答:
$a_1=1;a_2=1;a_{n}=a_{n-1}+a_{n-2};\;n>2$

Induction

for $n=1$ it is true because $a_1=a_2$ by definition

now suppose it is true for $n$

$a_1+a_3+\ldots+a_{2n-1}=a_{2n}$

and let's prove it for $n+1$, that is

$\left(a_1+a_3+\ldots+a_{2n-1}\right)+a_{2n+1}=a_{2n}+a_{2n+1}=a_{2n+2}$

for the definition of the Fibonacci sequence

Hope this helps

Body:
Let $(a_n)$ be a Fibonacci sequence , Prove that $a_1+a_3+a_5+\cdots+a_{2n-1} = a_{2n}$

In this question $a_n$ is a Fibonacci sequence, I tried solving it with the PMI , problem arose when i was solving it for $P(n)=k+1$ , I could not make both of the sides equal.


问:
Graph Automorphism
I'm trying to solve this question:
Show $\pi \in S_n$ is an automorphism over $G = (V, E)$ with $V = \{1,...,n\}$ if there is an edge-set $E'$ such that:
$$E = \bigcup_{i=0}^{\binom{n}{2}} \{ \pi^i(u) \pi^i(v)\ | \ uv \in E' \}$$
But I'm having a bit of trouble understanding how I would go about doing it!

Thanks for any help.

答:
Just some partial thoughts.

We have usually three steps when proving that something is an Automorphism.

1. It is endomorphism So you ask: if $uv$ edge, is $\pi(u)\pi(v)$ edge?
In decomposition of $E$, situation $i=1$ gives you this condition for edges in $E'$. Can you then extend it for the whole $E$?

2. It is surjective Decomposition gives you that any edge can be written as $\pi^i(u)\pi^i(v)$ for some $i$ and $uv\in E'$.  From this follows that each edge has a pre-image and moreover we know that this pre-image is of form $\pi^{i-1}(u)\pi^{i-1}(v)$

3. It is injective Once you have proven this, your proof is complete. (Sorry for not completing the proof)

ContentLicense: CC BY-SA 4.0

Title: Graph Automorphism


问:
Double factorials Combination
We know that a factorial is consecutive product from 1 upto n and we can used factorials to extend them into combination of k items from n or permutation of k items from n.so my question is using same method we can factorials to double factorials where  the double factorial or semifactorial of a number n, denoted by n‼,[1] is the product of all the integers from 1 up to n that have the same parity (odd or even) as n.[2] That is,For example, 9‼ = 9 × 7 × 5 × 3 × 1 = 945. The zero double factorial 0‼ = 1 as an empty product,so my question is their some analog for combination and permutations for double factorial ,since factorials permutes single item from a collection whereas a double factorial permutes a pair of two objects for a collection of n things where n is assumed even.</p>

答:
There is a nice interpretation of $(2n-1)!!$ as the number of ways to partition $2n$ distinct objects into $n$ pairs, where the order of the pairs does not matter.

A separate interpretation for $(2n-1)!!$ is the number of Stirling permutations of order $n$, which are rearrangements of the list $1,1,2,2,\dots,n,n$ such that for each $k\in \{1,\dots,n\}$, the numbers between the two copies of $k$ are greater then $k$. These are in turn equivalent to rooted plane trees with vertices numbered $0$ to $n$ with the property that every vertex has a label less than its child. See https://en.wikipedia.org/wiki/Stirling_permutation. These are also related to the Eulerian numbers of the second kind.

Finally, whereas $n!$ counts permutations, $(2n)!!=2^nn!$ counts signed permutations. These are bijections $\pi$ from $[-n,n]\cap \mathbb Z$ to itself which satisfy $\pi(-i)=-\pi(i)$ for $i\in \{1,\dots,n\}$.

Title: Double factorials Combination


问:
Using the triangle inequalities to prove a bound in the complex plane
Let $G$ be the interior of the circle in the complex plane given by $|z-(1+i)|=1$. Prove that if $z\in G$ then $\sqrt{5}-1<|z-3|<\sqrt{5}+1$.

Attempt so far: We know by the triangle inequality that $|a+b|\leq |a|+|b|$ and also $|a-b|\geq||a|-|b||$. Now, $G$ is a circle centred at $(1+i)$ with radius $1$. Using the inequality above, we have $|z-3|\leq |z|+3$, and so $|z-3|\leq 4$. For the lower bound, $|z-3|\geq ||z|-3|=|\sqrt{2}-3|$. This isn't the required inequality though. Any ideas?

Body: Use $|z-3| \leqslant |z-(1+i)| + |(1+i)-3|$ instead.


问:
Why does $\overline{z \,\overline{w}}=2 \text{Re}(z \,\overline{w})$?

why does $\overline{z \,\overline{w}}=2 \text{Re}(z \,\overline{w})$?

答:
What is true is that for a complex number $z$ we have
$$
z+\overline z = 2\operatorname{Re}(z).
$$
So starting from the right hand side of what you are claiming we have
$$
2\operatorname{Re}(z\,\overline w) = z\,\overline w + \overline{z\,\overline w} = z\,\overline w + \overline z\, w \neq \overline{z\,\overline w}.
$$


答:
If $\bigl|z-(1+i)\bigr|<1$, then

\begin{align}|z-3|&=\bigl|z-(1+i)+(1+i)-3\bigr|\\&\leqslant\bigl|z-(1+i)\bigr|+|-2+i|\\&<1+\sqrt5\end{align}
and\begin{align}|z-3|&=\bigl|z-(1+i)+(1+i)-3\bigr|\\&\geqslant\bigl|\bigl|z-(1+i)\bigr|-|-2+i|\bigr|\\&=\bigl|\bigl|z-(1+i)\bigr|-\sqrt5\bigr|\\&>\sqrt5-1.\end{align}Besides, the bounds $\sqrt5+1$ and $\sqrt5-1$ are the best ones, as the picture below suggests.

Title: Using the triangle inequalities to prove a bound in the complex plane

Body:
Let $G$ be the interior of the circle in the complex plane given by $|z-(1+i)|=1$. Prove that if $z\in G$ then $\sqrt{5}-1<|z-3|<\sqrt{5}+1$.

Attempt so far: We know by the triangle inequality that $|a+b|\leq |a|+|b|$ and also $|a-b|\geq||a|-|b||$. Now, $G$ is a circle centred at $(1+i)$ with radius $1$. Using the inequality above, we have $|z-3|\leq |z|+3$, and so $|z-3|\leq 4$. For the lower bound, $|z-3|\geq ||z|-3|=|\sqrt{2}-3|$. This isn't the required inequality though. Any ideas?


问:
Estimate the propotion of Republicans in a certain district
Hoel's Probability Book Problem Chapter 3: If you wished to estimate the propotion of Republicans in a certain district and wanted your estimate to be correct withing .02 unit with a probability of .90, how large a sample should you take (a) if you know the true proportion is near0.4 (b) if you have no idead what the true propotion is?
my solution to (a)
Let $p=0.9$, $q=0.1$ $\sigma=3$, then
$$p+3\sqrt{\frac{pq}{n}}=0.03$$
$$0.9+3\sqrt{\frac{0.9\times0.1}{n}}=0.03$$
$$3\sqrt{\frac{0.9\times0.1}{n}}=-0.87=\sqrt{\frac{0.9\times0.1}{n}}=-0.29$$
$$\frac{0.9\times0.1}{n}=0.0841$$
n=1.0702

Can you please help me with (a) and (b)?

答:
You are confusing notation slightly, but in general you're on the right track.

If you measure a proportion $\hat{p}$ out of a sample of size $n$, the confidence interval  of the true proportion $p$ is given by
$$ \hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}},$$
where $z$ is the appropriate quantile of the standard normal distribution.

You should interpret the question as: what $n$ do you need in order for the confidence interval to have a total length of $0.02$?

First, observe that $z = 1.65$ in our case, because $\Pr(-1.65 <Z < 1.65) \approx 0.9$, which is the level of precision we desire. (You can verify this in a $Z\text{-score}$ lookup table.)

Furthermore, because the total length of the interval (both plus and minus) should be $0.02$, we need that
$$  z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 0.01.$$

Now if we know that $\hat{p}$ will be approximately $0.4$, we can substitute to obtain

$$1.65\sqrt{\frac{(0.4)(0.6)}{n}}=0.01,$$
which, if you solve for $n$, yields $n=6534$. (part a)

So you need a sample size of $6534$ in order for the confidence interval to be about $0.02$ wide, if you know that the real probability is somewhere around $0.4$.

However, if you have no clue about the real probability, then you should assume a probability of $0.5$ (this is called the distribution with maximum entropy). Then, you can subsitute again, to obtain
$$1.65\sqrt{\frac{(0.5)(0.5)}{n}}=0.01,$$
which yields $n = 6807$. (part b)


问:
Intersection of a straight line and a convex polytope
I apoligize in advance if the question might appear trivial, but there is something unclear to me related to convex polytopes in high dimensions. Let $P$ be a convex polytope in $\mathbb{R}^d$ for $d\ge 2$, and let $\mathbf{p}$ an interior point of $P$. Let $L$ be straight line through $\mathbf{p}$.

When $d=2$, $L$ intersects $P$ at one of the its edges (more than one edge only if the intersection point is a vertex of $P$).

When $d=3$, $L$ intersects $P$ at one of of its $2$-dimensional faces. Hence, the set of all edges (which are ($d-2$)-dimensional objects) of $P$ does not contain all possible intersection points over the choice of $L$ given any $\mathbf{p}$ inside $P$.

Question: When $d\ge 4$, it is always the case that a subset of polygons (which are $d'$-dimensional objects for $d'\le d-2$) having as vertices the ones of $P$ and containing no interior point of $P$, contains all possible intersection points of $L$?

答:
A ray from an interior point $p$ of a $d$-dimensional convex polytope   will meet the boundary of $P$ at exactly one point $q$.

With probability $1$ the point will $q$ be in the interior of one of the $d-1$-dimensional facets of $P$. But it might be on the boundary of some facet, in which case it will be on several facets.

Any point on the boundary of $P$ will be the $q$ for some ray since you can always join $p$ to $q$.

Think about the cube in space for your intuition. A ray from the center might leave the cube at the interior of a face, at the interior of an edge (in which case it will be on two of the faces) or at a vertex (in which case it will be on three of the faces and three of the edges).


问:
Slater's Condition for Dual Problem
I am reading the paper "A Semidefinite Optimization Approach to Quadratic Fractional Optimization with a Strictly  Constraints" by Maziar Salahi & Saeed Fallahi.

I this paper, they tried to prove that if the Slater condition holds for the primal SDP problem:
\begin{equation*}
\begin{aligned}
\min_{X} \mathrm{tr}(C^\top X)\\
\text{s.t.}~\mathrm{tr}(A^\top X) = 1\\
\mathrm{tr}(B^\top X) \preceq 0 \\
X \succeq 0
\end{aligned}
\end{equation*}
as well as its dual problem:
\begin{equation*}
\begin{aligned}
\max_{\lambda,\eta} \eta\\
\text{s.t.}~C^\top -\eta A^\top + \lambda B^\top = Z\\
\lambda\geq 0 \\
Z \succeq 0
\end{aligned}
\end{equation*}
then both problems attain their optimal values and the duality gap is zero.

Because trace is a linear function, in order to show the strong duality, I know one has to show the Slater condition holds for the primal problem.

My Question is: why should I also have to show the Slater condition holds for the dual problem?

答: If either the primal or the dual satisfies Slater's condition, strong duality holds. However, the problem for which Slater's condition holds can still be unbounded, so you cannot conclude that "both problems attain their optimal values". To show that the primal problem is bounded you could give a feasible point for the dual or vice versa. That is not the only way. Instead, you can conclude that the primal is bounded if $C$ is positive semidefinite.

Title: Slater's Condition for Dual Problem


问:
What is the purpose to define central moment?
What is the purpose to define central moment?
I searched the google and all i could find is bunch of properties</p>

答: Not sure I got your question, one of the 'purposes' of central moment is to set $n=2$ and get the variance of the rv, which is a measure of dispersion around the mean. Higher central moments (e.g. kurtosis) can also be used for this purpose. </p>

ContentLicense: CC BY-SA 3.0


问:
Stopping time w.r.t Brownian filtration is predictable
In "Credit Risk: Modeling, Valuation and Hedging" they say, that any stopping time $\tau$ w.r.t the Brownian filtration is predictable.

Since predictable means that there exists a sequence of $(\tau_n)_n$ announcing $\tau$ i.e.

$\tau_n<\tau$ on $\{\tau>0\}$

$\lim_{n\rightarrow\infty}\tau_n=\tau$

I didn't manage to find such an announcing stopping time only based on the fact that $\{\tau<t\}\in\sigma(B_u:u\leq t)$ for all $t$.

Furthermore they say, that e.g. when $\tau$ is a stopping time w.r.t the Brownian Filtration $\mathbb{F}$, that it's not predictable w.r.t a strictly larger filtration containing $\mathbb{F}$. I don't get how one couldn't just use the announcing stopping time which has been found for $\mathbb{F}$.

答:
The predictability of Brownian stopping times follows from the fact that all (local) martingales in the filtration of a Brownian motion have continuous sample paths — a consequence of the Martingale Representation Theorem. (A nice discussion of such things can be found here: https://almostsuremath.com/2011/05/26/predictable-stopping-times-2/.)

A different way to see this is through Markov process theory: Brownian motion is a path continuous strong Markov process; as such all of its stopping times are predictable.

Your suspicion of the assertion made by "CR:M,VaH", reported in your last paragraph, is well-founded: any $\Bbb F$ predictable stopping time is $\Bbb G$ predictable, for the reason  you state.


问:
Why does $\frac{2P}{P}$ give $2$ and not $P$?
The formula for annual compounded interest (according to our lecturer) is  $$2P = P(1 + r)^n.$$
($P$: initial investment, $r$: interest rate, $n$: years).

To solve for $n$, he said the "$P$"s cancel out and you can then transform the equation to
$$n =\frac{\ln(2)}{\ln(1 + r)}.$$

But why is that? If there were an $x$ instead of $P$ (so $2x=x(1+r)^n$), you'd reduce it by dividing through by $x$ (so $2x/x$) which would leave $1x$. So, why is this not happening with $P$, and instead we keep the number $2$ but omit $P$?

答: Careful, remember that $2P$, or $2x$ for that matter, is shorthand for $2\times P$ or $2\times x$.  So, when you write $$\frac{2x}{x}$$ what you really mean is $$\frac{2\times x}{x}$$ and so the division by $x$ undoes the multiplication by $x$ and you are left with $$\frac{2x}{x} = 2.$$


问:
Obtain rate of change of an exponential without using numerical analysis
I though i was very good at math till i met this equation

$$Y=X\times (1.01)^X$$

If $X= [1,2,3,...]$ then $Y=[1.01, 2.0402, 3.0909,...]$

I expected that the rate of change should be $dY=[0.01,0.0402,0.0909,...]$ but when i differentiated it, the answer was extremly different. this can be clarified more without numbers.

The mathematical derivative of the above image is this
But the actual change is this.

I do not understand "what i do not understand".

Why is the rate of change (derivative...) does not give me the same as what i have done above (very similar to limits).

What kind of math branch do I need to study or see or understand to conceptualize the very big gap between what i understand and the reality.

Edit 1:

This is related to something called natural growth (e.g., half life time equation)

Edit 2:

The Equation was obtained from this sentence: write an equation that describes accumulative interest of 1%. In another words, AccInterest $= ((X(1.01)+X)(1.01)) +X)(1.01)...$

答:
The equation in the picture,
$$
Y = X \times 1.01 ^X
$$
is not the equation for constant growth at the rate of $1\%$ for each time interval. That equation is
$$
Y = s \times 1.01^X
$$
where $Y$ represents the amount at time $X$ and $s$ is the amount you started with at time $X=0$.

To find the half life you let $Y=2s$ and try to find $X$. The answer will be about $X=70$. To find it more precisely you need to experiment with various values or learn about logarithms.


问:
Why has the kernel of the 2x2 hessian 3 linear functions?
I have calculated something characteristic polynome and such things but i think thats the wrong way to see this. My try is :

On $\mathbb{R}^2$ the Hessian matrix can be relaized as
$$
\mathbf{H}_\mathbb{R^2}=
\begin{pmatrix}
\partial_{xx} & \partial_{xy}  \\
\partial_{yx} & \partial_{yy}
\end{pmatrix}
\Leftrightarrow
\begin{pmatrix}
\partial_{xx} & \partial_{xy}  \\
\partial_{xy} & \partial_{yy}
\end{pmatrix}
$$
The kernel, or null space, of a matrix is defined as all vectors that are mapped to the zero vector and the kernel is not trivial if the kernel not only contains the zero element. For the Hessian matrix  then the kernel is the set of functions that have all second partial derivatives mapped to zero. The only three functions for that this holds are $u(x,y) = x$, $u(x,y) = y$ and $u(x,y) = x + y$, so $\dim(\ker(\mathbf{H}_u)) = 3$.

Is this correct ?

答:
No, because the third function is a linear combination of the first and the second. And they are not the only ones. All their linear combinations belong to kernel.

The dimension is $3$, by the way. The kernel is the set
$$
\ker \mathbf{H} = \{\lambda x + \mu y + \varepsilon 1 : \lambda, \mu, \varepsilon \in \mathbb{R}\}.
$$


Title: Why does $\frac{2P}{P}$ give $2$ and not $P$?

Body:
The formula for annual compounded interest (according to our lecturer) is  $$2P = P(1 + r)^n.$$
($P$: initial investment, $r$: interest rate, $n$: years).

To solve for $n$, he said the "$P$"s cancel out and you can then transform the equation to
$$n =\frac{\ln(2)}{\ln(1 + r)}.$$

But why is that? If there were an $x$ instead of $P$ (so $2x=x(1+r)^n$), you'd reduce it by dividing through by $x$ (so $2x/x$) which would leave $1x$. So, why is this not happening with $P$, and instead we keep the number $2$ but omit $P$?

Body: $$\frac{2P}{P} = \frac{P + P}{P} = \frac{P}{P} + \frac{P}{P} = 1 + 1 = 2$$


答:
if you start with $P$ capital after $n$ years you have $P(n)=P(1+r)^n$. The question here is thus when you've doubled your starting capital, so when $P(n)=2P$, where $2P$ is short for $2 \times P$ of course. You get the equation (with $\times$) written out

$$2\times P = P \times (1+r)^n$$

The two $P$'s cancel out, just as when you solve $2x=6$ to conclude $x=3$.
so you're left with $2 = (1+r)^n$ which reduces to $\ln(2) = n\ln(1+r)$ and so $n= \frac{\ln(2)}{\ln(1+r)}$ etc.


问:
Show that if $K \leq H \leq G$ and $K \triangleleft G$ then $K \triangleleft H$
Show that if $K \leq H \leq G$ and $K \triangleleft G$ then $K \triangleleft H$

My try:

Let $a\in H$ then $a\in G $ because $H\leq G$

And how $K \triangleleft G$ then $aKa^{-1} \leq K \ \ \ \forall  \ a \in H$ (because $a$ is in $G$ too) then $K \triangleleft H$ is right?

答:
Your proof is fine.

It even works for equality, so you can say, if $K\le H\le G$ and $K\unlhd G$, then

$$K\unlhd H.$$


问:
$\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot ... \cdot \frac{995}{996} \cdot \frac{998}{999}$ and $0.1$ which is bigger?
$\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot ... \cdot  \frac{995}{996}  \cdot  \frac{998}{999}$ and $0.1$ which is bigger?

The expression seems a bit different from Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$

答:
Consider three products:

$$
P_1 = {1\over2}\cdot{4\over5}\cdot...\cdot{997\over998}\\
P_2 = {2\over3}\cdot{5\over6}\cdot...\cdot{998\over999}\\
P_3 = {3\over4}\cdot{6\over7}\cdot...\cdot{999\over1000}\\
$$
Obviously, $P_1<P_2<P_3$, because they all have equally many terms and the corresponding terms are all ordered the same way. Also, because of telescoping, $P_1P_2P_3={1\over1000}$, which means that $P_1<0.1$ and $P_3>0.1$. Pity it doesn't tell us much about $P_2$, which is what we need.

We'll have to make a better estimate. What about the geometric mean of the corresponding terms of $P_1$ and $P_3$? That would be
$$\sqrt{{3n-2\over3n-1}\cdot{3n\over3n+1}} = \sqrt{(3n-1)^2-1\over(3n)^2-1} = \sqrt{{(3n-1)^2\over(3n)^2}\cdot{1-{1\over(3n-1)^2}\over1-{1\over(3n)^2}}}<{3n-1\over3n}$$

That's it; $P_2>\sqrt{P_1P_3}$, hence $P_2>\sqrt[3]{P_1P_2P_3}=0.1$


问:
Arrangement without any monotone section

Here's the question:
"In how many ways can the sequence 45678 be arranged such that there aren't any three consecutive terms increasing or decreasing"

The first thing I thought of is to do complementary counting and then make cases for each number of consecutive terms that there could be (3, 4, 5).

However there's a lot of overlap between the three cases and I'm not sure how to deal with that in an efficient way. If someone could solve it and then tell me how they did it that would be great.

答:
We have 5 digits, so after filling the first digit, we have to decide 4 times if we want to increase ($i$) the next digit or decrease($d$).

We have to take $i$ and $d$ alternatively, because if two $i$ or two $d$ are together, then there will be 3 consecutive terms increasing or decreasing.

For that, there can only be 2 cases ($idid$) and ($didi$), in these two cases, we have only interchanged $i$ and $d$, so they will have the same number of cases.

For $idid$:

First digit 7, $78564$, $78465$.

First digit 6, $68475$, $68574$, $67485$, $67584$.

First digit 5, $58476$, $58674$, $57486$, $57684$, $56487$

First digit 4, $46587$, $47586$, $47685$, $48576$, $48675$

Number of ways: 16

$$\boxed{\text{FINAL RESULT}: \underbrace{16}_\text{idid} + \underbrace{16}_\text{didi} = 32}$$


问:
Simplifying matrix equation $A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}$
I want to simplify the inverse of a $2\times 2$ block matrix:
$$
\begin{bmatrix}
A_{n\times n} & B^T_{n\times m}\\
B_{m\times n} & 0_{m\times m}
\end{bmatrix}
$$
$A_{n\times n}$ is square and invertible (and possibly positive definite), and $B_{m\times n} (m < n) $ is not square but full row rank. Using this eq. (2.2), the inverse can be written as:
$$
\begin{bmatrix}
A & B^T\\
B & 0
\end{bmatrix} ^{-1}=
\begin{bmatrix}
A^{-1}-A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}  &  A^{-1}B^T(BA^{-1}B^T)^{-1} \\
(BA^{-1}B^T)^{-1}BA^{-1} & -(BA^{-1}B^T)^{-1}
\end{bmatrix}
$$
I want to see is there a way to simplify this matrix? For example, can we simplify this term $A^{-1}B^T(BA^{-1}B^T)^{-1}BA^{-1}$, or other terms?

答:
I believe that there is no way to significantly simplify it. You may rewrite it in different ways, but (as far as I see) nothing will cancel out completely.

One way of rewriting. Denote $ C = BA^{-1/2},\; D = A^{-1/2}$. Then the inverse is equal to
$$
\begin{bmatrix}
D(I-C^+C)D & DC^+ \\
C^{\rm T+}D & -(CC^{\rm T})^{-1}
\end{bmatrix}.
$$

Here $^+$ denotes the Moore-Penrose inverse.

Another way of rewriting. You can use the formula
$$
(BA^{-1}B^{\rm T})^{-1} = B^{\rm T+} A^{1/2} XA^{1/2}B^+,
$$
where
$$
X = I-A^{1/2}(I-B^{+}B) \big( A^{1/2}(I-B^{+}B)\big)^+.
$$
This formula opens $(BAB^{\rm T})^{-1}$ and takes some stuff out. It may be useful sometimes, but hardly simplifies something in your case.

Using limits. You may use Woodbury matrix identity to write
$$
\begin{align}
A^{-1}-A^{-1}B^{\rm T}(BA^{-1}B^{\rm T})^{-1}BA^{-1} & = \lim_{\varepsilon \to +0} \Big( A^{-1}-A^{-1}B^{\rm T}(\varepsilon I + BA^{-1}B^{\rm T})^{-1}BA^{-1} \Big) \\
& = \lim_{\varepsilon \to +0} \Big(A + \frac{1}{\varepsilon}B^{\rm T}B\Big)^{-1}.
\end{align}
$$

This looks nice, but again is not a simplification.


问:
Electrostatic Potential Energy
How is the boxed step , physically as well as mathematically justified and correct ?

Source:Wiki http://en.wikipedia.org/wiki/Electric_potential_energy

As work done = $- \Delta U $. for Conservative force and it shouldn't matter whether we take $ds$ or $-dr$ ?

And when $dr$ is just a notation to specify the variable and the real thing behind it , is a limit , why is it that $dr$ is so important here .

Image : http://www.artofproblemsolving.com/Forum/download/file.php?id=43358&mode=view

答:
Basically you $dr$ is negative , so $|dr|=-dr.$ like $|-2|=-(-2)$ , evaluate using this , the sign ambiguity will dissolve .

Also displacements are measured w.r.t. some origin and here , origin can't be placed at $\infty$. So you base your origin at the point where the charge '$Q$' is located .

Title: Electrostatic Potential Energy


问:
Why are finite groups linearly reductive?
Let $G$ be a linear algebraic group contained in $GL(n)$. $G$ is linearly reductive iff every regular representation is completely reducible. Among the examples of linearly reductive groups, there are finite groups: why is that?

答:
This is because for any such group $G$ and its (finite dimensional) representation $V$ we can define on $V$ a hermitian $G$-invariant inner product. Using this product we can easily check that an orthogonal compliment to any $G$-invariant subspace is $G$-invariant itself. Of course everything is in characteristic $c \neq \#G$.

To construct such inner product one can start with an arbitrary hermitian inner product $(\cdot,\cdot)$ and define
$$
\langle x \, | \, y\rangle = \sum_{g \in G}(gx, gy).
$$


问:
Solving $\tan(\frac\pi4 \left(4x-5\right))=-1$. Rewriting $-1$ as $\tan\frac\pi4$ vs $\tan\frac{3\pi}{4}$ gives different answers.
I'm solving this equation:
$$\tan \left(\frac{\pi \left(4x-5\right)}{4}\right)=-1$$

When I rewrite $-1$ as $-\tan\left(\frac{\pi }{4}\right)$, I get $x=1+n$,
but when I change $-1$ for $\tan \left(\frac{3\pi }{4}\right)$, I get $x=2+n$.

What is it I do wrong?

答: If $n$ are any integers then $x=1+n$ and $x=2+n$ define the same solution sets, just written differently.  You could even say $x = n$ and that would still define the same solution set.


问:
Are these two logical negation of each other
Are these two statements considered negation of each other?:

All  fruits are sweet.

Some fruits are not sweet.

I guess there is no relation between the two. I guess we can only say that statement 1 renders statement 2 "invalid" and vice versa. But nothing else can be said (in terms of "negation" and any other), am I right ?

答:
If you formalise these two statements in first order logic, they read something like:

$\forall f \cdot F(f) \to S(f)$

$\exists f \cdot F(f) \land \lnot S(f)$

where $F(f)$ asserts that $f$ is a fruit and $S(f)$ asserts that $f$ is sweet. The logical negation of 1 is $\lnot\forall f \cdot F(f) \to S(f)$ which (in classical logic) is logically equivalent to 2. Likewise the logical negation of 2 is equivalent to 1.


问:
Solving XAY = B for both X and Y
I have a problem where A and B are known in the above expression, and I'd like to solve for X and Y concurrently.

Dimensions:

X is a diagonal matrix (N x N)

A is square (N x N), and sparse if that helps

Y is a vector (N x 1)

B is a vector (N x 1)

For reference, this follows on from my question from yesterday (Solving a linear system of the form C .* (BX) = AX) which was solved, but unfortunately requires wading through "null space". I was not able to find the desired solution robustly using standard optimization approaches, which has led me to reformulate the problem as above (giving me, unfortunately, two unknowns to solve for).

Many thanks (again) in advance...

答:
Solution.
$$\begin{split}XAY&=B\\
\left(\begin{matrix}\frac 1 {x_1} & 0&... & 0\\0&\frac 1 {x_2}&...&0\\...&...&...&0\\0&0&0&\frac 1 {x_N}\end{matrix}\right)\left(\begin{matrix}A_1^T\\...\\A_N^T\end{matrix}\right)\left(\begin{matrix}Y_1\\...\\Y_N\end{matrix}\right)&=\left(\begin{matrix}B_1\\...\\B_N\end{matrix}\right)\text { write the matrix}\\
\left(\begin{matrix}A_1^T\\...\\A_N^T\end{matrix}\right)\left(\begin{matrix}Y_1\\...\\Y_N\end{matrix}\right)&=\left(\begin{matrix}x_1 & 0&... & 0\\0&x_2&...&0\\...&...&...&0\\0&0&0&x_N\end{matrix}\right)\left(\begin{matrix}B_1\\...\\B_N\end{matrix}\right)\text { left multiply by $X^{-1}$}\\
&=\left(\begin{matrix}B_1x_1\\...\\B_Nx_N\end{matrix}\right)\text { property of diagonal matrices}\\
\left(\begin{matrix}A_1^TY-B_1x_1+0x_2+...+0x_N\\...\\A_N^TY+0x_1+0x_2+...-B_Nx_N\end{matrix}\right)&=0\text { expand left hand side and subtract the right term from both sides}\\
\left(\begin{matrix}A_1^T&-B_1&0&...&0 \\...\\A_N^T&0&0&...&-B_N\end{matrix}\right)\left(\begin{matrix}Y_1\\...\\Y_N\\X_1\\...\\X_N\end{matrix}\right)&=\text { write in matrix form}\\
\left(\begin{matrix}A&-diag(B)\end{matrix}\right)\left(\begin{matrix}Y\\X\end{matrix}\right)&=0\text { write in block matrix form}\end{split}$$

I'm sorry, it seems like this solution requires null space too.

Title: Solving XAY = B for both X and Y


问:
Different complex structures of a torus.
Consider a complex $\mathbb C/\Lambda$, once we choose a lattice $\Lambda$, then the torus is uniquely determined.

For example, if we choose different lattices: $\Lambda_1=\mathbb Z\oplus i\mathbb Z$, $\Lambda_2=\mathbb Z\oplus 2i\mathbb Z$, and let $T_1=\mathbb C/\Lambda_1$, $T_2=\mathbb C/\Lambda_2$, then do $T_1$ and $T_2$ have different complex structures? I know they have the same diffeomorphism sturcture, but why they have different complex structures? Does there exist an intuitive explanation why they have different complex structures?

答:
Here is a general procedure used to determine if two tori are biholomorphic to each other. Consider a lattice $\Lambda$ in ${\mathbb C}$ generated by two complex numbers $\alpha, \beta$. I assume that the basis $(\alpha, \beta)$ in ${\mathbb R}^2$ is positively oriented, otherwise, swap $\alpha$ and $\beta$. Consider the ratio $\tau=\beta/\alpha$ (the Teichmuller parameter). Positive orientation implies that $\tau$ is in the upper half-plane $U$, $Im(\tau)>0$. Then, find an element $\gamma\in PSL(2, {\mathbb Z})$ sending $\tau$ to the modular fundamental domain
$$
F=\{z: - 1/2 \le Re(z)\le 1/2, |z|\ge 1\}.
$$

If $z=\gamma(\tau)$ is in the interior of $F$, then it is uniquely determined by $\tau$, otherwise, it is unique up to the action of the translation $z\mapsto z\pm 1$ (if $Re(z)=\pm 1/2$) or involution $z\mapsto -1/z$ (if $|z|=1$).

Up to this ambiguity, $z$ uniquely determines the conformal class of the quotient torus ${\mathbb C}/\Lambda$.

In the case of a "rectangular" torus, things are especially simple since $Re(\tau)=0$ and either $\tau\in F$ or $-1/\tau\in F$.

Now, to your specific example (from your comment): $\Lambda_2$ is generated by $\alpha=2$ and $\beta=i$. In this case, $\tau=i/2$, $-1/\tau=2i\in F$. Hence, the torus ${\mathbb C}/\Lambda_2$ is not conformal to the torus ${\mathbb C}/\Lambda_1$.

Regarding references: This Wikipedia article provides a nice proof of the fact that the moduli space of complex one-dimensional tori (aka nonsingular complex elliptic curves) is the quotient of the upper half-plane $U$ by  $\Gamma=PSL(2, {\mathbb Z})$. Concretely, this means that if $\Lambda_1, \Lambda_2$ are two lattices generated, respectively, by $\alpha_1, \beta_1$ and $\alpha_2, \beta_2$ with $\tau_k=\beta_k/\alpha_k, k=1, 2$, then the tori ${\mathbb C}/\Lambda_k$ are biholomorpbhic if and only if
$$
\Gamma \tau_1 =\Gamma \tau_2.
$$
The fact that this equality of two orbits is equivalent to
$$
\Gamma \tau_1 \cap F= \Gamma \tau_2 \cap F,
$$
i.e. that $F$ is a fundamental domain of $\Gamma$, is explained in many places. One reference is given by the linked Wikipedia article, Chapter VII in

Serre, Jean-Pierre, A course in arithmetic, Graduate Texts in Mathematics. 7. New York-Heidelberg-Berlin: Springer-Verlag. viii, 115 p. (1973). ZBL0256.12001.

Another is

Katok, Svetlana, Fuchsian groups, Chicago Lectures in Mathematics. Chicago: The University of Chicago Press,. x, 175 p. (1992). ZBL0753.30001.

Or, freely available Pete Clark's lecture notes on Shimura curves here.

Lastly, a true story: A math professor is flying in an airplane, reading Serre's book "A course in arithmetic." His next-seat neighbor notices the title and approvingly says "Very good, it is never too late to learn!"


问:
Is the fundamental lemma of Calculus of Variations wrong?
the title might be a bit absurd but this is how I currently feel just because I can't see where my reasoning is faulty.

So the lecturer presented the lemma as the following:

Let $M \in C[a, b]$. Then $\int_{a}^{b} M \cdot h \,dx = 0, \forall h \in C_0^1[a,b] \implies M \equiv 0$.

But how does the lemma still hold when $h(x) = 0$? The integral would still be zero but it would not imply that $M = 0$ since it could for example be $M = x$. Am I wrong to believe that $0 \in C_0^1[a,b]$?

He gives a proof by contradiction using a non-zero $h$ (the bump function ) somewhere in the lines of: Say $M$ is greater than zero and $h$ is greater than zero, then the integral must also be greater than zero which results in a contradiction since the integral must be equal to zero. Therefore $M$ must be zero. This is obviously true for the case when $h$ is non-zero but isn't this an obvious loss of generality? When $h$ is also allowed to be zero then one can not easily deduce which function is making the integral to be zero. If the lemma would require that it applies to $\forall h \in \{C_0^1[a,b] \setminus 0\}$ then everything would make sense but that is not how the lemma is generally expressed.

Any help would be appreciated.

答: The question “how does the lemma still hold when $h(x)=0$?” makes no sense, since the hypothesis is that the equality $\int_a^bM\times h=0$ for every compactly supported smooth function.


问:
If $K \leq G $ and $H\triangleleft G$ then $HK$ is a subgroup of $G$
If $K \leq G $ and $H\triangleleft G$ then $HK=\{hk : h \in H, k\in K\}$ is a subgroup of $G$

My try:

How $e \in H$ and $e \in K$ then $ee=e \in HK$

and let $h_1k_1, h_2k_2 \in HK$ then $h_1k_1h_2k_2=h_1h_3k_1k_2$ because $H$ is normal (that implies that $ha=ah'$ for all $a\in G$)

and how $h_1 $ and $h_3$ are in $H$ then $h_1h_3\in H$ and the same for $k_1$ and $k_2$ so $h_1k_1h_2k_2=h_1h_3k_1k_2 \in HK$

and $(hk)^{-1}=k^{-1}h^{-1}=h'k^{-1}$ again because $H$ is normal so $(hk)^{-1} \in HK$
so $HK \leq G$ is right?

答: You have used the two-step subgroup test. It is required that you show that $HK\subseteq G$. This is easy though: since $HK=\{hk\color{red}{\in G}\mid h\in H, k\in K\}$, we have $HK\subseteq G$.


答:
We notice if we rotate the digits of a number we get five different numbers. There is always $2$ good numbers (no three consecutive inc/dec terms) or $0$ good number among the five numbers. If there is only one inc/dec sequence in the "circular configuration" then there are two good numbers. If there are more than one such sequence then there is no good number.

It's easy to see there are $24$ such "rotational classes."

Among these five numbers of the same class there is always a number where the middle number is $6$. We notice that if the numbers to the left of $6$ and to the right of $6$ are both less than or greater than $6$. Then the number itself is good so it must be the "$2$ out of $5$" case. There are $2\times 2\times2=8$ such classes, calculated by grouping $(4,5)$ and $(7,8)$ together, switching each group's number order and switching the groups.

Out of the remaining $16$ classes, we again look at the number where $6$ is in the middle. Notice that we can have a bijection by switching the numbers before and after the middle $6$. If the original class is a "$2$ out of $5$" class then the class after switching is always "$0$ out of $5$", and vice versa. This comes from the fact that if the number is $ab6cd$ then the class is good only when $ad$ and $bc$ has reverse order so switching $b,c$ will turn a good class into bad and a bad class into good.

This means exactly $8$ classes out of the $16$ classes have "$2$ out of $5$" numbers being good.

This means there are totally $16$ good classes and $16\times2=32$ good numbers.

Title: Arrangement without any monotone section

Body:
Here's the question:
"In how many ways can the sequence 45678 be arranged such that there aren't any three consecutive terms increasing or decreasing"

The first thing I thought of is to do complementary counting and then make cases for each number of consecutive terms that there could be (3, 4, 5).

However there's a lot of overlap between the three cases and I'm not sure how to deal with that in an efficient way. If someone could solve it and then tell me how they did it that would be great.


问:
Finding length of arc intercepted by two tangent lines given radius and distance from circle
I'm in the process of writing a script for work but realized that I need to calculate the length of a circular arc intersected by two tangent lines, and when I try finding it I get lost in a mess of trig that seems to go nowhere.

Here's what I have so far:

Tangent lines intersect each other at point $P$

Tangent lines intersect circle at points $A,B$

Circle radius is $R$

Point on circle nearest to $P$ is $Q$

Length of $\vec{PQ}$ is $T$

Using some trig, I found that $\vec{PA}=\vec{PB} = R\arccos(\frac{R}{R+T})$

But that's where I'm stuck. How can I find $\widehat{AQB}$ ?

答:
From $ \Delta AOP$ subtended angle is directly (since $OP$ is hypotenuse)

$$ 2 \cos^{-1}\dfrac{R}{R+T}$$

$R$ times the above is the length of the arc subtended
by this angle.


问:
Help why counting but not conditional probability only gets the correct answer.
Let me setup my question with two very simple examples.

Example 1.

A group of 5 students, made up of 2 females and 3 males, are sat at random in a row. What is the probability the first two seats are occupied by females?

Solution by counting:

2* 3!/5!= 1/10

Solution via conditional probability:

2/5 * 1/4 = 1/10.

They agree.

Now let’s try the another problem:

Example 2.

Letters a,b,c,d,e,f are to be place in 3 distinct groups of two.  The assignments are made at random. What is the probability a,b are both in group 1.

Solution via counting:

4c2 / (6c2 *4c2)= 1/6c2= 1/15

Can’t be solved via conditional probabilities as follows:

P(a in group 1) p( b in group 1| a in group 1) =? 1/3 *1/3

Not the same answer.

Question:

Why does conditional probability work in example one and not( as how I have applied it) not work in problem 2.

I gather it’s how random is being a implied.

in example 2, random means assume all permutations are equally likely. Sometimes, but not all the time, that exact language is written.  So, if I apply conditional probability, as I have attempted, it fails because that would require each letter have equal probability for assignment and that is assuming too much.

Now, assume the same language, all permutations are equally likely, to example 1. Why does conditional probably, as I applied now work, because it seems It still assumes singleton assignments are equally likely, which is not assumed?

答: Your conditional probability in the second case is wrong. $a$ indeed has probability $\frac 13$ to end up in the first group, but with one of the two slots filled, the three groups are not equally probable for $b$, with only one of the five remaining available slots in the first group.


Body:
My simple Python script enumerates them as

(4, 6, 5, 8, 7)

(4, 7, 5, 8, 6)

(4, 7, 6, 8, 5)

(4, 8, 5, 7, 6)

(4, 8, 6, 7, 5)

(5, 4, 7, 6, 8)

(5, 4, 8, 6, 7)

(5, 6, 4, 8, 7)

(5, 7, 4, 8, 6)

(5, 7, 6, 8, 4)

(5, 8, 4, 7, 6)

(5, 8, 6, 7, 4)

(6, 4, 7, 5, 8)

(6, 4, 8, 5, 7)

(6, 5, 7, 4, 8)

(6, 5, 8, 4, 7)

(6, 7, 4, 8, 5)

(6, 7, 5, 8, 4)

(6, 8, 4, 7, 5)

(6, 8, 5, 7, 4)

(7, 4, 6, 5, 8)

(7, 4, 8, 5, 6)

(7, 5, 6, 4, 8)

(7, 5, 8, 4, 6)

(7, 6, 8, 4, 5)

(7, 8, 4, 6, 5)

(7, 8, 5, 6, 4)

(8, 4, 6, 5, 7)

(8, 4, 7, 5, 6)

(8, 5, 6, 4, 7)

(8, 5, 7, 4, 6)

(8, 6, 7, 4, 5)

so there are $32$


问:
Fundamental system of solutions for SLAE

Given a SLAE $\left\{\begin{matrix}
5x_{1}+x_{2}-2x_{3}+6x_{4}=0\\
x_{1}+3x_{2}-2x_{3}+4x_{4}=0\\
3x_{1}+2x_{2}-2x_{3}+5x_{4}=0\\
4x_{1}+5x_{2}-4x_{3}+9x_{4}=0
\end{matrix}\right.$

Check if the rows of matrices A and B given below form a fundamental solution of SLAE.

$A=\begin{pmatrix}
4 & 8 & 14 & 0\\
2 & 4 & 7 & 0
\end{pmatrix}$ $\hspace{0.5mm}$ $B=\begin{pmatrix}
-1 & -1 & 0 & 1\\
4 & 8 & 14 & 0
\end{pmatrix}$

I found the row reduced echelon form of the coefficient matrix for the homogeneous SLAE as the following:

$\begin{pmatrix}
1 & 0 & \frac{-2}{7} & 1\\
0 & 1 & \frac{-4}{7} & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{pmatrix}$

But am confused on how to relate it with the matrices A and B?

Can you please let me know the correct approach for the same?

答: Since the rang of your system is $2$ and the dimension of the space of variables is $2$, solutions of the system form a $4-2 = 2$-dimensional subspace. Fundamental solution is a basis of this subspace. So you should check whether rows of $A$ and $B$ provide solutions to the system or not. Then, in the first case, you should prove or disprove that they are linearly independent. If they actually are, they form a fundamental system by the dimensionality argument.


问:
Prove that $U_{n+1} + 1 \geqslant \tfrac{3}{4}(U_{n}+1)$
$U_{n}$ is a sequence where $n$ is natural.

$U_{n+1} = \tfrac{U_{n}^{2}+1}{U_{n}-1}$

$-3 \leqslant U_{n} < -1$

And the question is: Prove that $U_{n+1} + 1 \geqslant \tfrac{3}{4}(U_{n}+1)$

My attempt was:

$U_{n+1} + 1 = \tfrac{U_{n}}{U_{n}-1}(U_{n}+1)$

And $\tfrac{-1}{2} < \tfrac{1}{U_{n}-1} \leqslant \tfrac{-1}{4}$ ... (1)

And $-3 \leqslant U_{n} < -1$ ... (2)

And I wanted to multiply (1) by (2) but the signs are diffrent so I can't do that.

So how I solve it?

Body:
Since $-3 \le U_n < -1$, we have $U_n-1 <0$, then
$$U_{n+1} + 1 - \tfrac{3}{4}(U_{n}+1) = \frac{U_n^2+1}{U_n-1}+1 - \tfrac{3}{4}(U_{n}+1)\\
=\frac{(U_n+1)(U_n+3)}{4(U_n-1)} \ge 0.\blacksquare$$


问:
Plot Line that gets exponentially larger around a point
I'm building a program that calculates the cost of an item based on it's size (let's say a bamboo pole). As the customer requests a longer pole, it gets hard to find a bamboo, plus requires more resources to grow, therefore, I would want to charge more per inch for the piece of bamboo based on it's length approaching a particular length. Then after that point the cost per inch would really escalate.

I believe that log would be the function that I need to use, but I just can't figure out how to make my formula. I've tried log(-x), log(x-1), log(y-x), I can't figure out how to get the log to shoot up to infinity, nor target a specific point.

Referencing the example above, I would want the cost/inch of the bamboo to stay reasonable up to 72", but after that, the cost/inch should rapidly increase, until it gets to 100", where it would become ridiculously expensive. Before 72", it should rise in cost/inch, but at a slow rate (it costs a little bit more per inch to grow a 72" stick than a 6" stick). I'm looking for a uniform growth, not a split formula. No f(x) where x<72, g(x) where x>72.

I'm not necessarily looking for the formula to solve the above question. I am looking for the HOW to research and solve the above question.

Many Thanks,
Matt

答:
You are looking for functions that go to infinity, preferably in finite time (vertical asymptote) or something similar. you then construct a scale to get to your infinity point at the number of inches you want (say 100). so, for tan(x) as an example, pi/2 is infinity, so map 0-100 to pi/4-pi/2, and you have something that works (you might want to add a constant to formula to help you on on the low end).

A possibility: f(x) = Tan(x/200*pi/2+pi/4)+5

f(10) = 6.2

f(72) = 9.5

f(90) = 17.7

f(95) = 35.45

f(98) = 68.7

Feel free to massage the constants, or pick a different function to get the behavior that you want.


问:
Why is the direct product of two groups $G_1 \times G_2$ too small to be their coproduct?
We’re looking at the following definition in my class.

Definition. The coproduct $X_{1} \amalg X_{2}$ of $X_{1}$ and $X_{2}$, together with the morphisms $i_{j}: X_{j} \rightarrow X_{1}\amalg X_{2}$, is characterized by the following universal property: Given any object $Y$ with morphisms $f_{j}: X_{j} \rightarrow Y,$ there exists a unique $f: X_{1} \amalg X_{2} \rightarrow Y$ such that $f_{j}=f \circ i_{j}$.

In $G_1 \times G_2$, there exist two subgroups isomorphic to $G_1$ and $G_2$, namely the sets of elements $\{(g_1,e_2): g_1 \in G_1\}$ and $\{(e_1,g_2): g_2 \in G_2\}$, respectively. These two subgroups have the property that their members commute with each other: $(g_1,e_2) \cdot (e_1,g_2) = (g_1,g_2) = (e_1,g_2) \cdot (g_1,e_2)$.

The reason given for the direct product not being the coproduct is that, in a group $Y$ with maps $f_1: G_1 \to Y$ and $f_2: G_2 \to Y$, elements of $Y$ don’t commute with each other. I’m not yet able to derive this.

My understanding of the argument is that, given the property of the two subgroups above, I ought to be able to derive $f_1(g_1) \cdot_Y f_2(g_2) = f_2(g_2) \cdot_Y f_1(g_1)$, which would not be true for a general group $Y$ (unless $Y$ is abelian).

Now if I also assume $G_1 \times G_2$ to be the coproduct, I’d have:
$$f_1(g_1) \cdot_Y f_2(g_2) = (f \circ i_1)(g_1) \cdot_Y (f \circ i_2)(g_2) = f(g_1,e_2) \cdot_Y f(e_1,g_2)$$

It’s here that I don’t know how to keep going to get the relation above.

答:
Let $X_1, X_2$ be cyclic of order $2$ and let $x_i$ denote the non-trivial element of $X_i$. Let $Y$ be the set of permutations of $\Bbb Z$.
In particular, we have two elements $y_1,y_2\in Y$, given by $y_1(n)=-n$ and $y_2(n)=1-n$. Then $y_1\circ y_1=y_2\circ y_2=\operatorname{id}_{\Bbb Z}$, i.e., these are both of order $2$.
This allows us to define homomorphisms $f_i\colon X_i\to Y$ given by mapping $x_i\mapsto y_i$.

But $y_1( y_2(n))=n-1$ whereas $y_2(y_1(n))=n+1$, i.e., $y_1\circ y_2\ne y_2\circ y_1$ -- the elements do not commute in $Y$.
Hence neither may the elements $i_1(x_1)$ and $i_2(x_2)$ commute in $X_1\amalg X_2$. This shows that $X_1\amalg X_2$ cannot be just the direct product.

By the way, the subgroup of $Y$ that is generated by $y_1$ and $y_2$ (together with the above homomorphisms) does have the desired universal property. Can you show this?


问:
interpolating function with inclined asymptote
I would like to find an interpolating function that is a curve with an inclined asymptote. I already tried with the parabolic and with the homographic (y= (ax+b)/(cx+d)). Unfortunately they are both phisically inconsistent with the phenomenon that I am analysing. I need a function like the c) in the image attached, which is a curve that has an inclined asymptote. Can you help me ?

答:
I recommend looking at functions of the form polynomial/polynomial, where the top polynomial's degree is 1 larger than the bottom polynomial's degree. You'll also want to make sure the bottom polynomial never equals 0 when x is nonnegative. One family of examples (that go through the origin, as seems to be desired from your pictures) is
$$
y = \frac{ax^2+bx}{x+c} \tag{1}
$$
where $a,b,c$ are positive. (The inclined asymptote will then be the graph of $y=ax+b$, for that matter.)

If you don't insist that the graph go through the origin, then you can expand this family to
$$
y = \frac{ax^2+bx+d}{x+c}. \tag{2}
$$
Another way of writing this same family is
$$
y = ex+f+\frac g{x+c}
$$
(where the inclined asymptote is now $ex+f$). Whether $g$ is negative or positive controls whether the graph lies below or above the asymptote (equivalently, whether the curve is concave or convex). Both shapes are possible with different choices of the parameters—and the same is true for the forms (1) and (2).


问:
degree of polynomial satisfied by real part of complex algebraic number
If a complex number $z=x+iy$ is algebraic over $\mathbb Q$ of degree $n$ then its real part $x$ will also be algebraic over $\mathbb Q$. What will be the degree of polynomial satisfied by $x$? Can it be greater than the degree of $z$ over $\mathbb Q$?

答:
Let us assume $z\notin \mathbb{R}$, otherwise there is nothing to be said. Here is an elementary proof that $x$ has degree $\leq n(n-1)/2$ over $\mathbb{Q}$.

Consider the extension $\mathbb{Q}(z)\subseteq\mathbb{Q}(z,\overline{z})$. Let $f\in\mathbb{Q}[t]$ be the minimal polynomial of $z$ over $\mathbb{Q}$. Since $\overline{z}$ is a root of $f$, its minimal poynomial over $\mathbb{Q}(z)$ divides $f$, and so has degree $\leq n$. But, moreover, $f$ factorizes over $\mathbb{Q}(z)$ as \begin{equation}f=(t-z)g\end{equation} for some polynomial $g\in\mathbb{Q}(z)[t]$ of degree $n-1$. Then, since $z\neq\overline{z}$, we have $g(\overline{z})=0$ so actually $\overline{z}$ has degree $\leq n-1$ over $\mathbb{Q}(z)$.

We also have $[\mathbb{Q}(z,\overline{z}):\mathbb{Q}(x)]\geq 2$ simply because $z\notin\mathbb{R}$, so from

\begin{equation}
2[\mathbb{Q}(x):\mathbb{Q}]\leq [\mathbb{Q}(z,\overline{z}):\mathbb{Q}(x)][\mathbb{Q}(x):\mathbb{Q}]=[\mathbb{Q}(z,\overline{z}):\mathbb{Q}(z)][\mathbb{Q}(z):\mathbb{Q}]\leq (n-1)n
\end{equation}

it follows that $[\mathbb{Q}(x):\mathbb{Q}]\leq n(n-1)/2$.


问:
Finding explicit formula for this recurrence relation
This is the recurrence relation: $a_n=a_{n/2}+1,a_1=3$. I have never done this type of recurrence relation. I have been told, that I need to substitute $n/2$ with something else but I can't come up with it.

Body:
$a_n=a(n/2)+1,a_1=3 $

let's say $n/2=k  => n=2k$

which means n should be a power of two (because n/2 should be integer again due to $a_{n/2}$). Thus, $n =2^{k}$

$$ a_{2^{k}}=a_{2^{k-1}}+1 $$
$$ a_{2^{k-1}}=a_{2^{k-2}}+1 $$
$$ a_{2^{k-2}}=a_{2^{k-3}}+1 $$
$$.$$
$$.$$
$$.$$
$$a_{2}=a_{1}+1$$
$$a_{1}=3$$

Now, if we add both Left-hand side and Right-hand side, we'll get:

$$a_{2^k} = k*1 + 3 $$

Therefore:
$$ a_{n} = log_{2}^{n} + 3 $$


问:
Longest shortest path in an undirected unweighted graph
PROBLEM

I have an undirected unweighted graph and I would like to find the longest shortest path in that graph (in other words, for each two vertices I can calculate the minimal distance between them, and I want to find maximum over all those distances).

SOLUTION ATTEMPT

I've search for graph width, range, span -- but all those terms refer to something different.

答:
All considered graphs are finite, simple and undirected. Let $G=(V,E)$ be such a graph on $n$ vertices.

The minimum length of the paths connecting two vertices $v_x,v_y \in V$ is called the distance between $v_x$ and $v_y$ and is denoted by $d(v_x, v_y)$. If $G$ is disconnected and $v_x$ and $v_y$ are not in the same components, we define $d(v_x,v_y)=\infty$.

The $n \times n$ - graph distance matrix $D=(d_{xy})$ consists of all graph distances from vertex $v_x$ to vertex $v_y$. The maximum value of all entries of $D$ is called the diameter of $G$.

Example: Given $G=(V,E)$. Let $V=\{v_1,v_2,\cdots,v_n\}$ and $E=\{(v_1,v_2),(v_2,v_3), (v_2,v_5),(v_2,v_4)\}$. The $(x,y)^{th}$ entry of the distance matrix $D$ of $G$ is the distance $d(v_x,v_y)$, yielding in

$$
\begin{pmatrix}
    0 & 1 & 2 & 2 & 2 \\
    1 & 0 & 1 & 1 & 1 \\
    2 & 1 & 0 & 2 & 2 \\
    2 & 1 & 2 & 0 & 2 \\
    2 & 1 & 2 & 2 & 0
    \end{pmatrix}$$

Now, finding the diameter using the distance matrix $D$ is easy.

Question: How do we calculate $D$?

We could use Floyd-Warshall algorithm which runs in $O(n^3)$.

But since our graphs are undirected and unweighted, we can do better. There is an algorithm by Chan solving our problem for an unweighted and undirected graph with $n$ vertices and $m$ edges in $O(nm)$.

Yamane and Kobayashi implemented an algorithm for unweighted graphs in general. Their paper contains a small list of known algorithms solving the same problem. You can find a preprint on arXiv here.


问:
Big-$\mathcal{O}$ of the number of comparisons and multiplications needed
The number of comparisons and multiplications for the following algorithm should be:

$$\sum_{i=1}^{n}(n-i)\times 2$$

2 as the amount of operations for multiplication is same as for comparisons.

Question: But the solution I found claims that it's $n\times (n-1)\times 2$, could you pleas state why?

Edit: I found out the solution as follows:

$$\sum_{i=1}^{n-1}(n-i)\times 2$$
$$2\times \sum_{i=1}^{n-1} n-2\times \sum_{i=1}^{n-1}i$$
$$2\times (n)(n-1) -2\times \frac{(n-1)n}{2}$$
$$2\times (n)(n-1)$$

答: The product $a_ia_j$ and the maximum function are evalkuated exactly once per pair $(i.j)$ with $1\le i<j\le n$. Thenre are $n\choose 2$ such pairs, hence $n\choose 2$ multiplications and $n\choose 2$ comparisions (implicit in $\max$) givig a total of $n(n-1)$. (This assumes that we do not count the comparisons of index variable against end expressin that we implicitly have in the for loop).


问:
Time has dimension $2$ with respect to the Ricci flow scaling
Terence Tao in his lecture notes on Ricci flow has written:

If we are to find a scale-invariant (and diffeomorphism-invariant) monotone quantity for Ricci flow, it had better be constant on the gradient shrinking soliton. In analogy with $\frac{d}{dt}\mathcal{F}_m=2\int_M |Ric+Hess(f)|^2dm$, we would therefore like the variation of this monotone quantity with respect to Ricci flow to look something like

$$2\int_M|Ric+Hess(f)-\frac{1}{2\tau}g|^2dm (*)$$

where $\tau$ is some quantity decreasing at the constant rate $\dot{\tau}=-1.$

But the scaling is wrong; time has dimension $2$ with respect to the Ricci flow scaling, and so the dimension of a variation of a scale-invariant quantity should be $-2$, while the expression $(*)$ has dimension $-4$. (Note that $f$ should be dimensionless (up to logarithms), $\tau$ has the same dimension of time, i.e. $2$, and $\int_Mdm=1$ is of course dimensionless.

Question:

In the note what is the meaning of this sentence: time has dimension $2$ with respect to the Ricci flow scaling?

Thanks. (Sorry if the question is too trivial.)

答:
Let us start with the simple example of heat equation
$$ \partial_t u(t,x) = \triangle u(t,x) $$
we see that if $u(t,x)$ is a solution, then $u_\lambda(t,x) = u(\lambda^2 t,\lambda x)$ is also a solution, as
$$ (\partial_t u_\lambda)(t,x) = \lambda^2 (\partial_t u)(\lambda^2 t,\lambda x) = \lambda^2 (\triangle u)(\lambda^2 t,\lambda x) = (\triangle u_\lambda)(t,x)~.$$

This shows that if we change $x$ by a scale factor $\lambda$, we need to change $t$ by scale factor $\lambda^2$. So if $x$ has unit "Length", the units for $t$ should be "Length$^2$".

This fact is usually true for parabolic equations (though nonlinearities sometimes throw a wrench into the proceedings; in those cases it is also important to keep track of the solution $u$'s units; note that in the linear case $u$'s units do not matter, since by linearity they must appear in the same "strength" for all terms).

For the Ricci flow, consider the metric $g_{\mu\nu}$ as a function in local coordinates. Let the "scaled metric" $g^{(\lambda)}_{\mu\nu}(x) = g_{\mu\nu}(\lambda x)$ and compute the corresponding Christoffel symbols, you will see that $\Gamma^{(\lambda)} = \lambda \Gamma$ and the curvature scales like $\mathrm{Riem}^{(\lambda)} = \lambda^2 \mathrm{Riem}$. Taking the trace does not change the scaling, so we have that the same analysis as in the linear case gives that if $g_{\mu\nu}(t,x)$ solve, in a fixed coordinate system, the Ricci flow, then so would $g^{(\lambda)}_{\mu\nu}(t,x) = g_{\mu\nu}(\lambda^2 t,\lambda x)$.

Response to followup:

Again let us first start with the simple examples. Think back about classical physics. The distance $x$ an object has traveled has unit "Length". The time $t$ it took to travel has unit "Time". So the "variation of the distance with respect to time" $v = \frac{\mathrm{d}x}{\mathrm{d}t}$ has unit "Length / Time".

Now in the case of a parabolic equation, suppose $\mathcal{F}$ is something that has no units (a pure number, or in SI the unit of "one"), then the variation of $\mathcal{F}$ with respect to "time", which we recall has unit "Length$^{2}$", would have unit
$$ \frac{\text{Unit of } \mathcal{F}}{\text{Unit of time}} = \frac{1}{\text{Length}^2} = \text{Length}^{-2} $$


问:
Quadratic factoring
What is a general rule of thumb that we can use to check whether a quadratic expression can be factored without the use of the quadratic formula since using the quadratic formula can be time consuming in exam situations?

Is it even possible to tell whether an expression will require application of the quadratic formula even before beginning to factor it?

答:
All second degree equations can be factored but we might ask: Are the factors real or complex? Are they rational or irrational? Another answer mentioned the discriminant.

If $\quad b^2-4ac\lt0\quad $ the factors are complex (with an imaginary parts).

If $\quad b^2-4ac\ge 0\quad $ the factors are real numbers

If $\quad b^2-4ac\qquad$ is a perfect square, the factors are rational.

If you want to "find" the roots (factors) (if they exist) without the quadratic formula, you can apply the rational root theorem. In it, you just try all of the rational numbers composed of factors of the first and last terms, dividing using synthetic division, and seeing if they divide without remainder into the original quadratic polynomial.

Title: Quadratic factoring


问:
What is a way to solve for the argument of the absolute value of the natural exponential function?
I’d like to solve for $x$ in the equation:

$y = \lvert e^{-icx}\rvert$

where $x$ and $y$ are real constants and $c$ is a complex number of the form $a+ib$. Recognizing that the modulus of a complex number is the square root of the sum of the squares of the real and imaginary parts, I’ve tried to expand the equation in terms of sines and cosines but got stuck. Any ideas?

答:
$|e^{-ixc}|= |e^{-ix(a+ib)}|= |e^{-ixa} \cdot e^{xb}|=
|e^{-ixa}| \cdot |e^{xb}|= e^{xb}$, since
$|e^{iu}|=1 $ for all $u\in \mathbb R$ (Euler's formula). Therefore,
$x= \dfrac{\ln y}b$.


问:
$f(A\cap B)=f(A)\cap f(B)$ $\iff$ $f$ is injective.
Let $f:X\to Y$ where $X$ and $Y$ are nonempty. Prove that a sufficient and essential condition for any two subsets $A,B\subseteq X$ to fulfill $f(A\cap B)=f(A)\cap f(B)$ is that $f$ is injective. I sense there is some problem in my proof. I would be glad if you assisted me.

$Attempt:$ Let $f$ be injective and let $A,B\subseteq X$ be two subsets. If $A$ and $B$ are disjoint, then $A\cap B=\emptyset$ $\Rightarrow$ $f(A\cap B)=\emptyset$. Since $f$ is injective then there are no two elements with the same image and therefore $f(A)\cap f(B) =\emptyset =f(A\cap B)$. Now suppose $A\cap B\ne \emptyset.$ Let $ y\in f(A\cap B)$. There exists an $x\in A\cap B$ such that $f(x)=y$. Since $x \in A$ then $f(x)=y\in f(A)$ and since $x \in B$ then $f(x)=y\in f(B)$ $\Rightarrow$ $y\in f(A)\cap f(B)$. Therefore $f(A\cap B)\subseteq f(A)\cap f(B)$.

Now let $y\in f(A)\cap f(B)$. ( $f(A)\cap f(B)$ is not empty because if it were then $A$ and $B$ would have an empty intersection as well, which they don't.). Therefore $y\in f(A)$ and $y\in f(B)$ and therefore in $A$ there exists $x_1$ such that $f(x_1)=y$. The same with $B$, $f(x_2)=y$. By injectivity: $x_1=x_2 \Rightarrow x_1=x_2\in A\cap B\Rightarrow f(x_1)=y\in f(a\cap B)\Rightarrow f(A)\cap f(B)\subseteq f(A\cap B) \Rightarrow f(A\cap B)=f(A)\cap f(B).$

Necessary: Suppose it weren't necessary that $f$ is injective. Then there would exist a non-injective function $f$ such that for any two subsets $A,B\subseteq X$ we get $f(A\cap B)=f(A)\cap f(B)$. $f$ is non-injective and therefore there would be $x_1,x_2\in X$ such that $f(x_1)=f(x_2)$. Let us look at $\{x_1\},\{x_2\}\subseteq X$. $f(\{x_1\})\cap f(\{x_2\})=\{f(x_1)=f(x_2)\}\ne f(\{x_1\}\cap\{x_2\})=f(\emptyset)=\emptyset$. A contradiction.

答:
Let f : X → Y be a one-one, and A, B ⊆ X. We show that f(A ∩ B) =
f(A) ∩ f(B).
we need to prove that f(A) ∩ f(B) ⊆ f(A ∩ B).

Let y ∈ f(A) ∩ f(B). This implies y ∈ f(A) and y ∈ f(B). This means,
there exist x1 ∈ A and x2 ∈ B such that f(x1) = y = f(x2). Since f is one-one,
we have x1 = x2. Therefore, x1 ∈ A ∩ B, which means y = f(x1) ∈ f(A ∩ B).
Consequently, f(A) ∩ f(B) ⊆ f(A ∩ B).
Conversely, suppose for any two subsets A, B ⊆ X, f(A∩B) = f(A)∩f(B).
We show that f is one-one. Consider x1, x2 ∈ X, x1 6= x2. Let A = {x1}, B =
{x2}. Then A ∩ B = ∅. Therefore, {f(x1)} ∩ {f(x2)} = f(A) ∩ f(B) =
f(A∩B) = ∅. This implies that f(x1) 6= f(x2). Consequently, f is one-one.


问:
Minimizing costs to make a cylinder
A cylinder must have a volume $V=4000$ and its made from just one rectangle sheet, so its top and bottom must be cut from this sheet.

I have to find the dimensions of this sheet such that the parts I lost (when I cutoff the top and the bottom) are the lowest.

I thought about minimize the difference between the area of the original sheet and the parts I'll use to make my cylinder:

(I imagine this is the best way to don't waste sheet - the original problem doesn't have any image).

So the area of the original sheet is
$$(h+2R)\cdot4R=4hR+8R^2$$
and the area I'll use to make the cylinder is
$$4hR+2\cdot\pi R^2$$
so the difference between them is
$$8R^2-2\pi R^2=0$$
and derivating it:
$$16R-4\pi R$$
which the only root is $R=0$. What is wrong?

答:
Assume that you are free to use any or all the sheets stocked in hardware supply store whose total area is $S$.

From the wasted area  we have the useful area.

$$ A_{usable} = S-A_{wasted}= A $$

$$ V= \pi R^2 h, A= 2 \pi R h + 2 \pi R^2 $$

Eliminate one variable, of relation derivate and find its value. Using other relation find the second variable ( details not included... for the usual exercise).

We end end up with result of relative proportion

$$ h=2 R$$

$$ V= 4000 = \pi R^2 \cdot 2 R\to R_{opt}=\left(\frac{2000}{\pi}\right)^\frac13, h_{opt}=R_{opt}/2$$

Point is, there is no need to a priori optimize the area that is anyway later on going to enter into mathematical optimization. The remainder area is negative. Multiplying  by $-1,$ the minimization and optimal solution cannot be affected, the constants play no role here.

From Variational calculus we can see the irrelevance of sign,

$$ \frac{\partial V_{R}}{\partial V_{h}}=\frac{\partial A_{R}}{\partial A_{h}}=\quad \frac{\partial A_{uR}}{\partial A_{uh}}=\frac{\partial A_{wR}}{\partial A_{wh}}.$$

Title: Minimizing costs to make a cylinder


问:
How do we know that $y=e^{-\int P(x)dx}(c+\int e^{P(x)dx}f(x)dx)$ is the general solution of $\frac{dy}{dx}+P(x)y=f(x)$?
It is straightforward that any function of the form $y=e^{-\int P(x)dx}(c+\int e^{P(x)dx}f(x)dx)$  is a solution of the differential equation on the interval $I$ for which $P$ and $f$ are continuous. However, how are we sure that if $\frac{dy}{dx}+P(x)y=f(x)$ has a solution, then it must be of the form $y=e^{-\int P(x)dx}(c+\int e^{P(x)dx}f(x)dx)$ ?

The book used variation of parameters to arrive at this solution. How can we show that there is no other solution other than the one obtained using variation of parameters?

答:
That essentially follows from the linearity of the equation.

If you have a linear differential operator $L$ and the inhomogeneous equation $Lu=f$, then for any two solutions $u_1$ and $u_2$ that satisfy $Lu_1=f$ and $Lu_2=f$, you have by linearity
$$
L(u_1-u_2)=0
$$
which means, the difference $u_1-u_2$ is a solution to the homogeneous equation $Lu=0$.

In your example, $(Lu)(x)=\frac{du}{dx}+P(x)u(x)$.

Your solution is written as $u=u_h+u_p$, where

The part $u_h(x):=Ce^{-\int P(x)dx}$ is the general solution to $Lu=0$ and $u_p$ is a particular solution to $Lu=f$ found by some method (say, variation of parameters).

If you find another particular solution $u_q$ to the equation $Lu=f$, then $$
u_q-u_p=Ke^{-\int P(x)dx}
$$
for some constant $K$. So solution to $Lu=f$ is still of the form $u=u_h+u_p$.


问:
inequality:$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$

If $a,b,c,d>0$ such that $abcd=1$ prove that $$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$$

My first idea was to find counter examples and also check the bahaviour of inequality in the edge cases.Easy to see the equality occurs when $a=b=c=d=1$ However there is a big problem when $a\to 0^+ b\to 0^+,c\to 0^+,d\to \infty$ In this case the RHS $\to \infty$ and the LHS although appears  $\to 0$ we see that as $b+d\to \infty$ so we cannot say much.

Mixed variables doesnt help much as we cannot take WLOG $a\ge b\ge c\ge d$

I also tried Holder $$(a+c)(b+d)(bd+ac)\ge {(\sqrt[3]{ab^2d}+\sqrt[3]{ac^2d})}^3=ad{(b^{2/3}+c^{2/3})}^3$$ which  has only complicated it further

Looking forward to solutions without lagranges multiplier.

答:
Let $ac=x\geq1\geq\frac{1}x=bd.$ So if $a+c=2q$  and $b+d=2s,$ then $q\geq\sqrt x$  and $s\geq\frac{1}{\sqrt x}.$ The inequality writes as
$$s\left(2q\left(x+\frac{1}x\right)-3\right)+2-3q\geq0,$$
which we may regard as an affine function of varible $s,$ strictly increasing, for the rest of the elements fixed. It remains to prove
$$\frac{1}{\sqrt x}\cdot\left(2q\left(x+\frac{1}x\right)-3\right)+2-3q\geq0.$$
It's trivial to prove that this one is a strictly increasing affine function of variable $q.$ So we need to prove that
$$2\left(x+\frac{1}x\right)-3\left(\sqrt x+\frac{1}{\sqrt x}\right)+2\geq0,$$
which is obvious since $\sqrt x+\frac{1}{\sqrt x}\geq2.$ Done!


问:
lattice in a 3 dimension region
In a 3 dimension lattice, a random walker moves from origin. Let $a_n$ be the coordinate of the random walker in n-th steps. $a_0=(0,0,0)$

The random walker moves 6 directions with same probability in every steps with step size 1. It means, $a_1$ can be $(0,0,1),(0,0,-1),(0,1,0),(0,-1,0),(1,0,0),(-1,0,0)$ with probability of $1/6$

Let $a_n \equiv (X_n,Y_n,Z_n)$

Let $P(n) \equiv P(0<X_n,0<Y_n,0<Z_n,X_n<Y_n+Z_n)$

Prove or disprove that $P(n)$ is increasing when $n\geq3$.

Title: lattice in a 3 dimension region

Body:
False.

$P(3)=P[a_3=(1,1,1)]=\frac{6}{6^3}\approx 0.0277$

$P(4)=P[a_4=(1,2,1) \text{ or } a_4=(1,1,2)]=\frac{12+12}{6^4}\approx 0.0185$

$P(5)\approx 0.040$

$P(6)\approx 0.029$


问:
Adjoint problem for convection-diffusion-reaction equation
I have given the convection-diffusion-reaction equation
$$-\Delta u+\beta\cdot\nabla u+\gamma u=f\ \text{in}\ \Omega\\
\quad u=0\quad\quad\quad\quad\quad\quad\quad\quad \text{on}\ \partial\Omega
$$
with $\beta\in C^1(\Omega)^d$, $\nabla\cdot\beta=0$ and $\gamma\in L^\infty(\Omega)$.

Now I have to find a bounded linear operator $A:V\to V^*$ such that the the variational formulation correspondents to $Au=l$.

I was simply using $A=-\Delta\cdot+\beta\nabla\cdot+\gamma\cdot$.

To find the adjoint operator I was now doing integration by parts analog to Which means adjoint problem of a differential equation?.

$$
    \int_\Omega (Au)z\ \mathrm dx=\int_\Omega (-\Delta u+\beta\cdot\nabla u+\gamma u)z\ \mathrm dx\\
    =\int_\Omega (-\Delta uz+\beta\cdot\nabla uz+\gamma uz)\ \mathrm dx\\
    =\int_\Omega -\Delta uz\ \mathrm dx+\int_\Omega\beta\cdot\nabla uz\ \mathrm dx+\int_\Omega\gamma uz\ \mathrm dx\\
$$
integration by parts:
$$=u\nabla z-\nabla u z+\beta u z+\int_\Omega (-u\Delta z-\beta u\nabla z+uz)\ \mathrm dx\\
=u\nabla z-\nabla u z+\beta u z+\int_\Omega (-\Delta z-\beta \nabla z+z)u\ \mathrm dx$$
it would be perfect if this would lead to $$\int_\Omega (A^*z)u\ \mathrm dx$$

Did I choose $A$ and $l$ wrong?

答:
You are on the right track.

However, specifying the domain of such an operator is often crucial, and so I choose the domain of your operator $A$ to be $D(A)=H^2(\Omega) \cap H^1_0(\Omega)$ - I hope this is a reasonable assumption.

Indeed, you have to figure out $A^*$ such that
$$
\int_{\Omega} (A^*z)u \; dx=\int_{\Omega} (Au)z \; dx
$$

Integrating by parts is also the right idea, since boundary terms always vanish because of your imposed boundary conditions - but lets quickly split up the occuring terms:
$$
\int_{\Omega} (Au)z \; dx=\underbrace{\int_{\Omega} (-\Delta u)z\; dx}_{:=T_1}+\underbrace{\int_{\Omega} (\beta \cdot \nabla u)z\; dx}_{:=T_2}+\underbrace{\int_{\Omega} (\gamma u) z \; dx}_{:=T_3}
$$
For the first term $T_1$, integrating by parts (sometimes called Green's Formula) twice, which you can do because of the choice of $D(A)$, leads to
$$
\int_{\Omega} (-\Delta u)z\; dx=\int_{\Omega} u(-\Delta z)\; dx
$$
The second term $T_2$ is the tricky one. However, note that you can rewrite the integrand as
$$
z\beta \cdot \nabla u
$$
Then you integrate by parts again. But let us first simplify the following expression using the condition $\nabla \cdot \beta=0$:
$$
\nabla \cdot z \beta=\nabla z \cdot \beta + z \nabla \cdot \beta=\nabla z \cdot \beta
$$
You can now integrate by parts again:
$$
\int_{\Omega} (\beta \cdot \nabla u)z \; dx=\int_{\Omega} u(-\beta \cdot \nabla z)\; dx
$$
Note that we got a minus sign in the process. Therefore $T_2$ is ultimately the reason why $A$ is not symmetric (and therefore not self-adjoint).

The last term $T_3$ is just switching parentheses:
$$
\int_{\Omega} (\gamma u) z \; dx=\int_{\Omega} u(\gamma z)  \; dx
$$
Collection terms you have
$$
\int_{\Omega} (-\Delta z- \beta \cdot \nabla z+ \gamma z)u \; dx
$$
which gives you
$$
A^*=-\Delta - \beta \cdot \nabla + \gamma
$$


问:
Is the fundamental lemma of Calculus of Variations wrong?
the title might be a bit absurd but this is how I currently feel just because I can't see where my reasoning is faulty.

So the lecturer presented the lemma as the following:

Let $M \in C[a, b]$. Then $\int_{a}^{b} M \cdot h \,dx = 0, \forall h \in C_0^1[a,b] \implies M \equiv 0$.

But how does the lemma still hold when $h(x) = 0$? The integral would still be zero but it would not imply that $M = 0$ since it could for example be $M = x$. Am I wrong to believe that $0 \in C_0^1[a,b]$?

He gives a proof by contradiction using a non-zero $h$ (the bump function ) somewhere in the lines of: Say $M$ is greater than zero and $h$ is greater than zero, then the integral must also be greater than zero which results in a contradiction since the integral must be equal to zero. Therefore $M$ must be zero. This is obviously true for the case when $h$ is non-zero but isn't this an obvious loss of generality? When $h$ is also allowed to be zero then one can not easily deduce which function is making the integral to be zero. If the lemma would require that it applies to $\forall h \in \{C_0^1[a,b] \setminus 0\}$ then everything would make sense but that is not how the lemma is generally expressed.

Any help would be appreciated.

答: You seem to be reading it as$$\color{red}{\forall h\in C_0^1[a,\,b]\left(\int_a^bMhdx=0\implies M=0\right)},$$but it actually means$$\color{limegreen}{\left[\forall h\in C_0^1[a,\,b]\left(\int_a^bMhdx=0\right)\right]\implies M=0.}$$It's the equivalent of confusing$$\color{limegreen}{\text{If all people like me, I'm popular}}$$with$$\color{red}{\text{If you pick any one person, if they like me I'm popular}}.$$


问:
hypothesis testing problem
I'm stuck at the following problem

We have people picking one of A,B,C.

According to a theoretical model people pick in the ratio $p^2:2p(1-p):(1-p)^2$ respectively.

From a sample we have 42,52,22.

Is the model correct?

I know that I have to use chi-square test and since we know nothing about $p$
I should use MLE

When I do  that I find $\hat{p}=\frac{1}{2}$ but according to the book it should be $\hat{p}=\frac{68}{116}$

I would like to know what I'm doing wrong?

答:
You have the ratios $p^2:2p(1-p):(1-p)^2$, which sum to $1$

You are given the count data  $42,52,22$ summing to $116$.  This suggests A is more common than C so you might guess $\hat p > \frac12$.

The likelihood is proportional to $$\left(p^2\right)^{42}\left(2p(1-p)\right)^{52}\left(1-p)^2\right)^{22}$$ which $($ignoring powers of $2)$ is proportional to $$p^{2\times 42+52}(1-p)^{52+2\times 22} = p^{136}(1-p)^{96}$$  and this is maximised when $\hat p =\frac{136}{136+96}=\frac{136}{232}=\frac{68}{116}=\frac{17}{29} \approx 0.5862$ as your book says.

If that was correct, it might suggest expected values for the counts of about $39.86,56.28,19.86$ and you can perform your chi-squared test (with two degrees of freedom) in the usual way.  Intuitively these modelled numbers look close to the observed numbers so you will not reject the null hypothesis.  In R:

```<code>obs <- c(42, 52, 22)
phat <- (2*obs[1] + obs[2]) / (2*sum(obs))
chisq.test(obs, p=c(phat^2, 2*phat*(1-phat), (1-phat)^2))

#         Chi-squared test for given probabilities
#
# data:  obs
# X-squared = 0.66967, df = 2, p-value = 0.7155
</code>```

Title: hypothesis testing problem

Body:
You have the ratios $p^2:2p(1-p):(1-p)^2$, which sum to $1$

You are given the count data  $42,52,22$ summing to $116$.  This suggests A is more common than C so you might guess $\hat p > \frac12$.

The likelihood is proportional to $$\left(p^2\right)^{42}\left(2p(1-p)\right)^{52}\left(1-p)^2\right)^{22}$$ which $($ignoring powers of $2)$ is proportional to $$p^{2\times 42+52}(1-p)^{52+2\times 22} = p^{136}(1-p)^{96}$$  and this is maximised when $\hat p =\frac{136}{136+96}=\frac{136}{232}=\frac{68}{116}=\frac{17}{29} \approx 0.5862$ as your book says.

If that was correct, it might suggest expected values for the counts of about $39.86,56.28,19.86$ and you can perform your chi-squared test (with two degrees of freedom) in the usual way.  Intuitively these modelled numbers look close to the observed numbers so you will not reject the null hypothesis.  In R:

obs <- c(42, 52, 22)
phat <- (2*obs[1] + obs[2]) / (2*sum(obs))
chisq.test(obs, p=c(phat^2, 2*phat*(1-phat), (1-phat)^2))

#         Chi-squared test for given probabilities
#
# data:  obs
# X-squared = 0.66967, df = 2, p-value = 0.7155


问:
How can I find the domain of this sqrt log function?

Can someone help me to find the domain of this function step by steps,
I know how to find the domain for a log by making it >0 but I am unsure about the process doing it with 2 different logarithm and a square root.
$$f(x)=\sqrt{\log_{0.5}(5-x)+\log_2(2x-4)-1}$$

答:
Let's first define both the $\log$s, input of $\log$ should be a positive quantity.

$5 - x > 0 \implies x \in (-\infty, 5)$

$2x - 4 > 0 \implies x \in (2, \infty)$

Now we know that $\log_{a}{b} = -\log_{\frac{1}{a}}{b} = \log_{\frac{1}{a}}{\frac{1}{b}}$

$\implies\log_{0.5}(5-x) = \log_{2}{\frac{1}{5-x}}$

Now, we know that anything inside square root should be non negative.

$\implies \log_{0.5}(5-x)+\log_2(2x-4)-1 \geq 0$

$\implies \log_{2}{\frac{1}{5-x}+\log_2(2x-4)-1} \geq 0$

$\implies \log_2{\frac{2x-4}{5-x}} \geq 1$

$\implies \frac{2x-4}{5-x} \geq 2$

$\implies \frac{2x-4}{5-x} - 2 \geq 0$

$\implies \frac{4x-14}{5-x} \geq 0$

$\implies \frac{4x-14}{x-5} \leq 0$

$\implies x \in [\frac72, 5)$

Taking intersection of all 3 intervals:

$$\boxed{x \in [\frac72,5)}$$


问:
Get recurrent formula for integral $I_n=\frac{1}{\sin^n(x)}$
The first thing that I could notice is that we can do so

$$I_n - I_{n-2} = \int \frac{1 - \sin^2 x}{\sin^n x} dx = \int \frac{\cos^2 x}{\sin^n x} dx$$

(Sometimes I will skip putting $dx$ for contraction)

However when I was trying to continue integrating in this form, it seemed to me as even harder problem and I decided to change approach and try integrating by parts.

$$I_n = \int \frac{1}{\sin^n x} = \int \frac{1}{\sin x \cdot \sin^{n - 1}x}$$

This one was a bit despeate, I know. So, I decided on putting this idea aside either.

So, do you have any ideas on how to approach this problem? Thank you in advance.

Body:
$$
I_n = \int \dfrac{\sin x}{\sin^{n+1} x}dx = - \dfrac{\cos x}{\sin^{n+1} x} - (n+1)\int \dfrac{\cos^2 x}{\sin^{n+2} x}dx = - \dfrac{\cos x}{\sin^{n+1} x} - (n+1)(I_{n+2}-I_n)
$$


问:
asymptotic infinity
I am very very new to math as a whole, so please excuse my n00biness.

I read: An asymptote of a curve is a line that continually approaches the curve but never meets it at any finite distance. The distance between the line and the curve approaches zero as they tend to infinity. When we say variable n tends to infinity it means as n gets very very large. If we look at 1/n as n tend to infinity, then n gets very large and 1/n goes to zero

So my question is: Based on the above can 1/n be analogous to the distance between a line and a curve, and n thought of as the increase in the size of the two?

答:
If you look at the picture:

As you plug larger and larger values for $x$, the blue curve will get closer and closer to the horizontal black line.  The distance between the blue curve and black line is given by $\frac{1}{x}$, which gets smaller the bigger $x$ gets.  $x$ in this case represents how far to the right we go -- the further to the right, the closer the two are.

Title: asymptotic infinity


问:
can a parabolic curve have two turning points?
the curve $v= 0.00005(t-200)^2 - 1$ seems to have only $1$ minimum point to me at $(200,-1)$ as a minimum point.
(i completed the squares to find it)
but in my book it says it also has a maximum point at $(0,1)$

$0\leq t\leq800$

my question is how would you have found the maximum point? i learned the trigonometric graphs of sine and cosine and know that they have multiple turning points but how can a quadratic graph have multiple turning points?
could anyone please  explain why?

答: When looking for a curve's maxima or minima, you need to examine two things - inflection points and discontinuities. Inflection points are the turning points that you seem to be familiar with, which can be found through the typical use of derivatives. Derivatives don't exist at discontinuous points, however, so you need to check them individually. This parabolic function is discontinuous at the endpoints of t=0 and t=800, so you need to check values there. Basically, you can have maxima/minima at points where the slope is not equal to zero - these aren't "turning points", but rather endpoints or some other kind of discontinuity. You can see, for example, that the straight line x=y defined for [0,10] has no inflection points at all, but still has a maximum and minimum.</p>


问:
Broken Line Regression
$X = $Lot & $Y = $Cost

Give a broken line linear model with a breakpoint at $250$:
$$Y = B_0 + B_1X_1 + B_2X_2 + B_3X_3 + e$$
where $X_2 = 0$ or $1$ depending on whether the lot size is $\geq 250$ or $< 250$ and $X_3 = X_1\cdot X_2.$

Which hypothesis statement is equivalent to the statement: The two regression lines have the same intercept term?
$$H_0 : B_0 = 0\\
H_0 : B_1 = 0\\
H_0 : B_2 = 0\\
H_0 : B_3 = 0$$

$B_0$ is obviously the intercept for simple linear regression models, however, the broken line phrasing is causing me to be confused on this. My initial instinct was to assume $B_0$ hypothesis was appropriate, but now I'm wondering if $B_2 = 0$ makes more sense.

Any assistance or starting point would be very beneficial as I need to explain through the reasoning algebraically too !

答:
Presumably $X_1$ is the lot size

When $X_1 < 250$ you have $X_2=0$ and $X_3=X_1X_2=0$ so

$Y = B_0+B_1X_1$ with intercept $B_0$

When $X_1 \ge 250$ you have $X_2=1$ and $X_3=X_1X_2=X_1$ so

$Y = (B_0+B_2)+ (B_1+B_3)X_1$ with intercept $B_0+B_2$

so the intercepts are equal when $B_2=0$

Title: Broken Line Regression


问:
How to combine probabilities for guesses
Let 3 people guess an object A or B.
Each person have 80% chance of being correct.

What is the probability of object being an A, if:

all 3 of them say it is an A

only 2 say it is an A

Can you please also refer to generalised version of this question?

EDIT: chance for A or B is 50/50

答:
Assume guesses are independent:

First case - the probability that all three are wrong is $.2^3=.008$.

Second case - the probability that two are wrong and one is right is $.2^2\times .8$ for each, so total is $3\times .2^2\times .8=.096$.


问:
Is the non-homogeneous and inhomogeneous linear ODEs means the same?
Actually it's confusing me, somewhere it is written non-homogeneous and somewhere inhomogeneous.</p>

答:
First let me say that “in homogeneous” and “non homogeneous” means the same thing. But I suspect you already know this.

A homogeneous differential equation is one where every term involves derivatives of the dependent variable and (though not necessary) terms containing the dependent variable itself. For example

$$\frac{d^2 f}{dx^2} + \frac{d^2 f}{dy^2} + 7f =0$$

is called homogeneous.

A non homogeneous differential equation involves derivatives of the dependent variable with terms containing the independent variable(s). For example

$$\frac{d^2 f}{dx^2} + \frac{d^2 f}{dy^2} + f = x^2 + xy$$

is a non homogeneous differential equation.

So if you hear your teacher/lecturer say “in homogeneous” and “homogeneous” or even “non homogeneous” do not be confused since he/she means the distinction between “homogeneous” and “non homogeneous”.

OwnerDisplayName: Nandani Kumari

OwnerDisplayName: joseph h


问:
Find the ordered pair $(a,b)$ of real numbers for which $x^2+ax+b$ has a non-real root whose cube is $343$.
Because $x^2 + ax + b$ has all real coefficients, we know the roots must be complex conjugates. So let $\lambda e^{i \theta}$ be a root, so we know that $(x - \lambda e^{i \theta})(x - \lambda e^{-i \theta}) = x^2 - \lambda x (e^{i \theta} + e^{-i\theta}) + \lambda^2$ is our polynomial.

We also know that $(\lambda e^{i \theta})^3 = 7^3 \implies \lambda e^{i \theta} = 7$, and since $|\lambda e^{i \theta}| = |\lambda| = 7,$ we know that $\lambda = 7$.

This means that $e^{i\theta} = 1$ and so $\theta = 0$. But clearly this doesn't give the right answer, since our polynomial isn't supposed to have real roots. Where is the flaw in my reasoning?

答:
$(\lambda e^{i \theta})^3 = 7^3 \implies e^{3i \theta} = (\frac{7}{\lambda})^3e^{0}$

Using De Moivre's theorem

$\implies e^{i \theta} = (\frac{7}{\lambda}) e^{\frac{2k\pi}{3}i}, \text{k = 0,1,2}$

$\implies \lambda=7$

But as given in the question, roots are not real, so $k$ can not be $0$, therefore,

Roots are $7e^{\frac{2\pi}{3}i}$ and $7e^{\frac{4\pi}{3}i}$

$\implies 7(-\frac{1}{2} + \frac{\sqrt3}{2}i) \text{ and } 7(-\frac{1}{2} - \frac{\sqrt3}{2}i)$


问:
Prove $\sum_{n\le x} d(n)/n = \frac12 \log^2 x + 2e \log x + c + \mathcal O(1/\sqrt{x})$ where $d$ is the divisor function and $c$ is a constant
So I know that If $A(x) = \sum_{n \leq x} a(n)$, and $\phi(x)$ has a continuous derivative on $[X_1, X_2]$, where $0 \leq X_1 < X_2 \in \mathbb{R}$, then

$$\sum_{X_1 < n \leq X_2} a(n)\phi(n) = A(X_2)\phi(X_2) - A(X_1)\phi(X_1)- \int_{X_1}^{X_2} A(x)\phi'(x)\,\mathrm dx\tag1\label{eq1}$$

I also know that

$$\sum_{n \leq x} d(n) = x(\log x + 2e - 1) + \mathcal O(\sqrt{x})\tag2\label{eq2}$$

So from $\eqref{eq1}, \eqref{eq2}$ letting $a(n) = d(n)$ and $\phi(n) = 1/n$, I get

$$\begin{align}\sum_{n \leq x} d(n)/n &= \frac{1}{x} \sum_{n \leq x}d(n) + \int_{1}^x d(t)/t^2 \, \mathrm dt
\\&= (\log x + 2e - 1) + \mathcal O(\sqrt{x})/x + \int_{1}^x d(t)/t^2 \,\mathrm dt
\\&= (\log x + 2e - 1) + \mathcal O(1/\sqrt{x}) + \int_{1}^x d(t)/t^2 \,\mathrm dt\end{align}$$

but I am not sure where to go from here. I was thinking of rewriting $\int_{1}^x d(t)/t^2 \,\mathrm dt$ as

$$\begin{align}\int_{1}^x \frac{d(t)}{t^2} \, \mathrm dt &=\int_1^x \frac{\log(t)}{t}\,\mathrm dt + \int_{1}^x \frac{d(t) - t\log(t)}{t^2} \, \mathrm dt
\\&= \frac{1}{2}\log^2x +  \int_{1}^x \frac{d(t) - t\log(t)}{t^2} \, \mathrm dt
\\&= \frac{1}{2}\log^2x +  \int_{1}^\infty \frac{d(t) - t\log(t)}{t^2} \, \mathrm dt -  \int_{x}^\infty \frac{d(t) - t\log(t)}{t^2} \, \mathrm dt\end{align}$$
Would either of these two integrals converge to a constant?

答:
\begin{align}
\int_{1}^x \frac{d(t)}{t^2} \, \mathrm dt &=\int_1^x \frac{(\log(t) +2e -1)t+ \mathcal{O}(\sqrt{t})}{t^2}\ dt
\\&= \frac{1}{2}\log^2x + (2e-1) \log x + \int_{1}^x \mathcal{O}(t^{-\frac{3}{2}}) dt,
\end{align}
and $\int_{1}^\infty\mathcal{O}(t^{-\frac{3}{2}})dt$ is finite.


问:
Summing from m to n with initial modifier from 0 to 1
I am trying to find costs in a system, in the system going from Level 0 to Level 1 has a constant cost of $C$. As a result the cost to go from Level 1 to Level 2 is $C + i$ where $i$ is our cost increment for each increased Level. E.g. For $C = 10$ and $i = 5$ going from Level 0 to Level 1 would have a cost of $10$; similarly going from Level 1 to Level 2 would have a cost of $10 + 5 = 15$. How would I go about finding the costs for transitioning between two arbitrary Levels $m$ and $n$.

I know that if I have $C = 0$, my cost simply can be found out by using the formula $c =i\sum_{k=m}^n{k}$ where $c$ is the cost of transitioning from Level $m$ to Level $n$.

How would I incorporate $C$ into this formula?

答:
Following on from William Elliot's answer, if we let some $c(k)$ be the cost to go from Level $k$ to $k + 1$, we can express $c(k) = C + ki$, where $i$ is some constant increase factor. Let us express the cost to go from some Level $m$ to some Level $n$ as $c(m,n) = \sum_{k=m}^{n-1}{c(k)}$. Therefore to calculate the transition cost from Level $m$ to Level $n$, we can use summation laws to break $c(m,n)$ down into a simple $O(1)$ function:

\begin{align}
c(m,n) = \sum_{k=m}^{n-1}{c(k)} & = \sum_{k=m}^{n-1}{C + ki} \\
& = \sum_{k=m}^{n-1}{C} + i\sum_{k=m}^{n-1}{k} \\
& = \sum_{k=1}^{n-1}{C} - \sum_{k=1}^{m-1}{C} + i\left(\sum_{k=1}^{n-1}{k} - \sum_{k=1}^{m-1}{k}\right) \\
& = \dots \\
& = C(n - m) + \frac{1}{2}i(n^2 - m^2 + m - n)
\end{align}


问:
What is the purpose to define central moment?
What is the purpose to define central moment?
I searched the google and all i could find is bunch of properties</p>

答:
Quite often, you're interested in the characteristics of a distribution independent from where exactly its located on the $x$-Axis, or in other words independent from its mean.

The central moments do exactly that - they are the moments after shifting the random variable such that its mean becomes zero. I.e. the central $k$-th moment of $X$ is the $k$-th moment of $X-\mathbb{E}X$.


问:
Why does $\frac{2P}{P}$ give $2$ and not $P$?
The formula for annual compounded interest (according to our lecturer) is  $$2P = P(1 + r)^n.$$
($P$: initial investment, $r$: interest rate, $n$: years).

To solve for $n$, he said the "$P$"s cancel out and you can then transform the equation to
$$n =\frac{\ln(2)}{\ln(1 + r)}.$$

But why is that? If there were an $x$ instead of $P$ (so $2x=x(1+r)^n$), you'd reduce it by dividing through by $x$ (so $2x/x$) which would leave $1x$. So, why is this not happening with $P$, and instead we keep the number $2$ but omit $P$?

答:
Dividing a product by one of its factors gives the other factor: $(ab):b=a$.  On our case we have
$$(2P):P=2.$$

Similarly, subtracting an addend from a sum  gives the other addend: $(a+b)-b=a$.


问:
$\lim\limits_{n\to \infty}\frac{3^n+6^n-1}{6^n-1}$
$\lim\limits_{n\to \infty}\dfrac{3^n+6^n-1}{6^n-1} = 1$

I proved the statement above but now the exercise says that I have to find $n_0$ so that $\left\vert x_n-1\right\vert < \frac{1}{50}$ with $(x_n)$ as the given seqeunce above.

I wrote the sequence like this: $(x_n)=\frac{3^n}{6^n-1}+1$ and then I looked at $\left\vert \frac{3^n}{6^n-1}+1-1\right\vert < \frac{1}{50}$ which is equivalent to $\frac{\frac{1}{2^n}}{1-\frac{1}{6^n}} < \frac{1}{50}$. But now I don't know how to go on...

Can someone help me?

Body: $\frac{2^{-n}}{1-6^{-n}}<\frac{1}{50}$ is equivalent to $2^n-3^{-n}>50$. By $3^{-n}<1$ if $n>1$ you get the solution.


问:
Cohomology of a smoothly embedded space curve
Let $\mathbb{P}^3 = P(\mathbb{C}^4)$ and $\gamma:C\rightarrow \mathbb{P}^3$ be a smoothly embedded algebraic space curve. Then its total Chern class is $c(C) = c(TC) = 1+a\in H^*(C)$, where $a^2=0$. Since $C$ is compact and orientable we have Poincare duality and hence a well defined 'shriek - map' $$\gamma^!: H^*(C)\overset{PD_C}{\longrightarrow} H_{2-*}(M) \overset{\gamma_*}{\longrightarrow} H_{2-*}(\mathbb{P}^3)\overset{PD_{\mathbb{P}^3}^{-1}}{\longrightarrow}H^{*+4}(\mathbb{P}^3).$$

Let $g$ be the genus of $C$ and let $\chi(C):=2-2g$ and $d$ be the degree of $C$. Then $$\gamma^!(1) = d\cdot c_1(\mathcal{O}_{\mathbb{P}^3}(1))^2\quad\text{and }\quad \gamma^!(a) = \chi \cdot c_1(\mathcal{O}_{\mathbb{P}^3}(1))^3.$$

Does anyone know why these two relations hold and point me to some literature? My knowledge in algebraic geometry is very limited since I am more familiar with manifolds, homology/cohomolgy of topological spaces so Chern classes for me are always defined in that context. I know that there are Chern classes in algebraic geometry (i.e. for Chow rings) and that these notions are related, but my understanding is very superficial.

Still it would help me a lot if I could understand the relations above. Thanks!

答:
You can understand these facts using just a bare minimum of algebraic geometry.
All you need to know about $c_1(\mathcal{O}_{\mathbb{P}^3}(1))$ in this context is that it is the standard generator of $H^2(\mathbb{P}^3)$ (corresponding to the fundamental class of $\mathbb{P}^1=S^2$ when you restrict to $\mathbb{P}^1\subset\mathbb{P}^3$).  This can also be described as the Poincaré dual of the fundamental class of $\mathbb{P}^2\subset\mathbb{P}^3$, or of any hyperplane in $\mathbb{P}^3$ (since the inclusion maps of different hyperplanes $\mathbb{P}^2\to\mathbb{P}^3$ are all homotopic).  How you prove this depends on how you define Chern classes, but for some definitions it is essentially by definition: for instance the first Chern class of the tautological line bundle on $\mathbb{P}^\infty$ can be defined to be negative the standard generator of $H^2(\mathbb{P}^\infty)$, and then for other line bundles it can be defined by pulling back under the classifying map to $\mathbb{P}^\infty$.  The bundle $\mathcal{O}_{\mathbb{P}^3}(1)$ is just the dual of the tautological bundle on $\mathbb{P}^3$, so its first Chern class is the negative of the first Chern class of the tautological bundle, which is just the pullback of the tautological bundle under the inclusion $\mathbb{P}^3\to\mathbb{P}^\infty$.

Let me now say a bit about Poincaré duality on $\mathbb{P}^3$.  If you have two smooth oriented submanifolds $A$ and $B$ of an oriented manifold which intersect transversely, then the Poincaré dual of the fundamental class of the intersection $A\cap B$ is just the cup product of the Poincaré duals of the fundamental classes of $A$ and $B$.  This plays very nicely with smooth subvarieties of $\mathbb{P}^3$ via Bézout's theorem.  In particular, if $H\subset\mathbb{P}^3$ is a hyperplane, Bézout's theorem says that for any smooth curve $C\subset\mathbb{P}^3$ of degree $d$ which intersects $H$ transversely, $C$ and $H$ intersect at $d$ points.  So, writing $t$ for the Poincaré dual of the fundamental class of $H$ (which, as mentioned above, is $c_1(\mathcal{O}_{\mathbb{P}^3}(1))$) and $c$ for the Poincaré dual of the fundamental class of $C$, $tc=dt^3$.

Now to compute $\gamma^!(1)$, we first take the Poincaré dual of $1\in H^0(C)$, which is just the fundamental class $[C]\in H_2(C)$.  Then we push this forward to get the class $[C]\in H_2(\mathbb{P}^3)$, and take its Poincaré dual $c\in H^4(\mathbb{P}^3)$.  Since $tc=dt^3$, this class $c=\gamma^!(1)$ must be $dt^2$.

To compute $\gamma^!(a)$, we first take the Poincaré dual of $a=c_1(TC)\in H^2(C)$.  The top Chern class of a vector bundle is the same as its Euler class, and the Euler class of the tangent bundle of an oriented manifold is dual to the Euler characteristic of the manifold (that's why it's called the Euler class).  So the Poincaré dual of $a$ is $\chi\in H_0(C)$ (when you identify $H_0(C)$ with $\mathbb{Z}$ in the canonical way).  Pushing this forward to $\mathbb{P}^3$ we get $\chi\in H_0(\mathbb{P}^3)$, whose Poincaré dual is $\chi t^3$.


问:
Is Introduction to Calculus and Analysis by Courant a good book for a first Calculus course for a beginner?
I have Courant's Introduction to Calculus and Analysis and also have Spivak's Calculu's and Thommas-Finney Calculus with analytic Geometry, I'm finishing my precalculus studies, and I don't have much experience with calculus.

I want to become really good at math, and I want to know (Real) calculus, so I'd like to know if Courant is good for a first approach or should I start with Spivak?

答:
Courant and Spivak are classic excellent texts, from different eras. I suggest you use both, in parallel. You will learn a lot comparing the two approaches.

A word of warning: both these books work at a level of rigor that you may not be familiar with from your precalculus studies. Thomas will have a larger collection of relatively routine exercises.

Whatever route you choose, be sure to work lots of exercises.


问:
Given $1/2x^TQx+d^Tx$ s.t $a_1^Tx\leq b_1$ $a_2^Tx\leq b_2$ find optimal solution

Given min $1/2x^TQx+d^Tx$
s.t $a_1^Tx\leq b_1$

$a_2^Tx\leq b_2$

and $Q\succ0(n\geq3)$ $a_1,a_2,d\in\mathbb{R}^n$ and $b_1,b_2\in\mathbb{R}$,Assume that $a_1^TQ^{-1}a_1=a_2^TQ^{-1}a_2=2,a_2^TQ^{-1}a_1=0,a_1,a_2\neq0$

Find the optimal solution

I've set the lagrangian to be $L(x,\lambda,\eta)=1/2x^TQx+d^Tx+\lambda(a_1^Tx- b_1)+\eta(a_2^Tx-b_2)$

$\frac{\partial L}{\partial x}=Qx+d+\lambda a_1+\eta a_2$

I've tried setting $\lambda\neq0,\eta=0$ and solving but didn't get me anywhere,any hint please?

答:
So you got $L(x,\lambda,\eta)=1/2x^TQx+d^Tx+\lambda(a_1^Tx- b_1)+\eta(a_2^Tx-b_2)$ and derived the stationarity condition:
$$Qx+d+\lambda a_1+\eta a_2=0$$
You somehow need to use that $a_1^TQ^{-1}a_1=a_2^TQ^{-1}a_2=2,a_2^TQ^{-1}a_1=0,a_1,a_2\neq0$. If you pre-multiply the stationarity condition with $a_1^TQ^{-1}$ and $a_2^TQ^{-1}$ you get:
$$a_1^Tx+a_1^TQ^{-1}d+2\lambda=0 \quad \text{and} \quad a_2^Tx+a_2^TQ^{-1}d+2\eta=0$$
which means:
$$\lambda=-0.5(a_1^Tx+a_1^TQ^{-1}d) \quad \text{and} \quad \eta=-0.5(a_2^Tx+a_2^TQ^{-1}d)$$
Plugging that back into the stationarity condition, you get:
$$Qx+d-0.5(a_1^Tx+a_1^TQ^{-1}d) a_1-0.5(a_2^Tx+a_2^TQ^{-1}d) a_2=0$$
Since $(a_1^Tx)a_1 = a_1(a_1^Tx)=(a_1a_1^T)x$, the stationarity condition can be simplified to:
$$(Q-0.5a_1a_1^T-0.5a_2a_2^T)x+d-0.5a_1^TQ^{-1}d a_1-0.5a_2^TQ^{-1}d a_2=0$$
so
$$x = (Q-0.5a_1a_1^T-0.5a_2a_2^T)^{-1}(0.5a_1^TQ^{-1}d a_1+0.5a_2^TQ^{-1}d a_2-d)$$