问:
In a general eigenvector with eigenvalue having a multiplicity of 2, why is the first entry substituted instead of the other two for this example?
My question is about the computation of the eigenvector that corresponds to an eigenvalue with a multiplicity higher than 1. This is a pretty long example that establishes my question.

The matrix problem I am referring to can be seen in this link.

https://matrixcalc.org/en/vectors.html#diagonalize%28%7B%7B4,-2,-2%7D,%7B-2,4,-2%7D,%7B-2,-2,4%7D%7D%29

In determining the eigenvectors of the following matrix,

$$
    \begin{matrix}
    4 & -2 & -2 \\
    -2 & 4 & -2 \\
    -2 & -2 & 4 \\
    \end{matrix}
$$

The characteristic equation is
$$ -\lambda^3 + 12\lambda^2 -36\lambda = 0  $$

Solving for the eigenvalues,
$$ \lambda = 0 \ and\ 6 $$

The eigenvalue of 6 has a multiplicity of 2.

Substituting it to matrix $$ A - \lambda I $$ makes the matrix

$$
    \begin{matrix}
    -2 & -2 & -2 \\
    -2 & -2 & -2 \\
    -2 & -2 & -2 \\
    \end{matrix}
$$

Which can then be reduced to row-echelon form

$$
    \begin{matrix}
    1 & 1 & 1 \\
    0 & 0 & 0 \\
    0 & 0 & 0 \\
    \end{matrix}
$$

Where the equation
$$ x_1 + x_2 + x_3 = 0 $$

Can be transformed into
$$ x_1 = -x_2 + -x_3 $$

Which makes the eigenvector
\begin{pmatrix}
x_1  \\
x_2 \\
x_3 \\
\end{pmatrix}

Equal to
\begin{pmatrix}
-x_2 -x_3 \\
x_2 \\
x_3 \\
\end{pmatrix}

My question is, why was the first variable expressed in terms of the second and third variable? I got this problem right up until that step. Honestly, I don't have a good grasp of solving eigenvalues that have a multiplicity higher than one.

Do you always pick the first variable as the one being substituted? Why does it not work that way for other problems where you can choose what variable to express the eigenvector in? Does it have to do with the roots being 0 and 6 where 6 has a multiplicity of 2, which is why the first entry is substituted? What if the eigenvalue of a order 3 matrix had 2 distinct values but the lesser value had the multiplicity of 2? Would things change? What would also happen if it was an order 4 matrix with two distinct values each having a multiplicity of 2? Would both matrices have the first and second entries in the eigenvector be substituted?

Can people refer me to any reference regarding solution of eigenvectors corresponding to eigenvalues with multiplicity higher than 1? Honestly I tried searching but most that I have seen are so complicated. I mean call me stupid if I can't find them and they are easy to find, but I really couldn't get much of what I have seen. Most of them focus on the multiplicity part and not the eigenvectors.

This all leads to the bigger picture of, how do you solve for eigenvectors in general. I think I have a good grasp of solving because it was just really the substitution that I messed up in. More knowledge regarding this topic would help me. Thank you.

答:
For $\lambda=6$ with multiplicity 2 the eigenvector $(x_1,x_2,x_3)^T$ would satisfy $x_1+x_2+x_3=0$. Next choose two sets of $x_1,x_2,x_3$ such that the two eigenvectors $X_1, X_2$ are orthogonal: $X_1^T X_1=0$
So we get $X_1=\begin{bmatrix} 1 \\-2 \\ 1 \end{bmatrix}$ and $X_2=\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}.$


问:
Prove $\ln{(\frac {x}{y})} = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.
Prove $\ln (\frac{x}{y}) = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.

I am able to prove $\ln{xy} = \ln{x} + \ln{y}$, and $\ln{x^r} = r\ln{x}$, but with this one, I am stuck at $\ln\frac{x}{y} = \ln{x} + \ln{\frac{1}{y}}$, but is it okay to automatically assume $\ln{\frac{1}{y}}$ = $-\ln{y}$ from here?

答:
In case you don't like the $r=-1$ approach, here is a direct proof.

First write $\displaystyle\ln\frac{1}{y} = \int_1^\frac{1}{y}\frac{1}{t}\,\mathrm{d}t = -\int_\frac{1}{y}^1\frac{1}{t}\,\mathrm{d}t$. Now we change variables to $s = yt$. Then $\mathrm{d}s = y\mathrm{d}t$ and thus $\displaystyle\ln\frac{1}{y} = -\int_\frac{1}{y}^1\frac{1}{t}\,\mathrm{d}t = -\int_1^y\frac{1}{s}\,\mathrm{d}s = -\ln y$.

ContentLicense: CC BY-SA 3.0


问:
Is the halting problem still an open problem?
(Spoiler: Turing was right!)

I was thinking about Turing's proof that the halting problem is undecidable. But can the paradox, which is often given as a proof, be broken?

Proof that a single program cannot decide the halting problem

First, I want to repeat the paradox in my own words:

Let source be the source code of a program, args some arguments to that program, and eval(source, args) an instruction that executes the program with the given arguments. Assume we had a program will_halt(source, args) that returns True if and only if eval(source, args) halts. Consider the program:

```<code>def p(source):
    while will_halt(source, (source)):
        pass  # endless loop
    return 42
</code>```

p would run forever if will_halt(source of p, (source of p)) returns True, and halts otherwise. will_halt is faced with a conflict for the input (source of p, (source of p)). Turing concludes, will_halt cannot exist, because for any program will_halt a program p can be constructed that creates a conflict.

Resolving the conflict using two programs

will_halt must reason somehow whether loops terminate or not. It runs into a conflict for some inputs. However, we can define will_halt_or_true that returns True if and only if a conflict is encountered. Now, consider the following program:

```<code>def probe(source, args):
    while not will_halt_or_true(source, args):  # here we use 'not'!
        pass
    return eval(source, args)  # actually execute the program in question!
</code>```

probe(source, args) behaves as follows:

will_halt_or_true(source, args)
Conflict
eval(source, args)
probe(source, args)

True
No
Halts
Halts

False
No
Runs forever
Runs forever

True
Yes
Runs forever
Runs forever

We can use probe to distinguish the conflict from the non-conflict case. will_halt_or_true(probe, (source, (args))) returns True if and only if eval(source, args) halts. Note, that a single program running will_halt_or_true(probe, (source, (args))) is as powerful as running will_halt_or_true directly. So, the decision process always involves two steps and cannot be reduced to one.

Conclusion (false)

My main argument is that no single program can decide the halting problem but two might. Knowing how will_halt_or_true handles conflicts enables us to construct a probe to decide the halting problem. This is by no means a proof that will_halt exists, but I argue that the "common" proof is not sufficient to claim that the halting problem is undecidable.

Edit

I distinguished will_halt from will_halt_or_true to be more clear. However, I found the following counter example, which is quiet obvious:

```<code>def q(args):
    while will_halt_or_true(probe, (args, (args))):
        pass
    return 42
</code>```

Any value contributed to will_halt_or_true(probe, (source of q, (source of q))) will lead to a contradiction. So, will_halt_or_true(probe, ...) cannot be used. Yesterday it worked ;-)

答:
With your approach, the assumption on will_halt that you start out with "Assume we had a program will_halt(source, args) that returns True if eval(source, args) halts and False otherwise." does not hold anymore: there are cases where evaluating source runs forever, but your will_halt returns True. So, whatever you do afterwards does not address the halting problem anymore, because that starts out with your (original) assumptions.

(As an aside, note that it is trivial to write a will_halt function that returns True if the eval(source,args) terminates and without restriction on the return value if eval(source,args) does not terminate. It's useless, though.)

ContentLicense: CC BY-SA 4.0


答:
I am able to prove $\ln{xy} = \ln{x} + \ln{y}$, and $\ln{x^r} = r\ln{x}$, but with this one, I am stuck at $\ln{x/y} = \ln{x} + \ln{\frac{1}{y}}$, but is it okay to automatically assume $\ln{\frac{1}{y}}$ = $-\ln{y}$ from here?

Yes, absolutely. That's how proofs work in mathematics often work; certain things are proven, and than the results of those earlier proofs can be used in subsequent proofs.

So do as mathematicians do: just cite the problems in which you established both

$\,\ln(x\,y) = \ln x + \ln y,\,$

$\,\ln x^r = r \ln x,\,$

and you may argue as you suggest.

Title: Prove $\ln{(\frac {x}{y})} = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.

Body:
Prove $\ln (\frac{x}{y}) = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.

I am able to prove $\ln{xy} = \ln{x} + \ln{y}$, and $\ln{x^r} = r\ln{x}$, but with this one, I am stuck at $\ln\frac{x}{y} = \ln{x} + \ln{\frac{1}{y}}$, but is it okay to automatically assume $\ln{\frac{1}{y}}$ = $-\ln{y}$ from here?


问:
Find bases for a given linear map and matrix of linear map
One has given formula of linear map $T:\mathbb{R}^n \to \mathbb{R}^m$ and matrix of this linear map $ M(T)_\mathcal{A}^\mathcal{B} $ in (unknown) bases $\mathcal{A}$ of $\mathbb{R}^n$ and $\mathcal{B}$ of $\mathbb{R}^m$ and would like to find these bases. For example, get $ T:\mathbb{R}^3\to \mathbb{R}^2$ such that $T(x,y,z)=(x+y+z,2x-y+3z)$ with matrix $$M(T)_\mathcal{A}^\mathcal{B}=\begin{bmatrix} 1 & 3 & -1 \\ 4 & 0 & 7 \end{bmatrix}$$ and we would like to determine bases $\mathcal{A},\mathcal{B}.$

I would like to use a formula of base change: we have matrices $M(T)_\mathcal{A}^\mathcal{B}$ and $ M(T)_{st}^{st}.$ If I'll find $M(id)_\mathcal{A}^{st}$ and $M(id)_\mathcal{B}^{st}$, I have done. I tried by getting explicite form of these matrix in unknown elements but computations are so complicated, because we get a linear system with 13 coefficients... I would like to ask you for more satisfactory approach, more formal without such computations.

答:
Let $A$ be the $m\times n$ matrix of $T\colon \mathbb R^n\to\mathbb R^m$ with respect to the standard basis and $B=M(T)_{\mathcal A}^{\mathcal B}$ be the given matrix with respect to unknown bases $\mathcal A$ and $\mathcal B$.

The two matrices $A$ and $B$ can describe the same linear map $T$ if and only if they have the same rank. In this case, there are invertible matrices $X\in\mathbb R^{n\times n}$ and $Y\in\mathbb R^{m\times m}$ such that
$$
YAX = B.
$$
Comparing this with $M(\operatorname{id})_{\text{st}}^{\mathcal B}\ A\ M(\operatorname{id})_{\mathcal A}^{\text{st}} = B$, we might interpret $X$ and $Y$ as change of basis matrices. Hence, we can take the columns of $X$ as the basis vectors in $\mathcal A$ and the columns of $Y^{-1}$ as the basis vectors in $\mathcal B$.

Note that $X$ and $Y$ and hence $\mathcal A$ and $\mathcal B$ are far from being unique for given matrices $A$ and $B$. For example the identity map will have the same matrix with respect to any basis (taken in domain and codomain).

One algorithmic way to find such $X$ and $Y$ is to use row and column transformations to transform $A$ and $B$ into the rank normal form
$$
\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix} \in \mathbb R^{m\times n},
$$
where $r$ is the rank of both $A$ and $B$ and $I_r$ is the $r\times r$ identity matrix.

Keeping track of row and column operations seperately (by applying them simultaneously to $I_m$ and $I_n$, respectively) this yields
$$
X_1 A Y_1 = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix} = X_2 B Y_2,
$$
and hence
$$
\underbrace{X_2^{-1} X_1}_X\ A\ \underbrace{Y_1 Y_2^{-1}}_Y = B.
$$


问:
Proving unitarily equivalence using induction
Problem: A matrix B ∈ Mn,n(C) is said to be unitarily equivalent to A ∈ Mn,n(C) if there exists a unitary matrix U ∈ Mn,n(C) such that B= U†AU.
Show that every matrix A ∈ Mn,n(C) is unitarily equivalent to a triangular matrix.

Hint:This is obvious for 1×1 matrices.  Use this as induction hypothesis (A ∈ Mn,n(C)) and prove the statement for A ∈ Mn+1,n+1(C) by induction over n.  You can use that the product of two unitary matrices U,Q ∈ Mn,n(C) is a unitary matrix without having to prove it.

My attempt:

Let's prove this using induction.

Base case: As given in the hint, this is obvious for 1x1 matrices.

Induction hypothesis: There exists positive integer n such that nxn matrix A is unitarily equivalent to a triangular matrix.

Step case: Let A be an n+1 x n+1 matrix. ...

But I have no idea how to do the step case. Using induction over dimension confuses me, since n+1 x n+1 matrices are off course larger than nxn matrices.

Any help, whether it are hints or full answers, are appreciated.

答:
Have you seen inductive proofs of spectral theorem for Hermitian matrices? You find the eigenvector (which always exists, since characteristic polynomial has a root in $\mathbb{C}$), then look at orthogonal complement to the eigenvector, prove it's invariant and induct.

Here it's similar: find an eigenvector, make a basis using this eigenvector and a basis of its orthocomplement. Now see that in this basis your matrix has zeroes down the first column, then induct.

If you get stuck, google "Schur decomposition".


问:
Verifying that a computed density is a probability density
Question:

Let $n\in\Bbb N$. I want to show (or disprove) that for fixed $i\in \{1, \ldots ,n-1\}$ and $N\in\Bbb N$ we have

$$\sum_{m=i}^{n-1} \binom{m-1}{i-1} \sum_{k=i}^{m} (-1)^{i+k} \lambda_{k}^N \binom{m-i}{k-i} = \sum_{k=i}^{n-1} (-1)^{i+k}\lambda_k^N \binom{k-1}{i-1}\binom{n-1}{k}, $$

where $\lambda_\ell^N = \left(\frac{n-\ell}n \right)^N$ for $k\in\{1,\ldots , n-1\}$ and $\lambda_n^N = 1$.

Background:

The quantity above is supposed to be the sum over the probabilities of being in state $m \in \{i, \ldots , n -1\}$ started from $i$ after $N$ steps for the simple Markov chain on the state space $\{1, \ldots ,n\}$ with the transition matrix
$$p(i,m) = \begin{cases} \frac{n-i}n &: m=i \neq n ,\\ \frac i n&:m=i+1,\\ 1 &: m=i=n,\\ 0&: \text{else}\end{cases}$$

I calculated the probability of the $N$th step by computing the Jordan decomposition matrices. More precisely, a Jordan decomposition is then $D^N = \text{diag}( \left( \frac{n-1}n\right)^N , \ldots , \left( \frac{1}n\right)^N , 1)$ and $V = (v_{i,k})_{i,k}$ with $v_{i,k} = (-1)^{i-1} \binom{k-1}{i-1} 1_{i\leq k < n} +1_{k=n}$ as eigenvector matrix and $V^{-1} = \tilde V$ with $\tilde v_{i,k} = v_{i,k}1_{k<n} + (-1)^k \binom{n-1}{k} 1_{k=n}$.

To check this result I wanted to verify that the probabilities add up to $1$, but could not get it done yet. Here the left-hand side of the equation is $p^N (i, \{i, \ldots ,
 n-1\})$ and the right-hand side is $1-p^N (i,n)$.

答:
Let $(a_{k})_{k\in\Bbb N}$ be arbitrary numbers. We make the following claim: For $n\in\Bbb N$ holds for all $i\in\{1,\ldots , n\}$

$$\sum_{k=i}^n (-1)^{i+k} a_k \binom{k-1}{i-1} \binom{n}{k} = \sum_{m=i}^n \binom{m-1}{i-1} \sum_{k=i}^m (-1)^{i+k} a_k \binom{m-i}{k-i}$$

For the proof assume this holds for $n-1$. Then for $i\leq n-1$ holds

$$\sum_{k=i}^n (-1)^{i+k} a_k \binom{k-1}{i-1} \binom{n}{k} = \sum_{k=i}^{n-1} (-1)^{i+k} a_k \binom{k-1}{i-1} \binom{n}{k} + (-1)^{i+n} a_n \binom{n-1}{i-1}\\
= \sum_{k=i}^{n-1} (-1)^{i+k} a_k \binom{k-1}{i-1} \binom{n-1}{k} + \sum_{k=i}^{n-1} (-1)^{i+k} a_k \binom{k-1}{i-1} \binom{n-1}{k-1} + (-1)^{i+n} a_n \binom{n-1}{i-1}\\
=\sum_{m=i}^{n-1} \binom{m-1}{i-1} \sum_{k=i}^m (-1)^{i+k} a_k \binom{m-i}{k-i}+\sum_{k=i}^{n} (-1)^{i+k} a_k \binom{k-1}{i-1} \binom{n-1}{k-1}\\
=\sum_{m=i}^{n-1} \binom{m-1}{i-1} \sum_{k=i}^m (-1)^{i+k} a_k \binom{m-i}{k-i}+ \binom{n-1}{i-1}\sum_{k=i}^{n} (-1)^{i+k} a_k  \binom{n-i}{k-i}\\
=\sum_{m=i}^{n} \binom{m-1}{i-1} \sum_{k=i}^m (-1)^{i+k} a_k \binom{m-i}{k-i}
$$
Since for $n=1$ or $i=n$ the statement is clear the claim follows by induction.

From the claim the desired follows for $n$ by choosing $a_k = \lambda_k^N$. (I went the way with abitrary $a_k$ because the $\lambda_k^N$ depend on $n$.)


问:
Is it possible that a vector space can be a finite union of proper subspaces?
Let $V$ be a vector space, and let $V_i$ for $i=1,\ldots, n$ be non-zero subspaces of $V$. Is it possible that $V=\cup_{i=1}^n V_i?$

The underlying field is assumed to be infinite.

答:
Let $V$ be a vector space over an infinite field $k$.  Assume $V=\cup_{i=1}^nV_i$, where $V_i$ are all proper subspaces.  Let $u\in V_1$, and take any $v\in V\setminus V_1$.  There are infinitely many vectors of the form $u+cv$ for $c\in k^*$, none of which are in $V_1$.  Since there are infinitely many of them, and only finitely many subspaces, there must be some $V_j$ containing at least two of them.  It follows that both $v$ and $u$ are in $V_j$ so that $V_1\subset \cup_{i=2}^nV_i$.

Continuing this process, we can show that $V=V_n$, contradicting the assumption that the subspaces are proper.

I guess the assumption that $k$ is infinite is a bit overkill.  We only need that there are at least as many elements in the field as subspaces in the union to derive a contradiction.  This shows that even over finite fields, we need at least as many subspaces as elements in the field to obtain such a decomposition.


问:
Tricky Integral $\int_{0}^{\pi/2} \left[\frac{\sec^2 \theta}{2}\cdot \exp(\frac{\sec^2 \theta}{2})\right]^{-1} d\theta$ - Brain Slayer
Given the integral
$$
\int_{0}^{\pi/2}
\left[\frac{\sec^{2}\left(\theta\right)}{2}
\exp\left(\frac{\sec^{2}\left(\theta\right)}{2}\right)\right]^{\,-1} \mathrm{d}\theta
$$
How am supposed to evaluate it within few minutes $?$. Is there a quick trick solve the integral $?$.

Should you wish to see my attempt at doing so, then you should know that my solution is analyzed the author's one, because I could not solve it without seeing clues. So, I would like to get some ideas from; I think that I fail to solve this guy because I could not see something...

答:
I would be surprised if one can solve this integral within a few minutes without any prior knowledge. But there are some ideas that helps tackle this integral.

1. Write

$$ I = \int_{0}^{\frac{\pi}{2}} (2\cos^2\theta) e^{-\frac{1}{2}\sec^2\theta} \, \mathrm{d}\theta. $$

If you see an exponential function of something other than linear or quadratic polynomials inside the integral sign, then that integral is highly unlikely to be tackled directly except for some lucky cases. So, if we believe that this integral can be computed in a closed form, then we may first try to make the exponent simpler. Here, let us make the substitution $x = \tan\theta$. Then

$$ I = 2 \int_{0}^{\infty} \frac{1}{(x^2+1)^2} e^{-\frac{1}{2}(x^2+1)} \, \mathrm{d}x. $$

2. If we write $y = x^2 + 1$ for simplicity, we now that the integrand takes the form $\frac{1}{y^2} e^{-y/2}$. Both the factor $\frac{1}{y^2}$ and the exponential $e^{-y/2}$ suggest that we might possibly use the trick:

$$ \frac{1}{y^2} = \int_{0}^{\infty} s e^{-ys} \, \mathrm{d}s. $$

Plugging this to the integral $I$ above and interchanging the order of integration, we indeed confirm that the integral now assumes a much familiar form:

$$ I
= 2 \int_{0}^{\infty} \int_{0}^{\infty} s e^{-(x^2+1)(s+\frac{1}{2})} \, \mathrm{d}x\mathrm{d}s
= \sqrt{\pi} \int_{0}^{\infty} \frac{s}{(s+\frac{1}{2})^{1/2}} e^{-(s+\frac{1}{2})} \, \mathrm{d}s. $$

Although we can push this direction further to find the value of $I$, we will stop at here as it is already covered in another answer.

This trick can be thought as a generalization of integration by parts. As such, it serves as a versatile tool for computing many other tricky integrals. Here is another example of this sort:

Exercise. Using the above trick, show that
$$ \int_{0}^{\infty} \frac{\sin^2 x}{x^2} \, \mathrm{d}x = \frac{\pi}{2}. $$
(Hint: $\int_{0}^{\infty} e^{-sx} \sin^2 x \, \mathrm{d}x = \frac{2}{s(s^2+4)}$ for $s > 0$.)

3. Alternatively, we may ask if we can "cancel out" $\frac{1}{y^2}$ factor, since the integral will be much easier to compute if this factor were not there. This idea can be realized by invoking the Feynman's trick: Let

$$ f(s) := 2 \int_{0}^{\infty} \frac{1}{(x^2+1)^2} e^{-s(x^2+1)} \, \mathrm{d}x. $$

Then $I = f(1/2)$, and using the gaussian integral we check that

$$ f(\infty) = f'(\infty) = 0, \qquad f''(\alpha) = 2 \int_{0}^{\infty} e^{-s(x^2+1)} \, \mathrm{d}x = \sqrt{\frac{\pi}{s}} \, e^{-s}. $$

Then applying the "initial" condition $f'(\infty) = 0$ to $\int \sqrt{\frac{\pi}{s}} \, e^{-s} \, \mathrm{d}s = \pi \operatorname{erf}(\sqrt{s}) + C$ gives

$$ f'(s) = \pi (\operatorname{erf}(\sqrt{s}) - 1). $$

Similarly as before, using

\begin{align*}
\int \pi (\operatorname{erf}(\sqrt{s}) - 1) \, \mathrm{d}s
&= \pi s (\operatorname{erf}(\sqrt{s}) - 1) - \int \sqrt{\pi s} e^{-s} \, \mathrm{d}s \\
&= \pi s (\operatorname{erf}(\sqrt{s}) - 1) + \sqrt{\pi s} e^{-s} - \int \frac{\sqrt{\pi}}{2\sqrt{s}} e^{-s} \, \mathrm{d}s \\
&= \pi s (\operatorname{erf}(\sqrt{s}) - 1) + \sqrt{\pi s} e^{-s} - \frac{\pi}{2} \operatorname{erf}(\sqrt{s}) + C
\end{align*}

and applying the "initial" condition $f(\infty) = 0$,

$$ f(s) = \pi \bigl( s - \tfrac{1}{2} \bigr) (\operatorname{erf}(\sqrt{s}) - 1) + \sqrt{\pi s} e^{-s}. $$

Plugging $s = \frac{1}{2}$ gives the desired answer.

This is another versatile tool for computing many tricky integral. The key idea is to perturb the integrand by introducing an auxiliary parameter, and performing some calculus to the resulting integral regarded as a function of that parameter.

In the following example, both the idea in part 2 and Feynman's trick is utilized:

Exercise. For $a > 0$, consider the integral
$$ f(a) = \int_{0}^{\infty}\int_{0}^{\infty} a \cos(ax)\sin(ay)e^{-a xy} \, \mathrm{d}x\mathrm{d}y. $$
Computing this in two ways to show that
$$ f(a) = \int_{0}^{\infty} \frac{\cos(ax)}{x^2 + 1} \, \mathrm{d}x = \int_{0}^{\infty} \frac{y \sin(ay)}{y^2 + 1} \, \mathrm{d}y. $$
Using this, check that $f'(a) = -f(a)$ and $f(0) = \frac{\pi}{2}$. The unique solution of this equation for $a \geq 0$ is $f(a) = \frac{\pi}{2}e^{-a}$, hence it determines the value of the above integral.


问:
What does it mean to set the gradient of a multilinear function to $0$?
When you have a 1D function $f(x)$, the vector $x$ that satisfies $\nabla f(x) = 0$ is a stationary point which is a minimum when $f$ is convex.

Consider a multilinear function (linear in all arguments) that is defined on a compact set, so that it does attain minimum and maximum value.

$$f(x,y) = x(1-y)$$

The gradient of $f$ along $x$ is

$$\nabla_x f(x,y) = (1-y)$$

What does it mean to set $\nabla_x f(x,y) = 0$?

At this condition, we do not recover some $x$ that gives a minimum value to $f(x,y)$, but instead a function of $y$, which satisfies $y = 1$.

What does this condition tell us ?Does it tell us that $f(x,y)$ is minimized at $y = 1$ no matter what $x$? Or is it at the minimizer $x$, $y$ must be equal to $1$

答:
First you need to distinguish between stationary point and minimum. Stationarity is a necessary, but not sufficient, condition for minimum. The condition (or actually definition) of stationarity over a set $U$, is that for $x\in \text{int}(U)$ we have $\nabla f(x)=0$. If the set $U$ is open then every $x\in U$ is in the interior, however for closed convex sets, the condition for stationarity becomes $\nabla f(x)^T(y-x)\geq 0,
\; \forall y\in U$. This means that the directional derivative is nonnegative in all directions at a point $x$.

Second, for convex functions stationarity is also a sufficient condition for optimality. This is the case for $f(x)=x^2$ for example, but not for $f(x)=x^3$ as pointed out in a comment.

Lastly, the function you presented is indeed bilinear, however it is not linear or convex in general ("jointly"). Therefore even if the derivative w.r.t. $x, y$ equals zero, you will not necessarily have a minimum point. In this specific case, you will have a saddle point. What you do know is that in order for the derivative w.r.t. $x$ to equal zero, then we must have $y=1$. This is necessary for stationarity (but again, not sufficient). Notice that by ignoring the $y$ variable, you actually look at your function as it was $f_y(x)=x(1-y)$, where $y$ is just a parameter and not a variable. In that case the derivative will be zero iff the parameter $y=1$ and in fact you'll have a constant function $f(x)=0$. This also highlights the fact that linear functions have a zero derivative iff they are constant.


问:
Calculations with liars and sums of prime factors
Suppose you are given a number $n$ and told that the sum of its prime factors is $s$.  I'm looking for an efficient algorithm that checks the truth of the statement.

Obviously one can simply calculate the factorization of $n$, add, then check if the result is equal to $s,$ but it seems that the problem can be easier.  For example, if $s\approx n/100$ then $n$ must be contain at least one large prime which can be discovered by trial division.  Can this be formalized?  Are there other properties that can be used?

In my immediate case the sum is with multiplicity (so the sum of the prime factors of 1024 is 20) but I would be interested in the case where multiplicity is ignored (giving 2).  Similarly, in the problem at hand I am interested in small n ($n<10^{12}$) but approaches that apply to larger numbers are welcome.

答:
This isn't state of the art, but for your purposes it may be sufficient. You can implement Sieve of Eratosthenes using a dynamic programming approach.

It obviates the need to do any trial division and instead gives you trade off between upfront initialization and, roughly speaking, constant time look-up.

Let $T[i]$ denote the smallest prime, $p$, factor of $i$. To find the total factoring of $i$, start at $T[i]$ and record its value. Then, go to $T[ i / T[i] ]$ and record its value. If $i = T[i]$ terminate, otherwise continue until it does.

Take a second pass over the factoring table to evaluate the sum of factors in a similar way to build a sum of factors look-up table.

This process should take 3-5 secs (depending on your hardware) to build the sum of factors for all n between two and $10^8$.

Downside to this approach is that is not easily distributed or parallelized.

For a more concrete implementation, you can read my write-up of this approach from a couple years ago: http://antimatroid.wordpress.com/2009/04/01/integer-factorization-by-dynamic-programming-with-number-theoretic-applications/


问:
Find a POVM to distinguish the two states
Suppose $|\psi \rangle = |1\rangle$ and $| \phi \rangle = | + \rangle = \frac{1}{\sqrt{2}} (|0 \rangle + |1 \rangle )$. Write a POVM that allows for imperfect distinguishability between two states.

POVM stands for Positive Operator-Values Measures, it's a set of operators $\{E_m\}$ where the probability of measuring result $m$ is $Pr(m)=⟨ψ|E_m|ψ⟩$ and $∑_mE_m=I$. These operators allow for more general forms of measurement, such as distinguishing non-orthogonal states. One can do this if $⟨ψ|E_i|ψ⟩=0$ but $⟨ϕ|Ei|ϕ⟩>0$ but $<1$ for some $E_i$ and vice-versa for $E_j$. This allows one to imperfectly identify if the system is in state $| \psi \rangle$ or $| \phi \rangle$ since there some measurements are impossible (probability of 0).

答:
FYI this is probably better suited on the quantum computing stackexchange. But anyway, you can try taking $E_1 = c_1 |0\rangle \langle 0 |$, $E_2 = c_2 |-\rangle \langle - |$ and $E_3 = I - E_1 - E_2$. For some suitably chosen constants $c_1, c_2 > 0$ this should give you a POVM which satisfies
$$
\langle \psi | E_1 |\psi\rangle = 0 \qquad \langle \phi | E_1 |\phi \rangle >0
$$
and
$$
\langle \psi | E_2 |\psi\rangle > 0 \qquad \langle \phi | E_2 |\phi \rangle =0.
$$


问:
Show that : $\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$
I found it with help of Wolfram alpha .

$$\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$$

This integral admits an antiderivative (I can add it if so) so I can conclude that this result is true .I would like to see a proof with the use of complex integration or a 'real' proof . For that I think we can put $y=\frac{2}{x^2+1}$ and use integration by parts (perhaps?).For the proof with complex integration(if it exists I'm new for it )  can you detail the solution ?

Thanks!

答:
Here's a brute-force method:
$$
\begin{aligned}
 \int{\arctan}\left( \frac{2}{x^2+1} \right) \,dx&=\int{\mathrm{arccot}}\left( \frac{1}{2}+\frac{x^2}{2} \right) \,dx\\
 &=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) +\int{\frac{x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx\\
 &=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) -\frac{1}{2}\int{\frac{\sqrt{5}-x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx+\frac{1}{2}\int{\frac{\sqrt{5}+x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx\\
&=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) \\&+\frac{\displaystyle \int{\frac{\sqrt{2\left( -1+\sqrt{5} \right)}+2x}{-\sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x-x^2}}\,dx}{\sqrt{2\left( -1+\sqrt{5} \right)}}+\frac{\displaystyle \int{\frac{\sqrt{2\left( -1+\sqrt{5} \right)}-2x}{-\sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x-x^2}}\,dx}{\sqrt{2\left( -1+\sqrt{5} \right)}}\\&+\int{\frac{1}{\sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2}}\,dx+\int{\frac{1}{\sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2}}\,dx\\
&=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) \\&+\frac{\log \left( \sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2 \right)}{\sqrt{2\left( -1+\sqrt{5} \right)}}-\frac{\log \left( \sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2 \right)}{\sqrt{2\left( -1+\sqrt{5} \right)}}\\
&+\sqrt{\frac{2}{\sqrt{5}+1}} \arctan\left(\frac{2 x+\sqrt{2 \left(\sqrt{5}-1\right)}}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)\\&-\sqrt{\frac{2}{\sqrt{5}+1}} \arctan \left(\frac{\sqrt{2 \left(\sqrt{5}-1\right)}-2 x}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)
\\&\overset{\text{def}}{=} F(x)
\end{aligned}
$$
One can easily see that
$$
\underset{x\to \infty }{\text{lim}}x \text{ arccot}\left(\frac{x^2}{2}+\frac{1}{2}\right)=0
$$
and
$$
\begin{aligned}
&\lim_{x\to\infty} \frac{\log \left(x^2-\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}}-\frac{\log \left(x^2+\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}} \\&=\lim_{x\to\infty}\frac{\log \left(\frac{x^2-\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}}{x^2+\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}}\\
&=0
\end{aligned}
$$
Also it is easy to see that
$$
\underset{x\to \infty }{\text{lim}}\arctan\left(\frac{\sqrt{2 \left(\sqrt{5}-1\right)}-2 x}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)=-\dfrac{\pi}{2}
$$
and
$$
\underset{x\to \infty }{\text{lim}}\arctan\left(\frac{2 x+\sqrt{2 \left(\sqrt{5}-1\right)}}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right) = \dfrac{\pi}2
$$
Thus, combine the above results, from Newton-Leibniz formula, we get
$$
\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx= F(\infty) - F(0)=\sqrt{\frac{2}{\sqrt{5}+1}} \pi = \color{blue}{\pi\sqrt{\phi^{-1}}}
$$
As desired.


答:
Partial answer: When $y > 0$, we have that $\arctan (1/y)= \frac{\pi}{2} - \arctan y$, so the integral becomes
$$
\int_0^{+\infty} \frac{\pi}{2} - \arctan\left(\frac{1+x^2}{2}\right) dx
$$

Using integration by parts, we get to
$$
\int_{-\infty}^{+\infty} \dfrac{2 x^2}{4+(x^2+1)^2} dx
$$

This last integral can be computed by elementary means or complex integration.

Title: Show that : $\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$

Body:
I found it with help of Wolfram alpha .

$$\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$$

This integral admits an antiderivative (I can add it if so) so I can conclude that this result is true .I would like to see a proof with the use of complex integration or a 'real' proof . For that I think we can put $y=\frac{2}{x^2+1}$ and use integration by parts (perhaps?).For the proof with complex integration(if it exists I'm new for it )  can you detail the solution ?

Thanks!


问:
Given, (F,+,•) and (F,+,×) are fields; then • and × are same binary operations
(While reading about classification of simple groups, I was wondering if there could be something like simple fields or something... And this thought came in mind)

When we talk about field in abstract algebra of linear algebra,

It really feels like the second binary operation-multiplication is very much dependent the first binary operation-addition.

If there is no counter example to it then it implies that study of field is totally dependent on the group theory.

Also every finite dimensional vector space over $F$ is isomorphic to $F^k$.

答:
Since $(\mathbb R,+)$ and $(\mathbb C,+)$ are isomorphic groups (assuming the axiom of choice), there is a binary operation $\times$ on $\mathbb R$ such that $(\mathbb R ,+,\times)$ is isomorphic to the field of complex numbers. Therefore $\times$ is not the same binary operation as real multiplication.

The real numbers and the complex numbers have isomorphic additive groups because they are both vector soaces of (algebraic, i.e., Hamel) dimension $2^{\aleph_0}$ over the field of rational numbers.


问:
Existence of $x_0$ s.t. $\frac{f''( x_0 )}{2}=\frac{f( a )}{( a-b ) ( a-c )}+\frac{f( b )}{( b-a ) ( b-c )}+\frac{f( c )}{( c-a ) ( c-b )}$

Let $f$ be continuous on $[a,b]$, and $f''(x)$ exists on $(a,b)$. Prove that $\forall c\in(a,b), \exists x_0\in(a,b)$, s.t.
$$
\frac{f''\left( x_0 \right)}{2}=\frac{f\left( a \right)}{\left( a-b \right) \left( a-c \right)}+\frac{f\left( b \right)}{\left( b-a \right) \left( b-c \right)}+\frac{f\left( c \right)}{\left( c-a \right) \left( c-b \right)}
$$

My Approach: Assume $$\dfrac{K}{2} = \frac{f\left( a \right)}{\left( a-b \right) \left( a-c \right)}+\frac{f\left( b \right)}{\left( b-a \right) \left( b-c \right)}+\frac{f\left( c \right)}{\left( c-a \right) \left( c-b \right)}$$
Let
$$
g\left( x \right) =\frac{K}{2}\left( \left( a-x \right) \left( x-c \right) \left( c-a \right) \right) +f\left( a \right) \left( x-c \right) +f\left( x \right) \left( c-a \right) +f\left( c \right) \left( a-x \right)
$$
Then $g(a)=g(c)=g(b)=0$.
Since $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exist $x_1\in(a,c),x_2\in(c,b)$ s.t. $g'(x_1)=g'(x_2)=0$.
And since $g'$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $x\in(x_1,x_2)\subset (a,b)$, s.t. $g''(x_0)=0$.

From
$$
g''\left( x_0 \right) =\frac{K}{2}\cdot 2\left( a-c \right) +f''\left( x_0 \right) \left( c-a \right)
$$
We have $K=f''(x_0)$.

My Question is, is the proof right? And can we generalize the original claim(that is, change $f''(x)$ to $f^{(k)}(x)$)?

Other proofs of the original claim is also welcomed.

答:
Your proof is (now) correct. This is related to polynomial interpolation:
$$
 p(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)}f(a) + \frac{(x-a)(x-c)}{(b-a)(b-c)}f(b) + \frac{(x-a)(x-b)}{(c-a)(c-b)}f(c)
$$
is the Lagrange polynomial for the data points $(a, f(a))$, $(b, f(b))$, $(c, f(c))$, that is the (unique) second-degree polynomial with the property
$$
 p(x) = f(x) \text{ for } x= a, b, c \, .
$$
Repeated application of Rolle's theorem shows that $(p-f)''(x_0) = 0$ for some $x_0$, i.e.
$$
 \frac 12 f''(x_0) = \frac 12 p''(x_0) = \frac{f(a)}{(a-b)(a-c)} + \frac{f(b)}{(b-a)(b-c)} + \frac{f(c)}{(c-a)(c-b)} \, .
$$

Your function $g$ is essentially $f - p$ (up to some constant factor), so the above reasoning is equivalent to your proof. Writing it as an interpolation problem has the advantage that the generalization  to higher derivative becomes more apparent:

If $I \subset \Bbb R$ is an interval, $a_0, \ldots, a_n \in I$ are pairwise distinct, and $f: I \to \Bbb R$ is $n$-times differentiable, then
$$
\frac 1{n!} f^{(n)}(x_0) = \sum_{k=0}^n \frac{f(a_k)}{\prod_{j \ne k} (a_k-a_j)}
$$
for some $x_0 \in I$.

The proof is the same: We define $p$ as the Lagrange interpolation polynomial which interpolates $f$ at the $n+1$ points $x=a_0, \ldots, a_n$. $p$ has degree (at most) $n$ and $(p-f)^{(n)}(x_0) = 0$ for some $x_0 \in I$.


问:
linear transformations with f transpose
I am half confused with the following problem..

and

Then is equal to

and i have 4 choices:

I don't understand when it says Nu (f∘f), the nucleus of a linear transformation is not zero? I am quite confused. My idea was to get image and nucleus of f and then the associated matrix transposed. but I'm clearly confused.

答:
You have$$(f\circ f)(x_1,x_2,x_3,x_4)=(-2x_2-x_3-x_4,2x_2,x_3+x_4,x_3+x_4).$$So,$$\ker(f\circ f)=\{(a,0,b,-b)\mid a,b\in\Bbb R\}.$$For any element $(a,0,b,-b)$ of $\ker(f\circ f)$, you have $f(a,0,b,-b)=(a-b,0,a-b,-a+b)$, which belongs to $\Bbb H$ if and only if $a-b=-a+b$, which means that $a=b$.
So$$\ker(f\circ f)\cap f^{-1}(\Bbb H)=\{(a,0,a,-a)\mid a\in\Bbb R\}.$$

Title: linear transformations with f transpose


答:
$$I=\int_{0}^{\infty} \tan^{-1}\frac{2}{x^2+1} dx$$
Let us do it by parts taking 1 as second function:
$$I=x\tan^{-1}\frac{2}{1+x^2}|_{0}^{\infty}+4\int_{0}^{\infty}\frac{x^2}{x^4+2x^2+5}dx$$
$$\implies I=4\int_{0}^{\infty} \frac{dx}{x^2+5/x^2+2}$$ $$=2\int_{0}^{\infty} \left(\frac{(1+\sqrt{5}/x^2)}{(x-\sqrt{5}/x)^2+2+2\sqrt{5}}+\frac{(1-\sqrt{5}/x^2)}{(x+\sqrt{5}/x)^2+2-2\sqrt{5}}\right)dx$$
Let $x-\sqrt{5}/x=u$ and $x+\sqrt{5}/x=v$, then the limits in the second integral being identical it vanishes, we get
$$I=2\int_{-\infty}^{\infty} \frac{du}{u^2+2+2\sqrt{5}}=\frac{2}{\sqrt{2+2\sqrt{5}}} \tan^{-1}\frac{u}{\sqrt{2+2\sqrt{5}}}|_{-\infty}^{\infty}=\pi\sqrt{\frac{2}{1+\sqrt{5}}}=\pi \phi^{-1/2}.$$


问:
Set Notation (Axiom of Replacement)
This question is related to the one I asked yesterday here in that it's related to another one of the Zermelo-Fraenkel Axioms. After looking over the notation used to describe the axiom, that is:

$$ \forall \space x \space \forall \space y \space \forall \space z \space [\varphi (x,y,p) \wedge \varphi(x,z,p) \Rightarrow y = z] \Rightarrow \forall \space X \space \exists \space Y \space \forall \space y \space [y \in Y \equiv (\exists \space x \in X) \varphi(x, y, p) ]  $$

I believe I understand most of it, but I'm unsure of why we need to involve the variable z, so I thought I'd just write how I'm interpreting this and have someone correct me where it starts to get fuzzy.

Current Interpretation: For all the elements of the three sets $X, Y, Z,$ if the property $\varphi$ holds under some parameter $p$ for $x, y $ and $x, z$ conjointly implies that $y$ equals $z$ then for any set X there exists a set Y such that for any element of $Y$ there exists an element of X such that property $\varphi$ holds under both the element of $Y$ and the chosen element of $X$ for that property $p$.

What I'm confused about is the purpose of the extra parameter $p$ and the set $Z$ why couldn't you just say something like this:

$$ \forall \space x \space \forall \space y \space \varphi (x,y) \Rightarrow \forall \space X \space \exists \space Y \space \forall \space y \space [y \in Y \equiv (\exists \space x \in X) \varphi(x, y, p) ]  $$

What am I missing here? Also if someone could clear up my interpretation that would be awesome.

答:
Use $x=x$ for $\varphi(x,y)$. Then your version of the axiom gives us a universal set $Y$: every $y$ is in $Y$.  So it leads to an inconsistent theory.

The part that says that $\varphi(x,y,p) \land \varphi(x,z,p)$ implies $y=z$ says that the relation $\varphi$ is "function-like." It is a very strong set construction principle, but does not (one hopes) lead to inconsistency.

Title: Set Notation (Axiom of Replacement)


问:
Is it true that every element of $V \otimes W$ is a simple tensor $v \otimes w$?
I know that every vector in a tensor product $V \otimes W$ is a sum of simple tensors $v \otimes w$ with $v \in V$ and $w \in W$. In other words, any $u \in V \otimes W$ can be expressed in the form$$u = \sum_{i=1}^r v_i \otimes w_i$$for some vectors $v_i \in V$ and $w_i \in W$. This follows from the proof of the existence of $V \otimes W$, where one shows that $V \otimes W$ is spanned by the simple tensors $v \otimes w$; the assertion now follows from the fact that, in forming linear combinations, the scales can be absorbed in the vectors: $c(v \otimes w) = (cv) \otimes w = v\otimes (cw)$.

My question is, is it true in general that every element of $V \otimes W$ is a simple tensor $v \otimes w$?

答:
As has been pointed out, $V\otimes W$ is isomorphic to $Hom(V, W)$ once you pick a basis, with simple elements going to rank one matrices. Since (over any coefficients) as soon as $V$ and $W$ both have dimension at least 2 there is always a linear map from $V$ to $W$ with at least 2D image, we know that not every matrix is rank 1 and so not every tensor in $V\otimes W$ is simple.

I want to note that there is another proof that works for real or complex coefficients, or coefficients in a finite field. It works simply by comparing dimensions.

In the real and complex cases, the canonical map $V\times W/scalars \to V\otimes W$ is not linear, but it is smooth. And there is no surjective smooth map from a smaller dimensional space to a larger dimensional space, by say, Sard's theorem. This means there are non-simple elements in $V\otimes W$.

If $V$ and $W$  are  real vector spaces the dimensions or direct sum and tensor product are  $\dim V+\dim W$ and $\dim V\dim W$, respectively. (For complex vector spaces the real dimensions are $2\dim V+2\dim W$ and $2(\dim V\dim W)$.) So as long as $\dim V+\dim W-1< \dim V \dim W$ i.e. $(1-\dim V)(1-\dim W)<0$ i.e. one of $V$ and $W$ is at least 2-dimensional, we are done.

For coefficients in finite fields, we can see that there is no surjective map from a smaller dimensional space to a larger dimensional space simply by counting elements, and then the proof is the same.


问:
Determinants after Linear Algebra Done Right
I realize that Axler's Linear Algebra Done Right is a great textbook, and the removal of determinants is often pedagogically sound. However, I was doing some problems on a practice entrance exam, and I was surprised by the amount of times solutions would involve having to apply determinants to geometric situations.

Therefore, my question is this:
Where I can learn more about determinants, preferably without having to relearn all of linear algebra from another textbook?

答:
For an easy introduction to determinants there are a lot of online resources. See for example

Very basic intro to determinant

Another intro to determinant with graphical representation

For a very nice graphic representation have a look at this youtube video.

Hope this helps!


问:
Reference book for problem solving (Multivariable Calculus).
I am preparing for an entrance exam so I'm looking for a book that discusses and covers mainly techniques and methods to solve different kinds and types of problems on limits of multiple variables, multiple integrals etc.

Most of the books that were recommended to me in my undergraduate classes were theory books which were centered around proofs and well... theory.

Please note that I am NOT looking for a theory book (like Tom Apostol) which doesn't cover problem solving intensively. I am familiar with theory (at least most of it, I hope). I know how to go about doing proofs and everything but I'm extremely terrible at problem solving and hence this question. Any help is appreciated. Thanks!

答:
I think you may try once the following book

"Schaum's Outline of Advanced Calculus" by  Robert Wrede & Murray Spiegel

The book contains a large number of exercises with solutions. The essence of the theory is briefly repeated and then it goes to practice. Ideal for exam preparation - English is simple and easy to understand.


问:
Proving $\frac{1} {e^z -1} = \frac{1} {z} -1/2$ + 2z $\sum_{n=1}^{\infty} \frac{1} {z^2 + 4 {\pi}^2 n^2}$
This question is part of my complex analysis assignement and I am at loss of ideas on how should I approach it. So, I am posting it here.

Prove the identity $\frac{1} {e^z -1} = \frac{1} {z}$ -1/2 + 2z $\sum_{n=1}^{\infty} \frac{1} {z^2 + 4 {\pi}^2 n^2}$.

well ,only thing which comes to mind is using $e^z =1+z+z^2/2!+...$ but that doesn't seem to yield anything as RHS IS completely different to it.

I am kinda struck on how should i  manipulate LHS to get close to what is given on RHS.

Please shed some light on this.

答:
$$\frac1z (\frac1{e^z-1}+\frac12-\lim_{N\to \infty}\sum_{n=-N}^N \frac1{z-2i\pi n})$$ is entire and bounded and it vanishes at $+\infty$

The boundedness is shown in two steps: this expression (which is more or less $2i\pi$ periodic) gives that it is bounded near the poles and replacing $\lim_{N\to \infty}\sum_{n=-N}^N \frac1{z-2i\pi n}$ by your series gives that it is bounded everywhere else.


答: Is every real $m\times n$ matrix of the form $u \cdot v^T$ for some vectors $u\in \mathbb R^m$ and $v\in \mathbb R^n$?

Title: Is it true that every element of $V \otimes W$ is a simple tensor $v \otimes w$?

Body:
I know that every vector in a tensor product $V \otimes W$ is a sum of simple tensors $v \otimes w$ with $v \in V$ and $w \in W$. In other words, any $u \in V \otimes W$ can be expressed in the form$$u = \sum_{i=1}^r v_i \otimes w_i$$for some vectors $v_i \in V$ and $w_i \in W$. This follows from the proof of the existence of $V \otimes W$, where one shows that $V \otimes W$ is spanned by the simple tensors $v \otimes w$; the assertion now follows from the fact that, in forming linear combinations, the scales can be absorbed in the vectors: $c(v \otimes w) = (cv) \otimes w = v\otimes (cw)$.

My question is, is it true in general that every element of $V \otimes W$ is a simple tensor $v \otimes w$?


问:
Differentiability on normed spaces
My Advanced Calculus homework have the following problem: define differentiability on normed spaces $E,F$. Show that if $U \subset E$ is open and $f: U \to F$ is differentiable at $x \in U$ then continuity of $f$ at $x$ is equivalent to continuity of the derivative (linear transformation) $f'(x): E \to F$.

Now, the definition of differentiability on normed spaces should be the same as the one on euclidian spaces: $f: U \subset E \to F$ is differentiable at $x \in U$ if there exists a linear transformation $T: E \to F$ such that:
\begin{eqnarray}
f(x+h) = f(x) + T(h) + r(h) \tag{1}\label{1}
\end{eqnarray}
where:
$$\lim_{h\to 0}\frac{r(h)}{||h||} = 0,$$ provided $x+h \in U$.
Now, write $\rho(h) = r(h)/||h||$ for every $h \neq 0$ and set $\rho(0) := 0$. Then, (\ref{1}) becomes:
\begin{eqnarray}
f(x+h) = f(x) + f'(x)(h) + \rho(h)||h|| \tag{2}\label{2}
\end{eqnarray}
where $\lim_{h\to 0}\rho(h)=0$ (I wrote $f'(x)$ instead of $T$). Thus, $\lim_{h\to 0}\rho(h)||h|| = 0$ and we have:
\begin{eqnarray}
f(x+h)-f(x) = f'(x)(h) \tag{3}\label{3}
\end{eqnarray}
Now, if we write $x+h = y$, (\ref{3}) becomes
\begin{eqnarray}
f(y) - f(x) = f'(x)(y)-f'(x)(x) \Rightarrow \lim_{y\to x}f(y) - f(x) = 0 \iff \lim_{y\to x}f'(x)(y)-f'(x)(x) = 0 \tag{4}\label{4}
\end{eqnarray}

Question: Is (\ref{4}) sufficient to prove the second statement of the problem, i.e. that continuity of $f$ at $x$ is equivalent to continuity of the derivative $f'(x): E \to F$?

答:
It is sufficient. Linear operators $T: E \rightarrow F$ are continuous everywhere iff they are continuous in one point $x \in E$: Let $T$ be continuous in $x \in E$. Now let $z \in E$ be arbitrary and $z_n \rightarrow z$ for some sequence $(z_n)_{n \in \mathbb{N}} \subseteq Z$. Then $z_n+x-z \rightarrow x$. So:
$$
\lvert T(z_n) - T(z) \rvert = \lvert T(z_n +x-z) -T(x) \rvert \rightarrow 0
$$
We used linearity and then continuity in $x$.

Your other arguments are correct, too.

Title: Differentiability on normed spaces


问:
check if function one to one or onto
for example  f(x) = x^2
I know if we get same result from different value will be not one to one in other words if the power odd will be one to one

$(-2)^2=4$

$2^2=4$

But how do I check it is onto or not. The co domain not given in the exam

答:
It can be assumed in the context of the level of mathematics you seem to be at that if the codomain is not mentioned, then it is going to be the whole of $\mathbb R$

So, in order to prove a function is onto, you need to show that for all elements of $\mathbb R$, there exists an element of the domain (also $\mathbb R$ in this context) which maps to that element.

Knowing as you do that the square of a number is always non-negative, that tells you that negative numbers are never mapped to by the square function.

The answer follows directly from that.

Title: check if function one to one or onto


问:
Prove that $f(x) =∇f(0)·x$
For f - differentiable function, with the property  $ f(tx)=t f(x)$ for all real t and all $x$ in $\mathbb R^n$, prove that $f(x) =∇f(0)·x$, where $∇f$ is the gradient of $f$.

答:
Let $x$ be fixed .

Then we have $\frac{d}{dt}f(tx)= \frac{d}{dt}(tf(x))$.

Compute these derivatives and then plug in $t=0.$

Title: Prove that $f(x) =∇f(0)·x$


问:
Expected value of drawing bills
There are 2 $\$10$ bills, 2 $\$20$ bills, and 1 $\$50$ bill inside a box. You draw four bills at random. What is the expected value of amount that you have (assuming draws are without replacement)?

My approach: Expected value of bills from the box is $2/5*10 + 2/5*20 + 1/5*50 = 22$
I have a couple of questions at this point-

If draws are with replacement, then the expected value is simply 22*4 by linearity of expectations, am I correct?

How to approach this when the draws are without replacement?

Edit: I have corrected the arithmetic error as pointed out in the comments and the answer.

答:
If draws are with replacement, then the expected value is simply 18*4 by linearity of expectations, am I correct?

The principle of multiplying the expected value for one draw by four is correct.  That's a nice observation of the independence between the draws.  However, be careful with the arithmetic.

Let $\$X$ be the value of the bill drawn.

$$\text{expected value of the bill} = \$E[X] = \$10 \cdot P(X = 10) + \$20 \cdot P(X = 20) + \$50 \cdot P(X = 50) \\
= \$\left(10 \cdot \frac25+ 20 \cdot \frac25 + 50 \cdot \frac15\right) = \$(4 + 8 + 10) = \$22$$

Hence the expected value of drawing four bills with replacement is $\$4E[X] = \$4(22) = \$88$.

How to approach this when the draws are without replacement?

You have five bills in total.  When you draw four bills from them, only one bill is left, so the answer is simple: $\$(10\cdot2+20\cdot2+50\cdot1-E[X]) = \$(110 - 22) = \$88$.

Title: Expected value of drawing bills


答:
Start with a function $f:\mathbb{R}\to\mathbb{R}$ .

One one :

In order to check whether $f$ is one one or not, try to show that for any two points $x$ and $y$ , $$f(x) = f(y) \quad\iff\quad x=y$$ If so, then $f$ is one one. Otherwise not.

Onto :

In order to check whether $f$ is onto or not, you'll have to check if for any $y\in\mathbb{R}$ , there exists at least one $x\in\mathbb{R}$ such that $f(x)=y$ .

If so, then $f$ is onto. Otherwise, not.

Title: check if function one to one or onto

Body:
for example  f(x) = x^2
I know if we get same result from different value will be not one to one in other words if the power odd will be one to one

$(-2)^2=4$

$2^2=4$

But how do I check it is onto or not. The co domain not given in the exam


问:
Meaning of $S^{I}$
I have a measurable space $(S, \Sigma) $ and an index set I.
What does it mean by $S^I$. Can someone explain with the help of an example? I am studying stochastic processes and found this in the definition of Path space.

答: It is the space (or simply the set) of maps $I\to S$. Another way of saying the same thing in the case $I$ is countable is the set of sequences of elements of $S$ indexed by $I$, since any map $I\to S$ determines a sequence given by the images and any sequence indexed by $I$ determines the image of a map $I\to S$.

Title: Meaning of $S^{I}$


问:
Integration of radial functions?
Let $f(|x|)$ be a integrable radial function in $\mathbb{R}^n$ ($|\cdot|$ denotes the euclidean norm as in convention). The following identity is used to simplify computations

$$\int_{\mathbb{R}^n}f(|x|)\mathrm{d}x = \omega_{n-1}\int_0^\infty f(r) r^{n-1} \mathrm{d}r,$$
where $\omega_{n-1}$ denotes the surface area of the $(n-1)$-sphere of radius 1.

But I have never found any full proofs on such topic. Does anyone have some references in mind, better some source which I can actually refer to?

updates

Recently I found this subject has been mentioned by Stein in the appendix from his book "Fourier Analysis", and it further refers to Buck's "Advanced Calculus", Folland's "Advanced Calculus" and Lang's "Undergraduate Analysis".
I have gone through the later two. It seems to me that Lang's treatment is pretty self-contained, though it is given in exercise hence details omitted.

Mentioned by TCL, Evan's book on measure theory is decent but I'm not sure if it's too heavy. It let me feel like shooting mosquito with bazooka.

答:
Spherical coordinates in $\mathbb{R}^n$ is treated fully in Karl Stromberg's book "Introduction to Classical Real Analysis" p. 369-370. Note that on p. 370, it shows the Jacobian of the coordinate transformation as
$$r^{n-1} \prod_{j=1}^{n-2} (\sin\theta_j)^{n-j-1}$$ So if your function is radial, then
\begin{align*}\int_{\mathbb{R}^n}f(|x|)\mathrm{d}x &=\int_{-\pi}^\pi\int_0^\pi\int_0^\pi\cdots\int_0^\infty f(r)r^{n-1}\prod_{j=1}^{n-2} (\sin\theta_j)^{n-j-1}\,dr \, d\theta_1\cdots d\theta_{n-2}\,d\theta_{n-1}\\&= \omega_{n-1}\int_0^\infty f(r) r^{n-1} \mathrm{d}r,\end{align*}
where $$\omega_{n-1}=\int_{-\pi}^\pi\int_0^\pi\int_0^\pi\cdots\int_0^\pi \prod_{j=1}^{n-2} (\sin\theta_j)^{n-j-1}\,d\theta_1\cdots d\theta_{n-2}\,d\theta_{n-1}$$

Stromberg, Karl Robert, An introduction to classical real analysis, Providence, RI: AMS Chelsea Publishing (ISBN 978-1-4704-2544-9/hbk). xiv, 577 p. (2015). ZBL1331.00003.

Title: Integration of radial functions?


问:
Probability and Statistics using chess game board
There are 18,000 adults who belong to a nationwide chess club. Among the 18,000 adults, 8,000
prefer to play with white(W), whereas the rest prefer black(B). Among all members, 11,000 prefer
to play chess on the computer(C), the others prefer the regular board(RB). 3,000 members prefer
to play with white on the computer. Suppose an adult club member is selected at random.

(a) Find the probability that the person prefers to play with black. That is, find P(B).

(b) Find the probability that the person prefers to play with black on the regular board. That
is, find P(B ∩ RB).

(c) Find the probability that the person prefers the black pieces, given that the person prefers
to play on the regular board. That is, find P(B|RB).

(d) Are the two events “the person prefers black” and “the person prefers the regular board”
mutually exclusive? Are they independent? Give supporting calculations.

Can anyone help me solve this question.

答:
Perhaps this can help you get started
$$P(B)=\frac{\mbox{prefer to play black}}{\mbox{total chess club members}}$$
$$P(RB)=\frac{\mbox{prefer the regular board}}{\mbox{total chess club members}}$$
Study conditional probability to figure out the rest. Let me know if you have any questions.


答: By the Weierstrass factorization theorem, $\sin\tfrac{z}{2}=\tfrac{z}{2}\prod_{n\ge1}(1-\tfrac{z^2}{4\pi^2n^2})$. Taking log-derivatives, $\tfrac12\cot\tfrac{z}{2}=\tfrac1z-2z\sum_{n\ge1}\tfrac{1}{4\pi^2n^2-z^2}$. Replacing $z$ with $iz$, $\tfrac12\coth\tfrac{z}{2}=\tfrac1z+2z\sum_{n\ge1}\tfrac{1}{4\pi^2n^2+z^2}$. Subtracting $\tfrac12$ completes the proof.

Title: Proving $\frac{1} {e^z -1} = \frac{1} {z} -1/2$ + 2z $\sum_{n=1}^{\infty} \frac{1} {z^2 + 4 {\pi}^2 n^2}$

Body:
This question is part of my complex analysis assignement and I am at loss of ideas on how should I approach it. So, I am posting it here.

Prove the identity $\frac{1} {e^z -1} = \frac{1} {z}$ -1/2 + 2z $\sum_{n=1}^{\infty} \frac{1} {z^2 + 4 {\pi}^2 n^2}$.

well ,only thing which comes to mind is using $e^z =1+z+z^2/2!+...$ but that doesn't seem to yield anything as RHS IS completely different to it.

I am kinda struck on how should i  manipulate LHS to get close to what is given on RHS.

Please shed some light on this.


答:
$$I=\int_{0}^{\infty} \tan^{-1}\frac{2}{x^2+1} dx$$
Let us do it by parts taking 1 as second function:
$$I=x\tan^{-1}\frac{2}{1+x^2}|_{0}^{\infty}+4\int_{0}^{\infty}\frac{x^2}{x^4+2x^2+5}dx$$
$$\implies I=2\int_{-\infty}^{\infty} \frac{dx}{x^2+5/x^2+2}=2\int_{-\infty}^{\infty} \frac{dx}{(x-\sqrt{5}/x)^2+2+2\sqrt{5}}=2\int_{-\infty}^{\infty} \frac{dx}{x^2+2+2\sqrt{5}}$$ $$=\frac{2}{\sqrt{2+2\sqrt{5}}} \tan^{-1}\frac{x}{\sqrt{2+2\sqrt{5}}}|_{-\infty}^{\infty}=\pi\sqrt{\frac{2}{1+\sqrt{5}}}=\pi \phi^{-1/2}.$$

See the very interesting Glasser's property of integrals:
A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$


问:
Extreme values of a function without calculus
Given a function $f:[1,+\infty)\rightarrow\mathbb{R}$ with $f(x)=\frac{6x^2+x}{x^3+x^2+x+1}$
How do we determine whether the function has a maxima or a minima without using derivatives. It is easy to do it for quadratics since we can derive an inequality with $0$ on one side, however I am stuck when it comes to rational functions.

答: You know that $f(1)=\frac74$ and that $\lim_{x\to\infty}f(x)=0$. So, take $M\in(1,+\infty)$ such that $f(x)<\frac74$ when $x>M$ and then the point $x_0$ of $[1,M]$ at which $f|_{[1,M]}$ attains its maximum is that point of $[1,\infty)$ at which $f$ attains its maximum.


答:
In the simplest language possible: $S^I$ is the set of all functions from $I$ to $S$.

This avoids the possibly confusing term "space".

Title: Meaning of $S^{I}$

Body:
I have a measurable space $(S, \Sigma) $ and an index set I.
What does it mean by $S^I$. Can someone explain with the help of an example? I am studying stochastic processes and found this in the definition of Path space.


问:
Linearity of Conditional Expectation
I have read in several places that $E((\sum_{i=1}^{n} X_{i})|Y) = \sum_{i=1}^{n} E (X_{i}|Y)$, but I cannot seem to find a proof for it other than a rough sketch for the continuous case (Linearity of conditional expectation (proof for n joint random variables)). How will one show this result?

答:
What you want to show is that the mapping $X\mapsto \mathbb{E}[X|Y]$ is linear. To do this remember that the characteristic property of the conditional expectation is
$$\forall A\in\sigma(Y), \mathbb{E}[X\mathbf{1}_A]=\mathbb{E}\left[\mathbb{E}[X|Y]\mathbf{1}_A\right]$$

Now, if $X,X'$ are two random variables and $\lambda,\mu$ are scalars, the random variable $Z=\lambda \mathbb{E}[X|Y]+\mu\mathbb{E}[X'|Y]$ verifies this property due to the linearity of expectations. By unicity of the conditional expectation (up to sets of probability 0) we have $Z=\mathbb{E}[\lambda X + \mu X'|Y]$.

Once you have this you can simply apply well-known linear algebra properties to get the result for $n$ variable (or proceed by induction).

Title: Linearity of Conditional Expectation


答:
A few books comes to mind: (1) Multivariable Mathematics, by Theodore Shifrin has lot's of enlightening examples and exercise on multivariable limits. (2) Basic Multivariable Calculus, by Jerrold E. Marsden has quite a few exercises and examples on limits, integrals.

Some books that i can think right now; dedicated to problem solving is: (1) Solving Problems in Mathematical Analysis, Part II, by Tomasz Radożycki. Part II will specifically serve your interests. (2) Problems in Mathematical Analysis II: Continuity and Differentiation, by Kaczor & Nowak can be a handy reference too. Both (1) and (2) contains full solution to the problems.

Title: Reference book for problem solving (Multivariable Calculus).

Body:
I am preparing for an entrance exam so I'm looking for a book that discusses and covers mainly techniques and methods to solve different kinds and types of problems on limits of multiple variables, multiple integrals etc.

Most of the books that were recommended to me in my undergraduate classes were theory books which were centered around proofs and well... theory.

Please note that I am NOT looking for a theory book (like Tom Apostol) which doesn't cover problem solving intensively. I am familiar with theory (at least most of it, I hope). I know how to go about doing proofs and everything but I'm extremely terrible at problem solving and hence this question. Any help is appreciated. Thanks!


问:
Check if the function $f=x[\frac 1 x], x\ne 0$ is continuous.
Edit: I got carried away and forgot that the function is not defined at $x=0$, which means its not continuous at $x=0$, thanks for all the answers!

My attempt: check if the function is continuous at $x=0$.
I tried to apply the squeeze theorem:

$\frac 1 x -1<[\frac 1 x] \le \frac 1 x$ , Multiply all by $x\ne 0$, and i got:

$1-x<x[\frac 1 x] \le 1$, and this implies that the limit $x\rightarrow 0$ is 1.

and if the limit at $x=0$ exists and equal to 1, then the side limits when $x\rightarrow 0^+$ or $x\rightarrow 0^-$ both exist and equal, which implies that the function is continuos at $x=0$.

I feel like I have made some mistake, since the floor function can be tricky with limits, It would be appreciated if someone can approve if my work is right or wrong and tell me how to approach these kind of questions.
Thanks in advance!

答: Def of continuity is more severe in topology. Let $f: U \rightarrow V$ and put $f(x) = x * \left[\frac{1}{x}\right]$. Any function is said to be continuous at a point $x=a \in U$ if and only if for a neighborhood N of $f(x)$ in V, there exists the one S of the point such $f(S) \subset N$. I presume the idea has been caught. The function is continuous at the point $x_0$ if there exists neighborhood $f^{-1} (N)$ in V. You see that the function must reflect $U$ onto $V$ and vice versa. In the case, if $y(x) = x \cdot \left[\frac{1}{x}\right]$, then $\frac{y}{x} = \left[\frac{1}{x}\right]$. According to definition above, we must find a neighborhood $f^{-1} (N) $ where N is the neighborhood of $f(x)$ in V, so there exists a neighborhood of the point zero $S$ such that $f(S)$ is a subset of N. We see that we cannot find reciprocal in one-to-one manner. That is where you might think about discontinuity of the function.


问:
Equivalent inequality of $\left|\sin^{-1}(x) + \sin^{-1}(y)\right|\le\frac{\pi}{2}$ only in terms of x and y?
I was working with the inequality,
$$\left|\sin^{-1}(x) + \sin^{-1}(y)\right|\le\frac{\pi}{2} , \{|x|,|y|\le1\}.$$

I was trying to find an inequality only in terms of x and y without containing any trigonometric functions. Then after some hit and trial I found following beautiful inequality,
$$\biggl|x|x| + y|y| \biggr|\le 1 , \{|x|,|y|\le1\}.$$
But the problem is I couldn't derive this inequality and couldn't show their equivalence. I know they are equivalent by plotting them in Desmos(Function Plotting Program). Can you derive the proof showing the equivalence of these two inequalities ?

Note : The above inequality is the condition for the following relation to satisy,
$$\sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$$
I had thought this formula works for every value of $|x|,|y|\le1$ but turns out it doesn't. It only works when x and y satisfy above inequality.

Body:
We need $$\cos(\sin^{-1}x+\sin^{-1}y)\ge0$$

$$\implies\sqrt{(1-x^2)(1-y^2)}\ge xy$$

which is obvious $xy<0$

What if $xy=0?$

If $xy>0,$  $$(1-x^2)(1-y^2)\ge x^2y^2\iff 1\ge x^2+y^2$$


问:
Composition of functions that are onto or one-to-one
I found part of my answer here: If g(f(x)) is one-to-one (injective) show f(x) is also one-to-one (given that...); however I wanted to flesh out the last two statements I had in a proposition in my notes.

Proposition:
  Let $f: A \rightarrow B$ and $g: B \rightarrow C$.  Then:

(i) If $g \circ f$ is one-to-one, then $f$ is one-to-one.

(ii) If $g \circ f$ is onto, then $g$ is onto.

Proof:
(i)  Suppose $f(x)=f(y)$ for some $x,y$.

Since $g \circ f$ is one-to-one: $$g\circ f(x) = g\circ f(y) \Rightarrow x=y,\forall x,y \in A.$$

Therefore $f$ must be one-to-one.

(ii)  Since $g \circ f (x)$ is onto, then for every $c \in C$ there exists an $a \in A$ such that $c=g(f(a))$.  Then there exists a $b \in B$ with $b=f(a)$ such that $g(b)=c$.  Thus g is onto.

I wanted to confirm that these proofs are both correct for my peace of mind (as they weren't proven in class).

答: (i) If $f \circ g$ is one-to-one, then it is invertible. Its inverse is $g^{-1} \circ f^{-1}.$ Thus $f^{-1}$ exists. Therefore, $f$ is invertible, and thus is one-to-one. QED.


问:
What is the difference between the words "variables", "constants", "parameters" and "arguments"?
This has been nagging me for a while and then I saw a question on English SE asking what the difference is between variables and parameters.

In the context of mathematics, what is the difference between

variables

constants

parameters

arguments

I get 1. and 2. confused because, for example the general form of the quadratic function is often expressed $f(x)=ax^2+bx+c, a \neq 0$ and $x$ is the variable so what does that make a,b and c? I assume since math is so rigorous with precise definitions, these have one, but it is also a valid answer that as words they do not have a single precise definition.

答:
Variables, parameters and constants roughly differ in their degree of "fixedness".

In the context of equations as in the examples you gave:

A variable, in principle, is an atomic term that can take an arbitrary value.

On can distinguish between independent variables, whose value can be freely chosen within a given domain, and dependent variables, which can in principle take any value but have that value depend on the values of other terms.

A parameter is a special kind of variable whose value is arbitrary but fixed. This means that the value of a parameter is constant for the case currently under consideration but may vary between different contexts.

For instance, the braking distance $s$ is given by $s = f \cdot v^2$, where $v$ is a variable denoting the vehicle's speed and $f$ is a parameter denoting influencing factors such as the constitution of the pavement and the tires. Once this parameter is fixed, it remains constant in the equation for any value of the speed, but the parameter could be set to a different value in a different context, e.g. a different type of car or road.

A constant is a term which represents a specific value that is the same across all contexts.

For instance, $2$ is a constant which always means the number $2$, $\pi$ is a constant which always means $3.14 \ldots$, etc.

An argument is one of the inputs to a function. In a function definition, the arguments are represented by variables; when the function is applied, its arguments can be arbitrary terms, either unknown variables such as $x$ leaving the value partially unknown, atomic constants such as $2$, or complex terms such as $\sqrt{2+5}$.

Since this question has been linked from Variables vs. Constants, how are they defined in ZFC, and how they differ? , the concepts are a bit different in symbolic logic.

In propositional logic, a propositional variable is an atomic proposition which denotes a truth value, assigned by a valuation function.

In first-order logic, an individual variable is an atomic term which denotes an object from the domain, assigned by a variable assignment function. A variable may undergo different assignments and thus take different values within the same structure. A variable is like a pronoun; the referent of "it" depends on which object one is pointing at and which situation one is in.

In first-order logic, an individual constant is an atomic term which denotes an object from the domain, assigned by an interpretation function. The interpretation of a constant is fixed within a structure, but may vary between different structures. A constant is like a proper name; the referent of "John" is always the same person within one situation but may take different referents in different situations.

In formal logic, a parameter is a special kind of constant used in certain kinds of proof systems, such as natural deduction analytic tableaux for first-order logic. There is an unlimited number of so-called parameters available to choose in an instantiation of some quantified statement, which act as names for objects but without but don't have a concrete interpretation.

In first-order logic, an argument is an input to a function or predicate symbol. Arguments can be individual variables, individual constants or complex terms such as $+(4,x)$.

Thus "constants" in first-order logic perhaps come closer to what one would otherwise call a parameter, in the sense that their interpretation is fixed in a given context (structure) but may vary between different contexts, whereas there is no pendant of true constant terms that have the same interpretation across any structure in first-order logic.

And there are yet other kinds of variables, e.g. random variables in statistics.


问:
Find function for multiplication operator such that there are at least 2 eigenvalues
Consider $V=C([a,b])$ and let $A:V\to V: f(t)\mapsto u(t)f(t)$ be the multiplication operator with fixed $u(t)\in V$.

Let $E$ denote the set of eigenvalues of operator $A$.

Let $S$ denote the spectrum of operator $A$. The spectrum is defined as the set of non-regular points; i.e. the points $\mu$ for which the resolvent $(A-\mu\mathbf{1})^{-1}$ is not well-defined or bounded on $V$.

It is clear that $S= \operatorname{im}(u)$, the image of $u$. By choosing $u(t)$ wisely, we can create 3 different situations: 1) $E=\emptyset$, 2) $E = S$ and 3) $\emptyset\ne E \subset S$. My question is regarding this third situation.

A possible example (let's work on $[0,1]$), would be to consider $u(t) = 0$ outside of $[1/3,2/3]$ and have it form a little baseless triangle within $[1/3,2/3]$ (to ensure $u$ is continuous); and then we can choose $f$ to be $0$ inside $[1/3,2/3]$, but non-zero outside of it (but still continuous). Then $u(t)f(t)=0 = 0f(t)$, so this way we have found $E=\{0\}\subset S=[0,1]$ (albeit without explicitly writing $f$).

Now, I want to find an example (just by drawing, not necessarily expressions) of $u(t)$ such that $|E|>1$ but $E\subset S$ strictly. I'm having trouble finding continuous functions for this to work. Any ideas?

Thanks.

答:
You can show that, in general,  $E$ will be the set of $x\in\mathbb{R}$ such that $\operatorname{int}(u^{-1}(x))\neq\varnothing$ (use an argument similar to the one you did with the little triangle). So just take a function whose graph looks like a stair to get more eigenvalues.

It will also follow that $E$ is always countable.


问:
Find the factorization of the polynomial as a product of irreducible on rings R[x] and C[x]
Find the factorization of the polynomial $x^5-x^4+8x^3-8x^2+16x-16$ as a product of irreducible on rings $\Bbb R[x]$ and $\Bbb C[x]$

Testing with the simplest possible root in this case, $P(1) = 0$

Applying the schema of Ruffini

$
\begin{array}{c|lcr}
  & 1 & -1 & 8 & -8 & 16 & -16 \\
1 &  & 1 & 0 & 8 & 0 & 16 \\
\hline
  & 1 & 0 & 8  &0 &16 & 0 \\
\end{array}
$

$P(x)=(x-1).(x^4+8x^2+16)$

So, I need to make it irreducible in  $R[x]$, there's no more roots, but I don't know how to reduce it, I tried factoring:
$x^4+8.(x^2+2).(x-1)$, but I can not keep working it.

What I can do to solve it?

答: Remember $a^2+2ab+b^2=(a+b)^2$. Thus $x^4+8x^2+16=\,?$ Any polynomial $x^2+a$ with the number $a>0$ is irreducible over $\bf R$, but it should be clear how to reduce them over $\bf C$.</p>


问:
contrapositive prove if n and m are even integers then n-m are even
I solved it

n is odd and m is odd then n-m is odd

n= 2k+1

m=2b+1

(2k+1)-(2b+1)

2k+1-2b-1

2k-2b

2(k-b)

2r

How can this be may even. Is my answer wrong?

答:
It is easier to prove this directly. If $n$ and $m$ are even, then $n = 2r$ and $m = 2s$ for $r,s \in \mathbb{Z}$. Then
$$n - m = 2r - 2s = 2(r - s),$$
so $n-m$ is even.

If you insist on proving the contrapositive: if $n-m$ is odd, then $n-m = 2k + 1$ for some $k \in \mathbb{Z}$. Then $n = 2k + m + 1$. If $m$ were even, then $m = 2t$ for some $t \in \mathbb{Z}$, so
$$n = 2k + 2t + 1 = 2(k+t) + 1,$$
so $n$ is odd, so not both $n$ and $m$ are even. If $m$ were odd, the proof is complete already because not both $n$ and $m$ are even.


问:
Euler Charateristic of a surface of genus $g$
Using the definition of Euler charateristic from the theory of intersection numbers that is done in Hirsch's Differential Topology , I am trying to see that $\chi(G)=2-2g$, where $G$ is a closed surface of genus
$g$. Now my idea for this was to go by induction on $g$, and the case where $g=0$ it's true since we have that $\chi(S^2)=2$. Now I am having some trouble it the induction hypothesis. I am trying to use the following result

Suppose a compact $n$-manifold can be expressed by $A\cup B$ where $A$ and $B$ are compact $n-$dimensional submanfiolds and $A\cap B$ is a $(n-1)$-dimensional submanifold, then $\chi(A\cup B)=\chi(A)+\chi(B)-\chi(A\cap B)$.

I have tried using this but I can't seem to find what to use for the sets $A$ and $B$. Any help with this is aprecciated. Thanks in advance.

答:
We will first recall facts Hirch proves in the chapter.

$$ \chi\left(S^{n}\right)=1+(-1)^{n}=\left\{\begin{array}{ll} 2 &
\text { if }  n \text { is even } \\ 0 & \text { if }  n \text {
is odd. } \end{array}\right. $$

Let $M$ and $N$ be compact oriented manifolds without boundaries.
Then $\chi(M \times N)=\chi(M) \chi(N)$

So we can compute
$$\chi (S^1)=0$$
$$\chi(S^2)=2$$
$$\chi(T)=\chi(S^1 \times S^1)=0$$
We also need to compute $\chi(D)$. We have that $S^2=D_S \cup D_N$ where $D_N$ is the northern hemisphere and $D_S$ is the southern hemisphere. Thus by your formula we have
$$2=\chi(S^2)=\chi(D_S)+\chi(D_N)-\chi(S^1)=2\chi(D),$$
hence $\chi(D)=1$.

Let $S_g$ be the surface of genus $g$. We have $S_g=D\cup S_g\backslash \mathring{D}$ so $$\chi(S_g)=\chi(D)+\chi(S_g\backslash \mathring{D})+\chi(S^1).$$
The equation above implies that $\chi(S_g\backslash \mathring{D})=\chi(S_g)-1$.

Those computations set the stage for an induction arguments. We write $S_g=S_{g-1}\backslash \mathring{D}\cup T \backslash \mathring{D}$, plugging this to the formula we get
$$\chi(S_g)=\chi(S_{g-1}\backslash \mathring{D})+\chi(T \backslash \mathring{D})-\chi(S^1).$$
Useing the induction hypothisis we get
$$\chi(S_g)=(2-2(g-1)-1)+(-1)-0=2-2g+2-2=2-2g$$.
Note that we have allready shown the base case $T=S_1$ so this completes the proof.


问:
Probability that station select a different radio frequencies
Question: Two synchronous stations use frequency-hopping to select at random (with probability 1/n one of n frequencies. What is the probability that the stations select a different frequency?

My approach to solving this would be that the probability to select same frequency would be 1/n, so it for the probability of selecting a different frequency would be 1 - 1/n

Is my understanding correct? Or would it be n/(nC2) for same frequency?

答:
Yes, correct.

The probability that the 2 Stations select the same station is

$$n\frac{1}{n}\frac{1}{n}=\frac{1}{n}$$

thus the probability that the two stations select a different frequency is the complementary:

$$1-\frac{1}{n}$$


问:
$A$ and $B$ are positive self-adjoint matrix such that $AB$ is self-adjoint then $AB$ is positive
Question: $A$ and $B$ are positive self-adjoint matrix such that $AB$ is self-adjoint then prove that $AB$ is positive.

My first try: Here I can see that the same question was there but it has a little amount of typo. What I have to prove is that $(x, ABx)>0$ for all $x>0$.

Now, $(x,ABx)=(Ax,Bx)=(BAx,x)$ now we also have that $AB=BA$ because $AB=(AB)^*=B^*A^*=BA$ because $A,B,AB$ are all self-adjoint and we can also see that $(BAx,Ax)>0$ but I can't get rid of $A$ in the 2nd co-ordinate.

My second try: Then I tried using this result that "if $A$ and $B$ are positive self-adjoint matrix then there exist a basis $x_1,\cdots, x_n$ of $X$ satisfies an eqn of the form $$Ax_j=\lambda_j Bx_j.$$
where $\lambda_j$ is real."
Now $A,B$ are positive $\Rightarrow \lambda_j>0$ using Generalised Raleigh quotient then $$(x_j,ABx_j)=(Ax_j,Bx_j)=(\lambda_j Bx_j,Bx_j)=\lambda_j||Bx_j||^2>0.$$
Hence $AB$ is positive.
Here I am confused that what is the place where I used $AB$ is self-adjoint! Please help. You can also help me with different proofs.

答:
We can use a nice trick here given that both $A$ and $B$ are positive. As $A$ is positive, this implies that $A$ has a unique positive square root, $A^{\frac{1}{2}}$, which being positive has an inverse $A^{-\frac{1}{2}}$. Thus,
$$AB = A^{\frac{1}{2}}A^{\frac{1}{2}}BA^{\frac{1}{2}}A^{-\frac{1}{2}}$$
So we can see that $A^{\frac{1}{2}}BA^{\frac{1}{2}}$ is similar to AB. Thus showing $A^{\frac{1}{2}}BA^{\frac{1}{2}}$ is positive implies that $AB$ is positive.

For self-adjoint: $\left(A^{\frac{1}{2}}BA^{\frac{1}{2}}\right)^*=\left(A^{\frac{1}{2}}\right)^*B^*\left(A^{\frac{1}{2}}\right)^* = A^{\frac{1}{2}}BA^{\frac{1}{2}}$ because both $A^{\frac{1}{2}}$ and $B$ are self-adjoint.

For positive: For any $v\in V\setminus \{0\}$ as $A^{\frac{1}{2}}$ is invertible we have that $A^{\frac{1}{2}}v \neq 0$ and thus $\langle A^{\frac{1}{2}}BA^{\frac{1}{2}}v,v\rangle = \langle BA^{\frac{1}{2}}v, A^{\frac{1}{2}}v\rangle>0$ as $A^{\frac{1}{2}}$ is self-adjoint, and $B$ is positive.

Therefore $A^{\frac{1}{2}}BA^{\frac{1}{2}}$ is positive implying that $AB$ is positive as desired. In general when proving results given a positive operator it may be easier to prove the result for a similar operator using the above trick.


答:
Remember the form of a contrapositive; you want to prove something like
$$(P\land Q)\implies V,$$
the contrapositive of that is $\neg V\implies(\neg P\lor\neg Q )$.

With this you can now make a good contrapositive.

 Now, with the contrapositive, you have to prove that (n-m) is odd, then n is odd or m is odd

Title: contrapositive prove if n and m are even integers then n-m are even

Body:
I solved it

n is odd and m is odd then n-m is odd

n= 2k+1

m=2b+1

(2k+1)-(2b+1)

2k+1-2b-1

2k-2b

2(k-b)

2r

How can this be may even. Is my answer wrong?


问:
How to find the infimum and supremum of the set $\{x>0:3x^2-8x-3\le0\}$

Let $S=\{x>0:3x^2-8x-3\le0\}$. Find the supremum and infimum.

I want to solve this question, although I have no idea how to do it, I think I should guess the $\sup S$ and $\inf S$ first and try to prove it. I have solve the equation $3x^2-8x-3=0$ and get $x=3$. Then I found that if we have an equation $3x^2-8x-3=p$, for some $p\le0$, then we have $$x=\frac{8\pm \sqrt{64+12(3+p)}}{6}$$ and $$\frac{8\pm \sqrt{64+12(3+p)}}{6}\le3 \iff p\le0$$ Hence, I think I have proven that $3$ is a upper bound of $S$.

I know if I want to show that $3=\sup S$, I need to show that for any $\varepsilon>0$, there exists a number $t$ in $S$ such that $t>3-\varepsilon$, but I don't know how to prove that. And I have no idea how to find a lower bound of $S$ and $\inf S.$

答:
Rewrite the inequality as follows:

$$3x^2-8x-3 \le 0 \land  x>0 $$

$$\iff(3x+1)(x-3) \le 0 \land x > 0$$

$$\iff -\frac13 \le x \le 3 \land x > 0$$

$$\iff 0 < x \le 3$$

and it should be clear that the supermum is indeed $3$ and the infimum is $0$.


问:
Maximal solution of $x'+f\sin x - g\cos x + g =0$
Given an open interval $I$ and two smooth functions $f,g \colon I \to \mathbb{R}$, I would like to study the non-linear differential equation in $x \colon I \to \mathbb{R}$ defined by

\begin{equation}
x'(t)+f(t) \sin(x(t)) - g(t)\cos(x(t)) + g(t) =0.
\end{equation}

What can one say about the maximal domain of existence of the corresponding initial value problem? In particular, is there a simple condition on $f$ and $g$ guaranteeing global solvability – i.e., that the solution is defined on the entire $I$?

答:
A simple condition guaranteeing global solvability of our ODE is that $f$ and $g$ are bounded.

Let $F \colon I \times \mathbb{R} \to \mathbb{R}$ be defined by:
\begin{equation*}
F(t,x) = g(t)\cos(x) -f(t)\sin(x) - g(t).
\end{equation*}
We show that, if $f$ and $g$ are bounded, then $F$ is globally Lipschitz, i.e., for all $t\in I$ and all $x,y \in \mathbb{R}$, there exists $K>0$ such that:
\begin{equation*}
\left\lvert F(t,x)-F(t,y)\right\rvert \leq K \left\lvert x-y \right\rvert.
\end{equation*}
Computing
\begin{align*}
\left\lvert F(t,x)-F(t,y)\right\rvert &= \left\lvert g(t)\left(\cos(x)-\cos(y)\right) +f(t)\left(\sin(y)-\sin(x)\right) \right\rvert\\
&\leq \left\lvert g(t)\left(\cos(x)-\cos(y)\right)\right\rvert + \left\lvert f(t)\left(\sin(y)-\sin(x)\right) \right\rvert,
\end{align*}
we observe that, if $f$ and $g$ are bounded by $M$ and $N$, respectively, then
\begin{equation*}
\left\lvert F(t,x)-F(t,y)\right\rvert \leq 2(M+N).
\end{equation*}
Hence $F$ is globally Lipschitz, as desired.


问:
Proof by induction: $2 ^ {n - 1} \geq n$
I am trying to make a proof by induction but something is quite not adding up. The inequality to prove is like so.

Prove that:
$$ 2 ^ { n-1} \geq n $$ for $n \geq 0$

My inductive step is like below:

Suppose it works for $n=k$ what about $n = k + 1$
$$ 2^{k-1} \geq k$$
$$ 2^{k-1} + 1 \geq k + 1$$
So we just need to prove that $2^{k-1} \geq 1$ for $k\geq0$

The problem is that when $k=0$, the equation does not hold, i.e. $2^{-1} \geq 1$ is false.
But the inequality to prove works for $k=0$ i.e. $2^{-1}\geq 0$ is true. How can I reconcile this?

答:
Your inductive step works for $k\ge1$, so your base case must hold for $n=1$ (or more), which it does ($2^{1-1}\ge1$). Taking $n=0$ for the base case will not work.

Hence your inductive proof works only for $n\ge1$. But adding a complementary proof for $n=0$ (i.e. $2^{0-1}\ge 0$) is really nothing.


问:
Quantifier free L-rings sentence
So my question is as follows:
Let $\phi$ be the $L_{\text{rings}}$-sentence
$$\exists x(x^2+1=0\ \wedge x+1\neq0)$$
Give a quantifier-free $L_{\text{rings}}$-sentence $\psi$ which is $T_{\text{acf}}$ equivalent to  $\phi$, where $T_{\text{acf}}$ the theory of algebraically closed fields.
I don't even know how to start

答:
Think semantically. When $\varphi,\psi$ are sentences and $T$ is a theory, $\varphi$ is $T$-equivalent to $\psi$ iff the models of $T$ satisfying $\varphi$ are exactly the models of $T$ satisfying $\psi$. So if you want to find a "nice" $\psi$ which is $T$-equivalent to a given $\varphi$, the first step is to understand:

What are the models of $T$ which satisfy $\varphi$?

(More snappily, what are the models of $T\cup\{\varphi\}$?)

So in this particular case, what are the algebraically closed fields (= models of $T_{acl}$) where $-1$ has a square root other than $-1$ (= where $\exists x(x^2+1=0\wedge x+1\not=0)$)? Keep in mind that since we're looking at algebraically closed fields, we are guaranteed to have at least one square root of $-1$.

 Suppose $F$ is an algebraically closed field. Since we know that $-1$ has a square root, as long as $(-1)^2\not=-1$ we know $F$ satisfies the sentence in question. Conversely, it's not hard to show that if $(-1)^2=-1$ then $-1$ is in fact the only square root of $-1$ in $F$. So what fields satisfy $(-1)^2\not=-1$? Can you see a way to write down a quantifier-free first-order sentence which characterizes these?

Title: Quantifier free L-rings sentence


问:
Raising the floor function to a power
I've made a few plots and noticed that $\lfloor

Is it true that for positive $x > 1$ and $n \in \mathbb N,\quad n>=2$ the following holds:

$$(\lfloor x \rfloor + 1)^n >= \lfloor x^n \rfloor $$

If it is, how can it be proven?
If it is not, will that at least hold when $n=2$?
I am interested in the latter case, actually.

Thank you!

答: $\lfloor x\rfloor +1 \gt x$, so $(\lfloor x\rfloor +1)^n\gt x^n$.  On the other hand, $\lfloor x^n\rfloor\leq x^n$.</p>

Title: Raising the floor function to a power


问:
How many possibilites are there to draw n out of 4 balls (order matters, with putting back), if 2balls ogether are only allowed to occur n\2 times?
Given is a set S = {A, C, G, T}. We are interestend in the numbers of elements in the biggest subset from S^n (n natural number), which fulfills the following restriction: The amount of "G" and "C" together occuring in an Sequence is n\2 for even n, and (n-2)/2 for uneven n.

Example 1:
n = 6, even. {C,G} has to occur 3 times.
2 of the acceptet sequences are: (C,A,G,C,T,A) und (G,G,G,A,A,A)
2 of the NOT acceptet sequences are: (A,G,T,A,T,C) und (C,C,C,C,A,A)

Beispiel2
n = 3, not even, {C,G} has to occur 1 time.
2 of the acceptet sequences are: (C,A,T),(A,A,G)
2 of the NOT acceptet sequences are: (C,G,T),(T,A,A)

答:
Almost half of them will be filled from ${C,G}$.
Specifically in a sequence of $n$ elements,
$\lfloor{n \over 2}\rfloor$  will be filled from ${C,G}$ and
$\lceil{n \over 2}\rceil$ will be filled from ${A,T}$.
Since we should choose $\lfloor{n \over 2}\rfloor$ positions
from $n$ positions to be filled from ${C,G}$ we get

$$
S_n = {n \choose \lfloor{n \over 2}\rfloor}2^{\lfloor{n \over 2}\rfloor}2^{\lceil{n \over 2}\rceil} = {n \choose \lfloor{n \over 2}\rfloor}2^n
$$

(Hope, I understood your question)


问:
For primes $p,q$, what are the solutions to $x^2=(p+q)x\pmod{pq}$
For primes $p,q$, what are the solutions to $$x^2=(p+q)x\pmod{pq}$$

答:
Hint:

Use  the Chinese remainder theorem: for $p\ne q$,
$$\mathbf Z/pq\mathbf Z\simeq\mathbf Z/p\mathbf Z\times \mathbf Z/q\mathbf Z$$
so the given congruence is equivalent to the system of congruences
$$\begin{cases}
x^2\equiv qx\mod p \\
x^2\equiv px\mod q
\end{cases}\iff\begin{cases}
x\equiv 0 &\text{or }x\equiv q\mod p \\
x\equiv 0 &\text{or } x\equiv p \mod q
\end{cases}$$
Next, you can use the inverse isomorphism of the Chinese remainder theorem, which relies on a Bézout's relation $up+vq=1$:
\begin{align}
\mathbf Z/p\mathbf Z\times \mathbf Z/q\mathbf Z &\longrightarrow \mathbf Z/pq\mathbf Z \\
(x\bmod p,y\bmod q)&\longmapsto yup+xvq\bmod pq
\end{align}


答:
We can rewrite the modular equivalence as $$x^2-(p+q)x+kpq=0$$for some $k\in\mathbb Z$.

The discriminant of this quadratic is $(p+q)^2-4kpq$. To check whether there exists a square root of the discrimnant, we have to evaluate ${(p+q)^2-4kpq\choose pq}$, the legendre symbol. However, this is equal to ${(p+q)^2\choose pq}$ trivially, and this tells us that the discriminant is in fact a quadratic residue.

Hence, our solutions are of the form $$\frac{p+q\pm\sqrt{(p+q)^2-4kpq}}2$$for $k\in\mathbb Z$.

Title: For primes $p,q$, what are the solutions to $x^2=(p+q)x\pmod{pq}$

Body: For primes $p,q$, what are the solutions to $$x^2=(p+q)x\pmod{pq}$$


问:
Show that the following matrix is positive semidefinite
Problem: Let $0\leq a_1 \leq a_2 \leq \dots \leq a_n < \infty$. Show that the $n\times n$ matrix
\begin{align*}
A =
\begin{pmatrix}
a_1 & a_1 & \dots & a_1\\
a_1 & a_2 & \dots & a_2\\
\vdots & \vdots & \ddots & \vdots\\
a_1 & a_2 & \dots & a_n
\end{pmatrix}
\end{align*}
is positive semidefinite. (If it is not clear, the element in position $(i,j)$ of $A$ is $a_{\min\{i,j\}}$.)

My thoughts: I tried to use the definition of the determinant of a matrix in term of the determinants of submatrices but I did not get very far.

答:
You can write $A$ as the sum of rank 1 matrices:
$$
\begin{pmatrix}a_1 & \dots  & a_1 \\
\vdots & \ddots \\
a_1& & a_1 \end{pmatrix}
 +
\begin{pmatrix}0 & 0 & \dots & 0 \\
0 & a_2 - a_1 & \dots  & a_2-a_1 \\
\vdots & \vdots & \ddots \\
0 & a_2 - a_1& & a_2 - a_1 \end{pmatrix} + \text{etc}
$$
The $i^{th}$ term has eigenvector $(0, 0, \ldots, 0, 1, 1, \ldots 1)$, where the number of zeros is $i-1$ and eigenvalue $(n+1-i)(a_i - a_{i-i}) \geq 0$ (with $a_0=0$). Since the matrices have rank $1$, the remaining eigenvalues are $0$.

The sum of positive semidefinite matrices is positive semidefinite.


问:
Struggling to both understand and remember maths!
I hope it is OK to post this question here...

I'm currently in my 2nd year of maths at uni (doing cal 2, discrete, prob/stat) and hadn't studied maths formally for about 15 years beforehand. In my first year I did somewhat OK and scraped a pass in all of my modules (alg 1, calc 1, proofs) with thanks to distance exams. Otherwise, I would have absolutely failed.

I enjoy maths... the whole problem solving/challenging aspect is what drives me. However, I cry almost everyday because I'm so frustrated at how little I actually "get" it and it's actually starting to depress me (I dedicate at least 40 hrs p/week, 7 days a week to it). On the other hand, my boyfriend can look at a question, investigate it a little online and it makes sense to him with little effort.

In terms of learning methods I've found that notes do zilch. Perhaps my notetaking is just bad but I've found it wasted my time. My comp is full of ebooks as recommended by the staff too. So I have many sources of info at my hands but yet my intuition or core understanding is roughly the same as when I started.

Exams are in 5-6 months, I'm already getting myself stressed about it and I have to improve. I know maths is challenging, that's whats great about it! But when you find yourself staring at a question that involves bayesian theorem, wondering why the heck you have to even use division, then there's something missing.

Would it be most effective to simply do as many exercises as possible? How do you conceptualise theorems, etc and apply them to problems?

Any tips would be hugely appreciated :) Thank you in advance.

答:
Richard Feynman described his approach (in Surely You're Joking, Mr Feynman, my emphasis):

I can’t understand anything in general unless I’m carrying along in my mind a specific example and watching it go.  Some people think in the beginning that I’m kind of slow and I don’t understand the problem, because I ask a lot of these ‘dumb’ questions: “Is a cathode plus or minus?  Is an an-ion this way, or that way?”

But later, when the guy’s in the middle of a bunch of equations, he’ll say something and I’ll say, “Wait a minute!  There’s an error!  That can’t be right!”

The guy looks at his equations, and sure enough, after a while, he finds the mistake and wonders, “How the hell did this guy, who hardly understood at the beginning, find that mistake in the mess of all these equations?”

He thinks I’m following the steps mathematically, but that’s not what I’m doing.  I have the specific, physical example of what he’s trying to analyse, and I know from instinct and experience the properties of the thing.  So when the equation says it should behave so-and-so, and I know that’s the wrong way around, I jump up and say, “Wait!  There’s a mistake!”

I too find that very abstract stuff just leaks straight out of my brain without making much impact — but if I can visualise an example and follow it through, I have a much better chance of understanding and remembering.  (It's not always easy to find good examples, but it really helps.)


问:
Probability of senior citizens in a one million residence
In a city of over $1000000$ residents, $14\%$ of the residents are senior citizens. In a simple random sample of $1200$ residents, there is about a $95\%$ chance that the percent of senior citizens is in the interval [pick the best option; even if you can provide a sharper answer than you see in the choices, please just pick the best among the $5\%$ interval options ]

$N=1000000$ residents;

$p=14\%=0.14$ Senior Citizens

$n=1200$ residents simple random sample

$p=95\%=0.95$ chance that the $\%$ Senior Citizens is in the which interval?

$\quad\big[(9\%-19\%)\,$ or $\,(10\%-18\%)\,$ or $\,(11\%-17\%)\,$ or $\,(12\%-16\%)\,$ or $\,(13\%-15\%)\big]\,?$

答:
This is an application of a 95% confidence interval for a binomial proportion.

We calculate the confidence interval as such:

$$ p \pm z_{\frac{\alpha}{2}} \cdot SE $$

where the standard error is defined as: $$SE = \sqrt{\frac{p\cdot(1-p)}{n}}$$

It is quite straight-forward to "plug in" the data:

$p = 0.14$  (The probability of "success")

$n = 1200$  (The size of the sample)

$z = 1.96$  ("This value is the standard normal percentile having right-tail probability equal to $\frac{\alpha}{2}$." [1]) We get this value from the normal distribution, as we want a 95% Confidence Interval, so $\alpha=0.05$. $\frac{\alpha}{2}=0.025$ The corresponding z-score is 1.96.

The computation is easy to carry out:

$SE = \sqrt{\frac{0.14\cdot(1-0.14)}{1200}}$

$SE = \sqrt{\frac{0.14\cdot(1-0.14)}{1200}}$

$\therefore SE = 0.01002$

$\therefore Interval(p) = 0.14 \pm 1.96*0.01002$

$\therefore Interval(p) = (0.1203608,0.1596392)$

Rounded:

$Interval(p) = (0.12,0.16)$

So that will be the correct interval.

Let me know if you need any clarification.

[1] Cited from Alan Agresti's Introduction to Categorical Data Analysis, 2nd Ed.


问:
$f$ differentiable implies remainder differentiable at zero
Definition: A function $f: U\subset \mathbb{R}^{m} \to \mathbb{R}^{n}$ ($U$ open) is said to be differentiable at $x_{0} \in U$ if there exists a linear map $T: \mathbb{R}^{m}\to \mathbb{R}^{n}$ such that:
$$f(x+h) = f(x) + T(h) +r(h)$$
where $\lim_{h\to 0} r(h)/|h| = 0$. Here, $r$ is called the remainder. The above equality holds provided $x+h \in U$.

For convenience, let us write $T = f'(x_{0})$ if $f$ is differentiable at $x_{0}$.

Problem: Suppose $f$ is differentiable at $x_{0}$ and let $B_{\delta}(x_{0})$ be an open ball with center $x_{0}$ and radius $\delta > 0$. Define $r: B_{\delta}(x_{0}) \to \mathbb{R}^{n}$ by:
$$r(h) = f(x_{0}+h)-f(x_{0}) - f'(x_{0})(h)$$
Prove that $r$ is differentiable at zero.

My proof: Because $r(0) = 0$, it is enough to prove that there exists a linear map $T: \mathbb{R}^{m}\to \mathbb{R}^{n}$ such that:
\begin{eqnarray}
r(h') = T(h') + \tilde{r}(h') \tag{1}\label{1}
\end{eqnarray}
holds, where $\lim_{h'\to 0}\tilde{r}(h')/|h'| = 0$. Now, let us define:
$$\rho(h') := \frac{f(x_{0}+h')-f(x_{0})-f'(x_{0})(h')}{|h'|} $$
Then, because $f$ is differentiable at $x_{0}$, we must have $\lim_{h'\to 0}\rho(h') = 0$. Thus, let $\tilde{r}(h') = |h'|\rho(h')$ and $T \equiv 0$ be the identically zero linear transformation. Then $r(h')$ can obviously satisfies (\ref{1}), and this proves the claim.

Question: Is my proof correct?

答:
I see nothing wrong with your proof but you could simplify it. In this case you could simply write:
$$r(h) = f(x_0 +h) - f(x_0)-f'(x_0)h$$ and note by the definition of derivative for $f(x)$ that,
$$\lim_{|h|\to 0} \frac{r(h)}{|h|} = 0$$ which, as $r(0)=0$, is precisely the definition of the derivative of $r(h)$ at $h=0$.

Another way that should work is $f(x)$ differentiable at $x=x_0$ implies that $g(h):= f(x_0 +h)$ is differentiable at $h=0$ and thus $r(h)$ defined above is the sum of differentiable functions, making it differentiable.


问:
Let $H\lhd G$ and $\psi:G\to G/H,x\mapsto xH$ be a function. Show it is surjective
Let $H\lhd G$ and $\psi:G \to G / H$, $x \mapsto xH$ be a function. I am struggling to show that the map is surjective.

Is it just so simple to say let $y:=xH \in G / H$ than it is trivial that we get $x \in G$ s.t. $\psi(x)=xH$ and hence it is surjective? Many thanks for your help!

答: Take any element $y \in G/H$. By definition of $G/H$, there exists $x \in G$ such that $y=xH$. Then by definition of $\psi$, this can be rewritten as $y=\psi(x)$. So every element of $G/H$ is in the image of $\psi$, i.e. $\psi$ is surjective.


问:
How little o can verify for one example?
I have following facts:

$n^2log_2n+n^2$ = $o(n^3)$

how I can understand this is true fact? I'm not rain into a homework question, any intuitively idea to quickly find or doing by math?

答:
Just check if $\frac{n^2\log_2(n)}{n^3}+\frac{n^2}{n^3}$ tends to zero as $n$ goes to infinity. This can be easily checked. We know that
$$\frac{n^2\log_2(n)}{n^3}+\frac{n^2}{n^3}=\frac{\log_2{n}}{n}+\frac{1}{n}$$
$1/n$ tends to 0, it is a basic property of limits, which can be verfied computationally, and $\frac{\log_2(n)}{n}$ can be verified by L'Hôpital's rule.

In general, $f(x)=o(g(x))$ if
$$\lim_{n\to\infty}\frac{f(n)}{g(n)}=0$$
This expression means that as we put larger and larger values of $n$ in $f(n)/g(n)$, the result gets closer and closer to 0. This is the mathematical definition, but it is equivalent to the computer science definition.


答:
Let $X$ be the number of seniors in the sample. We are presumably sampling without replacement, but $1000000$ is very large compared to the sample size, so we can assume that the distribution of $X$ is binomial, $n=1200$, $p=0.14$. So $X$ has mean $(1200)(0.14)$ and standard deviation $\sqrt{1200(0.14)(0.86)}$.

Let $Y=\frac{X}{1200}$, the sample proportion of seniors.

Then $Y$ has mean $0.14$, and standard deviation $\frac{\sqrt{(0.14)(0.86)}}{\sqrt{1200}}\approx 0.0100$.

Note that the distribution of $Y$ is very close to normal. So with probability $0.95$, $Y$ is within $(1.96)(0.0100)$ of the mean $0.14$. The nearest in the given list is $12\%$ to $16\%$.

Title: Probability of senior citizens in a one million residence

Body:
In a city of over $1000000$ residents, $14\%$ of the residents are senior citizens. In a simple random sample of $1200$ residents, there is about a $95\%$ chance that the percent of senior citizens is in the interval [pick the best option; even if you can provide a sharper answer than you see in the choices, please just pick the best among the $5\%$ interval options ]

$N=1000000$ residents;

$p=14\%=0.14$ Senior Citizens

$n=1200$ residents simple random sample

$p=95\%=0.95$ chance that the $\%$ Senior Citizens is in the which interval?

$\quad\big[(9\%-19\%)\,$ or $\,(10\%-18\%)\,$ or $\,(11\%-17\%)\,$ or $\,(12\%-16\%)\,$ or $\,(13\%-15\%)\big]\,?$


答:
Yes, your argument is correct.

A very minor gripe: you don't "get" $x$ such that $\psi(x)=xH$. You actually start with one, and as you say, it trivially satisfies $\psi(x)=xH$.

Title: Let $H\lhd G$ and $\psi:G\to G/H,x\mapsto xH$ be a function. Show it is surjective

Body:
Let $H\lhd G$ and $\psi:G \to G / H$, $x \mapsto xH$ be a function. I am struggling to show that the map is surjective.

Is it just so simple to say let $y:=xH \in G / H$ than it is trivial that we get $x \in G$ s.t. $\psi(x)=xH$ and hence it is surjective? Many thanks for your help!