问:
How to motivate logarithm by breaking down $1/x$, into increasingly large segments, and proving that the area under the curve stayed the same?
James Stewart's Calculus 7e 2011 (not Early Transcendentals) presents the two ways to define a logarithm. First, on p 404 in § 6.3, Stewart defines $\log_ax$ as the inverse of $a^y$. Then on p 414 in § 6.4, he proves that $\frac d{dx} ln|x| = \frac 1x \iff \ln|x| + C = \int \frac 1x dx $.

Second, on p 421 in §6.2*, he first defines $\ln x = \int^x_1 \frac 1t \; dt, x > 0 $. Then on p 429 in §6.3*, he defines $e^x = y \iff \ln y = x$.

In either case, how does this comment add anything to the motivation or intuition behind $\log$?

I'm going to go back in time to the development of the logarithm. The log was a way of performing multiplication (in some base) by simply using addition. Meaning, every time you add one number, your output is proportional to the last output.

At some point in time shortly afterwards, mathematicians were working on integration techniques. They noticed that the basic methods of finding anti derivatives worked for everything except for 1/x. However it was realized, that if you break down the graph of 1/x, into increasingly large segments (1-2, 2-4, 4-8, 8-16...) That the area under the curve stayed the same. this is actually very easily provable! But now look what they found..... They found that the area under the curve increases by simple addition as you double the interval. (Meaning adding the area interval (1-2) + (2-4) + (4-8), is simply the same area added together 3 times).

This should stick out to you, because this is exactly how we defined a logarithm to work. Addition carrying out multiplication with some base number.

Because this logarithm seemed to be so "natural" (it just came about all by itself), it was called the "natural logarithm". So the relation was known that ln(x) = integral of 1/x, before it was known what that base number was. Euler was actually able to calculate the number, and the proof for that is very interesting as well! (See my paper for the proof).

答:
That comment you linked to is not a particularly meaningful one.  The reason is because any function of the form $f(x) = c/x$ for some constant $c > 0$ obeys the same property, namely that for $a > 0$, $$\int_{x=a}^{2a} \frac{c}{x} \, dx = \int_{x=2a}^{4a} \frac{c}{x} \, dx.$$  So this property in itself does not make the choice $c = 1$ "special" or "natural," although it is the choice for which the number $e$ satisfies $$\log_e x = \int_{t=1}^x \frac{1}{t} \, dt, \quad x > 0.$$

The question basically boils down to how $e$, the base of the natural logarithm, is defined.  And the "explanation" in the comment is inadequate motivation for the reason I described.

In reality, there are several equivalent definitions for $e$, each provable from any one of the others, as the Wikipedia article about $e$ states (see "Alternative characterizations").  Consequently, how one motivates the existence of this constant depends on what properties are of interest.


问:
Power Series with $a_{n}$ and $a_{k}$

If $P(x)= \sum_{n=0}^{\infty} a_{n} x^{n}$. It is known that $P$ satisfies: $$P^{\prime}(x)=2xP(x)$$ $\\$ for all $ x\in \mathbb{R}$ and $$P(0)= 1$$
\begin{split}\end{split} (1) Prove that $a_{2k+1}=0$ for every $ k\in \mathbb{\left \{1,2,3,4,...  \right \}}$

Its true that if two power series have the same coefficients then they are equal?

I can demonstrate the first equation transforming the series to an integral, but i cant prove (1) and (2) because $k$ confuses me, I was thinking that $k$ is for $P'(x)$, just as $n$ is for $P(x)$. Also i need orientation with the proof requested.

Thanks.

答:
It is true that if two power series have the same coefficients then they are equal. More is true actually. If two power series are equal on some interval then their coefficients must be equal.

Assuming that the series has a positive radius of convergence $R$, we can differentiate term by term to obtain a series for $P'(x)$.
$$P'(x)=\sum_{k=1}^{\infty}{ka_kx^{k-1}}$$
This series also has radius of convergence $R$, and since $P'(x)=2xP(x)\implies P'(0)=a_1=0$

Also, since $P'(x)=2xP(x)$
$$\sum_{k=1}^{\infty}{ka_kx^{k-1}}=\sum_{k=0}^{\infty}{2a_kx^{k+1}}$$
making a shift of index we see that,
$$\sum_{k=1}^{\infty}{(k+1)a_{k+1}x^k}=\sum_{k=1}^{\infty}{2a_{k-1}x^k}$$
Since these power series are equal on some interval with radius $R>0$, their coefficients must be equal. Thus we have,
$$(k+1)a_{k+1}=2a_{k-1}$$
Now you can probably finish the proof on your own using induction.

ContentLicense: CC BY-SA 4.0

Title: Power Series with $a_{n}$ and $a_{k}$


问:
Integration in a non-standard way: evaluating $ \int \frac 1 { x ^ 2 } \, \mathrm d x $ without applying the power rule - Is it nonsense?
I want to evaluate the integral
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x $$
in a non-standard way and without applying the power rule.

Let $ x \ne 0 $ and
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = f ( x ) + C \text . $$
For any $ a \in \mathbb R \setminus \{ 0 \} $ we have
$$ f ( a x ) = \int \frac 1 { a ^ 2 x ^ 2 } \, \mathrm d ( a x ) = \frac 1 a \int \frac 1 { x ^ 2 } \, \mathrm d x = \frac 1 a \bigl ( f ( x ) + C \bigr ) \text . $$

Then, we have
$$ f ( x ) = a f ( a x ) \text , $$
for any $ a \in \mathbb R \setminus \{ 0 \} $.

Putting $ x = 1 $, we have
$$ f ( a ) = \frac { f ( 1 ) } a \text . $$

This is the critical point for me. I'm not sure I'm not wrong here. I will just continue.

Here, I will replace the constant $ a $ with the variable $ x $. My argument for doing this is as follows: $ a $ covers all nonzero real numbers.

So, I will write the last equation like this (but, I'm not sure):
$$ f ( x ) = \frac { f ( 1 ) } x \text . $$

We continue.

Let $ u = x $ and $ v = f ( x ) $, so that $ \mathrm d u = \mathrm d x $ and $ \mathrm d v = \frac 1 { x ^ 2 } \, \mathrm d x $.

$$ \int x \cdot \frac 1 { x ^ 2 } \, \mathrm d x = x \cdot f ( x ) - \int f ( x ) \, \mathrm d x = \ln x + C _ 1 \text ; $$
$$ \implies x \cdot \frac { f ( 1 ) } x - \int \frac { f ( 1 ) } x \, \mathrm d x = \ln x + C _ 1 \text ; $$
$$ \implies f ( 1 ) - f ( 1 ) \left ( \ln x + C _ 2 \right ) = \ln x + C _ 1 \text ; $$
$$ \implies \bigl ( f ( 1 ) + 1 \bigr ) \ln x = f ( 1 ) ( 1 - C _ 2 ) - C _ 1 \text . $$

The right side is a constant. The left side must also be a constant.
$$ \implies f ( 1 ) + 1 = 0 \implies f ( 1 ) = - 1 \text . $$

Finally, we get
$$ f ( x ) = - \frac 1 x \text ; $$
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = - \frac 1 x + C \text . $$

Remark 1.

My goal was to find antiderivative of $ \frac 1 { x ^ 2 } $. Because,
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = \text {antiderivative of } \frac 1 { x ^ 2 } + \text{constant.} $$
In short, the function $ f ( x ) $ was an antiderivative.

Remark 2.

Initially $ a $ is a constant. However, it covers all real numbers except zero. Therefore, I replaced it with the variable $ x $.

How much of my work does math allow? Is the method I am using a nonsense? Otherwise, can the errors it contains be corrected?

答:
Task

To prove that if $f :(0,\infty) \to \mathbb R$ is a function which satisfies $f(1) = -1$ and for every $x,y>0$, $f(y)-f(x) = \int_{x}^y \frac{dt}{t^2}$, then $f(x) = \frac{-1}{x}$ for all $x$. We use monotonicity of the integral and the change of variables formula, but do not use FTC.

Idea

Create a functional equation for $f$ using a change of variables.

Solve the functional equation.

The functional equation

Let $x,y,b>0$ be arbitrary. Then, consider $f(y)-f(x) = \int_x^y \frac{dt}{t^2}$. Perform the change of variables $u=bt$, then $du = bdt$. This results in :
$$
f(y)-f(x) = \int_x^y \frac{dt}{t^2} = \int_{bx}^{by} \frac{b^2}{u^2}\frac{du}{b} = b\int_{bx}^{by} \frac{du}{u^2} = b(f(by)-f(bx))
$$

which leads to the equation : for all $b,x,y>0$, we have :
$$
f(y)-f(x) = b(f(by)-f(bx))
$$

Solving the equation

As Professor Vector shows here, there exists a constant $a$ such that $xf(x)=ax-a-1$ for all $x>0$. Note that no tools from calculus were used. That is, $f(x) = a - \frac{a-1}{x}$.

But then, $\lim_{x \to \infty} f(x) = a$. Therefore, $\lim_{x \to \infty} (f(x) - f(1)) = a+1$ i.e. $a = \int_{1}^\infty \frac{dt}{t^2} - 1$. We expect $\int_1^\infty \frac{dt}{t^2} = 1$.

But how do we show that? The most obvious substitution $u = \frac {-1}t$ is barred since we are not allowed to use the fact that this is the antiderivative of $\frac{1}{t^2}$, which we are trying to prove! Furthermore, if we attempt to express this as a Riemann sum and take the limit, we could have done the same for general $n$ instead of $\infty$ and landed with the expression of $\frac{-1}{x}$ earlier.

Conclusion

While our approach eventually turned into a roundabout, we showed that $f(x) = a - \frac{a-1}{x}$ for all $x$, where $a = \int_1^{\infty} \frac{dt}{t^2}$, a quantity that unfortunately cannot be calculated without going back to the theory of the integral itself.

However, our sojourn lead us to a few important observations about such $f$, which is all we can do without the power formula.

$0<\lim_{x\to \infty} \frac {dt}{t^2} = a < \infty$. The assertion that the integral is finite is quite surprising to those who have not seen the power law.

$f(x)$ decays exactly at the rate $\frac 1x$ as $x \to \infty$ i.e. $\lim_{x \to \infty} \frac{f(x)-a}{x} = a-1$. This is because $f$ is a strictly increasing function, so we must have $a-1<0$ i.e. $a<1$ so $a-1 \neq 0$.


问:
What is the expression $E[X_t dt]$ using stochastic calculus?
I am trying to finish a calculation using Itô calculus but do not know the final step.

What is the expression $E[X_t  dt]$ equal to?

Note that the $X_t$ is drifted Brownian motion $X_t = \mu_x t + \sigma_x W_t^x$.

答:
Well, $dt$ is "constant", so it is independent of $X_t$ and can come out of the expectation. Thus you get

$$\mathbb{E}[X_t  dt] = \mathbb{E}[X_t]dt $$

Assuming $\mu_x$ and $\sigma_x$ are deterministic, and that $W_t^x$ is a standard Brownian Motion, the expectation of $X_t$ is given by

$$\mathbb{E}[X_t] = \mathbb{E}[\mu_x t] + \mathbb{E}[\sigma_x W_t^x] = \mathbb{E}[\mu_x t] + 0 = \mu_x t $$

And finally

$$\mathbb{E}[X_tdY_t] = \mu_y \mu_x tdt$$


问:
How to solve $a^2 - 2b^2 - 3 c^2 + 6d^2 = 1$ over $\mathbb{Z}$?
I would like to find the integral elements int the quaternion algebra $D_{2,3}(\mathbb{Q})$.  These are $2 \times 2$ matrices with elements in $\mathbb{Z}[\sqrt{2}]$ with determinant $1$.  Observe these matrices are closed under multiplication.

$$ \det \left| \begin{array}{cc} x_1 + \sqrt{2} x_2 & x_3 + \sqrt{2} x_4 \\
3(x_3 - \sqrt{2} x_4) & x_1 - \sqrt{2} x_2\end{array}  \right| = x_1^2 - 2x_2^2 - 3x_3^2 + 6 x_4^2 = 1 $$
Books as recent as 2003, such on John Conway's On Quaternions and Octonions they collect fact about integral quaternions.

The usual mistake here is that I could have written an equation with no solutions, though I suspect that's unlikely.  This is similar to Pell's equation and we're looking for integer points on a conic section, a hyperboloid.

By exhaustive search (a standard problem-solving heuristic) I was able to show that:

$$ \det \left| \begin{array}{cc} 4 + 3 \sqrt{2} & 7 + 5 \sqrt{2}
 \\
3(7-5\sqrt{2}) & 4 - 3 \sqrt{2}
\end{array}\right| = 4^2 - 2 \times 3^2 - 3 \times 7^2 + 6 \times 5^2 = 1$$
These number theory problems are notorious for their brevity. In fact, they will get arbitrarily complex.  We also get an element of $D_{2,3}(\mathbb{Z})$.

There are books or papers on integer solutions to quadratic forms.  Cassels' book proves outlines and theory of qudratic forms over $\mathbb{Q}$.  For example,  there's no complete list and even writing down generators of this group could be complicated.

答:
From an elementary perspective:

Note that the equation is the same as $$a^2-2b^2-3(c^2-2d^2)=1.$$

For nonzero integers $p,q,r,s,t,u,v$:

Let $a=up^2+vq^2$, $b=2pqt$, $c=ur^2+vs^2$, $d=2rst$. Now let $2t^2=uv$. Then the equation is the same as $$(up^2-vq^2)^2-3(ur^2-vs^2)^2=1$$ which reduces to the form $m^2-3n^2=1$ for integers $m$ and $n$. If we let $f(m,n)=m^2-3n^2$ then we can find infinitely many solution pairs $(m,n)$ through $$f(m,n)=f(2m+3n, m+2n)=f(7m+12n, 4m+7n)=\cdots$$ Not sure if this counts as "solving" the equation, but it may help.


问:
Similar to inversion, what other geometric transformations have been used to prove theorems easily?
I just learned of the proof of Ptlomey's theorem and was just amazed by it (see here) and it made me wonder, are there other such transformations which gives out difficult geometrical results easily?</p>

答: One important transformation in projective geometry is changing points into lines and vice versa to get dual theorems. For example, once you've proved Ceva's theorem, its dual, Menelaus's theorem, follows.</p>


Title: A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$

Body:
The sequence is given by the formula $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and it is known that  $x_{2017} + x_{2023} = 990$, then what is $x_{2020}$ = ?

My little approch:

It is given that  $x_{2017} + x_{2023} = 990$ ----- (1)  and $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$

So,

$x_{2017} = x_{2016+1} = 3 x_{2016} + \sqrt{8x^2_{2016} + 2}$ -------(2)

$x_{2023} = x_{2022+1} = 3 x_{2022} + \sqrt{8x^2_{2022} + 2}$ -------(3)

Then from (1),(2),(3) =>

$3 x_{2016} + \sqrt{8x^2_{2016} + 2} + 3 x_{2022} + \sqrt{8x^2_{2022} + 2} = 990$

$3 (x_{2016}+x_{2022} ) + \sqrt{8x^2_{2016} + 2}  + \sqrt{8x^2_{2022} + 2} = 990$

$3 (x_{2016}+x_{2020} + x_{2021}) + \sqrt{8x^2_{2016} + 2}  + \sqrt{8x^2_{2020} + 2}+ \sqrt{8x^2_{2021} + 2}+ \sqrt{8x^2_{2022} + 2} = 990$

I stuck here and can't go further. Please help me with it.

Body:
We have
$$x_{n+1}-3x_n = \sqrt{8x_n^2+2}$$
$$(x_{n+1}-3x_n)^2 = 8x_n^2+2$$
$$x_{n+1}^2 -6x_{n+1} x_n + x_n^2 = 2   \tag{I}$$
And because
$$x_{n-1}^2 -6x_{n-1} x_n + x_n^2 = 2   \tag{II}$$
From $(I)$ and $(II)$, we have, for all $n$
$$(x_{n+1}-x_{n-1})(x_{n+1}+x_{n-1}-6x_{n})=0$$
Is it possible that $x_{n+1}=x_{n-1}$ ? No, because $x_n$ is an increasing series. Effectively, we have $x_{n+1}-x_n = \sqrt{8x_n^2+2}+2x_n > \sqrt{4x_n^2}+2x_n \ge 0$.

So, $x_{n+1}+x_{n-1}-6x_{n}=0$ for all $n$ or $x_n = \frac{x_{n+1}+x_{n-1}}{6}$

Now, we
$$
\begin{align}
x_n &= \frac{x_{n-1}+x_{n+1}}{6} \\
&= \frac{ \frac{x_{n-2}+x_{n}}{6} +\frac{x_{n}+x_{n+2}}{6}}{6} \\
&= \frac{x_{n-2}+x_{n+2}}{6^2} + \frac{2x_n}{6^2}  \\
&= \frac{\frac{x_{n-3}+x_{n-1}}{6} +\frac{x_{n+1}+x_{n+1}}{6}}{6^2} + \frac{2x_n}{6^2}  \\
&= \frac{x_{n-3}+x_{n+3}}{6^3}+\frac{x_{n-1}+x_{n+1}}{6^3} + \frac{2x_n}{6^2}  \\
&= \frac{x_{n-3}+x_{n+3}}{6^3}+\frac{x_n}{6^2} + \frac{2x_n}{6^2}  \\
\end{align}
$$

So,
$$(1-\frac{3}{6^2})x_n= \frac{x_{n-3}+x_{n+3}}{6^3}$$
Or
$$x_n = \frac{x_{n-3}+x_{n+3}}{6(6^2-3)}$$

Hence, if $x_{2017}+x_{2023} = 990$, then $x_n = \frac{990}{6(6^2-3)} = 5$.


问:
find solution of $\sum{\frac{1}{x_i-x}}=0$
Find $x$ that is a solution of $\sum_{i=1}^{n}{\frac{1}{x_i-x}}=0$.
I have no idea how to solve this problem. Taking gradient didn't help me also. Any hint?

答:
To avoid confusion, I shall change notations.

You are looking for the zero's of function
$$f(x)=\sum_{i=1}^n{\frac{1}{x-a_i}}$$ and they are $n$ solutions (one in each interval $(a_i,a_{i+1})$. The problem is that $f(x)$ is not very well conditioned because of these vertical asymptotes.

For a given interval, it is better to consider instead
$$g_k(x)=(x-a_k)(x-a_{k+1}) f(x)$$ which will write
$$g(x)=(x-a_k)(x-a_{k+1})\sum_{i=1}^{k-1}{\frac{1}{x-a_i}}+(2x-a_k-a_{k+1})+$$ $$(x-a_k)(x-a_{k+1})\sum_{i=k+2}^{n}{\frac{1}{x-a_i}}$$

Now, you see that
$$g(a_k)=a_k-a_{k+1} \qquad \text{and} \qquad  g(a_{k+1})=a_{k+1}-a_k$$ and the problem becomes simple. Use a combination of Newton and bisection steps (the lastest if Newton tends to take you out of the bounds).

May I confess that I spent decades working this type of equations ?

If you need more, just tell.


问:
Convergence of $\sum \frac{a_n}{(a_1+\ldots+a_n)^2}$
Assume that $0 < a_n \leq 1$ and that $\sum a_n=\infty$. Is it true that
$$ \sum_{n \geq 1} \frac{a_n}{(a_1+\ldots+a_n)^2} < \infty $$
?

I think it is but I can't prove it. Of course if $a_n \geq \varepsilon$ for some $\varepsilon > 0$ this is obvious.

Any idea? Thanks.

ContentLicense: CC BY-SA 3.0

Body:
Let $A_n=\sum\limits_{k=1}^na_n$. Then
$$
\begin{align}
\sum_{n=1}^\infty\frac{A_n-A_{n-1}}{A_n^2}
&\le\frac1{a_1}+\sum_{n=2}^\infty\frac{A_n-A_{n-1}}{A_nA_{n-1}}\\
&=\frac1{a_1}+\sum_{n=2}^\infty\left(\frac1{A_{n-1}}-\frac1{A_n}\right)\\
&\le\frac2{a_1}
\end{align}
$$


问:
Given N students each student sends K messages to any other student
There are N students, each student sends K different messages to any other student, 260 messages sent, how many students in the class?

Im struggling to solve this one, tried to solve it using the summation of in degree and out-degree using the 260 but it didn't work.

答:
Note that if the number of students is $N$, then every student sends $(N-1)K$ messages in total, as they send $K$ messages to every other student.

Thus, in total, $N(N-1)K=260$ messages are sent. As $N$ is a positive integer, we need that $N$, $N-1$ and $N(N-1)$ all divide 260. But the divisors of 260 are "1, 2, 4, 5, 10, 13, 20, 26, 52, 65, 130, 260". Thus, the only possibilities are that $N=5$, $K=13$ or that $N=2$ and $K=130$.


问:
Is $\omega \ncong \omega+1$ but $\omega \xrightarrow[\text{onto}]{1-1} \omega+1$
I am working my way through ordinal arithmetic and cardinality, so I wanted a quick reality check that I am correctly differentiating between structure preserving bijections (isomorphisms) and non-structure preserving bijections.

A consequence of this difference is the following statement:

$\omega \ncong \omega+1$ but $\omega \xrightarrow[\text{onto}]{1-1} \omega+1$, where $\omega$ refers to the set of all natural numbers

That is to say, there is no structure-preserving bijection between $\omega$ and $\omega+1$ (where the "structure" in question is $\in$ and its corresponding properties)...however $ \omega \approx \omega+1$ (i.e. there is a non-structure preserving bijection between $\omega$ and $\omega+1$).

Is this the right idea?

(There appears to be a similar post here: $\omega+1\sim\omega$ but $\omega+1\not\simeq\omega$ but I think the relationship $\sim$ defined by the author of the post presupposes that the individual is familiar with the Schroder-Bernstein theorem).

答:
Correct. There can't be a structure-preserving isomorphism $\omega \to \omega + 1$. The latter has a maximal element $\omega$ but the former does not, so what $x$ could satisfy $f(x) = \omega$? Or rather, if we did have such an $x$, where would $x+1$ go?

But they are of the same cardinality, as you can easily see by taking the maximal element $\omega$ from $\omega + 1$ and shoving it at the beginning. That particular bijection doesn't preserve the order structure, but it is a bijection.


问:
Symmetrical difference sets $A \bigoplus B$
Symmetrical difference for two sets I was taught it's defined as this way,

$A\bigoplus B = (x \in A \wedge  x\notin B) \vee (x \notin A \wedge x \in B)$

My question is shouldn't there be a 3rd case where, $x \in A \wedge x \in B$?

答: That would be the union of the two sets, which is the disjoint union of their intersection with their symmetric difference. The symmetric difference is a measure of how much the two sets are different. It is minimal (the empty set) when the two sets are equal and it is maximal (the union) when they are disjoint.</p>


Body:
$$x_{n+1} = 3x_{n} + \sqrt{8x_{n}^2+2}$$
$$x_{n+1}-3x_{n} = \sqrt{8x_{n}^2+2}$$
By squaring both sides,
$$x_{n+1}^2+x_{n}^2-6x_{n}x_{n+1}=2$$
Since this is a general expression, we can write -
$$x_{n+1}^2+x_{n}^2-6x_{n}x_{n+1}=x_{n}^2+x_{n-1}^2-6x_{n-1}x_{n}$$
Simplifying this we get,
$$x_{n+1}+x_{n-1}=6x_{n}$$
Now our task is to find $x_n$ for $n=2020$ when $x_{n+3}+x_{n-3}$ is given.
First we will find $x_{n+2}+x_{n-2}$.

Let $$x_{n+2}+x_{n-2}=y$$.

By adding $2x_{n}$ to both sides of the equation,
$$6(x_{n+1}+x_{n-1})=y+2x_{n}$$
So,
$$y=34x_{n}$$
Now let,
$$x_{n+3}+x_{n-3}=z$$
Add $x_{n+1}+x_{n-1}$ to both the sides,
$$6(x_{n+2}+x_{n-2})=z+6x_{n}$$
$$z=198x_{n}$$
So, we get -
$$x_{2020}=5$$


问:
Is this piecewise-defined function on $\mathbb{R}^2$ continuous at $(0,0)$? What about differentiable?
Is this function is a differentiable function, a continuous function at the point $(0,0)$?
How to show that ?

$$
f(x,y)=\begin{cases}
\frac{x^{3}y+xy^{3}}{x^{2}+y^{4}} &\text{if }(x,y)\neq (0,0),\\
0 & \text{if }(x,y)=(0,0)
\end{cases}
$$

Thanks a lot!

答: No it is not continuous. Hence also not differentiable. Express it in polar coordinates. Cancel common terms from numerator and denominator. Try approaching zero along different paths, you will get different values. For eg., along the path $\theta=0$ you get the value zero. Whereas along the path $\theta=\pi/2$ you will get infinity.</p>


问:
How $\sum\limits_{k=0}^n {n\choose k}\cdot 0^{n-k}\cdot 2^{k}=2^{n}$?
Note: I know this can be converted to $(0+2)^{n}=2^{n}$ according to the binomial theorem, I just got stuck after expanding the sum.

I expanded this sum and this is what I got:

$0+0+0+...+{n\choose n}\cdot 0^{0}\cdot 2^{n}$

Here, since $0^{0}$ is undefined how the sum will be equal to $2^{n}$ ?

Body:
Use $0^0=1, 0^k=0, k\ne 0$ In the series on ly the last term contibutes :
$$(0+2)^n=0+...+0+...+0+...+0+{n \choose n} 0^{n-n} 2^n=2^n.$$


答:
Indeed, it's necessary to define $0^0 = 1$ here.

Note that there are cases where it's not right to define $0^0 = 1$; for example, $\lim_{x \to 0} 0^x$ is $0$. For this reason, anywhere you have a formula which includes a term $0^0$, you need to give the value of $0^0$ alongside that formula.

Title: How $\sum\limits_{k=0}^n {n\choose k}\cdot 0^{n-k}\cdot 2^{k}=2^{n}$?

Body:
Note: I know this can be converted to $(0+2)^{n}=2^{n}$ according to the binomial theorem, I just got stuck after expanding the sum.

I expanded this sum and this is what I got:

$0+0+0+...+{n\choose n}\cdot 0^{0}\cdot 2^{n}$

Here, since $0^{0}$ is undefined how the sum will be equal to $2^{n}$ ?


问:
A proof that Superman does not exist
If Superman were able and willing to prevent evil, he would do so. If Superman were unable to prevent evil he would be impotent; if he were unwilling to prevent evil, he would be malevolent. Superman does not prevent evil. If Superman exists, he is neither impotent nor malevolent. Therefore, Superman does not exist

Read the argument above and encode it into symbolic sentences.

W: Superman is willing to prevent evil.

A: Superman is able to prevent evil.

P: Superman prevents evil.

I: Superman is impotent.

M: Superman is malevolent.

E: Superman exists

2. Show your process as you work through the logical steps of validating argument.

Superman were able and willing to prevent evil, he would do so ( A ∧ W ) → P

If Superman were unable to prevent evil he would be impotent   ¬A → I

if he were unwilling to prevent evil, he would be malevolent    ¬ W → M

Superman does not prevent evil  ¬P

If Superman exists, he is neither impotent nor malevolent   E  → (¬ I ∧ ¬ M )

Superman does not exist   ¬ E

------ the rest of the validation i do not understand ----------

¬ ( A ∧ W )

¬ A  ∨ ¬ W

A → ¬ W

A → W

A → M

A→ (M ∨ I)

¬ A → M ∨ I

M ∨ I

¬ ( ¬ I ∧ ¬ M )

¬ E

I found the answers online but i do not understand the other parts of number 2 can someone explain please

答: Determine whether this argument, taken from Kalish and Montague [KaMo64], is valid. If Superman were able and willing to prevent evil, he would do so. If Superman were unable to prevent evil, he would be impotent; if he were unwilling to prevent evil, he would be malevolent. Superman does not prevent evil. If Superman exists, he is neither impotent nor malevolent. Therefore, Superman does not exist...</p>

Title: A proof that Superman does not exist


答:
It's necessary to define $0^0 = 1$ only if you use $(a+b)^n=\sum_{k=0}^n {n\choose k}a^{n-k}b^{k}$.

The issue does not arise if you use $(a+b)^n=a^n+{n\choose 1}a^{n-1}b+\cdots +{n\choose n-1}ab^{n-1}+b^n$.


问:
$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$?
calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}.$$
Use Squeeze theorem we have
$$\frac{1}{n+1}\sum\limits_{k=1}^n(\frac{k}{n})^2<\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}<\dfrac{1}{n}\sum\limits_{k=1}^n(\frac{k}{n})^2$$
So $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\int_0^1x^2\mathrm{d}x=\frac{1}{3}.$$
Use $$\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}.$$
Hence $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right)=\frac{1}{2}.$$
If our method is correct, is there any other way to solve this problem? Thank you

答: Although $\sum\frac{k^2}{n^3+kn}$ and $\sum\frac{k^2}{n^3}$ have the same limit, they differ to $O(1/n)$. So when they are multiplied by $n$, they give different limits


问:
$\lim\limits_{x \to \infty} {x}(a^{1/x}-1)$, such that $a > 1$
$\lim\limits_{x \to \infty} {x}(a^{1/x}-1)$, such that $a > 1$

My Approach:

Since $x$ approaches infinity I presume we have to 'get rid' of the $x$ on the outside because the root $-1$ will be more than 0 but I have no idea how to reduce the $x$ so please help me out!

答:
Recall :  $\lim_{y \rightarrow 0} \dfrac{e^y-1}{y}=1;$

Set $y:=1/x,$ and consider $y \rightarrow 0^+$.

$\lim_{y \rightarrow 0^+}\log a (\dfrac{e^{y\log a-1}}{y\log a}) = \log a. $


问:
Why is the set of well formed formulas are defined as the smallest set of strings
Consider this definition of Well formed formulas in mathematical logic -
The set of all well formed formula is the smallest set of strings , WFF that satisfies

All Boolean variables are in WFF, and so are the symbols Τ and F. We call
such formulae atomic.

If A and Β are any strings in WFF, then so are the strings (not A), (A and B),
(A V β), (A —> β), (A = B)

Wouldn't there be exactly one such set of strings satisfying these properties of WFF ?

I don't understand the need for using the word smallest.

答:
Noah has already answered your question, but let me also add a side remark.

Instead of the "smallest set / intersection of all sets"-presentation of WFF, you sometimes also encounter definitions of the following form:

WFF is the set of strings which are generated in the following way:

All Boolean variables are in WFF, and so are the symbols Τ and F. We call such formulae atomic.

If A and Β are any strings in WFF, then so are the strings (not A), (A and B), (A V β), (A —> β), (A = B)

This "constructive" definition is also fine, and it is clear: I cannot see how two persons reading the above could possibly disagree about the object WFF which is being defined.

However, for some mathematicians nowadays the gold standard for rigorousness is formalizability in set theory; and here the "smallest set" definition is easier to state than the constructive one above.


答: It's best to use definite integrals; this not only distinguishes certain variables your reasoning risks conflating (though you've done your best to explain why the steps are legal), but also makes it shorter. Define $f(x):=\int_\infty^x\frac{dt}{t^2}$ so $f(ax)=\int_\infty^{ax}\frac{dt}{t^2}=\int_\infty^x\frac{du}{au^2}=\frac{f(x)}{a}$; in particular, $f(a)=\frac{f(1)}{a}$. For $x>0$, integrate by parts with $u:=t,\,v:=f(t)$ so$$\begin{align}\ln x&=\int_1^x\tfrac{dt}{t}\\&=[tf(t)]_1^x-\int_1^xf(t)dt\\&=\underbrace{xf(x)-f(1)}_0-\int_1^x\tfrac{f(1)}{t}dt\\&=-f(1)\ln x\\\implies f(1)&=-1.\end{align}$$

Title: Integration in a non-standard way: evaluating $ \int \frac 1 { x ^ 2 } \, \mathrm d x $ without applying the power rule - Is it nonsense?

Body:
I want to evaluate the integral
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x $$
in a non-standard way and without applying the power rule.

Let $ x \ne 0 $ and
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = f ( x ) + C \text . $$
For any $ a \in \mathbb R \setminus \{ 0 \} $ we have
$$ f ( a x ) = \int \frac 1 { a ^ 2 x ^ 2 } \, \mathrm d ( a x ) = \frac 1 a \int \frac 1 { x ^ 2 } \, \mathrm d x = \frac 1 a \bigl ( f ( x ) + C \bigr ) \text . $$

Then, we have
$$ f ( x ) = a f ( a x ) \text , $$
for any $ a \in \mathbb R \setminus \{ 0 \} $.

Putting $ x = 1 $, we have
$$ f ( a ) = \frac { f ( 1 ) } a \text . $$

This is the critical point for me. I'm not sure I'm not wrong here. I will just continue.

Here, I will replace the constant $ a $ with the variable $ x $. My argument for doing this is as follows: $ a $ covers all nonzero real numbers.

So, I will write the last equation like this (but, I'm not sure):
$$ f ( x ) = \frac { f ( 1 ) } x \text . $$

We continue.

Let $ u = x $ and $ v = f ( x ) $, so that $ \mathrm d u = \mathrm d x $ and $ \mathrm d v = \frac 1 { x ^ 2 } \, \mathrm d x $.

$$ \int x \cdot \frac 1 { x ^ 2 } \, \mathrm d x = x \cdot f ( x ) - \int f ( x ) \, \mathrm d x = \ln x + C _ 1 \text ; $$
$$ \implies x \cdot \frac { f ( 1 ) } x - \int \frac { f ( 1 ) } x \, \mathrm d x = \ln x + C _ 1 \text ; $$
$$ \implies f ( 1 ) - f ( 1 ) \left ( \ln x + C _ 2 \right ) = \ln x + C _ 1 \text ; $$
$$ \implies \bigl ( f ( 1 ) + 1 \bigr ) \ln x = f ( 1 ) ( 1 - C _ 2 ) - C _ 1 \text . $$

The right side is a constant. The left side must also be a constant.
$$ \implies f ( 1 ) + 1 = 0 \implies f ( 1 ) = - 1 \text . $$

Finally, we get
$$ f ( x ) = - \frac 1 x \text ; $$
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = - \frac 1 x + C \text . $$

Remark 1.

My goal was to find antiderivative of $ \frac 1 { x ^ 2 } $. Because,
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = \text {antiderivative of } \frac 1 { x ^ 2 } + \text{constant.} $$
In short, the function $ f ( x ) $ was an antiderivative.

Remark 2.

Initially $ a $ is a constant. However, it covers all real numbers except zero. Therefore, I replaced it with the variable $ x $.

How much of my work does math allow? Is the method I am using a nonsense? Otherwise, can the errors it contains be corrected?


问:
Proving prefix/suffix exists for finite automata
My task is to show that if a language $L \subseteq \{a, b\}^*$ is recognised by a finite automaton then there exists a finite automaton that recognise all the prefixes and suffixes of the word in $L$.

$$\text{Pref}(L) = \big\{x \in \{a,b\}^* \big| xy \in L\text{ for some }y \in \{a,b\}^*\big\}$$

and

$$\text{Suff}(L) = \big\{y \in \{a,b\}^* \big| xy \in L\text{ for some }x \in \{a,b\}^*\big\}$$

I've come so far to understand that:

A word $u$ is a prefix of a word $v$ if there is a word $w$ such that $v=uw$.
And intuitively I somehow believe that for a word $w$ a prefix/suffix can be all the partitions(?) in order for the word $w$, for example if the word is $abab$ then the possible prefixes/suffixes are

$$\big\{\emptyset, \{a\}, \{ab\}, \{aba\}, \{abab\}\big\}$$

However, I'm having a hard time actually proving this in a mathematical way. What I assume is that for a given finite automata, we'd have to turn all the states on a path to an accepting state into accepting states? Anyone care to help me prove these two?

答:
There are many ways to prove these kind of statements, but I think that it is useful to remember that a language $L \subseteq \{a, b\}^\ast$ can be recognized by a finite-state automaton if and only if it can be described as a regular expression.

We can then proceed by induction on the structure of regular expressions:

If $L$ is empty, $L = \{\varepsilon\}$ or $L = \{x\}$ for some $x \in \{a, b\}$ then result is trivial.

If L = $L_1.L_2$ (concatenation) then $Pref(L) = Pref(L_1) \cup L_1.Pref(L_2)$, both of which are regular languages by induction, and so their union is also a regular language.

If $L = L_1^\ast$ then ...

If $L = L_1|L_2$ then ...

In general, using a different characterization of regular languages gives another point-of-view, leading to some easier, clearer or more elegant proofs.


问:
Complex Integral : How can I evaluate $\int_{C_R} f(z) dz$?
I have to calculate $I$ ,using complex integral.
\begin{equation}
I:=\displaystyle\int_{-\infty}^{\infty} \dfrac{\text{log}{\sqrt{x^2+a^2}}}{1+x^2}.(a>0)
\end{equation}

Let $f(z)=\dfrac{\text{log}(z+ia)}{1+z^2}$.

$
C_R : z=Re^{i\theta}, \theta : 0 \to \pi.$

$C_1 : z=t, t : -R \to R.$

From Residue Theorem,
\begin{equation}
\displaystyle\int_{C_1} f(z) dz + \displaystyle\int_{C_R} f(z) dz = 2\pi i \text{Res}(f, i).
\end{equation}

If $R \to \infty$, $\displaystyle\int_{C_1} f(z) dz  \to \displaystyle\int_{-\infty}^{\infty} f(x) dx. $

And my mathematics book says that if $R \to \infty$, $\displaystyle\int_{C_R} f(z) dz \to 0.$ I cannot understand why this holds.

My attempt is following :

\begin{align}
\Bigg|\displaystyle\int_{C_R} f(z) dz \Bigg|
&=\Bigg|\displaystyle\int_0^{\pi} f(Re^{i\theta}) i Re^{i\theta} d\theta \Bigg| \\
&\leqq \displaystyle\int_0^{\pi} \Bigg| R \dfrac{\text{log} (Re^{i\theta} + ia)}{1+R^2e^{2i\theta}} \Bigg| d\theta \\
&\leqq \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (Re^{i\theta}+ia) \Bigg| d\theta \\
&= \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (R\cos \theta+i(R\sin \theta +a)) \Bigg| d\theta
\end{align}

I expect that
$ \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (R\cos \theta+i(R\sin \theta +a)) \Bigg| d\theta \to 0$,
but I cannot prove this.

I would like to give me some ideas.

Body:
$$I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx=
\int_0^{\infty} \frac{\ln(x^2+a^2)}{1+x^2}dx$$
$$\frac{dI}{da}=\int_0^{\infty} \frac{2a}{(1+x^2)(x^2+a^2)}dx$$
$$\int \frac{2a}{(1+x^2)(x^2+a^2)}dx=\frac{2}{a^2-1}\tan^{-1}(\frac{x}{a})-\frac{2a}{a^2-1}\tan^{-1}(x)$$
$$\frac{dI}{da}=\frac{2}{a^2-1}\frac{\pi}{2}-\frac{2a}{a^2-1}\frac{\pi}{2}$$
After simplification :
$$\frac{dI}{da}=\frac{\pi}{a+1}$$
$$I(a)=\pi\ln(a+1)+C$$
In order to determine $C$ we compute the integral for a particular value of $a$, for example $a=0$ :
$$I(0)=\int_0^{\infty} \frac{\ln(x^2)}{1+x^2}dx$$
Change of variable $\quad x=\frac{1}{t}$ :
$$I(0)=\int_{\infty}^0 \frac{\ln(\frac{1}{t^2})}{1+\frac{1}{t^2}}(-\frac{dt}{t^2})=
-\int_0^{\infty} \frac{-\ln(t^2)}{t^2+1}(-dt)=-\int_0^{\infty} \frac{\ln(t^2)}{t^2+1}dt$$
This implies
$$\int_0^{\infty} \frac{\ln(x^2)}{1+x^2}dx=-\int_0^{\infty} \frac{\ln(t^2)}{1+t^2}dt=0\quad\text{thus}\quad I(0)=0.$$
$$I(0)=\pi\ln(0+1)+C\quad\implies\quad C=0$$
$$\boxed{I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx=\pi\ln(a+1)}$$


问:
Rewriting equation through division
I have an equation $$\vec E = \frac{1}{4 \pi \epsilon_{0}} * \frac{Q}{x * \sqrt{x^2+a^2}}$$

where $\lambda = Q / 2a$

And this equation is rewritten into  $$\vec E = \frac{1}{2 \pi \epsilon_{0}} * \frac{\lambda}{x * \sqrt{(x^2/a^2)+1}}$$

I tried to divide it by $a$ but couldn't continue. I need some explanation on how it is rewritten.

答:
$$\begin{align}
\vec E & = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{x \cdot \sqrt{x^2+a^2}} \qquad [Q = 2a\lambda]\\[6pt]
& = \frac{1}{4 \pi \varepsilon_{0}} \frac{2a\lambda}{x \cdot \sqrt{x^2+a^2}} \\[6pt]
& = \frac{1}{2 \pi \varepsilon_{0}} \frac{\lambda}{\frac{x}{a} \cdot \sqrt{x^2+a^2}} \\[6pt]
& = \frac{1}{2 \pi \varepsilon_{0}} \frac{\lambda}{x \cdot \sqrt{\frac{x^2}{a^2}+\frac{a^2}{a^2}}}
\end{align}$$

Title: Rewriting equation through division


问:
Find inverse such that $A = A^{-1}$
I'm asked to find coefficients a, b and c for the following matrix $ A = \begin{pmatrix} a & b \\ c & 0 \end{pmatrix}$ such that it has an inverse $A^{-1}$ and such that $A^{-1} = A$. How would start on such a problem? I've tried calculating the inverse of the matrix but I have no idea where to go from there.

答:
$A^{-1}$ exists $\Rightarrow$ $Det(A)=-bc \neq 0$ which means both $b$ and $c$ can not be $0$

$A^{-1} = A \Rightarrow A^2=I$

$A = \begin{pmatrix} a & b \\ c & 0 \end{pmatrix} \Rightarrow A^2=\begin{pmatrix}a^2+bc &ab \\ ac & bc\end{pmatrix}=I$

$\begin{pmatrix}a^2+bc &ab \\ ac & bc\end{pmatrix}=I \Rightarrow bc=1$ and $a=0$

Title: Find inverse such that $A = A^{-1}$


问:
No. of ways of selecting 3 squares when they do not lie in same row, column, or diagonal

Total number of ways of selecting 3 small squares on a normal chess board so that they don’t belong to the same row, column or diagonal line, is equal to:

No. of ways of selecting 3 squares when they do not lie in same row or same column $=(64×49×36)×\frac{1}{3!}=18816$

Total ways of selecting the squares when they lie in the same diagonal line $=2({8 \choose 3} + {7 \choose 3} +{6 \choose 3}+{5 \choose 3} + {4 \choose 3}+ {3 \choose 3})=392$

But this only counts the cases when all three are in the same diagonal. How to count the cases when two of the three lie in the same diagonal?

答:
I will count the number of ways of selecting three named positions $X,Y,Z$ such that no two are in the same row or column under various conditions. The counts differ slightly according to the parity of $n$ and so I will assume that $n$ is even.

$A$. No other conditions.

$B$. $X$ and $Y$ are on the same diagonal.

$C$. $X,Y$ and $Z$ are on the same diagonal.

$D$. $XY$ and $XZ$ are on perpendicular diagonals.

Inclusion/exclusion

$B-C-2D$ counts the number of ways that $X$ and $Y$ are on the same diagonal and that $Z$ is not on the same diagonal as either.

$C+3D$ counts the number of ways that each of $X,Y,Z$ is on a diagonal with at least one other.

Then the total number of possibilities with no two of $X,Y,Z$ on the same diagonal is $$A-3(B-C-2D)-(C+3D)=A-3B+2C+3D.$$ Each configuration occurs $3!$ times because of the naming of the positions and so the answer requested in the post is given by $$\frac{1}{6}(A-3B+2C+3D).$$
A

$$A=n^2(n-1)^2(n-2)^2.$$
For $n=8$ this is $112896$.

B

The diagonals with one orientation are of lengths $1,2,...,n-1,n,n-1, ...,2,1$. Therefore the number of ways of choosing two positions on one of these diagonals is$${2 \choose 2} + {3 \choose 2} +...+{n-1 \choose 2}+{n \choose 2} + {n-1 \choose 2}+... {3 \choose 2}+{2\choose 2}=\frac{1}{6}n(n-1)(2n-1).$$ This must be multiplied by the number of orientations, $2$, the number of ways of naming $X/Y$, also $2$, and the number of ways of choosing the position of $Z$, $(n-2)^2$.
$$B=\frac{2}{3}n(n-1)(n-2)^2(2n-1) .$$
For $n=8$ this is $20160$.

C

The  number of ways of choosing three positions on one of the sets of parallel diagonals is$${3 \choose 3} +...+{n-1 \choose 3}+{n \choose 3} + {n-1 \choose 3}+... {3 \choose 3}=\frac{1}{12}n(n-1)^2(n-2).$$
This must be multiplied by the number of orientations, $2$, and the number of ways of naming $X,Y,Z$, $6$. $$C=n(n-1)^2(n-2).$$ For $n=8$ this is $2352$.

D

The condition for no two of $X,Y,Z$ to be on the same row or column is simply that the legs of the L-shape must be unequal. There are then $8$ different orientations of the L-shape and $2$ ways of choosing $Y$ and $Z$.

I have not found a neat formula for the number of L-shapes  with a particular orientation but it should be a reasonable challenge to find one. The shapes are however easy to count and for $n=8$ the number is $100$ (split into $49$ for one colour of squares  and $51$ for the other).
$$D=1600.$$

Grand total

$$\frac{1}{6}(112896-3\times 20160+2\times 2352+3\times 1600)=10320.$$

Addendum

The formula for $D$ appears to be $\frac{1}{6}n(n-2)(5n^2-16n+8)$.


问:
What is the limit of $ n \cdot 0 $ as n approaches infinity?
What is the limit of $ n \cdot 0 $ as $n$ approaches infinity?
Is it undefined or is it just $0$?

答:
$n\cdot 0 = 0$ for all naturals.
Therefore $$\lim_{n\to \infty} n\cdot 0 = \lim_{n\to \infty} 0 = 0$$


问:
Why is the square of a normal matrix as well a normal matrix?
Or is it? Be $A$ a normal matrix and my question is if $A^2$ is as well a normal matrix?</p>

答:
A matrix $A$ is normal iff $AA^*=A^*A$. If $A$ is normal, then
$$A^2(A^2)^*=A^2(A^*)^2=AAA^*A^*=AA^*AA^*=A^*AAA^*=A^*AA^*A=A^*A^*AA=(A^2)^*A^2$$
thus $A^2$ is normal as well.</p>


问:
Connected Subspace lies entrirely in $X$
Hi i am reading Topology by Munkres and i have one doubt at the last line of lemma 23.2. It is written that

Hence Y must lie entirely in C or in D.

I can't understand the above quoted statement. Everything else i was able to understand . I am attaching the screenshot where the lemma and its full proof is illustrated and i have highlighted the above statement(that i was not able to understand).The theorem's statement reads as: If the sets C and D form a separation of X, and if Y is a connected subspace of X, then Y lies entirely within either C or D.

答: If we assume that $C\cap Y$ and $D\cap Y$ are nonempty, then we conclude contradiction (contradiction with fact that $Y$ is conected). So it must be case that $C\cap Y=\emptyset$ or $D\cap Y=\emptyset$. Wlog, lets say that $D\cap Y=\emptyset$. Then $Y\subset C$ because $C\cup D = X$.


答:
Here's an answer to a slightly more general limit

$ \displaystyle L(m, n) = \lim_{x \to \infty} x^{m} ( a^{\frac{1}{x^{n}}} -1) \tag*{} $

Where $ a>1 $ and $(m, n) $ are natural numbers.

$ \displaystyle L(m, n) =  \lim_{x \to \infty} x^{m} ( e^{ \frac{\ln(a)}{x^n}} -1) =  \lim_{x \to \infty} \sum_{r=1}^{\infty} \frac{x^{m-rn}}{r!}(\ln(a))^{r} \tag*{} $

Where we have used the series expansion of the exponential function in the step above. It should be now straight forward to observe that the given limit is equal to $0$ whenever $ m<n$, Equal to $\ln(a) $ when $m=n$ and the limit diverges to infinity whenever $m>n$.

Body:
$\lim\limits_{x \to \infty} {x}(a^{1/x}-1)$, such that $a > 1$

My Approach:

Since $x$ approaches infinity I presume we have to 'get rid' of the $x$ on the outside because the root $-1$ will be more than 0 but I have no idea how to reduce the $x$ so please help me out!


问:
Why are functions with vanishing normal derivative dense in smooth functions?
Question

Let $M$ be a compact Riemannian manifold with piecewise smooth boundary. Why are smooth functions with vanishing normal derivative dense in $C^\infty(M)$ in the $H^1$ norm?

Here I define $C^\infty(M)$ to be those functions which have all orders of derivative continuous on $M$ and smooth in its interior. For example, $(x\mapsto \sin(\pi x))\in C^\infty([0,1])$ but $(x\mapsto \sqrt{x})\notin C^\infty([0,1])$.

I have crossposted this to MathOverflow.

Background

This is inspired by my belief that the form domain of the Friedrichs extension of the Neumann Laplacian on $M$ is equal to $H^1(M)$. If my belief is wrong, I would certainly accept as answer a counterexample, preferably with some discussion/references.

Here are the approaches I'm exploring.

Given $u\in C^\infty(M)$ construct a function that agrees away from an $\epsilon$ neighborhood of $\partial M$, but has been modified to have zero normal derivative. This is described below, but runs into some trouble at corners and with smoothing.

Given $u\in C^\infty(M)$, find some $\eta$ supported on an $\epsilon$ neighborhood of $\partial M$ such that $\nabla\eta|_{\partial M}$ is equal to the projection of $\nabla u|_{\partial M}$ onto the normal direction and $\eta$ and $|\nabla\eta|$ uniformly bounded in $\epsilon$. Then $u - \eta$ will be the desired approximation and uniform boundedness will imply $\|u - (u-\eta)\|_{H^1}\to 0$. I'm not sure if this is a search for an integrable harmonic vector field or if it's a constrained optimization problem.

Simply show that any $u\in C^\infty(M)$ that is perpendicular to all smooth functions with vanishing normal derivative must be zero. In order to do this, I think it still runs into the same fundamental difficulty as the others, which is constructing functions with vanishing normal derivative supported on an $\epsilon$-neighborhood of the boundary.

(I'm also happy to simply have a reference to follow up. A reference for domains in $\mathbb{R}^n$ should be fine, too, and just a partition of unity away from a Riemannian manifold.)

Current Approach

The approach I'm considering right now is an elaboration on possibility (2) from my list. The sketch is: set the boundary condition that $\nabla\eta$ on the boundary be equal to the normal component of $\nabla u$ and find an integrable harmonic vector field $E$ satisfying that boundary condition. Then cut off $E$ so it is only supported in a neighborhood of $\partial M$, and let $\eta$ be so that $E = \nabla\eta$.

Intuition is that the maximum principle will control $|E|$ and $|\eta|$, so that $\eta$ will be bounded above in the $H^1$ norm by a constant times the volume of the $\epsilon$-neighborhood of $\partial M$.

Another approach is inspired by zhw's comment below: approximate $\nabla u$ among $L^2$ vector fields, multiply it by a cutoff function so it is supported in the interior of $M$, and then integrate it to approximate $u$. This should work with some tweaking in $[0,1]$ but I'm not sure how well it will work in general.

Older work

My approach has been as follows. The intuition is to take an arbitrary smooth function, restrict it to the complement of a collar neighborhood, then extend the restriction to the collar neighborhood so that the value is constant on inward-normal geodesics. However I'm running into issues at corners.

Let $\epsilon > 0$ be such that $\{p\in M\ |\ d(p,\partial M) < \epsilon\}$ is a collar neighborhood of $\partial M$. Let $e_\epsilon(p)$ for $p\in\partial M$ be the smaller of $\epsilon$ or the greatest time parameter such that the inward normal geodesic collides with no other inward normal geodesic. Let $N$ be the set of inward normal vectors on $\partial M$ whose lengths are no greater than $e_\epsilon$. The interior of the set $\operatorname{exp}(N)$ is foliated by geodesics. Edit- Note this is not necessarily true if the manifold has inward corners.

Given a smooth, continuous function $u$ on $M$, define $\bar{u}$ to be the restriction of $u$ in the complement of $\operatorname{exp}(N)$ and on each geodesic leaf of the interior of $\operatorname{exp}(N)$ define $\bar{u}$ to take the value that $u$ takes on the inward limit of that leaf.

Here's trouble. As $\bar{u}$ need not be smooth, I'd like to mollify it. Edit- I had a detail incorrect. A standard mollifier produces a function defined on compact subsets of the interior of $M$. So the mollifier has to be modified. One idea I'm following up on is varying the support of the mollifying function based on distance to $\partial M$. I'm skeptical, as varying the mollifying function will add another component to the gradient, but if it works I'll post as an answer.

答:
I believe the following construction works. Analogous to the $n=1$ case from the comments, the idea is to approximate $\nabla u$ by compactly supported $\mathbf{g}_n$ and construct a suitable sequence $u_n$ such that each $\nabla u_n = \mathbf{g}_n.$ I will use some results from the following reference.

Maz’ya, Vladimir G., Sobolev spaces. With applications to elliptic partial differential equations. Transl. from the Russian by T. O. Shaposhnikova, Grundlehren der Mathematischen Wissenschaften 342. Berlin: Springer (ISBN 978-3-642-15563-5/hbk; 978-3-642-15564-2/ebook). xxviii, 866 p. (2011). ZBL1217.46002.

Definition (1.1.8): We say $\Omega \subset \Bbb R^n$ is starshapred with respect to a ball $B \subset \Omega$ if for all $x \in B$ and $y \in \Omega,$ we have the segment $[x,y] \subset \Omega.$

It is shown in 1.1.9 (Lemma 1) that if $\Omega$ satisfies the cone property, then it can be written as a union of finitely many domains which are startshared with respect to a ball. Note this also holds for $C^{0,1}$ domains, and it should include any reasonable definition of a piecewise smooth boundary.

Theorem (1.1.10): If $\Omega$ is bounded and starshapred with respect to a ball $B_{\delta}$ and $u \in W^{1,p}(\Omega),$ then we have
$$ u(x) = \delta^{-n} \int_{B_{\delta}} \varphi\left(\frac{y}{\delta}\right) u(y) \,\mathrm{d}y + \int_{\Omega} \frac{\mathbf{f}(x;r,\theta)}{r^{n-1}} \cdot \nabla u(y)\,\mathrm{d}y $$
for almost every $x \in \Omega,$ where $r = |y-x|,$ $\theta = (y-x)r^{-1},$ $\varphi \in C^{\infty}_c(B_1),$ and $\mathbf{f}$ is smooth and $\Bbb R^n$-valued such that
$$ |\mathbf{f}(x;r,\theta)| \leq c \left(\frac{\operatorname{diam}(\Omega)}{\delta}\right)^{n-1}.$$

I will omit a detailed proof (which is in the text), but the idea is to fix a mollifier $\varphi \in C^{\infty}_c(B_1)$ and set $\varphi_{\delta}(y) = \delta^{-n} \varphi(y/\delta).$ Then for $u$ smooth we have the Taylor representation
$$ u(x) = u(y) + \int_0^1 (1-t) \sum_{i=1}^n D^iu(y+t(x-y))(x_i-y_i) \,\mathrm{d}t, $$
for all $x \in \Omega$ and $y \in B_{\delta}.$ Multiplying by $\varphi_{\delta}(y)$ and integrating over $y \in B_{\delta}$ gives
$$ u(x) = \delta^{-n} \int_{B_{\delta}} \varphi\left(\frac{y}{\delta}\right) u(y) \,\mathrm{d}y + \int_{B_{\delta}} \varphi_{\delta}(y)\int_0^1 (1-t) \sum_{i=1}^n D^iu(y+t(x-y))(x_i-y_i) \,\mathrm{d}t \,\mathrm{d}y, $$
and the second term can be made to be of the right form by switching to $(r,\theta)$ coordinates. If $\delta=1$ one obtains
$$ f_i(x;r,\theta) = - \theta_i \int_r^{\infty} \varphi(x+ \rho\theta) \rho^{n-1} \,\mathrm{d}\rho, $$
which one can check satisfies the desired properties.

Using the above we can prove the reslt in the case when $M = \Omega$ is starsharped with respect to some ball $B_{\delta}.$ Let $\mathbf{g}_n \in C^{\infty}_c(\Omega;\Bbb R^n)$ such that $\mathbf{g}_n \to \nabla u$ in $L^2.$ Then define
$$ u_n(x) = \delta^{-n} \int_{B_{\delta}} \varphi\left(\frac{y}{\delta}\right) u(y) \,\mathrm{d}y + \int_{\Omega} \frac{\mathbf{f}(x;r,\theta)}{r^{n-1}} \cdot \mathbf{g}_n(y)\,\mathrm{d}y. $$
A calculation gives $\nabla u_n = \mathbf{g}_n,$ so we have $L^2$ convergence of the derivatives. For $L^2$ convergence this is essentially proved in 1.1.11, but since the text is light on the details I will include a proof. Applying Hölder and Fubini we have
$$\begin{split} \int_{\Omega} |u(x)-u_n(x)|^2 \,\mathrm{d} x &=\int_{\Omega} \left| \int_{\Omega} \frac{\mathbf{f}(x;r,\theta)}{r^{n-1}} \cdot (\nabla u(y) - \mathbf{g}_n(y))\,\mathrm{d}y \right|^2 \,\mathrm{d} x  \\
&\leq\int_{\Omega}\int_{\Omega} \frac1{r^{1-n}} \,\mathrm{d} y \int_{\Omega} \left|\mathbf{f}(x;r,\theta) \cdot (\nabla u(y) - \mathbf{g}_n(y))\right|^2 \frac1{r^{n-1}}\,\mathrm{d}y \,\mathrm{d} x \\
&\leq C \int_{\Omega} \left(\int_{\Omega} |x-y|^{1-n} \,\mathrm{d} x\right)^2 |\nabla u(y) - \mathbf{g}_n(y)|^2 \,\mathrm{d}y,
\end{split}$$
which converges to zero as $n \to \infty$ noting that $\int_{\Omega} |x-y|^{1-n} \,\mathrm{d} x \leq C(\Omega)$ uniformly in $y.$ Hence it follows that $u_n \to u$ in $H^1(\Omega).$

For the general case we can using a patching argument; let $\{\Omega_i\}$ be a finite covering of $M$ by subsets which can be identified by means of a diffeomorphism with a starshaped domain in $\Bbb R^n.$ Let $\{\psi_i\}$ be a subordinate partition of unity, approximate $u$ on each $\Omega_i$ by $u_n^{(i)}$ as above and set $u_n = \sum_i \psi_i u_n^{(i)}.$ Note this is only a sketch as it depends on the precise formulation of a piecewise smooth boundary, but I will leave you to fill in the details.


问:
What would be the answer to $\int (\cos(\frac1 x)+ \frac1 x\sin(\frac1 x))dx$ for $x \neq 0$?
Well, that pretty much wraps up my question. But to give you some clues, I tried Geogebra and it said this Integral is undefined (I'm assuming since $x$ cannot be zero?). However, Symbolab does give me an answer $(x\cos(\frac1 x)+2C)$. However, the app fails to show the steps. Does anyone know if this answer is correct and, if so, how you go about solving the said integral? At first, I tried substituting $\frac1 x$, but that didn't seem to help much.

Thank you in advance for any help you can provide.

答:
You can always check indefinite integrals solutions by differentiating them. So, Symbolab gives you right answer.

If you want to solve it by yourself, then first make substitution $u=1/x$, and then solve each of summands by partial integrations. Or, you can recognizine derivate of product under integral.


答: With $u=1/x$ it becomes arguably easier to spot:$$\int(\tfrac{-1}{u^2}\cos u+\tfrac{1}{u}\cdot-\sin u)du=\tfrac1u\cos u+C.$$Then there's integration by parts:$$\int\cos\tfrac1xdx=x\cos\tfrac1x-\int\tfrac1x\sin\tfrac1xdx.$$Either way though, you pretty much have to spot the same thing regardless of technique.

Title: What would be the answer to $\int (\cos(\frac1 x)+ \frac1 x\sin(\frac1 x))dx$ for $x \neq 0$?

Body:
Well, that pretty much wraps up my question. But to give you some clues, I tried Geogebra and it said this Integral is undefined (I'm assuming since $x$ cannot be zero?). However, Symbolab does give me an answer $(x\cos(\frac1 x)+2C)$. However, the app fails to show the steps. Does anyone know if this answer is correct and, if so, how you go about solving the said integral? At first, I tried substituting $\frac1 x$, but that didn't seem to help much.

Thank you in advance for any help you can provide.


问:
Spectrum of operator $Ax(t)=x(-t)$ on $L_1(-1,1)$
My teacher gave me the homework with the following exercise:

Find the spectrum of an operator $Ax(t)=x(-t)$ on $X = L_1(-1,1)$.

According to the definition: spectrum is $\lambda$ such that

$Im(A-I\lambda)\neq X$

$Ker(A-I\lambda)\neq 0$

I can find $\lambda$ for the 2. condition:

$$
        x(-t) - \lambda x(t) = 0 \\
        x(-t) = \lambda x(t)\\
        A^2x(t) = x(t) = \lambda Ax(t) = \lambda^2x(t)
$$
So $\lambda = \pm 1$. For the $\lambda=1$,  $x(t)$ is an even function, for the $\lambda=-1$, $x(t)$ is
an odd function. So $\lambda = \pm 1$ in the spectrum of $A$.

But I can't find the solution of
$$
    x(-t)-\lambda x(t) = y(t)
$$
From which we can find the $\lambda$ for 1. condition.

答:
An operator that satisfies some minimal polynomial has spectrum consisting of its roots.

This follows from the spectral mapping theorem: If $p(T)=0$ then $$\{0\}=\sigma(p(T))=\{p(\lambda):\lambda\in\sigma(T)\}$$

In this case, $A^2x(t)=x(t)$, that is, $A^2-I=0$, so $\sigma(A)=\{\pm1\}$.

Concerning $\mathrm{im}(A-\lambda I)$, note that $$(A-\lambda I)(A+\lambda I)=(1-\lambda^2)I$$ So, if $\lambda\ne\pm1$, then both $A\pm\lambda I$ must be surjective and injective.

If $\lambda=\pm1$, then $A-\lambda I$ is not surjective; their images are the odd/even functions.


答:
In fact you are very close to the right answer. I would suggest you check Euler - Maclaurin formula first - plus, not minus in the second term, though it does not change the result that you got $(+\frac{1}{2})$: $$\sum\limits_{k=1}^nf(\frac{k}{n})=n\int_1^nf(t)dt+\frac{1}{2}\left(f(\frac{1}{n})+f(1)\right)+\sum\limits_{k=2}^{\infty}(\frac{1}{n})^{k-1}\frac{B_k}{k!}\left(f^{(k-1)}(1)-f^{(k-1)}(\frac{1}{n})\right)$$
Next, the integral in fact is $$n\int_0^1\frac{t^2}{1+\frac{1}{n}t}dt=n\int_0^1t^2dt-\int_0^1t^3dt+O(\frac{1}{n})$$ the second term here gives you additionally $-\frac{1}{4}$.

All together,
$$ \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$

Title: $\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$?

Body:
calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}.$$
Use Squeeze theorem we have
$$\frac{1}{n+1}\sum\limits_{k=1}^n(\frac{k}{n})^2<\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}<\dfrac{1}{n}\sum\limits_{k=1}^n(\frac{k}{n})^2$$
So $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\int_0^1x^2\mathrm{d}x=\frac{1}{3}.$$
Use $$\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}.$$
Hence $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right)=\frac{1}{2}.$$
If our method is correct, is there any other way to solve this problem? Thank you


问:
$(P\Lambda P^{-1}=T^2)~\implies~(\exists \Lambda'~\text{s.t.}~T=R\Lambda' R^{-1})$: $\;P,R\;$ Unitary Matrices
Let $T$ be a linear operator such that the operator $T^2$ is diagonalizable. Is $T$ necessarily diagonalizable?</p>

答:
No. Counterexamples are plentiful, but the mother of them all has to be the non-diagonalisable Jordan block
$$
T := \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix},
$$
which squares to $T^2 = 0$.</p>


问:
On Langrange Multipliers
I don’t understand why in the Lagrange multiplier formula the constraint has to have a gradient too.
$\nabla f(x_1,\dots,x_n)= \lambda \nabla g (x_1, \dots, x_n)$

Isn’t $g(x_1, \dots, x_n)=0 $ just a curve?

答:
First, note that even $g\left(x_1,\dots,x_n\right)=0$ that doesn't mean that  $\nabla g\left(x_1,\dots,x_n\right)=0$.

$\nabla g$ is a vector, and with parametar $\lambda$ we can check when this vector is proportional to the vector $\nabla f$. If you remember that $\nabla f$ represent direction in which function $f$ grows, then proportionality of $\nabla g$ and $\nabla f$ tell us that curve $g=0$ is tangent to the levels of $f$ and that means that we have the stationary point of $f$ on the curve $g=0$.

Reading Wikipedia page on Lagrange multiplier should give you more motivation for the $\lambda$.

Title: On Langrange Multipliers


问:
Functions $f:\mathbb{R}\to\mathbb{R}$ with a local minimum at every point
We say that $f:\mathbb{R}\to\mathbb{R}$ has a local minimum at every point, if for any point $x_0$ we can find an open interval containing $x_0$ such that $f(x)\ge f(x_0)$ for every point in the interval. Suppose $f$ is any function $\mathbb{R}\to\mathbb{R}$ satisfying this condition (so, in particular, it is not necessarily continuous).

It does not follow that $f$ is constant. For example, take $f$ to be 0 on the integers and 1 elsewhere.

So the question is prove that $f$ is constant except at (at most) countably many points, or find a counter-example.

答: Hint: Your example where $f$ is $0$ on the integers and $1$ elsewhere does not use the fact that the integers are a countable set, only that they are a closed set (i.e. that its complement is open)


问:
Operator from vector space to Hilbert space
Let X be a linear space and Y be a Hilbert space.$ A:X \rightarrow Y$ be a linear operator.

Let $ S_{y} = \{x\in X : \left\Vert Ax-y\right\Vert  \leq \left\Vert Au-y\right\Vert  \forall u\in X\} $.

Show that $ S_{y}$ is non empty if and only if y$\in R(A)+ R(A)^\perp$.

If y$\in R(A)+ R(A)^\perp$ , $\exists$ v such that y=Av+k where k$\in R(A)^\perp$ its easy to show that $v\in S_{y}$.

If $S_{y}$ is non empty , Im not sure on how to proceed to show that $ y\in $ $R(A)+R(A)^\perp$

答: Hint: $\|Ax-y\|^{2} \leq \|A(u+tx)-y\|^{2}$ for al $t$, for all $u \in X$. Expanding this and  choosing  a suitable $t$ show that $ \langle (Ax-y), Au \rangle=0$. Since this is true for each $u$ we see that $Ax-y \in R(A)^{\perp}$. Finally, $y=Ax+(y-Ax)$ finisihes the proof.


答:
How about $f(x) = -\lfloor x \rfloor$.  (see definition .)
This function has a local minimum at every point, but $\lim_{x \to +\infty} f(x) = -\infty$.

So the function is not "constant" except at countably many points.  It is "locally constant" except at countably many points.

As maritsm remarked, a "closed set" will do, instead of the countable set of the OP.  Let $C$ be the Cantor set, and let $f(x) = 0$ on $C$ and, for each interval of the  complement $\mathbb R \setminus C$, let $f(x)$ be a negative constant; possibly different constants for each such interval.
Now we could have $\inf f = -\infty$ and $f$ is locally constant except on the uncountable set $C$.

Title: Functions $f:\mathbb{R}\to\mathbb{R}$ with a local minimum at every point

Body:
We say that $f:\mathbb{R}\to\mathbb{R}$ has a local minimum at every point, if for any point $x_0$ we can find an open interval containing $x_0$ such that $f(x)\ge f(x_0)$ for every point in the interval. Suppose $f$ is any function $\mathbb{R}\to\mathbb{R}$ satisfying this condition (so, in particular, it is not necessarily continuous).

It does not follow that $f$ is constant. For example, take $f$ to be 0 on the integers and 1 elsewhere.

So the question is prove that $f$ is constant except at (at most) countably many points, or find a counter-example.


问:
Vector bundles of rank $q+k$ in $K(Y)$, where $Y$ is a projective scheme of dimension $q$ over some field.

Given a vector bundle $G$ of rank $q+k$ on a projective scheme $Y$ of dim $q$ over some field,show that there exists an injection $0\rightarrow O_Y^k\rightarrow G(n)$ for some sufficiently large $n$.

I met this question when considering $[G(n)]$ in $K(Y)$.Thanks for your help.

答:
Here we will be almost copying the proof of Theorem 8.18 (Bertini's theorem), Chapter II from Algebraic Geometry by Hartshorne.

Let us assume that the base field is $k$ and it is algebraically closed. Let $V_n = \Gamma(Y, G(n))$. Consider the evaluation map

$$\varphi_x : \wedge^p V_n \rightarrow \wedge^p G(n)_y,$$

where $y \in X$ and $G(n)_y$ is the notation for $G(n)_y/m_yG(n)_y$. We choose $n$ large enough so that $V_n$ generates $G(n)$, this implies in particular that $\varphi_x$ is surjective. Hence $\dim(\ker \varphi_x) = \dim \wedge^p V_n - \text{rk} \wedge^p G$. Define

$$B := \lbrace (x, v_1 \wedge \dots \wedge v_p) \in Y \times \mathbb{P}(V_n) \big| \text{ such that } \varphi_x(v_1 \wedge \dots \wedge v_p) = 0 \rbrace.$$

It is not difficult to see that these are the closed points of a Zariski closed set to which we give the induced reduced subscheme structure. The composite map

$$B \hookrightarrow Y \times \mathbb{P}(V_n) \rightarrow Y$$

is surjective with fibers having dimension $\dim(\ker \varphi_x) - 1$. Hence the dimension of the reduced noetherian scheme whose closed points is $B$ can be seen to be

$$\dim(\wedge^p V_n) - \binom{q+p}{p} - 1 + p.$$

This is strictly less than $\dim(\mathbb{P}(\wedge^p V_n))$ for any $p > 0$. We get that there exists $v_1, v_2, \dots, v_p \in V_n$ such that $\varphi_x(v_1 \wedge \dots \wedge v_p) \neq 0$ for any $x \in X(k)$. In particular it implies that $v_1, v_2, \dots, v_p$ are linearly independent sections of $G(n)$. This directly implies the existence of an injective map $\mathcal{O}^p_X \rightarrow G(n)$.


问:
show that $a_{n+874}=a_{n}$,if such $a_{n+2}=\left\lceil \frac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$

Let the sequence  $\{a_{n}\}$ be such that $a_{1}=1, a_{2}=100$, and  $$a_{n+2}=\left\lceil \dfrac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$$

Prove that the sequence $\{a_{n}\}$ is periodic.

I have used a computer and found the periodic is $T=874$, but how to prove it?

答:
N.B. This is not complete answer, but just extrapolation of the clue provided by @SangchulLee.

First of all we can rewrite the recurrence as:

$$a_{n+2}=\frac{4a_{n+1}}{3}-a_{n}+v_{n+1}$$
where
$$v_{n+1}=\begin{cases}
  1 & a_{n+1} = 0 \bmod 3\\
  \frac{2}{3} & a_{n+1} = 1 \bmod 3 \\
  \frac{4}{3} & a_{n+1} = 2 \bmod 3
\end{cases}$$

Now we define
$$A_{n}  =    \begin{bmatrix}
  a_{n+1} & a_{n} \\
  a_{n+2} & a_{n+1}
  \end{bmatrix}$$
$$B  =  \begin{bmatrix}
    0&1  \\
    -1&\frac{4}{3}
    \end{bmatrix}$$
$$U_1 = \begin{bmatrix}
    0&0  \\
    0&1
    \end{bmatrix}$$

$$U_2 = \begin{bmatrix}
    0&0  \\
    1&0
    \end{bmatrix}$$
This means
$$A_{n+1}  =  BA_{n}+v_{n+1}U_1+v_{n+2}U_2$$
Then we can write
$$\det(A_{n+1}-v_{n+2}U_2) = \det(A_{n}+v_{n+1}B^{-1}U_1) $$

Simplifying this one obtains
$$a_{n+2}^2-\frac{4a_{n+2}a_{n+1}}{3}+a_{n+1}^2-v_{n+1}(a_{n+2}+a_{n+1})  =  a_{n+1}^2-\frac{4a_{n+1}a_n}{3}+a_n^2-v_{n+1}(a_{n+1}+a_n)$$

Notice that the equation $x^2-4xy/3+y^2-v(x+y)=C$ represents an ellipse for a given constant $C$ and $v\in\{1,2/3,4/3\}$

I was not able to proceed from here! Hope this helps although its a little late!


问:
How to find the average compound interest for all years, knowing the starting and ending values, as well as the number of years
For example:

Company X shares in 2016 - 38 USD

5 years later, in 2021 - 66.95 USD

38 + 12% (4.56 USD) + 12% (5.10 USD) + 12% (5.72 USD) + 12% (6.40 USD) + 12% (7.17 USD) = 66.95 USD

What formula should I use to know that the average compound interest is 12% for each year?

答:
We have:

$38(1+0.12)^5=66.96$

The general formula is:

$S=S_1(1+i)^n$

Where $S_1$ is initial investment, i is interest rate and n is number of years for investment. Total compound interest rate is:

$I=(1+0.12)^5-1=0.762$

The average rate of interest is:

$\frac{0.762}5=0.152=15.2$%


问:
expectation of cauchy distribution when power less than 1
How do I prove that $E|X|^p$, p<1 is less than infinity when $X$ is Cauchy distributed?

The pdf is $f(x) = 1/\pi(1+x^2)$.

When $p=1$, EX integrates nicely to $\frac{1}{2\pi} \ln(1+x^2)$, but its not so nice when eg $p=1/2$. Then we get something like $\ln \left|2x+2^{1+\frac{1}{2}}x^{\frac{1}{2}}+2\right| - arctan(...)$

Body: $E|X|^{p}=\int\frac {|x|^{p}} {\pi (1+x^{2})}dx\leq \int_{-1}^{1} \frac 1 {\pi} |x|^{p}dx+\frac 1{\pi} \int_{|x|>1} \frac 1  {|x|^{2-p}}dx<\infty$.


问:
Angle subtended by circular segment on external point
I'm trying to analytically calculate the angle subtended by a circular segment on an external point (external to the circle corresponding to the circular segment). Could someone please help me out here?

AngleSubtendedByCircularSegment

What I've figured out currently is that the angle subtended by any external point on the circular segment will either

Go through both the cut off points of the circular segment

Go through one cutoff point and a tangent to the circle

Go through tangents to the circle.

I'm kinda lost after this step. Any help will be much appreciated.

(In the problem I have at hand, the external points lie on a circle as well. So I need to figure out the angle subtended by a circular segment on points on an external circle. A more detailed image is here AngleSubtendedByCircularSegmentDetailed)

AnotherExample

More content added for context -

This problem comes from another problem I'm trying to solve where I have two concentric spheres, with a fixed outer and inner radius. There are photons that start on the outer sphere in an isotropic direction inwards (i.e., the photons only radiate inwards into the outer sphere, not outwards. Think of every point on the inner side of the outer sphere as a point source radiating in all directions inwards. Image attached for reference here).

The inner sphere is filled with water to a certain height (so the chord is indeed parallel to the x-axis). The problem is to find the fraction of photons that reach the water.

The case where the water completely fills the inner sphere is relatively straightforward to solve. Image here. In this case, I find the solid angle subtended by an arbitrary point on the inner sphere (any arbitrary point works since it is symmetric) and divide that by 2$\pi$ to get the fraction (dividing by $2\pi$ cause the photons only go inside). Using some geometry (solid angle formula for a cone given the angle), I get a value of $1 - \frac{\sqrt{r_2^2-r_1^2}}{r_2}$. So this fraction of photons starting on the outer sphere reaches the inner sphere.

Now the same problem is to be solved but for a partially filled inner sphere. It becomes slightly more complicated because the cone subtended by the inner sphere on the outer sphere has a solid angle that varies with theta. It's symmetrical in phi, so the problem can be reduced to a 2D problem. But the problem still persists.

答:
Say, $R$ and $r$ are radii of the outer and inner circles resp. and
$OE = FH = h$.

We want to find angle subtended by point $C$ on a horizontal chord $AB$, which is $\angle BCA$.

$\angle BCA (\alpha + \beta) = \angle BCH (\alpha) + \angle ACH (\beta)$

$AE = BE = \sqrt{OA^2 - OE^2} = \sqrt{r^2-h^2}$

If $\angle COD = \theta, $ then $\angle OCF = \theta$

$CF = R\cos\theta,$ so $CH = R\cos\theta - h$

$OF = R\sin\theta = EH, $

So $AH = \sqrt{r^2-h^2} + R\sin\theta, BH = \sqrt{r^2-h^2} - R\sin\theta$

$\alpha = \tan^{-1} \big(\frac{BH}{CH}\big) = \tan^{-1}\bigg(\frac{\sqrt{r^2-h^2} - R\sin\theta}{R\cos\theta - h}\bigg)$

$\beta = \tan^{-1} \big(\frac{AH}{CH}\big) = \tan^{-1}\bigg(\frac{\sqrt{r^2-h^2} + R\sin\theta}{R\cos\theta - h}\bigg)$

$\angle BCA = \alpha + \beta$


问:
linear space W which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$
Give an example of a linear space W of finite dimension $d \geq 4$ which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$.

Is that possible? I mean the only possibility of $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$ that I can think of is $V_1=V_2=V_3$ but then $V_1⊕V_2=V_1⊕V_3=V_2⊕V_3 = \emptyset$. It's probably something basic and I wasted way to much time thinking of it.

答:
Does this work? Let $\{e_1,e_2,e_3,e_4\}$ be a basis for a 4-dimensional $V$, then take

$V_1=\langle e_1+e_2,e_1+e_3\rangle$

$V_2=\langle e_4+e_2,e_4-e_3\rangle$

$V_3=\langle e_1,e_4\rangle$.

It's not difficult to show that $V_i\oplus V_j=V$ for $i\neq j$. Also not difficult to show that $V_1\cap V_3=V_2\cap V_3=\emptyset$. The only difficult bit is to show that $V_1\cap V_2=\emptyset$, which I haven't done here.


答:
If $V_1\oplus V_2 = W$ then $\dim V_1 +\dim V_2 = \dim W$. Similarly, $\dim V_2 +\dim V_3 = \dim V_3 +\dim V_1 = \dim W$, and we can conclude that $\dim V_1 =\dim V_2 =\dim V_3 = n$ and $\dim W = 2n$.

We can easily solve problem when $n=1$: we can take $V_1=\mathcal L\{(1,0)\}$, $V_2=\mathcal L\{(1,1)\}$ and $V_3=\mathcal L\{(0,1)\}$. You can easily see that $V_1\oplus V_2 = V_2 \oplus V_3 = V_3 \oplus V_1 = \mathbb R^2$.

Now we can solve problem for $n=2$, by taking $V'_1=V_1\times V_1$, $V'_2=V_2\times V_2$ and $V'_3=V_3\times V_3$. Then $V'_1\oplus V'_2 = V'_2 \oplus V'_3 = V'_3 \oplus V'_1 = \mathbb R^4$ holds.

Title: linear space W which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$

Body:
Give an example of a linear space W of finite dimension $d \geq 4$ which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$.

Is that possible? I mean the only possibility of $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$ that I can think of is $V_1=V_2=V_3$ but then $V_1⊕V_2=V_1⊕V_3=V_2⊕V_3 = \emptyset$. It's probably something basic and I wasted way to much time thinking of it.


问:
Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $
Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $

I did these steps:

Basis step

$$P(10): \ \ 1024 \geq 1000 \ (True)$$

Inductive step
$$P(n) \implies P(n+1) \\P(n+1) = 2^{n+1} \geq (n+1)^3$$
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3  \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.

does this demonstration work?

I know there's a similar question but the solution is different, I want to know if my demonstration is valid as well.

答:
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3  \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.

does this demonstration work?

No, your analysis (in and of itself) hasn't shown that

$(n+1)^3 \leq 2^{n+1}$, assuming that I am not overlooking anything.  This is what you are trying to prove.

The easy way, is:

after noting that $2^{10} > 10^3$, and assuming that $2^n > n^3$, simply note that for $n \geq 10$,

$$\left(\frac{n+1}{n}\right)^3 < 2. \tag1$$

Note that $2^{1/3} > 1.25$, and for $n \geq 10, \frac{n+1}{n} < 1.25$.

This justifies equation (1) above.

Thus:

$$2^{n+1} = 2 \times 2^n > 2 \times n^3 > \left(\frac{n+1}{n}\right)^3 \times n^3 = (n+1)^3.$$


问:
$f'(a) \gt 0$ and $f'$ is continuous at $a$, then there $I_a$ where f is increasing.
I have a little problem with a proof, I managed to get to a result, but I think I got there committing some mistakes.

So, i have to prove that if

$f'(a) \gt 0$

$f'$ is continuous at $a$

Then

$f$ is increasing in some interval containing $a$

So i can say there is an interval where $f'$ is positive using the intermediate value theorem for continuous functions, as it is also said in this other post:

But i don't know how to rigorously say $f$ is increasing in the interval:

I think i can use the definition of derivative:

$$\lim_{h\to 0} \frac{f(a+h)-f(a)}{h} \gt 0$$

Then it follows:

$$\lim_{h\to 0} \frac{f(a+h)}{h} \gt \lim_{h\to 0}\frac{f(a)}{h}$$

Case $h \gt 0$

$$\lim_{h\to 0} f(a+h) \gt \lim_{h\to 0}f(a)$$

Case $h \lt 0$

$$\lim_{h\to 0} f(a+h) \lt \lim_{h\to 0}f(a)$$

I choose $b = a +h$ and then I have $$\lim_{b\to a} f(b) \gt \lim_{b\to a}f(a)$$

Is that sufficient to end the proof?

As last point I have to prove that the continuity of $f'$ at $a$ is essential, but I can't understand that... If I proceed with the defintion of derivative alone, can't I show that there is at least an interval (for
how small it could be) containing $a$ where "f" is increasing proceeding as I did before?

That's the reason because I think that the way I approached the problem is not correct.

答:
I will give a hint that if $f'(a)=u > 0 $ then using the same continuity argument you used to find an interval where the derivative is positive you can restrict the interval you choose a little more so that $f'(x)>\frac{u}{2}>0$ on the interval.

The next steps will depend on what techniques you are allowed to use without proof.

Certainly if you can use the Fundamental Theorem of Calculus and comparison results the solution will follow quickly from there.


答:
Knowing there is an interval $I_a$ surrounding $a$ where $f' > 0$.  To show $f$ is increasing on $I_a$.  The rigorous way to do this is the Mean Value Theorem.  Let $u, v \in I_a$, $u<v$.  Note $f$ is continuous on $[u,v]$, differentiable on $(u,v)$.  By the MVT there is $w$ with $u < w < v$ where
$$
\frac{f(v)-f(u)}{v-u} = f'(w)
$$
thus
$$
\frac{f(v)-f(u)}{v-u} > 0\qquad\text{so}\qquad f(v)-f(u) > 0 .
$$

Title: $f'(a) \gt 0$ and $f'$ is continuous at $a$, then there $I_a$ where f is increasing.

Body:
I have a little problem with a proof, I managed to get to a result, but I think I got there committing some mistakes.

So, i have to prove that if

$f'(a) \gt 0$

$f'$ is continuous at $a$

Then

$f$ is increasing in some interval containing $a$

So i can say there is an interval where $f'$ is positive using the intermediate value theorem for continuous functions, as it is also said in this other post:

But i don't know how to rigorously say $f$ is increasing in the interval:

I think i can use the definition of derivative:

$$\lim_{h\to 0} \frac{f(a+h)-f(a)}{h} \gt 0$$

Then it follows:

$$\lim_{h\to 0} \frac{f(a+h)}{h} \gt \lim_{h\to 0}\frac{f(a)}{h}$$

Case $h \gt 0$

$$\lim_{h\to 0} f(a+h) \gt \lim_{h\to 0}f(a)$$

Case $h \lt 0$

$$\lim_{h\to 0} f(a+h) \lt \lim_{h\to 0}f(a)$$

I choose $b = a +h$ and then I have $$\lim_{b\to a} f(b) \gt \lim_{b\to a}f(a)$$

Is that sufficient to end the proof?

As last point I have to prove that the continuity of $f'$ at $a$ is essential, but I can't understand that... If I proceed with the defintion of derivative alone, can't I show that there is at least an interval (for
how small it could be) containing $a$ where "f" is increasing proceeding as I did before?

That's the reason because I think that the way I approached the problem is not correct.


问:
Show that $P$ lies inside an even number of triangles with vertices among $A_1 , . . . , A _{2m}$

Assume a convex $2m$-gon $A_1 , . . . , A _{2m}$ . In its interior we choose a point $P$ , which
does not lie on any diagonal. Show that $P$ lies inside an even number of triangles
with vertices among $A_1 , . . . , A _{2m}.$

This is from Problem solving strategies, Invariance chapter. I tried to find the invariance, but couldn't get one till now.

Any hints would be helpful.

答:
Let the point be outside. In this case, it lies inside even number of triangles i.e., zero.
Now move that point through every region in such a way its locus does not cross any intersection point. Suppose that point crosses $A_iA_j$ then it goes out of all triangles of form $A_iA_jA_k$ and moves inside all triangles of form $A_iA_jA_m$ where $k$ and $m$ are on opposite sides of $A_iA_j$. As the polygon has even number of vertices, parity of number of triangles on each side is same, so no change in parity of number of triangles that have that point inside them. As initially it was even, it will always remain even.


答:
You already proved the base case

Now suppose that $n^3\le 2^n$  is true and

let's prove it for $n+1$

$(n+1)^3=n^3+\left(3n^2+3n+1\right)\tag{1}<\ldots$

for $n\ge 4$ we have $3n^2+3n+1 < n^3$ indeed, adding $n^3$ to both sides we get

$n^3+3 n^2+3 n+1<n^3+n^3\to (n+1)^3<2n^3\to n+1<n\sqrt[3]{2}\to n(\sqrt[3]{2}-1)>1\to n>\frac{1}{\sqrt[3]{2}-1}\approx 3.8$

continue $(1)$ $$\ldots<n^3+n^3\le 2^n+2^n=2^{n+1}$$

Body:
Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $

I did these steps:

Basis step

$$P(10): \ \ 1024 \geq 1000 \ (True)$$

Inductive step
$$P(n) \implies P(n+1) \\P(n+1) = 2^{n+1} \geq (n+1)^3$$
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3  \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.

does this demonstration work?

I know there's a similar question but the solution is different, I want to know if my demonstration is valid as well.


问:
True or false, if $A^2 = \mathbb{I}$ and $B^2 = \mathbb{I}$, then $(AB)^{-1} = BA$
The title says it all, is $A^2 = \mathbb{I}$ and $B^2 = \mathbb{I}$, then $(AB)^{-1} = BA$ valid or not. The question is more per say on how I would solve this question (rather than just being spoon fed)

I have no idea how I should tackle this problem.

答:
$A^{-1}$ is a matrix for such $AA^{-1} = A^{-1}A = \mathbb{I}$. That's the definition of inverse matrix. If you are given that $A^2=\mathbb{I}$, you can also write it as $AA=\mathbb{I}$ From here, you see that $A=A^{-1}$. The same holds for $B$.

You can also figure out what is $(AB)^{-1}$ from this definition.$$
(AB)(AB)^{-1} = \mathbb{I}\\
ABB^{-1}A^{-1} = A\mathbb{I}A^{-1}=AA^{-1}=\mathbb{I}
$$
making these equal:$$
(AB)(AB)^{-1} = (AB)(B^{-1}A^{-1})
$$
And you get that $(AB)^{-1}=B^{-1}A^{-1}$. And we know that $A=A^{-1}$ and $B=B^{-1}$, so we know that $B^{-1}A^{-1} = BA$. And you're done.


问:
If there is a 30% chance of rain for both Saturday and Sunday, what is the probability there would be rain on both days?
Can someone give me a formula for this question?

Wanda said there was a 30% chance of rain for both Saturday and Sunday. It rained on both days.

If her calculations were correct and there was a 30% chance of rain each day, what is the probability there would be rain on both days?

Please help!!!

答:
Here is a formula:
$P(A\text{ and }B)=P(A)P(B)$ for independent events $A$ and $B$.</p>


问:
How to motivate logarithm by breaking down $1/x$, into increasingly large segments, and proving that the area under the curve stayed the same?
James Stewart's Calculus 7e 2011 (not Early Transcendentals) presents the two ways to define a logarithm. First, on p 404 in § 6.3, Stewart defines $\log_ax$ as the inverse of $a^y$. Then on p 414 in § 6.4, he proves that $\frac d{dx} ln|x| = \frac 1x \iff \ln|x| + C = \int \frac 1x dx $.

Second, on p 421 in §6.2*, he first defines $\ln x = \int^x_1 \frac 1t \; dt, x > 0 $. Then on p 429 in §6.3*, he defines $e^x = y \iff \ln y = x$.

In either case, how does this comment add anything to the motivation or intuition behind $\log$?

I'm going to go back in time to the development of the logarithm. The log was a way of performing multiplication (in some base) by simply using addition. Meaning, every time you add one number, your output is proportional to the last output.

At some point in time shortly afterwards, mathematicians were working on integration techniques. They noticed that the basic methods of finding anti derivatives worked for everything except for 1/x. However it was realized, that if you break down the graph of 1/x, into increasingly large segments (1-2, 2-4, 4-8, 8-16...) That the area under the curve stayed the same. this is actually very easily provable! But now look what they found..... They found that the area under the curve increases by simple addition as you double the interval. (Meaning adding the area interval (1-2) + (2-4) + (4-8), is simply the same area added together 3 times).

This should stick out to you, because this is exactly how we defined a logarithm to work. Addition carrying out multiplication with some base number.

Because this logarithm seemed to be so "natural" (it just came about all by itself), it was called the "natural logarithm". So the relation was known that ln(x) = integral of 1/x, before it was known what that base number was. Euler was actually able to calculate the number, and the proof for that is very interesting as well! (See my paper for the proof).

答:
One of the most important properties of logarithms is that
$$
\log a+ \log b = \log ab \, .
$$
In fact, if we try to solve the functional equation
$$
f(a)+f(b) = f(ab) \tag{*}\label{*} \, ,
$$
then the solution is $f(x)=c\log x$, where $c$ is an arbitrary constant and $\log$ is the natural logarithm. Now consider the task of finding the area under the curve of $1/x$. As Stewart notes, 'if you break down the graph of $1/x$ into increasingly large segments ... the area under the curve stayed the same'. Symbolically,
$$
\int_{1}^{2} \frac{1}{t} \, dt \, = \int_{2}^{4}\frac{1}{t} \, dt = \int_{4}^{8}\frac{1}{t} \, dt = \ldots
$$
If we play with this equation a little, we find an intriguing way of computing areas. For example,
\begin{align}
\int_{1}^{4} \frac{1}{t} \, dt &= \int_{1}^{2}\frac{1}{t} \, dt +\int_{2}^{4}\frac{1}{t} \, dt \\[6pt]
&= \int_{1}^{2}\frac{1}{t} \, dt + \int_{1}^{2}\frac{1}{t} \, dt
\end{align}
More generally, let $n$ be a positive integer. Then,
\begin{align}
\int_{1}^{2^n} \frac{1}{t} \, dt &= \int_{1}^{2} \frac{1}{t} \, dt + \int_{2}^{4} \frac{1}{t} \, dt + \int_{4}^{8} \frac{1}{t} \, dt + \ldots + \int_{2^{n-1}}^{2^n} \frac{1}{t} \, dt \\[6pt]
&= \underbrace{\int_{1}^{2} \frac{1}{t} \, dt + \int_{1}^{2} \frac{1}{t} \, dt + \int_{1}^{2} \frac{1}{t} \, dt + \ldots + \int_{1}^{2} \frac{1}{t} \, dt}_{\text{$n$ terms}}
\end{align}
What does this have to with logarithms? Well, if we let
$$
f(x) = \int_{1}^{x}\frac{1}{t} \, dt \, ,
$$
then the above equation translates to
$$
f(2^n) = \underbrace{f(2) + f(2) + f(2) + \ldots + f(2)}_{\text{$n$ terms}}
$$
meaning that it appears that these integrals do have the property
$$
f(a) + f(b) = f(ab) \, !
$$
Further investigation suggests that this property is not unique to powers of $2$—it works for powers of $3$, and all other powers for that matter. It is therefore sensible to conjecture that
$$
f(a) + f(b) = f(ab) \, ,
$$
in other words that the integral
$$
\int_{1}^{x}\frac{1}{t} \, dt
$$
is a logarithm. And we can prove this in the following way:
$$
\int_{1}^{ab}\frac{1}{t} \, dt = \int_{1}^{a}\frac{1}{t} \, dt + \int_{a}^{ab}\frac{1}{t} \, dt
$$
Let $z=t/a$, meaning that $dz=\frac{1}{a}dt=\frac{z}{t}dt$. Then we have
\begin{align}
\int_{1}^{ab}\frac{1}{t} \, dt &= \int_{1}^{a}\frac{1}{t} \, dt + \int_{1}^{b}\frac{1}{z} \, dz \\[4pt]
f(ab) &= f(a) + f(b) \, .
\end{align}
Since the solution to $\eqref{*}$ is $f(x)=c\log x$,
$$
\int_{1}^{x}\frac{1}{t} \, dt = c\log x
$$
for some constant $c$. It is only a matter of time before we discover that $c=1$, providing us with the motivation behind defining
$$
\boxed{
\;\\[4pt]
\quad \log x = \int_{1}^{x} \frac{1}{t} \, dt \, . \quad
\\
}
$$
As indicated by the length of this post, the easiest way to make sense of this definition is to pick another starting point, and then discover that
$$
\log x = \int_{1}^{x} \frac{1}{t} \, dt
$$
as a theorem. Once we have done that, it makes sense to use the integral as an alternative way of defining the exponential function.


答:
Start with,
$$A^2 = \mathbb{I}$$
$$A\cdot{A} = \mathbb{I}$$
$$A\cdot\mathbb{I}\cdot{A} = \mathbb{I}$$
$$A\cdot{B^2}\cdot{A} = \mathbb{I}$$
$$(AB)\cdot{BA} = \mathbb{I}  $$
$$ BA = (AB)^{-1}$$

This is assuming, $AB$ is invertible i.e. $|AB|\ne0$.

Title: True or false, if $A^2 = \mathbb{I}$ and $B^2 = \mathbb{I}$, then $(AB)^{-1} = BA$

Body:
The title says it all, is $A^2 = \mathbb{I}$ and $B^2 = \mathbb{I}$, then $(AB)^{-1} = BA$ valid or not. The question is more per say on how I would solve this question (rather than just being spoon fed)

I have no idea how I should tackle this problem.


问:
Using generalized logistic curve to create a mathematical model from data.

The first row is time and the second row is height of a plant. We need to use generalized logistic curve to model the behavior of the plant. The equation of the logistic curve is :
$$N = \frac{N_*}{1+(N_*/N_0-1)e^{-a_0t}}$$ where $N_*$ is the maximum height of the plant or in other words, the supremum of the logistic function.

$a_0$ is how fast the function increases. My question is the following:

I know how to determine $N_*$, we just make up a number that is above $251$ and less than $251 + (251 - 247) $. But how to exactly determine $a_0$. I am quite lost here. How I would do it is to calculate every $a_0$ for every time and height, with equation $a_0 = \frac{1}{t}\ln(\frac{N}{N_0})$. In this equation $N$ would be the difference between heights, so let's say between $18$ and $33$ the $N$ for time equals $1$ would be $33 - 18$. Is the process here of creating a table of $\sum_{k = 1}^{10}(10 - i)$ column elements of differences, so that we take multiple $N_0$ until we run out of data from the primary table in the picture (so until $N_0 = 9$). Then we can use a formula for $a_0 = \frac{\sum a_{0l}*t_l}{\sum t_l}$ where $l$ is the number of elements.

Is my procedure for creating a logistic out of the data provided correct?

答:
Method 1: One degree of freedom

$\newcommand{\eqd}{\triangleq}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}$
$\newcommand{\pderiv}[2]{{\frac{\partial#1}{\partial#2} }}$
Have you considered using the Method of Least Squares to find $a_0$? For that method you can define an error function $$e(t_n)\eqd N(t_n)-y_n\eqd \frac{N_{*}}{1+\brs{N_{*}/N_0-1}e^{-a_0t_n}}-y_n$$ and a cost function (the error cost) as the norm squared of $e(t_n)$ as $$cost(a_0)\eqd\norm{e}^2\eqd\sum_{n=0}^{n=10}e^2(t_n)$$ You want to find the $a_0$ that minimizes that cost; that is, you want to find where (with respect to $a_0$) the cost function "goes to the lowest point". Suppose you set $N_0\eqd N(0)=18$ and $N_*\eqd252$ $\ldots$From the plot, it appears that $cost(a_0)$ is minimized around $a_0=0.65$.

To get a more accurate optimal $a_0$, we can differentiate $cost(a_0)$ with respect to $a_0$ and set the expression to $0$. To do that, this (lemma) may come in handy:
$$\begin{align}
\boxed{N'(t)}
  &\eqd \pderiv{}{a_0}N(t)
\\&\eqd \pderiv{}{a_0}\brs{\frac{N_{*}}{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0t}}}
  && \text{by definition of $N(t)$}
\\&= \frac{0-N_*\brs{\frac{N_*}{N_0}-1}e^{-a_0t}(-t)}{\brp{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0t}}^2}
  && \text{by Quotient Rule}
\\&= \frac{N_*^2}{\brp{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0t}}^2}
\brs{\frac{1}{N_0}-\frac{1}{N_*}}te^{-a_0t}
\\&\eqd \boxed{\brs{\frac{1}{N_0}-\frac{1}{N_*}}N^2(t)te^{-a_0t}}
&& \text{by definition of $N(t)$}
\end{align}$$
Then$\ldots$
$$\begin{align}
\boxed{0}&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\cdot0
\\&=\frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\norm{e}^2
\\&\eqd \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\sum_{n=0}^{n=10}e^2(t_n)
  && \text{by definition of $\norm{\cdot}$}
\\&\eqd \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\sum_{n=0}^{n=10}\brs{N(t_n)-y_n}^2
  && \text{by definition of $e$}
\\&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\sum_{n=0}^{n=10}2\brs{N(t_n)-y_n}N'(t_n)
  && \text{by Chain Rule}
\\&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\sum_{n=0}^{n=10}2\brs{N(t_n)-y_n}\brs{\frac{1}{N_0}-\frac{1}{N_*}}N^2(t_n)te^{-a_0t_n}
  && \text{by (lemma)}
\\&= \boxed{\sum_{n=0}^{n=10}N^2(t_n)\brs{N(t_n)-y_n}t_ne^{-a_0t_n}}
\\&\eqd Dcost(a_0)
  && \text{(call the sum $Dcost(a_0)$)}
\end{align}$$
Plotting $Dcost$ with respect to $a_0$, it appears that $Dcost(a_0)$ crosses $0$ at around $a_0=0.66$:

The uniroot function from the R stats package indicates that $Dcost(a_0)$ crosses $0$ at $a_0=0.6631183$ with estim.prec=6.103516e-05.

Using $N_0\eqd18$, $N_*\eqd252$, and $a_0\eqd0.6631183$, $N(t)$ seems to fit the 11 data points fairly well ($cost(0.6631183)=31.32307$) $\ldots$

Some R code supporting Reproducible Research:

```<code>#---------------------------------------
# packages
#---------------------------------------
#install.packages("stats");
 require(stats);
 rm(list=objects());

#---------------------------------------
# Data
#---------------------------------------
 tn = c(0:10)
 yn = c(18,33,56,90,130,170,203,225,239,247,251)
 t  = seq( from=min(tn), to=max(tn), length=1000 )

#---------------------------------------
# Estimate Function N(t)
#---------------------------------------
 N0=18
 Nh=252
 N = function(t,a) Nh / (1 + (Nh/N0-1)*exp(-a*t))

#---------------------------------------
# Cost Function
#---------------------------------------
 cost = function(x,tn,yn)
 {
  #sum((N(tn,x)-yn)^2)
   summ = 0;
   for (i in c(1:11))
   {
     summ = summ + (N(tn[i],x)-yn[i])^2
   }
   result = summ
 }

#---------------------------------------
# Partial derivative of Cost Function
#---------------------------------------
 Dcost = function(x)
 {
   summ = 0;
   for (i in c(1:11))
   {
     summ = summ + (N(tn[i],x))^2 * (N(tn[i],x)-yn[i]) * tn[i]*exp(-x*tn[i])
   }
   result = summ
 }

#---------------------------------------
# Find 0 crossing
#---------------------------------------
 x=seq(from=0.3, to=0.9, length=1000)
 aRoot = uniroot( Dcost, c(0.3, 0.9) )
 a0 = aRoot$root

#---------------------------------------
# Graphics
#---------------------------------------
 colors = c( "red" , "blue", "orange", "green"    );
 traces = c( "N(t)", "data", "cost"  , "Dcost"    );
 plot ( t , N(t, a0)      , col=colors[1], lwd=2, type='l', xlab="t", ylab="y", ylim=c(0,max(yn)+10) )
 lines( tn, yn            , col=colors[2], lwd=3, type='p' )
 lines( t , Dcost(t)      , col=colors[4], lwd=2, type='l' )
 lines( t , cost(t,tn,yn) , col=colors[3], lwd=2, type='l' )
 legend("topleft", legend=traces, col=colors, lwd=3, lty=1:1)
 grid()
</code>```

Body:
Method 1: One degree of freedom

$\newcommand{\eqd}{\triangleq}$
$\newcommand{\brp}[1]{{\left(#1\right)}}$
$\newcommand{\brs}[1]{{\left[#1\right]}}$
$\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}$
$\newcommand{\pderiv}[2]{{\frac{\partial#1}{\partial#2} }}$
Have you considered using the Method of Least Squares to find $a_0$? For that method you can define an error function $$e(t_n)\eqd N(t_n)-y_n\eqd \frac{N_{*}}{1+\brs{N_{*}/N_0-1}e^{-a_0t_n}}-y_n$$ and a cost function (the error cost) as the norm squared of $e(t_n)$ as $$cost(a_0)\eqd\norm{e}^2\eqd\sum_{n=0}^{n=10}e^2(t_n)$$ You want to find the $a_0$ that minimizes that cost; that is, you want to find where (with respect to $a_0$) the cost function "goes to the lowest point". Suppose you set $N_0\eqd N(0)=18$ and $N_*\eqd252$ $\ldots$From the plot, it appears that $cost(a_0)$ is minimized around $a_0=0.65$.

To get a more accurate optimal $a_0$, we can differentiate $cost(a_0)$ with respect to $a_0$ and set the expression to $0$. To do that, this (lemma) may come in handy:
$$\begin{align}
\boxed{N'(t)}
  &\eqd \pderiv{}{a_0}N(t)
\\&\eqd \pderiv{}{a_0}\brs{\frac{N_{*}}{1+\brp{\frac{N_*}{N_0}-1}e^{-a_0t}}}
  && \text{by definition of $N(t)$}
\\&= \frac{0-N_*\brs{\frac{N_*}{N_0}-1}e^{-a_0t}(-t)}{\brp{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0t}}^2}
  && \text{by Quotient Rule}
\\&= \frac{N_*^2}{\brp{1+\brs{\frac{N_*}{N_0}-1}e^{-a_0t}}^2}
\brs{\frac{1}{N_0}-\frac{1}{N_*}}te^{-a_0t}
\\&\eqd \boxed{\brs{\frac{1}{N_0}-\frac{1}{N_*}}N^2(t)te^{-a_0t}}
&& \text{by definition of $N(t)$}
\end{align}$$
Then$\ldots$
$$\begin{align}
\boxed{0}&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\cdot0
\\&=\frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\norm{e}^2
\\&\eqd \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\sum_{n=0}^{n=10}e^2(t_n)
  && \text{by definition of $\norm{\cdot}$}
\\&\eqd \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\pderiv{}{a_0}\sum_{n=0}^{n=10}\brs{N(t_n)-y_n}^2
  && \text{by definition of $e$}
\\&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\sum_{n=0}^{n=10}2\brs{N(t_n)-y_n}N'(t_n)
  && \text{by Chain Rule}
\\&= \frac{1}{2\brp{\frac{1}{N_0}-\frac{1}{N_*}}}\sum_{n=0}^{n=10}2\brs{N(t_n)-y_n}\brs{\frac{1}{N_0}-\frac{1}{N_*}}N^2(t_n)te^{-a_0t_n}
  && \text{by (lemma)}
\\&= \boxed{\sum_{n=0}^{n=10}N^2(t_n)\brs{N(t_n)-y_n}t_ne^{-a_0t_n}}
\\&\eqd Dcost(a_0)
  && \text{(call the sum $Dcost(a_0)$)}
\end{align}$$
Plotting $Dcost$ with respect to $a_0$, it appears that $Dcost(a_0)$ crosses $0$ at around $a_0=0.66$:

The uniroot function from the R stats package indicates that $Dcost(a_0)$ crosses $0$ at $a_0=0.6631183$ with estim.prec=6.103516e-05.

Using $N_0\eqd18$, $N_*\eqd252$, and $a_0\eqd0.6631183$, $N(t)$ seems to fit the 11 data points fairly well ($cost(0.6631183)=31.32307$) $\ldots$

Some R code supporting Reproducible Research:

#---------------------------------------
# packages
#---------------------------------------
#install.packages("stats");
 require(stats);
 rm(list=objects());

#---------------------------------------
# Data
#---------------------------------------
 tn = c(0:10)
 yn = c(18,33,56,90,130,170,203,225,239,247,251)
 t  = seq( from=min(tn), to=max(tn), length=1000 )

#---------------------------------------
# Estimate Function N(t)
#---------------------------------------
 N0=18
 Nh=252
 N = function(t,a) Nh / (1 + (Nh/N0-1)*exp(-a*t))

#---------------------------------------
# Cost Function
#---------------------------------------
 cost = function(x,tn,yn)
 {
  #sum((N(tn,x)-yn)^2)
   summ = 0;
   for (i in c(1:11))
   {
     summ = summ + (N(tn[i],x)-yn[i])^2
   }
   result = summ
 }

#---------------------------------------
# Partial derivative of Cost Function
#---------------------------------------
 Dcost = function(x)
 {
   summ = 0;
   for (i in c(1:11))
   {
     summ = summ + (N(tn[i],x))^2 * (N(tn[i],x)-yn[i]) * tn[i]*exp(-x*tn[i])
   }
   result = summ
 }

#---------------------------------------
# Find 0 crossing
#---------------------------------------
 x=seq(from=0.3, to=0.9, length=1000)
 aRoot = uniroot( Dcost, c(0.3, 0.9) )
 a0 = aRoot$root

#---------------------------------------
# Graphics
#---------------------------------------
 colors = c( "red" , "blue", "orange", "green"    );
 traces = c( "N(t)", "data", "cost"  , "Dcost"    );
 plot ( t , N(t, a0)      , col=colors[1], lwd=2, type='l', xlab="t", ylab="y", ylim=c(0,max(yn)+10) )
 lines( tn, yn            , col=colors[2], lwd=3, type='p' )
 lines( t , Dcost(t)      , col=colors[4], lwd=2, type='l' )
 lines( t , cost(t,tn,yn) , col=colors[3], lwd=2, type='l' )
 legend("topleft", legend=traces, col=colors, lwd=3, lty=1:1)
 grid()


问:
There are pawns at each of two ending cells of 1 × 20 strip
There are pawns at each of two ending cells of 1 × 20 strip. In a single step a player can shift one of the pawns towards the second one by one or two cells. It is forbidden to jump over a pawn. The player who cannot make a step loses the game. Who has a winning strategy - the player who starts the game or his opponent?</p>

答: The losing positions for the player to move are those in which the number of empty squares between the pawns is a multiple of $3$. So the second player wins.


问:
How to calculate limit $\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$ without using L'Hospital rule
I have to calculate next limit without using L'Hospital rule:
$$\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$$
I came this far with that $$1+\frac{1}{e}\lim_{t\rightarrow1}\frac{\ln(t)}{t-1}$$
I know that result for this limit using the L'Hospital rule is $1+\frac{1}{e}$
Any suggestions how to finish this?

答: If $f(x)=x+\log(x+e)$, then $f(0)=\log(e)=1$. Therefore, your limit is $f'(0)$, which is $1+\frac1e$.


答:
There are four possibilities:

Rain Saturday, Rain Sunday

Rain Saturday, No Rain Sunday

No Rain Saturday, Rain Sunday

No Rain Saturday, No Rain Sunday

You can calculate the probability of each of these four cases using @Zev's formula; note that their sum will be 100%.

Title: If there is a 30% chance of rain for both Saturday and Sunday, what is the probability there would be rain on both days?

Body:
Can someone give me a formula for this question?

Wanda said there was a 30% chance of rain for both Saturday and Sunday. It rained on both days.

If her calculations were correct and there was a 30% chance of rain each day, what is the probability there would be rain on both days?

Please help!!!


Title: How to calculate limit $\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$ without using L'Hospital rule

Body:
I have to calculate next limit without using L'Hospital rule:
$$\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$$
I came this far with that $$1+\frac{1}{e}\lim_{t\rightarrow1}\frac{\ln(t)}{t-1}$$
I know that result for this limit using the L'Hospital rule is $1+\frac{1}{e}$
Any suggestions how to finish this?

Body:
Using your approach, $$\lim_{t\to 1}\frac{\ln t}{t-1}=\lim_{t\to 1}\frac{\ln (1+(t-1))}{t-1}= \lim_{t\to 1}\ln\left(1+(t-1)\right)^{1/(t-1)}$$
Now,  recall that $\lim_{x\to 0}(1+x)^{1/x}=e$


问:
Finding the solution to $\frac{d^2 h(y)}{dy^2}-2y\frac{dh(y)}{dy}+(e-1)h(y)=0$
I have this diff equation:   $\frac{d^2 h(y)}{dy^2}-2y\frac{dh(y)}{dy}+(e-1)h(y)=0$

I am having trouble understanding the method of solution given by the textbook:

This is only part of the solution but i having trouble understating the step that led to (9.27).

How are we able to change the $s->s+1$ for just one term and not the whole term.

Why is $s=0$ in that term and not $s=1$

Is there a general rule for relabelling the dummy summation index that you could list

答:
The name of summing index is always dummy this means that
$$\sum_{s=0}^{\infty} 2s(2s-1) a_s y^{2s-2}= \sum_{t=0}^{\infty} 2t(2t-1) a_t y^{2t-2}$$
Now let the first term in (9.26) be
$$F=\sum_{t=0}^{\infty} 2t(2t-1) a_t ~y^{2t-2}$$
$$F= 0 a_0 y^{-2}+ \sum_{t=1}^{\infty} 2t(2t-1) a_t~ y^{2t-2}$$
Now put $t=s+1 \implies t-1=s $ to get
$$F=\sum_{s+1=1}^{\infty} 2(s+1)(2s+1) a_{s+1}~ y^{2s}=\sum_{s=0}^{\infty} 2(s+1)(2s+1) a_{s+1}~ y^{2s}$$

The main motivation was to have similar power of $y$ as $y^{2s}$in both terms of 9.26. Now replace the first term of 9.26 as above and get 9.27. Yhis is usually done in the series solution of ODEs.