问:
Find the pointwise limit
The question is:

Consider the sequence $$ f_k(x):(0,1)\to \Bbb R\ $$ given by $$f_k(x)=kx^{k}$$ and find its pointwise limit.

Without the (0,1), I'd say the sequence diverges but now we have the interval (0,1) I tried to say that $|x^{k}|<1$ converges, but I am not sure how to then find the pointwise limit.

Any help would be appreciated.

答:
Let $k>\dfrac x{1-x}$. We have

$$\frac{f_{k+1}(x)}{f_k(x)}=\frac{k+1}{k}x=r<1$$

so that by induction

$$f_{k+n}(x)<f_k(x)r^n$$ where the RHS is a decreasing geometric sequence.

Title: Find the pointwise limit


问:
Determine the base b in
I have doubts about how to solve this issue about bases in numbering systems:

"Determine the base b in $(104)_b = 8285$"

can anybody help me?
Thanks.

答:
Hint: In the number $104_b$, the final digit represents the units, the middle digit represents the number of "$b$"s, and the leftmost digit represents the number of "$b^2$"s. (Compare to base $10$, where these would be units, tens, and hundreds.) In other words, you have
$$b^2+4 = 8285.$$

Now solve for $b$.

ContentLicense: CC BY-SA 4.0

Title: Determine the base b in


答:
Since $(104)_b = 1\cdot b^2 + 0 \cdot b + 4$ you get the equation
$$b^2+4 = 8285,$$
which you should be able to solve.

Title: Determine the base b in

Body:
I have doubts about how to solve this issue about bases in numbering systems:

"Determine the base b in $(104)_b = 8285$"

can anybody help me?
Thanks.


问:
relation according to height factor between two specific trees
I have $2$ trees as follows:

$First$

$Second$

I want to find a relation between these two trees, I need this relation: if we get height of the second three as $H$ what is the relation between $height$ of these two example according to $H$?

答: The height of the second tree is $n$, the length of the path from the vertex $2^0$ to the root. Problem $5$ in the PDF specifies that $n+1=2^N$, so $\log_2(n+1)=N$, and the claim is that the height of the first tree is $N$. To see this, let $h$ be the height of this tree; it has $2^N$ leaves, and they are on level $h$. Clearly level $h-1$ has half as many nodes as level $h$, so it has $2^{N-1}$ nodes. Similarly, level $h-2$ has $2^{N-2}$ nodes, and in general level $h-i$ has $2^{N-i}$ nodes for $i=0,\ldots,N$. Thus, level $h-N$ has $2^{N-N}=1$ node; clearly that node is the root of the tree, so it is on level $0$, i.e., $h-N=0$. But then $h=N$, as claimed.


问:
How many ways to form a binary sequence with 0 and 1, where no 3 consecutive digits are 1? (without recurrence)
The straight forward question is,

How many ways to form a binary sequence with 0 and 1, where no 3 consecutive digits are 1?

As it is a same question as this one or so like these. But everywhere I found recurrence. For this problem there is also a recurrence $$f_n = f_{n-1} +f_{n-2} +f_{n-3} $$

But I was wondering and couldn't find out a way that Is it possible to get any solution without recurrence ? Also Is there any combinatorial solution as this seems to be like a counting problem? If yes, how to reach the solution for problems like this? If no, how to prove that this is not possible?

答:
Any such sequence with a $0$ appended after it must be made out of the blocks $0, 10, 110$, by looking at where the $1$s occur in the sequence. If there are $i$ copies of the block $0$, $j$ copies of the block $10$ and $k$ copies of the block $110$ then there are ${i + j + k \choose i, j, k} = \frac{(i + j + k)!}{i! j! k!}$ ways to arrange these blocks and the total length of the sequence is $i + 2j + 3k$, which gives

$$f_n = \sum_{i + 2j + 3k = n+1} {i+j+k \choose i, j, k}.$$

There's a similar but simpler formula for the Fibonacci numbers. Whether this counts as a "solution" is subjective because there is still a summation with an unbounded number of terms involved. It's also hard to analyze the growth rate. Whereas the recurrence relation immediately tells you the growth rate, namely that it's $r^n$ where $r$ is the largest positive root of $r^3 = r^2 + r + 1$. It's an unintuitive technique at first but it gives very powerful and precise results. This argument in terms of blocks only works in some special cases and it doesn't give an answer that's very easy to work with.


答:
Hint: The number in base $10$ is $8285$. Remember that this means the number can be expressed as
$$5\cdot 10^0 + 8\cdot 10^1 + 2\cdot 10^2 + 8\cdot 10^3 = 8285$$
You are told that, in base $b$, the number can be expressed as
$$4\cdot b^0 + 0\cdot b^1 + 1\cdot b^2 = 8285$$
The left side is just $b^2 + 4$. So you want to solve
$b^2 + 4 = 8285$ for $b$.

Title: Determine the base b in


问:
Bhaskara approximation of cos(x)
In my knowledge the best aproximation of $\sin$ and $\cos$ is Bhaskara approximation which is :
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
and
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$

what is the best approximation of $\cos(x)$ for $ -\pi\leq x\leq 0$ (by a regular function ) ?

答:
The approximation for $\sin(x)$ on $[-\pi/2, \pi/2]$ that fits Bhaskara’s spirit the best is

$$\frac{22 \pi^2 x - 24 x^3}{\pi (7 \pi^2 + 4 x^2)}.$$

It is exact for $\lvert x \rvert \in\{0, \pi/6, \pi/2\}$ and its derivative is zero at the end points.  Shift left over $\pi/2$ to get an approximation of $\cos(x)$ on $[-\pi, 0]$.

This is a better approximation in terms of maximum absolute error than Padé’s of the same degree. (Remez’ algorithm can find an even better one.  See section 6 of this exposition.) Scaling to the interval $[-1,1]$ gives the nicer looking approximation $$\frac{11 x - 3 x^3}{7 + x^2} \approx \sin \left(\tfrac{\pi}2 x\right).$$

Title: Bhaskara approximation of cos(x)


问:
Prove that there is no rational $x$ such that $x-1 = 1/x$
As said in title,

I need to prove that there doesn't exist a rational number $x$ that satisfies $x-1 = 1/x$.

I remember doing something like this a while back at school but I can't recall how to do it. Am I right in starting out by simplifying it to:

$x^2-x=1$?

答:
Let $x=\frac{p}{q}$ be a solution in lowest terms, i.e., $\gcd(p,q)=1$, with $q>0$. By substituting this into the equation for $x$, we get

$$\left(\frac{p}{q}\right)^2-\frac{p}{q}=1\Leftrightarrow p^2-pq=q^2\Leftrightarrow p^2=q(p+q).$$

That is, $q$ divides $p^2$. But we assumed $\gcd(p,q)=1$ and thus $q$ must be $1$. Hence, we get

$$p^2=p+1\Leftrightarrow p(p-1)=1,$$

which means that $p$ divides $1$, i.e., $p=\pm 1$, but obviously neither $p=1$ nor $p=-1$ satisfies $p^2=p+1$. We have thus reached a contradiction and $x$ cannot be rational.


问:
To prove the interior of a circle is a path connected region
I'm trying to prove that the interior of a circle is a path connected region.

To prove that, we take two random points $A \neq B$ in the interior of a circle we don't have any point of the circumference in line segment $AB$ and I come to the following result which I can't prove:

Given $\triangle ABC$ and a point $X \in \overline{BC}$ then $AX < \max(b,c)$.

I'm taking a point $P \in \overline{AX}, AP = \max (b,c)$ and trying to use triangular inequality on some triangle, but it is not working so far

答:
This is an application of greater angle, greater side (Euclid I-19).

Since $\angle AXB + \angle AXC = 180^\circ$, one of $\angle AXB, \angle AXC \ge 90^\circ$.

If $\angle AXB \ge 90^\circ$, it follows that $\angle ABX < 90^\circ$.

By greater angle, greater side, we have $AB > AX$. The case for $\angle AXC \ge 90^\circ$ is similar.

Hence $AX < \max(AB, AC)$.


问:
What is the limit of $(3n^2 + n)^{1/n}$ and why?
I am really unsure of how to answer this as I have tried taking out n and using the ratio lemma and am still confused!</p>

答:
Yes taking out $n$ is a good idea but $3n^2$ is a better one to obtain

$$(3n^2 + n)^{1/n}=(3n^2)^\frac1n\left(1 + \frac1{3n}\right)^{1/n}$$

and since

$$\left(1 + \frac1{3n}\right)^{1/n} \to 1^0=1$$

we reduce to evaluate the limit for $(3n^2)^\frac1n=3^\frac1n \,\left(n^\frac1n\right)^2$.

As an alternative we can use that

$$1=1^\frac1n \le (3n^2 + n)^{1/n}\le (4n^2)^{1/n} =4^\frac1n \,\left(n^\frac1n\right)^2$$


问:
Do we have $ U \bigoplus W$ under these circumstances?
let $V$ be a finite dimensional inner product space with subspaces $W$ and $U$ such that $W \cap U = \{ 0\}$

suppose $W$ and $U^{\perp}$ are isomorphic, is it true that $  V = U \bigoplus W$ ?

I know for one that if $  V = U \bigoplus W$ then necessarily $W=U^{\perp}$

I'm positive it is not true, but I'm struggling to construct a practical counter example, any ideas ?

答:
NO. Consider the case that $V=\mathbb{R}^{2}$ with the usual inner
product. Let $e_{1}=(1,0)$ and $e_{2}=(0,1)$. Let $U=\mbox{span}(e_{1})$
and $W=\mbox{span}(e_{1}+e_{2})$, then $U^{\perp}=\mbox{span}(e_{2})$.
Clearly $U\cap W=\{0\}$ and $W$ and $U^{\perp}$ are isomorphic
(because both are one dimensional). However we do not have $V=U\oplus W$.

Although it is true that for each $x\in V$, there exist uniquely
$u\in U$ and $w\in W$ such that $x=u+w$. However, the condition
$\langle u,w\rangle=0$ is NOT fulfilled. For example, take $x=2e_{1}+e_{2}$,
then we have decompostion $x=u+w$, where $u=e_{1}$ and $w=e_{1}+e_{2}$.
However $\langle u,w\rangle=1$.

Remark: If the symbol $V=U\oplus W$ just means: For any $v\in V$,
there exist uniquely $u\in U$ and $w\in W$ such that $v=u+w$, then
the answer is Yes.

Reason: Let $n=\mbox{dim}(V)$, $k=\mbox{dim}(U)$, then $\mbox{dim}(W)=\mbox{dim}(U^{\perp})=n-k$.
Since $U\cap W=\{0\}$, it is not difficult to see that $\mbox{dim}(U\oplus W)=k+(n-k)=n$,
from which we conclude that $U\oplus W=V$.

I work out the details in below: Let $e_{1},e_{2},\ldots,e_{k}\in U$
be linearly independent. Let $f_{1},f_{2},\ldots,f_{n-k}\in W$ be
linearly independent. We asssert that $e_{1},\ldots,e_{k},f_{1},f_{2},\ldots,f_{n-k}$
are linearly independent. Let $\alpha_{1},\ldots,\alpha_{k},\beta_{1},\ldots,\beta_{n-k}\in\mathbb{R}$
be such that $\alpha_{1}e_{1}+\ldots+\alpha_{k}e_{k}+\beta_{1}f_{1}+\ldots+\beta_{n-k}f_{n-k}=0$,
then $\alpha_{1}e_{1}+\ldots\alpha_{k}e_{k}=-(\beta_{1}f_{1}+\ldots+\beta_{n-k}f_{n-k})$.
It follows that $\alpha_{1}e_{1}+\ldots\alpha_{k}e_{k}\in U\cap W\Rightarrow\alpha_{1}e_{1}+\ldots\alpha_{k}e_{k}=0$.
Since $e_{1},\ldots,e_{k}$ are linearly independent, we further have
$\alpha_{1}=\ldots=\alpha_{k}=0$. Now, $\alpha_{1}e_{1}+\ldots+\alpha_{k}e_{k}+\beta_{1}f_{1}+\ldots+\beta_{n-k}f_{n-k}=0$
reduces to $\beta_{1}f_{1}+\ldots+\beta_{n-k}f_{n-k}=0$, from which
we conclude that $\beta_{1}=\ldots\beta_{n-k}=0$ because $f_{1},\ldots,f_{n-k}$
are linearly independent.

Since $U+W=\mbox{span}(e_{1},\ldots,e_{k},f_{1},\ldots,f_{n-k})$,
we have $\mbox{dim}(U+W)=n$. On the other hand, $U+W$ is a vector
subspace $V$. Recall that the only vector subspace of $V$ that has
dimension $n$ is $V$ itself. Therefore $V=U+W$. Lastly, we show
that the decomposition $v=u+w$ is unique. For, suppose that $v=u_{1}+w_{1}=u_{2}+w_{2}$
for some $u_{1},u_{2}\in U$ and $w_{1},w_{2}\in W$, then $u_{1}-u_{2}=w_{2}-w_{1}\in W\Rightarrow u_{1}-u_{2}\in U\cap W\Rightarrow u_{1}-u_{2}=0$.
It follows that $u_{1}=u_{2}$, then $w_{1}=w_{2}$.


问:
Cardinality of sets and union
I ve been thinking about this lately, but couldn t come up with an answers.Let A and B be two infinite  sets of the same cardinality, is it true that the union of A and B has the same cardinality as A and B?</p>

答: Yes, assuming the axiom of choice it is true. Without the axiom of choice there can be counterexamples. In particular, if $A$ is an amorphous set, let $A_0=A\times\{0\}$ and $A_1=A\times\{1\}$. Clearly there is a bijection between $A_0$ and $A_1$, but if there were a bijection between $A_0\cup A_1$ and $A$, $A$ would be the disjoint union of two infinite sets and therefore not amorphous.

Title: Cardinality of sets and union


问:
Construct any solid with identical plane shapes and the resulting solid must be a fair die?
Take a bunch of identical shapes and stick them together to form a convex 3-d solid. Of course, this won't be possible for most shapes. But when it is possible, one should always get a fair die. All platonic solids are examples of this, so are two tetrahedrons pasted together along a face. So is a Tetratoid where the faces are not even regular polygons. This result seems to make intuitive sense. But is it possible to prove it beyond doubt?</p>

答:
This statement is not true. At least, there are shapes for which it no longer makes "intuitive sense" that they would be fair dice, though rigorously proving unfairness would require a more formal specification of what it means to roll a die.

As an example, consider the snub disphenoid:

The general issue here is that intuitive "fairness" comes from a polyhedron being face-transitive, or isohedral, so that we cannot distinguish any two faces from one another. Thus, if spun with enough random noise that any orientation of the shape is equally likely, we shouldn't expect any one face to be favored over another, since we can't even write down a property possessed by one face and not another without making reference to a fixed orientation or location on the solid.

But "having all congruent faces", or being monohedral, is not in general enough to guarantee face-transitivity; one can find other counterexamples among the non-uniform convex deltahedra, like the triaugmented triangular prism and the gyroelongated square bipyramid. I believe these are the only counterexamples with regular polygons as faces, but using non-regular faces we can find examples like the pseudo-deltoidal icositetrahedron, the rhombic icosahedron, and the Belinski dodecahedron.


问:
Don’t really want an answer, but I don’t understand it so can someone explain this a little bit to get me going
A parent has washed some nappies in a strong bleach solution and wishes to rinse them so that they contain as weak a bleach solution as possible. By wringing out, the nappies can be made to contain just half a litre of solution. Show that two thorough rinses, such that the solution strength is uniform, the first using 12 litres of water and the second using 8 litres of water, reduces the strength of the bleach solution to 1/425 of its original value
If 20 litres of clean water is all that is available and the parent is prepared to do only two rinses, how best should the water be divided between the two rinses?</p>

答: After the wash and wringing, the nappies contain $\frac 12$ litre of strong solution.  In the first rinse you add $12$ litres of water, mix that with what was in the nappies to start, and wring them out.  That leaves $\frac 12$ litre of the solution in the nappies.  What is the concentration of that solution compared to the one they were washed in?  The second step is the same with $12$ replaced by $8$.  You are asked for the concentration after the second step compared to the start.


问:
Closed-form for $ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $ where $p\in\mathbb{N}$
Is there a closed form to the sum $$ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $$
where $p$ is a positive integer? I know that for $p = 1$ the series sums to $xe^x$, but not sure if there's a simple expression for $p > 1$. If there isn't a closed form, is this series related to any special functions? Thanks!

答:
Note that in general $$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n} f(x) = \sum_{k=1}^n {n \brace k} \, x^k f^{(k)}(x)$$
which can be derived by noting the general form $$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n} f(x) = \sum_{k=1}^n a_k(n) \, x^k f^{(k)}(x)$$
and then proceed inductively i.e.
$$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n+1} f(x) = xf^{(1)}(x) + \sum_{k=2}^n \left\{k \, a_k(n) + a_{k-1}(n)\right\} x^k f^{(k)}(x) + x^{n+1} f^{(n+1)}(x) \\
= \sum_{k=1}^{n+1} \left\{ k \, a_k(n) + a_{k-1}(n) \right\} x^k f^{(k)}(x) \stackrel{!}{=} \sum_{k=1}^{n+1} a_k(n+1) \, x^k f^{(k)}(x) $$
where $a_1(n)=1$, $a_n(n)=1$ and $a_0(n)=a_{n+1}(n)=0$ with $n\in \mathbb{N}$.

Finally check that the Stirling Numbers of Second kind $${n\brace k}=\frac{1}{k!} \sum_{i=0}^k (-1)^{k-i} \binom{k}{i} i^n$$ solve the recurrence $$a_k(n+1)=k \, a_k(n) + a_{k-1}(n) \, .$$

Your special case is $f(x)=e^x$.


问:
$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof
Does anyone know hot to prove this inequality?

Having: $a, b, c \gt 0$

$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$

I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I don't know how to continue.

Furthermore, I don't know how to get rid of the known term. I tried with the AM-HM inequality but I'm still not getting any results.

Also, just some hints would be appreciated, thanks

答:
I am assuming $a,b,c>0$. Clear denominators, expand, cancel like terms and obtain

$$a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2\geq 0$$

However, this is easily proved by cyclically adding up the obvious inequalities $\frac{1}{2}a^4+\frac{1}{2}b^4\geq a^2b^2$.


问:
Show that $3n^4+3n^2+1$ is never a perfect square
I am looking for a proof for the fact that $3n^4+3n^2+1$ can never be a perfect square for a natural number $n>0$.

I know for a fact that the statement must be true as it came up as one of the cases in a solution of the diophantine equation $y^2=x^3-1$ using the LTE lemma and, according to two different solutions I have come across, this equation has no solutions apart form $(x,y)=(1, 0)$.

I have spent a considerable amount of time looking for a suitable attack strategy, however, I was not able to make any progress. Can anyone help?

答: $3n^4 + 3n^2 + 1 - x^2=0$ is an elliptic curve with Weierstrass form $s^3 + t^2 - 15s + 22=0$.  According to Sage, the rank is $0$ and the torsion group has order $2$, consisting of a point at infinity (which corresponds to $n=0, x=-1$) and $s=2,t=0$ (which corresponds to $n=0,x=1$).  Thus $n=0$, $x = \pm 1$ are the only rational points.


问:
Show that $C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$
Given $$C=1+r\cos{\theta}\ +\frac{r^2\cos{2\theta}}{2!}\ + \frac{r^3\cos{3\theta}}{3!}\ + \dotsb$$ and $$S = r\sin{\theta}\ + \frac{r^2\sin{2\theta}}{2!}\ + \frac{r^3\sin{3\theta}}{3!}\ + \dotsb$$
Show the following$$C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$$

I am currently solving the problems given in Differential Calculus for Beginners by Joseph Edwards. As a beginner I am completely clueless about the approach to the above question. I tried to find the answer keys to this book, but sadly non exist on the internet.

答:
We have
$$\begin{align}
C+iS&=1+r(\cos\theta+i\sin\theta)+\frac{r^2}{2!}(\cos2\theta+i\sin2\theta)+\frac{r^3}{3!}(\cos3\theta+i\sin3\theta)+\cdots\\
&=1+re^{i\theta}+\frac{1}{2!}(re^{i\theta})^2+\frac{1}{3!}(re^{i\theta})^3+\cdots\\
&=e^{re^{i\theta}}
\end{align}$$
using De Moivre's Theorem and Euler's relation in the second line and the Maclaurin series expansion for $e^x$ in the final line. Using $C+iS$ is a common approach when finding closed form formulae for trigonometric summations (where $C$ is the series for $\cos$ and $S$ is a similar series for $\sin$), whether they be infinite or finite. I can make the working more complete if you like.

ADDED SECTION
This is also equivalent to
$$\begin{align}
(e^{e^{i\theta}})^r &=(e^{\cos\theta+i\sin\theta})^r\\
&=(e^{\cos\theta}\cdot e^{i\sin\theta})^r\\
&=e^{r\cos\theta}(\cos(\sin\theta)+i\sin(\sin\theta))^r\\
&=e^{r\cos\theta}(\cos(r\sin\theta)+i\sin(r\sin\theta))\\
&=e^{r\cos\theta}\cos(r\sin\theta)+ie^{r\cos\theta}\sin(r\sin\theta)
\end{align}$$
So we have
$$C=e^{r\cos\theta}\cos(r\sin\theta)$$
$$S=e^{r\cos\theta}\sin(r\sin\theta)$$

FINAL SECTION
So, if
$$C=e^{r\cos\theta}\cos(r\sin\theta)$$
$$S=e^{r\cos\theta}\sin(r\sin\theta)$$
then $$\frac{dC}{dr}=e^{r\cos\theta}\cos\theta\cos(r\sin\theta)-e^{r\cos\theta}\sin\theta\sin(r\sin\theta)=e^{r\cos\theta}\cos(\theta+r\sin\theta)$$
$$\frac{dS}{dr}=e^{r\cos\theta}\cos\theta\sin(r\sin\theta)+e^{r\cos\theta}\sin\theta\cos(r\sin\theta)=e^{r\cos\theta}\sin(\theta+r\sin\theta)$$
So
$$\begin{align}
C\frac{dC}{dr}+S\frac{dS}{dr}&=e^{2r\cos\theta}\cos(r\sin\theta)\cos(\theta+r\sin\theta)+e^{2r\cos\theta}\sin(r\sin\theta)\sin(\theta+r\sin\theta)\\
&=e^{2r\cos\theta}\cos\theta.
\end{align}$$
on using the compound angle formulae for Cosine.

And finally,
$$(C^2+S^2)\cos\theta=(e^{2r\cos\theta}\cos^2(r\sin\theta)+e^{2r\cos\theta}\sin^2(r\sin\theta))(\cos\theta)=e^{2r\cos\theta}\cos\theta$$
as required!!

Thanks so much for giving me the opportunity to answer this question, I have hugely enjoyed it!


答:
Since the expression is cyclic, we can, WLOG, reduce it into two cases:

Suppose $a\ge b\ge c > 0$. We have

$$\frac{a^2-c^2}{b+c} = \frac{a^2-b^2}{c+b} + \frac{b^2-c^2}{c+b} \ge \frac{a^2-b^2}{c+a} + \frac{b^2-c^2}{a+b}$$

Now suppose $a \ge c \ge b > 0$. We then have

$$\frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge \frac{c^2-b^2}{a+c} + \frac{a^2-c^2}{a+c} =\frac{a^2-b^2}{c+a}$$

Title: $\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof

Body:
Does anyone know hot to prove this inequality?

Having: $a, b, c \gt 0$

$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$

I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I don't know how to continue.

Furthermore, I don't know how to get rid of the known term. I tried with the AM-HM inequality but I'm still not getting any results.

Also, just some hints would be appreciated, thanks


答:
$$L=\lim_{n \to \infty}(3n^2 + n)^{1/n}$$
Apply Cauchy-D'Alembert criterion:
$$L=\lim_{n \to \infty}\dfrac {3n^2+7n+ 4}{3n^2+n}=\lim_{n \to \infty}\dfrac {3n^2}{3n^2}$$
$$\implies L=1$$

Title: What is the limit of $(3n^2 + n)^{1/n}$ and why?

Body: I am really unsure of how to answer this as I have tried taking out n and using the ratio lemma and am still confused!</p>


问:
Sequence of functions defined by recursion
let $ b>0 $ and let $ f_{1}:[0,b]\to\mathbb{R} $ be an integrable functions.

Prove that the sequence of functions, that defined recursivly by:

$ f_{n+1}\left(x\right)=\int_{0}^{x}f_{n}\left(t\right)dt $

is uniformly convergent to the function $ f=0 $

Actually, I have no idea where to start, I cant really see anything that would justify that the values of the functions really approaches to $ 0 $ (I cant even see why $ f_n(x) $ is getting smaller for a constant $ x $ ).

What I did notice is that for $ n>3 $ the function $ f_n $ is the anti-deriviative of $ f_{n-1} $ (except for $ f_2  $ and  $f_1 $ ), and ofcourse they are all continious functions (except for maybe $ f_1 $ ). Any hints would help, thanks in advance.

答:
Since $f_1$ is Riemann integrable then by definition $f_1$ is bounded, i.e, $|f_1(x)|\leq M$.

Then $|f_2(x)|=|\int_{0}^x f_{1}(t)dt|\leq \int_{0}^x |f_{1}(t)|dt\leq Mx$.

So $|f_3(x)|=|\int_{0}^x f_{2}(t)dt |\leq \int_{0}^x |f_{2}(t)|dt\leq  \int_{0}^x M tdt=M\frac{x^2}{2}.$

By induction we obtain $|f_n(x)|\leq M\frac{x^{n-1}}{(n-1)!}$.

Thus $|f_n(x)|\leq M\frac{x^{n-1}}{(n-1)!}\leq M\frac{b^{n-1}}{(n-1)!}\xrightarrow{n\rightarrow \infty}0$.

The result follows.


问:
If $A\subset B$ and $\mu^*(A)=\mu^*(B)<\infty$, then $\mu^*(B\setminus A)=0$
I feel like this should be an easy problem, but I've been banging my head against a wall on this.

Let $X$ be a non-empty set and $\mu^*\colon 2^X\to \left[0,\infty\right]$ an outer measure. Assume that $A\subset B\subset X$ with $\mu^*(A)=\mu^*(B)<\infty$. Then $\mu^*(B\setminus A)=0$.

Is that true? Without the assumption $\mu^*(A)=\mu^*(B)<\infty$ this is easily shown to be false, but the counterexample relies on the fact that $\infty-\infty$ is undefined. In this case, however, that is no longer possible. I've tried to prove it without getting really anywhere. Any help is appreciated.

On my notation: $X\subset Y$ includes the case $X=Y$.

答: This is false even with the finiteness assumption. For example, take $X=\{a,b\}$, and set $\mu^*(\emptyset)=0, \mu^*(\{a\})=\mu^*(\{b\})=\mu^*(X)=1$: then $\mu^*$ is an outer measure on $X$, but $\mu^*(\{a\})=\mu^*(X)$ and $\mu^*(X\setminus \{a\})=1$.


答:
$$
C\frac{dC}{dr}+S\frac{dS}{dr}=C\sum_{n=1}^\infty\frac{nr^{n-1}\cos n\theta}{n!}+S\sum_{n=1}^\infty\frac{nr^{n-1}\sin n\theta}{n!}\\
=C\sum_{n=1}^\infty\frac{ r^{n-1}\cos n\theta}{(n-1)!}+S\sum_{n=1}^\infty\frac{ r^{n-1}\sin n\theta}{(n-1)!}=C\sum_{n=0}^\infty\frac{ r^{n }\cos (n+1)\theta}{ n !}+S\sum_{n=0}^\infty\frac{ r^{n }\sin (n+1)\theta}{ n !}\\
=C\sum_{n=0}^\infty\frac{ r^ n }{ n !}(\cos n\theta\cos\theta-\sin n\theta\sin\theta )+S\sum_{n=0}^\infty\frac{ r^ n }{ n !}(\sin n\theta\cos\theta+\cos n\theta\sin\theta )\\
= C^2 \cos\theta-C S\sin\theta+SC\sin\theta+S^2 \cos\theta
=(C^2+S^2)\cos\theta
$$

Title: Show that $C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$

Body:
Given $$C=1+r\cos{\theta}\ +\frac{r^2\cos{2\theta}}{2!}\ + \frac{r^3\cos{3\theta}}{3!}\ + \dotsb$$ and $$S = r\sin{\theta}\ + \frac{r^2\sin{2\theta}}{2!}\ + \frac{r^3\sin{3\theta}}{3!}\ + \dotsb$$
Show the following$$C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$$

I am currently solving the problems given in Differential Calculus for Beginners by Joseph Edwards. As a beginner I am completely clueless about the approach to the above question. I tried to find the answer keys to this book, but sadly non exist on the internet.


问:
Tychonoff theorem motivation
I am studying for a master's degree and I have to choose a topic and finally I decided to choose. Tychonoff Theorem in General Topology.
Here I studied about Tychonoff Theorem and also I found some easy proof of this theorem. But I am looking for motivation behind this theorem and what is the idea of  this theorem. If I would found some interesting application regarding masters level then it'll be great.
Currently I am following Munkres Topology , Ryszard Engelking and Armstrong basic topology.

Any help will be appreciated.

Thanks in advance!

答:
Hmm, well I have interesting memories of Tychonoff theorem, dating back to when I took Spanier's point-set topology and Intro to functions of a real variable at Berkeley.  He said he knew I didn't get the proof right, because I didn't use the axiom of choice.

But, I can tell you that the Tychonoff theorem, along with Urysohn metrization, is one of the two premiere theorems in general topology.  It says, as you know, that the product of compact spaces is compact.  This distinguishes the product topology from the box topology, since the theorem would not be true in the latter.

Munkres is a good reference.  And there's Kelley.

Title: Tychonoff theorem motivation


答:
When you add $1/2 \,\mathrm{L}$ bleach solution to $12 \,\mathrm{L}$ of water (and mix thoroughly), you have a new solution of $25/2 \,\mathrm{L}$.  The concentration of bleach is that solution is $\frac{1/2}{25/2} = 1/25$ of the original concentration and that is the concentration carried by the nappies from the first rinse to the second rinse.  You should be able to determine the concentration of bleach after the second rinse by following a similar procedure.  After the second rinse, you should determine that the nappies carry $1/2 \,\mathrm{L}$ of a solution at $1/425$ the concentration of the original nappies.

The question asks you to generalize.  You have $20 \,\mathrm{L}$ of water which you may partition in any way you like into two rinses.  Suppose you put quantity $x$ (in liters) into the first rinse.  They you may put quantity $y$ water into the second rinse, with $y \leq 20 \,\mathrm{L} - x$.  The question is, what values of $x$ and $y$ make the final bleach concentration the least?

Things to think about:

Is it ever a good idea to use $y < 20 \,\mathrm{L} - x$?

Can you compute the final concentration in terms of $x$ and $y$?

Is there any way to eliminate one of these variables so that you are writing in terms of $x$ only or $y$ only?

Can you minimize the resulting expression in terms of a single independent variable?

Title: Don’t really want an answer, but I don’t understand it so can someone explain this a little bit to get me going

Body:
A parent has washed some nappies in a strong bleach solution and wishes to rinse them so that they contain as weak a bleach solution as possible. By wringing out, the nappies can be made to contain just half a litre of solution. Show that two thorough rinses, such that the solution strength is uniform, the first using 12 litres of water and the second using 8 litres of water, reduces the strength of the bleach solution to 1/425 of its original value
If 20 litres of clean water is all that is available and the parent is prepared to do only two rinses, how best should the water be divided between the two rinses?</p>


问:
Confused about types of variable substitutions
I am confused about (dummy?) variable substitutions in comparison with change of variable type of substitutions.

On the one hand, I can write $f(x) = \dfrac{1}{1+x}$ or $f(u) = \dfrac{1}{1+u}$ and the functions produce same result for equal values of $x$ and $u$.

On the other hand, if $x$ and $u$ are related, say $x = 2u$, I get $f(x) = f(2u) = \dfrac{1}{1 + 2u}$. Now if I refer to $f(u)$, would I be talking about $\dfrac{1}{1 + u}$ or $\dfrac{1}{1 + 2u}$? Obviously in this case, $f(x=5)$, for example, would not product the same result as $f(u=5)$ without knowing which $f(\cdot)$ we are talking about.

Is there a more formal way to think about this? Should one of the functions $f(\cdot)$ actually be called something else?

答: When you write $f(x) = \frac{1}{1+x}$ you are referring to the function $f\colon x\mapsto \frac{1}{1+x}$.  Writing $f(u)$ means the same thing, except you've relabeled your variable.  If you make the substitution $x = 2u$ so that you have $f(x) = f(2u) = \frac{1}{1+2u}$, you are still referring to the same function $f$, but you are inputting $2u$.  If you want to refer to the function $u\mapsto \frac{1}{1+2u}$ in the same context then you should call it something other than $f$.


问:
Quasi-finite surjective holomorphic map is proper?
Let $X$ and $Y$ be connected complex manifolds. Let
$f:X \to Y$ be a surjective holomorphic map such that pre-image of every $y \in Y$ is a finite set.

Then can we say that $f$ is a proper map? I feel that this is not true but I am unable to think of a counterexample.

If not counterexample, hints towards $f$ is proper are welcome!

答:
Such a map need not be proper. A simple example is given by the map $f \colon z \mapsto z^2$ with $Y = \mathbb{C}\setminus \{0\}$ and $X$ any connected open subset of $Y$ such that $f(X) = Y$. Standard choices for $X$ are $\mathbb{C} \setminus (-\infty,0]$ or a three-quarter plane. A maximal $X$ is for example $Y \setminus \{1\}$.

Then if $z_0$ is a boundary point of $X$ in $Y$, the preimage of a compact neighbourhood of $f(z_0)$ isn't compact.


问:
Partitions of an interval and convergence of nets
Let $\mathscr{T}$ be the set of partitions $\tau = (\tau_0 = 0 < \tau_1 < \dots < \tau_N = 1)$ of the interval $[0,1]$ (where $N$ is not fixed). This becomes a directed set by setting $\tau < \tau^\prime$ iff $\tau^\prime$ is a subdivision.

Now we can look at the nets
$$ s_\tau := \sum_{j=1}^N (\tau_j - \tau_{j-1})^2$$
or
$$ r_\tau := N \sum_{j=1}^N (\tau_j - \tau_{j-1})^3$$

We should have both $s_\tau \longrightarrow 0$ and $r_\tau \longrightarrow 0$ in the sense of nets, but I don't know how to prove it. Does anybody have an idea?

答:
A general hint and a partial illustration.

As I understood, you must show that for each $\varepsilon>0$ there are nets $\tau_1$  and $\tau_2$ such that $s_{\tau_1’}<\varepsilon$ and $r_{\tau_2’}<\varepsilon$ for each $\tau_1’>\tau_1$ and $\tau_2’>\tau_2$.

Since $(a+b)^2\ge a^2+b^2$, provided $a$ and $b$ are non-negative,  $s_\tau$ is monotone. So
it is enough to show that for each real $\varepsilon>0$ there is a net $\tau$ such that $s_\tau<\varepsilon$.

ContentLicense: CC BY-SA 3.0


问:
Free group on a set B. If the free group is finitely generated, then B is finite.
Let B be a set. For simplicity, assume B contains all the formal inverses of its elements. Let W(B) be the set of words created from elements in B, and let F(B) be the set of equivalence classes [w] where w is a word in W(B). Any two words in the same equivalence class are related by a finite number reductions (insertions or deletions) by elements in B.

Here's my question: If F(B) is finitely generated, how do we prove that B is finite? In another problem I proved that F(B) is freely generated by the set of equivalence classes of elements arising from B; that is, equivalence classes of the form [b] where b is an element of B. Call this set S(B). My math prof. told me to use the universal extension property (where each map from S(B) to a group G extends to a unique homomorphism from F(B) to G), but I'm not sure how to do it.

答: Hint: Assume $F(B)$ is generated by the words $w_1,\dots,w_n$, and take the set of letters occuring in these words $w_i$.


问:
Are the sum and direct sum of vector subspaces associative and commutative?

Definition

If $S$ is a subset of a vector space $\mathscr{V}$ then the subspace $L(S)$ generated by $S$ is the smallest vector subspace of $\mathscr{V}$ containing $S$.

Proposition

The subspace $L(S)$ generated by $S$ is the set of all finite linear combination of vectors of $S$.

Definition

If $W_1$ and $W_2$ are vector subspaces of $\mathscr{V}$ then their sum $W_1+W_2$ is the subspace $L(W_1\cup W_2)$ generated by $(W_1\cup W_2)$.

Proposition

If $W_1$ and $W_2$ are vector subspaces of $\mathscr{V}$ then
$$
W_1+W_2=\{w_1+w_2:w_1\in W_1,\,w_2\in W_2\}
$$

Definition

The sum of two vector subspaces $W_1$ and $W_2$ of $\mathscr{V}$ is direct if $W_1\cap W_2={0}$. In particular the finite sum of a collection $\mathcal{W}:=\{W_i: i\in\ I\}$ vector subspace is said direct if $W_i\cap W_j=\{0\}$ for each $i,j\in I$.

So with the previous formalism I’m asked to prove that the sum of subspaces is commutative and associative and the direct sum too. In the case of the simple sum I thought that this could trivially follow by the fact that
$$
W_1+W_2=\{w_1+w_2:w_1\in W_1,\,w_2\in W_2\}=\{w_2+w_1:w_2\in W_2,\,w_1\in W_1\}=W_2+W_1
$$
and
$$
W_1+(W_2+W_3)=\{w_1+(w_2+w_3):w_i\in W_i, i=1,2,3\}=\{(w_1+w_2)+w_3:w_i\in W_i, i=1,2,3\}=(W_1+W_2)+W_3
$$
but I am not sure about the correctedness of the arguement. Then I think that I cant implement the first argument to show that
$$
W_1\oplus W_2=W_2\oplus W_1
$$
but I still am not sure about this and finally if the second argument shows that the sum is associative I would have only prove that $(W_1\oplus W_2)\cap W_3=\{0\}$ but I don't be able to do this unfortunately. So could someone help me, please?

答:
The fact that $W_1+W_2=W_2+W_1$ is fairly obvious, because
$$
W_1+W_2=L(W_1\cup W_2)=L(W_2\cup W_1)=W_2+W_1
$$
by the very definition.

What about associativity? In this case you use the proposition: if $W_1,W_2,W_3$ are subspaces, $X=W_1+W_2$ and $Y=W_2+W_3$, you want to prove that
$$
X+W_3=W_1+Y
$$
Let $x\in X,w_3\in W_3$; then, by the proposition, $x=w_1+w_2$, with $w_1\in W_1$, $w_2\in W_2$; then
$$
x+w_3=(w_1+w_2)+w_3=w_1+(w_2+w_3)\in W_1+Y
$$
because $w_2+w_3\in Y$. Thus $X+W_3\subseteq W_1+Y$. The reverse inclusion follows similarly.

About direct sums there is a big misunderstanding. While the definition of “direct sum” in the case of two subspaces is correct, it is incorrect to say that the sum of more than two subspaces is direct when $W_i\cap W_j=\{0\}$ for $i\ne j$.

The condition is stricter, namely that
$$
W_i\cap\sum_{j\ne i}W_j=\{0\},\qquad i=1,2,\dots,n
$$
at least if one wants to stick with the common terminology and one of the most important properties of direct sums, namely that
$$
\dim(W_1\oplus W_2\oplus\dots\oplus W_n)=\dim W_1+\dim W_2+\dots+\dim W_n
$$
in case of finite dimensional spaces.

For instance, the enclosing vector space being $\mathbb{R}^3$, if $W_1$ is generated by $(1,0,0)$, $W_2$ by $(0,1,0)$ and $W_3$ by $(1,1,0)$, it is true that $W_1\cap W_2=\{0\}$, $W_1\cap W_3=\{0\}$, $W_2\cap W_3=\{0\}$, but
$$
\dim(W_1+W_2+W_3)=2\ne\dim W_1+\dim W_2+\dim W_3
$$

In any case, since a direct sum is a sum of subspaces to begin with, proving associativity (once the definition is fixed) and commutativity is not a problem, because it has already been done.

Let's tackle associativity of direct sum. Suppose $W_1,W_2,W_3$ are independent subspaces (meaning that their sum is direct). Then, by definition,
$$
W_1\cap(W_2+W_3)=\{0\}=(W_1+W_2)\cap W_3
$$
Therefore also $W_1\cap W_2=\{0\}=W_2\cap W_3$; hence $W_1+W_2=W_1\oplus W_2$ and $W_2+W_3=W_2\oplus W_3$. Hence
$$
W_1+(W_2+W_3)=W_1\oplus(W_2+W_3)=W_1\oplus(W_2\oplus W_3)
$$
$$
(W_1+W_2)+W_3=(W_1\oplus W_2)+W_3=(W_1\oplus W_2)\oplus W_3
$$
But these are equal by the previous argument.


答:
$$3n^4+3n^2+1-k^2 =0\implies \Delta = 9-12(1-k^2)$$

Since $\Delta$ is divisible by $3$ and it is a perfect square we have $$3d^2 = 4k^2-1$$

Now $2k-1$ and $2k+1$ are relatively prime so we have $$2k-1=3a^2\;\;\;\wedge \;\;\;\; 2k+1=b^2 $$ or $$2k-1=a^2\;\;\;\wedge \;\;\;\; 2k+1=3b^2 $$

In first case we get $2 =b^2-3a^2$ and in second $2=3b^2-a^2$ (which has no solution, see by modulo 3). So we are left with $2 =b^2-3a^2$ which has something to do with the Pell equation...

Title: Show that $3n^4+3n^2+1$ is never a perfect square

Body:
I am looking for a proof for the fact that $3n^4+3n^2+1$ can never be a perfect square for a natural number $n>0$.

I know for a fact that the statement must be true as it came up as one of the cases in a solution of the diophantine equation $y^2=x^3-1$ using the LTE lemma and, according to two different solutions I have come across, this equation has no solutions apart form $(x,y)=(1, 0)$.

I have spent a considerable amount of time looking for a suitable attack strategy, however, I was not able to make any progress. Can anyone help?


答:
Suppose that $3n^4+3n^2+1=m^2$, so that $$3n^4+3n^2+(1-m^2)=0$$  This is a quadratic in $n^2$.  $$n^2=\frac{-3\pm\sqrt{9-12(1-m^2)}}{6}=-\frac12\pm\frac{\sqrt{12m^2-3}}6$$

So if $n$ is an integer, there is an integer $k$ such that $$12m^2-3=k^2$$  Clearly, $3|k$, so let $k=3t$ and let $s=2m$.  Then we have $$3s^2-9t^2=3\\s^2-3t^3=1$$ a Pell's equation.

This is as far as I've carried it, but presumably, solving for $s$ and $t$ and putting it back in terms of $m$ and $k$ will show that you can't have an integer solution to the original equation.


问:
Set difference of an open set and a closed set
Suppose $A$ is an open set and $B$ a closed set. Is it necessarily true that $A \backslash  B$ open? I have thought about a few counter examples but could not come up with any, so I am inclined to say it is true. For example, $\mathbb{R} \backslash \phi$ is both open and closed but this does not impose any problem.

If $A \subset B$ then $A \backslash B = \phi$ which is open. If $A$ and $B$ do not intersect each other then $A \backslash B = A$ which is of course open. The only edge case is when $B \subset A$, which again I think should make $A \backslash  B$ open. Am I right?

答:
ANSEWR ACCORDING COMMENT:

Take $A=(-2,2)$ and $C_n=[-1+1/n,1-1/n]$.

Then $B=\bigcup C_n=(-1,1)$ and $A\setminus B=(-2,-1]\cup[1,2)$.

FIRST ANSWER:

You're not right. You missed the genereal case when $A\cap B\ne\varnothing$ but neither $A\subset B$ nor $B\subset A$.

A better approach:
Recall that $A\setminus B=A\cap B^c$ and take into account that intersection of finite number of open set is always open.


问:
Alternate expression for a Taylor Series
The exercise goes as follows: Represent $f(x)=\sin(x)$ as the sum of its Taylor series centered at $\pi/3$.

This is how I solved the exercise:
$$
f(x) = f(\pi/3) +\frac{f'(\pi/3)}{1!}(x-\pi/3)+\frac{f''(\pi/3)}{2!}(x-\pi/3)^2+\frac{f'''(\pi/3)}{3!}(x-\pi/3)^3+\cdots
$$
$$
= \frac{\sqrt{3}}{2}+ \frac{1}{2\cdot 1!} (x-\pi/3)-\frac{\sqrt{3}}{2\cdot 2!}(x-\pi/3)^2 - \frac{1}{2\cdot 3!}(x-\pi/3)^3+\cdots
$$

Rearranging all terms in two groups, the ones with $\sqrt(3)/2$ as a common factor and $1/2$, we have:
$$
\left(\frac{\sqrt{3}}{2}+ -\frac{\sqrt{3}}{2\cdot 2!}(x-\pi/3)^2 +\cdots\right)+\left(\frac{1}{2\cdot 1!} (x-\pi/3)- \frac{1}{2\cdot 3!}(x-\pi/3)^3+\cdots\right)
$$
$$
=\frac{\sqrt{3}}{2}\left(1+ -\frac{1}{ 2!}(x-\pi/3)^2 +\cdots\right)+\frac{1}{2}\left(\frac{1}{1!} (x-\pi/3)- \frac{1}{ 3!}(x-\pi/3)^3+\cdots\right)
$$
$$
\frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{(-1)^n(x-\pi/3)^{2n}}{(2n)!}+\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n(x-\pi/3)^{2n+1}}{(2n+1)!}
$$

It seemed simple enough to find the pattern that led to the answer.

To corroborate that my answer was correct, I typed the problem in WolframAlpha and I found this elegant, succinct and perplexing answer:
$$
\sum_{n\ge 0}\frac{(-1)^n\cos(1/6(\pi +3n\pi)) (x-\pi/3)^n}{n!}
$$

Does anyone know how this alternate answer was obtained?
Thank you very much for your time and consideration.

答:
Hint: if you use angle-addition we have
$$
\cos(\pi/6 + n\pi/2) = \cos(\pi/6)\cos(n\pi/2) +\sin(\pi/6)\sin(n\pi/2)
$$Then look at what happens if you take $n$ modulo $4$; can you take it from here?


问:
Propositional calculus logical proof help
I am asked to prove:

(Y => Z) V X    ≡    (Y V X) =>(Z V X)

This is what I have so far,

(Y V X) =>(Z V X)

(Definition of => X,Y := Y V X, Z V X)

(Y V X) V (Z V X)≡ (Z V X)

(V SYMMETRY twice )

(X V Y) V (X V Z)≡ (Z V X)

(V/V X,Y,Z := X,Y,Z)

X V (Y V Z) = (Z V X)

How do I get rid of the (Z V X) so that I have the full proof Picture of the proofs I am using

答:
$$\begin{align} (Y \to Z) \lor X &\equiv (Y\lor X) \to (Z\lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv \lnot (Y \lor X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \land \lnot X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \lor Z \lor X) \land \underbrace{(\lnot X \lor Z \lor X)}_{true}\\ \\\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \lor Z \lor X)\end{align}$$

I leave it to you to justify the steps.  If you are unsure of any of them, feel free to comment below the answer.  I've made use of the equivalence between $(P\to Q)\equiv (\lnot P \lor Q)$ (and more than once).  I've also used DeMorgan's law, and the fact that $\lnot X \lor X$ is always true ($\top$).  I used the fact that $\top \lor Z$ is always true, and the fact that $\top \land P \equiv P$.


问:
Showing $\lim_{x\rightarrow0}\frac{1}{x^4}=\infty$ via $\epsilon$-$\delta$

Show that $\lim_{x\rightarrow0}\frac{1}{x^4}=\infty$.

Consider our preliminary work:
$$\begin{align}
|\frac{1}{x^4}-0|<\epsilon &\implies|x^4-0|<\frac{1}{\epsilon}  \tag1\\
&\implies|x^2-0||x^2+0|<\frac{1}{\epsilon} \tag2\\
&\implies|x-0||x+0||x^2+0|<\frac{1}{\epsilon} \tag3
\end{align}$$
Choose $\delta=1$, then
$$|x-0|<\delta=1\implies -1<x<1\Rightarrow|x-0|<1 \tag4$$
Thus,
$$|x-0||x+0||x^2+0|<\frac{1}{\epsilon}\implies|x-0|<\frac{1}{\epsilon} \tag5$$ Now, let $\epsilon>0$ and choose $\delta= \min\{1,\frac{1}{\epsilon}\}$.

What should I do next in writing this proof? I'm confused how to choose $\delta$ so that  $\frac{1}{\epsilon}$ becomes $\epsilon$. Also, do I need to pick multiple upper bounds since I broke my polynomial into three separate absolute values?

答:
Let $M$ be given. Then, take $\delta = \dfrac{1}{\sqrt[4]{M}}$. Thus,

$$0< |x| < \delta \implies |x| < \dfrac{1}{\sqrt[4]{M}} \implies |x^4| < \dfrac{1}{M} \implies |\dfrac{1}{x^4}| > M $$


答:
Completing the proof of A-level Student:
$$C\frac{dC}{dr}+S\frac{dS}{dr}=\frac12\frac{d}{dr}(C^2+S^2)=\frac12\frac{d}{dr}(C+iS)(C-iS)\\
=\frac12\frac{d}{dr}(e^{re^{i\theta}}\cdot e^{re^{-i\theta}})=\frac12(e^{i\theta}+e^{-i\theta})(e^{re^{i\theta}}\cdot e^{re^{-i\theta}})=(C^2+S^2)\cos\theta
$$
where I have used the fact that for any power $p$, $\bar{ z^p }=(\bar{z})^p$


答:
Clearly
$$ C+iS = e^{re^{i\theta}} $$
and hence
$$ C^2+S^2=(C+iS)(C-iS)=e^{re^{i\theta}}e^{re^{-i\theta}}=e^{2r\cos\theta}. $$
So
$$ C\frac{dC}{dr}\ + S\frac{dS}{dr}=\frac12\frac{d}{dr}(C^2+S^2)=\frac12\frac{d}{dr}e^{2r\cos\theta}=e^{2r\cos\theta}\cos\theta=(C^2+S^2)\cos\theta. $$


问:
Need help with transforming a binomial distribution
This week, Bob heads to the roulette table. When betting on red, if the spin lands on red then you win the value of the bet, otherwise you win nothing. Bob will use the Martingale strategy: starting with a bet of $\$1$, he will double his bet after each loss, and return his bet to $\$1$ after each win. Bob will bet on red every time.

(a) Let $Y$ be the net winnings from playing 3 spins using Strategy B,

i) [2] What is the range and pmf of $Y$ ?

The probably the roulette wheel will be red is $\frac{18}{38}$.

So I found the $R(X)=0,1,2,3$

and the $p(x)=\binom3x\cdot (9/19)^x\cdot(10/19)^x$

I am completely slumped with finding the range and pmf of $Y$.

I think the range of $R(Y)$ might $=\{-6,-2,0,1,2,3\}$, (based on the money factored) but I might be wrong.

If anyone can help me answer this question that would be fantastic!

答:
You should be able to do this with a tree diagram, for instance, listing all 8 possibilities.

Another way to help compute the winnings would be to code Bob's outcomes, as W for win, L for loss, to give sequences like $WLL$ (for a win on the first throw, and losses on the second and third throw). Then notice, by the Martingale strategy, substrings of type $L^n W$ ($n \ge 0$) always win a dollar (regardless of "n"). Finally, if Bob finishes his gambling with a string of $n$ losses, $L^n$ amounts to the loss $-(2^n-1)$.

This gives, for the range:

$WWW = 3$; $WWL = 1$; $WLW = W + LW = 1+1 = 2$; $WLL = -2$; $LWW = LW + W = 2$; $LWL = LW + L = 1-1 = 0$; $LLW = 1$; $LLL = -7$ (poor Bob!)

The tricky part is these win/loss amounts are not commutative with respect to the string ($WLL$ does not equal $LWL$).

For the pmf of your discrete RV, $Y$, the probabilities are commutative. You'd just have your odds calculation as above: $(9/19)^{\#wins} (10/19)^{\# losses}$--so eg $P(WLW) = (9/19)^2 (10/19)^1$. At any value in the range of $Y$, you'd add up all the probabilities (for instance $P(Y=-7)=(10/19)^3$).


答:
We can prove it without expending by SOS and the Tangent Line method.

Indeed, $$\sum_{cyc}\frac{b^2-a^2}{a+c}=\frac{1}{2}\sum_{cyc}\frac{2b^2-2a^2}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{2b^2-2a^2}{a+c}+a-c\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{b^2-c^2-(a^2-b^2)}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{a^2-b^2}{c+b}-\frac{a^2-b^2}{a+c}\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{2(a+c)(b+c)}\geq0.$$


问:
Why would an ill-conditioned problem not be solvable with much accuracy when using floating-point arithmetic?
Could anyone explain using an example why an ill-conditioned problem would not be solvable with much accuracy when using floating-point arithmetic to compute an approximate solution?</p>

答:
Consider the common problem of computing $y = f(x)$ where $f :\mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function. If the algorithm is relative backward stable, then the computed value $\hat{y}$ of $y$ satisfies $$ \hat{y} = f(\hat{x})$$ where $$\left|\frac{x-\hat{x}}{x}\right| \leq C u.$$
Here $u$ is the unit round off and $C>0$ is a constant independent of $u$. A good algorithm has a small value of $C$. This is as good as it gets. Now if the problem is ill-conditioned, then small changes in the input can cause large changes in the output. Specifically, if $\bar{x}$ is an approximation of $x$, then we cannot hope to do better than $$\left| \frac{f(x)-f(\bar{x})}{f(x)} \right| \approx \kappa_f(x) \left|\frac{x-\bar{x}}{x}\right|,$$ where $ \kappa_f(x)$ is the relative condition number of $f$ at the point $x$ given by $$\kappa_f(x) = \left| \frac{xf'(x)}{f(x)} \right|.$$ A rigorous deriviation of this relation from an abstract definition of the condition number can be found in this answer to a related question.

In particular, we have the following bound for the forward relative error
$$\left| \frac{ y - \hat{y} }{y} \right| = \left| \frac{f(x)-f(\hat{x})}{f(x)} \right| \approx \kappa_f(x) \left|\frac{x-\hat{x}}{x}\right| \leq C \kappa_f(x) u.$$
In summary, the best we can hope for is a small relative backward error, but this is not enough to guarantee a small relative forward error when the problem is ill-conditioned, i.e., when $\kappa_f(x)$ is large relative to $u$. Conversely, if $C\kappa_f(x)u$ is tiny then all is well and the forward relative error is always small.


问:
How to expand the Fourier series for $f(x)=\max \{0, \frac{\pi}{2}-\lvert x\rvert \} $?
My Question: My Goal is to determine the Fourier series for $f(x)=\max \{0, \frac{\pi}{2}-\lvert x\rvert \} \quad$ for $x \in [-\pi, \pi ]$ This function is $2\pi$-periodic.

My Approach: i found out, $f(x)$ is an even function, because its graph is symmetric to the y-axis. The Expression: $\frac{\pi}{2}-\lvert x \rvert$ has to "zeros":

$\frac{\pi}{2}-\lvert x \rvert = 0$

$\frac{\pi}{2}= \lvert x \rvert  $

$x_{0}=-\frac{\pi}{2}$

$x_{1}=+\frac{\pi}{2}$

the graph must look somehow like that.. so the function is linear all over the Intervall:

But now Comes the hard stuff. I'm stuck in building the Fourier series for $f(x)$
The integral should be builded maybe this way, but i am not sure if if am right:

$\int (\frac{\pi}{2}-x)\cdot \cos (nx)\ dx = \frac{\pi}{2n}\sin (nx) - \frac{x}{n}\sin (nx) - \frac{1}{n^2}\cos (nx)$

What do you think, is that right? And what would be the next step? My textbook doesn't describes the further calculation.

p.s. edits for improving language and latex

答:
Hint: Consider the $2\pi$ continuation of $$f(x)=\begin{cases} 0 & x\in\left[-\pi, -\frac{\pi}{2}\right] \\
  \frac{\pi}{2}+x & x\in \left[-\frac{\pi}{2}, 0\right] \\
\frac{\pi}{2}-x & x\in\left[0,\frac{\pi}{2}\right] \\
 0 & x\in\left[\frac{\pi}{2}, \pi\right]\end{cases}$$

Edit: As your function is even, each $b_n=0$. Also, the integral can be computed by integrating from $0$ to $\pi$ and multiplying the result by 2.


问:
What is the maximum number of marked squares on a 10x10 table s.t. every square has at most one marked neighbour?
There is a $10 \times 10$ table of $100$ unit squares which squares can be marked or unmarked. We want to mark as many squares as possible s.t. every square (marked or unmarked) has at most one marked neighbour. Two squares are adjacent iff they have a common side or vertex.

What is the maximum number of marked squares under these conditions?

答:
Here is a solution with $16$ marked squares (in blue).

The lines $x=2$, $x=5$, $x=8$, $y=2$, $y=5$, $y=8$ divide the grid into $16$ boxes in each of which at most one square can be marked, so this is best possible.


问:
Describe the quotient space that results from the partition of $\mathbb{R}$ into the equivalence classes on $\mathbb{R}$ given by $x-y\in \mathbb{Z}$

Consider the equivalence relation defined in $\mathbb{R}$ defined by $x \sim y$ if $x-y\in \mathbb{Z}$.
Describe the quotient space that results  from the partition of $\mathbb{R}$ into the equivalence classes result from the equivalence relation.

First assume that $\sim$ is a relation of equivalence, following tha definition of class of equivalence
$[x]=\lbrace y\in \mathbb{R}:x \sim y\rbrace = \lbrace x-n:n\in \mathbb{Z} \rbrace$ this for each $x\in \mathbb{R}$
For find the quotient space i need to know the open sets in the quotient space, fix $[x]\in \mathbb{R/ \sim}$ then $p^{-1}([x])$ where $p:\mathbb{R}  \longrightarrow \mathbb{R/ \sim}$ is just
$p^{-1}([x])=\lbrace y\in \mathbb{R}:p(y)\in [x]\rbrace= \lbrace x-n:n\in \mathbb{Z} \rbrace$ therefore since $<\mathbb{Z}>=\mathbb{Z}$ we can change of representant of the class and claim that the quotient is homeomorphic to the $[0,1)$ but i can prove it.
Someone that can give me a proof or contraexample, any help wasvery useful.

答:
The quotient space can be naturally identified with $[0,1)$, true, but not as a group and not as a topological space. The quotient space is $\mathbb R/\mathbb Z$, which is isomorphic (as a group) and homeomorphic to the circle $S^1$. You can map from $[0,1)$ via $t\mapsto e^{2\pi i t}$, and this actually gives you the quotient map from $\mathbb R$ that induces the topology.

If you're not looking for any kind of structure-preserving map, then identifying it with $[0,1)$ is easy and fine. But you mention isomorphism, for which $[0,1)$ would not be appropriate algebraically or topologically.


问:
Formula or upper bound for $\sum_{i=1}^n i^3 {n\choose i} a^i b^{n-i}$
I want to upper bound (or find a closed form for):
$$ \sum_{i=1}^n i^3 {n\choose i} a^i b^{n-i} $$
This seems really close to the binomial formula $\sum_{i=1}^n {n\choose i} a^ib^{n-i} = (a+b)^n$, so I was wondering if there's an easy way to deal with sums of this type.

答: Since $\sum_{i=0}^n\binom{n}{i}(ax)^ib^{n-i}=(ax+b)^n$, you can apply $x\tfrac{d}{dx}$ three times, then set $x=1$. The $i=0$ lower limit has the same effect as taking $i=1$, because $0^3=0$.


问:
Lebesgue measurable simple function and its Borel measurable counterpart
Q) Suppose that $f$ is a simple Lebesgue measurable function on $R^n$. Show that
there exists a Borel measurable function $g$ such that $f(x) = g(x)$ for almost every $x \in R^n$.

I know that a simple Lebesgue measurable function takes finitely many values on Lebesgue measurable sets. But I'm not sure how I can find an appropriate Borel measurable function $g$, i.e. function wherein the pre-image of every open set is Borel measurable, which is the same as $f$ except atmost on a countable set?

答:
The problem boils down to: For each Lebesgue measurable set $A$,
there exists a Borel set $B$ such that $m(A\Delta B)=0$. For simplicity, I demonstrate the proof for the case $n=1$.

Proof: Let $A$ be Lebesgue measurable. Firstly, consider the case
that $m(A)<\infty$. For each $n$, there exists an open set $U_{n}$
such that $A\subseteq U_{n}$ and $m(U_{n})-m(A)<\frac{1}{n}$. Define
$B=\cap_{n}U_{n}$, which is a Borel set (Actually, we can say more:
it is a $G_{\delta}$-set. However, we do not need this fact.) Clealry
$A\subseteq B$. Moreover $B\setminus A\subseteq U_{n}\setminus A$
implies that $m(B\setminus A)\leq m(U_{n}\setminus A)=m(U_{n})-m(A)<\frac{1}{n}$.
Since $n$ is arbitrary, we have $m(B\setminus A)=0$. Next, we drop
the assumption that $m(A)<\infty$. For each $n\in\mathbb{Z}$, let
$A_{n}=A\cap(n,n+1]$. For each $n$, choose a Borel set $B_{n}$ such
that $A_{n}\subseteq B_{n}$ and $m(B_{n}\setminus A_{n})=0$. Let
$B=\cup_{n}B_{n}$, which is a Borel set. Clearly $A\subseteq B$.
Moreover, $B\setminus A=\cup_{n}(B_{n}\setminus A)\subseteq\cup_{n}(B_{n}\setminus A_{n})$,
so $m(B\setminus A)=0$.


问:
Problem related with a linear transformation and its rank,nullity...
I am stuck on the following problem:

Suppose $W$ be  a vector space of dimension $m \geq 2.$ Let $F \colon W \to W$ be a linear transformation such that $F^{n+1}=0$ and $F^n \ne 0$ for some $n \ge 1.$ Then I have to check which of  the following statement(s) is/are true?

1.$\text{trace}(F) \ne 0 $

2.$\text{Rank}(F^n) \le \text{Nullity}(F^n)$

3.$n=m$

4.$F$ is diagonalizable.

I do not know how to progress with the problem. Any kind of hints with some explanation will be helpful.

答:
To start with (and indeed to answer some of the points) try writing down the simplest (who knows, maybe a $2\times 2$) nonzero matrix you can think of which squares to 0.

For the second point, think about what happens to a vector $v$ as we map $v\mapsto Fv \mapsto F^n F v$.


问:
Sets and the principle of inclusion and exclusion
I found an interesting problem regarding the principle of inclusion and exclusion.

Given $1985$ sets, each containing $45$ items, find their total union, if the union of any two is $89$.

This is what I did:
First, I found out the intersection of any two. Using the principle of inclusion and exclusion, I managed to find out that |$A_{1}$|$\cup$|$A_{2}$|=|$A_{1}$|+|$A_{2}$|-|$A_{1}$|$\cap$|$A_{2}$|. From this, we know that $89=45+45-x$, where $x=1$. The first thing that came to my mind was that this intersection could be the common intersection of all the sets:

A solution I found to the entire problem is that there is one element that every set has in common, and 44 elements that each set does not share with no other set. To clarify, one can imagine a flower, whose petals are individual sets, their only common item being the center of the flower. Each and every pair of sets therefore contains the required 89 elements ($44$ elements on one petal $+$ $44$ elements on the other petal $+ 1$ element in the centre), each set also contains $45$ elements. Therefore, the total sum of all the elements, or the union of all the sets, is $1985\times44+1$. But I do not know how to prove this with formulas and expressions. I am also not sure whether this is correct, because of what I found when I tried to solve a similar problem but only with three sets:

The rules of this problem are exactly the same as before, only now we have $3$ sets instead of $1985$. I found out two ways to arrange the elements of the sets which satisfy the rules, each of them having a different total union.

If we had a three-set Venn diagram, one solution is this:

1 element in the intersection of them all ($A \cap B \cap C$)

44 elements in each individual set but not in any intersection (in $A$,$B$,$C$)

The other solution is this:

1 element in each intersection of every pair, but not in the intersection of them all

43 elements in each individual set.

Their unions are different, because while in the first solution, the union is $44\times3+1$, while in the second solution, the union is $43\times3+3$. These are different numbers and therefore I am unsure about my solution to the original problem. My question is whether there is an analytical way to solve this problem, or whether I am missing something.

答:
Claim: There is one element that every set has in common.

Proof by contradiction. Suppose there isn't such an element.

Fix a set $A_1$.

For each element $a_{1,i} \in A_1$, let $ A_{1,i}$ denote that sets (not including $A_1$) which contain $a_{1,i}$.

The $A_{1,i}$ are disjoint from each other, so $\sum |A_{1,i}| = 1985 - 1$.

Fix an element $a_{1,i} \in A_1 $.

By the assumption, $|A_{1,i} | < 1984$, and so there is another $j\neq i$ such that $ a_{1,j} \in A_1$ and $|A_{1,j}| > 0 $.

Let $B_k \in A_{1,j}$, where $B_k$ is one of the original sets with 45 elements.

We will prove by contradiction that $|A_{1,i}| \leq 44$.

Suppose not, so$ |A_{1,i}| \geq 45$. Then $B_j \backslash \{ a_{1,j}\} $ has 44 elements, and doesn't contain $a_{1,i}$.

So $B_k$ cannot intersect the 45+ sets in $A_{1,i}$, which are distinct sets after excluding $a_{1,i}$, which is a contradiction.

This shows that $ |A_{1,i} | \leq 44$.

Coming back to the original claim, we have
$$1984 = \sum_{i=1}^{45} |A_{1,i} | \leq 45 \times 44 = 1980,$$
which is a contradiction.


问:
Derivative of different variables and using chain ruleb

Suppose $\phi(x)$ is defined by a formula in which x occurs in several
places (for example, there are three $x$’s in $\phi(x)=
> \frac{x^2e^x}{x+3}$).

Show that the derivative $\phi'(x)$ is obtained by differentiating
with respect to each of the $x$’s in turn, treating the others as
constants, and adding the results.

Hint: Let $F(x_1, . . . , x_n)$ be the function of several variables
obtained by replacing each of the $x$’s in the formula for $\phi(x)$
by a different variable.

I need to Express $\phi$ in terms of F and use the chain rule. Now i am having trouble in proving that $\phi'(x)$ is actually the sum of partial derivatives of the function. And therefore cant really reach to a conclusion for this question. The answer in the text book was given the following

```<code>E“_  E“?ŽY“?ŽLŽY“_```
```

Body:
Hint

By the quotient rule,
$$
\frac{d}{dx}\left(\frac{x^2e^x}{x+3}\right)=\frac{(x^2e^x)'(x+3)-(x^2e^x)(x+3)'}{(x+3)^2} \, \text{.}
$$


问:
$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$ Proof
Having $c\gt0$

$a, c \in \Bbb{R}$

$$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$$

I don't know how to solve this inequality, I tried with the AM-GM inequality and I reached this point so far.

$$(a+b)^2\le a^2(c+\frac{1}{c})+b^2(c+\frac{1}{c})$$

I don't know how to continue, and I don't even know if this is the right path.

If you have any hints, those would be much appreciated.
Thanks

答:
We have that

$$(a+b)^2\le(1+c)a^2+\left(1+\frac{1}{c}\right)b^2 $$

$$a^2+b^2+2ab\le a^2+b^2+ca^2+\frac1cb^2$$

$$2ab\le ca^2+\frac1cb^2 \iff \frac{ca^2+\frac1cb^2}2 \ge ab$$

which is true by AM-GM.


问:
Taylor expansion of matrix product
Suppose we have two real $n \times n$ matrices $A$ and $B$. Consider the linear operator $T = A^{-1} B$ which maps $\mathbb{R}^n$ to itself.

I'm trying to calculate a Taylor expansion of the function $T(A)$ about $A = B$,

$$
T(A) = I + \dfrac{\partial T}{\partial A}|_{A=B} (A - B) + \dfrac{1}{2} \dfrac{\partial^2 T}{\partial A^2}|_{A=B} (A - B)^2 + \dots
$$
but I'm stuck at calculating $\partial T / \partial A$.

I know that $\partial A^{-1} / \partial A = -A^{-2}$, and that $\partial T / \partial A^{-1} = B^T$. In what order do these derivatives need to be multiplied to obtain $\partial T / \partial A$?

I'm also not sure whether the order of multiplication in my Taylor series is correct. Any advice on this would be much appreciated!

答:
Hint: For $ \frac{\partial T}{\partial A} $ you can use the fact that

$$ \partial (XY) = (\partial X)Y + X(\partial Y) $$

Now substitute $ X \mapsto A^{-1} $ and $ Y \mapsto B $.

Title: Taylor expansion of matrix product


问:
What is the limit of $(3n^2 + n)^{1/n}$ and why?
I am really unsure of how to answer this as I have tried taking out n and using the ratio lemma and am still confused!</p>

Body:
Try to factor $3n^2$ out.

$$(3n^2 + n)^{\frac{1}{n}} = (3n^2)^\frac{1}{n}(1 + \frac{1}{3n})^{\frac{1}{n}}$$

Now notice that

$$\lim_{n \rightarrow \infty} (3n^2)^\frac{1}{n} = 1$$
$$\lim_{n \rightarrow \infty} (1 + \frac{1}{3n})^{\frac{1}{n}} = 1$$

Hence

$$\lim_{n \rightarrow \infty} [(3n^2 + n)^{\frac{1}{n}}] = \lim_{n \rightarrow \infty} [(3n^2)^\frac{1}{n}(1 + \frac{1}{3n})^{\frac{1}{n}}] = 1 $$


问:
$\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$
This problem was proposed by Cornel and he showed that

$$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)$$

here is my approach.

we know that $\quad\zeta(2)-H_{2n}^{(2)}=\psi^{(1)}(2n+1)=-\displaystyle\int_0^1\frac{x^{2n}\ln x}{1-x}\ dx$

then
$$S=-\int_0^1\frac{\ln x}{1-x}\sum_{n=1}^\infty\frac{x^{2n}H_n}{n}\ dx=-\int_0^1\frac{\ln x}{1-x}\left(\frac12\ln^2(1-x^2)+\operatorname{Li}_2(x^2)\right)\ dx$$
and by applying IBP , we get $$S=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\ln(1-x^2)}{x(1-x^2)}\ dx$$
I applied the dilogarithmic identity $\quad\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)$

but it was not that helpful. any idea?

答:
A solution by Cornel Ioan Valean

If we set $a_n=H_n/n$ and $b_n=\zeta(2)-H_{2n}^{(2)}$, and then apply Abel's summation, we arrive at
$$\sum_{n=1}^{\infty}\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$$
$$=\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(2n+1)^2}}_{\displaystyle A}+\underbrace{\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{(2n+1)^2}}_{\displaystyle B}+\frac{1}{8}\underbrace{\sum_{n=1}^{\infty}\frac{H_n^2+H_n^{(2)}}{n^2}}_{\displaystyle C}-\frac{1}{4}\underbrace{\sum_{n=1}^{\infty}\frac{H_n}{n^3}}_{\displaystyle 5/16\zeta(3)}.\tag1$$

For the sum $A$, we consider a result from the book (Almost) Impossible Integrals, Sums, and Series, page $291$,
\begin{equation*}
A=\sum_{k=1}^{\infty} \frac{H_k^2-H_k^{(2)}}{(k+1)(k+n+1)}
\end{equation*}
\begin{equation*}
=\frac{(\psi(n+1)+\gamma)^3+3(\psi(n+1)+\gamma)(\zeta(2)-\psi^{(1)}(n+1))+2(\zeta(3)+1/2\psi^{(2)}(n+1))}{3n},
\end{equation*}
where if we multiply both sides by $n$, differentiate once with respect to $n$, and set $n=-1/2$, we arrive at
\begin{equation*}
\sum_{n=1}^{\infty} \frac{H_n^2-H_n^{(2)}}{(2n+1)^2}=\frac{15}{4}\zeta (4)-7 \log(2) \zeta(3)+3\log^2(2)\zeta(2).
\end{equation*}

For the sum $B$, we apply Abel's summation with $a_n=1/(2n+1)^2$ and $H_n^{(2)}$ that gives
\begin{equation*}
B=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{(2n+1)^2}=\frac{15}{8}\zeta(4)-\frac{7}{4}\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}+2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^{(2)}}{n^2}
\end{equation*}
\begin{equation*}
=\frac{1}{3}\log^4(2)-2\log^2(2)\zeta(2)+7\log(2)\zeta(3)-\frac{121}{16}\zeta(4)+8 \operatorname{Li}_4\left(\frac{1}{2}\right),
\end{equation*}
where
\begin{equation*}
\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2}=\frac{1}{6}\log^4(2)-\log^2(2)\zeta(2)+\frac{7}{2}\log(2)\zeta(3)-\frac{51}{16}\zeta(4)+4\operatorname{Li}_4\left(\frac{1}{2}\right),
\end{equation*}
may be found calculated in (Almost) Impossible Integrals, Sums, and Series, pages $505$-$506$.

Finally, for the sum $C$ we may elegantly exploit the fact that $\displaystyle\int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x =\frac{H_n^2+H_n^{(2)}}{n}$, you may also find nicely calculated in (Almost) Impossible Integrals, Sums, and Series, page $60$. In general, such integrals may also be viewed in terms of Beta function and hence a solution is pretty easily extracted. Then,
$$C=\sum_{n=1}^{\infty}\frac{H_n^2+H_n^{(2)}}{n^2}=-\int_0^1 \frac{\log^3(1-x)}{x}\textrm{d}x=-\int_0^1 \frac{\log^3(x)}{1-x}\textrm{d}x=6\zeta(4). $$

Collecting all sums in $A$, $B$, $C$ and plugging them in $(1)$, we arrive at the desired result
$$\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac{1}{3}\log^4(2)-\frac{1}{2}\log^2(2)\zeta(2)+\frac{7}{2}\log(2)\zeta(3)-\frac{21}{4}\zeta(4)+8\operatorname{Li}_4\left(\frac{1}{2}\right).$$

COOL BONUS

Combining the sums $A$ and $B$ we obtain that
$$\sum_{n=1}^{\infty} \frac{H_n^2}{(2n+1)^2}=\frac{1}{3}\log^4(2)+\log^2(2)\zeta(2)-\frac{61}{16}\zeta(4)+8 \operatorname{Li}_4\left(\frac{1}{2}\right).$$


答:
The multivariable chain rule says that if you assume each variable $x_i = x_i(t)$ is a function of another variable $t$, then the total derivative is

$$ \frac{dF}{dt} = \sum_i \frac{\partial F}{\partial x_i} \frac{dx_i}{dt} $$

If you take $x_i(t) = t$ for all $i$, then $\frac{dx_i}{dt} = 1$, and so you get that $\frac{dF}{dt}$ is the sum of the partial derivatives (with $x_i = t$ substituted).

Title: Derivative of different variables and using chain ruleb

Body:
Suppose $\phi(x)$ is defined by a formula in which x occurs in several
places (for example, there are three $x$’s in $\phi(x)=
> \frac{x^2e^x}{x+3}$).

Show that the derivative $\phi'(x)$ is obtained by differentiating
with respect to each of the $x$’s in turn, treating the others as
constants, and adding the results.

Hint: Let $F(x_1, . . . , x_n)$ be the function of several variables
obtained by replacing each of the $x$’s in the formula for $\phi(x)$
by a different variable.

I need to Express $\phi$ in terms of F and use the chain rule. Now i am having trouble in proving that $\phi'(x)$ is actually the sum of partial derivatives of the function. And therefore cant really reach to a conclusion for this question. The answer in the text book was given the following

```<code>E“_  E“?ŽY“?ŽLŽY“_```
```


Title: $(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$ Proof

Body:
Having $c\gt0$

$a, c \in \Bbb{R}$

$$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$$

I don't know how to solve this inequality, I tried with the AM-GM inequality and I reached this point so far.

$$(a+b)^2\le a^2(c+\frac{1}{c})+b^2(c+\frac{1}{c})$$

I don't know how to continue, and I don't even know if this is the right path.

If you have any hints, those would be much appreciated.
Thanks

Body:
By C-S $$(1+c)a^2+\left(1+\frac{1}{c}\right)b^2=$$
$$=\left(\frac{1}{1+c}+\frac{c}{1+c}\right)\left((1+c)a^2+1+\frac{(1+c)b^2}{c}\right)\geq(a+b)^2.$$


答:
hint

Put $$c=d^2, A=ad \;and\; B=\frac bd$$

By AM-GM, we have

$$2AB\le A^2+B^2$$

or

$$2ab\le a^2c+\frac{b^2}{c}$$


问:
Help with the notation of the definition of vector spaces
My textbook gives the following statement as of what a vector space is:

Let $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$. A vector space onto $\mathbb{F}$ consists of a non empty set $V$, a representation $V$x$V \rightarrow V, (u_1,u_2) \mapsto u_1 + u_2$, which is called addition and a representation $(\mathbb{F}$x$V) \rightarrow V, (λ,u) \mapsto λu$, which is called scalar multiplication of an element of $\mathbb{F}$ with an element of V, for which the following axioms are true: ...

Can someone explain what the notation that we used for the addition and multiplication mean? e.g. what does $V$x$V$ mean, and why is this arrow $\mapsto$ used? What does it mean in this context? Thank you!

答:
The Cartesian product $V\times V$ is the set of ordered pairs of vectors of $V$.

Addition is a binary operation, or a function from $V\times V$ to $V$;

this can be written as $+:V\times V\to V $.

The "maps to" symbol "$\mapsto$" shows where a function takes an individual element;

e.g., $+:(v_1,v_2)\mapsto v_1+v_2$.

Scalar multiplication is a function from $\mathbb F\times V$ to $V$,

mapping the ordered pair $(\lambda,v)$ to $\lambda v$, where $\lambda\in\mathbb F$ and $v\in V$.


答:
$$L=\underset{n\to \infty }{\text{lim}}\left(3 n^2+n\right)^{1/n}$$
Take the logarithm of the limit
$$\log L=\log \underset{n\to \infty }{\text{lim}}\left(3 n^2+n\right)^{1/n}=\underset{n\to \infty }{\text{lim}}\frac{\log \left(3 n^2+n\right)}{n}=0$$
If $\log L=0$ then $L=1$


问:
$A^X+B^Y=C^Z\pm 1$ Beal's conjecture "almost" solutions
Beal's conjecture is a generalization of Fermat's last theorem. Fermat's last theorem states that there are no solutions to the equation $A^N+B^N=C^N$ where $A,B,C,N\in \Bbb{N}\space |\space N\ge 3$ In Beal's conjecture the exponents are allowed to vary independently, but there is an added restriction that the bases are coprime. So Beal's conjecture states that there are no solutions to the equation $A^X+B^Y=C^Z$ where $A,B,C,X,Y,Z\in\Bbb{N}\space |\space X,Y,Z \ge 3$ and $gcd(A,B)=gcd(B,C)=gcd(A,C)=1$

(As far as I'm aware this has not been proven or disproven yet.)

At one point Ramanujan looked into Fermat's last theorem he created generating functions which gave an infinite number of near miss solutions to Fermat's last theorem they are of the form $A^3+B^3=C^3\pm 1$ here are the generating functions and a few examples

My question: is there at least one near miss solution to Beal's conjecture? $$A^X+B^Y=C^Z\pm 1 \quad \text{{1}}$$
where $A,B,C,X,Y,Z\in\Bbb{N}\space |\space X,Y,Z \ge 3$ and $gcd(A,B)=gcd(B,C)=gcd(A,C)=1$

To be clear one near miss solution is sufficient. A generating function isn't necessary. (Although it would be extremely cool if someone finds one.)

Every equation which consists only of integer terms has an even number of odd terms. There can't be zero odd terms in {1} because $\pm 1$ is odd. There can't be two odd terms in {1} because that implies that there are two even terms and therefore one of the three gcd pairs isn't one. This leaves the only possibility that all terms in {1} are odd. Only odd bases can generate odd power numbers so $A$,$B$, and $C$ are all odd.

I have found two cases where all conditions are met except for coprime condition
$3^3+15^4=37^3-1$ and $7^4+19^3=21^3-1$

I also found another case where the power numbers meet all conditions except for the
$\pm 1$ and was only two off $31^3+63^3=23^4-3$

If there is a counter example one of the terms in {1} must be above $10^6$

Edit: I Realize that I should have stated this from the beginning. I want to avoid really trival counter examples so $A\neq 0, B\neq 0, C\neq 0, C\neq 1$. One of $A$ or $B$ can equal one but if this is the case the equation of the form $A^X+B^Y=C^Z+1$ cannot be used. If one of $A$ or $B$ is one then only the form $A^X+B^Y=C^Z-1$ is allowed. (If one of $A$ or $B$ is one then this would be equivalent to finding a pair of  odd power numbers where both of the exponents are greater than two and the difference between the numbers in the pair is two. I don't think such a pair exists.)

答:
Among the sixty solutions of $A^3 + B^3 = C^3 - 1$ tabulated in OEIS sequences A050787, A050788 and A050789, there are three that also meet the coprimality condition:
$$
25765^3 + 33857^3 = 38239^3 - 1,
$$
$$
54101^3 + 56503^3 = 69709^3 - 1,
$$
and
$$
51293^3 + 64165^3 = 73627^3 - 1.
$$
In the second example, each of $A,B,C$ is actually a prime.

Similarly, $368$ solutions of $A^3 + B^3 = C^3 + 1$ are tabulated in OEIS sequences A050791, A050792 and A050793.  In this much longer list, $18$ triples are pairwise coprime; here are the first three:

$$
35131^3 + 76903^3 = 79273^3 + 1,
$$
$$
190243^3 + 219589^3 = 259495^3 + 1,
$$
and
$$
466727^3 + 867679^3 = 910541^3 + 1.
$$
Only one of the $18$ solutions in this set involves three primes:
$$
457047089^3 + 552546031^3 = 641644319^3 + 1.
$$


问:
X-sided dice probability
I have an $x$-sided dice and I roll it $y$ times.

What is the probability that I will see at least $z$ of the $x$ sides?

e.g. If I rolled a $6$-sided dice $8$ times, what is the probability I see at least $4$ of the sides?

But I want to do with much bigger numbers

Specifically, I want to know  what is probability I see at least $z$ of the $160,000$ sides if I rolled a $160,000$-sided dice $25,000$ times.

Thanks in advance!

答:
Possible practical approach (without necessarily needing to approximate):

Define $P(a,b)$ to be the probability of seeing at least $z$ distinct sides after $a$ additional rolls having seen already $b$ distinct sides. Then we have a simple recurrence
$$P(a,b)=\frac{b}{x}P(a-1,b)+(1-\frac{b}{x})P(a-1,b+1)$$
Base cases: $P(a,b)=0$ if $a+b<z$ and $P(a,b)=1$ if $b \geq z$.

You want $P(y,0)$.

This is around three hundred million subproblems for $y=25000$.

For example, for $z = 23145$, we get about $0.504$ (using double-precision floating point with no concern for numerical issues, took maybe a minute to compute by computer).

($23145$ chosen as an "interesting" value of $z$ using the coupon collector problem to calculate how many distinct coupons to collect from $160000$ to give expected number of draws of about $25000$, ie $160000 (H_{160000}-H_{160000-23145}) \approx 25000$ where $H_n$ is the $n$th harmonic number; can also get $23145$ from Empy2's answer.)

Title: X-sided dice probability


答:
The symbol $A \to B$ is general set theory shorthand for "a function with domain $A$ and range $B$". Often functions come with names, in which case one would put the name first, like this: $f : A \to B$, which means "a function, named $f$, with domain $A$ and range $B$". So without the name, one might also read $A \to B$ as "an unnamed function with domain $A$ and range $B$.

The notation $A \times B$ is general set theory notation for "the cartesian product of a set $A$ and a set $B$".

You might benefit from finding an account of basic set theory notation and usage. Undergraduate textbooks on the "foundations of mathematics" will contain these kinds of things, and sometimes more advanced textbooks will contains these in an introductory section. I quite like the introduction of Munkres "Topology" for that purpose, although that's quite an investment if all you want is to learn some linear algebra.

Title: Help with the notation of the definition of vector spaces

Body:
My textbook gives the following statement as of what a vector space is:

Let $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$. A vector space onto $\mathbb{F}$ consists of a non empty set $V$, a representation $V$x$V \rightarrow V, (u_1,u_2) \mapsto u_1 + u_2$, which is called addition and a representation $(\mathbb{F}$x$V) \rightarrow V, (λ,u) \mapsto λu$, which is called scalar multiplication of an element of $\mathbb{F}$ with an element of V, for which the following axioms are true: ...

Can someone explain what the notation that we used for the addition and multiplication mean? e.g. what does $V$x$V$ mean, and why is this arrow $\mapsto$ used? What does it mean in this context? Thank you!


问:
prove the following using the triangle inequality
QUESTION: https://imgur.com/gpaUg3V

can anyone check my MY ANS:

let a , b, c ∈ R. then

| |a| + b - c| = ||a|| + (b - c)|

| |a| + b - c| ≤ ||a|| + |b - c| ---- by triangle inequality and x=|a| and y= b - c

≤ ||a|| + |b| + |-c| ---- triangle again on |b - c| [and -c = +(-c)]

≤ |a| + |b| + |c| ---- |-c| = |c| and ||a||=|a|

答:
Your answer looks OK

By the triangle inequality:
$$|a|+|b|+|c|=||a||+|b|+|-c|\geq||a||+|b-c|\geq||a|+b-c|.$$


答:
$V \times V$ is a notation for the cartesian product. The elements of $V \times V$ are basically ordered the pairs $(v_1,v_2)$ with $v_1,v_2 \in V$. In general when you have two sets $A,B$, $A \times B$ denotes the the set containing all the ordered pairs $(a,b)$ with $a \in A, b \in B$.

The $\mapsto$ symbol denotes what your map does with the elements. Its the same thing as writing $\varphi((v_1,v_2))=v_1+v_2$ in your example.


问:
There are 7 boys and 8 girls sitting in a row. In how many different ways can they sit if certain 3 girls always sit together?
question There are 7 boys and 8 girls sitting in a row. In how many different ways can they sit if certain 3 girls always sit together? Do you to 15! and 3! then minus them.Im very confused how to get this answer.</p>

答: Hi I cannot comment now but I will give you a hint. First create a three-girl seat, then the problem reduces to 7 boys and 6 girls sitting in a row. Then within that three-girl seat you have 3! ways of arranging the three girls.</p>


问:
Real world uses of homotopy theory
I covered homotopy theory in a recent maths course. However I was never presented with any reasons as to why (or even if) it is useful.

Is there any good examples of its use outside academia?

答: Homotopy methods can be used for solving nonlinear equations (algebraic and differentials) where these equation appear in different problems in engineering and science. One example of these equations is a system of nonlinear algebraic equations which is a model of electrical circuit. This mathematical model consists of two equations which can be easily solved using Newton homotopy to find the value of the current in this circuit. For more information, please see my PhD thesis entitled "Newton homotopy algorithms for solving nonlinear systems" and my papaers presented in some conferences and published in some journals upon my name: Talib hashim Hasan.</p>

ContentLicense: CC BY-SA 2.5

Title: Real world uses of homotopy theory


问:
Prove: $f(x) = \sin(x) + 2x$ is monotonically increasing

Prove, without using derivatives, that

$$f(x) = \sin(x) + 2x$$

is monotonically increasing.

My intuition says it has something to do with the fact that $|\sin(x)| \leq |x|$ but I couldn't figure it out.

答:
Note : $|\sin x- \sin y|\leq |x-y|$ for all $x,y \in \mathbb{R}$.

For the function $f(x)=\sin x+2x$    $\quad$  and let $x<y$ then

$\sin x-\sin y \leq |\sin x- \sin y|\leq |x-y|$

$\implies \sin x-\sin y +2(x-y) \leq |\sin x- \sin y|+2(x-y) \leq |x-y|+2(x-y)$

$\implies f(x)-f(y) \leq |x-y|+2(x-y)$

$\implies f(x)-f(y)\leq -x+y+2(x-y)$

$\implies f(x)-f(y) \leq x-y<0$

$\implies f(x)<f(y)$.


问:
Why does this function and contour I chose not work in solving this integral?
I'm trying to solve

$$ \int_{-\infty}^\infty \frac{\cos x}{1+x^2}\, \mathrm{d} x,$$

which has been done to death and widely available. What I'm confused with is a particular function I chose and why it won't produce the correct answer $\pi/e$.

For a complex variable $ z $, consider the integral

$$ \int_C \frac{\cos z}{1+z^2}\, \mathrm{d} z, $$

where $ C $ is the usual infinitely large semicircle in the upper-half of the complex plane. There is one pole in this region at $ z = i $, so the integral is equal to $ 2 \pi i \mathrm{Res}(\cos z/(1 + z^2), i) = \pi \cosh(1) $. Now, we can split the integral into 2 portions, one along $ z = x $ and another along $ z = r \mathrm{e}^{i \theta} $:

$$
\begin{align}
    \int_C \frac{\cos z}{1+z^2}\, \mathrm{d} z
    &= \int_{-\infty}^\infty \frac{\cos x}{1+x^2}\, \mathrm{d} x
       + \int_0^{\pi} \frac{\cos\left(r\mathrm{e}^{i \theta} \right)}{1+r^2 \mathrm{e}^{2 i \theta}}\, \mathrm{d} \theta
    = \int_{-\infty}^\infty \frac{\cos x}{1+x^2}\, \mathrm{d} x
    = \pi \cosh 1,
\end{align}
$$

since when $ r \to \infty $, the 2nd integral vanishes. But, this isn't correct, it should be $ \pi/e $. I'm not understanding why this is wrong.

答:
The problem is that your function does not vanish on the semicircle. The reason is that $\cos(z)$ is not bounded by $\pm 1$, and can go to infinity. So you need to express your function is a different way. Say $$\cos(z)=\frac{e^{iz}+e^{-iz}}2$$
Once you split the integral into two parts, you notice that one of the integral will vanish on the upper contour, one will vanish on the lower one. $$\int_{-\infty}^\infty\frac{\cos x}{x^2+1}=\frac12\int_{-\infty}^\infty\frac{e^{ix}}{x^2+1}+\frac12\int_{-\infty}^\infty\frac{e^{-ix}}{x^2+1}$$
We continue the first integral on the upper side, where it will vanish, and the second on the lower side. For the first you need to calculate the residue at $i$, for the second at $-i$.
$$\int_{-\infty}^\infty\frac{\cos x}{x^2+1}=\pi i\mathrm{Res}\left(\frac{e^{iz}}{z^2+1},i\right)+\pi i\mathrm{Res}\left(\frac{e^{-iz}}{z^2+1},-i\right)=\pi i\frac{e^{-1}}{2i}+\pi i\frac{e^{-1}}{2i}=\frac\pi e$$


问:
Surjectivity of natural morphism between multiplicative modulo groups
Let $n$ be an integer and $d$ be a divisor of $n$. Then there is a natural group homomorphism:
\begin{align}\varphi:(\mathbb{Z}/n\mathbb{Z})^{\times} &\to (\mathbb{Z}/d\mathbb{Z})^{\times}\\ [x] &\mapsto [x] \end{align}
I want to know whether $\varphi$ is surjective. This may be a silly question, but I don't know where to look for the answer. Any help would be appreciated.

答: $\varphi$ is surjective. By the Chinese remainder theorem it suffices to show this when $n$ and $d$ are powers of the same prime, say $p^a$ and $p^b$ where $b \le a$. In this case it's clear because every lift of a unit $\bmod p^b$ is a unit $\bmod p^a$ (since it's not divisible by $p$).


问:
When does an optimization problem have a unique solution, no solution, or infinitely many?
What conditions are required for an optimization problem (e.g., given  $f(x) = a \cdot x^2 + b \cdot x + c $, find $\min f(x)$) to have a unique solution, no solution, or infinitely many?

I tried to analyze the function by taking its derivative(s) and to look for minima (or maxima). But is that the only way to approach such a problem? Or is there a faster – and a more efficient – way?

答:
Generalities

Let us introduce $X \subseteq \mathbb{R}$, the set of admissible solutions of your optimization problem.

If $X$ is closed and convex ($X = \mathbb{R}$ or $X = [a, b]$), if $f$ is strictly convex on $X$ ($\forall x \in X, \; f''(x) > 0$), then you have existence and uniqueness of a global minimum.

When $X = \mathbb{R}$, if $f$ is coercive ($f(x) \to \infty$ when $\vert x \vert \to \infty$), then there exists a minimum (but you have not uniqueness).

When $X$ is compact (bounded and closed, e.g. a closed interval $X = [a, b]$), if $f$ is continuous, then there exists a minimum (but you have not uniqueness).

If $X$ is open ($X = \mathbb{R}$ for example), any minimizer $x_0 \in X$ verifies $f'(x_0) = 0$ (it is a necessary condition: you need convexity at $x_0$ to ensure it to be a local minimum, i.e $f''(x_0) > 0$). In general, the minimum is not ensured to be global.

All these properties can be generalized when optimizing in $\mathbb{R}^n$ or a subset of $\mathbb{R}^n$.

In general, apart for strict convexity, it may be hard to prove uniqueness of a minimum, or to check that a minimum is global.

Answer to your question

On your example $X = \mathbb{R}$, you can divide cases depending on $a$ and $b$:

when $a > 0$, you have a unique global minimizer because of strict convexity

when $a = 0$, $b \neq 0$, you don't have any minimizer, since $f'(x) = 0$ has no solution and $X$ is open (see point 4)

when $a = 0$, $b = 0$, you have an infinite number of minima as $f$ is identically equal to $c$

when $a < 0$, $f$ is strictly concave and goes to $-\infty$ when $\vert x \vert \to \infty$: you can check that no points verify $f'(x) = 0$ in this case apart at some $x_0$ which is a global maximizer


答: Think 3 girls as one girl.  Then we have 7 boys and 6.       (8-3+1) girls so 13 people are here. We know 13 people sit in 13! different ways. But don't forget we get three girls as one and they can sit in 3! different ways. It means those all people can sit in 13! * 3! different way</p>

Body: question There are 7 boys and 8 girls sitting in a row. In how many different ways can they sit if certain 3 girls always sit together? Do you to 15! and 3! then minus them.Im very confused how to get this answer.</p>


问:
Measure theory and existence of set integral
Let $(X,S,\mu)$ be a measure space. Suppose $f\in \mathcal{L}^p$. Let $\epsilon>0$. Show that there exists a set $E \in S$ with $\mu(E)<\infty$ such that if $F\in \mathcal{S}$ and $F\cap E=\varnothing$ then $||f\chi_{F}||_p<\epsilon$

My attempt was to do something like:

For $f\in \mathcal{L}^p$, clearly $|f|^p \in \mathcal{L}^1$. Put $E_n=\{ x\in X$ $:$ $|f|^p(x)\geq \frac{1}{n}\}$

Then, $|f|^p\chi_{E_n}\leq |f|^p$ . Since $|f|^p\chi_{E_n} \rightarrow |f|^p$  as $n\rightarrow \infty$ , we get by DCT that

$\lim_{n\rightarrow \infty}\int |f|^p \chi_{E_n} d\mu= \int |f|^p d\mu <\infty$.

How do I complete the proof? Is it possible from this stage?

答:
Once you have
$$
\lim\int_{E_n}|f|^p=\int |f|^p
$$
choose $N$ big enough such that
$$
\int |f|^p < \int_{E_N}|f|^p + \epsilon^p.
$$
Then for $E=E_N$ and for any $F \cap E=\emptyset$
$$
\int_F |f|^p+\int_{E_N}|f|^p \leq \int |f|^p < \int_{E_N}|f|^p + \epsilon^p.
$$
Then
$$
\int_F |f|^p<  \epsilon^p.
$$


问:
Proving that if $n$ is odd and $\gcd(m, n) = 1$, then $\gcd(2m + n, 2n) = 1$
I've been trying to crack this one for the last little while. I've tried a few approaches, but none have bore any fruit.

Let $n > 0$ be an odd integer. Prove that if $\gcd(m, n) = 1$, then $\gcd(2m + n, 2n) = 1$.

答: Hint: Suppose $k\mid 2n,2m+n$. Could $k$ be even? What can you deduce?


答:
Let
$$f(x) = ax^2 + bx +c.$$
Suppose $a \ne 0.$
Then
$$f(-\frac{b}{2a})=-\frac{b^2}{4a}+c.$$
Hence we have
$$f(x) = ax^2 +bx + f(-\frac{b}{2a})+\frac{b^2}{4a} =a(x+\frac{b}{2a})^2 + f(\frac{-b}{2a}).$$
If $a > 0$ then we have

$f(x) \ge f(\frac{-b}{2a})$  for all  $x.$
$f(\frac{-b}{2a})$ is the global minimum.

Same way for $a < 0.$

Title: When does an optimization problem have a unique solution, no solution, or infinitely many?

Body:
What conditions are required for an optimization problem (e.g., given  $f(x) = a \cdot x^2 + b \cdot x + c $, find $\min f(x)$) to have a unique solution, no solution, or infinitely many?

I tried to analyze the function by taking its derivative(s) and to look for minima (or maxima). But is that the only way to approach such a problem? Or is there a faster – and a more efficient – way?


问:
How is transitive closure of a relation related to the cardinality of the set?
This is my question, and I don't know how to approach it.

$R$ is a binary relation on set $S$. $R_0$, $R_1$, $R_2$…. are defined as below:

$R_0 := I = \{(x,x) : x ∈ S\}$

$R_{n+1} := R_n ∪ (R;R_n)$ for $n >= 0$

There exists $i ∈ N$ such that $R_i = R_{i+1}$

If |S| = u, show that $R_{u^2} = R_{u^2 + 1}$ and explain the reason.

答: HINT: $\langle a,b\rangle\in R_n$ if and only if there are $a_0=a,a_1,\ldots,a_n=b\in S$ such that $\langle a_k,a_{k+1}\rangle\in R$ for $k=0,\ldots,n-1$. If $n>u^2$, the pigeonhole principle says that there are distinct $k,\ell$ such that $\langle a_k,a_{k+1}\rangle=\langle a_\ell,a_{\ell+1}\rangle$.


答:
In this context, $V\times V$ means the space created in 2 dimensions by all possible combinations of $V$ with itself.

For example, the cartesian plane is a space in two dimensions which could be described as $\mathbb{R} \times \mathbb{R}$ or $\mathbb{R}^2$. In fact, we can construct every ordered pair in the cartesian plane by choosing any $a, b \in \mathbb{R}$ such that $(a, b) \subset \mathbb{R}^2$. Note that $a$ is in $\mathbb{R}$ but $(a, b)$ is in $\mathbb{R}^2$. If you want an ordered tern you would also choose $c \in \mathbb{R}$ such that $(a, b, c) \subset \mathbb{R} \times \mathbb{R} \times \mathbb{R}$ or $\mathbb{R}^3$.

The arrow means its an application from V into V. Recall that you used that same notation for functions: $f : \mathbb{D} \rightarrow \mathbb{R}$ where $\mathbb{D}$ is the domain of $f$.
Well, a function is a particular case of an application from the set $\mathbb{D}$ to the set $\mathbb{R}$. Instead of a function one could have another application from a set to another, for example, a Transformation from $\mathbb{R}^2$ to $\mathbb{R}^3$. Or $T: \mathbb{R}^2 \rightarrow \mathbb{R}^3$.


问:
Conditions for equivalent definition of absolutely continuous measures
Let $\lambda, \mu$ be two real valued measures in the measurable space $(X, \mathcal{A})$. Bartle's "Elements of Integration" Lemma 8.8 says that, if both are finite measures, then $\lambda \ll \mu$ if, and only if, for every $\varepsilon > 0$, there exists $\delta > 0$ such that, for all $E \in \mathcal{A}$, if $\mu(E)< \delta$, then $\lambda(E)< \epsilon$.

The $(\Leftarrow)$ side of the proof is obvious, because if $\mu(E) = 0$, then $\mu(E)< \delta$ for each $\delta > 0$, which implies $\lambda(E) < \varepsilon$ for each $\varepsilon > 0$. Then we have $\lambda(E) = 0$. This proves that $\lambda \ll \mu$.

The other side of the proof goes as follows: suppose the conclusion of $(\Rightarrow)$ is false. Then there is $\varepsilon > 0 $ such that for each $n \in \mathbb{N}$, there is $E_n \in \mathcal{A}$ with $\mu(E_n)< 2^{-n}$ and $\lambda(E_n) \geq \varepsilon$. Define $F_n = \bigcup_{k=n}^\infty E_n$. Then $F_n$ is a decreasing sequence of sets, $\mu(F_n) \leq 2^{-n + 1}$, $\mu(F_1) \leq 1 < \infty$ and $\lambda(F_1)< \infty$, because $\lambda$ is finite. Then $\mu(\bigcap F_n) = \lim \mu(F_n) = 0$ and $\lambda(\bigcap F_n) = \lim \lambda(F_n) \geq \varepsilon$.

It seems to me that the hypothesis that $\mu$ is finite can be dropped. Is this the case?

答:
Note that after you prove Radon-Nikodym theorem the equivalence is still true.
That is, if $\lambda$ is finite and $\mu$ is $\sigma$-finite with $\lambda\ll \mu$, then there is an
measurable function $g$ s.t
$$\lambda(A)=\int_Ag\ d\mu$$
for all measurable $A$. Now, by finiteness of $\lambda$ it follows that $g$ is integrable. So, by the continuity of the integral you have the direction ($\implies$).

As it seems by your observation the equivalence is still true in the special case where $\lambda$ is finite and $\mu$ is an arbitraty positive measure whereas Radon-Nikodym is not since you can take as $\lambda$ the Lebesgue measure on $(0,1)$ and $\mu$ the counting measure on $(0,1)$.


答: To say that $V$ is a vector space over $\mathcal{F}$ means that there is an "addition" function and a "scalar multiplication" function. Addition is a function $a:V\times V\rightarrow V$, where $V\times V$ is the set of ordered pairs of elements in $V$, i.e. an element of $V\times V$ is an ordered pair $(v,w)$ with $v$ and $w$ in $V$. We often write $a(v,w)=v+w$, or alternatively we say that the function $a$ is given by $(v,w)\mapsto v+w$ (the arrow $\mapsto$ is used in this way throughout all of mathematics). Scalar multiplication is a function $m:\mathcal{F}\times V\rightarrow V$, where as above $\mathcal{F}\times V$ is ordered pairs of elements $(\lambda, v)$ where $\lambda$ is in $\mathcal{F}$ and $v$ is in V. We often write $m(\lambda,v) = \lambda v$, or alternatively we say that $m$ is given by $(\lambda, v)\mapsto \lambda v$. We further require that these functions $a$ and $m$ respect each other in certain ways, but hopefully this clears up some of the notational confusion.


问:
$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$ Proof
Having $c\gt0$

$a, c \in \Bbb{R}$

$$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$$

I don't know how to solve this inequality, I tried with the AM-GM inequality and I reached this point so far.

$$(a+b)^2\le a^2(c+\frac{1}{c})+b^2(c+\frac{1}{c})$$

I don't know how to continue, and I don't even know if this is the right path.

If you have any hints, those would be much appreciated.
Thanks

答:
Hi others have given you solutions. I only want to comment on your partial results but since I cannot comment now I'll type it in the answer box.

If you have already proved $(a+b)^2 \leqslant (c+\frac{1}{c}) (a^2+b^2)$ then you are done because $RHS\geqslant 2(a^2+b^2) \geqslant (a+b)^2=LHS$.

However I believe you made a mistake because the following inequality cannot always hold:

$(1+c)a^2 + (1+\frac{1}{c})b^2 \geqslant (c+\frac{1}{c})(a^2+b^2)$.

For example if $c=2$, it reduces to $3a^2+\frac{3}{2}b^2 \geqslant \frac{5}{2} (a^2+b^2) \iff a^2\geqslant 2b^2$ so if $a=1, b=1$ it's reversed.


问:
finite neighborhood base
My question might seem obvious but still I want to sure if the result I shall state below is true:

if the neighborhood base in each point of a space say X is finite. Then we have a discrete topology.

Many thanks and apologies in case I just stated some insanity above :)

答: Am I misreading this question? It seems like it's obviously not true, for example any finite non-discrete space.</p>

Title: finite neighborhood base


问:
What is the meaning of the parentheses in $\phi^{-1}\left[\{\phi(g)\}\right]=gH=Hg$?
I am studying homomorphisms is groups and i saw a theorem saying:

For $g$ in a group $G$, the cosets $gH$ and $Hg$ are the same, and collapsed onto the single element $\phi(g)$ by $\phi$. That is, $\phi^{-1}\left[\{\phi(g)\}\right]=gH=Hg$.

I am confused about the parentheses here. I think $\{\phi(g)\}$ denotes the set, formed by the elements of the mapping, but i am not sure what $[\ ]$ represent. Can anyone tell how should i interpret those parentheses? Or explain the theorem in other words?

Thank you

答:
It is never helpful to express symbolically what can be simply stated.  The mathematical sentence $$\Phi^{-1}\left[\{\Phi(g)\}\right]=gH=Hg$$
is equivalent to the much easier sentence $$gH=Hg\text{ is the preimage of }\Phi(g)\text{ under }\Phi.$$
@amWhy's answer explains what all the various parenthesis mean.  I will elaborate on the group theoretic meaning.  In this context, $H$ is normal in $G$, and $\Phi$ is the homomorphism $$\Phi:G\rightarrow G/H\hspace{20pt}:\hspace{20pt}\Phi(g)=gH.$$
The point the text is trying to make is that any element $g^\prime$ in the coset $gH$ other than $g$ will also map to the coset $gH$ under $\Phi$.  In other words, $$gH=g^\prime H\Leftrightarrow \Phi(g)=\Phi(g^\prime).$$
This is a tautological statement, but intuitively this demonstrates that $\Phi$ represents the division of $G$ into a set of $H$-cosets (which are then imbued with a group law).  When the text refers to $gH$ being "collapsed into a point," they mean that the subset $gH$ of $G$ is only one element of $G/H$, whereas it is $|gH|$ elements in $G$.  Under the coset multiplication group law, it is sometimes possible to visualize this geometrically, for example the canonical homomorphism $\mathbb{R}^2\rightarrow \mathbb{R}^2/\left(\mathbb{R}\oplus 0\right)$ may be viewed as taking lines to points.</p>


问:
Find the coefficient of $x^{12}$ in $(1+x)^{-1}$
My work:

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

So we do $C(1,0) \cdot C(-1,6) = 0$ to find all the ways to get $x^{12}$

But the final answer is $1$ not $0$.

What am I doing wrong?

答: Remember that $$\frac{1}{1-t}=1+t+t^2+t^3+\cdots$$ for all $\left| t\right| <1$. Substitute $t=-x$.


问:
Finding the equation real root.
where $c$ is a positive real number;

The equation $c .\ln(x) = x^2$ has a single real root. So what is the value of the number $c$?

How should I go about it? Anyone have a solution?

答:
Let consider

$$f(x)=c\ln x -x^2 \implies f'(x) =\frac c x -2x \implies f''(x)=-\frac c{x^2}-2$$

then find $x$ such that $f'(x)=0$ and $c$  such that $f(x)=0$.

$$f'(x) =\frac c x -2x =0 \implies x=\sqrt{\frac c 2}$$

with

$$f\left(\sqrt{\frac c 2}\right)=c \ln\left(\sqrt{\frac c 2}\right)-\frac c 2=0 \iff \sqrt{\frac c 2}=e^\frac12 \iff c=2e$$

and since $f(x)$ is convex this is the unique value such that $f(x)=0$.

Title: Finding the equation real root.


问:
Proof of identity involving vector field, unit tangent vector and a smooth curve.
Let $\textbf{t}$ be a continuous vector field which is parallel to the unit tangent vector at each point of a smooth curve C.

Prove that,
$$
\int_{C} \textbf{t} \cdot d\textbf{r}=\int_{C} ||\textbf{t}|| d\textbf{s}\space.
$$

$C$ is parametrised by $$ and I have been given the points $(0,0,0)$ and $(6,18,36)$

答:
Assuming we're working with three dimensions here, any curve can be parameterized in one variable (called a parameter), with some bounds on the parameter. (Usually the bounds are specified in an interval, e.g. $a \leq s \leq b$.) Let us assume that $\mathbf{r}(s)$ is a parameterization of $C$, and define $\mathbf{r}'(s)$ to be the derivative (tangent) vector at the point $\mathbf{r}(s)$ (the point on the curve corresponding to when the parameter is equal to $s$). What the left-hand side denotes is then is the vector line integral, which is by definition
$$\int_a^b \mathbf{t}(\mathbf{r}(s)) \cdot \mathbf{r}'(s) ds$$
However, it is also given that for any point $(x, y, z) = \mathbf{r}(s)$ which lies on the curve $C$, $\mathbf{t}(x, y, z)$ returns a vector parallel to the tangent vector at $\mathbf{r}(s)$, meaning that $ \mathbf{t}(\mathbf{r}(s))$ is actually always parallel to $\mathbf{r}'(s)$. We know that the dot product of two parallel vectors is just the product of their magnitudes, so it would follow that
$$\mathbf{t}(\mathbf{r}(s)) \cdot \mathbf{r}'(s) = \|\mathbf{t}(\mathbf{r}(s))\| \cdot \| \mathbf{r}'(s) \|$$
(The dot on the right-hand side is just normal multiplication.) Thus, the integral we wrote down is equivalent to
$$\int_a^b \|\mathbf{t}(\mathbf{r}(\mathbf{s}))\| \cdot \|\mathbf{r}'(s)\| ds$$
which is exactly the definition of the scalar line integral
$$\int_C \|\mathbf{t}\| ds$$
on the right-hand side of the identity you wanted to verify.

One key takeaway here: when evaluating line integrals (scalar or vector), you will most likely need to find a parameterization of the curve you are evaluating across.


问:
a≤b, b-a<ε => unique C which is A's upper bound, and B's lower bound
$$A,B\subseteq \mathbb R \quad a\in A\quad b\in B\quad \forall a,b \quad a\leq b
\\
\forall \varepsilon >0 \quad b-a<\varepsilon
$$
Prove there's a unqiue $c$, which is the upper bound of $A$, and the lower bound of $B$.

I've found a way to prove that $\inf(B) = \sup(A)$. But that's it, I have no idea how to prove that there's a unique $c$ which is both the upper bound of A, and that lower bound of B.

答: I can't comment but here is a hint: use the property of inf and sup. You have shown $C =\inf B=\sup A$. Will anything less than $C$ be a lower bound?