d}x &=&\frac{1}{2}x\sqrt{x^{2}+c}-%
\frac{c}{2}\int \frac{1}{\sqrt{x^{2}+c}}\;\mathrm{d}x \\
&=&\frac{1}{2}x\sqrt{x^{2}+c}-\frac{c}{2}\ln \left( x+\sqrt{x^{2}+c}\right)
\end{eqnarray*}$$

because

$$\int \frac{1}{\sqrt{x^{2}+c}}\;\mathrm{d}x=\ln \left( x+\sqrt{x^{2}+c}%
\right)+\text{ constant}.$$
Thus

$$\int_{0}^{2}\frac{p^{2}}{\sqrt{p^{2}+3}}\;\mathrm{d}p=\sqrt{7}-\frac{3}{2}%
 \ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right).$$

And finally

$$I=4\pi \left( \sqrt{7}-\frac{3}{2}\ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right) \right) .$$

ContentLicense: CC BY-SA 3.0

Title: Trig substitution for a triple integral


问:
least squares minimum
it's the age old least-squares problem:

Find the value of b for which the following sum has the minimum value (assuming that x and y are one-dimensional vectors of the same length):

$$ \sum_{i=1}^n (y_i - bx_i)^2 = min $$

This must have been asked before but I can't find it, would appreciate a redirect from somebody thanks

答:
If you know linear algebra, the LHS can be written as
$$
(y-bx)^T(y-bx)
$$
with $y=(y_0,y_1,\cdots y_n)^T$, $x=(x_0,x_1,\cdots x_n)^T$.

The minimization solution is given by
$$
b=(x^Tx)^{-1}x^Ty
$$

See this article for the general case.

Title: least squares minimum


问:
Two fair dice roll. Lower score discarded. In case of tie, either die is discarded. Compute the mean value of the remaining number
"Two fair dice are thrown and the one with a lower score is discarded. In case of a tie, either one of them is discarded. Compute the mean value of the remaining number."

I know how to calculate the expected mean which would be 7. I'm stuck on the "remaining value" part when the lower score die is discarded. How would I calculate the mean value of the remaining number? Any clarity is appreciated!

答: What has to be calculated here is just the maximum of the two dice. I don't see how to speed up the computation, either. Just get the distribution of the maximum number and you're good to go.</p>


问:
What's the first-order logical expression of this statement?
Definition: a set $X$ is finite if and only if it has cardinality $n$ for some natural number $n$

"Let $X$ and $Y$ be finite sets. Then $X \cup Y$ is finite and $\#(X \cup Y) \le \# X + \# Y $" ($\#X$ means the cardinality of $X$)

What's the first-order logical expression of this statement?

答:
Asaf Karaglia's question in comments is very relevant: what predicates and constants do you have in your language?

Meaning, can I write $\#X=n$ in a formula to mean that the cardinality of $X$ is $n$? Can I write $\mathbb{N}$ for the set of natural numbers? Can I use the order of natural numbers as a given? I assume I can use all of this, although it would not be specially problematic to assume both the cardinality equality and the order as not given (things are not as simple if $\mathbb{N}$ is not available).

So what I would write is
$$\forall X\forall Y\Big(\exists n\exists m(m\in\mathbb{N}\wedge n\in\mathbb{N}\wedge \#X=m\wedge\#Y=n)\rightarrow \exists p(p\in\mathbb{N}\wedge \#(X\cup Y)=p\wedge p\leq m+n)\Big).$$
That is:

for any two sets corresponds to $\forall X\forall Y$;

the assumption that both $X$ and $Y$ are finite corresponds to $\exists n\exists m(m\in\mathbb{N}\wedge n\in\mathbb{N}\wedge \#X=m\wedge\#Y=n)$, meaning there exist ($\exists m\exists n$) natural numbers ($m\in\mathbb{N}\wedge n\in\mathbb{N}$) that equal the cardinalities of $X$ and $Y$ ($\#X=m\wedge\#Y=n$);

the conclusion that $X\cup Y$ is finite and has its cardinality bounded by $m+n$ corresponds to $\exists p(p\in\mathbb{N}\wedge \#(X\cup Y)=p\wedge p\leq m+n)$, meaning there exists ($\exists p$) a natural number ($p\in\mathbb{N}$) which equals the cardinality of $X\cup Y$ ($\#(X\cup Y)=p$) and is bounded by $\#X+\#Y$ ($p\leq m+n$).


问:
The length of projection of the line segment joining the points $(1, –1, 0)$ and $(–1, 0, 1)$ to the plane $2x + y + 6z = 1$ is equal to?

The length of projection of the line segment joining the points $(1, –1, 0)$ and $(–1, 0, 1)$ to the plane $2x + y + 6z = 1$ is equal to?

I am confused if i have to find the length of perpendicular from line segment to plane or the length of projection

答:
Point $A(1,-1,0)$ lies on the plane $2 x+y+6 z=1$  thus
$A\equiv A'$

From $B(-1,0,1)$ we find the perpendicular line parametric equations
$$(x=-1+2t,y=t,z=1+6t)$$
$$2 (-1+2t)+t+6(1+6t)=1\to t=-\frac{3}{41}\to B'=\left(-\frac{47}{41},-\frac{3}{41},\frac{23}{41}\right)$$
and finally
$$A'B'=\sqrt{\frac{237}{41}}\approx 2.4$$


问:
Is the cylinder homotopic to the circle without using real multiplication?
Let $X$ be a nice topological space (for example $X = \mathbb{S}^1$). It is possible to show by hand that the cylinder $X \times [0,1]$ is homotopic to $X$. All the proof I know rely on the multiplication of two reals.
Since being homotopic has an intuitive meaning that do not need multiplication, I would like to see a method that do not use this fact.

答:
Here is a proof using cubical Agda, which is an environment allowing you to do synthetic homotopy. This means that certain principles are directly assumed that characterize homotopy (and make the reasoning itself homotopical), rather than building it out of e.g. real numbers. In the case of Agda, this means one can define spaces in terms of being generated by points and paths between them, and one can talk about points along a path via formal, continuously varying parameters.

A key lemma that the interval is contractible is then:

```<code>[0,1]-lemma : ∀ i → Path [0,1] zero i
[0,1]-lemma zero = λ k → zero
[0,1]-lemma one  = λ k → seg k
[0,1]-lemma (seg j) = λ k → seg (j ∧ k)
</code>```

Which involves only definition by (coherent) cases, the generating points/paths, and certain operations on the formal parameterization of paths (like being able to take the 'minimum' of two parameters).

If you're interested in this sort of synthetic homotopy, there's the Homotopy Type Theory book.

For a more general explanation of the 'synthetic' method (with an example other than homotopy), there was recently a nice talk on the subject by Andrej Bauer.

Body:
Here is a proof using cubical Agda, which is an environment allowing you to do synthetic homotopy. This means that certain principles are directly assumed that characterize homotopy (and make the reasoning itself homotopical), rather than building it out of e.g. real numbers. In the case of Agda, this means one can define spaces in terms of being generated by points and paths between them, and one can talk about points along a path via formal, continuously varying parameters.

A key lemma that the interval is contractible is then:

[0,1]-lemma : ∀ i → Path [0,1] zero i
[0,1]-lemma zero = λ k → zero
[0,1]-lemma one  = λ k → seg k
[0,1]-lemma (seg j) = λ k → seg (j ∧ k)

Which involves only definition by (coherent) cases, the generating points/paths, and certain operations on the formal parameterization of paths (like being able to take the 'minimum' of two parameters).

If you're interested in this sort of synthetic homotopy, there's the Homotopy Type Theory book.

For a more general explanation of the 'synthetic' method (with an example other than homotopy), there was recently a nice talk on the subject by Andrej Bauer.


问:
Indirect Lyapunov method for exponential stability
Le the following non linear system:

$\begin{cases}
\dot x_1=x_2\\
\dot x_2=-x_1^3-x_2
\end{cases}$

with $V(x)=\frac{1}{4}x_1^3+\frac{1}{2}x_2^2$ by Lyapunov Theorem I have proven that the origin is an asymptotically stable point.

Since the jacobian of the function $f$ (let's rewrite the system in vectorial form as $\dot x=f(x)$) has an eigenvalue equal to $0$, I would say that surely the origin cannot be exponentially stable. Do you think it is right my intuition?

答:
If we understand exponential stability as the existence of a lyapunow function $V(x)$ such that $V(x) = -\lambda \dot V(x)$ then from

$$
\begin{cases}
\dot x_1=x_2\\
\dot x_2=-x_1^3-x_2
\end{cases}\Rightarrow\begin{cases}
x_1^3\dot x_1=x_1^3x_2\\
x_2\dot x_2=-x_1^3x_2-x_2^2
\end{cases}
$$
after addition

$$
\frac 12\frac{d}{dt}\left(\frac 12x_1^4+ x_2^2\right)= -x_2^2\ge-\left(\frac 12x_1^4+ x_2^2\right)
$$

so

$$
V(x)\ge -\lambda \dot V(x)
$$


问:
Why does $|[\alpha]^{< \kappa}|=|\alpha^{< \kappa}|$?
Why does $|[\alpha]^{< \kappa}|=|\alpha^{< \kappa}|$?
I already know that $|[\alpha]^\kappa|=|\alpha^\kappa|$ if the sets are infinite, but why does $|[\alpha]^{< \kappa}|=|\alpha^{< \kappa}|$?</p>

答:
If $\kappa=\lambda^+$ then $|[\alpha]^{<\kappa}|=|[\alpha]^\lambda|$, and $|\alpha^{<\kappa}|=|\kappa\cdot\alpha^\lambda|=|\alpha^\lambda|$, so we reduced this to an equality you already know.

If $\kappa$ is singular, then we can write this as the union of $|[\alpha]^\lambda|$ for a cofinal sequence of $\lambda<\kappa$, and again we reduce this to the previous situation, only now we have two infinite sums over the same index set (the cofinality of $\kappa$), and the summands are the same. Equality ensues.


问:
Change in the Closure, Interior, and Boundary of a set when the topology is changed
Let $A=(0,1], B=[0,1], C=(0,1)$, and $D=\{0,1\}$. I know that in the usual topology we have
$\operatorname{closure}(A) = B, int(A) = C$ and $\sigma(A)=D$. However, when we consider the set of real numbers as a  topological space where the topology is the power set of the reals, then it is given that $\operatorname{closure}(A) = (0,1], \operatorname{int}(A) = (0,1]$ and $\sigma(A)=\emptyset$. What is the reason?

答:
Because, since that topology consists of all subsets of $\Bbb R$, then, with respect to that topology, every set is open. And, since every set is open, every set is closed.

Now, since the closure of a set is the smallest subset that contains it, if a set is closed, then it is equal to its own closure. And, since the interior of a set is the largest of its subsets, if a set is open, then it is equal to its own interior.

Finally, since the boundary of a set $S$ is its closure minus its interior, if a set is both closed and open, then its boundary is the empty set.


问:
How to prove that $\int_{\partial A}\frac{x}{\lVert x\rVert^n}\cdot \nu(x)d\sigma = \int_{\partial B_1(0)}1d\sigma$?
I am having some difficulty finding a proof idea for this question I found in an old exam:

Suppose $A\subset\mathbb R^n$ is a bounded $C^1$ set with outer unit normal $\nu_A(x)$ and $0\in A$. Prove that $$\int_{\partial A}\frac{x}{\lVert x\rVert^n}\cdot \nu_A(x)d\sigma = \int_{\partial B_1(0)}1d\sigma.$$
Hint: Consider the set $A_\epsilon:=A\setminus B_\epsilon(0)$ for sufficiently small $\epsilon$.

答:
Thanks @TedShifrin

Following the hint and comment recommendations, consider $A_\epsilon\subset A$ and note that $B_\epsilon(0)\subset A$ for sufficiently small $\epsilon$, using the quotient rule
$$\begin{aligned}\int_{A_\epsilon}\text{div} \frac{x}{\lVert x\rVert^n}dx
&= \int_{A_\epsilon}\sum_{i=1}^n \frac{\partial_i}{\partial x_i} \frac{x}{\lVert x\rVert^n}dx\\
&= \int_{A_\epsilon}\sum_{i=1}^n \frac{\lVert x\rVert^n - x_in\lVert x\rVert^{n-1}x_i\lVert x\rVert^{-1}}{\lVert x\rVert^{2n}}dx\\
&= \int_{A_\epsilon}\sum_{i=1}^n\frac{\lVert x\rVert ^2 - x_i^2 n}{\lVert x\rVert^{n+2}}dx\\
&=n\int_{A_\epsilon}\frac 1 {\lVert x\rVert^n}(1 -\sum_{i=1}^n\frac{x_i^2}{\lVert x\rVert^{2}})dx\\
&= 0.
\end{aligned}$$

So by the divergence theorem, keeping in mind the orientation of the outer normals, we have
$$0 = \int_{\partial A} \frac{x}{\lVert x\rVert^n}\cdot \nu_A(x)d\sigma - \int_{\partial B_\epsilon(0)}\frac{x}{\epsilon^n}\cdot \frac x {\epsilon}d\sigma.$$
Taking a closer look at the second term,
$$\int_{\partial B_\epsilon(0)}\frac{x}{\epsilon^n}\cdot \frac x {\epsilon}d\sigma = \int_{\partial B_\epsilon(0)}\frac{\lVert x\rVert^2}{\epsilon^{n+1}}d\sigma = \int_{\partial B_\epsilon(0)}\frac{1}{\epsilon^{n-1}}d\sigma = \int_{\partial B_1(0)}1d\sigma$$


问:
Strengthening of the axiom of Dependent Choice
I have in mind the following statement:

For every non empty set $X$ and for every entire binary relation $R$ over $X$, there exists a function $f:X\rightarrow X$ such that, for all $x \in X, \ \ x \  R \ f(x)$

This statement obviously implies $\text{DC}$ in ZF, but how much stronger is it? Is it a proper strengthening? Is it equivalent to other known choice axioms?

Thanks

答:
The axiom you propose is equivalent to the axiom of choice. It can be used to produce choice functions on $V_\alpha\setminus\{\emptyset\}$ for arbitrary $\alpha$:

Let $X=V_\alpha$ and say $x\ R\ y $ iff $y\in x$ or $x=\emptyset=y$. The last clause is only there to make sure that $R$ is entire. If now $f: V_\alpha\rightarrow V_\alpha$ satisfies $x\ R\ f(x)$ for any $x$ then $f(x)\in x$ for $x\neq\emptyset$.

This of course implies that any $X$ (with $\emptyset\notin X$) admits a choice function as $X$ is included in some $V_\alpha$ for $\alpha$ large enough.


问:
What is the Definition of Linear Algebra?
For introducing a field of science, usually we require a definition which summarizes the goal which is intended to be achieved by this field. Linear Algebra is one of the most important parts of mathematics which has not only interesting pure mathematical ideas but also a lot of applications in physics and engineering. So it seems reasonable to have a Complete, Clear, Brief, and Delicate definition for this important branch of mathematics.

I am wondering that what is the best fit for such a definition. For example, the very first sentence that the Linear Algebra Done Right by Sheldon Axler starts with is

Linear Algebra is the study of linear maps on finite dimensional vector spaces.

What is your definition of Linear Algebra in few sentences?

答:
I would actually add what user Samasambo comments and say

Linear Algebra is the study of linear maps on finite dimensional vector spaces; the original motivation and first application of this is solving systems of linear equations.

Beyond that, I agree with a sentiment voiced in comments that such "definitions" might be handy for grant applications and TED talks, but are virtually useless for the actual pursuit of study, research, or teaching.


问:
which vectors are perpendicular to each other?
which vectors are perpendicular to each other?

$\vec a = (1, -2, 3)$, $\vec b = (5, 4, 1)$, $\vec c = (1, 0, -5)$

Do i just take the dot product of 2 of them. If the dot product they are at $90^\circ$? But how do i know if there perpendicular?

答: Two vectors are perpendicular if the angle between them is $\frac{\pi}{2}$, i.e., if the dot product is $0$. This follows from the fact that for two vectors $\vec{v}, \vec{w}$, we have $\vec{v}\cdot\vec{w} = |\vec{v}||\vec{w}|\cos(\theta)$, where $\theta$ is the angle between $\vec{v}$ and $\vec{w}$. </p>


问:
What does it mean to equip a subset with the subspace topology?
This goes for equipping any set with any topology. Suppose that I want to equip a subset $A \subseteq X$ where $X$ is a topological space, with the subspace topology. I know that by definition of the subspace topology that $\tau_A = \{U \cap A | U \in \tau\}$. But what does this do to $A$?

答:
Nothing is altered about the set $A$ itself. Instead, the language "to equip with a topology" means to spawn a topological space $(A, T_A)$ with the topology $T_A$ being the topology in question. The topological space $(A, T_A)$ and the set $A$ are not the same thing. $A$ forms a part of it, but it is not identical to $A$, nor does it alter the definition of $A$.

How much differentiation there is between these two things depends on how rich our formalizing foundations are. With ordinary, "ZFC" set theory, $(A,\ T_A)$ is just an ordered pair of sets, which in turn is the set $\{ \{ A\}, \{ A, T_A \} \}$. Note this is not equal to $A$, nor does it have $A$ as a direct member. A richer system of foundations, though, would incorporate a typed language and say that $(A, T_A)$ is an object of a different "data type", the type of topological spaces, and thus could not be interconverted to a set or other type of ordered pair unless explicit type casting rules are defined.


问:
A line parallel to the altitude of a triangle
The altitude $CH$ of a triangle $ABC$ divides $AB$ into two parts $-$ $AH=14$ $cm$ and $BH=36$ $cm$. A line $a$ parallel to $CH$ divides the area of the triangle in half and intersects $AB$ at $P$. Find $AP$ and $BP$.

Since $a$ is parallel to $CH$ and $CH\perp AB\Rightarrow a\perp AB.$ From here $\triangle CHB \sim \triangle NPB$ and $$\dfrac{CN}{NP}=\dfrac{BH}{BP}=\dfrac{BC}{BN}.$$ This doesn't help much. I don't see how we can use the fact that the line divides the area of triangle $ABC$ in half. Thank you in advance! Any help would be appreciated.

答:
On the contrary, the similar triangles you found is the key.

Let $BP = x$ and $CH = h$. By the similar triangles, we have $NP = \dfrac {hx}{36}$.

The area of $\triangle BPN$ is $\dfrac {hx^2}{2\times 36}$, and the area of $\triangle ABC$ is $25h$, which is twice that of $\triangle BPN$.

Hence $25h = \dfrac {hx^2}{36}$, which gives $x=30$.


问:
Birthday Paradox with Leap Year
I looked online, and found more than one and inconsistent answers to the Birthday Paradox when we throw the leap year into the mix. None of the answers I saw match with my own. I am posting my solution to see if it is correct, or if I am missing something.

Question:

Assume that the leap year occurs every four years. (i.e. ignore the 100 and 400 year rule). Also assume that the number of people born each day is the same. What is the probability that in a group of $n$ people (each one selected randomly), no two people share the same birthday?

My Solution:

Let $\mathcal{D}$ be the set of all possible dates in an year. (Thus $\mathcal{D}$ contains 366 elements. Note that these possibilities are not equally likely, since a person is four times as likely to be born on (say) Jan 1 than on Feb 29. This is true for any given day other than Feb 29, and it is encoded in the probability assignments given below.)

Now probability that a randomly selected person is born on Feb 29 is $\frac{1}{1 + 4*365} = \frac{0.25}{365.25}$.

Also, the probability that a randomly selected person is born on a given day other than Feb 29 is $\frac{1}{365.25}$.

Now, for a group of $n$ randomly selected people, the Sample Space of birthdays is $\mathcal{D}^n$. Let $\mathcal{A} \subset \mathcal{D}^n$ be the subset such that no two people share the same birthday.

Divide $\mathcal{A}$ into two disjoint sets $\mathcal{A}_1$ and $\mathcal{A}_2$ such that

$\mathcal{A}_1 = \{\xi: \xi \in \mathcal{D}^n \text{ and no two people have same birthday, and none is born on Feb 29} \}$, and

$\mathcal{A}_2 = \{\eta: \eta \in \mathcal{D}^n \text{ and no two people have same birthday, and exactly one is born on Feb 29} \}$

Now, $\mathbb{P}(\xi) = \frac{1}{(365.25)^n}$ for each $\xi \in \mathcal{A}_1$, and $\mathbb{P}(\eta) = \frac{0.25}{(365.25)^n}$ for each $\eta \in \mathcal{A}_2$.

Also, $|\mathcal{A}_1| = \; ^{365}P_{n}$, and $|\mathcal{A}_2| = \; n \; \cdot \; ^{365}P_{n-1}$

Finally, $\mathcal{A}_1$ and $\mathcal{A}_2$ being disjoint, it follows that

$$\mathbb{P}(\mathcal{A}) = \mathbb{P}(\mathcal{A}_1) + \mathbb{P}(\mathcal{A}_2) = \frac{^{365}P_{n}}{(365.25)^n} + \frac{0.25 \; \cdot \; n \; \cdot \; ^{365}P_{n-1}}{(365.25)^n}$$

PS:

As usual, if you want to find out the probability that at least two people share the same birthday then you would calculate $1 - \mathbb{P}(\mathcal{A})$. Interestingly, the number of people required so that this probability is more than 0.5 is still $n = 23$, same as the birthday paradox without leap year.

Please let me know if the solution above looks accurate.

答:
The two earlier answers are correct both in their analysis and when they say the effect of including a leap-day is small.

It worth checking that is not big enough to change the standard answer of $23$ to the question of smallest number of people for which the probability of no matches is below $\frac12$.   Using the following R code to calculate this for $365$ days and $22,23,24$ people, we get

```<code>probnomatch <- function(people, daysinyear){
  (product((daysinyear %/% 1) - (0:(people-1))) +
   product((daysinyear %/% 1) - (0:(people-2))) *people* (daysinyear %% 1)) /
     daysinyear ^ people
  }
probnomatch(22, 365)
# 0.5243047
probnomatch(23, 365)
# 0.4927028
probnomatch(24, 365)
# 0.4616557
</code>```

which is the standard birthday problem result, with the probability falling below $\frac12$ when there are $23$ people.

Increasing the average number of days in a year to $365.25$ gives

```<code>probnomatch(22, 365.25)
# 0.5247236
probnomatch(23, 365.25)
# 0.493135
probnomatch(24, 365.25)
# 0.4620987
</code>```

which is similar, and leaves $23$ as the median.  Using the current estimate of $365.24217$ mean solar days in a tropical year, or the average of $365.2425$ across a $400$-year cycle of the Gregorian calendar would also leave the $23$ unchanged.  This is not surprising as a year of $366$ days also leaves it unchanged:

```<code>probnomatch(22, 366)
# 0.5252494
probnomatch(23, 366)
# 0.493677
probnomatch(24, 366)
# 0.4626536
</code>```

The switch would happen when there were about $372.34695$ days in a year:

```<code>probnomatch(23, 372.34695)
# 0.5
</code>```

These calculations do not take into account seasonal or other effects on particular dates of birth in the year, which would tend to reduce the median number of people or leave it the same.

Title: Birthday Paradox with Leap Year

Body:
The two earlier answers are correct both in their analysis and when they say the effect of including a leap-day is small.

It worth checking that is not big enough to change the standard answer of $23$ to the question of smallest number of people for which the probability of no matches is below $\frac12$.   Using the following R code to calculate this for $365$ days and $22,23,24$ people, we get

probnomatch <- function(people, daysinyear){
  (product((daysinyear %/% 1) - (0:(people-1))) +
   product((daysinyear %/% 1) - (0:(people-2))) *people* (daysinyear %% 1)) /
     daysinyear ^ people
  }
probnomatch(22, 365)
# 0.5243047
probnomatch(23, 365)
# 0.4927028
probnomatch(24, 365)
# 0.4616557

which is the standard birthday problem result, with the probability falling below $\frac12$ when there are $23$ people.

Increasing the average number of days in a year to $365.25$ gives

probnomatch(22, 365.25)
# 0.5247236
probnomatch(23, 365.25)
# 0.493135
probnomatch(24, 365.25)
# 0.4620987

which is similar, and leaves $23$ as the median.  Using the current estimate of $365.24217$ mean solar days in a tropical year, or the average of $365.2425$ across a $400$-year cycle of the Gregorian calendar would also leave the $23$ unchanged.  This is not surprising as a year of $366$ days also leaves it unchanged:

probnomatch(22, 366)
# 0.5252494
probnomatch(23, 366)
# 0.493677
probnomatch(24, 366)
# 0.4626536

The switch would happen when there were about $372.34695$ days in a year:

probnomatch(23, 372.34695)
# 0.5

These calculations do not take into account seasonal or other effects on particular dates of birth in the year, which would tend to reduce the median number of people or leave it the same.


答:
If the dot product two vectors is $0$, they are orthogonal; in other words, they are perpendicular.

The dot product between two vectors $\vec u, \vec v$ is given by $\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos(\theta)$, so $$\vec u \cdot \vec v = 0 \implies \cos \theta  = 0 \implies \theta = \pi/2 \;\;(90^\circ).$$

(Recall:  two vectors that are orthogonal (perpendicular) form a right angle $\theta = \pi/2 = 90^\circ$.)

Algebraic definition would be $\vec{u}\cdot\vec{v} = \sum_{i=1}^n\: a_i b_i.$

Body:
which vectors are perpendicular to each other?

$\vec a = (1, -2, 3)$, $\vec b = (5, 4, 1)$, $\vec c = (1, 0, -5)$

Do i just take the dot product of 2 of them. If the dot product they are at $90^\circ$? But how do i know if there perpendicular?


问:
Integration with trig substitution
Trying to evaluate this using trig substitution:

$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$

Here's how I'm going about it, using $x = 5/7(\tan\theta)$

$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$
$$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}\right)} \mathrm{d}\theta$$
$$=\int \frac {1}{(25\tan^2\theta + 25)} $$
$$=\int \frac {1}{25(\tan^2\theta + 1)}\mathrm{d}\theta $$
$$=\int \frac {1}{25\sec^2\theta}\mathrm{d}\theta $$
$$=\int \frac {\cos^2\theta}{25}\mathrm{d}\theta $$
$$=\frac{1}{50}(\theta + \sin\theta + \cos\theta) $$
To generalize for $x$, $\theta = \arctan(7x/5)$
$$\frac{1}{50}\left(\arctan\left(\frac{7x}{5}\right) + \sin\left(\arctan\left(\frac{7x}{5}\right)\right) + \cos\left(\arctan\left(\frac{7x}{5}\right)\right)\right) $$
$$\frac{1}{50} \left(\frac{7x}{5\left(\frac{49x^2}{25}+1\right)} + \arctan\left(\frac{7x}{5}\right)\right)$$
But taking the derivative of this gets me:
$$ \frac{35}{(49x^2 +25)^2}$$
Where is my mistake?

答:
Welcome to MSE. The issue is that you left off $dx$:
$$\int \frac1{49x^2 + 25} dx = \int \frac1{49(5/7\tan(\theta))^2 + 25} d(5/7 \tan(\theta) = \frac57\int \frac1{25\tan^2(\theta) + 25} \sec^2(\theta)d\theta = \cdots $$

Title: Integration with trig substitution


问:
Math derivative problem
Consider
$$\mathcal C: \quad g(x)=x^3-5x^2+1$$
and $\mathcal C$ its curve.

Show that $\mathcal C$ has two tangents parallel to a line with equation $y=13x$.

Find the points of tangency $E$ and $F$.

I can't think of a way to solve the first part but for part b I found the derivative of $g(x)$ and put it equal to $13$ and got $x=-1$ and $x=13/3$ (I don't know if it is correct though).

答:
Your work for this part is correct so far. The derivative is $g'(x) = 3x^2 - 10x$. Recall that the equation $y = 13x$ has the slope-intercept form, which is a straight line with slope $13$.
We solve for the value of $x$ such that $g'(x) = 13$. We have
\begin{align}
3x^2 - 10x &= 13 \\
3x^2 - 10x - 13 &= 0 \\
x &= -1 \text{ or } \frac{13}{3}.
\end{align}
This implies that at $x = -1$ and $x = \frac{13}{3}$ the tangents have a slope of $13$ which is parallel to $y = 13x$.
If required by your course, you can check that these two values are on two different tangents indeed, though that will also answer part 2.

To solve for the points of tangency $E$ and $F$ simply substitute the values of $x$ solved in part 1. into $g(x)$, that is,

$$ E = (-1, g(-1)) \quad F = \left(\frac{13}{3}, g\left(\frac{13}{3}\right)\right).$$

Title: Math derivative problem


答:
We can calculate a more general integral of the form
$$I =\int\frac{1}{a^2x^2+b^2}\,\mathrm{d}x.$$
In your example $a=7$ and $b=5$. First of all do the non-trigonometrical substitution $u=ax/b$. That will give you
$$\int\frac{b}{a(b^2u^2+b^2)}\,\mathrm{d}x=\frac{1}{ab}\int\frac{1}{u^2+1}\,\mathrm{d}u.$$
You should be familiar with this integral. It's equal to $\arctan u$. Substituting back yields
$$I = \dfrac{\arctan(ax/b)}{ab}+C.$$

Title: Integration with trig substitution

Body:
Trying to evaluate this using trig substitution:

$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$

Here's how I'm going about it, using $x = 5/7(\tan\theta)$

$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$
$$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}\right)} \mathrm{d}\theta$$
$$=\int \frac {1}{(25\tan^2\theta + 25)} $$
$$=\int \frac {1}{25(\tan^2\theta + 1)}\mathrm{d}\theta $$
$$=\int \frac {1}{25\sec^2\theta}\mathrm{d}\theta $$
$$=\int \frac {\cos^2\theta}{25}\mathrm{d}\theta $$
$$=\frac{1}{50}(\theta + \sin\theta + \cos\theta) $$
To generalize for $x$, $\theta = \arctan(7x/5)$
$$\frac{1}{50}\left(\arctan\left(\frac{7x}{5}\right) + \sin\left(\arctan\left(\frac{7x}{5}\right)\right) + \cos\left(\arctan\left(\frac{7x}{5}\right)\right)\right) $$
$$\frac{1}{50} \left(\frac{7x}{5\left(\frac{49x^2}{25}+1\right)} + \arctan\left(\frac{7x}{5}\right)\right)$$
But taking the derivative of this gets me:
$$ \frac{35}{(49x^2 +25)^2}$$
Where is my mistake?


问:
Why are homeomorphisms the isomorphisms of topological spaces?
I'm having trouble understanding why the statement in the question is true. My current understanding is as follows.

In category theory, an isomorphism between two objects $A$ and $B$ in the same category is defined as a map between them that has an inverse map.

A homeomorphism is defined as a continuous function between two topological spaces that has a continuous inverse.

Why are homeomorphisms exactly the isomorphisms of the topological space category, i.e. why can we not have discontinuous maps (given that we can probably find inverse maps, which would fit the bill for an isomorphism)?

There is a similar question, but with a slightly broader scope.

答: The definition of the category of topological spaces demands that the morphisms are the continuous functions between the underlying sets. So, for a morphism to be an isomorphism in that category it must itself be continuous and its inverse must be in the same category, namely must also be continuous.</p>


问:
Prove that the number of admissible $L$-tuples is the $L$'th Fibonnaci number
I find this exercise in a very old looking number theory book and I've been working on it for awhile. Anyone got any ideas?? Thanks!

Fix a length $L$ and consider an $L$-tuple of $a$ and $b$ that obey the following rule:

$Rule:$ If there is a consecutive string of $b$'s in the $L$-tuple, then there must be an even number of $b$ in that consecutive string

Non-examples:

For $L=6$ the $6$-tuple $(b,b,b,a,a,a)$ breaks the rule because there are $3$ consecutive $b's$.

For $L=12$, the $12$-tuple $(b,b,b,b,a,a,a,a,b,b,b,a)$ breaks the rule because there are $3$ consecutive $b's$

Examples:

For $L=7$, the $7$-tuple $(a,b,b,a,a,a,a)$ is an admissible tuple.

Claim: Fixing $L$, the total number of $L$-tuples of $a$'s and $b$'s that obey this rule is the $L$'th fibonnaci number

Bonus: Fixing a length $L$ and natural number $n$, how many admissible $L$ tuples are there where $a$ appears $n$-times?

答:
Let $L_n$ be the nunmber of good words of length $n$.

We remark that a good word of length $n$ must either be of the form $aW_1$ or $bbW_2$  where $W_i$ is a good word of length $n-i$.  It follows that $$L_n=L_{n-1}+L_{n-2}$$

Checking initial values finishes the proof.


问:
How to prove $a^2+b^2+c^2+d^2+ac+bd\ge\sqrt3$?
We have $ad-bc=1$ . prove

$$a^2+b^2+c^2+d^2+ac+bd\ge\sqrt3$$

To solve it, I multiplied the inequality by $2$ and added $2(ad-bc)=2$ to it :

$$2a^2+2b^2+2c^2+2d^2+2ac+2bd+2ad-2bc\ge2+2\sqrt3$$
$$(a+c)^2+(b+d)^2+(a+d)^2+(b-c)^2\ge 2+2\sqrt3$$
From here I don't know how to prove the minimum value of sum of these squares is $2+2\sqrt3$ .

答:
by AM-GM $$a^2+b^2+c^2+d^2\ge 2\sqrt{(a^2+b^2)(c^2+d^2)}=2\sqrt{{(ac+bd)}^2+{(ad-bc)}^2}$$ $$=2\sqrt{{(ac+bd)}^2+1}$$ Let $ac+bd=t$ we have to prove $$2\sqrt{t^2+1}+t\ge \sqrt{3}$$ Can you end it now?

 we have to prove $$2\sqrt{t^2+1}\ge \sqrt{3}-t$$ if $\sqrt{3}-t\le 0$ the inequality is obvious other wise squaring both sides of inequality it suffices to prove $$4(t^2+1)\ge t^2-2\sqrt{3}t+3\iff {(\sqrt{3}t+1)}^2\ge 0$$ which is true


问:
How might I prove or disprove this conjecture: $7p_{k-2} < 8p_{k-4}$ for all $k \ge 35$?
Patterns observed while working with recursive prime factorizations (arXiv:2102.02777) lead me to the following conjecture (which is a specific case of a much more general conjecture that I am not presently prepared to state):

Let $k$ be an integer greater than or equal to 33.  Then $7p_{k} < 8p_{k-2}$, where $p_i$ is the $i$th prime number.

@Lulu and @Peter have been very helpful in pointing me to a solution for that conjecture.  Having experienced so much success, I venture to ask about another special case of the generalized conjecture (please forgive me if I do not immediately see that the answers already given are sufficient to address it):

Let $k$ be an integer greater than or equal to 50.  Then

$p_k < 2p_{k-1} < 3p_{k-1} < 4p_{k-2} < 5p_{k-1} < 6p_{k-2} < 7p_{k-1} < 8p_{k-3} < 9p_{k-2} < 10p_{k-2}$.

Since @lulu was kind enough to subsequently point out that Bertrand's Postulate is sufficient to prove that latter conjecture, I am now ready to present the generalization of the conjecture (note that I begin with $p_{k-1}$ rather than $p_{k}$ to show successive terms in OEIS sequence A073093 as the subtrahend in the subscript):

Let $n \in \mathbb{N}_{+}$.  Then there exists a positive integer $k_{\text{min}}$ such that for all integers $k \ge k_{\text{min}}$,

$p_{k-1} < \cdots < np_{k-a_{n}},$

where $a_{n}$ is the number of prime power divisors of $n$, i.e., $a$ is Sequence A073093 in the Online Encyclopedia of Integer Sequences (OEIS).

By the way, I cannot figure out what the pattern is in the sequence of $k_{\text{min}}$ for successive values of $n$; that sequence is $(2,3,8,8,13,13,35,35,10,14,\ldots)$.  Can anyone identify a formula for the sequence?

答:
EDIT: As has been pointed out in the comments, there is an error in this proof (basically from copying the bounds incorrectly from wikipedia) which means the details of the argument are wrong. However, the overall gist of the argument is correct so I think it is worth it to leave this answer up. Just note that the bounds should be

$$k\ln(k)+k\ln(\ln(k))-k<p_k<k\ln(k)+k\ln(\ln(k))$$

where I had accidentally copied

$$k\ln(k)+k\ln(\ln(k))-1<p_k<k\ln(k)+k\ln(\ln(k))$$

With this fix, the $K$ described below should be $611$ instead of the $27$ that I originally got.

As has been mentioned in the comments, this conjecture seems true but case checking might be unavoidable. To prove its true for some $k\in\mathbb{N}$, I use the following bounds

$$k\ln(k)+k\ln(\ln(k))-1<p_k<k\ln(k)+k\ln(\ln(k))$$

for $k\geq 6$. Then

$$\frac{7p_k}{8p_{k-2}}<\frac{7}{8}\cdot \frac{k\ln(k)+k\ln(\ln(k))}{(k-2)\ln(k-2)+(k-2)\ln(\ln(k-2))-1}$$

We must now find some $K\in\mathbb{N}$ such that $k\geq K$ implies

$$\frac{7}{8}\cdot \frac{k\ln(k)+k\ln(\ln(k))}{(k-2)\ln(k-2)+(k-2)\ln(\ln(k-2))-1}<1$$

Of course, such a $K$ must exist since

$$\lim_{k\to\infty}\frac{7}{8}\cdot \frac{k\ln(k)+k\ln(\ln(k))}{(k-2)\ln(k-2)+(k-2)\ln(\ln(k-2))-1}=\frac{7}{8}$$

the only difficulty being finding such a $K$. Okay, here is a method for finding such a $K$. Define

$$f(x)= \frac{x\ln(x)+x\ln(\ln(x))}{(x-2)\ln(x-2)+(x-2)\ln(\ln(x-2))-1}$$

Then differentiating gives two awful functions in the numerator and denominator

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

(I haven't written them here since they are rather large and unweildy and don't really provide any insight into the problem). Since $h(x)^2>0$, it is sufficient to find $X$ such that $x\geq X$ implies $g(x)<0$. Again, after lots of simplification, we find that

$$g(x)<-2\ln(\ln(x))^2$$

(with the minor stipulation that $x\geq 3^3$ in order for some of the inequalities we used to work). That is, the derivative of $f(x)$ is always negative for $x\geq 3^3$. This basically completes the proof since

$$\frac{7}{8}f(27)=0.975<1$$

Thus, a sufficient bound is $K=27$ which is already less than what you need. This checks well with the actual bound, since the final prime the conjecture is not true for is $k=25$. That is

$$\frac{7p_{25}}{8p_{23}}=\frac{679}{664}>1$$

and the actual bound is $K=26$.

FINAL EDIT: This edit will prove the existance of $K(n)$ such that

$$p_k<2p_{k-1}<3p_{k-2}<...<np_{k-a_n}$$

where $a_n$ is the number of non-distinct prime factors in $n$, is true for all $k\geq K(n)$

By the prime number theorem, we have that

$$\lim_{k\to\infty}\frac{ap_{k}}{bp_{k-r}}=\lim_{k\to\infty}\frac{ak\ln(k)}{b(k-r)\ln(k-r)}=\frac{a}{b}$$

(note that $r\in\mathbb{Z}$). Thus, $K(2)$ exists since

$$\lim_{k\to\infty}\frac{p_k}{2p_{k-1}}=\frac{1}{2}$$

(in fact, $K(2)=2$ follows from Bertrand's postulate). Now, to find $K(n+1)$ from $K(n)$, define $K(n+1)$ as the maximum of $K(n)$ and $K^{'}$. Here, $K^{'}$ is defined as the smallest natural such that $k\geq K^{'}$ implies

$$\frac{np_{k-a_n}}{(n+1)p_{k-{a_{n+1}}}}<1$$

Again, we know such a $K^{'}$ must exist from the prime number theorem. In this manner, all $K(n)$ can be defined. Of course, finding these $K(n)$ and proving that they are certain values is a much more difficult challenge.


问:
Show that $g(x)=\frac{f(x)}{x} (x>0)$ is increasing
Let $f$ a continuous function for $x\ge 0$, $f'(x)$ exists for $x>0$, $f(0)=0$ and $f'$ increasing. Show that $g(x)=\frac{f(x)}{x} (x>0)$ is increasing.

I would like to know if my proof holds, please.

As $f'$ exists and increasing, $f$ is convex. Consider the interval $0<x<b$ and by convex function property we have the following inequality:

$\dfrac{f(x)}{x}\le \dfrac{f(b)}{b}$

which holds $\forall \ 0<x<b$. And as $f$ is continuous for $x\ge 0$ and $b$ is an arbitrary point (respecting the inequality), we can conclude that $g(x)$ is increasing.

答:
What you did is correct.

You could improve readability by stating that for a convex map the slope of the line joining a fixed point with abscissa $a$ to a point with abscissa $x \gt a$ increases with $x$. Applying that in your case with $a=0$ and knowing that

$$\frac{f(x) - f(0)}{x- 0} = \frac{f(x)}{x}$$ you get that $g$ is increasing.

The introduction of $b$ seems useless to me.


答: The assumptions implies that $f$ is convex. Therefore if $a<b $ we have $$f(a)=f(b\frac{a}{b} ) \leq \left(1-\frac{a}{b}\right) f(0) +\frac{a}{b} f(b) $$

Body:
Let $f$ a continuous function for $x\ge 0$, $f'(x)$ exists for $x>0$, $f(0)=0$ and $f'$ increasing. Show that $g(x)=\frac{f(x)}{x} (x>0)$ is increasing.

I would like to know if my proof holds, please.

As $f'$ exists and increasing, $f$ is convex. Consider the interval $0<x<b$ and by convex function property we have the following inequality:

$\dfrac{f(x)}{x}\le \dfrac{f(b)}{b}$

which holds $\forall \ 0<x<b$. And as $f$ is continuous for $x\ge 0$ and $b$ is an arbitrary point (respecting the inequality), we can conclude that $g(x)$ is increasing.


答:
You could also evaluate the integral in this way:

$$\int \frac {1}{49x^2 + 25}dx=\frac1{49}\int\frac1{x^2+(\tfrac57)^2}dx$$
By knowing that $\int\frac{1}{x^2+a^2}dx=\frac1a\tan^{-1}(\frac xa)$ we have:

$$\frac1{49}\int\frac1{x^2+(\tfrac57)^2}dx=\frac1{49}\times\frac75\tan^{-1}\left(\frac{7x}{5}\right)+C=\frac1{35}\tan^{-1}\left(\frac{7x}5\right)+C$$

Title: Integration with trig substitution


问:
Are there implicit parentheses around fractions?
I was just looking at a simple math question that came across my social media feed and noticed something I must have always ignored in the past.

In this equation $x = 3 \div \frac{1}{3}$ there seems to be an implicit set of parentheses around $\frac{1}{3}$ which yields a result of $x= 9$.

This seems strange to me because I look at $\frac{1}{3} = 1 \div 3$ and wonder why we can't rewrite it as $x = 3 \div \frac{1}{3} = 3 \div 1 \div 3 = 1$

Is omitting the the parentheses just shorthand that I never realized I was using; $\frac{1}{3} \neq 1 \div 3$, but instead $\frac{1}{3} = (1 \div 3)$?

Should fractions come at the front of (F)PEMDAS/(F)BEDMAS?

Or is it just that the definition of a fraction entails that it is an individual unit and expressing      $\frac{1}{3}$ as $(1 \div 3)$ doesn't really make sense either?

答:
Since $\frac13=1\div3$, we can write $3\div\frac13$ as $3\div(1\div3)$. But this is not the same thing as $(3\div1)\div3$, since division is not associative. In other words, you don't get the same thing if you compute first $b\div c$ and then you compute $a\div(b\div c)$ and if you compute first $a\div b$ and then you compute $(a\div b)\div c$.

It's just like subtraction:$$12-(5-2)=12-3=9,$$but$$(12-5)-2=7-2=5.$$


问:
Proof without using truth tables
I need some further explanation, as I feel I am just missing some critical piece of information. I have shown my work thus far (I excluded the truth tables as they take up so much space but included the pertinent values). Can someone shed some light on what I'm overlooking here?

Which of the following formulas are equivalent? Without showing truth tables, explain why which of the following five statements are equivalent. Hint – analyze the possible values of P, Q, and R that make each of the statements false.

a. P → (Q → R).

b. Q → (P → R).

c. (P → Q) ∧ (P → R).

d. (P ∧ Q) → R.

e. P → (Q ∧ R).

My work:

So using truth tables I was able to determine that a,b,d are all equivalent to each other AND c and e are equivalent to each other.
A, B, D are all false ONLY when p and q are true and r is false, as that is the  only time that the conditional statement becomes true implies false.
C and E are false for several instances, when p=T q=T r=F; p=T q=F r=T; p=T q=F r=F.

What has me stumped is how to write a proof from that information. I have tried using the logical equivalence identities, but am unable to  come up with the correct answer, so I feel like that is not the proper way to do it. So I'm wondering, is there somethin I'm missing? Is this supposed to be a proof by implication, or and indirect proof, and if so, how am I supposed to start that? I understand that these are conditional statements, but again, I feel as if I am missing something.

答:
(Posted after a previous answer was accepted)

Using a form natural deduction, we can prove as follows:

Similarly, we can prove:

Then, we have the required logical equivalence:

Title: Proof without using truth tables


问:
Midpoints of diagonals
Let $ABCD$ be convex quadrilateral such that $AB=CD$. And $E\neq F$ where $E, F$ is midpoint of $AC, BD$ respectively. Then prove that angle between$(AB, EF)$ and $(CD, EF)$ are equal.

I'll prove $\vec{AB}\cdot\vec{EF}=\vec{CD}\cdot\vec{EF}$. Using $\vec{EF}=\frac{1}{4}(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})$, it's sufficient to prove that $$(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})\cdot\vec{AB}=(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})\cdot\vec{CD}\iff (\vec{CB}+\vec{AD})\cdot\vec{AB}=(\vec{CB}+\vec{AD})\cdot\vec{CD}\iff BC\cdot CDcos\angle C+AD\cdot CDcos\angle D=BC\cdot ABcos\angle B+BA\cdot ADcos\angle A\iff BCcos\angle C+ADcos\angle D=CBcos\angle B+ADcos\angle A$$. I can't continue after this. Can anyone help me?

答:
It seems that my solution is to advanced, nobody want's to read it.

Translate $D$ for vector $\vec{CB}$ in to new point $G$. Then $BCDG$ is paralelogram, so $$\angle (EF,CD) = \angle (EF,BG)$$
Also, since diagonals in parallelogram halves each other we see that $F$ is on $CG$. But then $EF$ is midlline in triangle $ACG$ parallel to $AG$ and since triangle $ABG$ is isosceles we are done.

Title: Midpoints of diagonals


答:
Assume the point $P$ is some proportion, $r$ of along the path of $HB$. That is to say $HP = r\cdot HB$.

Then the area of $\triangle PNB$ is $r^2\times$ the area of $\triangle HCB=r^2\times \frac 12 HB\times HC=r^2\times 18 HC$

So we have the area of $\triangle PNB = \frac 12\times$ the area $\triangle ABC = \frac 12 \times \frac 12 AB\times HC = \frac 14\times (14+36)HC = 12.5HC$.

SO we have $r^2 \times 18 HC = 12.5 HC$ and so E$r^2 = \frac {12.5}{18}= \frac {25}{36}$ and $r = \frac 56$.

So $HP = \frac 56 36=30$ and $BP = 36 -30 = 6$.  And $AP = HP +AH = 14 + 30 = 44$.

Title: A line parallel to the altitude of a triangle

Body:
The altitude $CH$ of a triangle $ABC$ divides $AB$ into two parts $-$ $AH=14$ $cm$ and $BH=36$ $cm$. A line $a$ parallel to $CH$ divides the area of the triangle in half and intersects $AB$ at $P$. Find $AP$ and $BP$.

Since $a$ is parallel to $CH$ and $CH\perp AB\Rightarrow a\perp AB.$ From here $\triangle CHB \sim \triangle NPB$ and $$\dfrac{CN}{NP}=\dfrac{BH}{BP}=\dfrac{BC}{BN}.$$ This doesn't help much. I don't see how we can use the fact that the line divides the area of triangle $ABC$ in half. Thank you in advance! Any help would be appreciated.


Title: Demonstrate $3+9+27+\cdots +3^{n}=(\frac{3}{2})\left ( 3^{n}-1 \right )$

Body:
Good day! I have this for this one, but I cannot get it. What am I doing wrong?

$3+9+27+\cdots +3^{n}=(\cfrac{3}{2})\left ( 3^{n}-1 \right )$

$3^{1}=\left ( \cfrac{3}{2} \right )\left ( 3^{1} -1\right )$

$3=\left ( \cfrac{3}{2} \right )\left ( 2\right )$

$3 = 3$

Then the second part:

$3+9+27+\cdots +3^{k}=(\cfrac{3}{2})\left ( 3^{k}-1 \right )$

$3+9+27+\cdots +3^{k}+3^{k+1}=(\cfrac{3}{2})\left ( 3^{k+1}-1 \right )$

In the left side I have:

$3+9+27+\cdots +3^{k}+3^{k+1}=(3+9+27+\cdots +3^{k})+3^{k+1}$

Substituting in:

$3+9+27+\cdots +3^{k}=(\cfrac{3}{2})\left ( 3^{k+1}-1 \right )$

$=(\cfrac{3}{2})\left ( 3^{k+1}-1 \right )+3^{k+1}$

Body:
You can use the fact that $3^i=\frac{3^{i+1}-3^i}{2}$ whenever $i\in\Bbb N$
$$
\begin{aligned}\sum_{i=1}^n 3^i&=\frac{1}{2}\sum_{i=1}^n 3^{i+1}-3^i\\&=\frac{1}{2}\left(\sum_{i=1}^n 3^{i+1}-\sum_{i=1}^n3^i\right)\\&=\frac{1}{2}\left(\sum_{i=2}^{n+1} 3^{i}-\sum_{i=1}^n3^i\right)\\&=\frac{1}{2}\left(3^{n+1}+\sum_{i=2}^{n} 3^{i}-\left[3+\sum_{i=2}^n3^i\right]\right)\\&=\frac{1}{2}(3^{n+1}-3)\\&=\frac{3}{2}(3^{n}-1)
\end{aligned}$$

In general, we have that $a^i=\frac{a^{i+1}-a^i}{a-1}$, and, therefore
$$\sum_{i=1}^n a^i=\frac{1}{a-1}\sum_{i=1}^n a^{i+1}-a^i=\frac{1}{a-1}(a^{n+1}-a)=\frac{a(a^{n}-1)}{a-1}$$


问:
Elements and Conjugacy Classes of a group
Let

$G=(C_{p_1} : C_{3}) \times(C_{p_2} : C_{3})$

where $p_1,p_2\equiv{1}\pmod{3}$.

How many elements does the group $G$ have of each order?  Furthermore, what is the total number of conjugacy classes?

I assumed that G contains exactly p-1 elements of order p, 2(p-1) elements of order 3p and 2(3p+1) elements of order 3, for each p_{i}. But I could be wrong.

Similarly, can I ask the same question for $G= A_{4} \times(C_{p} : C_{3})$ where $p\equiv{1}\pmod{3}$.

I am trying to adapt a current proof where $G= C_{3} \times(C_{p} : C_{3})$ and $p\equiv{1}\pmod{3}$.  The proof is shown above and the authors claim that similar arguments can be used to prove the two cases I have presented above.

答:
For a direct product $G \times H$, the conjugacy classes are of the form $C \times D$, where $C$ and $D$ are conjugacy classes of $G$ and $H$, and the order of the elements in $C \times D$ is the least common multiple of the orders of elements in $C$ and in $D$. This makes it straightforward to answer your questions in direct products provided that we can answer them in the factors.

So lets apply that to the case when both $G$ and $H$ are nonabelian groups with structure $C_p:C_3$.

The factors have one element of order $1$, $p-1$ of order $p$ in $(p-1)/3$ classes, and two classes of elements of order $3$, both of size $p$.

So in $(C_{p_1}:C_3) \times (C_{p_2}:C_3)$ with $p_1 \ne p_2$, we have (if I have counted correctly):

one element of order $1$,

$p_1-1$ of order $p_1$ in $(p_1-1)/3$ classes,

$p_2-1$ of order $p_2$ in $(p_2-1)/3$ classes,

$2p_1 + 2p_2 + 4p_1p_2$ of order $3$ in $2 + 2 + 4 = 8$ classes,

$2p_2(p_1-1)$ of order $3p_1$ in $2(p_1-1)/3$ classes,

$2p_1(p_2-1)$ of order $3p_2$ in $2(p_2-1)/3$ classes, and

$(p_1-1)(p_2-1)$ of order $p_1p_2$ in $(p_1-1)(p_2-1)/9$ classes.

The only significant difference when $p_1=p_2$ is that the elements in the final class have order $p_1$.


问:
Interesting Normality Condition (topology)
I just got through an inquiry-based course in introductory topology, and I enjoyed the experience a lot. If you're unfamiliar, this roughly means that we were given a text full of theorems and the class consisted of providing as many proofs as possible, with student presentations in lieu of lectures.

There was one theorem which I got a lot of use out of, but was never able to prove. It wasn't assigned, and I talked to the professor about it a couple times, but he didn't have any advice for me that I hadn't figured out myself.

The text calls it the "Normality Lemma" if that helps, but it doesn't appear to be a standard term.

Let $A$ and $B$ be subsets of a topological space $X$ and let $\{U_i\}_{i\in\mathbb N}$ and $\{V_i\}_{i\in\mathbb N}$ be two collections of open sets such that:

$A\subseteq \bigcup_{i\in\mathbb{N}} U_i$ and $B\subseteq \bigcup_{i\in\mathbb{N}} V_i$

$\overline{U_i}\cap B$ and $\overline{V_i}\cap A$ are empty for all $i\in\mathbb N$.

Then there are disjoint open sets $U$ and $V$ such that $A\subseteq U$ and $B\subseteq V$.

The thing that really puzzles me about this is the countability condition. Is this doing anything more than saying the collections need to be infinite?

Another interesting quirk about this theorem is that it doesn't show that $X$ is normal explicitly. Rather it guarantees disjoint open sets around a specific pair of (sufficiently ''cushioned'') arbitrary sets for any space.

Can anyone help me through a proof of this?

答:
$\newcommand{\cl}{\operatorname{cl}}$This lemma is actually the basis of the usual proof that every regular, Lindelöf space is normal. The countability condition isn’t to make the collections of open sets infinite: it’s to make sure that they aren’t uncountable.

Let $G_0=U_0$ and $H_0=V_0\setminus\cl U_0=V_0\setminus\cl G_0$. Suppose that for some $n\in\Bbb N$ you’ve defined $G_n$ and $H_n$ so that $G_n\subseteq\bigcup_{k\le n}U_k$, $H_n\subseteq\bigcup_{k\le n}V_k$, and $G_n\cap H_n=\varnothing$. Let

$$G_{n+1}=(G_n\cup U_{n+1})\setminus\cl H_n$$

and

$$H_{n+1}=(H_n\cup V_{n+1})\setminus\cl G_{n+1}\;;$$

clearly $G_{n+1}\subseteq\bigcup_{k\le n+1}U_k$, $H_{n+1}\subseteq\bigcup_{k\le n+1}V_k$, and $G_{n+1}\cap H_{n+1}=\varnothing$, so the recursive construction can continue.

Now let $G=\bigcup_{n\in\Bbb N}G_n$ and $H=\bigcup_{n\in\Bbb N}H_n$. Clearly $G$ and $H$ are open. I’ll let you see if you can show that $A\subseteq G$, $B\subseteq H$, and $G\cap H=\varnothing$; feel free to ask for help if you need it.

Added: You might be interested in an old paper of mine in The American Mathematical Monthly: A ‘More Topological’ Proof of the Tietze-Urysohn Theorem, Vol. $85$, No. $3$ (Mar., $1978$), pp. $192$-$193$, JSTOR; the proof in question uses a lemma that is proved by the same ‘climbing a chimney’ technique that I used here.


问:
Does representability of the subfunctor coherent with taking quotients?
I'm currently learning the deformation of Galois representations by reading Gouvêa's note.

Let $\mathsf{CNL}$ be the category of complete noetherian local rings with (fixed) residue field $\mathbb{F}$. The morphisms are local homomorphisms fixing the residue field. Let $\Lambda \in \mathsf{CNL}$ and denote the full subcategory of $\Lambda$-algebras in $\mathsf{CNL}$ by $\mathsf{CNL}_{\Lambda}$. The full subcategory of artinian objects are denoted by $\mathsf{Ar}_{\Lambda}$.

Let $\mathbf{F}: \mathsf{Ar}_{\Lambda} \rightarrow \mathsf{Set}$ be a functor and $\mathbf{F}_1$ a subfunctor of $\mathbf{F}$. By a "subfunctor", we mean for all $A \in \mathsf{Ar}_{\Lambda}$, $\mathbf{F}_1(A) \subseteq \mathbf{F}(A)$. (More precise definitions here: Subfunctor in nLab)

Now suppose $\mathbf{F}$ and $\mathbf{F}_1$ are pro-representable, i.e. they are represented by object in
$\mathsf{CNL}_{\Lambda}$. Let $R, R_1 \in \mathsf{CNL}_{\Lambda}$  represent  $\mathbf{F}$ and $\mathbf{F}_1$ accordingly, then one can show that $R_1$ is a quotient of $R$. Say $R_1 = R/I$.

Let $A \in \mathsf{Ar}_{\Lambda}$ and $a \in \mathbf{F}_1(A) \subseteq \mathbf{F}(A)$ fixed. Then for $a$,

By representability of $\mathbf{F}_1$, there exists a corresponding homomorphism $\phi_{1,a}: R_1 \rightarrow A$.   (Via $\mathbf{F}_1(A) \cong \mathrm{Hom}(R_1, A)$.)

By representability of $\mathbf{F}$, there exists a corresponding homomorphism $\phi_{a}: R \rightarrow A$.   (Via $\mathbf{F}(A) \cong \mathrm{Hom}(R, A)$.)

My main question is: Are the two homomorphisms $\phi_{1,a}$ and $\phi_{a}$ "coherent"?

To be more specific, let $\pi: R \rightarrow R_1=R/I$ be the natural homomorphism. Is the composition
$$ R \stackrel{\pi}{\longrightarrow} R/I \stackrel{\phi_{1,a}}{\longrightarrow} A $$
EXACTLY the homomorphism $\phi_{a}$?

I feel like this is correct but I have spent quite a long time struggling for a neat proof but failed. So I'm sitll doubting whether this is really true or not?

Thank you all for commenting and answering!

答:
The Yoneda embedding associates to any natural transformation $\text{Hom}(R_1,-)\to \text{Hom}(R,-)$ a ring morphism $\pi:R \to R_1$ such that for any $T$
$$\text{Hom}(R_1,T)\to \text{Hom}(R,T),\quad\phi\mapsto \phi\circ \pi$$
is the morphism induced by the natural transformation. Choose an arbitrary $a\in F_1(A)$
given by $\phi_{1,a}\in \text{Hom}(R_1,A).$ Then its image under the inclusion of the subfunctor, by definition of the induced ring morphism $R\to R_1$, is given by the image of $\phi_{1,a}$ under $\text{Hom}(R_1,T)\to \text{Hom}(R,T)$, i.e its $\phi_{1,a}\circ \pi$. Hence $\phi_{1,a}\circ \pi=\phi_a$ to use your notation.


Body:
Here is what you were looking for.

$$(3/2)(3^k - 1) + 3^{k+1} = (3/2)(3^k - 1 + 2\cdot 3^k)
= (3/2)(3\cdot 3^k - 1) =(3/2)(3^{k+1} - 1)$$


答:
Hint:

You should know from high school that the sum of consecutive terms of a geometric series with common ratio $q$ is given by the formula:
$$a_m+a_{m+1}+\dots +a_n=\frac{a_{n+1}-a_m}{q-1}\qquad (n>m).$$


问:
Super-Gaussian random number generation
Is there any simple method to generate random numbers, which follow super-Gaussian distribution?

$$f(x,y)=A\exp \left(-\left({\frac {(x-x_{o})^{2}}{2\sigma _{X}^{2}}}+{\frac {(y-y_{o})^{2}}{2\sigma _{Y}^{2}}}\right)^{P}\right)$$

One idea was rejection sampling, but drawing a bounding box would cut off the tails.

答:
This'll be for the case $P>1$.

With a slight change in notation your density $f$ is proportional to $\exp(-\phi^P(x, y))$ where $\phi(x, y) = (x - m_X)^2/{2\sigma_X^2} + (y - m_Y)^2/2\sigma_Y^2$. You have a bivariate Gaussian for $P=1$.

First the easy part. If we knew there to be a constant $a$ for which
$a \exp(-\phi(x, y))/\exp(-\phi^P(x)) \geq 1$ for all $x$ and $y$, a simple acceptance-rejection procedure would be:

Generate $X \sim N(m_X, \sigma_X^2)$, $Y \sim N(m_Y, \sigma_Y^2)$.

Generate $U$ uniform on $[0, 1]$.

If $a U \exp(-\phi(X, Y)) \leq \exp(-\phi^P(X, Y))$ accept $X, Y$, otherwise go back to Step 1.

There is such an $a$. For $P > 1$ the smallest value of $\exp(-u)/\exp(u^P)$ for $u \geq 0$ occurs at $u_0 = P^{-1/(P -1)}$, so we can choose $a = \exp(u_0 - u_0^P)$ as the rejection constant.

I doubt this generator will have uniform performance as $P \rightarrow \infty$, you typically have to do a bit more work for that, but maybe this is a start.

Title: Super-Gaussian random number generation


问:
Independent and uniformly distributed on $(\frac{1}{2},1]$
I have two random variables $X,Y$ which are independent and uniformly distributed on $(\frac{1}{2},1]$. Then I consider two more random variables, $D=|X-Y|$ and $Z=\log\frac{X}{Y}$. I would like to calculate both, the disitrbution functions $F_D(t), F_Z(t)$ and the the density functions $f_D(t),f_Z(t)$

To do that I think the first thing we need to do is to evaluate the density of the common distribution of $X$ and $Y$, but I do not know how to do that.

The only thing which is clear to me is the density and distribution function of $X$ and $Y$ because we know that they are uniform.

EDIT: Please read my own answer to this question. I need someone who can show me my claculation mistakes.

答:
The area of a $\color{red}{\mathrm{right\ triangle}}$ being half the product of its legs is all that one needs, really...

For example, $D\geqslant x$ with $0\leqslant x\leqslant\frac12$ means that $(X,Y)$ is either in the triangle $\color{red}{T_x}$ with vertices $(\frac12+x,\frac12)$, $(1,\frac12)$, $(1,1-x)$ or in the triangle $\color{red}{S_x}$ symmetric of $\color{red}{T_x}$ with respect to the first diagonal. Both legs of $\color{red}{T_x}$ are $\frac12-x$, the triangles $\color{red}{T_x}$ and $\color{red}{S_x}$ are disjoint and with the same area, and the domain of $(X,Y)$ is the full square $(\frac12,1)\times(\frac12,1)$ with area $\frac14$, hence
$$
P[D\geqslant x]=4\cdot2\cdot|\color{red}{T_x}|=(1-2x)^2=1-4x(1-x).
$$
From here, one gets for every  $0\leqslant x\leqslant\frac12$,
$$
F_D(x)=4x(1-x),\qquad f_D(x)=4(1-2x).
$$
Likewise, consider $R=\frac{Y}X$, then $\frac12\leqslant R\leqslant2$ and $R$ and $\frac1R$ are identically distributed. For every $\frac12\leqslant x\leqslant 1$, $R\leqslant x$ means that $(X,Y)$ is in the triangle $\color{red}{U_x}$ with vertices $(\frac1{2x},\frac12)$, $(1,\frac12)$, $(1,x)$. The legs of $\color{red}{U_x}$ are $1-\frac1{2x}$ and $x-\frac12$ hence
$$
P[R\leqslant x]=4\cdot|\color{red}{U_x}|=\frac1{2x}(2x-1)^2.
$$
Likewise, for every $1\leqslant x\leqslant2$,
$$
P[R\geqslant x]=\frac1{2x}(2-x)^2.
$$
(This can be proved either by considering the triangle $\color{red}{V_x}$ which corresponds to the event $R\geqslant x$, or directly using the equidistribution of $R$ and $\frac1R$.)

Now, if $Z=\log R$ then, for every $-\log2\leqslant z\leqslant0$, $[Z\leqslant z]=[R\leqslant\mathrm e^z]$ hence
$$
P[Z\leqslant z]=\frac12\mathrm e^{-z}(2\mathrm e^z-1)^2,
$$
and for every $0\leqslant z\leqslant\log2$, $[Z\geqslant z]=[R\geqslant\mathrm e^z]$ hence
$$
P[Z\geqslant z]=\frac12\mathrm e^{-z}(2-\mathrm e^z)^2.
$$
From here, one gets, for every $-\log2\leqslant z\leqslant0$,
$$
F_Z(z)=\tfrac12\mathrm e^{-z}(2\mathrm e^z-1)^2,\qquad f_Z(z)=2\mathrm e^z-\tfrac12\mathrm e^{-z},
$$
and, for every $0\leqslant z\leqslant\log2$,
$$
F_Z(z)=1-\tfrac12\mathrm e^{-z}(2-\mathrm e^z)^2,\qquad f_Z(z)=2\mathrm e^{-z}-\tfrac12\mathrm e^{z}.
$$
The invariance of $f_Z(z)$ by the symmetry $z\mapsto-z$, which follows from the fact that $R$ and $\frac1R$ are identically distributed, should be apparent.


问:
$X^5-6X+3$ has two complex roots
Show that the polynomial $X^5-6X+3$ has exactly two non-real roots.

I am struggling to solve this question for some time now and I wonder whether there is a theorem that helps to answer it.

Any help is appreciated.

答: $$P'(x) = 5x^4-6$$ Equation $P'(x)=$ has exactly 2 real solutions so $P$ has exactly two extremal points at $x_{1,2} = \pm\sqrt[4]{6\over 5}$. Since $P(x_1)\cdot P(x_2)<0$ we have a conclusion.

Title: $X^5-6X+3$ has two complex roots


问:
Error propagation of a variable for an integral
I have an integral that depends on two parameters $a\pm\delta a$ and $b\pm \delta b$. I am doing this integral numerically and no python function can calculate the integral with uncertainties.

So I have calculated the integral for each min, max values of a and b.
As a result I have obtained 4 values, such that;

$$(a + \delta a, b + \delta b) = 13827.450210 \pm 0.000015~~(1)$$
$$(a + \delta a, b - \delta b) = 13827.354688 \pm 0.000015~~(2)$$
$$(a - \delta a, b + \delta b) = 13912.521548 \pm 0.000010~~(3)$$
$$(a - \delta a, b - \delta b) = 13912.425467 \pm 0.000010~~(4)$$

So it is clear that $(2)$ gives the min and $(3)$ gives the max. Let us show the result of the integral as $c \pm \delta c$. So my problem is what is $c$ and $\delta c$ here?

The integral is something like this

$$I(a,b,x) =C\int_0^b \frac{dx}{\sqrt{a(1+x)^3 + \eta(1+x)^4 + (\gamma^2 - a - \eta)}}$$

where $\eta$ and $\gamma$ are constant.

Note: You guys can also generalize it by taking $\eta \pm \delta \eta$ but it is not necessary for now.

I have to take derivatives or integrals numerically. There's no known analytical solution for the integral.

$\eta = 4.177 \times 10^{-5}$, $a = 0.1430 \pm 0.0011$, $b = 1089.92 \pm 0.25$, $\gamma = 0.6736 \pm 0.0054$, $C = 2997.92458$

答:
```<code>from numpy import sqrt
from scipy import integrate
import uncertainties as u
from uncertainties.umath import *

#Important Parameters
C = 2997.92458  # speed of light in [km/s]
eta = 4.177 * 10**(-5)
a = u.ufloat(0.1430, 0.0011)
b = u.ufloat(1089.92, 0.25)
gama = u.ufloat(0.6736, 0.0054)

@u.wrap
def D_zrec_finder(gama, a, b):
    def D_zrec(z):
        return C / sqrt(a * (1+z)**3 + eta * (1+z)**4 + (gama**2 - a - eta))
    result, error = integrate.quad(D_zrec, 0, b)
    return result

print((D_zrec_finder(gama, a, b)).n)
print((D_zrec_finder(gama, a, b)).s)
</code>```

This works

Body:
from numpy import sqrt
from scipy import integrate
import uncertainties as u
from uncertainties.umath import *

#Important Parameters
C = 2997.92458  # speed of light in [km/s]
eta = 4.177 * 10**(-5)
a = u.ufloat(0.1430, 0.0011)
b = u.ufloat(1089.92, 0.25)
gama = u.ufloat(0.6736, 0.0054)

@u.wrap
def D_zrec_finder(gama, a, b):
    def D_zrec(z):
        return C / sqrt(a * (1+z)**3 + eta * (1+z)**4 + (gama**2 - a - eta))
    result, error = integrate.quad(D_zrec, 0, b)
    return result

print((D_zrec_finder(gama, a, b)).n)
print((D_zrec_finder(gama, a, b)).s)

This works


问:
Solve nonlinear ordinary differential equation?
I have no idea how to solve the following ODE:
$$
f'(t) + \gamma\left[ r+\frac{\sigma^{-2}B^2}{2(1-\gamma)} \right]f(t)+(1-\gamma)e^{-\frac{\rho t}{1-\gamma}}\left( f(t) \right)^{\frac{\gamma}{\gamma-1}} = 0,
$$
$$
f(T) = e^{-\rho T},
$$
where all the letters except $f$ and $t$ can be regarded as constant. Any solution or hint will be appreciated!

答:
$$
f'(t) + \gamma\left[ r+\frac{\sigma^{-2}B^2}{2(1-\gamma)} \right]f(t)+(1-\gamma)e^{-\frac{\rho t}{1-\gamma}}\left( f(t) \right)^{\frac{\gamma}{\gamma-1}} = 0,
$$
$$
f(T) = e^{-\rho T},
$$

This ODE can be re-written as
$$f'(t)+af(t)+b e^{-ct} f^q=0=f^{-q}f'+a f^{1-q}=-be^{-ct}$$
which is a Bernoulli IDE as pointed out in the comment above.
Take $f^{1-q}=z \implies (1-q) f^{-q} f'=z'$, then
$$\frac{z'}{1-q}+az=-be^{-ct} \implies z'+(1-q)az=-b(1-q)e^{-ct}$$
This is linear ODE with integrating factor as $I=e^{(1-q)at}$.
Then $$z=f^q= e^{-(1-q)at} \int e^{(1-q)at} (-b(1-q)e^{-ct}) dt+D e^{-(1-q)at}$$
$$\implies f(t)=\left(\frac{-b(1-q)}{a-aq-c} e^{-ct}+ D e^{(q-1)at}\right)^{1/q}.$$


答:
Yes, you can certainly say that there are implicit parentheses, since $3 \div \frac{1}{3}$ equals $3 \div (1 \div 3)$, not $(3 \div 1) \div 3$. It helps to think of $\frac{1}{3}$ as a number in its own right, an individual unit as you have suggested. It can be visualised as the proportion of the circle below that has been shaded:

When you divide $3$ by $\frac{1}{3}$, you are asking 'what number, when multiplied by $\frac{1}{3}$ gives you $3$?'
$$
\frac{1}{3} \times \, ? \, = 3
$$
Here, the $?$ must be equal to $9$, since
$$
\frac{1}{3} \times 9 = 3 \, .
$$
and so $3 \div \frac{1}{3}=9$. The number $3$ can be visualised as shading in $3$ circles. Then, you are trying to think of how many sectors of area $\frac{1}{3}$ can be fit into $3$. If we count them up, we get $9$. Alternatively, note that $\frac{1}{3} \times 3 = 1$, and $1 \times 3 = 3$. Hence,
$$
\left(\frac{1}{3} \times 3\right) \times 3 = 3 \, .
$$
With multiplication, we are allowed to regroup the brackets*:
$$
\frac{1}{3} \times \left(3 \times 3 \right) = 3 \, ,
$$
and so
$$
\frac{1}{3} \times 9 = 3 \, .
$$

*The technical term for an operation where you are allowed to regroup the brackets is associative, as outlined in José Carlos Santos' answer. For instance, addition is associative:
$$
(3+2)+7 = 3+(2+7) \, .
$$
In general,
$$
(a+b) + c = a+(b+c) \, .
$$
Multiplication is also associative. Be careful, though: division and subtraction are not associative.

Title: Are there implicit parentheses around fractions?

Body:
I was just looking at a simple math question that came across my social media feed and noticed something I must have always ignored in the past.

In this equation $x = 3 \div \frac{1}{3}$ there seems to be an implicit set of parentheses around $\frac{1}{3}$ which yields a result of $x= 9$.

This seems strange to me because I look at $\frac{1}{3} = 1 \div 3$ and wonder why we can't rewrite it as $x = 3 \div \frac{1}{3} = 3 \div 1 \div 3 = 1$

Is omitting the the parentheses just shorthand that I never realized I was using; $\frac{1}{3} \neq 1 \div 3$, but instead $\frac{1}{3} = (1 \div 3)$?

Should fractions come at the front of (F)PEMDAS/(F)BEDMAS?

Or is it just that the definition of a fraction entails that it is an individual unit and expressing      $\frac{1}{3}$ as $(1 \div 3)$ doesn't really make sense either?


问:
Show that the product of bounded functions converges uniformly
If $f_n$ and $g_n$ are bounded sequences of functions that converge uniformly on a set $E$, show that $f_ng_n$ converges uniformly on $E$.

What I've tried

Call the functions that $f_n,g_n$ converge to $f,g$, respectively.

My intuition is that normalizing the functions here will help with the argument.

To simplify the notation, I will refer to $f(x)$ as $f$.

We need to show that:

$$\sup_{x\in E} \left| f_ng_n - fg \right| \rightarrow 0$$

We will deal with two cases, first assume that $f,g$ are bounded away from $0$ so that the above is equivalent to:

$$\sup_{x\in E} \left| \frac{f_n}{f}\frac{g_n}{g} - 1 \right| \rightarrow 0$$

factoring out $\frac{f_n}{f}$ we have that the term inside the $\sup$ is:

$$\left |\frac{f_n}{f} \left(\frac{g_n}{g} - 1\right) + \left(\frac{f_n}{f} - 1\right)\right |$$

By the triangle inequality it is sufficient to show that each of these converges to $0$ uniformly. The second term converges uniformly to $0$ from the hypothesis. The first term is:

$$O(1)\left(\frac{g_n}{g} - 1\right)$$

since $f_n$ is bounded and $f$ is bounded away from $0$. This converges uniformly to $0$ by the hypothesis on $g_n$.

Now I'm stuck figuring out how to handle the case where $f,g$ may be arbitrarily close to $0$ for some $x$. Is there some way to complete this argument, or does splitting it into these two cases not make sense here?

答: Hint: $f_ng_n -fg = (f_ng_n -fg_n) + (fg_n -fg).$


问:
NP hard/complete
I have never been very clear on this concept. Please help:

At the end of the day, we should want to identify useful problems for which we don't have polynomial solution so far and only have exponential solutions.We want to keep finding if we can find a polynomial solution to these problems and the use of reductions is only to be able to identify new only exponential solvable problems with help of existing known exponential problems.??

Why is concept of determinism and non determinism brought into this whole concept of exponential and polynomial solvable problems?? How is this notion useful?

What do we call the problems for which we have only exponential solution so far, but we haven't been able to find reduction to a known NP - hard/complete problems..??

Thanks,

答:
The NP problems are those for which we have a (deterministic) algorithm to verify that a proposed solution really is a solution. Thus, the question of whether a given logical expresion is satisifiable is NP, since we can correctly verify whether a proposed truth assignment is satisfying or not in polynomial time. The questions of whether a given salesman travel map has a route visiting every city within a fixed travel budget is NP, because we can verify whether any proposed route fits the requirements in polynomial time. The difficult part of an NP problem, rather, is to come up with the proposed solution, to come up with the satisfying truth assignment or the efficient travel route. </p>

<p>These NP problems are all exponential time solvable, because the silly algorithm of just trying out all the possible combinations in turn takes exponential time. Non-determinism comes into the picture because one can view the procedure of "guessing the right answer" and then verifying it as a nondeterministic algorithm. Thus, NP problems are those which have a non-deterministic polynomial time algorithm. </p>

Title: NP hard/complete

Body:
The NP problems are those for which we have a (deterministic) algorithm to verify that a proposed solution really is a solution. Thus, the question of whether a given logical expresion is satisifiable is NP, since we can correctly verify whether a proposed truth assignment is satisfying or not in polynomial time. The questions of whether a given salesman travel map has a route visiting every city within a fixed travel budget is NP, because we can verify whether any proposed route fits the requirements in polynomial time. The difficult part of an NP problem, rather, is to come up with the proposed solution, to come up with the satisfying truth assignment or the efficient travel route.

These NP problems are all exponential time solvable, because the silly algorithm of just trying out all the possible combinations in turn takes exponential time. Non-determinism comes into the picture because one can view the procedure of "guessing the right answer" and then verifying it as a nondeterministic algorithm. Thus, NP problems are those which have a non-deterministic polynomial time algorithm.


答:
The function is a polynomial of odd degree and positive coefficient. So you know it goes down on the left and up on the right, so it has to have one real root at least. A distinct real root is one where the function changes sign.  Try a few values:

$$f(-2) = -17$$

$$f(-1) = 8$$

$$f(0) = 3$$

$$f(1) = -2$$

$$f(2) = 23$$

So not only are there $3$ real roots, the first one is in the interval $(-2,-1)$, the second is in $(0,1)$, and the third is in $(1,2)$.
It only remains to check if that is all of them.
Taking the derivative, we get $f'(x) = 5x^4 - 6$, which has two real roots, meaning only two places where the function can turn around, and we have already accounted for those.  QED.

Title: $X^5-6X+3$ has two complex roots

Body:
Show that the polynomial $X^5-6X+3$ has exactly two non-real roots.

I am struggling to solve this question for some time now and I wonder whether there is a theorem that helps to answer it.

Any help is appreciated.


问:
How to find matrix of linear map sample of vectors and their projection?
I would like to find matrix of linear transformation $f$, which projects vectors like this:

$$f((x, y, z)^T)= (a, b, c, d, e)^T$$

How Can I find this matrix ? I am confused that vectors have different amount of elements.

Can you provide please minimal explanation / proof why your solution works ?

Thank you for help

答:
the fact that the vectors have different amounts of elements is not a problem. The linear map simply maps from $K^3$ to $K^5$ where $K$ is any field (probably $K= \mathbb R$ or $K = \mathbb Q$ is meant in this case). Now we know that the columns of the matrix of the transformation consist of the values of $f(e_1), f(e_2), f(e_3)$ where $e_i$ is the $i$-th canonical basis vector (or in general: Let $f: V \rightarrow W$ be a linear transformation of finite dimensional vector spaces. The $i$-th column of the matrix representing the transformation consists of the coordinates of $f(b_i)$ in the chosen basis of $W$ where $b_i$ is the $i$-th basis vector of the chosen basis of $V$).

In this case we use the canonical bases of $K^3$ and $K^5$ and we need to know $f(\begin{pmatrix} 1 \\ 0 \\0 \\ \end{pmatrix})$, $f(\begin{pmatrix} 0 \\ 1 \\0 \\ \end{pmatrix})$ and $f(\begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix})$ which we can figure out because $f$ is linear and we know the given three values. We just write the three values as columns of the matrix and we are done.

Hope this helps you.


答: Hint: You could prove that ist has three real solution, by considering , the part, where $f(x)>0$ and $f<0$  ($x$ real)  finding the the two extrema with $f'= 0$ helps.

Title: $X^5-6X+3$ has two complex roots


问:
Establish the inequality
Establish the inequality $2^n < \binom{2n}{n} < 2^{2n}$, for $n > 1$.

Hint: Put $x = 2\cdot4\cdot6\cdots(2n)$ ; $y = 1\cdot3\cdot5\cdots(2n - 1)$ and $z = 1\cdot2\cdot3\cdots n$ , then show that $x > y > z $ and subsequently $x^2 > xy > xz$.

Now I am not able to understand that how can I equate $x^2 > xy > yz $ with $ 2^n < \binom{2n}{n} < 2^{2n}$, any sort of help would be appreciated.

Title: Establish the inequality

Body:
Note that $x=2^nn!$:

$$\binom{2n}{n}=\frac{(2n)!}{(n!)^2}=\frac{xy}{x^22^{-2n}}<2^{2n},$$

since $x^2>xy$. Also,

$$\binom{2n}{n}=\frac{xy}{(x2^{-n})z}=2^n\frac{xy}{xz}>2^n,$$

since $xy>xz$.


问:
Beginner question on conditional probability
Three soccer players doing penalty kicks.

Player $X$ has a hit ratio of $90$%, Player $Y$ has a hit ratio of $80$%, Player $Z$ has a hit ratio of $70$%.

Player $X$ takes $70$% of penalty kicks, Player $Y$ takes $20$% of penalty kicks, Player $Z$ takes $10$% of penalty kicks.

You hear on the radio that a penalty has just been missed. What's the probablity of $X,Y,Z$ being the one who missed the shot?

Do we look for ("$M$ = player missed penalty", "$C$ = player got chosen")

$$P(M|C)=\frac{P(M\cap C)}{P(C)}$$

?

Because if so, i always get wrong values according to the solutions.

答:
A straightforward application of Bayes' Theorem:

$P(X) = .7$, $P(Y) = .2$, and $P(Z) = .1$.

$P(M|X) = .1$, $P(M|Y) = .2$, and $P(M|Z) = .3$.

$$P(X|M) = \frac{P(X) P(M|X)}{P(X) P(M|X) + P(Y) P(M|Y) + P(Z) P(M|Z)}$$

and likewise for $P(Y|M)$ and $P(Z|M)$.  Just plug in the numbers.


问:
Category theory- Historical aspect of universal property
In many fields, category theory is an important tool for problem solving, especially in algebraic topology. When I studied algebraic topology, one of the most useful and important things was the universal property of objects (in $Grp$), i.e free product, free product with amalgamation, abelianization and etc'.

My question is, in historical aspect, what was the first thing to get discovered, and what was the first need? the need of universal property, or the need of these objects?

Thanks in advance.

答: As Eilenberg and Mac Lane state in their original article introducing category theory, the reason to define categories is to define functors, and the reason for defining functors is to define natural transformations. So, the need was the inherent 2-categorical computations in homological algebras. The need was to develop the 2-categorical language. As far as I know, the characterisation of objects via their universal property, and the usage of that in practice, e.g., the Van Kampen theorem, came later.</p>


Body:
$3+9+27+\cdots +3^{n}=$

$3(1 + 3 + 9 + ... + 3^{n-1}) =$

$\frac 32\cdot [2\times (1 + 3 + 9 + ... + 3^{n-1})]$

So we now have to prove that $3^n-1 = 2(1 + 3 + 9 + ... + 3^{n-1})$

There's a trick and an insight and as a hint I will ask you to think about how on earth the $2$ could be related to $3$.

  $2 = (3-1)$

So we have to prove:

 $(3-1)(1+3+9 + .... + 3^{n-1}) = 3^n-1$.

Which try it out.....

 $(3-1)(1+3+9 + .... + 3^{n-1}) = 3^n-1$

 $3(1+3+9 + .... + 3^{n-1}) - 1(1+3+9 + .... + 3^{n-1}) =$

 $(3 + 9 + 27 + ...... + 3^n) - (1 + 3+ 9 + ..... + 3^{n-1}) = $

 $(3+9+27 + 81 + ..... + 3^{n-2}+3^{n-1} + 3^n) - (1 + 3 + 9 + 27+81 + .... +3^{n-2}+3^{n-1}) =\require{cancel}$

 $(\cancel 3+\cancel 9+\cancel{27} + \cancel{81} + ..... + \cancel{3^{n-2}}+\cancel{3^{n-1}} + 3^n) - (1 + \cancel 3+\cancel 9+\cancel{27} + \cancel{81} + ..... + \cancel{3^{n-2}}+\cancel{3^{n-1}}) =$

 $3^n -1$.


问:
Finding the fourth side given other three and diagonals.
Say I had a quadrilateral and was given sides of length $a,b,c$ and diagonals $d,e.$ How can I find the fourth side?

Let the fourth side be $x$.

I though of using Ptolemy's first of all to get
$$ac + cx \ge de.$$
We can solve for $x$ to get
$$\frac{de-ac}{c} \le x.$$

But, that doesn't really help much, other than giving a bound. I can't really think of another good way to do this. I also thought of using the Law of Cosines somehow. Can someone help?

答:
Label the vertices $P, Q, R, S$ so that $PQ = a, QR = b, RS = c, SP = x, PR = d$, and $QS = e$.
Since you know all three sides of $\Delta PQR$ you can find the angles using the Law of Cosines.  Likewise for $\Delta QRS$. So you can find $\angle PQR$ and $\angle SQR$. Subtracting gives you $\angle PQS$. Using the Law of Cosines one more time gives you $x$.


问:
Uniqueness of ODE solution with absolute function
I have an ODE   $dy/dx = 2\sqrt{|y|}$.
Suppose I was given initial conditions $y(1) = 1$ for $y > 0$ and $y(-3) = -1$ for $y < 0$, and from these combining with y(x) = 0, I could find a solution.
But, for example, if I keep the same initial condition for y > 0 and change the one for $y < 0$ to $y(-4)=-1$, I will have a different solution.
In this case, for the same initial condition $y(-3) = -1$, we have multiple solutions.

I want to ask is this solely because that the absolute function is not differentiable at 0, thus it fails the conditions of the Picard existence and uniqueness theorem? Or is the answer more subtle and I need more explanation for this non-uniqueness?

Thank you

答:
Picard-Lindelof doesn't apply because $f(x,y)=2\sqrt{|y|}$ is not Lipschitz as a function of $y$, due to an infinite slope at $y=0$. (Differentiability is not actually an issue, it is just used in the common "elementary" form of Picard-Lindelof to avoid explaining the term Lipschitz. As a concrete example to support this remark, the IVPs of $y'=|y|$ always have unique solutions.)

However, the failure of the hypotheses of Picard-Lindelof does not immediately imply that the solutions are not unique.

In this specific problem, you see that $y \equiv 0$ is a solution, but you can also see that there are solutions with $y(x_0)=0$ for some $x_0$ but $y \not \equiv 0$, such as $y=\operatorname{sign}(x) x^2$. So the solutions aren't unique.

I can't really understand what you mean about initial conditions for $y>0$ or $y<0$. In actuality none of the IVPs of this particular equation have unique solutions; if $y(x_0)>0$ then you can go backward in time until you hit $y=0$, and then you have non-uniqueness after that. Similarly if $y(x_0)<0$ then you can go forward in time and eventually uniqueness will break down.


问:
Expressing third covariant derivatives in terms of second covariant derivatives
I'm following the tutorial at this link, where the author states:

These follow from the various way one can iterate covariant derivative

$$\nabla^3_{xyz}s = \nabla^2_{xy}(\nabla_zs) - \nabla_{\nabla^2_{xy}z}s$$
and
$$\nabla^3_{xyz}s = \nabla_x (\nabla^2)_{yz}s + \nabla^2_{yz}(\nabla_{x}s)$$

I'm unable to derive these equations myself.

My Attempt

For the second covariant derivative, I seem to be able to derive this. I used the product rule and the fact that we can commute the covariant derivative with contractions to show that:

$$\nabla_x\nabla_y s= \nabla_x C(\nabla s \otimes y) $$
$$\nabla_x\nabla_y s=  C(\nabla_x \nabla s \otimes y) + C(\nabla s \otimes \nabla_x y ) $$
$$\nabla_x\nabla_y s=  (\nabla_x\nabla s)(y) + \nabla_{\nabla_x y} s $$

I then assume that $(\nabla_x\nabla s)(y)$ is the second covariant derivative, so:

$$ \nabla^2_{xy}s = \nabla_x\nabla_y s - \nabla_{\nabla_x y } s $$

Now at this point we have one expression for the second covariant derivative. To get the next one, I just used the product rule to get:

$$ \nabla_x\nabla_y s = (\nabla_x\nabla)_y s + \nabla_{\nabla_xy}s + \nabla_y(\nabla_x s) $$
$$ \nabla_x\nabla_y s - \nabla_{\nabla_xy}s = (\nabla_x\nabla)_y s  + \nabla_y(\nabla_x s) $$

$$ \nabla^2_{xy}s = (\nabla_x\nabla)_y s  + \nabla_y(\nabla_x s) $$

But any attempts to do the same thing for the third covariant derivative seem to be failing for me. Is there some straightforward way to get to the results I quoted above from here that I'm not seeing?

答:
My recommendation is to work out your own definitions and notation for everything. When you read someone else's writing, use their notation and proof as a guide to how to write everything including the proof in your own notation. Don't worry about understanding their notation literally.

I find higher covariant derivatives to be very confusing. The way I deal with it is that I view the covariant derivative of a higher order covariant derivative to be just a special case of the covariant derivative of a tensor. For example, the covariant derivative of a second order tensor $T$ is defined to be
$$
(\nabla T)(X,Y,Z) = \partial_X(T(Y,Z)) - T(\nabla_XY,Z) - T(Y,\nabla_XZ)
$$
So the second order covariant derivative of $T$ is
\begin{align*}
(\nabla^2T)(X,Y,Z,W) &= (\nabla(\nabla T))(X,Y,Z,W)\\
&=\partial_X(\nabla T(X,Y,Z)) - \nabla T(\nabla_XY,Z,W)\\
&\quad - \nabla T(Y,\nabla_XZ,W) - \nabla T(Y,Z,\nabla_XW)
\end{align*}
And so on.

Therefore, $\nabla^3T = \nabla(\nabla(\nabla T))) = \nabla^2(\nabla T) = \nabla(\nabla^2T)$. Now skew-symmetrization and the Ricci identity should give you what you want.

Note that my personal convention is to never write $\nabla_XT(Y,Z)$. I find that notation difficult to work with, even though I like the way the chain rule identity looks using that notation:
$$
\partial_X(T(Y,Z)) = \nabla_XT(Y,Z) + T(\nabla_XY,Z) + T(Y,\nabla_XZ)
$$


答:
Here is one way to approach it.

Say we know $a, b, c, d, e$ and we need to find $x$.

Apply law of cosine in $\triangle BAD$ to find $\angle A$.

Then apply law of cosine in $\triangle CAD$ to find $\angle CAD$.

Now you know $\theta$. Apply law of cosine in $\triangle BAC$ to find $x$.

Title: Finding the fourth side given other three and diagonals.

Body:
Say I had a quadrilateral and was given sides of length $a,b,c$ and diagonals $d,e.$ How can I find the fourth side?

Let the fourth side be $x$.

I though of using Ptolemy's first of all to get
$$ac + cx \ge de.$$
We can solve for $x$ to get
$$\frac{de-ac}{c} \le x.$$

But, that doesn't really help much, other than giving a bound. I can't really think of another good way to do this. I also thought of using the Law of Cosines somehow. Can someone help?


问:
Surjective holomorphic map from a domain to $\mathbb{C}$
Let $U\subset \mathbb{C}$ be a domain (i.e. open and connected). Is it always the case that there exists a holomorphic map $f:U\to \mathbb{C}$ such that $f(U)=\mathbb{C}$?

I managed to prove it for the upper half plane ($(z-i)^2$) (and so for every simply connected set by the Riemann mapping theorem)
and for $\mathbb{C}-\{z_0,\dots,z_n\}$ (just take $f(z)=\prod_{i=0}^{n+1} (z-z_i)$, where $z_{n+1}$ is a point in $\mathbb{C}-\{z_0,\dots,z_n\}$) but I do not seem to be able to prove it in general.

答:
Either there exists an open disc $\Delta\subseteq U$ which gives $\#(\partial\Delta\cap\partial U)\geq 2$ or there isn't.

If no such $\Delta$ exist, then $U$ misses at most one point and so $U$ contains a half-plane.

If such $\Delta$ exists, pick two points in $\partial\Delta\cap\partial U$ and send by Mobius to $0,\infty$ so you have a $U$ contains a half-plane.

So you have $U\twoheadrightarrow\text{something containing upper half-plane}\twoheadrightarrow\mathbb{C}$.


答: Hint. You can use coordinate geometry to find the endpoints of the third side, each of which is determined by the triangles whose sides are $a, b, d$ and $a, c, e$ respectively. The $x$ is the distance between the two points. Clearly you should be able to do this once you've chosen an origin, a unit and an orientation of axes, since the sides with lengths $b, d$ and $c, e$ respectively in each triangle determine intersecting lines, which give you the needed points.


问:
Riemann zeta function is analytic for $\operatorname{Re}(z)>1$
I want to show that, Riemann zeta function is analytic on the domain $\operatorname{Re}(z)>1$. I know that it is absolutely and uniformly convergent on the right half of the line $Re(z)=1$. Now, I am trying to invoke Morera's theorem to prove its analyticity, but I am not quite sure how  to do that. Can anyone help?

In general, if anyone can point me to some other direction then that is also welcome.

答:
From analyticity of $\sum_{n=1}^k \frac{1}{n^z}$, we get that
$
\int_{\mathfrak{D}}\sum_{n=1}^k \frac{1}{n^z}=0
$, from Goursat's Theorem. Now, from uniform convergence of the partial sums, $\sum_{n=1}^k \frac{1}{n^z}$, to $\zeta(z)$ in $Re(z)>1$, we get

$$
lim_{k\rightarrow\infty}\int_{\mathfrak{D}}\sum_{n=1}^k \frac{1}{n^z}=\int_{\mathfrak{D}}\zeta(z)=0
$$
Therefore, analytic.


问:
discrete random variable with cumulative distribution function
Consider a discrete random variable X with CDF (Cumulative
Distribution Function) () specified below:

I am just new to the course of statistics and wonder if only Pr(X=1), Pr(X=2), Pr(X=3) and Pr(X=5)are all the probabilities that can be drawn out from this cdf and other remaining probabilities like Pr(X=4)etc are equal to 0. Am I correct?

Based on my understanding, I have tried to compute the following, are they correct?
(a)For the value of Pr(2 <  ≤ 3) and Pr( = 3), I think that both Pr(2 <  ≤ 3) and Pr( = 3)are equal to find the value of Pr( = 3) and the value is 0.2.
(b)For the value of Pr( = 4)and Pr( = 6), both of the answers are 0.
(c)Pr(3 ≤  ≤ 5)= 1-0.6=0.4
(d)Pr(3 <  ≤ 6)=0.2
(d)Is the population mean  = E() equal to the sum of product of all possible values (which are X= 1, 2, 3, 5)with their corresponding probabilities?

Thanks for your precious time solving my problem!
Thank you!!!!

答:
You should begin working such a problem by plotting the CDF.
In this problem the CDF should look a lot like the empirical CDF (ECDF) of my sample (using R) simulated from your distribution. With a million observations, simulated
values should be accurate to two or three significant digits.

```<code>set.seed(220)
x = sample(c(1,2,3,5), 10^6, rep=T, p = c(.3,.3,.2,.2))
</code>```

Approximate probability distribution:

```<code>table(x)/10^6
x
       1        2        3        5
0.300511 0.298717 0.200321 0.200451
</code>```

Approximated values of $E(X), Var(X), SD(X):$

```<code>mean(x)
[1] 2.501163   # aprx E(X) = 2.5
sd(x)
[1] 1.433084   # aprx SD(X) = 1.432
</code>```

Approximate CDF:

```<code>plot(ecdf(x), lwd=2, col="blue")
 abline(h=0:1, col="green2")
</code>```

Using R as a calculator, here are exact values of
$E(X)=2.5,\, Var(X)=2.05,\, SD(X)=1.431782\,:$

```<code>k = c(1,2,3,5);  p = c(.3,.3,.2,.2)
mu = sum(k*p);  mu
[1] 2.5
vr = sum((k-mu)^2 * p); vr
[1] 2.05
sg = sqrt(vr);  sg
[1] 1.431782
</code>```

Body:
You should begin working such a problem by plotting the CDF.
In this problem the CDF should look a lot like the empirical CDF (ECDF) of my sample (using R) simulated from your distribution. With a million observations, simulated
values should be accurate to two or three significant digits.

set.seed(220)
x = sample(c(1,2,3,5), 10^6, rep=T, p = c(.3,.3,.2,.2))

Approximate probability distribution:

table(x)/10^6
x
       1        2        3        5
0.300511 0.298717 0.200321 0.200451

Approximated values of $E(X), Var(X), SD(X):$

mean(x)
[1] 2.501163   # aprx E(X) = 2.5
sd(x)
[1] 1.433084   # aprx SD(X) = 1.432

Approximate CDF:

plot(ecdf(x), lwd=2, col="blue")
 abline(h=0:1, col="green2")

Using R as a calculator, here are exact values of
$E(X)=2.5,\, Var(X)=2.05,\, SD(X)=1.431782\,:$

k = c(1,2,3,5);  p = c(.3,.3,.2,.2)
mu = sum(k*p);  mu
[1] 2.5
vr = sum((k-mu)^2 * p); vr
[1] 2.05
sg = sqrt(vr);  sg
[1] 1.431782


问:
Factoring a polynomial in $\mathbb Q[x]$.
I have a question regarding the factorisation of $f(x) = x^4 - 3x^2 + 2x + 1 \in \mathbb Q[x]$. Because of Theorem I, the only rational roots of $f(x)$ are $\pm 1$. By substituting them we see that they are no roots. Thus $f(x)$ has no linear factors.
So if $f(x)$ could be written as a product of irreducible factors, then they must have degree two. Because of Theorem II (here is where I don't understand why Theorem II applies) we could choose the factors such that they have integer coefficients. We suppose
\begin{align*}
 x^4 - 3x^2 + 2x + 1 & = (x^2 + ax + b)(x^2 + cx + d) \\
                     & = x^4 + (a+c)x^3 + (b+d+ac)x^2 + (bc + ad)x + bd.
\end{align*}
By comparing coefficients whe find
$$
  a + c = 0, \quad b + d + ac = -3, \quad bc + ad = 2, \quad bd = 1.
$$
These equations have no common integer solution, because of $bd = 1$ it must $b = d = \pm 1$ and then $b(a+c) = \pm 2$ in contradiction to $a+c=0$. Hence the polynomial $f(x)$ is irreducible in $\mathbb Z[x]$ and therefore also in $\mathbb Q[x]$.

Theorem I: Let $R$ be a UFD and let $K$ be his field of quotients. Let $p(x) = a_0 + a_1x + \ldots + a_n x^n \in R[x]$. If $\frac{r}{s} \in K$ with $r,s \in R, s \ne 0$, $r$ and $s$ coprime and $p(\frac{r}{s}) = 0$, then $r \mid a_0$ and $s \mid a_n$.

Theorem II: Let $R$ be a UFD and let $K$ be his field of quotients. Then

(i) For every $g(x) \in K[x] \setminus \{ 0 \}$ there exists $a \in K$ with $g_1(x) = ag(x) \in R[x]$ and $g_1(x)$ is primitive.

(ii) If $f(x), g(x) \in R[x]$ and $g(x)$ primitive, the following implication holds
$$
 f(x) = ag(x) \textrm{ with $a \in K$} \Rightarrow a \in R
$$

In the proof, where Theorem II is applied, I don't understand why we can suppose that the coefficients are integer, because when I look at Theorem II it must be the case that by changing the coefficients to an integer we multiply every coefficient (by the common denominator), but then a normed polynomial is no longer normed in general, but the polynomials in the factorisation are assumed to be normed?

答:
If $f=ab$, and $\mu_a$ resp. $\mu_b$ is the leading coefficient of $a$ resp. $b$, then $1=\mu_a \mu_b$. It follows $f=(\mu_a^{-1} a) (\mu_b^{-1} b)$ is a factorization into monic, in particular primitive polynomials. Does this answer your question?

See also Gauss's Lemma.


问:
Show there exists differentiable $g : (0, \infty) \to \mathbb{R}$ s.t $f(\vec{x}) = g(||\vec{x}||)$ for $f : \mathbb{R}^3 \to \mathbb{R}$

Let $f : \mathbb{R}^3 \setminus \left \{0 \right \} \to \mathbb{R}$ be a differentiable function s.t $\nabla f \neq 0$ and:

$y \frac{\partial f}{\partial x} -  x \frac{\partial f}{\partial y} =0 \\
z \frac{\partial f}{\partial y} -  y \frac{\partial f}{\partial z} =0$

Show there is a differentiable $g : (0, \infty) \to \mathbb{R}$ s.t $f(x,y,z) = g(||(x,y,z)||)$

I tried using spherical coordinates to show that the gradient is only determined by $r$ which I think should help, but I had some trouble with the algebra, help appreciated.

答:
Define $g(r,\theta,\phi) := f(r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta)$.
Our assumptions say
$$\frac{\partial f}{\partial x}r\sin\theta\sin\phi = \frac{\partial f}{\partial y}r\sin\theta\sin\phi$$
$$\frac{\partial f}{\partial y}r\cos\theta = \frac{\partial f}{\partial z}r\sin\theta\sin\phi$$
We have
$$\frac{\partial g}{\partial \phi} = -\frac{\partial f}{\partial x}r\sin\theta\sin\phi + \frac{\partial f}{\partial y}r\sin\theta\sin\phi = 0$$
so $g$ in fact does not depend on $\phi$ and we can fix $\phi = \frac\pi2$ so
$$g(r,\theta) = f(0, r\sin\theta, r\cos\theta).$$
We have
$$\frac{\partial g}{\partial \theta} = \frac{\partial f}{\partial y}r\cos\theta - \frac{\partial f}{\partial z}r\sin\theta = 0$$
so $g$ doesn't depend on $\theta$ either. Therefore $g(r,\theta,\phi) = g(r)$ and is differentiable.

Taking $(x,y,z)$ from a dense subset of $\Bbb{R}^3$ we can define $$(r,\theta,\phi) = \left(\sqrt{x^2+y^2+z^2}, \arctan\frac{\sqrt{x^2+y^2}}z, \arctan\frac{y}x\right)$$
and notice
$$f\left(x,y,z\right) = f(r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta) = g(r,\theta,\phi) = g(r)= g(\|(x,y,z)\|).$$
By continuity the result extends to all of $\Bbb{R}^3$.


问:
Random varible defined from another ($Y:=\left | X \right |-1$)
Given a PMF of $X$

$$p(x)=\left\{\begin{matrix}
0.1, & x=-4, \\
0.2 & x=-1\\
0.6 & x=1\\
0.1 & x=2\\
0 & else
\end{matrix}\right.$$
Find $\mathbb{P}(X \geq 0.8)$, $\mathbb{E}X$, var$(X)$ and sd$(X)$. Furthermore, find the mean and pmf for $Y:=\left | X \right |-1$.

I have found the firs 4 solutions $\mathbb{P}(X \geq 0.8)=0.7$, $\mathbb{E}X=1$, The varians to be $17/5$ and thus the sd to be roo square root of that answer.

Now for the last 2: Would this mean that the PMF's entries of $Y$ would just correspond to the aboslute value of the discrete r.v $X$ minus 1? Would you have to compute the mean from the start even though we just calculated the expected value?

答:
For the pmf question: basically yes. But you should notice that $X$ takes 1 and -1 which will be the same after taking absolute value. So $P(Y=0)=p(1)+p(-1)=0.8$.

You have to calculate the mean from start. Mean is a linear functional of random variables, so it couples pretty well with linear transformations. However, the absolute value is not linear, so you can not tell directly from mean of $X$ the mean of $|X|$. Actually this can be even not well-defined: see Cauchy distribution. Intuitively we should have $``\mathbb{E}X=0"$(and we do, in the sense of Cauchy principal value), but the integral $\mathbb{E}|X|$ diverges.


问:
test for convergence in this series $(-1)^n\frac{2^{2n+1}}{n!}$
I'm testing for convergence in this series:

$(-1)^n\frac{2^{2n+1}}{n!}$

I approached it with the ratio-test, although is there a better way to test for convergence in alternating series?

This is what I've done so far:

$$(-1)^{n+1}\frac{2^{2n+1+1}n!}{(-1)^n(n+1)!2^{2n+1}}$$

Then cancelling the terms:
$$-\frac{1}{n+1}\frac{1}{2}$$ The series converges to $-\frac{1}{2}$?

答:
No. That last sequence converges to zero, which means the convergence radius of the absolute value of the sequence's series is infinite...and thus the alternating series (what you have) converges. You also cancelled incorrectly:

$$\frac{2^{2(n+1)+1}}{(n+1)!}\cdot\frac{n!}{2^{2n+1}}=\frac4{n+1}\xrightarrow[n\to\infty]{}0\implies R=\infty$$

and if you want to know the series sum:

$$\sum_{n=0}^\infty(-1)^n\frac{2^{2n+1}}{n!}=2\sum_{n=0}^\infty\frac{(-4)^n}{n!}=2e^{-4}$$


答:
Since I prefer a more conceptual approach, note that you can deduce from the two equations that
$$\nabla f(x,y,z) = \lambda(x,y,z)(x,y,z)$$
for some function $\lambda$. Since the gradient of a differentiable function is everywhere orthogonal to the level surfaces of that function, this tells you that the function $f$ is constant on spheres centered at the origin. Thus, $f(x,y,z) = g(\|(x,y,z)\|)$ for some function $g$.

(If you prefer, you can see that the directional derivative is $0$ in any direction tangent to such a sphere, because that directional derivative is given by the dot product of the gradient and the tangent vector, and we all know that $(x,y,z)$ is the normal to the sphere centered at the origin passing through the point $(x,y,z)$.)

Title: Show there exists differentiable $g : (0, \infty) \to \mathbb{R}$ s.t $f(\vec{x}) = g(||\vec{x}||)$ for $f : \mathbb{R}^3 \to \mathbb{R}$

Body:
Let $f : \mathbb{R}^3 \setminus \left \{0 \right \} \to \mathbb{R}$ be a differentiable function s.t $\nabla f \neq 0$ and:

$y \frac{\partial f}{\partial x} -  x \frac{\partial f}{\partial y} =0 \\
z \frac{\partial f}{\partial y} -  y \frac{\partial f}{\partial z} =0$

Show there is a differentiable $g : (0, \infty) \to \mathbb{R}$ s.t $f(x,y,z) = g(||(x,y,z)||)$

I tried using spherical coordinates to show that the gradient is only determined by $r$ which I think should help, but I had some trouble with the algebra, help appreciated.