答:
You've touched on one of the reasons why a careful axiomatization of measure theory is important. How you answer your question depends on how you are defining your notion of "volume"/"area" (what we generally call "measure").

Let's say in your setup you give line segments of length $1$ a nonzero volume $c$. And let's suppose that your notion of volume is additive (in the sense that the volume of the union of disjoint regions is equal to the sum of the volumes of the individual regions). One might think of a $1 \times 1$ square as stacking infinitely many (actually, an uncountable infinitude of) copies of this line  segment. Then your $1 \times 1$ square would have infinite volume since each segment has volume $c$ and you are combining infinitely many of these segments.

On the other hand, if we say that these line segments have zero volume, then this seems to suggest that the $1 \times 1$ square has zero volume as well, since it is made up of infinitely many segments each having zero volume.

The "solution" to this paradox that is established in measure theory is to restrict the additivity property to only countable unions. That is, if $E_1, E_2,\ldots$ are a countable collection of disjoint sets, then $\text{volume}\left(\bigcup_{i =1}^\infty E_i \right) = \sum_{i=1}^\infty \text{volume}(E_i)$; we do not require that this property holds for uncountable collections of disjoint sets. Then, in the case of Lebesgue measure in two-dimensional space, we assign "two-dimensional" regions volume in the usual sense ("area"), and we assign "one-dimensional" regions (like line segments) zero volume. The "paradox" in the previous bullet-point does not occur simply because we do not require our notion of volume to be additive across uncountable collections of disjoint sets (like stacking uncountably many line segments to make a square). This is just a description of Lebesgue measure, however; you can assign measures in other non-intuitive ways.

Title: Is the area of a line really $0$?

Body:
Let's take a square with side $A$. The area of this is defined as $A\times A$.
The way I explained this to myself is by reapeatedly deviding the square till I reached a single line.
Then stacking up lines one on top of the other $A$ times.

But if the area of a line was really $0$ no matter how many lines were stacked up, they would always have an area of $0$.(would exist in one dimension only)

Another way to see this is defining a function $V(F)$ which gives the volume of two dimensional figure $F$.
If this volume became $0$ , no matter how many such Figures I stack up on each other they would never have a volume.

When this is combined with physics it makes a little more sense.
Even if I were to chop up a cube infinitely I could never reduce its height to less than the Planck length. This means that the volume would be $Area \times  ℓP$ (Planck length)

So the question is for a three dimensional object to exist must every two dimensional object have a non zero volume?


问:
finding terms in ODE
Consider the equation

$\frac{\partial{y}}{\partial{t}}+y\frac{\partial{y}}{\partial{x}}=\Gamma \frac{\partial^2y}{\partial{x}^2}$

where $\Gamma$ is a constant term.

Given that $y=\frac{y_0}{2}-\frac{y_0}{2}tanh(\frac{x-ct}{L})$  satisfies the equation above, find $L$ and $c$ in terms of the constants $y_0$ or $\Gamma$

My attempt:

$\frac{\partial{y}}{\partial{x}}=\frac{-y_0}{2L}sech^2(\Phi)$  where $\Phi=\frac{x-ct}{L}$.

$\frac{\partial{y}}{\partial{t}}=\frac{y_0c}{2L}sech^2(\Phi)$

$\frac{\partial^2{y}}{\partial{x}^2}=\frac{y_0}{L^2}sech^2(\Phi)tanh(\Phi)$

Hence plugging them into the equation above we get

$1-\frac{y}{c}=\Gamma \frac{2}{cL}tanh(\Phi)$, so $1-\frac{y_0}{2c}=\frac{4\Gamma-y_0L}{2cL}tanh(\Phi)$

What else should I do?

答:
$$1-\frac{y_0}{2c}=\frac{4\Gamma-y_0L}{2cL}\tanh(\Phi)$$
must hold for any $t,x$ so must be
$$
\begin{cases}
1-\frac{y_0}{2c}=0\\
4\Gamma-y_0L=0\\
\end{cases}$$
Solution is $\left(L= \frac{4 \Gamma}{y_0},c= \frac{y_0}{2}\right)$

ContentLicense: CC BY-SA 4.0

Title: finding terms in ODE


问:
A differentiable function is onto or not in higher dimension
Suppose we have a differentiable function $ F : \mathbb R^n \rightarrow \mathbb R^n$ and it's derivative 'function(linear function)' at each point always has non zero determinent then is the function onto? I know it is one-one and onto in small domain but can we extend it over all to whole domain?

答:
No. For instance, the function $f: \Bbb R \to \Bbb R$ defined by $$f(x) = e^x$$

has non-zero derivative but is not onto.


问:
Counterexamples in Double Integral
I need to:

$a.$ Give an example of function $f:\mathbb{R\times R}$ $\to$
$\mathbb{R}$ with domain in $[0,1]^2$ so that double integral exists but the function is not Riemann integrable.

$b.$ Give an example (if any) for a non-integrable function $f:\mathbb{R\times R}$ $\to$
$\mathbb{R}$ with domain in $[0,1]^2$ such that both iterated integrals exists(i.e. in both order of integration.)

Here is what I think about $(a)$:

$$
f(x,y) =
\begin{cases}
1, & \text{if }x=1/3, \text {and } y\in \mathbb{Q} \\
0, & \text{otherwise }
\end{cases}
$$

According to me, this function should work. As irrationals are dense in $[0,1]$, points of discontinuities are not finite, thus Jordan measure is not zero. This implies function is not Riemann integrable.
As the infimums are zero for all partions of the $[0,1]^2$, lower riemann integral is zero. But refinement of partitions can also make the volume of the each little "cube" approach zero. Hence double integral exists.

Please correct this if need be.

About $(b)$, intuitively, I don't think such function exists. But I cannot give a rigorous proof for this. Again, I look forward for your help. Thanks.

答:
b) There actually exists such a function. Sierpinski gave an example of a nonmeasurable set in $\mathbb{R}^2 $ which intersects any line in at most two points. It is discussed on MathOverflow. The characteristic function of this set, restricted to $[0,1] \times [0,1]$, is not measurable, but the iterated integrals are equal to 0.

Edit: And here is Sierpinski's article in French.

ContentLicense: CC BY-SA 3.0

Title: Counterexamples in Double Integral


答: No: a counterexample already for $n=1$ is $f(x) = \arctan x$.

Title: A differentiable function is onto or not in higher dimension

Body: Suppose we have a differentiable function $ F : \mathbb R^n \rightarrow \mathbb R^n$ and it's derivative 'function(linear function)' at each point always has non zero determinent then is the function onto? I know it is one-one and onto in small domain but can we extend it over all to whole domain?


问:
If every normal section of a surface is a geodesic, then for every point of the surface, the curvature is the same in any direction.
How should I show that if every normal section of a surface in $\mathbb{R}^3$ is a geodesic,  then for every point of the surface, the curvature is the same in any direction?

I would like some hints. Thank you!

答:
Sketch: Assume $p=(0,0,0)$ is a point on the surface $S$ and the unit normal vector to $S$ at $p$ is $(0,0,1)$. Take a ball of radius $r>0$ small enough on the tangent plane at $S$ (the $X,Y$-plane). Note that if you go $r$ distance from $p$, along $S$, in any direction, to say $x$, the normal vector at $x$ is in the plane passing through $Z$-axis and $x$ (uniqueness of geodesics given a direction and at a point).

Now consider the curve on $S$ which is the set of all points at Riemannian distance $r$ from $p$. Consider the function which assigns to each point on this curve its euclidean distance from the $X,Y$-plane. Note that the vectors tangent to this curve are perpendicular to planes containing the base point and the $Z$-axis by the observation above. This means that the function considered above is a constant. This is true for any $r$. Check that this gives local symmetry (rotations in the $X,Y$-plane send geodesics from $p$ to geodesics at $p$). The claim follows from this.


问:
Non-commutativity of the group ring of a non-commutative group
Let $G$ be a group and $R$ be a unital ring. Let's define $R\left( G \right)$ to be the group ring of $G$ over $R$ with underlying set $\left\{ f:G\to R\left| {{\forall }^{\infty }}x :f\left( x \right)=0 \right. \right\}$. Then $R\left( G \right)$ is a ring with identity element.

I'm struggling to prove that if $G$ is a noncommutative group then $R\left( G \right)$ is a noncommutative ring (even if $R$ is a commutative ring).

Any help would be greatly appreciated.

答:
Assume that $R$ is a non-trivial ring (ie $0\neq 1$).

Take two non-commuting elements $u,v\in G$ (with $uv\neq vu$). Then define $f,g:G\to R$ to be the characteristic functions of $u$ and $v$ respectively. Technically: $f(x):=\delta_{ux}$ and $ g(x):=\delta_{vx}$.

Then $$(f\cdot g)(uv)=\sum\limits_{xy=uv}f(x)g(y)=f(u)g(v)=1$$
but
$$ (g\cdot f)(uv)=\sum\limits_{xy=uv}g(x)f(y)=0$$
because $f(x)\cdot g(y)=1$ if and only if $x=v$ and $y=u$, but $vu\neq uv$ by our initial assumption.


问:
Convolution with mollifier $f\star \rho_n\not\to f$? for $f\in L^{\infty}$
I just learned that if $\rho_n$ is a sequence of molifiers and $f$ a function in $L^p$ with $p\neq \infty$ than the convolution $f\star \rho_n\to f$. But what about $p=\infty$? I there an easy example $f\in L^{\infty}$ such that $f\star \rho_n\not\to f$?

答:
Yes, take any $f \in L^{\infty}$ which is not continuous.

To see this, observe that each $f \ast \rho_n$ is continuous and $f \ast \rho_n \to f$ in $L^{\infty}$ implies the limit $f$ must also be continuous by the uniform limit theorem.

Note that you will still have convergence in $L^p_{\mathrm{loc}}$ for all $p < \infty$ and convergence almost everywhere (including at all points of continuity), but the convergence can never be uniform if $f$ is discontinuous.


问:
if there are three strings, randomly pair the 6 ends together, what is the probability of getting a complete circle
I try to use recursive formula to solve it, but I do not know whether it is right or not. I let Pn be the probability of getting a complete circle with n strings. Then I consider the situation when I add one more string. The first end of the string can be attached to any other ends, but the second end of the new string can only attached to one end.(Imaging inserting the new string into the circle). It gives the recursive formula Pn=Pn-1/2n-1.</p>

答:
We have $P_1=1$.

Assume that there are $n\geq2$ strings. We select a random pair of the $2n$ ends. This can be done in ${2n\choose2}$ ways. When the two chosen ends belong to the same string we have a failure. There are $n$ failure pairs, so that with probability
$$1-{n\over{2n\choose2}}={2n-2\over2n-1}$$
we are successful. Conditioned on this happening we now have $n-1$ strings, not all of the same length. The probability that a random pairing of the $2n-2$ ends of these strings leads to a success is $P_{n-1}$. It follows that we have the recursion
$$P_n={2n-2\over2n-1}\,P_{n-1}\qquad(n\geq2)\ .$$
One can easily find an expression of $P_n$ in terms of factorials and powers of two. The numerical values are
$$(P_n)_{n\geq1}=\left(1,{2\over3},{8\over15},{16\over35},{128\over315},{256\over693},\ldots\right)\ .$$


问:
Prove arithmetic formula for the number of bits in $j=2^k-1$
If $k$ is an integer with $0\le k<155$, and $j=2^k-1$, then it holds:
$$k\,=\,5\left(\left\lfloor\frac j{31((j\bmod31)+1)}\right\rfloor\bmod 31\right)-\left\lfloor\frac{12}{(j\bmod31)+3}\right\rfloor+4$$

Why does this equation hold for such a large interval of $k$? How can we prove that?

This is a shortened version of more complex formula# posted by Hallvard B. Furuseth in a comp.c post on 2003-11-03, which reportedly works to $k$ of at least $3\cdot10^{10}$.

Update (not part of the bountied question): per this source, the same author gave that formula working for $0\le k<2040$:
$$k\,=\,8\left(\left\lfloor\frac j{255((j\bmod255)+1)}\right\rfloor\bmod 255\right)-\left\lfloor\frac{86}{(j\bmod255)+12}\right\rfloor+7$$

Try it online with Python or with Wolfram Mathematica! The code computes the upper limits for both formulas.

# Full formula, for reference (not in the above code or part of the bountied question):
$$\begin{align}k\,=\,&30\left(\left\lfloor\frac j{((j\bmod(2^{30}-1))+1)\,(2^{30}-1)}\right\rfloor\bmod (2^{30}-1)\right)+\\&5\left(\left\lfloor\frac{j\bmod(2^{30}-1)}{31((j\bmod31)+1)}\right\rfloor\bmod 31\right)-\left\lfloor\frac{12}{(j\bmod31)+3}\right\rfloor+4\end{align}$$

答:
Looking into the binary representation, and starting from $j=({\underbrace{111\ldots 1}_k})_2$, notice that $j$ can be split (from the left) into groups of five $1$'s, with less than five $1$'s remaining at the right end. So, let's write $k=5q+r$ where $0\le r \lt 5$, and look at:

$$j=(\underbrace{11111,11111,11111,\ldots,11111}_q,\underbrace{11\ldots 1}_r)_2$$

(commas added for clarity). Notice also that $(11111)_2=31$ so $j\bmod{31}=(\underbrace{11\ldots 1}_r)_2$ as the rest of the number is obviously divisible by $31$.

First part of the formula

For the first part of the formula, let's try to calculate $\left\lfloor\frac{j}{31((j\bmod{31})+1)}\right\rfloor$ first. In fact, we want to first divide $j$ by $(j\bmod{31})+1$, which is $(1\underbrace{00\ldots 0}_r)_2$, and then divide by $31$ - but when shall we take the floor function? We will use a helping Lemma below:

Lemma: Let $a\in\mathbb R$ and $b,c\in\mathbb N\setminus\{0\}$. Then $\left\lfloor\frac{a}{bc}\right\rfloor=\left\lfloor\frac{\lfloor\frac{a}{b}\rfloor}{c}\right\rfloor$.

Proof: See here: https://math.stackexchange.com/a/6793/700480 . $\blacksquare$

Back to division: to calculate $\left\lfloor\frac{j}{31((j\bmod{31}+1)}\right\rfloor$, we will take the "floor" function twice (as per above Lemma) - first after division by $(j\bmod{31})+1$, which is just the right shift by $r$ places, and then after division by $31$, which already produces the whole number $(\underbrace{1,00001,00001,\ldots,00001}_{q\text{ }1\text{'s}})_2$. (Each group of binary digits $11111$ turns into $00001$.)

Now, this is a sum of numbers of the form $2^{5i}=(2^5)^i=32^i\equiv 1^i=1\pmod{31}$, so $\bmod{31}$ every single one of those $1$'s (at its place value) is equivalent to just $1$, and so the whole number is equivalent to $q$. Now, assuming the total number $k$ of "ones" to start with was $<155$, we have $q<31$ and therefore the whole number $\bmod{31}$ is $q$.

Altogether, this means $5\left(\left\lfloor \frac{j}{31((j \bmod 31)+1)}\right\rfloor  \bmod 31\right)=5q$. This explains the first part of the formula.

Second part of the formula

The second part of the formula: $4-\left\lfloor\frac{12}{(j \bmod 31)+3}\right\rfloor$ is easily explainable, because $j\bmod{31}=(\underbrace{11\ldots 1}_r)_2$ and so it depends only on the remainder $r, 0\le r\lt 5$ rather than on the "whole" $j$. We can do the legwork and check the cases $r=0,1,2,3,4$ and in every case verify that $4-\left\lfloor\frac{12}{(j \bmod 31)+3}\right\rfloor=r$. It just so happens that, when $12$ is divided by numbers $0+3,1+3,3+3,7+3,15+3$, you get $4,3,2,1\frac{1}{5},\frac{2}{3}$, the integral part of those numbers are $4,3,2,1,0$, respectively, so subtracting them from $4$ produces the number $r$ we started from.

Adding it all together

One can now easily verify that $(\text{the first part})+(\text{the second part})=5q+r=k$, as expected.

The formula with $255$'s

I have not looked in detail, but this second formula looks just like the first one, where $31=(11111)_2$ is replaced by $255=(11111111)_2$. One could expect that $j=2^k-1$ would be broken up into blocks of eight $1$'s, that the proof for the first part would go as previously, while, for the second part, the number $86$ was a lucky choice and plays the same role as $12$ played above. (Numbers $0+12, 1+12, 3+12, 7+12, 15+12, 31+12, 63+12, 127+12$, when $86$ is divided by them and then truncated, yield $7, 6, 5, 4, 3, 2, 1, 0$ and subtracting them from $7$ recovers the remainder (of $k$ divided by $8$, this time).

The formula with three terms

Your two-term formula does not work for $k\ge 155$ because, when $k\ge 155$, $q\ge 31$ and instead of $k=5q+r$ you get $5(q\bmod 31)+r$.

To patch that, the third term is added:

The additional term is based on breaking up the number into groups of $30$ "ones", while the number of those groups should be less than $2^{30}-1$. Thus, it will work for $k<30(2^{30}-1)\approx 3.2212\cdot 10^{10}$.

The remaining part has up to $30$ "ones", which is less than $155$, so the original formula finishes it off.


问:
Approximation of alternating series $\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$
$\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$

I am asked to find the no. of terms needed to approximate the partial sum to be within  0.0000001 from the convergent value of that series. Hence, I tried find the remainder, $R_n = S-S_n$ by the improper integral

$\int_{n}^{\infty} \frac{ (-1)^n(0.55)^{2n-1}} {(2n-1)!} dn $

However, I don't know how to integrate this improper integral so is there other method to solve this problem?

答:
Since this is an alternating series, the "alternating series estimation theorem" states that the "remainder" $R_n = | s - s_n | $ is no larger than the absolute value of the next term in the series $| a_{n+1} |$.  The desired precision ($10^{-7}$ , in your case) can be set as greater than this term.

The way your summation is written, the general term is

$$ \ a_n = \frac{(-1)^{n-1}  \cdot x^{2n-1}}{(2n-1)!} \ ,$$

so the inequality based on the "next term" will be

$$| \frac{(-1)^{(n+1)-1} \cdot x^{2(n+1)-1}}{(2[n+1]-1)!} | \ \le 10^{-7}  \ \Rightarrow \  \frac{x^{2n+1}}{(2n+1)!}  \ \le 10^{-7} \ .$$

This is not too convenient to think about, so we can take reciprocals in the inequality (remembering, of course, to reverse the sign!):

$$ \ \Rightarrow \  \frac{(2n+1)!}{x^{2n+1}}  \ \ge \ 10^7 \ .$$

We are evaluating the series at $x = 0.55$, but it will not introduce any significant disturbance in the result to solve this for $ \ x = 0.5 \ $:

$$ \frac{(2n+1)!}{(\frac{1}{2})^{2n+1}} \ = \ 2^{2n+1}  \cdot  (2n+1)!\ \ge \ 10^7 \ .$$

Now, an inequality with a factorial in it is not going to be something we can solve "algebraically", but we can get an estimate for  $ \ n \ $ pretty quickly if we know some factorial values.  Since $ \ 2n + 1 \ $ is odd, we note that $ \ 9! \approx 360,000 \ $ and $ \ 11! \approx 40,000,000 \ $ , so starting from $ \ n = 4 \ $ would be a good first attempt:

$$2^{2 \cdot 4 \ + \ 1}  \cdot  (2 \cdot 4 \ + \ 1)! \ = \ 2^9 \cdot 9! \ \approx \ 500 \cdot 360,000 \ \approx \ 1.8·10^8 \ \ge \ 10^7  \ , $$

indicating that $ \ n = 4 \ $ certainly does the job.  The only remaining question is whether we could cut the finite series back by one term more.  Setting $ \ n = 3 \ $ yields

$$2^{2 \cdot 3 \ + \ 1}  \cdot  (2 \cdot 3 \ + \ 1)! \ = \ 2^7 \cdot 7! \ \approx \ 130 \cdot 5000 \ \approx \ 660,000 \ \le \ 10^7  \ , $$

so we cannot spare the $ \ n = 4 \ $ term.  Your summation has an initial index of $ n = 1 $ , but the exponent in the general term is $ \ 2n - 1 \ $ , so we have need of five terms to reach the desired level of precision.  (Indeed, we find that the fifth term produces $ \ + \frac{0.55^9}{9!} \ \approx \ 1.27·10^{-8} \ $ , so we have attained entirely adequate precision; the fourth term is only $ \ - \frac{0.55^7}{7!} \ \approx \ -3.0·10^{-6} \ $, which does not affect the seventh decimal place.)

[Warning to students: a "popular" (with exam-writers -- not necessarily exam-takers) multiple-choice question is to ask how many terms are needed for a given level of precision for a series.  Always check what the initial index of the series is and which terms in the series have non-zero coefficients.  It may be that a series has all the terms up to $ \ n = 8 \ $, but the count begins at $ \ n = 0 \ $, so in fact nine terms are required.  In the problem here, the series runs to the ninth-power term, and the initial index is $ \ n = 1 \ $, but only five terms are used.  So it is important to have an awareness of the specificiations for the series being evaluated.]


问:
Classification of countably infinite Abelian groups?
There’s a pretty simple classification of finitely generated Abelian groups, and there’s a relatively understandable classification of countably infinite Abelian $p$-groups for any prime $p$.  But my question is, what is the general classification upto isomorphism of countably infinite Abelian groups?

How complicated is it to state?  Is there a sentence we can write, like “Two countably infinite Abelian groups are isomorphic if and only if ...”?

答:
There is no simple classification theorem for countably infinite Abelian groups.

There are many different theorems in the literature that make the sentence above precise in various ways, but here's one that I think is pretty compelling: Hjorth proved that the isomorphism relation for countable abelian groups (even torsion-free ones) is non-Borel.

Another result along the same line is a result of Thomas, which says that if you look at torsion-free countable abelian groups of finite rank, there is no Borel map that takes invariants that work for rank $n$ groups and produces invariants that work for rank $n+1$ groups.  Thus the complexity problems gets strictly harder as the rank increases.

The results I've mentioned here show that the classification problem is hard even for torsion-free countable abelian groups, so you can imagine the situation certainly isn't any better for arbitrary ones.


问:
Is the area of a line really $0$?
Let's take a square with side $A$. The area of this is defined as $A\times A$.
The way I explained this to myself is by reapeatedly deviding the square till I reached a single line.
Then stacking up lines one on top of the other $A$ times.

But if the area of a line was really $0$ no matter how many lines were stacked up, they would always have an area of $0$.(would exist in one dimension only)

Another way to see this is defining a function $V(F)$ which gives the volume of two dimensional figure $F$.
If this volume became $0$ , no matter how many such Figures I stack up on each other they would never have a volume.

When this is combined with physics it makes a little more sense.
Even if I were to chop up a cube infinitely I could never reduce its height to less than the Planck length. This means that the volume would be $Area \times  ℓP$ (Planck length)

So the question is for a three dimensional object to exist must every two dimensional object have a non zero volume?

答:
The answer to your question is no.

You have the right idea that decomposing areas into arealess constituents provides no help in finding areas.  Of course, no one does that.  One finds a reference area and uses that to measure other areas.  For instance, one declares a particular figure (commonly a square or rectangle) to have area $1$.  Then other areas are determined to be multiples of that area.  In short, you divided too far if you divided all the way down to lines.

The same thing can be done for volumes: declare a unit volume and measure other volumes in terms of that volume.

If you read the old Greek geometers they a constantly measuring with respect to some other figure, for instance, this line segment is twice as long as that line segment.  They let the initial geometry dictate the scale of the magnitudes.

You may wonder how we can deal with areas and volumes that are not unions of unit squares and unit volumes.  The underlying method is to proceed iteratively through dimensions.  Start by declaring a unit length.  Then, declare that a rectangle of side lengths $m$ and $n$ has area $mn$.  Then declare that a rectangular parallelpiped of side lengths $m$, $n$, and $p$ has volume $mnp$.  So now we have reference volumes of any size.

Decompose a region containing a figure into rectangles.  If there are any rectangles in the region that do not meet the figure, just delete those from the region and continue with this reduced region.  The area of the region is the sum of the areas of the rectangles and is an upper bound for the area of the figure.  A lower bound on the area of the figure is the sum of the areas of the rectangles contained in the figure.  Call the lower bound $m$ and the upper bound $M$.  We know that the area of the figure is in the interval $[m,M]$.  Suppose the longest edge of any rectangle is $L$.  Now make a new rectangulation where the length of the longest edge is $\leq L$.  We expect this finer partition to have more rectangles which we discard, so the upper bound decreases, and to include additional area interior to the figure, so the lower bound increases.  If the lower bound and upper bound approach the same value, $a$, in the limit as the longest edge goes to zero (for every sequence of refining rectangulations), then the area of the figure is $a$.  If the two limits do not approach the same value or the value depends on details of the rectangulation, then the area does not exist.

Notice that this method does partition the figure into pieces and we sum up the contributions of these pieces, but we never actually have pieces of zero area.  We contemplate what would happen in the limit of having all pieces have zero area, but we never actually work with such pieces.

An analogous process, partitioning to produce an inner volume (lower bound) and outer volume (upper bound) is used for volumes (and this process can be continued to higher dimensions).  Again, if the two bounds approach the same constant as the length of the longest edge goes to zero for any sequence of decompositions into rectangular parallelpipeds, that constant is the volume.

(This diagram (by Blacklemon67 at the English Wikipedia, used unmodified under CC-BY-SA 3.0)

shows a variant of this method using the analogue of rectangular parallelpipeds in circular polar coordinates to produce an inner volume.  Finer partitioning in the radial direction will allow the circular cylindrical shells to more closely follow the contour of the surface bounding the volume and hence increase toward the desired volume.)

So what happens to a line (segment)?  This is a figure in the plane, so we can ask for its area.  For any enclosing region, we get some finite outer area, but there are no rectangles contained in the line, so we always get zero inner area.  As we refine and refine the rectangulation, we gradually discard the area outside the line.  In the limit, we would discard all the area outside the line, and the two bounds for the area converge to zero.  So the area of a line (segment) is zero.

So what causes the jump between "this collection of line segments has zero area because each line segment has zero area" and "this figure has positive area even though we can imagine decomposing it into zero area parts"?  Fundamentally, this depends on what the word "area" means.  We don't define an area based on the number of line segments a figure can be decomposed into.  Line segments are too small (in area) to support any nonzero notion of area.  We choose (not just in this Answer) to define area in terms of rectangles.  Decomposition into rectangles (however many and however large or small) is a process that can arrive at an area.  Decomposition into line segments immediately leaves the arena of things with areas so immediately cuts off access to the notion of area.

One might object that one should always be able to decompose into line segments and should get a sensible notion of area when doing so.  But this doesn't always happen.  Consider the function $f(x) = \frac{x}{x}$.  We might (incorrectly) simplify this to $1$.  However, $f$ is defined by the expression we gave and that expression is undefined when $x = 0$.  As a consequence, $f$ has no value when $x = 0$.  $f$ has a sensible value, $1$, all the way down to $x = 0$, but is not sensible at $x = 0$.  A similar phenomenon occurs with area : we get sensible areas (bounds) by dividing into rectangles with bounded longest side length as that bound decreases all the way down to zero, but we do not get a sensible result when the bound on the length of the longest side is set to zero.

This can be generalized.  One takes some properties of areas (and analogues in higher dimensions) and calls anything which satisfies these properties a "measure": areas are nonnegative.  The empty figure contains no area.  A finite or countably infinite collection of disjoint figures have collective area equal to the sum of their individual areas.  (Note that an infinite collection of areas need not have a finite combined area, so we should be careful about summing up infinitely many areas.)  One then can define an inner measure (our lower bounds, above) and an outer measure (our upper bounds, above) and then use the same approach as above to determine whether a figure has an area and if so, what it is.

Title: Is the area of a line really $0$?


问:
Logic and implication negation
I am new to predicate logic and there is something I can't seem to understand.

Let's have an implication:
If I have a sister, I have a sibling.

Now, it's negation should be:
I have a sister and I do not have a sibling.

But I can't understand why the following is not a valid negation:
If I have a sister, I do not have a sibling.

It should be a negation because the first part of the implication happened and the second did not.

答:
A statement $\neg A$ is the negation of a statement $A$ if and only if whenever $A$ is true, $\neg A$ is false and whenever $A$ is false, $\neg A$ is true. So to find out which is the negation of the original statement, you just need to investigate all the possible cases and verify that the two statements have "opposite" truth values.

Remember that "If $A$ then $B$" is true whenever $A$ is false or $B$ is true -- that's just how material implication is defined. The problem is the former case: When "I have a sister" is false, then "If I have a sister, I have a sibling" and "If I have a sister, I don't have a sibling" are both true, so they do not have opposing truth values in all cases.

In contrast, "I have a sister and I don't have a sibling" is false whenever "If I have a sister, I have a sibling" is true (namely in those cases wher "I have a sister" is false or "I have a sibling" is true), and "I have a sister and I don't have a sibling" is true whenever "If I have a sister, I have a sibling" is false (namely in that case when "I have a sister" is true but "I have a sibling" false), as you can easily verify with a truth table.

Title: Logic and implication negation


问:
How many values can $y/n_2-x/n_1$ take given $n_2, n_1$ are positive integers and coprime?
Suppose $X \in \{0,\dots,n_1\}$ and $Y \in \{0,\dots,n_2\}$, with $n_1,n_2 \in \mathbf{N}$ and coprime. Then form the random variable $$\frac{Y}{n_2}-\frac{X}{n_1} \ .$$ How many values can it take?

What does coprime imply in this specific case?

答:
Consider the following equation with the numerator in the required ranges:
$$\frac{Y_1}{n_2}-\frac{X_1}{n_1}=\frac{Y_2}{n_2}-\frac{X_2}{n_1}$$
Then, we have:
$$(Y_1-Y_2)n_1=(X_1-X_2)n_2$$
This is where the condition of co-primality comes into play. We must have $n_1$ dividing $X_1-X_2$. From the fact that $0 \leqslant X \leqslant n_1$, this is only possible if $X_1=X_2$ (which in turn forces $Y_1=Y_2$), or if one of $X_1$ and $X_2$ is $n_1$ and the other is $0$. In the latter case, it is similarly forced that one of $Y_1$ and $Y_2$ is $n_2$ and the other is $0$ (they cannot be equal as the values of $X$ are unequal). We can indeed see that the pairs $(X,Y)=(n_1,n_2),(0,0)$ give the same output. The significant fact is that we have proven this to be the only pair to give the same output. The total number of pairs is $(n_1+1)(n_2+1)$. Hence,  excluding one of the two pairs which give the same output yields the answer:
$$(n_1+1)(n_2+1)-1=n_1n_2+n_1+n_2$$


问:
Proof verification and understanding needed
Prove :if A is infinite and B finite and B is a finite subset of A then A\B is infinite by using Exercise 1.

Exercise 1
Let A,B be disjoint finite sets. and A≈m. and B≈n,then. A ∪ B ≈ m + n. Conclude that the union of two finite sets is finite.

Note: problem comes from A book of set
Theory by Pinter

Attempted proof
(Caveat Lector: let the reader beware...
My knowledge of infinite set is shaky
I can use induction and mapping)

I proved exercise 1.
(Complete rewrite)

Write A=(A\B)$\cup$  B (1)

Using $A \cup B $ from exercise 1 we get
A\B=($A\cup B)\cap B^{c}$ (2)

Now suppose that A has a denumerable subset B and A is finite; that is, A ≈ n, B ⊆ A, and B ≈ ω. So B$\subset$(A\B)$\cup$ B.

A\B can’t be finite since A is infinite
If a$\in$A\B then a$\in B^{c}$ then $B^c$
is infinite which is contradiction since
B is finite

Hence A/B is infinite

Help

答:
A few things:

$A\setminus B = \{x \in A: x \notin B\}$. Thus $$A\setminus B = A\cap B^\complement$$
There is no reason to union in all the elements of $B$ before you remove them by intersecting with $B^\complement$.

You deduce

$A\setminus B= ((A\setminus B)\cup B)\cup B)\cap B^\complement$

So $A\setminus B$ and $B$ are disjoint.

Any argument by which you could get "$A\setminus B$ and $B$ are disjoint" from $A\setminus B= ((A\setminus B)\cup B)\cup B)\cap B^\complement$ would work far more easily from your statement (2): $A\setminus B= (A\cup B)\cap B^\complement$. Or more easily yet from (what I assume is the definition Pinter gives for $A\setminus B$): $A\setminus B = A\cap B^\complement$. You were quite clearly heading in the wrong direction and evidently just decided to fake it, hoping your reader would be equally lost and assume that you actually had demonstrated something.

That $A\setminus B$ and $B$ are disjoint is something so obvious that it is questionable whether it needed to be demonstrated at all. By the set-builder definition I gave, it is provable by noting that $x \in A\setminus B \implies x \notin B$, therefore there is no $x$ which is in both $A\setminus B$ and $B$. If you insist on a "set-algebraic" proof, then $$(A\setminus B) \cap B = (A \cap B^\complement)\cap B = A\cap(B^\complement\cap B) = A\cap\varnothing = \varnothing$$

You are not keeping track of your own assumptions:

Now suppose that $A$ has a denumerable subset $B$ and $A$ is finite; that is, $A \approx n, B \subseteq A$, and $B \approx \omega$. So $B\subset (A\setminus B)\cup B$.

$A\setminus B$ can’t be finite since A is infinite ...

Further, you make no use of any of the items above in the rest of your argument, so why did you mention them? The only thing you used was that $A$ is infinite, which is a hypothesis of the theorem.

If $a\in A\setminus B$ then $a\in B^\complement$ then $B^\complement$ is infinite which is contradiction since $B$ is finite.

I assume you are showing that $A\setminus B \subseteq B^\complement$, which would indeed imply $B^\complement$ is infinite (assuming that it has already been proven that a class with an infinite subclass is itself infinite). But $B^\complement$ being infinite does not in anyway contradict $B$ being finite. In fact the complement of every finite set is infinite. Complements of sets are not sets under Pinter's set theory. They are proper classes, and proper classes are always infinite.

If you want to use exercise 1 to prove this, proof by contradiction is needed. But what you are trying to prove is "$A\setminus B$ is infinite", so the assumption you need to make is the opposite: "$A\setminus B$ is finite". When you arrive at a contradiction, it means that the assumption that led you to it is false, and if "$A\setminus B$ is finite" is false, then its opposite "$A\setminus B$ is infinite" will be true.

So you have the hypotheses of the theorem:

$A$ is infinite.

$B$ is finite.

And the assumption you are trying to disprove:

$A\setminus B$ is finite.

You also have the already-proven theorem:

If $C$ and $D$ are both finite, then so is $C\cup D$.

Can you see how to combine these to arrive at a contradiction?


答:
$$z = \frac{Y}{n_2} - \frac{X}{n_1} = \frac{n_1Y - n_2X}{n_1n_2}$$

Since $n_1$ and $n_2$ are coprime, there is no factor you can pull out of the $n$ terms on the numerator and denominator to cancel things out into a smaller-looking fraction with the same value - any change in how the fraction looks will be entirely due to $X$ and $Y$.

It makes the counting easier, because otherwise there are multiple ways to get a particular value $z$ and the problem becomes more complicated. This way the only "tricky" case is when $z = 0$ (which can happen in two ways: When $X=Y=0$, and when $X=n_1, Y=n_2$) - every other value is achieved uniquely.

Title: How many values can $y/n_2-x/n_1$ take given $n_2, n_1$ are positive integers and coprime?

Body:
Suppose $X \in \{0,\dots,n_1\}$ and $Y \in \{0,\dots,n_2\}$, with $n_1,n_2 \in \mathbf{N}$ and coprime. Then form the random variable $$\frac{Y}{n_2}-\frac{X}{n_1} \ .$$ How many values can it take?

What does coprime imply in this specific case?


问:
Graphs in which the degree of every vertex is at least 3
I have a trouble in proving the following:

If $G$ is a graph such that every vertex in it has degree greater than or equal to 3, then for any tree $T$ of $G$ the following is true:

$$|E(T)| \leq 2|E(G-T)|-3,$$

provided that $|E(G-T)| \geq 3.$

I tried to use the fact that the sum of degrees is equal to $2|E(G)|$ but couldn’t get a useful result.

答:
Let $n_d$ be the number of vertices of degree $d$.  Then
$$|V(G)|=\sum_{d \ge 3} n_d$$
and
$$2|E(G)|=\sum_{d \ge 3} d n_d.$$
So
$$3|V(G)| = 3 \sum_{d \ge 3} n_d = \sum_{d \ge 3} 3 n_d \le \sum_{d \ge 3} d n_d = 2|E(G)|.$$
Subtract 3 from both sides to obtain
$$3(|V(G)|-1) \le 2|E(G)|-3.$$
Now $|E(T)|=|V(G)|-1$ implies that
$$3|E(T)| \le 2|E(G)|-3.$$
Finally, $|E(G-T)|=|E(G)|-|E(T)|$ yields
$$|E(T)| \le 2|E(G-T)|-3.$$


问:
Finding the orthogonal projection on a subspace $K \in \ell^2(\mathbb{N})$
I'm trying to find the orthogonal projection of $\ell^2(\mathbb{N})$ on $K$ and then determining $p_K(x)(2n)$ for $x\in \ell^2(\mathbb{N})$ and $n\in \mathbb{N}$. The subspace K is defined as
$$ K = \{x\in \ell^2(\mathbb{N})|x(2n+1) = x(2n) \text{ for all } n \in \mathbb{N} \} $$
and $\ell^2(\mathbb{N}) = \{ x: \mathbb{N} \rightarrow \mathbb{C} \mid\sum_{n=0}^\infty |x(n)|^2 < \infty \}$.

I'm sure the problem is much simpler than what I think, I am just completely lost in this topic at the moment. Can anyone help?

答:
Hint: Consider $e_0=\left(\frac1{\sqrt2}, \frac1{\sqrt2}, 0,0,\dots\right),\ e_1=\left(0,0,\frac1{\sqrt2},\frac1{\sqrt2},0,0,\dots\right),\,\dots$, prove that it's an orthonormal basis of $K$, and then verify that
$$p_K(x)=\sum_n\langle x,e_n\rangle e_n\,.$$

 You should receive that $p_K(x)(2n)=\dfrac{x_{2n}+x_{2n+1}}2$.


问:
Calculating the limit $ \lim_{x \to 0} \frac{e^{\sin^2(x)} - e^x}{\sin(2x)}$ without L'Hospital's rule
I calculated this limit before using L'Hospital and got a result of $-1/2$, but I was wondering if we could somehow calculate this without resorting to L'Hospital and derivatives in general. Any tips and ideas will be appreciated.

答:
You can simply use the equivalence of the functions sinus and exponential at $x=0$. Thus $\sin(x)$ is equivalent to $x$ at $x=0$, and $(\sin(x))^2$ is equivalent to $x^2$ at $x=0$. In addition, $\exp(x)$ is equivalent to $1+x$ at $x=0$.
Bearing this in mind, you find that $\exp( (\sin(x))^2 ) - \exp(x)$ is equivalent to $(1+x^2) - (1+x) = x(x-1)$ at $x=0$. Hence, since $\sin(2x)$ is equivalent to $2x$ at $x=0$, you find that your fraction is equivalent to $\frac{(x-1)}{2}$ at $x=0$, that is the limit is $\frac{-1}{2}$.


答:
As shown in this answer, for any sum of non-negative numbers to make sense, only countably many terms can be non-zero. So if we chop up the unit square into lines, each one indexed by the real number which is its $y$-coordinate, we cannot expect the sum of the uncountably many $0$ areas to give a meaningful result.

There is a similar problem when trying to assign a uniform probability measure on the integers. To divide the total measure of $1$ among the infinitely many integers, we need to assign a measure of $0$ for each integer. However, summing the countably many zeros, we get $0$. So we cannot have a uniform probability measure on the integers.

This indicates that we cannot divide up a square into uncountably many lines of $0$ area and add those areas to get the area of the square.

Title: Is the area of a line really $0$?


答: You want$$\underbrace{\lim_{x\to0}e^x}_{1}\cdot\underbrace{\lim_{x\to0}\frac{e^{\sin^2x-x}-1}{\sin^2x-x}}_{1}\cdot\lim_{x\to0}\frac{\sin^2x-x}{\sin(2x)}$$(the second limit uses $\lim_{y\to0}\frac{e^y-1}{y}=1$). The last limit is$$\underbrace{\lim_{x\to0}\frac12\tan x}_0-\underbrace{\lim_{x\to0}\frac{x}{\sin(2x)}}_{1/2}=-\frac12.$$

Title: Calculating the limit $ \lim_{x \to 0} \frac{e^{\sin^2(x)} - e^x}{\sin(2x)}$ without L'Hospital's rule

Body: I calculated this limit before using L'Hospital and got a result of $-1/2$, but I was wondering if we could somehow calculate this without resorting to L'Hospital and derivatives in general. Any tips and ideas will be appreciated.


问:
How is the derivative calculated with respect to w^2?
I have the following function.

\begin{array}{l}
\phi(w)=e^{b w}: \\
\frac{\partial \phi}{\partial t}=0, \quad \frac{\partial \phi}{\partial w}=b \phi, \quad \text { and } \quad \frac{\partial^{2} \phi}{\partial w^{2}}=b^{2} \phi
\end{array}

I have trouble with calculating the derivative with respect to $w^2$. When I do reverse engineering with integrals, it makes sense. I would like to understand how to calculate this derivative. I find it difficult that you have to take the derivative with respect to $w^2$, but I don't see any $w^2$ in the expression.

Nadine

答:
First of all, the standard notation for the second (and higher) derivatives is misleading, because it is non-algebraic (i.e., the ratios don't cancel).  It's even more problematic when using partial differentials, because the symbols themselves are equivocal within the same expression.

In any case, what happens when you take a derivative is that you first take a differential and then divide by the differential of the variable you are taking the derivative "with respect to".  When you take a derivative twice, you are dividing by that differential twice, which is why it is squared.  Note that the proper interpretation of that notation is not $d(w^2)$ but rather $(d(w))^2$.

If you want an algebraically manipulable form of the second derivative, it is:

$$y'' = \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2} = \frac{d(d(y))}{(d(x))^2} - \frac{d(y)}{d(x)}\frac{d(d(x))}{(d(x))^2}$$

I've expressed it above in two ways.  The first is in more standard notation, the second is more explicit, so you can see what is being multiplied vs. what is being differentiated.  Just note that the differential, in absence of parentheses, usually very tightly associates with the variable next to it.

So, the short answer, it is actually $(d(w))^2$.  The long answer - if you are wanting something that is algebraically manipulable, and not mere symbology, you actually have to write it a different way.  More information about this is available in my paper "Extending the Algebraic Manipulability of Differentials."


问:
What is the dimension of the eigenspace corresponsing to eigenvalue lambda = 9?
Another homework problem is asking me to find the dimension of the eigenspace corresponding to lambda = 9. From what I understand, I need to subtract original matrix A from 9 times a 4x4 identity matrix, but I'm seeking to understand what to do after that because my matrix has the number 9 in every column:

$$
    \begin{pmatrix}
    3-\lambda & 0 & 0 & 0 \\
    3 & 9-\lambda & 0 & 0 \\
    9 & 0 & 9-\lambda & 0 \\
    3 & 3 & 9 & 3-\lambda \\
    \end{pmatrix}
$$

What should I do now? Am I even on the right track?

答:
Well...you're sort of on the right track. Plug in $9$ for $\lambda$ in your matrix and you get
$$
A - 9I =
    \begin{bmatrix}
    -6 & 0 & 0 & 0 \\
    3 & 0 & 0 & 0 \\
    9 & 0 & 0 & 0 \\
    3 & 3 & 9 & -6 \\
    \end{bmatrix}
$$
And now you have to compute the dimension of the nullspace of that matrix (or you could compute the rank of that matrix and use the rank-nullity theorem).


问:
How to integrate $f(x|\theta)=\int_{0}^\infty \frac{3x\theta^3}{(x+\theta)^4} dx$
I've been struggling for quite some time to solve this integration:

$f(x|\theta)=\int_{0}^\infty \frac{3x\theta^3}{(x+\theta)^4} dx$

Can anyone help me try to solve this please? Thanks in advance!

Body:
\begin{align*}
\int_{0}^{\infty}\frac{3x\theta^{3}}{(x+\theta)^{4}}\mathrm{d}x & = 3\theta^{3} \int_{0}^{\infty}\frac{x}{(x+\theta)^{4}}\mathrm{d}x\\\\
& = 3\theta^{3}\int_{0}^{\infty}\frac{(x+\theta) - \theta}{(x+\theta)^{4}}\mathrm{d}x\\\\
& = 3\theta^{3}\left[\int_{0}^{\infty}\frac{1}{(x+\theta)^{3}}\mathrm{d}x - \int_{0}^{\infty}\frac{\theta}{(x+\theta)^{4}}\mathrm{d}x\right]\\\\
& = 3\theta^{3}\left[-\frac{1}{2(x+\theta)^{2}} + \frac{\theta}{3(x+\theta)^{3}}\right]\Biggr|_{0}^{\infty}\\\\
& = 3\theta^{3}\left[\frac{1}{2\theta^{2}} - \frac{1}{3\theta^{2}}\right] = \frac{\theta}{2}
\end{align*}


问:
Simplify binomial sum
It is well-known that for any integer $j$
$$
\sum _{i=0}^k (-1)^{i} \binom{k}{i} (j-i)^k = k!
$$
But I wonder if there's a closed form for the following sum
$$
\sum _{i=j}^k (-1)^{i} \binom{k}{i} (j-i)^k
$$

The following table gives the value for this $k = 1,\cdots, 6$ and $j = 0, \cdots k$.

```<code>1   0
2   1   0
6   5   1   0
24  23  12  1   0
120 119 93  27  1   0
720 719 662 360 58  1   0
</code>```

答:
Suppose we seek an alternate representation of

$$\sum_{p=q}^k (-1)^p {k\choose p} (q-p)^k.$$

This is

$$\sum_{p=0}^k (-1)^p {k\choose p} (q-p)^k
- \sum_{p=0}^{q-1} (-1)^p {k\choose p} (q-p)^k.$$

We get for the first piece

$$k! [z^k] \sum_{p=0}^k (-1)^p {k\choose p} \exp((q-p)z)
\\ = k! [z^k] \exp(qz)
\sum_{p=0}^k (-1)^p {k\choose p} \exp(-pz)
\\ = k! [z^k] \exp(qz) (1-\exp(-z))^k.$$

Now $(1-\exp(-z))^k = z^k + \cdots$ so this evaluates to $k!.$
We thus have

$$k!- \sum_{p=0}^{q-1} (-1)^p {k\choose p} (q-p)^k.$$

Using an Iverson bracket we get for the sum component

$$[w^{q-1}] \frac{1}{1-w}
\sum_{p\ge 0} (-1)^p {k\choose p} (q-p)^k w^p
\\ = k! [z^k] [w^{q-1}] \frac{1}{1-w}
\exp(qz) (1-w\exp(-z))^k
\\ = k! \;\underset{z}{\mathrm{res}}\; \frac{1}{z^{k+1}}
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^q} \frac{1}{1-w} \exp(qz)
(1-w\exp(-z))^k.$$

We now apply Jacobi's Residue Formula.  We put $w=v \exp((1-v)u)$ and $z
= (1-v)u$. The scalar to obtain a non-zero constant term in $u$ and $v$
for $z$ and $w$  is $u$ for $z$ and $v$ for $w.$
Using the determinant of the Jacobian we obtain

$$ \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}^{-1}
\begin{vmatrix}
1-v & -u \\
v (1-v) \exp((1-v)u) & \exp((1-v)u) - u v \exp((1-v)u) \\
\end{vmatrix}
\\ = \exp((1-v)u)
\begin{vmatrix}
1-v & - u \\
v (1-v) & 1 - uv \\
\end{vmatrix}
\\ = \exp((1-v)u) (1 - uv - v + uv^2 + uv - uv^2)
\\ = \exp((1-v)u) (1 - v).$$

Doing the substitution we find

$$k! \;\underset{u}{\mathrm{res}}\;
\frac{1}{u^{k+1}} \frac{1}{(1-v)^{k+1}}
\;\underset{v}{\mathrm{res}}\; \frac{1}{v^q} \frac{1}{\exp(q(1-v)u)}
\\ \times \frac{1}{1-v\exp((1-v)u)} \exp(q(1-v)u)
(1-v\exp((1-v)u)\exp(-(1-v)u))^k
\\ \times \exp((1-v)u) (1-v)
\\ = k! \;\underset{u}{\mathrm{res}}\;
\frac{1}{u^{k+1}} \frac{1}{(1-v)^{k+1}}
\;\underset{v}{\mathrm{res}}\; \frac{1}{v^q}
\frac{1}{1-v\exp((1-v)u)} (1-v)^k
\\ \times \exp((1-v)u) (1-v)
\\ = k! \;\underset{u}{\mathrm{res}}\; \frac{1}{u^{k+1}}
\;\underset{v}{\mathrm{res}}\; \frac{1}{v^q}
\frac{1}{1-v\exp((1-v)u)} \exp((1-v)u)
\\ = k! \;\underset{u}{\mathrm{res}}\; \frac{1}{u^{k+1}}
\;\underset{v}{\mathrm{res}}\; \frac{1}{v^q}
\frac{1}{\exp((v-1)u)-v}.$$

Consider on the other hand the quantity

$$\sum_{p=0}^{q-1} \left\langle k \atop p \right\rangle.$$

This is

$$k! [z^k] \sum_{p=0}^{q-1} [w^p] \frac{1-w}{\exp((w-1)z)-w}
\\ = k! [z^k] [w^{q-1}] \frac{1}{1-w} \frac{1-w}{\exp((w-1)z)-w}
\\ = k! [z^k] [w^{q-1}] \frac{1}{\exp((w-1)z)-w}
\\ = k! \;\underset{z}{\mathrm{res}}\; \frac{1}{z^{k+1}}
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^q} \frac{1}{\exp((w-1)z)-w}.$$

This is the same as the sum term and we conclude the argument having
shown that

$$\sum_{p=q}^k (-1)^p {k\choose p} (q-p)^k
= k! - \sum_{p=0}^{q-1} \left\langle k \atop p \right\rangle$$

which is

$$\bbox[5px,border:2px solid #00A000]{
\sum_{p=q}^k (-1)^p {k\choose p} (q-p)^k
= \sum_{p=q}^k \left\langle k \atop p \right\rangle.}$$

Reference, as per request. The Jacobi Residue Formula is Theorem 3
in  the paper  A Combinatorial  Proof of  the Multivariable  Lagrange
Inversion Formula by I. Gessel.

Title: Simplify binomial sum


问:
Calculate $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\prod_{k=1}^{n}\left(n+3k-1\right)\right)^{\frac{1}{n}}$
I'm need of some assistance regarding a homework question:

$$
\mbox{"Calculate the following:}\quad \lim_{n \to \infty}
\frac{1}{n}\left[%
\prod_{k = 1}^{n}\left(n + 3k -1\right)\right]^{1/n}\
\mbox{"}
$$

Alright so since this question is in the chapter for definite integrals ( and because it is similar to other questions I have answered ) I assumed that I should play a little with the expression inside the limit and change the product to some Riemann sum of a known function.

$0$K so I've tried that but with no major breakthroughs$\ldots$

Any hints and help is appreciated, thanks $!$.

答:
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 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}\braces{{1 \over n}
\bracks{\prod_{k = 1}^{n}\pars{n + 3k - 1}}^{1/n}\,}} =
\lim_{n \to \infty}\braces{{3 \over n}
\bracks{\prod_{k = 1}^{n}\pars{k + {n - 1 \over 3}}}^{1/n}\,}
:\ {\Large ?}}$

\begin{align}
&\bbox[5px,#ffd]{\prod_{k = 1}^{n}\pars{k + {n - 1 \over 3}}} =
\pars{n + 2 \over 3}^{\overline{n}} =
{{\Gamma\pars{\bracks{n + 2}/3 + n} \over
\Gamma\pars{\bracks{n + 2}/3}}}
\\[5mm] = &\
\left.{{\Gamma\pars{m + n + 1} \over \Gamma\pars{m + 1}}}
\,\right\vert_{\,m\ = \pars{n - 1}/3}\ =\
{\pars{m + n}! \over m!}
\\[5mm]
\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&
{\root{2\pi}\pars{m + n}^{m + n + 1/2}\,\,\expo{-m - n}
\over
\root{2\pi}m^{m + 1/2}\,\,\expo{-m}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &
{n^{m + n + 1/2}\,\,\,\,\pars{1 + 1/3}^{4n/3}
\over
3^{-m - 1/2}\,\,\,n^{m + 1/2}\,\,
\pars{1 - 1/n}^{n/3}}\,\expo{-n}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,\,\, &
3^{n/3}\,n^{n}\,\pars{4 \over 3}^{4n/3}
\expo{1/3}\expo{-n} =
\bracks{3^{1/3}\,n\,\pars{4 \over 3}^{4/3}
\expo{1/\pars{3n}}\,\expo{-1}}^{n}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,\,\, &
\pars{{2^{8/3} \over 3\expo{}}n}^{n}
\end{align}
Finally,
$$
\bbox[5px,#ffd]{\lim_{n \to \infty}\braces{{1 \over n}
\bracks{\prod_{k = 1}^{n}\pars{n + 3k - 1}}^{1/n}\,}} =
\bbx{2^{8/3} \over \expo{}} \approx 2.3359 \\
$$


答:
Just to give an alternative to J.G.'s approach, we have

$$\begin{align}
{e^{\sin^2x}-e^x\over\sin2x}
&={(e^{\sin^2x}-1)-(e^x-1)\over2\sin x\cos x}\\
&={1\over2\cos x}\left({e^{\sin^2x}-1\over\sin^2x}\sin x-{e^x-1\over x}\cdot{x\over\sin x} \right)\\
&\to{1\over2\cdot1}\left(1\cdot0-1\cdot1 \right)\\
&=-{1\over2}
\end{align}$$


答:
$$(A-\lambda I)v=0$$
$$\begin{pmatrix}
    -6& 0 & 0 & 0 \\
    3 & 0 & 0 & 0 \\
    9 & 0 & 0 & 0 \\
    3 & 3 & 9 & -6 \\
    \end{pmatrix}\pmatrix {v_1 \\v_2 \\v_3 \\v_4}=\pmatrix {0\\0\\0\\0}$$
You deduce that:
$$
\begin{cases}
v_1=0 \\
v_2+3v_3-2v_4=0
\end{cases}
$$
So that we have :
$$v=\left(0,v_2,v_3,\dfrac 12 v_2+\dfrac 32 v_3\right)^t$$
$$v=c_1\left(0,2,0,1\right)^t+c_2\left(0,0,2,3\right)^t$$

You can deduce the dimenson of the eigenspace now.

Title: What is the dimension of the eigenspace corresponsing to eigenvalue lambda = 9?

Body:
Another homework problem is asking me to find the dimension of the eigenspace corresponding to lambda = 9. From what I understand, I need to subtract original matrix A from 9 times a 4x4 identity matrix, but I'm seeking to understand what to do after that because my matrix has the number 9 in every column:

$$
    \begin{pmatrix}
    3-\lambda & 0 & 0 & 0 \\
    3 & 9-\lambda & 0 & 0 \\
    9 & 0 & 9-\lambda & 0 \\
    3 & 3 & 9 & 3-\lambda \\
    \end{pmatrix}
$$

What should I do now? Am I even on the right track?


答:
Anyway, no need for L'Hospital's rule here.

Just use Maclaurin's expansions at the relevant orders and compose them where required:

$\mathrm e^x=1+x+o(x)$;

$\sin^2x=x^2+o(x^2)$, so $\mathrm e^{\sin^2x}=1+x^2+o(x^2)=1+o(x)$;

$\sin 2x=2x+o(x)$.

Therefore
$$\frac{\mathrm e^{\sin^2x}-\mathrm e^x}{ {\sin2x}}=\frac{-x+o(x)}{2x+o(x)}=-\frac12+o(1).$$


问:
Problem 7.3 I. Martin Isaacs's Character Theory Of Finite Groups

Let $H\subseteq G$ and $\xi\in Irr(H)$. Suppose $(\xi-\xi(1)1_H)^G=\theta$ and $[\theta,\theta]=1+\xi(1)^2$. Show that there exists $N\lhd G$ with $N\cap H=\ker\xi$ and every $x\in G-N$ conjugate to some element of $H$.

My attempt:

I am trying to imitate the proof of the Frobenius theorem(Theorem 7.2).

Obviously $ \xi\ne 1_H $ since $ [\theta,\theta]>1 $.

Let $ \chi=\theta+\xi(1)1_G $, it is a generalized character of $ G $, we claim that $ \chi\in Irr(G) $ by showing that $ [\chi,\chi]_G=1 $:

\begin{align*}
  [\chi,\chi]_G&=[\theta,\theta]+2\xi(1)[\theta,1_G]+\xi(1)^2\\
  &=1+\xi(1)^2+2\xi(1)[\xi^G-\xi(1)(1_H)^G, 1_G]+\xi(1)^2\\
  &=1+2\xi(1)^2+2\xi(1)[\xi^G, 1_G]-2\xi(1)^2[(1_H)^G, 1_G]\\
  &=1+2\xi(1)^2+0-2\xi(1)^2\\
  &=1.
 \end{align*}
Also, $ \chi(1)=\theta(1)+\xi(1)=\xi(1)>0 $, it is indeed an irreducible character of $ G $.

Suppose we have shown that $ \chi|_H=\xi $.

Let $N=\ker\chi$.

We have $ N\lhd G $ and we claim that $ N\cap H=\ker\xi $. Since if $ h\in N\cap H $, then $ \xi(h)=\chi(h)=\chi(1)=\xi(1) $, we know that $ h\in\ker\xi $. Conversely, if $ \xi(x)=\xi(1) $, then of course $ x\in H $ and $ x\in\ker\xi\subseteq N $. Hence $ N\cap H=\ker\xi $.

To show that every $ x\in G-N $ conjugate to some element of $ H $, it suffices to show that $ G\setminus N\subseteq\bigcup_{g\in G} H^g $ which is equivalent to show that $ G\setminus\bigcup_{g\in G} H^g\subseteq N $. For any $ y\in G\setminus\bigcup_{g\in G} H^g  $, and $ \chi|_H=\xi\in Irr(H)\setminus\{1_H\} $, we have $ \chi(y)-\chi(1)=\theta(y)=(\xi-\xi(1)1_H)^G(y)=0 $ by the definition of induced characters.

But the problem is, I do not know how to show that $ \chi|_H=\xi $. We do not have any TI set here and I have tried to compute $[\chi|_H,\xi]=[\xi^G,\xi^G]-\xi(1)[(1_H)^G,\xi^G]$. And it suffices to show that $[\chi|_H,1_H]=0$.

答:
I figured it out.

Obviously $ \xi\ne 1_H $ since $ [\theta,\theta]>1 $.

Let $ \chi=\theta+\xi(1)1_G $, it is a generalized character of $ G $, we claim that $ \chi\in Irr(G) $ by showing that $ [\chi,\chi]_G=1 $:

\begin{align*}
  [\chi,\chi]_G&=[\theta,\theta]+2\xi(1)[\theta,1_G]+\xi(1)^2\\
  &=1+\xi(1)^2+2\xi(1)[\xi^G-\xi(1)(1_H)^G, 1_G]+\xi(1)^2\\
  &=1+2\xi(1)^2+2\xi(1)[\xi^G, 1_G]-2\xi(1)^2[(1_H)^G, 1_G]\\
  &=1+2\xi(1)^2+0-2\xi(1)^2\\
  &=1.
 \end{align*}
Also, $ \chi(1)=\theta(1)+\xi(1)=\xi(1)>0 $, it is indeed an irreducible character of $ G $.

Note that
\begin{align*}
[\chi_H,\xi]_H&=[\chi,\xi^G]\\
&=[\xi^G-\xi(1)(1_H)^G+\xi(1)1_G,\xi^G]\\
&=[\xi^G,\xi^G]-\xi(1)[(1_H)^G,\xi^G]+\xi(1)[1_G,\xi^G]\\
&=[\xi^G,\xi^G]-\xi(1)[(1_H)^G,\xi^G]
\end{align*}
since $ [1_G,\xi^G]=[1_H,\xi]=0 $.

Note that $ [(1_H)^G,1_G]=[1_H,1_H]=1 $. Let $ (1_H)^G=1_G+\sum_{i=1}^l a_i\varphi_i+\sum_{i=1}^{m} b_i\psi_i $, $ \xi^G=\sum_{i=1}^{l} a_i'\varphi_i+\sum_{i=1}^{m'} c_i\lambda_i $ where $ b_i,c_i,l,m,m'\in\mathbb Z_{\ge 0} $, $ a_i,a_i',i\in\mathbb{Z}_{>0} $ and $ \varphi_i,\psi_i,\lambda_i\in Irr(G) $ are distinct irreducible characters of $ G $ for all $ i $. So $ \varphi_1,...,\varphi_l $ are the common irreducible constituents of $ (1_H)^G $ and $ \xi $. When $ l=0 $, there are no common irreducible constituents.

Therefore $$ [\chi_H,\xi]=\sum_{i=1}^{l}a_i'^2+\sum_{i=1}^{m'}c_i^2-\xi(1)\sum_{i=1}^l a_ia_i'. $$

On the other hand,
\begin{align*} \chi&=\xi^G-\xi(1)(1_H)^G+\xi(1)1_G\\&=\sum_{i=1}^{l} a_i'\varphi_i+\sum_{i=1}^{m'} c_i\lambda_i-\xi(1)\left(\sum_{i=1}^l a_i\varphi_i+\sum_{i=1}^{m} b_i\psi_i\right) .
\end{align*}

Since $ \chi\in Irr(G) $, only one of the irreducible character above will survive. But $ \xi(1)>0 $ forces all $ b_i $ are zero. Hence $$ \chi=\sum_{i=1}^{l} (a_i'-\xi(1)a_i)\varphi_i+\sum_{i=1}^{m'} c_i\lambda_i .$$

The first case is $ \chi=\lambda_i $ for some $ i\in [m'] $, we suppose W.L.G. that $ c_1=1 $ and $ c_i=0 $ for all $ i=2,...,m' $ and $ a_i'=\xi(1)a_i $ for all $ i\in [l] $. Then $$ [\chi_H,\xi]=\sum_{i=1}^l\xi(1)^2a_i^2+1-\xi(1)\sum_{i=1}^la_i(\xi(1)a_i)=1 $$
which implies $ \chi_H=\xi $ since $ \xi\in Irr(H) $ and $ \chi_H(1)=\chi(1)=\xi(1) $.

The other case is $ \chi=\varphi_i $ for some $ i\in l $. We suppose W.L.G. that $ \chi=\varphi_1 $. Thus, $ c_i=0 $ for all $ i\in [m'] $, $ a_1'=\xi(1)a_1+1 $ and $ a_i'=\xi(1)a_i $ for all $ i=2,...,l $. We compute that
\begin{align*}
[\chi_H,\xi]&=(\xi(1)a_1+1)^2+\sum_{i=2}^l (\xi(1)a_i)^2-\xi(1)a_1(\xi(1)a_1+1)-\xi(1)\sum_{i=2}^la_i(\xi(1)a_i)\\
&=\xi(1)^2a_1^2+2\xi(1)a_1+1-\xi(1)^2a_1^2-\xi(1)a_1\\
&=\xi(1)a_1+1.
\end{align*}

Note that $ \xi(1)\ge 1 $ and $ a_1\ge 1 $ so $ [\chi_H,\xi]\ge 2 $. But this can never happen since $ \chi(1)=\xi(1) $ and $ \xi\in Irr(H) $, $ \chi\in Irr(G) $.

So we conclude that $ \chi_H=\xi $.

Let $ N=\ker\chi $.

We have $ N\lhd G $ and we claim that $ N\cap H=\ker\xi $. Since if $ h\in N\cap H $, then $ \xi(h)=\chi(h)=\chi(1)=\xi(1) $, we know that $ h\in\ker\xi $. Conversely, if $ \xi(x)=\xi(1) $, then of course $ x\in H $ and $ x\in\ker\xi\subseteq N $. Hence $ N\cap H=\ker\xi $.

To show that every $ x\in G-N $ conjugate to some element of $ H $, it suffices to show that $ G\setminus N\subseteq\bigcup_{g\in G} H^g $ which is equivalent to show that $ G\setminus\bigcup_{g\in G} H^g\subseteq N $. For any $ y\in G\setminus\bigcup_{g\in G} H^g  $, and $ \chi|_H=\xi\in Irr(H)\setminus\{1_H\} $, we have $ \chi(y)-\chi(1)=\theta(y)=(\xi-\xi(1)1_H)^G(y)=0 $ by the definition of induced characters. We are done.


问:
Differential generators of 4D rotations and $so(4)\sim so(3)\times so(3)$
Rotations in $\mathbb{R}^3$ are generated by differential operators
$$ O_{yz}=y\frac{d}{dz}-z\frac{d}{dy}, O_{zx}=z\frac{d}{dx}-x\frac{d}{dz}, O_{yx}=y\frac{d}{dx}-x\frac{d}{dy},$$
which satisfy the commutation relations
$$[O_{yz},O_{zx}]=O_{yx}$$ and $[O_{ab},O_{bc}]=O_{ac}$ in general, with $O_{ab}=-O_{ba}$.

I suppose rotations in $\mathbb{R}^4$ must be generated by those three generators plus three others $O_{xw}$, $O_{yw}$, $O_{zw}$, defined analogously. This is consistent with the dimension of algebra $so(4)$ being 6.

Now I have seen that $so(4)\sim so(3)\oplus so(3)$, so there should be possible to construct linear combinations of the six $O_{ab}$ generators in $\mathbb{R}^4$ to produce two separate $so(3)$. How is this done?

Body:
$$L_{xy}=\tfrac12(O_{xy}+O_{zw}),\quad L_{yz}=\tfrac12(O_{yz}+O_{xw}),\quad L_{zx}=\tfrac12(O_{zx}+O_{yw})$$

$$R_{xy}=\tfrac12(O_{xy}-O_{zw}),\quad R_{yz}=\tfrac12(O_{yz}-O_{xw}),\quad R_{zx}=\tfrac12(O_{zx}-O_{yw})$$


问:
Is this a standard type of graph?
I have some nodes linked together in this graph fashion. The top node has a left and a right "child", but as you can see, when you go a level further, left and right children share common children and this pattern continues. The question is, is this a standard type of graph?

答: One might think of it as a truncated dependency graph for Pascal's Triangle:

Title: Is this a standard type of graph?


问:
Why General Leibniz rule and Newton's Binomial are so similar?
The binomial expansion:

$$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$

The General Leibniz rule (used as a generalization of the product rule for derivatives):

$$(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}$$

Both formulas can be obtained simply by induction; Newton's binomial also has a combinatorial proof (here's the relevant wikipedia page).
It's striking how these formulas are similar; is there a possible connection between them?
I was thinking that maybe the General Leibniz rule could be obtained using a combinatorial argument as well (hence the binomial coefficients)...

答:
The main feature of the Leibniz law $\left[f(x)g(x)\right]' = f'(x)g(x) + f(x)g'(x)$ is that it turns a product into a sum. Another way to write it to show that more explicitly is

$$
\frac{d}{dx}\left(f(x)g(x)\right)=\left(\left.\frac{\partial}{\partial s}\right\rvert_{s,t=x}+\left.\frac{\partial}{\partial t}\right\rvert_{s,t=x}\right)\left(f(s)g(t)\right)=\frac{df}{dx}g(x)+f(x)\frac{dg}{dx}
$$

or if you don't like the contrivance of introducing distinct dummy variables only to then sub them out for $x$ again, you could write with a tensor product like

$$
\frac{d}{dx}\left(f(x)\otimes g(x)\right) = \left(1\otimes \frac{d}{dx} + \frac{d}{dx}\otimes 1\right)\left(f(x)\otimes g(x)\right).
$$

Either way we think of it, the point is that the derivative of a product is really a sum of two differential operators, applied to that product.

And how do you compute the $n$th power of a sum of two commuting variables? Why with the binomial theorem of course. The formula

$$
(x+y)^n = \sum_{i=0}^n{n \choose i}x^iy^{n-i}
$$

makes sense for any formal commuting variables. So it makes sense for $\frac{\partial}{\partial s}$ and $\frac{\partial}{\partial t}$. So we have

$$
\left(\frac{\partial}{\partial s}+\frac{\partial}{\partial t}\right)^n = \sum_{i=0}^n{n \choose i}\frac{\partial^i}{\partial s^i}\frac{\partial^{n-i}}{\partial t^{n-i}}
$$
which, when applied to the product $f(s)g(t)$ and evaluated at $s=t=x$, gives you the generalized Leibniz formula you desire. So we see that formula as just a special case of the binomial theorem. After all, the binomial theorem is just a bookkeeping of the combinatorics of applying the distributive law $n$ times. And differential operators distribute over addition, just as much as multiplication does. That is why an $n$th derivative product rule is just $n$ applications of the distributive law. The combinatorics of it is nicely explained in Tad's answer.

So the question here isn't why an $n$-fold derivative that's actually a sum has the same form as the $n$th power of a sum. That's now obvious: the former is just a special case of the latter. The question instead is: why is the derivative of a product a sum in the first place? What is the intuition behind the non-generalized first derivative Leibniz law? The graphic on wikipedia is helpful. This thread has some nice answers. I will offer this: a derivative is the linear response of a function to a small change of input. It makes sense that when you compute the linear response of a product of two functions, you end up with a product of sums, which you distribute. But mostly that's probably a topic for a separate question.

I will add that in physics and signal processing and other places where you spend your days in Fourier transforms, it is very natural to think of differentiation and multiplication by our independent variable as two sides of the same coin. Every function admits dual representation in configuration space and in momentum space. Multiplying in configuration space is the same as differentiating the momentum space representation, and vice versa. So of course $n$ applications of one is the same as $n$ applications of the other. I bet this point of view could be fleshed out into a complete answer to the question.


问:
Finding an explicit formula for $a_n$ defined recursively by
The sequence $a_n$ is defined recursively by $a_0=0$, $a_n=4a_{n-1}+1$. I must use generating functions to solve this. $n\geq1$.

I have found a pattern:
$$\sum_{n=1}^\infty(4a_{n-1}+1)x^n = x+5x^2+21x^3+85x^4+341x^5+\ldots$$

If we subtract 1 from each term, respectively (ignoring the first term) we would have:

$$x+4x^2+20x^3+84x^4+340x^5+\ldots$$

As you can see, the pattern is the second term increases by 4, the third term increases by 4^2, the fourth term increases by 4^3, and the fifth term increases by 4^4. I wanted to use the formula for the geometric series, but the common ratio isn't 4, but $4^n$. Is there any way I can use this pattern to help me find the explicit formula, using generating functions?

Using $$\sum_{n=0}^\infty 4^nx^n = \frac{1}{1-4x}$$ Isn't helping me too much. But I have a feeling I can manipulate this generating function to solve this.

However, this might seem helpful:

$$\sum_{n=0}^\infty4^nx^{n+1} = 4^0x+4^1x^2+4^2x^3+4^3x^4+\ldots = x+4x^2+16x^3+64x^4+\ldots$$

After some work I have gotten this:

$$A(x)=\frac{x^2}{(1-x)(x-4)}$$ Where $A(x)$ is our unknown generating function (which we have been looking for). But I don't think this is correct because it does not match the answer down below.

Edit: I have finally figured it out, I will be back with my LaTex code to present it.

Final edit:
Find an explicit formula for $a_n$ using the generation function technique.
\newline We will begin by multiplying each side by $x^n$ and summing over $n\geq1$. We will also let $A(x)=\sum_{n=0}^\infty a_nx^n$ be the unknown generating function which we are looking for.
$$\begin{align*}
\implies \sum_{n=1}^\infty a_nx^n&=\sum_{n=1}^\infty (4a_{n-1}+1)x^n
\end{align*}
$$
Because
$a_0=0$ is given, we can rewrite $A(x)$:
$$
\begin{align*}
A(x)=a_0+\sum_{n=1}^\infty a_nx^n=\sum_{n=1}^\infty a_nx^n
\end{align*}
$$
Now, we continue to manipulate our equality.
$$
\begin{align*}
\implies \sum_{n=1}^\infty a_nx^n&=\sum_{n=1}^\infty (4a_{n-1}+1)x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n + x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n+\sum_{n=1}^\infty x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n+ \frac{x}{1-x} \\[2mm]
&=\sum_{n=0}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=a_0+\sum_{n=1}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=\sum_{n=1}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=4x\sum_{n=1}^\infty a_{n}x^{n}+ \frac{x}{1-x}. \\[2mm]
\end{align*}$$
It is clear now, that we have a multiple of our previously stated $A(x)$ on the left and right hand side.
$$
\begin{align*}
\implies&\sum_{n=1}^\infty a_{n}x^{n}=4x\sum_{n=1}^\infty a_{n}x^{n}+ \frac{x}{1-x} \\[2mm]
\implies&A(x)=4xA(x)+\frac{x}{1-x} \\[2mm]
\implies&A(x)-4xA(x)=\frac{x}{1-x}\\[2mm]
\implies& A(x)(1-4x)=\frac{x}{1-x}\\[2mm]
\implies& A(x)=\frac{x}{(1-x)(1-4x)}=\frac{x}{4x^2-5x+1} \\[2mm]
\end{align*}$$
Thus, the explicit formula for $a_n$ is given by $A(x)=\dfrac{x}{4x^2-5x+1}$ through the use of generating functions.

Finally, it follows that we have $$A(x)=\sum_{n=0}^\infty\frac{4^n-1}{3}x^n=\frac{x}{4x^2-5x+1}.$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
A particular solution is $\ds{-\,{1 \over 3}}$.

Then
$\ds{\pars{~\mbox{with}\ a_{0} = 0~}}$,
\begin{align}
a_{n} + {1 \over 3} & =
4\pars{a_{n - 1} + {1 \over 3}} =
4^{2}\pars{a_{n - 2} + {1 \over 3}} =
4^{3}\pars{a_{n - 3} + {1 \over 3}}
\\[5mm] & = \cdots =
4^{n}\pars{a_{0} + {1 \over 3}} = {4^{n} \over 3}
\implies
\bbx{a_{n} = {4^{n} - 1 \over 3}} \\ &
\end{align}


问:
Why aren't the homotopy group functors a generalized homology theory?
I am currently enrolled in an introductory course on algebraic topology.

In a recent research seminar at my home institution, it was stated that the homotopy group functors $\pi_1, \pi_2, \pi_3, ...$ are not a generalized homology theory, and I'd like to know why.

I know that $\pi_n(X)$ consists of the homotopy classes of basepoint-preserving maps from $S^n$ into $X$, and I know of the four axioms (homotopy, exactness, excision, additivity) that need to be satisfied to define a generalized homology theory. What axiom isn't satisfied when trying to define a generalized homology theory via the homotopy group functors ?

Thanks!

答:
First, I'll note that homotopy groups do not even solely take values in groups since $\pi_0$ is a set. Even if we ignore $\pi_0$ they do not take values in abelian groups since $\pi_1$ may be noncommutative.

Let's suppose we ignore both of those issues. If the homotopy groups were a generalized homology theory, then they would have a suspension isomorphism. This would mean that the homotopy groups of $S^1$ would be the same as the shifted homotopy groups of $S^2$. However, $\pi_2(S^1)=0$ while $\pi_3 (S^2)=\mathbb{Z}$.

When one forces the suspension isomorphism to work for homotopy groups, one arrives at the stable homotopy groups $\pi_* ^s$ which do form a homology theory. This has the result of making the stable homotopy groups of $S^0$ very complicated, even though the homotopy groups of $S^0$ are trivial.


问:
Do mathematicians really use infinity in their computations?
I say that mathematicians use the words infinity and infinite, but in their computations they change the meanings of infinity and infinite, so that they are no longer with the meanings of without end, but with end, and it is because they cannot manipulate something whatever it is that is literally without end i.e. without ‘finis’.

Am I correct?

答:
Actually, you are not correct.  Mathematicians have actually developed several ways of approaching the infinite.  The nice thing about having a mind is that we can approach problems that seem impossible, and do amazing things with them.

As an example, let's look at the ratio: $\frac{x^2 + x + 1000}{3x^2 - 3x + 200}$.  Let's say we want to know what this looks like at $x = \text{infinity}$.  Well, we can see that as $x$ grows, the importance of the term with the highest exponent ($x^2$) becomes more and more important.  likewise, as $x$ gets larger, the importance of the other terms get less and less important.  After $x$ gets infinitely larger, the other terms get infinitely less important, or, not important at all.  Therefore, at $x = \text{infinity}$ then fraction then reduces to $\frac{x^2}{3x^2} = \frac{1}{3}$.  So, even though we can't actually count to infinity, we can use the properties of infinity to reason about things in the infinite.

Now, there are multiple different conceptions of infinity, so, to deal with infinity rigorously, you have to state which conception of infinity you are using.  Cantor's "cardinal numbers" deal with absolute set sizes, and shows that there are infinities of different absolute sizes.  Cantor's "ordinal numbers" deal with the ordering of different ways of expressing infinite sets.  My favorite way of dealing with infinity is the "hyperreal numbers", which basically give a unit value for a "base infinity", and then you can treat infinity as an algebraic unit.

For an example of how to use infinitely practically, think about the infinitely small.  If we take $\epsilon$ to be an infinitely small value, we can do some interesting things with it.  Let us say we want to know the value of $\frac{x^2 - 25}{x - 5}$ at $x = 5$.  If you plug it in, you get $\frac{0}{0}$.  However, if you nudge it slightly, you get an actual number.  What if we nudged it an infinitely small amount, by $\epsilon$, so that $x = 5 + \epsilon$?.  This yields: $$\frac{(5 + \epsilon)^2 - 25}{5 + \epsilon - 5} = \frac{5^2 + 2\cdot 5\cdot\epsilon  + \epsilon^2 - 25}{\epsilon} = \frac{10\epsilon + \epsilon^2}{\epsilon} = 10 + \epsilon$$

In other words, if $x$ is infinitely close to $5$, the result of the expression is infinitely close to $10$.

As you can see, infinite can be used in a number of ways by mathematicians.  We can do infinite sums of infinitely small numbers.  We can express a calculation as a polynomial with an infinite number of terms.  We can operate on series that diverge to infinity.

The key part is recognizing what infinity means in a particular context, and how the rules of that sort of infinity affect the problem you are working on.


问:
Showing that the symmetric subspace of N copies of the fundamental representation space is the space of the n-th irrep for SU(2).
I would like to have a clear explanation of why the symmetric subspace of N copies of the fundamental representation space is the space of the n-th irrep for SU(2) (and also more generally I don't know?).

I can see for two spin half objects $\frac{1}{2}\otimes \frac{1}{2}$ one can choose the multiplet and singlet as a basis $\{|00>,|01>+|10>,|11>\}$ and $\{|01>-|10>\}$. So one can say that the largest of the spaces in the Clebsch-Gordan decomposition ($1\oplus 0$) is indeed the space $sym(H^2\otimes H^2)$

I'm not clear however how to move beyond this. In an intuitive sense I can see that if I take some symmetric subspace of  $sym(H^{2n})$ I could view it in terms of a basis made up of sums of vectors where one has 1 in every possible position i.e |000>,|010>+|100>+|001>,|110>+|101>+|011>,|111>. However in this case I can't quite see how to link this to being the space of the nth (spin 3/2 in my example) irrep as determined by the Clebsch-Gordan decomposition (which would guarantee it was an irrep at least).

I suppose another way of phrasing the question is why is the largest space in the Clebsch-Gordan decomposition of n copies of the fundamental representations the symmetric subspace of those copies? (And is this just a property of SU(2)?)

答:
In this case there is no need of the Clebsch-Gordan decomposition, you can really do things more concretely with basis elements.

Let $V = \mathbb C^2$ be the fundamental rep of $\mathfrak{su}_2$ with basis $e_1,e_2$. You should have elements $x,y \in \mathfrak{su}_2$ such that $x e_1 = e_2, xe_2 = 0$ and $ye_2 = e_1, ye_1 = 0$.

A basis of $Sym^n(V)$ is given by $e_1^n,e_1^{n-1}e_2, \dots, e_2^n$. Then, it's easy to see that from $e_1^n$ you can get the whole $Sym^n(V)$ by applying $x$ several times, for example $x \cdot e_1^n = ne_1^{n-1}e_2, x^2e_1^n = x(ne_1^{n-1}e_2) = n(n-1)e_1^{n-2}e_2^2$, etc ...

It follows that $Sym^n(V)$ is irreducible.


问:
If $x, y \in \overline{B}(z,r)$ such that $\|x-y\| \ge \epsilon r,$ then $ \|z-\frac{x+y}{2}\|\le r \sqrt{1-\frac{\epsilon ^ 2}{4}}$
Let $z\in \mathbb{R}^n, r > 0$, and $\epsilon \in (0,2]$. Prove that if $x, y \in \overline{B}(z,r)$ such that $\|x-y\| \ge \epsilon r,$ then $ \|z-\frac{x+y}{2}\|\le r \sqrt{1-\frac{\epsilon ^ 2}{4}}$. So far I have:

$$\|z - \frac{x+y}{2}\| = \sqrt{\langle z-\frac{x+y}{2},z-\frac{x+y}{2}  \rangle}$$

Then:

$$\langle z - \frac{x+y}{2}, z-\frac{x+y}{2}\rangle = \langle z - x, z-\frac{x+y}{2}\rangle + \langle \frac{x-y}{2}, z-\frac{x+y}{2}\rangle
$$$$=\langle z-x, z- y \rangle + \langle z - x, \frac{y-x}{2}\rangle + \langle \frac{x-y}{2}, \frac{y-x}{2}\rangle + \langle \frac{x-y}{2}, z-y\rangle $$
$$=\langle z-x,z-y\rangle + \langle \frac{x-y}{2}, x-y\rangle-1/4\|x-y\|^2$$
$$=\langle z-x, z-y\rangle+1/2\|x-y\|^2-1/4\|x-y\|^2$$
$$\le\|z-x\|\|z-y\|-\frac{(\epsilon r)^2}{4}+1/2\|x-y\|^2$$
$$\le r^2-\frac{(\epsilon r)^2}{4}+1/2\|x-y\|^2=r^2(1-\frac{\epsilon^2}{4})+1/2\|x-y\|^2$$

I think I'm on the right track but I keep getting the $1/2\|x-y\|^2$ and I don't know what to do. Help?

答:
By translating by $z,$ you can assume $z = \vec{0}$ for simplicity. Then,

$$\|x+y\|^2+\|x-y\|^2 = 2\|x\|^2+2\|y\|^2\leq4r^2\implies \|x+y\|^2\leq 4r^2-\epsilon^2 r^2$$

and you are done.


问:
Find the solution to cauchys problem
I'm learning differential equation and not sure how to solve the following problem:

Find the solution for the cauchy's problem:

$y''-6y'+9y=e^{3x}\frac{x^2+2}{x^2+1}$

$y(0)=0$

$y'(0)=1$

What I tried so far:

$\lambda^2-6\lambda+9=0 \Leftrightarrow (\lambda-3)(\lambda-3)=0$

$y(x)=c_1*e^{3x}+c_2*x*e^{3x}+\varphi(x)$

I'm not sure how to continue.

Could you help me?

Thanks

Title: Find the solution to cauchys problem

Body:
hint

It is easier to begin by putting

$$\boxed{y=z.e^{3x}}$$

with
$$y'=(z'+3z)e^{3x}$$
$$y''=(z''+6z'+9z)e^{3x}$$

thus

$$y''-6y'+9y=$$
$$\color{red}{z''}e^{3x}=\color{red}{(1+\frac{1}{1+x^2})}e^{3x}$$
and
$$z'=c_2+x+\arctan(x)$$
$$z=(c_1+c_2x)+\frac{x^2}{2}+x\arctan(x)-\frac 12\ln(1+x^2)$$

With initial conditions, you will find $$c_2=1 \;\; and \;\;c_1=0$$


问:
Given a Geodesic square what is the area of the square?
I need help with this question

Given a Geodesic square on unit ball, The interiors angle are 2 radians each,
What is the area of the square?

答:
$$\text{Area (n-gon)}=\sum_{i=1}^n \alpha_i - (n-2)\pi = 2\pi - \sum_{i=1}^n (\alpha_i - \pi)$$

In the case of a triangle, for example, the area will be the difference between the sum of the angles and $pi.$ In the question it is $8-2\pi.$

This follows from the Gauss-Bonnet theorem:

$$\int_T K \text{d}A + \sum_{i=1}^3 \int_{E_i} k_g \text ds + \sum_{i=1}^3 \alpha_{\text{ext}_i}= 2 \pi \chi (T)\tag 1$$

where $T$ is the spherical triangle; $K$ is the Gaussian curvature; $E_i$ are the three edges of the triangle; $k_g$ is the geodesic curvature of the edges; $\alpha_{\text{ext}_i}$ the exterior angles, which are related to the interior angles by $ \alpha_{\text{ext}_i}= \pi-\alpha_i.$

Since $\int_{E_i} k_g \text ds=0$ because the triangle is bounded by geodesics (zero curvature); and $\chi (T)=1,$ because the spherical triangle is topologically a disc, equation $(1)$ becomes

$$\int_T K \text{d}A +  \sum_{i=1}^3 \pi - \alpha_i= 2 \pi $$

Since the curvature of the sphere is constant, and $K=\frac 1 {r^2},$

$$\begin{align}
A &=\frac 1 K \left( \sum_{i=1}^3  \alpha_i - \pi \right)\\
A &= r^2 \left( \sum_{i=1}^3  \alpha_i - \pi \right)\\
A &=  \sum_{i=1}^3  \alpha_i - \pi
\end{align}$$

when $r=1.$

The general formula for $n$-gons can be deduced inductively by splitting $n$-gons into spherical triangles.


问:
Funny integral inequality
Assume $f(x) \in C^1([0,1])$,and $\int_0^{\frac{1}{2}}f(x)\text{d}x=0$,show that:
$$\left(\int_0^1f(x)\text{d}x\right)^2 \leq \frac{1}{12}\int_0^1[f'(x)]^2\text{d}x$$

and how to find the smallest constant $C$ which satisfies
$$\left(\int_0^1f(x)\text{d}x\right)^2 \leq C\int_0^1[f'(x)]^2\text{d}x$$

Title: Funny integral inequality

Body: write $g=f'$ and observe that $f(0)=-\int_0^{1/2}(1-2t)g(t)dt$ from $\int_0^{1/2}f(x)dx=0.$ Therefore $$(\int_0^{1}f(x)dx)^2=(\int_0^1g(t)\min(t,1-t)dt)^2\leq \int_0^{1}g(t)^2dt\times \int_0^1(\min(t,1-t)^2dt$$ from Schwarz.


问:
Prove by using the inverse functions, $f$ is surjective if and only if $h \circ f \circ g$ is surjective
Let $f: A \to B$ be a function and $g: A \to A$, $h: B \to B$ be bijections. Prove by using the inverse functions, $f$ is surjective if and only if $h \circ f \circ g$ is surjective.

I can see a very clear proof for the first direction, however it doesn't use the inverse functions. Maybe it is acceptable to use the inverse functions on only the second direction.

$\Rightarrow$: Suppose $f$ is surjective

Since $g$ is bijective, then $f \circ g$ is surjective

Similarly, since h is a bijection, then $h \circ (f \circ g)$ is surjective

$\Leftarrow$: Suppose $h \circ f \circ g$ is surjective.

答: Since $h^{-1}$ and $g^{-1}$ are bijective and thus surjective we have $$(h \circ f \circ g)\circ g^{-1}= h \circ f$$ is surjective and so $$h^{-1} \circ (h \circ f) = f$$ is surjective.


答:
To begin with, I would notice that
\begin{align*}
y'' - 6y' + 9y & = (y'' - 3y') - (3y' - 9y) = (y' - 3y)' - 3(y' - 3y)
\end{align*}

Thus if we make the change of variable $u = y' - 3y$, one gets that
\begin{align*}
u' - 3u = e^{3x}\left(\frac{x^{2}+2}{x^{2}+1}\right) & \Longleftrightarrow (ue^{-3x})' = \frac{x^{2}+2}{x^{2}+1} = 1 + \frac{1}{x^{2}+1}\\\\
& \Longleftrightarrow ue^{-3x} = x + \arctan(x) + c_{1}\\\\
& \Longleftrightarrow (ye^{-3x})' = x + \arctan(x) + c_{1}\\\\
& \Longleftrightarrow ye^{-3x} = \frac{x^{2}}{2} + x\arctan(x) - \frac{\ln(x^{2}+1)}{2} + c_{1}x + c_{2}\\\\
& \Longleftrightarrow y(x) = e^{3x}\left(\frac{x^{2}}{2} + x\arctan(x) - \frac{\ln(x^{2}+1)}{2} + c_{1}x + c_{2}\right)
\end{align*}

Title: Find the solution to cauchys problem

Body:
I'm learning differential equation and not sure how to solve the following problem:

Find the solution for the cauchy's problem:

$y''-6y'+9y=e^{3x}\frac{x^2+2}{x^2+1}$

$y(0)=0$

$y'(0)=1$

What I tried so far:

$\lambda^2-6\lambda+9=0 \Leftrightarrow (\lambda-3)(\lambda-3)=0$

$y(x)=c_1*e^{3x}+c_2*x*e^{3x}+\varphi(x)$

I'm not sure how to continue.

Could you help me?

Thanks


问:
Find every quadratic polynomial of the form $p(x)=x^2+bx+c$ such that if $r$ is a root then $r^2 − 4$ is also a root of $p(x)$.

Consider polynomials $p(x)$ with the following property, called $(†)$.
$(†)$: If r is a root of $p(x)$, then $r^2 − 4$ is also a root of $p(x)$. We want to find every quadratic polynomial of the form $p(x) = x^2 +bx+c$ such that $p(x)$ has two distinct roots, has integer coefficients, and has property $(†)$. Prove that there are exactly two such polynomials and list them.

If $r^2-4$ is a root then it would imply $(r^2-4)^2-4$ is a root which would again imply $((r^2-4)^2-4)^2-4$ is a root and so on. But since we can have only two roots, we must have all the roots except $r$ and $r^2-4$ to be equal to either of them. Is my reasoning correct? How to explicitly find the equations?

Body:
$$p(x)=x^2- x-4$$
indeed solving the equation we get
$$x_1=\frac{1}{2} \left(1-\sqrt{17}\right);\;x_2=\frac{1}{2} \left(\sqrt{17}+1\right)$$
$$\left(\frac{1}{2} \left(1-\sqrt{17}\right)\right)^2-4=\frac{1}{2} \left(1-\sqrt{17}\right)$$
$$\left(\frac{1}{2} \left(1+\sqrt{17}\right)\right)^2-4=\frac{1}{2} \left(\sqrt{17}+1\right)$$


问:
Is $(1+x^2,x^3) \in \mathbb{Q}[x]$ a principal ideal?
Is this line of reasoning correct?

If $\big(1+x^2,\:x^3\big) \in \mathbb{Q}[x]$ were a principal ideal, then there would be a polynomial $f\in \mathbb{Q}[x]$ such that $\big(f(x)\big) = \big(1+x^2,x^3\big)$, and so we could conclude that there exist polynomials $a,b,g,h\in \mathbb{Q}[x]$ such that
\begin{align}
f(x) &= (1+x^2) a(x) + x^3 b(x) \\
1+x^2 &= f(x)g(x) \\
x^3 &= f(x)h(x) .
\end{align}
But that would imply that $\mbox{deg}(f)\leq 2$ and hence that $x^3 = (1+x^2) a(x) h(x)$, where $\mbox{deg}(ah) \leq 1$, that is, $a(x) h(x) = A + Bx$ for some $A,B \in \mathbb{Q}$. This, however, contradicts the fact that there exist no rational numbers $A,B$ such that $(1+x^2)(A + Bx) = A+Bx+Ax^2+Bx^3=x^3$. Therefore, $\big(1+x^2,\:x^3\big)$ is not a principal ideal.

答:
The issue in your reasoning is the implication from $\deg(f)\leq 2$ to $x^3=(1+x^2)a(x)h(x)$. You've lost the $x^3b(x)$ term, and what you get is actually that
$$x^3=(1+x^2)a(x)h(x)+x^3b(x)h(x),$$
which factors to
$$(1+x^2)a(x)h(x)=x^3(1-b(x)h(x)),$$
not a contradiction if $1+x^2$ divides the $1-b(x)h(x)$ factor.

In fact, this ideal (like all ideals in $\mathbb Q[x]$) is principal, and its single generator can be found using a sort of $\gcd$ argument. To get you started, you can find a polynomial in the ideal of degree smaller than $2$, since
$$x=x(x^2+1)-x^3\in (1+x^2,x^3).$$


问:
Pythagoras Theorem Proof
Is my logic correct below to prove the Pythagoras Theorem? Thanks.

Area Rectangle R
\begin{align*}
R &= WL\\
  &=(2a+b)(2b+a)\\
  &=4ab+2a^2+2b^2+ab\\
  &=5ab+2a^2+2b^2
\end{align*}

Total Area Yellow Triangles T
\begin{align*}
T &= 10(\frac{ab}{2})\\
&=5ab
\end{align*}

Calculate Area $c^2$

\begin{align*}
c^2 &= R-T-a^2-b^2 \\
   &=5ab+2a^2+2b^2-5ab-a^2-b^2\\
   &=a^2+b^2
\end{align*}
∎

Proof by rearrangement

答: Yes. This is correct provided you have shown that all yellow triangles are congruent.</p>

Title: Pythagoras Theorem Proof


问:
Triple integral/tetrahedron
Let the Tetrahedron be bounded by the planes

$x+2y+z=2$, $x=2y$, $y=0$, $z=0$

so the limits for z are easy and would be $0 \leq z \leq  2-x-2y$
I have calculated the intersection point in the $(x,y)$ plane and I get $(1, \frac{1}{2})$.
So $y=0$ means that the lower boundary of $y$ should start with zero. I drew a sketch, but I am unsure about the x limits and y limits.

what would the functions $h_1$ and $h_2$ be?

$\int_{0}^{1} \int_{h_1(y)}^{h_2(y)} \int_{0}^{2-2y-x} dzdxdy$

on the other hand, if we had $x=0$ it would be easy to write because I could write it as
$\int_{0}^{1} \int_{x/2}^{1-x/2} \int_{0}^{2-2y-x} dzdydx$

but since we have $y=0$ I need to determine the $h_1(x)$, $h_2(x)$, where I am confused how to do it

]1

答:
y goes from 0 to 0.5

And for a fixed y we know that x goes from $2y$ to $2 - 2y$ because the x values are to the right of the line $x = 2y$ and to the left of the line $x = 2 - 2y$

So we end up with:

$\int_0^{0.5}\int_{2y}^{2-2y}\int_{0}^{2 - x -2y}dzdxdy$

Title: Triple integral/tetrahedron


问:
Proving a theorem using its converse
Converse of Pythagoras' theorem: If the lengths of the sides of a triangle $T$ are $a$, $b$, and $c$, and if $a^2+b^2=c^2$, then the triangle is a right triangle and the side opposite to the right angle is the one whose length is $c$.

Proof: Construct a line segment $XY$ whose length is $a$. Then construct a line segment $YZ$ whose length is $b$ which is perpendicular to $XY$. By construction, the triangle $XYZ$ is a right triangle, and therefore, by Pythagoras' theorem and because we are assuming that $a^2+b^2=c^2$, the length of $XZ$ is equal to $c$. So, the triangle $XYZ$ is similar to the original triangle $T$. Since the triangle $XYZ$ is a right triangle, then so is $T$.
What I find peculiar about this proof is the fact that it uses Pythagoras' theorem in order to prove its converse.

It is not the only situation that I am aware of in which this occurs. For instance, there is a proof of the converse of Ceva's theorem  which uses that theorem. But I am not aware of any example outside Euclidean Geometry.

Can anyone provide an example of a theorem of the type $A\implies B$ outside Geometry with a proof which uses the fact that $B\implies A$?

答:
Let $f: A \to B$ be a function and $g: A \to A$, $h: B \to B$ be bijections.

Prove that $f$ is surjective if and only if $h \circ f \circ g$ is surjective.

Proof:  Suppose $f$ is surjective .... we get $ h \circ f \circ g$ is surjective.

Now the converse: Suppose $h \circ f \circ g = :e$ is surjective. Since $g^{-1}$ and $h^{-1}$ are surjective then, by already proven part, we have $h^{-1} \circ e \circ g^{-1}=f$ is surjective.

Title: Proving a theorem using its converse


答:
$$y''-6y'+9y=e^{3x}\frac{x^2+2}{x^2+1}$$
The differential equation is equivalent to:
$$(ye^{-3x})''=1+\frac{1}{x^2+1}$$
Integrate twice.

Title: Find the solution to cauchys problem


问:
In intuitionistic first-order logic, can the "nothing" quantifier define either of the universal or existential quantifiers?
In classical first-order logic, the nothing quantifier $Nx$, which means "there are no x such that", can define both the universal and existential quantifiers. I suspect this is not the case in intuitionistic first-order logic. I don't think that quantifier can define either of the universal or existential quantifiers. Is this true, and what is the proof?

答:
Your quantifier, which I'll denote $И$, cannot be used to define either the universal or the existential quantifiers in intuitionistic logic. I offer a proof-theoretic argument: needless to say, one needs to know the sequent calculus and the cut-elimination theorem to understand what follows.

I. Frankly, I was not entirely sure why my right and learned colleague grilled you on the inference rules that the $И$ quantifier should satisfy, since there is exactly one sensible way to assign rules to the $И$ quantifier. What's more, it's easy to verify that both possible classical formalizations of $Иx.P(x)$, as either $\forall x. \neg P(x)$ or $\neg \exists x. P(x)$, yield intuitionistically equivalent quantifiers.

Let me start by proving this claim, as it plays a crucial role in the proof I give below.

II. The formulae $\forall x. \neg P(x)$ and $\neg \exists x. P(x)$ are equivalent in intuitionistic logic. An informal natural deduction proof follows.

First, assume that $\forall x. \neg P(x)$ holds, and assume for a contradiction that $\exists x. P(x)$ also holds. By this second assumption we have $P(s)$ for some term $s$, so we can set $x$ to $s$ in the first assumption to obtain $\neg P(s)$, a contradiction. This proves the implication $(\forall x. \neg P(x)) \rightarrow \neg \exists x. P(x)$.

Now assume that $\neg \exists x. P(x)$ holds. Take any $y$, and assume for a contradiction that $P(y)$ holds. Then $\exists x. P(x)$ also holds, contradicting our first assumption. Thus $\neg P(y)$ holds. But we had an arbitrary $y$, so we get $\forall x. \neg P(x)$, and we can conclude $(\neg \exists x. P(x)) \rightarrow \forall x. \neg P(x)$.

So we can define the meaning of $Иx.M(x)$ as either $\forall x. \neg M(x)$ or equivalently as $\neg \exists x. M(x)$, whichever we find more convenient in any given context.

III. From here on, we will use the following general proof-theoretic facts about intuitionistic logic. All three are easily provable by induction on cut-free derivations.

Existence-disjunction property (see Theorem 4.2.4. of [1], we could use much weaker results): if $\Gamma$ does not contain any existential quantifiers, and $\Gamma \vdash \exists x. M(x)$, then we can find terms $t_1,t_2,\dots t_n$ such that $\Gamma \vdash M(t_1) \vee M(t_2) \vee \dots \vee M(t_n)$.

Disjunction property: if $\Gamma$ does not contain any disjunction connectives or existential quantiifers, and $\Gamma \vdash M \vee N$, then we can find a proof of either $\Gamma \vdash M$ or $\Gamma \vdash N$.

Universal independence: If $\Gamma$ contains no universal quantifier, and $\Gamma \vdash \forall x. M(x)$, then $\Gamma \vdash N$ for any formula $N$.

III. Let us prove that $\exists$ cannot be defined in terms of the quantifier $И$ and the usual connectives $\rightarrow, \neg, \vee, \wedge$.

Take a unary atomic predicate symbol $P(x)$ and suppose that we could find a formula $Q$ in the $\{\rightarrow, \neg, \vee, \wedge, И \}$ fragment of Intuitionistic Logic such that $Q \vdash \exists x. P(x)$ and $\exists x. P(x) \vdash Q$ both hold. Given the result of Section II, we can interpret the quantifier $Иx.M$ as an abbreviation for $\forall x. \neg M$. Thus, $Q$ belongs to the $\{\rightarrow, \neg, \vee, \wedge, \forall \}$ fragment of intuitionistic logic.

Since $Q \vdash \exists x. P(x)$, the existence-disjunction property allows us to find $t_1,t_2,\dots t_n$ such that $Q \vdash P(t_1) \vee P(t_2) \vee \dots \vee P(t_n)$. Choose a variable symbol $y$ that does not occur in any of the terms $t_1,t_2,\dots t_n$. We have $P(y) \vdash \exists x. P(x)$. From $\exists x. P(x) \vdash Q$ and the cut rule we get a proof that $P(y) \vdash Q$. From $Q \vdash P(t_1) \vee P(t_2) \vee \dots \vee P(t_n)$ and the cut rule we get a proof that $P(y) \vdash P(t_1) \vee P(t_2) \vee \dots \vee P(t_n)$. Using the disjunction property we could then have $P(y) \vdash P(t_m)$ for some $m$, which is impossible since $y$ does not occur in any of the terms $t_1,t_2,\dots t_n$.

Hence our supposed formula $Q$ cannot exist.

IV. Let us prove that $\forall$ cannot be defined in terms of the quantifier $И$ and the usual connectives $\rightarrow, \neg, \vee, \wedge$.

Take a unary atomic predicate symbol $P(x)$ and suppose that we could find a formula $Q$ in the $\{\rightarrow, \neg, \vee, \wedge, И \}$ fragment of Intuitionistic Logic such that $Q \vdash \forall x. P(x)$ and $\forall x. P(x) \vdash Q$ both hold. This time, given the result of Section II, we interpret the quantifier $Иx.M$ as an abbreviation for $\neg \exists x. M$. Thus, $Q$ belongs to the $\{\rightarrow, \neg, \vee, \wedge, \exists \}$ fragment of intuitionistic logic: in particular $Q$ contains no universal quantifiers. From $Q \vdash \forall x. P(x)$, universal independence gives that $Q \vdash \neg \forall x. P(x)$. Cutting this against our assumption $\forall x. P(x) \vdash Q$ would yield $\forall x. P(x) \vdash \neg \forall x. P(x)$, but we can immediately see that the latter admits no proof.

Hence our supposed formula $Q$ cannot exist.

[1] A. S. Troelstra, H. Schwichtenberg: Basic Proof Theory, 2nd edition, web


问:
Opposite of Dirac delta distribution
The multivariate Dirac delta distribution can be - more or less intuitively - be expressed as

\begin{align}
\delta(\mathbf x) = \begin{cases} \lim\limits_{a\rightarrow0} \quad \dfrac{1}{a^n} & \forall  x_i \in [-\frac a2,\frac a2], 1\le i\le n \\[6pt]
\quad 0 & \text{otherwise}
\end{cases}
\end{align}

where

$$
\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \delta(\mathbf x) \text{ d}\mathbf x = 1
$$

Is there an "opposite" of that, which can be expressed as

\begin{align}
\epsilon(\mathbf x) = \begin{cases} \lim\limits_{a\rightarrow\infty} \quad \dfrac{1}{a^n} & \forall  x_i \in [-\frac a2,\frac a2], 1\le i\le n \\[6pt]
\quad 0 & \text{otherwise}
\end{cases}
\end{align}

where also

$$
\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \epsilon(\mathbf x) \text{ d}\mathbf x = 1
$$

?

Is there a name for this distribution and/or a symbol?

For context: I am planning to use them in convolutions and I am treating them as probability densities.

答: Both limits $$\lim_{a\to 0} a^{-n} 1_{x\in [-a/2,a/2]^n}, \qquad \lim_{a\to \infty} a^{-n} 1_{x\in [-a/2,a/2]^n}$$ are perfectly rigorous definitions of distributions, the first one converges in the sense of distributions to $\delta$ and the second one to $0$.

Title: Opposite of Dirac delta distribution


问:
Integrating left to right versus right to left.
OK, I understand that when integration is done left to right with respect to x increasing left to right (dx is positive), that the answer is positive, and vice versa when integrating right to left. But, as a reality check, there is something that I have never understood (or, understanding, I find so counterintuitive as to want to question it): Interpreting the result as area, one and the same area is either positive or negative not depending on what it is, but on how it is counted. I recognize that it is a logical extension of, say, -1 versus +1 on the number line, but at least -1 and +1 occupy different positions! Whenever I've brought this up, I tend to get blank looks, like, "how could you be so stupid", or "huh", but, clearly it does not keep me from being bothered by it, so I'll ask here ... once again! To me, apart from what the mechanics of integration require, a more logical "positioning" of the negative area would be its reflection across the x-axis.</p>

答:
Although integrals can often be explained as "area", you are going wrong when you assume that integral is area. That is not true. Yes, they are related and numbers sometimes coincide.

But take a look at sine integral - it's zero, although the circumscribed areas are clearly non-zero. Does that seem O.K. to you? You have to accept that integral is not area. It can behave differently and it can have different signs.

Have you had physics in school? If you are familiar with potential energies, this might help.

Take a look at such situation. Object with mass $m$ is at height $a$. It is lifted up to a higher height $b$. Force of gravity is $F=mg$. The object-Earth system has gained additional potential energy:
$$\Delta U_1 = \int\limits_a^b mg \mathrm{d}h = mg(b-a)$$

If the object was instead lowered from b to a then
$$\Delta U_2 = \int\limits_b^a mg \mathrm{d}h = mg(a-b) = -\Delta U_1$$

Wither way the change in potential energy is the familiar $mg\Delta h$, but the change in height $\Delta h$ is opposite and so is the change in potential energy. Hope that's understandable and alright to you.

What I want to show, that integral is not the area function. It's more like accumulating operator. In the gravity case it accumulates energy by overcoming gravitational force while the argument $x$ (height) is changed. If you change it in other direction, the situation becomes obviously opposite.

Yes, if you take positive function and integrate it from left to right, you accumulate area and therefore you get the area. But if you go backwards, you are more like subtracting that area, therefore you get the minus sign.


问:
Is there a way to prove that the 2-norm of a matrix is greater than the largest element?
I would like to prove

$$\max_{ij} |A_{ij}| \leq \| A \|_2$$

Any help would be greatly appreciated!

答:
Hint: since the two-norm $\|A\|_2$ is defined as the maximum of some function over all vectors, all you need to do is find some vector for which that function is at least any given matrix entry; more concretely, you need some vector $x$ for which
$$\frac{|Ax|}{|x|}\geq A_{ij},$$
where $|\cdot|$ denotes the standard (Euclidean) norm on vectors. For scaling reasons, you might as well pick $x$ for which $|x|=1$; can you find some unit vector for which $A_{ij}$ becomes an entry of $Ax$?