问:
Making Exponent the Subject When Equation Contains Multiple Unknowns
** Sorry if some of the terminology is wrong here, I am but a novice **

I want to make $g$ the subject of the following equation

$x_T=x_1 e^{g(T-1)}$

I have the answer, which is given as

$g=\cfrac{\ln(x_T)-\ln(x_1)}{T-1}$

But I would like to understand how this answer was obtained.

Unfortunately, if you Google something like 'exponential into log form,' you will just run into rudimentary explanations of how to convert exponentials into logs, something like:

$3^x=9\rightarrow x=\log_3 9$

I cannot seem to find any examples of how to make the exponent the subject when the equation contains multiple variables. Hoping someone can point me in the right direction!

答:
If $x_1 \ne 0$we can write the equation as
$$
\frac{x_T}{x_1}=e^{g(T-1)}
$$
so, using the definition of logarithm, we have
$$
\log \frac{x_T}{x_1}=g(T-1)
$$
and using the formula
$$
\log \frac{a}{b}=\log a -\log b
$$
you can find the result.


问:
Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

I did this by creating 3 sets:

$S\,\,=\{a,b,c,d\}$

$T\,\,=\{c,d,e,f\}$

$T'=\{a,b,c,d\}$

And then I performed the operations and checked they are the same. Is there something else I should do or only this will suffice?

答:
Checking one example is insufficient in order to conclude a general proof.

You need to do this for abstract sets. Pick an element from $S\cap (T\cup T')$, using the definitions of $\cap,\cup$ conclude that it has to be in $(S\cap T)\cup (S\cap T')$, and then do the same thing from the other direction.

That would be a proof.

ContentLicense: CC BY-SA 3.0


问:
About Euler totient function $\underbrace{\varphi( \varphi( \dots \varphi(}_{k\ \text{times}} n) \dots ))$
Let $\varphi^k(n)=\underbrace{\varphi( \varphi( \dots \varphi(}_{k\ \text{times}} n) \dots ))$. Define $f:\mathbb{N}/ \{1\}\rightarrow \mathbb{N}$ so that $f(n)$ is the number of iterations of the Euler function required to obtain $2$ starting from $n$ (it is obvious that this number exists and is unique as $\varphi (k)$ is even and $\varphi(k)<k$ for $k\ge 3$).
How can I show that $f(ab)=f(a)+f(b)$ if at least one of $a$ and $b$ is odd and $f(ab)=f(a)+f(b)+1$ otherwise.

答:
It will be convenient for stating results to define $f(1)=0$.

Let $n$ be minimal such that $n$ can be expressed as a product which is a counterexample to this result. We shall first prove a couple of Lemmas.

Let $PQ\le n$, where $P$ and $Q$ are coprime. Then $f(PQ)=f(P)+f(Q)$.

Proof

$\phi(PQ)=\phi(P)\phi(Q)$ and so $f(PQ)=1+f\left (\phi(P)\phi(Q)\right)$, where $\phi(P)\phi(Q)<n$.

If either $P$ or $Q$ is $1$ or $2$, then the result is trivial so we can suppose that both $\phi(P)\text { and }\phi(Q)$ are even.

Then  $f(PQ)=1+f (\phi(P))+f(\phi(Q))+1=f(P)+f(Q)$.

If $a\ge1, f(2^a)=a-1$ and, if  $p$ is an odd prime such that $2^a\le n$, then $f(p^a)=af(p).$

Proof

Since $\phi(2^a)=2^{a-1}$ we have $f(2^a)=a-1$.

Since $\phi(p^a)=p^{a-1}(p-1)$ we have $$f(p^a)=1+f\left (p^{a-1}(p-1)\right )=1+f\left (p^{a-1}\right)+f\left (p-1\right )=f\left (p^{a-1}\right)+f(p).$$ Therefore $f(p^a)=af(p).$

Solution

Let $n=uv$ and let a prime $p$  occur to power $a$ in $u$ and to power $b$ in $v$.

If $p$ is odd, this prime contributes the same amount, $(a+b)(p-1)$, to both$f(uv)$ and $f(u)+f(v)$.

If $p=2$ and occurs in only one of $u$ and $v$ then it also contributes the same to $f(uv)$ and $f(u)+f(v)$. Therefore $f(uv)=f(u)+f(v)$.

If $p=2$ and if $ab>0$, then it contributes $a+b-1$ to $f(uv)$ and $a-1+b-1$ to $f(u)+f(v)$. Therefore  $f(uv)=f(u)+f(v)+1$.

ContentLicense: CC BY-SA 4.0


问:
Good partial differential equation books.
I have good command in real and complex analysis as well i had studied ordinary differential equation. Now i am trying to study partial differential equation but facing a lot of problems. Please suggest me some books which are suitable for me. I start with Strauss book but it is not comfortable for me. Also tell me books in which Cauchy problem for PDE is given in very simple and nice form. I studied ODE from tenenbaum dover book which is very nice to digest. I want same type of books for PDE. Please suggest me . Thanks a lot. </p>

答: If you need a bit of physics to keep the rigor interesting, I'd recommend Guenther and Lee's Partial Differential Equations of Mathematical Physics and Integral Equations. I used it in grad school, suited me fine. Looks like Dover publishes it now.</p>


问:
How to check if other points are on the geodesic between 2 points on a sphere?
Given any 2 points $p_1, p_2 \neq p_1$ on a sphere, how can we check if $p_3$ is on the geodesic from $p_1$ to $p_2$?

I think of checking the $||\text{arc } p_1 p_3|| + ||\text{arc } p_3 p_2|| = ||\text{arc } p_1 p_2||$. Is this condition sufficient? Any other suggestions?

答:
Evaluate the following double integral
$$
\int_{0}^{\frac{\pi}{2}} \int_{0}^{\pi}\left(x^{2} \cos y-\sin x\right) d x d y
$$


问:
Matrix equation with diagonal matrix
Let $A \in\mathbb{R}^{n\times n}$ be a given Positive-Semi Definite matrix ,$\theta\in\mathbb{R}^n$ a vector, $x\in\mathbb{R}$ an unknown variable, and $a\in\mathbb{R}^+$ a positive constant. I have the following equation
$$
\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta  = a
$$
where $I$ is the identity matrix.

Is it possible to solve this equation for $x?$

What I have tried:

$$
\text{trace}(\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta)  = \text{trace}(a)
\\\theta^TAA\theta  \cdot\text{trace}((A+xI)^{-1}(A+xI)^{-1})  = a
\\ \sum_{i = 1}^n\frac{1}{(x+s_i)^2}=\frac{a}{\|\theta\|_{A^2}^2}
$$

where $s_i$ are the singular values of the $A$. Is it correct? How can i proceed from here?

答:
$\def\T#1{\operatorname{tr}(#1)}$I would suggest using Newton's Method, i.e.
$$\eqalign{
f(x) &= 0, \quad
f_k &= f(x_k), \quad
f'_k &= f'(x_k) \quad\implies\quad
x_{k+1} &= x_k - \frac{f_k}{f'_k} \\
}$$
For typing convenience, define the matrices
$$\eqalign{
B &= A+Ix \quad\implies\quad
\frac{dB^n}{dx} = nB^{n-1} \\
M &= A\,\theta\theta^T\!A \\
}$$
Set up an appropriate function and apply Newton's Method
$$\eqalign{
f  &= a - \T{MB^{-2}} \\
f' &= +\T{2MB^{-3}} \\
B_k &\doteq A + Ix_k \\
x_0 &= 1, \qquad
x_{k+1}
  = x_k - \frac{a-\T{MB_k^{-2}}}{\T{2MB_k^{-3}}} \\
}$$
NB:  The iteration will converge to different roots depending upon the choice of $x_0$

Also note that if $A$ is diagonal, then $B$ is also diagonal and the matrix inversion required by the iteration can be computed very efficiently.

Title: Matrix equation with diagonal matrix


问:
Approximation of beam
Assume that there is a simply supported beam subjected to concentrated moments $M_0$ at each end. The governing equation is
$$EI\frac{d^2y}{dx^2}-M(x)=0$$
with the boundary conditions $y(0)=0$ and $y(H)=0$. I know that there is an exact solution in the form of $y(x)=\frac{M_0\,x}{2EI}(x-H)$ but I must find an approximate solution for the deflection of the beam using trigonometric functions via Least Squares Method. The least squares method is known to me in general; but how to use it in this case?</p>

答:
Suppose that $\hat y(x)$ is an approximate solution then it follows that
$$EI\frac{d^2 \hat y}{dx^2}-M(x)=R(x)\ne 0$$
since the approximate solution does not satisfy the equation. In LSM the error term is defined such as
$$I=\int_a^b R(x)^2 dx$$
and minimized wrt to coefficients in $\hat y(x)$. To satisfy the boundary constraint the approximation function can be taken as (Note that you can have more parameters and satisfy the boundary constraints later on)
$$\hat y(x)=a\sin\bigg(\frac{\pi\,x}H\bigg)$$
By noting that $M(x)$ has a constant value $M_0$
$$I=\int_0^H \bigg(EI\frac{d^2 \hat y}{dx^2}-M_0\bigg)^2 dx=\int_0^H \bigg(-EI\frac{a\pi^2}{H^2}\sin\bigg(\frac{\pi\,x}H\bigg)-M_0\bigg)^2 dx$$
Now use Wolfram Alpha to integrate (or use below result)
$$I=\frac{E^2\,I^2\,\pi^4}{2\,H^3}a^2+\frac{4\,M_0\pi\,E\,I}Ha+M_0^2\,H$$
For minimization use the derivative wrt $a$
$$\frac{d\,I}{d\,a}=\frac{E^2\,I^2\,\pi^4}{H^3}a+\frac{4\,M_0\pi\,E\,I}H=0\Rightarrow a=-\frac{4\,M_0\,H^2}{\pi^3\,E\,I}$$
and the approximate function becomes
$$\hat y(x)=-\frac{4\,M_0\,H^2}{\pi^3\,E\,I}\sin\bigg(\frac{\pi\,x}H\bigg)$$</p>

Title: Approximation of beam


问:
Why is $\int 2 \cos x \sin x$ $\neq$ $\int \sin 2x$?
So I had this school assignment with this question

$\int (\cos x - \sin x)^2 \ dx$

And through computations I came up with this:

$\int (\cos x + \sin x )\ dx - \int 2 \cos x \sin x$

And this is where I got stuck, that when I integrate $2 \cos x \sin x$ right away, I would actually get $\color{blue}{x - \sin^2 x + C}$, which is different if I applied the double angle trigonometric identities in $2 \cos x \sin x$ , rewrite it to $\sin 2x$ , and it will give me a different answer, $\color{red}{x + \cos(2x)/2 + C}$

I went to Symbolab, and it really shows the difference, whether I integrate it right away or apply the trigonometric identity first before integrating

I tried to plot the points in Desmos, two answers seems close enough that it's just one constant away (particularly $0.5$), but still mathematically not equal.

So, does this mean that this trigonometric identity changed the actual answer that I should avoid using it as much as possible. Or this is just an exception? Which one is right? Is it both of the two answers?

I already passed my answer to the teacher, but what I actually did before passing it is integrating right away. But this problem haunts me in sleep that I actually opened my pc to ask this question here. Still learning integration techniques that I got introduced to this kind of problem, that overall it's pretty fun and stressing.

答:
Hint:

$$x-\sin^2(x)+\color{blue}C=x+\frac{\cos(2x)}2+\color{blue}{C- \frac12}$$


问:
Using $\sec(x)$ for integral.

Find the undefined integral $\int \frac{\sqrt{x^2+4x}}{x^2}\mathrm{dx}$

$$\displaystyle\int \dfrac{\displaystyle\sqrt{x^2+4x}}{x^2}\mathrm{dx}$$

I tried to rearrange the square root and I got:

$$\sqrt{(x+2)^2-4}$$ and I substitute the $x+2$ with $2\sec(u)$ so indeed I got these two:

$$x+2=2\sec(u) \\
\sqrt{x^2+4x}=2\tan(u) \\
\mathrm{dx}=2\tan(u)\sec(u)\mathrm{du}$$

And the integral turns out like this:

$$\displaystyle \int \dfrac{\tan^2(u)\sec(u)}{(\sec(u)-1)^2}\mathrm{du}$$

And I continued to rearrange the integral:

$$\displaystyle\int \dfrac{1-\cos^2(u)}{\cos(u)(1-\cos(u))^2}\mathrm{du}$$

I apply partial fractions method, saying $\cos(u)=u$ without integral sign and I ended up with:

$$\displaystyle\int \dfrac{1}{\cos(u)}\mathrm{du}+2\displaystyle\int \dfrac{1}{1-\cos(u)}\mathrm{du}$$

One can easily integrate those integrals se I skip the calculating part. In the end I get:

$$\ln\left|\sec(u)+\tan(u)\right|+2(-\cot(u)-\csc(u))+\mathrm{C}$$

I drew a triangle:

Knowing $\sec(u)=\dfrac{x+2}{2}$ says us $\cos(u)=\dfrac{2}{x+2}$

As I calculate I got:

$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{2x-8}{\sqrt{x^2+4x}}+\mathrm{C}$$

Bu the answer key is:

$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{8}{x+\sqrt{x^2+4x}}+\mathrm{C}$$

I have been thinking where my wrong is for five hours but I couldn't find anything. Please if you see any gap tell me. Thanks.

答:
The error occurred in partial decomposition. Instead, it should be

$$
\dfrac{1-\cos^2(u)}{\cos(u)(1-\cos(u))^2}=
\dfrac{1}{\cos(u)}+\dfrac{2}{1-\cos(u)}$$

Title: Using $\sec(x)$ for integral.


问:
general form of summation $N$ to $N!$
Is there a general form of summation of $N$ to $N!$? Let's say $N = 5$:

$$5 + (5 \cdot 4) + (5 \cdot 4 \cdot 3) + (5 \cdot 4 \cdot 3 \cdot 2) + (5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)$$

答:
Like I already wrote in my comment: this question was already answered here Expression for $n+n(n-1)+n(n-1)(n-2)+...+n!$

I'll write out the answer from that post here if you just want the answer. For an explanation why it is true look into the aforementioned thread. The closed form for your summation is:

$$n+n \times (n-1) \times...\times2\times1+n!=\displaystyle n! \sum_{k=0}^{n-1} \frac1{k!} = \lfloor n!\, e\rfloor -1$$

For your example ($n=5$) the answer is: $$\lfloor 5!\, e\rfloor -1 = \lfloor 120\times 2.7182818...\rfloor -1 = \lfloor 326.1938\rfloor -1 =326-1=325$$

Title: general form of summation $N$ to $N!$


答:
The difference between the two answers is:

\begin{align}
&\quad\frac {2x\color{red}{+}8}{\sqrt{x^2+4x}}-\frac {8}{x+\sqrt{x^2+4x}}
\\&=\frac {(2x+8)(x+\sqrt{x^2+4x}) - 8\sqrt{x^2+4x}}{(x+\sqrt{x^2+4x})\sqrt{x^2+4x}}
\\&=\frac {2x^2+8x+2x\sqrt{x^2+4x}}
{(x+\sqrt{x^2+4x})\sqrt{x^2+4x}}
\\&=\frac {2(x^2+4x+x\sqrt{x^2+4x})}
{x\sqrt{x^2+4x}+x^2+4x}
\\&=2
\end{align}

which is a constant. So both answers are correct.

Title: Using $\sec(x)$ for integral.

Body:
Find the undefined integral $\int \frac{\sqrt{x^2+4x}}{x^2}\mathrm{dx}$

$$\displaystyle\int \dfrac{\displaystyle\sqrt{x^2+4x}}{x^2}\mathrm{dx}$$

I tried to rearrange the square root and I got:

$$\sqrt{(x+2)^2-4}$$ and I substitute the $x+2$ with $2\sec(u)$ so indeed I got these two:

$$x+2=2\sec(u) \\
\sqrt{x^2+4x}=2\tan(u) \\
\mathrm{dx}=2\tan(u)\sec(u)\mathrm{du}$$

And the integral turns out like this:

$$\displaystyle \int \dfrac{\tan^2(u)\sec(u)}{(\sec(u)-1)^2}\mathrm{du}$$

And I continued to rearrange the integral:

$$\displaystyle\int \dfrac{1-\cos^2(u)}{\cos(u)(1-\cos(u))^2}\mathrm{du}$$

I apply partial fractions method, saying $\cos(u)=u$ without integral sign and I ended up with:

$$\displaystyle\int \dfrac{1}{\cos(u)}\mathrm{du}+2\displaystyle\int \dfrac{1}{1-\cos(u)}\mathrm{du}$$

One can easily integrate those integrals se I skip the calculating part. In the end I get:

$$\ln\left|\sec(u)+\tan(u)\right|+2(-\cot(u)-\csc(u))+\mathrm{C}$$

I drew a triangle:

Knowing $\sec(u)=\dfrac{x+2}{2}$ says us $\cos(u)=\dfrac{2}{x+2}$

As I calculate I got:

$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{2x-8}{\sqrt{x^2+4x}}+\mathrm{C}$$

Bu the answer key is:

$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{8}{x+\sqrt{x^2+4x}}+\mathrm{C}$$

I have been thinking where my wrong is for five hours but I couldn't find anything. Please if you see any gap tell me. Thanks.


问:
References for De Rham’s cohomology and De Rham’s theorem
I’m looking for a reference (preferably lecture notes or a book) that introduces De Rham’s cohomology and gives a proof of De Rham’s theorem. It would be nice if they included some applications too. I have some background in differential geometry and calculus on manifolds, and a bit of background in algebraic topology (simplicial homology mostly) and very basic category theory.

I’ve found the book ‘From calculus to cohomology: De Rham cohomology and characteristic classes’ by Ib Madsen and Jørgen Tornehave which seems quite helpful but it doesn’t include a discussion of De Rham’s theorem/singular (co)homology and its relation to De Rham’s cohomology.

答:
You can find a treatment of De Rham cohomology and De Rham's theorem in chapters 17 and 18 of Lee's Introduction to Smooth Manifolds respectively.

I took an undergraduate course on differential geometry not long ago where this book was the main reference. I believe the categorical prerequisites - if any - should be very light.

In any case, as Jyrki says you should probably wait for an expert's opinion.


答:
The "part of the antiderivative without the $+C$" is not unique, it can always be adjusted by a constant. Confusingly, there are expressions that don't appear to involve constants that nevertheless differ by constants, such as $\cos(x)^2$ and $-\sin(x)^2$. This is what is happening here:

$$\int 2 \sin(x) \cos(x) dx = \sin(x)^2 + C \\
\int \sin(2x) dx = -\frac{1}{2} \cos(2x) + C = -\frac{1}{2} \cos(x)^2 + \frac{1}{2} \sin(x)^2 + C = -\frac{1}{2} + \sin(x)^2 + C$$

where in that last step I used $\cos(x)^2=1-\sin(x)^2$.

These have to be the same thing, because the functions being integrated are the same...and they are the same thing, because $-1/2 + C$ in the second version is just another arbitrary constant. I find this is intuitive after you've "isolated" the constant like I did above, whereas the idea that $\sin(x)^2+C$ and $-\frac{1}{2} \cos(2x) + C$ are the same thing is a bit puzzling on its face.

Title: Why is $\int 2 \cos x \sin x$ $\neq$ $\int \sin 2x$?

Body:
So I had this school assignment with this question

$\int (\cos x - \sin x)^2 \ dx$

And through computations I came up with this:

$\int (\cos x + \sin x )\ dx - \int 2 \cos x \sin x$

And this is where I got stuck, that when I integrate $2 \cos x \sin x$ right away, I would actually get $\color{blue}{x - \sin^2 x + C}$, which is different if I applied the double angle trigonometric identities in $2 \cos x \sin x$ , rewrite it to $\sin 2x$ , and it will give me a different answer, $\color{red}{x + \cos(2x)/2 + C}$

I went to Symbolab, and it really shows the difference, whether I integrate it right away or apply the trigonometric identity first before integrating

I tried to plot the points in Desmos, two answers seems close enough that it's just one constant away (particularly $0.5$), but still mathematically not equal.

So, does this mean that this trigonometric identity changed the actual answer that I should avoid using it as much as possible. Or this is just an exception? Which one is right? Is it both of the two answers?

I already passed my answer to the teacher, but what I actually did before passing it is integrating right away. But this problem haunts me in sleep that I actually opened my pc to ask this question here. Still learning integration techniques that I got introduced to this kind of problem, that overall it's pretty fun and stressing.


答:
Depending on when and what you do the u-substitution, there are at least three ways to calculate $\int 2\sin(x)\cos(x)dx$:

$$
\int 2\sin x\cos x\ dx=\int \sin(2x)\ dx=-\frac12 \cos(2x)+C\tag{1}
$$

$$
\int 2\sin x\cos x\ dx = (\sin x)^2+C\tag{2}
$$

$$
\int 2\sin x\cos x\ dx = -(\cos x)^2+C\tag{3}
$$

These answers are all correct.

In general, if you get (correctly via different methods)
$$
\int f(x)\ dx=F(x)+C,\quad
\int f(x)\ dx=G(x)+C
$$

then there is some constant $K$ such that $G(x)=F(x)+K$.

Recall that $\int f(x)\,dx=F(x)+C$ really means $F'(x)=f(x)$ (on some interval $I$). The solution $F$ is not unique.

To see another example, both the following are correct:
$$
\int \cos(x)\,dx=\sin x+C,\quad \int \cos(x)\,dx=\sin x+1+C
$$


问:
How to calculate $\lim_{n\to\infty}\left(n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}} \right)$?
This is an admission question of Tokyo University：
For all natural number $n\geqslant2$, we always have

$$n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}}\geqslant\cfrac{i}{10}$$
in which $i$ is integer, please calculate the maximum 0f $i$.

I have worked out this problem, furthermore, how to calculate

$$\lim_{n\to\infty}\left(n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}}\right)$$
is the value a transcendental number？

答:
First note that
$$
\frac{k}{{\sqrt {k^2  - 1} }} = 1 + \frac{1}{{(k + \sqrt {k^2  - 1} )\sqrt {k^2  - 1} }}.
$$
Thus,
$$
\mathop {\lim }\limits_{n \to  + \infty } \left( {n - \sum\limits_{k = 2}^n {\frac{k}{{\sqrt {k^2  - 1} }}} } \right) = \mathop {\lim }\limits_{n \to  + \infty } \left( {1 - \sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2  - 1} )\sqrt {k^2  - 1} }}} } \right).
$$
Observe that our sequence is decreasing, so we indeed need to estimate this limit efficiently. Now
$$
\sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2  - 1} )\sqrt {k^2  - 1} }}}  \le \frac{1}{2}\sum\limits_{k = 2}^n {\frac{1}{{k^2  - 1}}}  \to \frac{1}{2}\sum\limits_{k = 2}^\infty  {\frac{1}{{k^2  - 1}}}  = \frac{3}{8}
$$
and
$$
\sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2  - 1} )\sqrt {k^2  - 1} }}}  \ge \frac{1}{2}\sum\limits_{k = 2}^n {\frac{1}{{k^2 }}}  \to \frac{1}{2}\sum\limits_{k = 2}^\infty  {\frac{1}{{k^2 }}}  = \frac{1}{2}\left( {\frac{{\pi ^2 }}{6} - 1} \right) \ge 0.322.
$$
Consequently,
$$
0.625 \le \mathop {\lim }\limits_{n \to  + \infty } \left( {n - \sum\limits_{k = 2}^n {\frac{k}{{\sqrt {k^2  - 1} }}} } \right) \le 0.678,
$$
and the answer is $i=6$.


问:
Proving that a solution to the "Cat behind 7 doors" puzzle is optimal

The "Cat behind 7 Doors" Puzzle.

There are 7 doors on a corridor, and a cat is behind one of them. We are trying to find the cat. Interesting thing is that, whenever the door we open is empty, we must close it after, and the cat must move to the door adjacent to him (1 right or 1 left). The cat can move to the door we have just closed.

I was able to find that the solution for the number of trials for $n$ number of doors (when $n$ is larger than 3) is $2(n-2)$. And the order in which we do the trials is starting with 2nd door and going one by one until the $(n-1)$th door, repeating $(n-1)$ and going back to the 2nd door. For example for $5$ doors, the trials are $2$, $3$, $4$, $4$, $3$, $2$.

I understand why this solution is correct. However, I am having a hard time proving if/why this solution is optimal, and I'm not even sure how to make a start on it. Do I use proof by induction? Do I use the formula? Any answers/help will be very much appreciated!

答:
Note: The original poster said "I understand why this solution is correct". What he wanted was a proof that the solution is optimal. I find it strange that none of the responses here give any proof of the optimality for the original 7 doors puzzle. The proof of optimality for 5 doors in this response seems quite convoluted.

I will now sketch a proof of the minimality of $2(n-2)$ moves to catch the cat.

First, we can draw out the situation in a grid such that the $i^\text{th}$ row represents the doors at the $i^\text{th}$ step and plot the doors opened by the strategy on this grid by a slash. We also color the cells red or blue in a checkerboard pattern. For example, one of the optimal solutions are:

time \ door
1
2
3
4
5
6
7

1

̸

2

̸

3

̸

4

̸

5

̸

6

̸

7

̸

8

̸

9

̸

10

̸

Now note that if there is a path from the top to the bottom that always goes from the previous cell to the cell diagonally left-and-down or right-and-down, then the strategy misses such a cat. The intuition is that these paths are all of one color, red or blue, so if the path starts on a red cell, it will never be blocked by a slash on a blue cell and vice versa.

Suppose there are less than $2(n-2)$ slashes. Then one of the colors, call this $C$, will have less than $n-2$ slashes. Consider the middle $n-2$ columns. One of these columns $x$ will have no slashes on color $C$. Note that a row cannot have more than one slash because you cannot open two doors at the same time. In every row, pick a cell of color $C$ which is in one of the columns $\left\{x-1,x,x+1\right\}$ and has no slash (this is clearly always possible). Together, these cells form a path for the cat to escape.

In the following example, the red cells have fewer than $n-2$ slashes, and the asterisks mark a path for the cat to follow.

time \ door
1
2
3
$x$=4
5
6
7

1

̸
*

2

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答:
Hint:

Another way:

For $x>0,$

$$I=\int\dfrac{\sqrt{x^2+4x}}{x^2}\ dx=\int\dfrac{\sqrt{x+4}}{x^{3/2}}\ dx$$

Integrating by parts,

$$I=\sqrt{x+4}\int\dfrac{dx}{x^{3/2}}-\int\left(\dfrac{d\sqrt{x+4}}{dx}\int\dfrac{dx}{x^{3/2}}\right)dx$$

$$=-\sqrt{x+4}\cdot\dfrac{2}{\sqrt x}+\int\dfrac{dx}{\sqrt{(x+2)^2-2^2}}$$

Now observe that $\dfrac8{\sqrt{x^2+4x}+x}=\dfrac{8(\sqrt{x^2+4x}-x)}{4x}=2\sqrt{\dfrac{x+4}x}-2$

Title: Using $\sec(x)$ for integral.


问:
what is running yield?
Suppose an investor who pays £120 on 15 July 2007 for £100 nominal of a six-year stock with half-yearly coupon payments of 13% pa and is not subject to tax. The stock is redeemable at par. Calculate the running yield?

I understood entire question but what actually is running yield?

答:
Running yield is the annual income from an investment divided by its current market value.  Unless the market price is equal to the nominal price, it will not be the same as the nominal yield or the redemption yield.

So in this case  $\dfrac{13}{120}$ or about 10.83%.

Title: what is running yield?


问:
Improper integrals involving logarithm in denominator
Discuss/comment on the convergence/divergence of the following improper integrals:

(A)
$$\int_0^{+\infty}\dfrac{e^{-x}}{x^{\alpha}(|\log x|)^{\beta}},\text{ with }\alpha,\beta>0$$
(B)
$$\int_0^{+\infty}\dfrac{xe^{-x}}{\log^2(x)}$$

I find it difficult to deal with improper integrals when logarithms are in the denominator. Could you suggest some method to tackle them?

I understand that we'll first divide the domain of integrals so that we have only one type of improper integral to deal with in that domain. Then we try to come up with some function using which we apply the limit comparison test and proceed further.
I face difficulties in finding that function.

答:
The second integral you wrote is
$$\int_0^{+\infty}\dfrac{xe^{-x}}{\log^2(x)}dx.$$
We have to study the integral in a neighbourhood of the points $0,1$ and for $x\to +\infty$.

Let $x\in\mathcal U(0)$ we have that $\dfrac{xe^{-x}}{\log^2(x)}\sim\dfrac{x}{\log^2(x)}\underset{x\to 0^+}{\longrightarrow}0\implies$ the integrand is defined in a neighbourhood of zero and the integral is clearly convergent.

Let now $x\in \mathcal U(1)$, then the integrand is asymptotic to $\dfrac{c}{\log^2(x)}$ and since a function of the form $\dfrac{1}{|x|^p|\log x|^q}$ has improper integrand convergent iff $q>1$ we can state that the integral is divergent in a neighbourhood of $x=1$.

Finally if $x\in \mathcal U(+\infty)$, then $\dfrac{xe^{-x}}{\log^2(x)}\sim \tilde c/e^x\le\dfrac{1}{x^2}$, whose integral is convergent.

So the integral will be convergent only in the set of the form $[0,1-\delta]\cup[1+\delta,+\infty)$.

Let $f(x)=\dfrac{1}{|x|^p|\log x|^q}$ we have the following properties
$$x\to 0\implies \begin{cases}\text{convergent integral}&&\text{for }p>1,\forall q\in\mathbb R\text{ or }p=1,q>1\\\text{divergent integral}&& \text{otherwise} \end{cases}$$

$$x\to 1\implies \begin{cases}\text{convergent integral}&&\text{for }p<1,\forall q\in\mathbb R\text{ or }p=1,q>1\\\text{divergent integral}&& \text{otherwise} \end{cases}$$

$$x\to \infty\implies \begin{cases}\text{convergent integral}&&\text{for }\forall p\in\mathbb R,q<1\\\text{divergent integral}&& \text{for }\forall p\in \mathbb R,q>1 \end{cases}$$


问:
Claim on Wikipedia in connection with integrability and Risch's algorithm. Any references?
The Wikipedia article on Risch's algorithm makes this interesting statement:

the following algebraic function has an elementary antiderivative: $$f(x) = \frac{x}{\sqrt{x^4 + 10 x^2 - 96 x - 71}}$$ [...] But if the constant term 71 is changed to 72, it is not possible to represent the antiderivative in terms of elementary functions.

For the first part of the statement the antiderivative is given, together with a (quite cool) reference to a post by the late Bronstein. But for the second part of the statement no reference is given (as often happens in Wikipedia; so the second part is just a claim). I tried to check the book by Geddes & al but didn't manage to find that statement there.

Does anyone know any reference to a proof that if 71 is changed to 72 no antiderivative exists that can be expressed in terms of elementary functions?

答:
Comment:May be this idea helps:

$x^4+10x^2-96x-71=(x^2+5)^2-96(x+1)=$

Let:

$x^2+5=u\Rightarrow 2xdx=du\Rightarrow dx\frac {du}{2x}=\frac{du}{2\sqrt{u-5}}$

Putting this in integrand you get:

$$ \frac{dx}{x^4+10x^2-96x-71}=\frac{du}{[u-96(\sqrt{u-5}+1)](2\sqrt{u-5})}$$

Which can be transformed to two fractions.This is not possible if you replace 71 by 72.


问:
Is a perfect complexes locally quasi-isomorphic to a bounded complex of free modules of finite type?
I'm a little confused by the definition of perfect complex on a scheme $X$, namely a complex $E^{\cdot}$ of $\mathcal{O}_X$-modules such that for every $x\in X$ there exists $U\ni x$ open such that $E^{\cdot}|_U$ is quasi-isomorphic to a bounded complex of locally free $\mathcal{O}_X$-modules of finite type.

My question could be silly, but is that true that one could just define a perfect complex as a complex locally quasi-isomorphic to a bounded complex of free $\mathcal{O}_X$-modules of finite type?

Given a perfect $E^{\cdot}$, I would argue that one can consider $U\ni x$ such that $E^{\cdot}|_U$ is quasi-isomorphic to $F^{\cdot}$, which is a bounded complex of locally free sheaves over $U$, hence find for every $k$ an open $U_k\ni x$ such that $F^k|_{U_k}$ is free, and taking finite intersection over those $k$ such that $F^k\neq 0$, I find another open $V=\bigcap U_k$ neighborhood of $x$ such that $F^{\cdot}|_V$ is in fact a complex of free $\mathcal{O}_X$-modules

Is this reasoning correct or am I wrong somewere? It sounds wierd to me because the definition is usually given talking abound locally free modules. If I'm wrong, could you suggest a counterexample? Thanks in advance

答: Yes, this is correct.  The use of "locally free" in the definition may stem from the fact that for modules over a ring, you can define a perfect complex simply as one that is quasi-isomorphic to a bounded complex of locally free modules of finite type.  That is, in the affine case, there's no need to restrict to an open set $U$ if you say "locally free" instead of "free".


问:
Find an equation of the tangent line to a graph
Consider the graph defined by

$$y = \dfrac{x - 3}{x - 5}$$

(a) Use the definition of the derivative to find the slope of the tangent line to the graph at the point $(6, 3)$.

(b) Find an equation of the tangent line to the graph at the point $(6, 3)$.

答:
The definition of the derivative , which gives the slope is $$y_0'=\lim\limits_{x\to x_0}\dfrac{y-y_0}{x-x_0}$$
The equation of the tangent line is $$y-y_0=y'_0(x-x_0)$$
Can you now continue?</p>


问:
statement logic, consequence
I need to formalize this sentence and decide whether the logical consequence applies

Tom drove either by bus (b) or tram (t). If he drove by tram (t) or by car (a) then he arrived late (p) and did not hear the lecture (v). He did not arrive late (non p). It follows that he rode the tram (t) or missed the lecture (v)

I create this:

$T \vee B$, $(T \vee A)\Rightarrow(P \wedge V)$, $\neg P$, ? $T \vee V$

with the help of a semantic tree, I decided that is not consequence. Is that right?

答:
Your formalization of the problem sounds correct to me, although such formalizations depend heavily on your interpretation of the text.

Now, consider the case in which $B$ is true and $A$, $T$, $V$ and $P$ are false:

$T\vee B$ is true, since $B$ is true;

$(T\vee A)\rightarrow(P\wedge V)$ is true, since $T\vee A$ is false (since $T$ and $A$ are false);

$\neg P$ is true, since $P$ is false;

$T\vee V$ is false, since both $T$ and $V$ are false.

This shows one can have all premises true while the consequence is false, meaning that the consequence does NOT follow logically from the premises.

Title: statement logic, consequence


问:
Prove $t^{1\over t-1}$ is rational only when $t=2$ in the context of solving $x^y=y^x$ where $x\neq y$
Prove $t^{1\over t-1}$ is rational only when $t=2$ where $t$ is any real number.

This problem arises from me watching the blackpenredpen video solving $x^y=y^x$ where $x\neq y$. https://www.youtube.com/watch?v=PI1NeGtJo7s, in which he gives the solution $x=t^{1\over t-1}$.

First I considered the integer solutions where $x\neq y$ other than $2^4=4^2$ but realized it is trivial because $1<t^{1\over t-1}<2$ for any $t\geq 2$ and $-1>t^{1\over t-1}>-2$ for any $t\leq-2$.

The next question is then, are there any rational solutions other than the $2,4$ pair. It does not seem trivial to me as inequalities do not help.

My guess is that $t^{1\over t-1}$ cannot be rational other than when $t=2$ but I wan't able to prove.

答:
We have,

$$t-1=a^{t-1}-1, ~ \text{where} ~ a\in\mathbb {Q^{+}}$$

$$\implies a^x=x+1, x=t-1$$

Substitute $a^x=t \implies x\ln a=\ln t $

$$\implies  t=\dfrac {\ln t}{\ln a}+1$$

Substitute $\ln t=u$

$$\implies e^u=\dfrac{u+\ln a}{\ln a}$$

$$\implies e^{u+\ln a}=\dfrac {a}{\ln a}(u+\ln a)$$

$$\implies (u+\ln a)^{-1} e^{u+\ln a}=\dfrac {a}{\ln a}$$

$$\implies  \left((u+\ln a)^{-1} e^{u+\ln a}\right)^{-1}=\dfrac {\ln a}{a}$$

Substitue $z=u+\ln a$

$$\implies -ze^{-z}=-\dfrac {\ln a}{a}$$

$$\implies W(-ze^{-z})=W\left(-\dfrac {\ln a}{ a}\right)$$

$$\implies z=-W\left(-\dfrac {\ln a}{a}\right)$$

If we take the substitutes backwards

$$t=\dfrac{e^z}{a}$$

We have:

$$\implies t=\dfrac{e^{-W\left(-\frac {\ln a}{a}\right)}}{a}$$

So, $a \in\mathbb {Q^{+}}$, where $$\color {gold}{\boxed {\color{blue}{t=\dfrac{e^{-W\left(-\frac {\ln a}{a}\right)}}{a}}}}$$

Thus, your conjecture is not true.

Here $W(x)$ is a Lambert function


问:
Computation of laurent series and convergence test
I have a few questions about Taylor and Laurent series expansions.

Is it possible to compute a laurent series of a real function with poles?

If $f(x)=\frac{1}{(x^{2}-1)^{2}}$ is a real function with a singularity at $x=\pm 1$.

How do I compute its Taylor series expansion and also determine whether the series converges?

答:
A Taylor series for $f$ is

$$f(x) = \sum_{k=0}^\infty a_k(x-x_0)^k,$$ so we need to specify $x_0$.  Here is one approach to compute the Taylor series centered at $x_0=0:$

Consider the following identity involving a geometric series:
$$ \sum_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}, \quad (r \ne 0).$$
When $|r|<1,$ the limit of this series as $n\to\infty$ converges.

$$\frac{1}{1-x} = 1+x+x^2 +x^3 + \cdots, \quad (|x|<1).$$

$$\frac{1}{(1-x^2)^2} = 1+x^2+x^4 +x^6 + \cdots, \quad (|x|<1).$$

$$\begin{aligned} \frac{1}{(1-x^2)^2} &= [1+x^2+x^4 +x^6 + \cdots]\cdot[1+x^2+x^4 +x^6 + \cdots], \quad (|x|<1)\\
&= 1 + 2x^2 + 3x^4 + 4x^6 + 5x^8+ \cdots , \quad (|x|<1)
\end{aligned}$$

This particular series converges because it is the product of convergent series.  Each of these converge by comparison with the geometric series.  There are other tests for convergence of a series including the ratio test and the root test.  See here (and the associated links) for more info:

https://mathworld.wolfram.com/RadiusofConvergence.html

A Laurent series has also terms with negative powers of $x-x_0$.  There are a number of $\textit{obvious}$ choices for $x_0$, for example $x_0\in\{0,1,-1\}.$

Let's write down the Laurent series for $x_0=0$ and $|x|>1.$

$$\begin{aligned}
\frac{1}{(1-x^2)^2}&=\frac{1}{x^4} \cdot \frac{1}{(1-\frac{1}{x^2})^2}\\
&=\frac{1}{x^4}\left[1+\frac{2}{x^2}+\frac{3}{x^4}+\frac{4}{x^6}+\frac{5}{x^8}+\cdots\right]\\
&=\frac{1}{x^4}+\frac{2}{x^6}+\frac{3}{x^8}+\frac{4}{x^{10}}+\frac{5}{x^{12}}+\cdots, \quad (|x|>1).
\end{aligned} $$


问:
Find the probability that the product of 2 dice throws is a multiple of 3
I am a retired person who studies maths as a hobby.
This is the problem I am struggling with:

2 unbiased dice are thrown. Find the probability that the product of the scores is a) odd, b) a multiple of 3, c) a multiple of 12.

I know the total number of possibilities is $6 \times 6 = 36$

As regards a) an odd number x an odd number = an odd number
so the probability is $\frac{1}{2} \times\frac{1}{2} = \frac{1}{4}$

Now for b) I would have thought if one dice rolls a 3 or a 6 then any  score on the other dice would give a product which is a multiple of 3. Since there is a $\frac{1}{3}$ chance of this happening on any one dice I would have thought the probability is $\frac{1}{3}$.

But the book says $\frac{5}{9}$

As regards c) I was able to work out this as $\frac{7}{36}$, but only by actually counting the possible outcomes. I don't know the method.

答:
For second, the favorable outcomes are as follows -

If you take $3$ on first dice, $6$ favorable outcomes

If you take $6$ on first dice, again $6$ favorable outcomes

If you take $3$ on second dice, only $4$ additional outcomes as two were counted with first being $3$ or $6$.

Similarly for $6$ on second dice.

So total favorable outcomes $ = 6 + 6 + 4 + 4 = 20$.

But there is an easier way to do this. As we know, $3$ or $6$ on any dice would give product as multiple of $3$. So for product not to be multiple of $3$, we have to get from $(1, 2, 4, 5)$ on both dice. That probability is $\frac{4}{6} \times \frac{4}{6} = \frac{4}{9}$.

So probability that the product is multiple of $3 = 1 - \frac{4}{9} = \frac{5}{9}$.

For third as you rightly said, we count the favorable outcomes as below.

$12 = 3 \times 4$ or $2 \times 6$ so $4$ favorable outcomes - $(3,4), (4,3),(2,6),(6,2)$

$24 = 4 \times 6$ so $2$ favorable outcomes $(4,6), (6,4)$.

$36 = 6 \times 6$ so $1$ favorable outcome $(6,6)$.

That leads to $\frac{7}{36}$.


问:
Convergence of $\sum_{k = 1}^{\infty} \cos (\log k)$ and $\sum_{k = 1}^{\infty} \sin (\log k)$
I was trying to determine if

$$\sum_{k = 1}^{\infty} \cos (\log k)$$

(with $k = 1, 2, \ldots$) converges. $\log$ is the natural logarithm.

I tried the ratio test,

$$\lim_{k \to \infty} \frac{\cos \left[ \log ( k + 1 ) \right]}{\cos \left[ \log ( k ) \right]} = 1$$

but it is not useful to determine the convergence.

According to Wolfram Alpha, it diverges. How to prove it?

And what about $\displaystyle \sum_{k = 1}^{\infty} \sin (\log k)$?

What I can empirically observe is that, regardless of how big $\log k$ can become, its cosine will always be a number $x$ such that $-1 \leq x \leq 1$. But I can't immediately see that this leads to an infinite sum.

Trivially, the term test is:

$$\lim_{k \to \infty} \cos (\log k)$$

and this limit is not defined, due to the cosine. Maybe this is sufficient to state that the series diverges.

答: For $k$ sufficiently big, $\log(k+1)-\log k<\frac\pi2$. On the other hand, $\log k$ is unbounded. Therefore, $\log k$ "walks" to $\infty$ in steps so small that it will hit every interval $[2m\pi-\frac\pi4,2m\pi+\frac\pi4]$ with $m\gg0$. That is, we will have $\cos \log k\ge \frac{\sqrt 2}2$ infinitely often.


答:
I do not think you need "a method". Simply drawing a table like the following it is very easy to see the favourable outcomes

total 20 favourable on 36 possible.

Title: Find the probability that the product of 2 dice throws is a multiple of 3

Body:
I am a retired person who studies maths as a hobby.
This is the problem I am struggling with:

2 unbiased dice are thrown. Find the probability that the product of the scores is a) odd, b) a multiple of 3, c) a multiple of 12.

I know the total number of possibilities is $6 \times 6 = 36$

As regards a) an odd number x an odd number = an odd number
so the probability is $\frac{1}{2} \times\frac{1}{2} = \frac{1}{4}$

Now for b) I would have thought if one dice rolls a 3 or a 6 then any  score on the other dice would give a product which is a multiple of 3. Since there is a $\frac{1}{3}$ chance of this happening on any one dice I would have thought the probability is $\frac{1}{3}$.

But the book says $\frac{5}{9}$

As regards c) I was able to work out this as $\frac{7}{36}$, but only by actually counting the possible outcomes. I don't know the method.


答:
A common mistake.

$$\int(2\sin x\cos x-\sin 2x)\,dx=C,$$ so

$$\int2\sin x\cos x\,dx=\int\sin 2x\,dx+C$$ for some $C$.


Title: Find an equation of the tangent line to a graph

Body:
Consider the graph defined by

$$y = \dfrac{x - 3}{x - 5}$$

(a) Use the definition of the derivative to find the slope of the tangent line to the graph at the point $(6, 3)$.

(b) Find an equation of the tangent line to the graph at the point $(6, 3)$.

Body:
$$\dfrac {dy}{dx}=\dfrac{(x-5)\dfrac {d}{dx}(x-3)-(x-3)\dfrac {d}{dx}(x-5)}{(x-5)^2}$$

$$\dfrac {dy}{dx}=\dfrac{(x-5)-(x-3)}{(x-5)^2}$$
$$\dfrac {dy}{dx}_{(6,3)}=\dfrac{-2}{(x-5)^2}$$
$$\dfrac {dy}{dx}_{(6,3)}=\dfrac{-2}{(6-5)^2}$$
$$\dfrac {dy}{dx}_{(6,3)}=-2$$

slope of tangent at $(6,3)$=$-2$

b)

equation of tangent :  $y-y_0=m(x-x_0)$

(where $(x_0,y_0)$ is a point on line and m is slope)

$$y-3=-2(x-6)$$
$$2x+y=15$$


答:
Hint: Whenever you have positive and negative exponents, factor out the negative exponent. I'm also factoring out the fractions so we could see a perfect square a bit more easily.

$\sqrt{\frac{1}{2}+x^{\frac{2}{3}}+\frac{x^\frac{-2}{3}}{16}}=\sqrt{\frac{x^{\frac{-2}{3}}}{16}}\sqrt{8x^{\frac{2}{3}}+16x^{\frac{4}{3}}+1}$

Taking the square root of $\frac{x^{\frac{-2}{3}}}{16}$ is easy. And for the other square root, try factoring $16y^2+8y+1$, where we have $y=x^{\frac{2}{3}}$ here.

Body:
The length of the curve $ =\dfrac{3}{4}x^{\frac{4}{3}}-\dfrac{3}{8}x^{\frac{2}{3}}+7$ from $x=1$ to $x=8$ equals.

$(A)=\ \dfrac{99}{8} \ \ \ \ \ \ \ \ (B)=\dfrac{117}{8} \ \ \ \ \ \ \ \ (C)=\dfrac{99}{4} \ \ \ \ \ \ \ \ (D)=\dfrac{117}{4}$

$\int_{1}^{8}\sqrt{1+(y')^2}dx$

$y'=x^{\frac{1}{3}}-\frac{1}{4}x^{\frac{-1}{3}}$

$(y')^2=(x^{\frac{1}{3}}-\frac{1}{4}x^{\frac{-1}{3}})^2=\sqrt{\frac{1}{2}+x^{\frac{2}{3}}+\frac{x^{\frac{-2}{3}}}{16}}$

$\int_{1}^{8}\sqrt{\frac{1}{2}+x^{\frac{2}{3}}+\frac{x^{\frac{-2}{3}}}{16}}dx$

Which is hard to compute for the exam I am preparing for. This question came for $2$ marks(short type questions) and there is something I missing. Help me here finding the solution.


答:
I=∫2cosx sinx dx=2∫cosx sinx dx   .....1)

integration by parts

I=2{cosx∫sinx dx-∫ [d/dx(cosx)]* [ ∫sinx dx ] }dx

I=2{cosx (-cosx)- ∫ [-sinx][-cosx)]dx }

I=-2cos²x - 2∫[sinx][cosx]dx

from.....1)

I=-2cos²x -I

2I=-2cos²x

I=-cos²x  + constant

I=-(1+cos2x)/2 + constant

I=-1/2-cos2x/2 + constant

I=-cos2x/2 + constant-1/2........(-1/2 is also constant)
I=-cos2x/2 +C

∫sin2x dx= -cos2x/2 +C


答:
$$u=x^{1/3}\Rightarrow du=\frac13x^{-2/3}dx\therefore dx=3x^{2/3}du=3u^2du$$
and so:
$$\sqrt{\frac12+x^{2/3}+\frac1{16}x^{-2/3}}dx\Rightarrow3\sqrt{\frac12+u^2+\frac1{16}u^{-2}}u^2du$$
then actually you could make the substitution:
$$v=u^2\Rightarrow du=\frac{dv}{2u}\therefore u^2du=\frac{udv}{2}=\sqrt{v}\frac{dv}{2}$$
now sub in:
$$\frac32\sqrt{\frac12+v+\frac1{16}v^{-1}}\sqrt{v}dv=\frac32\sqrt{v^2+\frac v2+\frac1{16}}dv$$
now take out the factor of $16$ so:
$$\frac38\sqrt{16v^2+8v+1}dv$$

If we try and factor this we get:
$$16v^2+8v+1=(4v+1)^2$$
and now your integral is simply:
$$\frac38\int_1^4(4v+1)dv$$

Title: The length of the curve $ =\dfrac{3}{4}x^{\frac{4}{3}}-\dfrac{3}{8}x^{\frac{2}{3}}+7$ from $x=1$ to $x=8$ equals.


问:
Are the following real numbers constructible?
1) $\sqrt[4]{5+\sqrt2}$

2)$\sqrt[6]{2}$

3) $3/(4+\sqrt13)$

4) $3+\sqrt[5]{8}$

From what I know, a number is constructible if it can be converted in a finite number of steps using only the operations addition, subtraction, multiplication, division, and square roots.

This in mind, I would think #1 is constructible number because it is calculated using addition and square roots, with the 4th being a square root of a square root.

I would say #2 is not constructible as it is not possible to take a third root.

I would say #3 is constructible because it makes use of the addition, division, and square root operations.

Finally, #4 I would say is not constructible because it is not possible to generate the 5th root.

Can anyone weigh in on whether these conjectures are correct/incorrect, or maybe comment on the rationale and whether this method (if correct) could still be used on harder examples?

Thank you in advance!

答:
While all these answers are correct, just having a higher root does not guarantee a failure of constructibility.  The number $\sqrt[3]{2+\sqrt5}\approx 1.618$ (big hint there) has been known to be constructible since Euclid, even if he would not have recognized the number in that form.

Suppose we are given a number $\sqrt[n]{a+\sqrt{b}}$, where $n$ is an odd natural number, $a$ is any integer and $b$ is any positive integer.  We can check such a number for cobstructibility with a two-step process.

First, if $\sqrt[n]{a+\sqrt{b}}$ is to be constructible then so is the conjugate $\sqrt[n]{a-\sqrt{b}}$.  Thus so is their product $\sqrt[n]{a^2-b}$ and thus, $a^2-b$ must be an $n$th power.  If this passes, define $\sqrt[n]{a^2-b}=R$ and move on to step 2.

In step 2, propose that

$(x+\sqrt{x^2-R})^n=a+\sqrt{b}.$

We also have the conjugate relation:

$(x-\sqrt{x^2-R})^n=a-\sqrt{b}.$

Clearly their product matches up with our defining equation $R^n=a^2-b$.  If you now expand the left sides by the binomial theorem and average the two results, the radicals cancel and you are left with

$x^n+(\frac{n(n-1)}{2})(x2-R)x^{n-2}+(\frac{n(n-1)(n-2)(n-3)}{24})(x^2-R)^2x^{n-4}+...+n(x^2-R)^{(n-1)/2}x=a$

If this has a rational root for $x$, then that rational root may be substituted for $x$ and we then have $x+\sqrt{x^2-R}$ as a constructibke $n$th root of $a+\sqrt{b}$.  Failure at either this step or step 1, above, means the numbervis not constructible.

Plug in $\sqrt[3]{2+\sqrt5}$ and see what happens.  First off $2^2-5=-1$ is a perfect cube, and accordingly we will set $R=-1$ for step 2.

For step 2, with a cube root the binomial expansion after canceling radicals will generate two terns, to wit:

$x^3+3(x^2+1)=2$

$4x^3+3x-2=0$

We seek a rational root, using the Rational Root Theorem to identify candidates.  In this case $x=1/2$ succeeds, thus our cube root is constructible and  we identify what we might have suspected from that decimal approximation I quoted:

$\sqrt[3]{2+\sqrt5}=(1/2)+\sqrt{(1/2)^2+1}=(1+\sqrt5)/2.$


问:
Sylow theorem in fraighle's text
I've been struggling with the first Sylow theorem.
Then γ−1[K] is a subgroup of N[H] and hence of G. This subgroup contains H and is of order pi+1 How come it has order pi+1, Couldn't understand that part. I've came across this in the proof of the First Sylow Theorem

enter image description here

答:
You have $\gamma : N[H] \rightarrow N[H]/H$.

Now $K$ is a subgroup of order $p$ of $N[H]/H$, and $H$ has order $p^i$, so $\gamma^{-1}(K)$ has order $p \times p^i = p^{i+1}$.

Title: Sylow theorem in fraighle's text


问:
Faithful state and associated measure.
Let $A= C^*(1,a)$ where $a \geq 0$, that is $A$ is a unital $C^*$-algebra generated by a positive element $a$ and the identity. Let $f: A \to \mathbb{C}$ be a state on $A$. We have a canonical identification $X: A \cong C(\sigma(a))$ where $\sigma(a) \subseteq \Bbb{R}$ is the spectrum of $a$. By Riesz' theorem, there is a Radon Borel measure $\mu$ on $\sigma(a)$ such that
$$f(a) = \int_{\sigma(a)} X(a) d \mu.$$
Is it true that
$$\|a \| = \sup (\operatorname{supp} \mu)$$?

Attempt:

Clearly, $\operatorname{supp}\mu \subseteq \sigma(a)$ so $\sup \operatorname{supp} \mu \leq \sup \sigma(a) = r(a)= \|a\|$. However, I can't justify why the other inequality is true.

答:
This is not true.  Suppose $x\in\sigma(a)$, with $x\neq\sup\sigma(a)$, and let $f$ be the state such that $f\circ X^{-1}\in C(\sigma(a))^*$ is evaluation at $x$, that is, $f(b)=[X(b)](x)$ for $b\in A$.  The corresponding measure is $\delta_x$, the Dirac mass at $x$, and $\operatorname{supp}\delta_x=\{x\}$, so
$$\sup(\operatorname{supp}\delta_x)=x<\sup\sigma(a)=\|a\|.$$

Title: Faithful state and associated measure.


问:
"Stars and Bars" summation identity
Can anybody help me solve this problem?
I don't even know where to start or how to interpret this.

Proof for all $1 \leq k \leq n$ that the folowing identity holds:

\begin{equation*}
\begin{pmatrix}
n+k-1 \\
n-1
\end{pmatrix} =
\sum_{i=1}^k \begin{pmatrix}k-1 \\ i-1 \end{pmatrix} \begin{pmatrix} n \\ i \end{pmatrix}
\end{equation*}

This problem is in fact a homework assignment, so it would be kind if the solution isn't spoiled right away. There is one hint given in the assignment that says that for k identical objects and n different colors ($k \geq n$) the amount of different color combinations (you may repeat colors) when using every color at least once, is equal to: $\begin{pmatrix} k-1 \\ n-1 \end{pmatrix}$

I hope that anybody can assist me.

Thanks in advance!

答: Rewrite $\binom{n}{i}$ as $\binom{n}{n-i}$ and use a combinatorial proof of Vandermonde's identity.

Title: "Stars and Bars" summation identity


Body:
It is not hard to compute:
$$\sqrt{1+(y')^2} = \sqrt{1+x^{\frac{2}{3}}+\frac{x^{\frac{-2}{3}}}{16}-\frac{1}{2}} = \sqrt{x^{\frac{2}{3}}+\frac{x^{\frac{-2}{3}}}{16}+\frac{1}{2}} = \sqrt{(x^{\frac{1}{3}}+\frac{1}{4}x^{\frac{-1}{3}})^2} = x^{\frac{1}{3}}+\frac{1}{4}x^{\frac{-1}{3}}$$ Integral is easy to calculate now (just power rule).


问:
Alcuin's problem of inheritance.
A certain father died and left as an inheritance to his three sons $30$ glass flasks, of which $10$ were full of oil, another $10$ were half full, while another $10$ were empty. Divide the oil and flasks so that an equal share of the commodities should equally come down to the three sons, both of oil and glass

One could find the answer just by finding coefficient of $x^{10}$ in
$$\frac{1}{(1-x^2)(1-x^3)(1-x^4)}.$$

So could anyone give me a reason why this generating function gives the solution to this problem.

答:
EDIT: I've thought of an argument which I think conveys the idea better.

That generating function counts the partitions of $n$ into parts $2$, $3$, and $4$. I'll give a bijection between such partitions of $n$ and divisions of the inheritance. Note that among the sons, we can order them by who gets the most full flasks of oil. Call the sons $A$, $B$, and $C$ where $A$ gets the fewest half flasks and $C$ gets the most (some of them may get the same number of half flasks, but that's fine).

Say you have some way of fairly dividing the inheritance, and let $A_1, B_1$, and $C_1$ be the number of half flasks given to $A$, $B$ and $C$ respectively. Since $A$ gets the fewest half flasks, every time we give $A$ a half flask, we must give one to $B$ and $C$ as well. Thus, we can distribute some half flasks in groups of $3$, giving each son one half flask, a total of $A_1$ times. Now that we've given $A$ all of his half flasks, we turn to $B$. Observe that $A_1$, $B_1$, and $C_1$ must all have the same parity (that is, they are either all even or all odd), so $B_1-A_1$ is even. Hence, we can give $B$ the rest of his half flasks in groups of $2$, and since $C$ has at least as many half flasks as $B$, we must give $C$ the same number of half flasks. This allows us to distribute more half flasks in groups of $4$, giving $B$ and $C$ each two half flasks, a total of $(B_1-A_1)/2$ times. Finally, we give $C$ the rest of his half flasks, and since $C_1$ and $B_1$ have the same parity, we can give $C$ his in groups of $2$, a total of $(C_1-B_1)/2$ times. This is a partition of $n$ into parts $2$, $3$, and $4$.

Now we show that this correspondence is bijective by giving its inverse. Say we have a partition of $n$ into parts $2, 3,$ and $4$. Each $2$, $3$, and $4$ will tell us how to divide that many full flasks, half flasks, and empty flasks among the sons. For each $2$, give $A$ and $B$ one full flask and one empty flask each, and give $C$ two half flasks. For each $3$, give each son one full flask, one half flask, and one empty flask. For each $4$, give $A$ two full flasks and two empty flasks, and give $B$ and $C$ one full flask, two half flasks, and an empty flask. In each case, we divide the flasks and oil evenly, so this division will satisfy the desired conditions. Furthermore, this is the inverse of our original correspondence, since it distributes the half flasks using the same groups of $2$, $3$, and $4$. Hence, the number of fair divisions of $3n$ flasks is the number of partitions of $n$ into parts $2$, $3$, and $4$, so we have the desired generating function.

Title: Alcuin's problem of inheritance.


问:
Tetration and its inverse to various exponents
I've recently seen in my studies tetration, or the next operation in the addition, multiplication, exponentation... series. I've also heard much discussion about how to extend this operation to "heights" (that is, the number of times to which the exponentation is repeated) other than the natural numbers.
Now given that I've had not much higher math experience (pretty not much farther than real analysis) and I might very well be oversimplifying things, but my question is: why can't we define fractional 'heights' like we do with exponents?

If $x$ is any base, then $\displaystyle x^{1\over 2}$ is ${\rm sqrt}(x)$. Similarly, why can't $\displaystyle {}^{1\over 2}x$ be its inverse, ${\rm ssrt}(x)$ (I will use  ${\rm ssrt}(x)$ as its inverse and  ${\rm scrt}(x)$ for the inverse of tetration with height $3$). It also works for expressions like $5\over 3$: ${}^{5\over 3}x = {}^{5}{\rm scrt}(x)$. As for irrational numbers, well I guess that's another story, just like it is for exponents.

答:
If you write an operator-notation for your function $f(x)$ say the operator $E$ for the exponential-function, such that $$ \exp(x) = E \circ x  \qquad \qquad \text{and} \\ \exp(\exp(x)) = E \circ E \circ x \qquad \qquad \text{ and } \ldots $$ then tetration of a certain height $h$ must be written as $$ \exp^{\circ h}(x) = E^h \circ x $$  and the question is, whether or how we could invent/understand an operator and even a power of an operator.

1.

Now one idea to implement an operator for analytic functions (which have a power series with nonzero range of convergence) there is the concept of "Carleman"-matrices (see wikipedia) which is construed for the manipulation of the coefficients of the (formal) power series.

With that you need thus the concept of powers and even of fractional powers of matrices.

For instance for rational/linear functions $f(x)$ the Carleman-matrix is nearly trivial: a 2x2-matrix, of which you can then define a fractional power using diagonalization or logarithmization very simply. And this can even be understood and done by any amateur.

However, the inverse function, say to find the height of a tetrations needs to find the exponent for the Carleman-matrix which would have iterated $x$ to $x_h$ (the h'th iteration of the function). Here the logarithm of a matrix becomes involved...

For general functions the matrices become of infinite size and it is non-trivial to define meaningfully fractional powers of such matrices. Accepted solutions are so far when the infinite Carleman-matrix is lower triangular (which means the powerseries of the function $f(x)$ has no constant term and has thus a fixpoint at zero).

For the exponentiating this is exceptionally difficult because also the inverse (the logarithm) is multivalued and the exponential-function has no real fixpoint.

2.

Another perspective is that of the "Schröder-function" which can intuitively be seen as an analogon to the log/exp-function for fractional iteration of the multiplication. If we ask for the fractional height of iterated multiplication with a certain base $b$, say $$f: x_1 = b \cdot x_0 \qquad \qquad \text{ and } \\ f°^2 : x_2 = b \cdot b \cdot x_0   \qquad \qquad \text{ and } \\ f°^h : x_h = \underset{ h \text{-times}}{\underbrace{b \cdot b \cdot b \cdot \ldots b \cdot} } x_0$$ then $h$ is the "height" of iterated multiplication. To find the iterated multiplication with a fractional height
$ x_h = f°^h(x)  = b^h \cdot x_0$ we use the pair of log/exp-function in the following way. First there is a constant $u_b$ defined with the "base" $b$ (in the context of multiplication ) such that $u_b=\log(b)$. Then we write:
$$ x_h = \log°^{-1}( u_b \cdot h + \log(x_0))  $$

$  \qquad \qquad \qquad $ (where of course $\log°^{-1}(x)$ means $\exp(x)$)

For the question of the inverse: determine, which height $h$ is required to iterate from $x_0$ to $x_h$ by multiplication with base $b$ ? we  employ the $\log$ for both $x_0,x_h$ and thir height-difference is then linear in h:
 $$ \operatorname{hgh} : h = {\log(x_h) - \log(x_0) \over u_b }$$

Similarly there exists the "Schröder-function" which is used in a roughly analoguous way. For a base $b$ in connection with the function-iteration there is some constant $u_b$ defined, which after transformation of the initial value $x_0$ the h'th power of $u$ leads to the $h$'th iterate. We have for the h'th iteration
$$ x_h = \operatorname{Schröder}°^{-1} ( u_b^h \cdot \operatorname{Schröder}(x_0))  $$

(we might say, the $\log$ is the Schröder-function for the iteration of multiplication, while the Schröder-function for the iterated exponentiation has not yet a name).

You can read more on Schröder-functions in the wikipedia; possibly this concept is even nearer to the focus of your question than that of the Carleman-matrices although I found the Carleman-matrices easier to understand (even I arrived at this incidentally by my own fiddling with powers of matrices...)


问:
closed set of closed subspace topology
Let $\left(X,\tau\right)$ be a topological space and $M\subseteq X$ be closed and equip with the subspace topology $\tau_M$. I am trying to show that if $A\subseteq M$ is closed in $\tau_M$ then $A$ is closed in $\tau$. However, I've come to a roadblock.

Here is my attempt: $A$ is closed in $\tau_M \iff A^c$ is open in $\tau_M \iff \exists U\in\tau: A^c = M\cap U.$ So, carrying out De Morgan gives, $A = M^c\cup U^c$. Clearly, $M^c\in\tau$ however, $U^c\notin\tau$. So, how do am I suppose to conclude that $A$ is open in $\tau$ hence $A^c$ is closed in $\tau$?

答: If $A$ is closed in $M$, then $M\setminus A$ is open in $M$, and therefore $M\setminus A=M\cap O$, for some $O\in\tau$. So, $A=M\cap O^\complement$. But, since both $M$ and $O^\complement$ are closed subsets of $(X,\tau)$, $M\cap O^\complement$ is also a closed subset of $(X,\tau)$; in other words, $A$ is a closed subset of $(X,\tau)$.

Title: closed set of closed subspace topology


问:
Punctual convergence of $f_n(x+1/n)$
Let $(f_n)$ be a sequence of functions (with real values) that converges uniformly to a continuous function $f$. I have to proof that $f_n(x+1/n)\to f(x)$ por each $x\in\mathbb{R}$.

So I know that I have to prove punctual convergence but I have trouble understanding the problem since it is evaluated in $(x+1/n)$. Any hints would be appreciated.

答:
Hint:$$\left|f_n\left(x+\frac{1}{n}\right)-f(x)\right|\leq \left|f_n\left(x+\frac{1}{n}\right)-f\left(x+\frac{1}{n}\right)\right|+\left|f\left(x+\frac{1}{n}\right)-f(x)\right|$$
Now try to make the two terms on the right side small enough by taking large enough $n$.

Title: Punctual convergence of $f_n(x+1/n)$


问:
Proving strong continuity of semigroup in $L^2(\mu)$ ($\mu$ invariant measure) from non-cadlag Markov process
I am trying to derive the definition of a Markov semigroup from Section 1.1 Bakry, Gentil, Ledoux Analysis and Geometry of Markov Diffusion Operators from a Markov process. I am unsure about my proof of strong continuity of the semigroup on $L^2(\mu)$ for an invariant measure $\mu$. Here is my work:

Suppose we are given a Markov process $X_t, t \geq 0$ on a Polish space $\mathcal X$. The semigroup is defined as

\begin{equation}
        \mathbf{P}_{t}f(x) = \mathbb E [f(X_t)|X_0=x].
\end{equation}
on bounded measurable functions $\mathcal M_b (\mathcal X)$.

The semigroup property, mass conservation $\mathbf{P}_{t}  1 = 1$ and positivity preservation  $f \geq 0 \Rightarrow \mathbf{P}_{t} f \geq 0$ are straightforward to show from the definition.

Also, these properties allow to show a Jensen's inequality for the semigroup, for convex functions $\Phi$
\begin{equation}
    \mathbf{P}_{t}(\Phi(f)) \geq \Phi\left(\mathbf{P}_{t} f\right)
\end{equation}

A $\sigma$-finite measure $\mu$ is invariant under the semigroup if
\begin{equation*}
  \int_\mathcal X \mathbf{P}_{t}f \: d \mu = \int_\mathcal X f \: d \mu, \quad \forall f \in \mathcal M_b (\mathcal X).
\end{equation*}

With this, we can extend the semigroup to a contraction of $L^p(\mu), 1 \leq p\leq\infty$

\begin{align*}
   \| f \|_{L^p(\mu)}^p= \int_\mathcal X| f|^p \:d \mu =\int_\mathcal X \mathbf{P}_{t} |f|^p \:d \mu \geq \int_\mathcal X |\mathbf{P}_{t} f|^p \:d \mu  = \|\mathbf{P}_{t} f \|_{L^p(\mu)}^p
\end{align*}

Finally, for any bounded, continuous function $f \in \mathcal C_b(\mathcal X), 1 \leq p< \infty$,

\begin{equation*}
\int_\mathcal X |P_t f(x)- f(x)|^p \: d \mu(x) \to 0 \quad \text{as}\quad  t \to 0
\end{equation*}
by the DCT since $|P_t f- f|^p(x) \leq 2^{p-1}(\|P_t f\|_\infty +\|f\|_\infty)\leq 2^{p}\|f\|_\infty< \infty$. Since $C_b(\mathcal X)$ is dense in $L^p(\mathcal X, \mu)$ we are done.

Now this argument with the DCT seems only to work for finite measures. For $\sigma$-finite measures, should I split the state-space into finite regions and apply the argument?

Also I am not sure what licenses me to say that $P_t f(x) \to f(x)$ as $t \to 0$ for $\mu$-a.e. $x$ and bounded continuous $f$. This works when the process is right continuous but I am not sure about the general case.

答: The strong continuity (in $L^2(\mu)$) of $({\bf P}_t)$ is part of the definition of Markov semigroup in BGL (condition (vi) on page 11). On the other hand, the strong continuity on $L^p(\mu)$ for $1\le p<\infty$ then follows automatically because $\|{\bf P}_t\|_p\le 1$: By a well-known result, strong continuity of a semigroup like  $({\bf P}_t)$ follows from weak continuity (see, for example, Theorem IX.1 in Yosida'a Functional Analysis). The weak continuity, that is the continuity of $t\mapsto \int g\cdot {\bf P}_tf d\mu$ for $g\in L^q(\mu)$ and $f\in L^p(\mu)$, follows because (i) $g$ and $f$ can be taken to be in $L^1(\mu)\cap L^\infty(\mu)\subset L^2(\mu)$ by the $L^p(\mu)$ contraction property of $({\bf P}_t)$ and the denseness of $L^1(\mu)\cap L^\infty(\mu)$ in $L^p(\mu)$ ($\mu$ is $\sigma$-finite!) and (ii) continuity of $t\mapsto   \int g\cdot {\bf P}_tf d\mu$ for $f,g\in L^1(\mu)\cap L^\infty(\mu)$ follows from $L^2(\mu)$ continuity of the semigroup.


问:
Solving $\tan(x) = \cos(x)$
I've been trying to solve the following trigonometric equation unsuccessfully. My intuition was to reduce the degree of the equation from $4$ to $2$ so I could solve it as a quadratic equation, but my attempt so far have proved unsuccessful. Any help would be appreciated.

$$\begin{align}
\tan(x) &= \cos(x) \tag1\\[4pt]
\frac{\sin(x)}{\cos(x)} &=\cos(x) \tag2\\[4pt]
\frac{\sin^2(x)}{\cos^2(x)} &=\cos^2(x) \tag3\\[4pt]
\frac{1-\cos^2(x)}{\cos^2(x)} &=\cos^2(x) \tag4\\[4pt]
1-\cos^2(x) &=\cos^4(x) \tag5\\[6pt]
\cos^4(x) + \cos^2(x) - 1 &= 0  \tag6
\end{align}$$

It's at this point that I get stuck

答: It is much simpler to deal with this as follows:\begin{align}\frac{\sin(x)}{\cos(x)}=\cos(x)&\iff\sin(x)=\cos^2(x)\\&\iff\sin(x)=1-\sin^2(x)\\&\iff\sin^2(x)+\sin(x)-1=0.\end{align}So, solve the equation $y^2+y-1=0$.

Title: Solving $\tan(x) = \cos(x)$


问:
Let x and y be positive numbers. Which of the following implies $x^y >= y^x.$
Let x and y be positive numbers. Which of the following implies $x^y \geq y^x$.
$(a)x\leq e\leq y$ $(b) y\leq e\leq x$ $(c) x\leq y\leq e$ or $e\leq y\leq x$ $(d) y\leq x\leq e$ or $e\leq x\leq y$.

I've proceeded with taking log on both sides but is kinda stuck over there, help!
Thanks in advance.

答:
Hint :

$$x^y \geq y^x \Longleftrightarrow y \ln(x) \geq x \ln(y) \Longleftrightarrow \frac{\ln(x)}{x} \geq \frac{\ln(y)}{y}$$

Now study the function
$$t \mapsto \frac{\ln(t)}{t}$$


问:
$f$ bounded on $[a,b]$ with one or finite discontinuities implies $f$ Riemann-integrable.
I have two problems:

Prove that if $f$ is bounded on $[a, b]$ and has exactly one discontinuity in $[a, b]$ then $f$ is Riemann-integrable on $[a, b]$.

Prove that if $f$ is bounded on $[a, b]$ and $f$ has only finitely many discontinuities in $[a, b]$ then $f$ is Riemann-integrable on $[a, b]$.

I have a possible proof for the first one which I would like checked:

Theorem 6.1 states that $f$ is Riemann-integrable on $[a, b]$ iff given any $\varepsilon > 0$ there exists a partition $P$ of $[a, b]$ with $U(P, f) - L(P, f) < \varepsilon$.

Theorem 6.2 states that if $f$ is continuous on $[a, b]$ then it is Riemann-integrable on $[a, b]$.

Suppose $f$ is bounded on $[a, b]$. Then by definition there exists $M > 0$ such that $\vert f(x) \vert \leq M$ for all $x \in [a, b]$. Suppose $f$ has exactly one discontinuity in $[a, b]$ call it $c$. Without loss of generality suppose $c \in (a, b)$. Let $\varepsilon > 0$. Choose $\delta > 0$ such that $\displaystyle \delta = \frac{\varepsilon}{12M}$. Observe that $f$ is continuous on $[a, c - \delta]$ and $[c + \delta, b]$. By Theorem 6.2 $f$ is Riemann-integrable on $[a, c - \delta]$ and $[c + \delta, b]$. By Theorem 6.1 there exists a partition $P_1$ of $[a, c - \delta]$ with $U(P_1, f) - L(P_1, f) < \varepsilon/3$. Again, by Theorem 6.1 there exists a partition $P_2$ of $[c + \delta, b]$ with $U(P_2, f) - L(P_2, f) < \varepsilon/3$. Define $P = P_1 \cup P_2$. Now observe that
          \begin{align*}
   U(P, f) &=  U(P_1, f) + 2\delta \cdot \sup_{x \in [c - \delta, c+ \delta]} f(x) + U(P_2, f) \\
   &\leq U(P_1, f) + 2\delta \cdot M + U(P_2, f)
  \end{align*}
  and
          \begin{align*}
   L(P, f) &= L(P_1, f) + 2\delta \cdot \inf_{x \in [c - \delta, c+ \delta]} f(x) + L(P_2, f)\\
   & \geq L(P_1, f) + 2\delta \cdot (-M) + L(P_2, f)
  \end{align*}
  which is implies $$-L(P, f) \leq - L(P_1, f) + 2\delta \cdot M - L(P_2, f) $$ Hence we have
          \begin{align*}
   U(P, f) - L(P, f) &\leq U(P_1, f) + 2\delta \cdot M + U(P_2, f) - L(P_1, f) + 2\delta \cdot M - L(P_2, f)\\
   &= \big[U(P_1, f) - L(P_1, f) \big] + 4 \delta \cdot M + \big[U(P_2, f) - L(P_2, f)\big]\\
   &< \frac\varepsilon3 + 4M\frac{\varepsilon}{12M} + \frac\varepsilon3\\
   &= \frac\varepsilon3 +\frac\varepsilon3 +\frac\varepsilon3 \\
   &= \varepsilon
  \end{align*}
          Thus by Theorem 6.1 we can conclude that $f$ is Riemann-integrable on $[a, b]$.

My question namely is: is it sufficient to check that $c \in (a, b)$ or should I explicitly show that it holds when the discontinuity is on the endpoints?

For the second problem, I am thinking of using induction on the first problem, but I haven't seen an induction problem done like that so I don't know how to get started. Any push in the right direction would be greatly appreciated!

答:
Suppose f is a bounded real valued function on [a, b] such that f
2 ∈ R[a, b]. Does it follow that
f ∈ R[a, b]? Does the answer change if we assume f
3 ∈ R[a, b]? Either give a proof or provide a counter
example in each of the two cases.</p>


问:
Is there a smarter way to integrate?
My textbook uses the following technique to integrate. Let us take the following as an example,  evaluate the integral of:

$$f(x) = 6x(x^2 + 1)^5$$

Notice that $[x^2+1]' = 2x$, so $6x\space dx = 3 \space d (x^2+1)$

$$6x(x^2+1)^5 \space dx = \dfrac{6}{2} (x^2+1)^5 \space d(x^2+1) = 3u^5 \space du = d \dfrac{1}{2} u^6 = d \dfrac{1}{2} (x^2+1)^6$$

So $$F(x) = \dfrac{1}{2} (x^2 +1)^6 + c$$

This method seems very confusing (what is the meaning of the $d$ for example?) and besides, it seems like it would work only in textbook-cases. Is there a 'better' way of integrating these types of basic functions?

p.s. - I looked in my textbook for reference, but the toughest types of functions we have to evaluate are the integrals of for example $h(x) = \dfrac{2x-3}{x^2+6x+10}$ and $k(x) = \dfrac{\arcsin^2(x)}{\sqrt{1-x^2}}$. So the method wouldn't of course have to work for integrals tougher than these types.

答: There are many dedicated methods to solve different integration problem.The method you have shown here is fundamental method which is not directly used in complex problem but it will use in many complex problem when that problem reduces to easier form. d comes from differentiation.This is substitution example.we differentiate it to reduce it easier form</p>

Title: Is there a smarter way to integrate?


问:
Minimal boolean matrix coverage
Given a boolean matrix $M \in {\{ 0,1 \}}^{n \times m}$

A pseudo-rectangle of $M$ is defined as a pair of rows and columns of $M$. I.e. a pair $(I_n, I_m)$ such that  $I_n \subseteq \{1,..,n\}, I_m \subseteq \{1,..,m\}$

Find the minimal set of pseudo-rectangles such that each $1$ cell in $M$ is covered by exactly 1 pseudo-rectangle and no pseudo-rectangle covers any $0$ cell

答:
You can solve this as a set partitioning (also known as exact cover) problem as follows.  For each pseudo-rectangle $R$ that does not contain a $0$ cell ($M_{i,j}=0$), let binary decision variable $x_R$ indicate whether $R$ is selected.  The problem is to minimize $\sum_R x_R$ subject to
$$\sum_{R: (i,j) \in R} x_R = 1 \quad \text{for all $(i,j)$ where $M_{i,j}=1$}$$

Title: Minimal boolean matrix coverage


问:
How can we integrate a function for surface area of revolution?

This is the equation (with my values plugged in) for surface area of revolution which I found online and I understand the derivation, thus I wanted to use it for a math project I am currently working on. I attempted to use the quadratic function (outside the radical) and inside the radical is the first derivative squared. However, I am stuck in evaluating the integral. I tried u substitution where u is equal to -1/12x+2/3 but was unable to finish it.

Is trigonometry required and does the u substitution I attempted not work in this case? What is the Surface area (answer to the integral)?

答:
Have you tried u-sub with $u=(-\frac{1}{24}x^2+\frac{2}{3}x)$?

edit: I just realized you made these values up. My suggestion may be wrong, but I'm curious to see if this is integrable because I can't think of anything.


问:
How to find the volume of a solid bounded by x which is equal to an arbitrary constant
(solved)

I stumbled on this question today and I'm extremely curious on how to solve it.

My main concern is that x = b (b>1). I don't understand how can I bound this if x= b and b is any value greater than 1. If I ignore that last restriction I can do it.

This is what I got ignoring the last restriction. Anyhow, how do you actually solve the problem for a and b? Please let me know.

答:
First of all, you have the wrong region; you get a ring or stand-like object when you revolve around the Y axis, not the X axis. You should take a rectangle-ish region to the right of the one you selected, going from 1 out to b, and then swing that in a horizontal circle.
Second, I don't know if you need to do this by one particular method, washers versus cylinders. For washers, you need to solve the function for x, and integrate in two pieces...easier to use cylinders centered on the y axis with radii from 1 to b.  The extra factor of 2$\pi$ x makes the integral more doable.


问:
uniform integrability does not imply $L_p$ boundednes
Let $\mathcal H$ denote a family of random variables on ($\Omega,\mathcal A ,P$) . We all know that If $\mathcal H$ be $L_p$ bounded then $L_p$ is uniformly integrable but the converse is not true.
How can I give an example for this and how can I show that this is not uniformly integrable.

答: Let $X \in L_1$ be an integrable random variable that is not $L_p$ integrable, and set $\mathcal H := \{X\}$.  Then $\mathcal H$ is uniformly integrable because it is a singleton, but is not $L_p$ bounded because ${\sup_{Y \in \mathcal H} \|Y\|_p = \|X\|_p = \infty.}$


答:
Use Cylindrical method. The volume is found by integrating  (2 pi x f(x)) from x=1 to x=b, where f(x) is the given function.
Your answer depends on b of course.</p>

Title: How to find the volume of a solid bounded by x which is equal to an arbitrary constant

Body:
(solved)

I stumbled on this question today and I'm extremely curious on how to solve it.

My main concern is that x = b (b>1). I don't understand how can I bound this if x= b and b is any value greater than 1. If I ignore that last restriction I can do it.

This is what I got ignoring the last restriction. Anyhow, how do you actually solve the problem for a and b? Please let me know.


问:
Finding a vector perpendicular to $n-1$ vectors in $\mathbb{R}^n$
I was trying to find the cross-product, or a vector perpendicular to vectors in $\mathbb{R}^n$

Then I recalled the formula for the cross product in $\mathbb{R}^3$ and then tried to generalize it.

So, I let the vectors be $\left(a_{1, 1}, a_{1, 2},\cdots , a_{1, n}\right), \left(a_{2, 1}, a_{2, 2},\cdots , a_{2, n}\right) \cdots \left(a_{n-1, 1}, a_{n-1, 2}, \cdots , a_{n-1, n}\right)$

Then using the formula to find the cross product of two vectors in $\mathbb{R}^3$ and generalizing it I get the perpendicular vector to be $$\det\pmatrix{e_1&e_2&\cdots &e_n\\a_{1, 1}&a_{1, 2}&\cdots & a_{1, n}\\ a_{2, 1}&a_{2, 2}&\cdots & a_{2, n}\\ \vdots&\vdots & \ddots & \vdots\\a_{n-1, 1}&a_{n-1, 2}&\cdots & a_{n-1, n}}$$where $e_1, e_2,\cdots, e_n$ is like $i, j, k...$

So, I tried this with $3$ vectors in $\mathbb{R}^4$ then took the dot product of the resultant vector with each of the original vectors and the answer came as $0$ in every case, so I did confirm that it worked, but I do not know why. Is there any proof of this?

答:
Consider a scalar product of

$$\mathbf{v}=\det\begin{pmatrix}\mathbf{e}_1&\cdots&\mathbf{e}_n\\
a_{1,1}&\cdots&a_{1,n}\\
\vdots&\ddots&\vdots\\
a_{n-1,1}&\cdots&a_{n-1,n}\\
\end{pmatrix} = A_1 \mathbf{e}_1 + \ldots + A_n \mathbf{e}_n
\\
\text{with}
\qquad
\mathbf{a}_i=\begin{pmatrix}a_{i,1}\\ \vdots\\ a_{i,n}\end{pmatrix} = a_{i,1} \mathbf{e}_1 + \ldots + a_{i,n} \mathbf{e}_n
$$

Since the basis $\mathbf e_i$ is orthonormal:

$$
\mathbf v\cdot \mathbf a_i = A_1a_{i,1} + \ldots+a_{i,n}A_n =
\det\begin{pmatrix}a_{i,1}&\cdots&a_{i,n}\\
a_{1,1}&\cdots&a_{1,n}\\
\vdots&\ddots&\vdots\\
a_{n-1,1}&\cdots&a_{n-1,n}\\
\end{pmatrix}
$$

In this matrix two rows are exactly the same, so the determinant is zero.


问:
Natural partial order of Mitsch on Natural numbers
So according to Mitsch, the natural partial order $\leq$ of any semigroup $S$ is given by
$$a \leq b \iff a = xb = by, xa = a, \quad \text{for some } x, y \in S^1$$
But obviously in $(\mathbb N, \cdot)$ we have $2 \leq 3$, and yet there is no $x \in \mathbb N$ that satisfies the above condition.

Similarly in $(\mathbb N, +)$ we have $2 \leq 3$ and the same problem.

Is there something that I'm missing? Is regular "less than or equal" not a natural partial order on $\mathbb N$?

答: Mitsch is defining one particular partial order on any semigroup and calling it "the natural partial order".  This is nothing more than a name; it is not an assertion that any other partial orders are "unnnatural" in any sense.  As you have observed, in the case of $(\mathbb{N},\cdot)$ or $(\mathbb{N},+)$, it is not the same as the usual partial order of natural numbers.


答:
In general:

$L^p$ bounded $\Longrightarrow$ uniformly integrable $\Longrightarrow$ $L^1$ bounded.

So if your question is for a fixed $p$, Then we can choose $s$ with $1 < s < p$ and find a family $\mathcal H$ that is $L^s$ bounded but not $L^p$ bounded.  Then $\mathcal H$ will be uniformly integrable.  On the probability space $[0,1]$, such an example can be found of the form $\{f_n : n = 1,2,3,\cdots\}$ where
$f_n(x) = \min\{n,x^\alpha\}$ for an appropriate choice of $\alpha$.  Choose $\alpha$ so that
$$
\int_0^1x^{\alpha p}\;dx = +\infty\quad\text{but}\quad\int_0^1x^{\alpha s}\;dx < +\infty
$$

We can insert another space here.  Define $\mathcal H$ is $L\log L$ bounded iff
$\sup_{X \in \mathcal H}\mathbb E[|X| \log^+|X|] < +\infty$.  Where $\log^+(t) = \max\{0,\log t\}$.  Then

$L^p$ bounded $\Longrightarrow$ $L\log L$ bounded $\Longrightarrow$ uniformly integrable.

So for a more powerful counterexample, choose $\mathcal H$ to be $L\log L$ bounded, but $L_p$ unbounded for all $p>1$.
On the space $[0,1/2]$, such an example can be found of the form $\{f_n : n = 1,2,3,\cdots\}$ where
$f_n(x) = \min\{n,x^\alpha\;|\log(x)|^\beta\}$ for an appropriate choice of $\alpha, \beta$.

Choose $\alpha, \beta$ so that
$$
\int_0^{1/2}|x^{\alpha}\log^\beta(x)|^p\;dx = +\infty\quad \text{ for all } p>1\\ \qquad\text{but}\quad
\int_0^{1/2}|x^\alpha\log^\beta(x)|\log^+|x^\alpha\log^\beta(x)|\;dx< +\infty
$$
A choice that works is $\alpha=-1, \beta=-3$.

Title: uniform integrability does not imply $L_p$ boundednes

Body:
Let $\mathcal H$ denote a family of random variables on ($\Omega,\mathcal A ,P$) . We all know that If $\mathcal H$ be $L_p$ bounded then $L_p$ is uniformly integrable but the converse is not true.
How can I give an example for this and how can I show that this is not uniformly integrable.


问:
System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$
This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\
3^z + 4^z + 5^z = 2^z \cdot 3^{z -1} \cdot x
\end{array}
\right.
$$
So, the first thing I notice was $4^x = 2^{2x} = a$ so we get a system of quadratic equations. This proved to be quite hard to deal with, so I gave it up. I also tried dividing by $2^x \cdot 3^{x - 1} \cdot y$ and the others, but it was weird to deal with. Do you have any ideas?

答:
This is mostly a variant of NN2's answer, deriving the unique solution from an argument about the middle of the three variables rather than the minimum. I'm posting it mainly because it occurs to me that proving that each variable must be "between" the other two might be easier to follow than eyeballing the directions that a bunch of inequalities point.

Let

$$f(u)=3\left(\left(1\over2\right)^u+\left(2\over3\right)^u+\left(5\over6\right)^u\right)$$

and rewrite the system as

$$\begin{align}
f(x)&=y\\
f(y)&=z\\
f(z)&=x
\end{align}$$

Now since $1\over2$, $2\over3$, and $5\over6$ are all less than $1$, $f$ is a decreasing function of $u$. Hence if $x$ is between $y$ and $z$ (i.e., greater than or equal to one and less than or equal to the other), then $f(x)=y$ is between $f(y)=z$ and $f(z)=x$, which in turn implies $f(y)=z$ is between $f(z)=x$ and $f(x)=y$ (which in turn implies $f(z)=x$ is between $f(x)=y$ and $f(y)=z$). Since at least one variable is between the other two, we conclude that each is between the others, i.e., $x=y=z$.

Finally, since $f$ is decreasing, the equation $f(u)=u$ can have at most one solution. By inspection, we see that the unique solution is $u=3$. So the OP's system of equations has $(x,y,z)=(3,3,3)$ as its unique solution.


问:
Nash equilibria - Why can we calculate a player's strategy without reference to their payoffs?
Consider the following calculation of Nash equilibria from Wikipedia:

```<code>                  Player B plays H  Player B plays T
Player A plays H    −1, +1          +1, −1
Player A plays T    +1, −1          −1, +1

To compute the mixed strategy Nash equilibrium, assign A the probability p
of playing H and (1−p) of playing T, and assign B the probability q of
playing H and (1−q) of playing T.

E[payoff for A playing H] = (−1)q + (+1)(1−q) = 1−2q
E[payoff for A playing T] = (+1)q + (−1)(1−q) = 2q−1
E[payoff for A playing H] = E[payoff for A playing T] ⇒ 1−2q = 2q−1 ⇒ q = 1/2
</code>```

We have calculated the strategy for player 2, but we haven't used their payoffs, only those for player 1! How is this possible?

答:
The reasoning is as follows. Player 2 should choose $q$ such that player 1 will be indifferent between playing $H$ and playing $T$. We ex ante know that this reasoning will work out because we know that
1) there exists no pure equilibrium in the game therefore there will be a mixed equilibrium
2) According to the following theorem: The strategies in the support of Nash equilibrium (a strategy which is played with positive probability) always yield the same payoff. </p>

<p>[one can showthis as follows. assume that one strategy in the support of a NE yields more payoff than the other strategy in the support of a NE, then one would 'transfer' probability from the strategy which gives lower payoff than to the one which gives higher payoff.]</p>

Body:
The reasoning is as follows. Player 2 should choose $q$ such that player 1 will be indifferent between playing $H$ and playing $T$. We ex ante know that this reasoning will work out because we know that
1) there exists no pure equilibrium in the game therefore there will be a mixed equilibrium
2) According to the following theorem: The strategies in the support of Nash equilibrium (a strategy which is played with positive probability) always yield the same payoff.

[one can showthis as follows. assume that one strategy in the support of a NE yields more payoff than the other strategy in the support of a NE, then one would 'transfer' probability from the strategy which gives lower payoff than to the one which gives higher payoff.]


问:
Product expression for roots of polynomial
I am working slowly through "Riemann's zeta function", HM Edwards, Dover Publications, 1974. At the top of page 18, I read (I have altered the notation from $p(s)$ to $P(x)$)

"any polynomial $P(x)$ can be expanded as a finite product $P(x) = P(0) \prod_\rho \bigl(1-\frac{x}{\rho } \bigr)$ where $\rho$ ranges over the roots of the equation $P(\rho) = 0$.

Really? Surely, $P(x) = \prod_\rho \bigl(x-\rho  \bigr)$? I wondered if this was something to do with extracting the factor $P(0)$ in order to give the roots of a monomial, but of course this cannot be the case, since this would require division by the coefficient of highest order. And besides, a simple counter-example establishes that Edwards' assertion is incorrect. Define

$$
\begin{aligned}
P(x) :=& (x+1) (x+2) (x+3) (x+4) (x+5)
\\
=& x^5 + 15x^4 + 85x^3 + 225x^2 + 274x + 120
\end{aligned}
$$

(I have chosen a polynomial with real roots for simplicity, but I see no reason why the same definitions should not apply to finite polynomials with complex coefficients.)

Applying $P(x) = P(0) \prod_\rho \bigl(1-\frac{x}{\rho } \bigr)$ gives

$$
\begin{aligned}
P(x)=&120 (1-x) \left(1-\frac{x}{2}\right)
   \left(1-\frac{x}{3}\right)
   \left(1-\frac{x}{4}\right)
   \left(1-\frac{x}{5}\right)
\\
=& -x^5 + 15x^4 - 85x^3 + 225x^2 - 274x + 120
\end{aligned}$$

So, in this case, the correct assertion would be $P(x) = P(0) \prod_\rho \bigl(1+\frac{x}{\rho } \bigr)$.

Could someone please explain what is going on here? Weierstrass and Hadamard's factorisation theorems don't seem to help, but perhaps I have understood them only poorly.

How is Edwards' original assertion arrived at, and why does the counter-example not work?

答:
Your simple example establishes that your assertion is incorrect.

The roots of

$$P(x) := (x+1) (x+2) (x+3) (x+4) (x+5)$$

being $-1,-2,-3,-4,-5$ we do have

$$P(x)=120\left(1+\dfrac x1\right) \left(1+\dfrac x2\right) \left(1+\dfrac x3\right) \left(1+\dfrac x4\right) \left(1+\dfrac x5\right).$$

The proof is very trivial by transforming all factors with

$$\rho-x=\rho\left(1-\dfrac x\rho\right).$$


问:
How many non-redundant periodic sequences exist from a given set of terms?
I'll try to clarify the question using an example, using a set of 4 elements, $A, B, C, D$.

The number of non-redundant sequences of period 1 would be 4, namely $"AAA...", "BBB...", "CCC...", "DDD..."$.

The number of non-redundant sequences of period 2 would be 10, namely $"AAA...", "BBB...", "CCC...", "DDD...", "ABAB...", "ACAC...", "ADAD...", "BCBC...", "BDBD...", "CDCD..."$.

By "redundant," I mean that "ABAB..." is the same as "BABA...", since they're essentially the same sequence but shifted.

So far so good.

I'm running into trouble finding a good way to calculate the number of non-redundant sequences with period 3 or greater though, and would appreciate any help with this.

答:
Let $k$ be fixed, and let $a_n$ be the number of cyclic sequences of length $n$ which are non-periodic. Then
$$
a_n=\frac1n\sum_{d|n} \mu(n/d)k^d,\tag1
$$
where $\mu$ is the number-theoretic Möbius function, and $d$ ranges over all divisors of $n$.  In your example, the number of words of length $2$ is $\frac12\left(\mu(2/1)4^2+\mu(2/2)4^1\right)=\frac12(16-4)=6$.

This can be proven with Möbius inversion. Note that
$$
k^n=\sum_{d|n}da_{d},\tag2
$$
since the $k^n$ linear words of length $n$ are uniquely specified by choosing a non-periodic sequence of length $d$, repeating it $n/d$ times, and then rotating it between $0$ and $d-1$ places. Applying Möbius inversion to $(2)$ yields $(1)$.

Edit: I misunderstood your question. $a_n$ is the number of sequences of length $n$ which cannot be decomposed into any smaller sequences. To get the total number of sequences, you need to sum $a_m$ over all divisors of $n$:
$$
\text{# sequences}=\sum_{m|n}a_m=\sum_{m|n}\frac1m\sum_{d|m}\mu(m/d)k^d
$$
There might be a way to simplify this by reducing the order of summation, I am not sure...


问:
Hungarian Algorithm with different metric
I have a modified Assignment Problem, that can almost be solved using the Hungarian Algorithm.

Instead of trying to minimize the sum of costs of assignments, I want to minimize the cost of the costliest assignment. Using the explanation of Hungarian algorithm at wikipedia , one could say that I want to find a solution that minimizes the amount paid to the highest salaried worker.

One suggestion I've received is to simply use a modified cost matrix, with every value in the cost matrix raised to a biggish exponent (say, 10), and then use the standard Hungarian Algorithm.

But perhaps there is a better way?

答:
Actually, this change of metric makes the problem a little easier to compute. You can find the objective value by dichotomy. To answer the question "is it possible to find a matching whose highest edge costs $M$ ?", you just remove the edges costing more than $M$ and search for a perfect matching (which is easier than the minimum-cost matching).

This simple idea, with a Hopcroft-Karp algorithm, already gives a $O(|E| \sqrt{|V|} \log(|E|))$ algorithm ; but actually you could make it run faster, because the subgraphs are included one in the other so you can reuse the information from one to the other. If I remember correctly, you can completely remove the $\log(n)$ factor by doing it in the right order ; but I'm not sure about it.


问:
using pigeonhole principle for a hand of thirteen cards
Say I shuffle and deal a hand of thirteen cards. How can I apply the pigeonhole principle in these cases:

The hand has at least four cards in the same suit

The hand has at exactly four cards in some suit

The hand has at least five cards in some suit

The hand has exactly one suit containing four or more cards

I have problems in which to assign holes and pigeons to, can anyone help?

答: You can apply the pigeonhole principle to case 1: can you fit thirteen pigeons into four holes each with space for three pigeons? </p>


答:
First of all, the following boils down to exactly the same as your textbook does, but it might make more sense to you. Make the substitution $u = x^2 +1$. Then $du= 2xdx$. Hence $\int 6x(x^2+1)^5dx=\int 3 u^5du$. Now integrate and substitute back for $u$.

Notice that it was a very plausible idea to make this substitution because we see the the integral involves a product of a certain function ($u$) multiplied by the derivative of this function ($du$).

Title: Is there a smarter way to integrate?

Body:
My textbook uses the following technique to integrate. Let us take the following as an example,  evaluate the integral of:

$$f(x) = 6x(x^2 + 1)^5$$

Notice that $[x^2+1]' = 2x$, so $6x\space dx = 3 \space d (x^2+1)$

$$6x(x^2+1)^5 \space dx = \dfrac{6}{2} (x^2+1)^5 \space d(x^2+1) = 3u^5 \space du = d \dfrac{1}{2} u^6 = d \dfrac{1}{2} (x^2+1)^6$$

So $$F(x) = \dfrac{1}{2} (x^2 +1)^6 + c$$

This method seems very confusing (what is the meaning of the $d$ for example?) and besides, it seems like it would work only in textbook-cases. Is there a 'better' way of integrating these types of basic functions?

p.s. - I looked in my textbook for reference, but the toughest types of functions we have to evaluate are the integrals of for example $h(x) = \dfrac{2x-3}{x^2+6x+10}$ and $k(x) = \dfrac{\arcsin^2(x)}{\sqrt{1-x^2}}$. So the method wouldn't of course have to work for integrals tougher than these types.


问:
Find the coefficients of this polynomial.

Given $f(x)=x^4 +ax^3 +bx^2 -3x +5$ , $  a,b\in \mathbb{R}$ and knowing that $i$ is a root of this polynomial. what are $a,b$?

My work until now: since all the coefficients are real, I now know that $i$ and $-i$ are roots of this polynomial, there must be two more roots. I tried to use vietta but it didn't really help, I get that $x_3+x_4=-a$ and $i(-i)x_3x_4=5$. What am I missing?

Appreciate any help.

答:
$$f(x)=x^4 +ax^3 +bx^2 -3x +5$$

Since $i$ is a root, we have
$$f(i) = 0$$
$$\implies 1 -ia -b -3i + 5 = 0$$
$$\implies (6-b) + i (-a -3) = 0$$

Since $a, b \in \mathbb{R}$, the real and imaginary part of the above expression must separately vanish.

Hence $$-a -3 = 0$$ and $$6 - b = 0$$

Finally
$$a = -3, b = 6$$

Note

In general, if $a = a_x + i a_y$ and $b = b_x + i b_y$, then we may have many solutions that satisfy $a_y - b_x + 6 = 0, a_x + b_y + 3 = 0$. Since $a, b \in \mathbb{R}$, we must have $a_y = b_y = 0$


问:
"Stars and Bars" summation identity
Can anybody help me solve this problem?
I don't even know where to start or how to interpret this.

Proof for all $1 \leq k \leq n$ that the folowing identity holds:

\begin{equation*}
\begin{pmatrix}
n+k-1 \\
n-1
\end{pmatrix} =
\sum_{i=1}^k \begin{pmatrix}k-1 \\ i-1 \end{pmatrix} \begin{pmatrix} n \\ i \end{pmatrix}
\end{equation*}

This problem is in fact a homework assignment, so it would be kind if the solution isn't spoiled right away. There is one hint given in the assignment that says that for k identical objects and n different colors ($k \geq n$) the amount of different color combinations (you may repeat colors) when using every color at least once, is equal to: $\begin{pmatrix} k-1 \\ n-1 \end{pmatrix}$

I hope that anybody can assist me.

Thanks in advance!

答: Hint: Well, the RHS is choosing some color ($i$ of them) and coloring $k$ objects with those $i$ colors. Using the hint given, every of the $i-$th colors chosen is used at least once. But notice that you choose the colors, that means the other $n-i$ were not used. So, if you were to consider the $n$ colors at a whole, then you are relaxing the condition of at least once to no restriction at all. What is, then, the LHS?

Title: "Stars and Bars" summation identity


问:
Uniqueness of coefficients in plane wave representation of wave equation's solution
3D wave equation
$$
\Delta p(\boldsymbol{r},t) - \frac{1}{c^{2}} \frac{\partial^{2} p}{\partial t^{2}}(\boldsymbol{r},t) = 0 \tag{1}
$$
By applying the spatial Fourier transform to this PDE, we obtain the following ODE
$$
c^{2} |\boldsymbol{k}|^{2} \hat{p}(\boldsymbol{k},t) - \frac{\partial^{2} \hat{p}}{\partial t^{2}}(\boldsymbol{k},t) = 0 \tag{2}
$$
Since $(2)$ is ODE of harmonic oscillator, the solution can be written as
$$
\hat{p}(\boldsymbol{k},t) = A(\boldsymbol{k}) e^{ic|\boldsymbol{k}|t} + B(\boldsymbol{k}) e^{-ic|\boldsymbol{k}|t} \tag{3}
$$
Since $A=0$ by physical interpretation, apply the inverse Fourier transform to $(3)$.
$$
p(\boldsymbol{r},t) = \int_{\mathbb{R}^{3}} B(\boldsymbol{k}) e^{i(\boldsymbol{k}\cdot \boldsymbol{r} - c|\boldsymbol{k}| t)} d\boldsymbol{k} \tag{4}
$$
This is plane wave representation of 3D wave equation's solution.

Now if sound pressure $p(\boldsymbol{r},t)$ is known, can the coefficients $\{B(\boldsymbol{k})\}$ be uniquely determined?

In other words, I want to know invertibility of following integral operator $\mathcal{G}$ and its inverse operator if $\mathcal{G}$ is invertible.
$$
\mathcal{G}\hat{f} (\boldsymbol{r},t) = \int_{\mathbb{R}^{3}} \hat{f}(\boldsymbol{k}) e^{i(\boldsymbol{k}\cdot \boldsymbol{r} - c|\boldsymbol{k}| t)} d\boldsymbol{k} \tag{5}
$$
In the first place, it is difficult for me to calculate multiple integrals with polar coordinate component and I'm not familiar with integral operator theory.

Thank you for your advice.

答:
You can write
$$ p(\boldsymbol{r},t) = \int \limits_{\mathbb{R}^3} B(\boldsymbol{k}) \,\mathrm{e}^{\mathrm{i} (\boldsymbol{k} \cdot \boldsymbol{r} - c \lvert \boldsymbol{k} \rvert t)} \, \mathrm{d} \boldsymbol{k} = \int \limits_{\mathbb{R}^3} B(\boldsymbol{k}) \mathrm{e}^{-\mathrm{i} c \lvert \boldsymbol{k} \rvert t} \,\mathrm{e}^{\mathrm{i} \boldsymbol{k} \cdot \boldsymbol{r}} \, \mathrm{d} \boldsymbol{k} = [\mathcal{F}^{-1} b (\cdot, t)] (\boldsymbol{r}) \, , $$
where $b (\boldsymbol{k}, t) = B(\boldsymbol{k}) \mathrm{e}^{-\mathrm{i} c \lvert \boldsymbol{k} \rvert t}$. Since the Fourier transform $\mathcal{F}$ is invertible (on suitable function spaces), $B$ is uniquely determined by
$$ B (\boldsymbol{k}) = b (\boldsymbol{k}, t) \mathrm{e}^{\mathrm{i} c \lvert \boldsymbol{k} \rvert t} = [\mathcal{F} p(\cdot,t)] (\boldsymbol{k}) \mathrm{e}^{\mathrm{i} c \lvert \boldsymbol{k} \rvert t} = \frac{\mathrm{e}^{\mathrm{i} c \lvert \boldsymbol{k} \rvert t}}{(2 \pi)^3} \int \limits_{\mathbb{R}^3} p (\boldsymbol{r},t) \, \mathrm{e}^{-\mathrm{i} \boldsymbol{k} \cdot \boldsymbol{r}} \, \mathrm{d} \boldsymbol{r} \, .$$


问:
Ordinary generating function for $\binom{3n}{n}$
The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$

can be derived by using the duplication formula for the gamma function and the generalized binomial theorem.

But what about the ordinary generating function for $ \displaystyle  \binom{3n}{n}$?

According to Wolfram Alpha,   $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{2\cos \left(\frac{1}{3} \arcsin \left(\frac{3 \sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \, , \quad |x| < \frac{4}{27}. $$

Any suggestions on how to prove this?

EDIT:

Approaching this problem using the fact that $$ \text{Res} \Big[ \frac{(1+z)^{3n}}{z^{n+1}},0 \Big] = \binom{3n}{n},$$

I get $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = -\frac{1}{2 \pi i x} \int_{C} \frac{dz}{z^{3}+3z^{2}+3z - \frac{z}{x}+1},$$

where $C$ is a circle centered at $z=0$ such that every point on the circle satisfies $ \displaystyle\Big|\frac{x(1+z)^{3}}{z} \Big| < 1$.

Evaluating that contour integral would appear to be quite difficult.

答:
The following is a proof of the hypergeometric identity $$_2F_1\left(a,1-a;\frac{1}{2};\sin^2(x)\right)=\frac{\cos[(2a-1)x]}{\cos x} \, , \quad - \frac{\pi}{2} < x < \frac{\pi}{2} \tag{1}.$$

This identity is used in R. J. Mathar's answer.

Similar to my answer here, I will first use the generalized binomial theorem to show that $$_2F_1\left(a,1-a;\frac{1}{2};-z^{2}\right)  = \frac{1}{2 \sqrt{1+z^{2}}} \left[\left(\sqrt{1+z^{2}}+z \right)^{2a-1} + \left(\sqrt{1+z^{2}}-z \right)^{2a-1} \right] \, , \quad |z| <1. $$

Identity $(1)$ then follows if $z$ is replaced with $i \sin (x)$.

\begin{align} &\frac{1}{2\sqrt{1+z^{2}}} \left[\left(\sqrt{1+z^{2}}+z \right)^{2a-1} + \left(\sqrt{1+z^{2}}-z \right)^{2a-1}\right] \\ &= \frac{1}{2\sqrt{1+z^{2}}} \left[\sum_{k=0}^{\infty} \binom{2a-1}{k} z^{k}\left(\sqrt{1+z^{2}} \right)^{2a-1-k} + \sum_{j=0}^{\infty} \binom{2a-1}{j} (-z)^{j} \left(\sqrt{1+z^{2}} \right)^{2a-1-j}\right ] \\ &= \sum_{k=0}^{\infty} \binom{2a-1}{2k} z^{2k} (1+z^{2})^{a-1-k}\\ &= \sum_{k=0}^{\infty} \binom{2a-1}{2k} z^{2k} \sum_{l=0}^{\infty} \binom{a-1-k}{l}z^{2l}\\ & \stackrel{(2)}= \sum_{n=0}^{\infty} \sum_{m=0}^{n}\binom{2a-1}{2m}  \binom{a-1-m}{n-m} z^{2n}  \\ &=\sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)}\sum_{m=0}^{n} \frac{\Gamma(a-m)}{\Gamma(2m+1)\Gamma(2a-2m) \Gamma(n-m+1)} \, z^{2n} \\ & \stackrel{(3)}= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \sum_{m=0}^{n} \frac{\pi}{2^{2a-1}} \frac{1}{\Gamma \left(m+\frac{1}{2}\right) \Gamma(m+1) \Gamma(a-m+\frac{1}{2}) \Gamma(n-m+1)} \,  z^{2n}  \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \frac{\pi}{2^{2a-1}} \frac{1}{\Gamma(n+ \frac{1}{2}) \Gamma(a+ \frac{1}{2})}  \sum_{m=0}^{n} \binom{n-\frac{1}{2}}{n-m} \binom{a-\frac{1}{2}}{m}  \, z^{2n} \\ & \stackrel{(4)}= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \frac{\pi}{2^{2a-1}}\frac{1}{\Gamma(n+ \frac{1}{2}) \Gamma(a+ \frac{1}{2})}  \binom{a+n-1}{n} \, z^{2n}  \\ & \stackrel{(5)} = \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a-n)} \frac{\sqrt{\pi}}{\Gamma (n+\frac{1}{2})}  \, \frac{z^{2n}}{n!} \frac{\Gamma(a)}{\Gamma(a)}  \\  & \stackrel{(6)} = \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a)} \frac{\Gamma(1-a+n)}{\Gamma(1-a)} \frac{\Gamma(\frac{1}{2})}{\Gamma(n+ \frac{1}{2})} \frac{(-z^{2})^{n}}{n!}  \\ &= \, _2F_{1} \left(a, 1-a; \frac{1}{2}; -z^{2} \right) \end{align}

$(2)$: Cauchy product

$(3)$: Duplication formula for the gamma function

$(4)$: Chu-Vandermonde identity

$(5)$: Duplication formula again

$(6)$:  $\frac{\Gamma(a)}{\Gamma(a-n)}= (-1)^{n} \frac{\Gamma(1-a+n)}{\Gamma(1-a)}$


Title: Find the coefficients of this polynomial.

Body:
Given $f(x)=x^4 +ax^3 +bx^2 -3x +5$ , $  a,b\in \mathbb{R}$ and knowing that $i$ is a root of this polynomial. what are $a,b$?

My work until now: since all the coefficients are real, I now know that $i$ and $-i$ are roots of this polynomial, there must be two more roots. I tried to use vietta but it didn't really help, I get that $x_3+x_4=-a$ and $i(-i)x_3x_4=5$. What am I missing?

Appreciate any help.

Body:
$$f(i)=i^4 + a i^3 + b i^2 - 3 i + 5=-i a-b+(6-3 i)=0$$
$$6-b-i(3+a)=0$$
as $a,b\in\mathbb{R}$ we have $6-b=0,3+s=0$

$a=-3,b=6$


问:
Rigorous definition of derivative of a real-valued function
I am learning Real Analysis through the book Analysis 2 of Terence Tao and there is a notation of Partial derivative (in Definition 6.3.7 below).

In the definition, the author uses the notation $\quad\frac{d}{dt} f(x_0 + t.e_j)|_{t=0} \quad$ without stating its definition in the previous chapters. (Please note that $e_j = (0,0,0,..1,0,0)$ ($i.e.$ the row vector with all zeros except for the $j^{th}$ element).

I understand the meaning of that notation is to take the $derivative$ of each element of the vector $f(x_0 + t.e_j)$ with respect to $t$.

But I am wondering whether there is a rigorous definition of derivative of a vector-valued function with respect to a single variable ?

答:
I am wondering whether there is a rigorous definition of derivative of a vector-valued function with respect to a single variable ?

Check Definition 6.2.2 in the book. What you want is a special case of differentiability of function $f:\mathbb{R}^n\to\mathbb{R}^m$ with $n=1$.