问:
Arc length equation question
If you were given the following and were told to find $\dfrac{\text{d}y}{\text{d}x}$.

$y^2=2(x+4)^3$

How do you solve the problem when there is a $y^2$? When I tried solving the problem I got

$\dfrac{\text{d}y}{\text{d}x}= \dfrac{3x^2+24x+48}{y}$

I was trying to put the answer into the equation for finding the length of a curve which doesn't make sense because the equation is $\displaystyle \int_{a}^{b} \sqrt{1+\left(\dfrac{\text{d}y}{\text{d}x}\right)^2}\text{d}x$

My answer is a fraction with a $y$ variable on the bottom, but the formula at the end has a $\text{d}x$? How do you solve for $\dfrac{\text{d}y}{\text{d}x}$ so that there isn't a $y$ on the bottom fraction?

答:
Easy enough to square root both sides:
$$y=\pm\sqrt2(x+4)^{3/2}$$
Then differentiate:
$$y'=\pm\frac32\sqrt2(x+4)^{1/2}=\frac32\sqrt{2(x+4)}$$
where the sign depends on which branch you want to compute the arc length over.

Title: Arc length equation question


问:
Movie Earnings and Relative Ticket Sizes
Given that movie A made 308 million and a 1 by 3 cm ticket represented A and movie B made 609 million, assuming that prices have been adjusted for inflation and ticket B is the same shape as ticket A, what would the size of ticket B be? </p>

<p>I said if the area of A was 3 cm which represented 308 million than ticket B would just need to be about 1.97 $cm^2$ larger i.e. have an area of about 5.91. I got this because since its a graphical representation then ticket B should be proportional to ticket A. Not really sure if it's right so if anyone would check it out and let me know, thanks! </p>

答:
Movie A: $308$ Units, $1cm$ by $3$$cm$: Resultant Area: $3cm^2$

Movie A: $609$ Units

$\frac{609}{308} = 1.977$

$\therefore$ $Area\;B = 1.977 \cdot Area\;A$

$\therefore$ $Area\;B = 5.931$

You can then work out the dimensions of $Area\;B$. You know that the rectangles are proportional ("same shape"), so one side is three times as long as the other.  $x$ will be the length of one side, so $3x$ will be the length of another.

$x \cdot 3x = 5.931$

$3x^2 = 5.931$

$x = \sqrt{1.977}$

$x = \sqrt{1.977}$

$x = 1.4061$

$3x = 4.218$

$\therefore$ the dimensions of Ticket B will be $1.4061cm$ by $4.218cm$.

Graphical Representation in this way is a weird idea, but that should be right.

ContentLicense: CC BY-SA 3.0


问:
Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$
I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.

The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$

So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so that'd be a go-to. Also, since the degree of th enumerator is one less than the denominator we can maybe treat this as $du/u$ which would imply a natural log, though we don't quite know that yet.

So we decompose: I know that $x^2+3x-4$ can be factored as $(x-1)(x+4)$.

That gets me

$$\int{\frac{x}{(x-1)(x+4)}dx}$$
so I can do this:
$$\frac{Ax}{x-1}+\frac{B}{x+4} = \frac{x}{(x-1)(x+4)}$$
Which implies
$$A(x+4)+B(x-1) = x$$
and since the roots are at $x=1$ and $x=-4$, I can set it up like this:
$$5A = 1;-5B=1 \text{  and } A = \frac{1}{5}, B=-\frac{1}{5}$$

Leading to:

$$\frac{x}{5(x-1)}-\frac{1}{5(x+4)}$$
Which I can set up the integral like so:
$$\int{\frac{x}{5(x-1)}-\frac{1}{5(x+4)}dx}=\int{\frac{x}{5(x-1)}dx-\int{\frac{1}{5(x+4)}dx}}$$
I can now integrate by addition here  $$\int{\frac{x}{x-1}}dx=\int{\frac{x-1+1}{x-1}}dx=x+\int{\frac{1}{x-1}}dx=x-\ln(x-1)$$
And doing the same thing for the second term and bringing back my $\frac{1}{5}$
$$\frac{1}{5}(x-\ln(x-1)+\ln(x+4))$$

I suspect there is a further simplification I could do. On a problem like this I also saw it integrated as an arctangent, but that seemed needlessly complex? In any case I was curious if I did this correctly.

答:
Its wrong ! the step $$A(x+4)+B(x-1) = x$$ is not true instead you should have $$Ax(x+4)+B(x-1)=x$$ As you can see we would then have to take $A=0$ and the purpose of partial fractions is gone

I prefer $$\frac{x}{x^2-3x+4}=\frac{1}{5}(\frac{1}{x-1})+\frac{4}{5}(\frac{1}{x+4})$$

another way is to write $x=l(2x-3)+m$ (because derivative of $x^2-3x+4=2x-3$) then we have $$\int \frac{x}{x^2-3x+4}=\int \frac{2x-3}{2(x^2-3x+4)}+\int \frac{3}{2(x^2-3x+4)}$$ which is easy

ContentLicense: CC BY-SA 4.0


问:
Area of paraboloid on top of a triangle

Calculate the area of a paraboloid $z=y^2$ that lies on top of the triangle whose tips are located at the origin and $(0,1)$ & $(1,1)$.

The situation looks as follows:

However, I'm not entirely sure how I can integrate this to find the area of the "patch" that the $xy$-triangle bounds. Should I define equations for the different lines that the points on the triangle defines or what? Any hints are appreciated.

Title: Area of paraboloid on top of a triangle

Body:
$f(x,y)=y^2$

$$S=\iint_T\sqrt{f'_x{^2}+f'_y{^2}+1}\,dA$$
where $f'_x$ and $f'_y$ are the partial derivatives
$$\int _0^1\int _x^1\sqrt{(2 y)^2+0^2+1}\,dydx=\frac{1}{12} \left(5 \sqrt{5}-1\right)$$


Title: Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$

Body:
I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.

The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$

So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so that'd be a go-to. Also, since the degree of th enumerator is one less than the denominator we can maybe treat this as $du/u$ which would imply a natural log, though we don't quite know that yet.

So we decompose: I know that $x^2+3x-4$ can be factored as $(x-1)(x+4)$.

That gets me

$$\int{\frac{x}{(x-1)(x+4)}dx}$$
so I can do this:
$$\frac{Ax}{x-1}+\frac{B}{x+4} = \frac{x}{(x-1)(x+4)}$$
Which implies
$$A(x+4)+B(x-1) = x$$
and since the roots are at $x=1$ and $x=-4$, I can set it up like this:
$$5A = 1;-5B=1 \text{  and } A = \frac{1}{5}, B=-\frac{1}{5}$$

Leading to:

$$\frac{x}{5(x-1)}-\frac{1}{5(x+4)}$$
Which I can set up the integral like so:
$$\int{\frac{x}{5(x-1)}-\frac{1}{5(x+4)}dx}=\int{\frac{x}{5(x-1)}dx-\int{\frac{1}{5(x+4)}dx}}$$
I can now integrate by addition here  $$\int{\frac{x}{x-1}}dx=\int{\frac{x-1+1}{x-1}}dx=x+\int{\frac{1}{x-1}}dx=x-\ln(x-1)$$
And doing the same thing for the second term and bringing back my $\frac{1}{5}$
$$\frac{1}{5}(x-\ln(x-1)+\ln(x+4))$$

I suspect there is a further simplification I could do. On a problem like this I also saw it integrated as an arctangent, but that seemed needlessly complex? In any case I was curious if I did this correctly.

Body:
HINT

Here it is an alternative way to solve it:

\begin{align*}
\frac{x}{(x-1)(x+4)} & = \frac{(x-1) + 1}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{5}\times\frac{(x+4) - (x-1)}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{5}\times\left[\frac{1}{x-1} - \frac{1}{x+4}\right]\\\\
& = \frac{1}{5(x-1)} + \frac{4}{5(x+4)}
\end{align*}

Can you take it from here?


问:
$\epsilon$-$\delta$ proof
I found an interesting problem in my textbook, it asks to prove the following statement:

If $f'(x_{0}) >0$, then there is a $\delta >0$ such that $f(x)<f(x_{0})$ if $x_{0}- \delta < x <x_{0}$, and $f(x) > f(x_{0})$ if $x_{0} <x < x_{0} + \delta$.

How would one prove this from the definitions?

答:
Since $f'(x_0) > 0$, we can find $\delta > 0$ so that:
$$
|x - x_0| < \delta \implies \frac{f(x) - f(x_0)}{x - x_0} > 0
$$
(why?)

Now, the quotient above is positive iff the numerator and denominator have the same sign. Split this into two cases and handle each case separately to get the desired result.

Title: $\epsilon$-$\delta$ proof


问:
A problem in geometry(tangents) to show that $CD= 2AB \cdot (AC^2−AB^2)/(AC^2+AB^2)$
Suppose that $O$ is the circumcentre of Triangle $BCD$. Draw the tangent to the circumcircle of $BCD$ at $B$. Extend $CO$ meet this tangent at $A$. Show that

$$CD= 2AB \,\frac{AC^2−AB^2}{AC^2+AB^2}$$

What I thought was to apply Cevas theorem or Menelaus theorem. But things are not working out.I need some little help. I can't show what I have done as absolutely other than construction I could not do anything.
So I would be obliged if anyone can help.

答:
Showing why this is not necessarily true.

$\angle OCB = \angle OBC = \theta$

In $\triangle ABC,$

$\displaystyle \frac{AB}{\sin \theta} = \frac{AC}{\sin (90^0 + \theta)} = t$

$AB = t \sin \theta, AC = t \cos \theta$

That gives us

$ \frac {AC^2 − AB^2}{AC^2+AB^2} = \cos 2 \theta$

Now please note $\angle D = \frac{1}{2} \angle BOC = 90^0 - \theta$

Also $AB = R \tan 2 \theta \,$ (as $\angle AOB = 2 \theta \, , \,R$ is the circumradius)

$2 AB \cos 2 \theta = 2R \sin 2 \theta = 4 R \sin \theta \cos \theta = 4 R \sin D \cos D = 2 BC \cos D $


问:
Topology of 3D rotations
Wikipedia chapter about topology of 3D rotation group says:

Consider the solid ball in $\mathbb {R} ^{3}$ of radius $\pi$ (that is, all points of $\mathbb{R} ^{3}$ of distance $\pi$ or less from the
origin). Given the above, for every point in this ball there is a
rotation, with axis through the point and the origin, and rotation angle equal to the distance of the point from the origin. The identity rotation corresponds to the point at the center of the ball.

followed by:

consider the path running from the "north pole" straight through the
interior down to the south pole [...] this loop represents a
continuous sequence of rotations about the z-axis starting and ending at the identity rotation (i.e. a series of rotation through an angle $\varphi$
where $\varphi$ runs from $0$ to $2\pi$).

I do not understand why "north pole" (and also south pole) is the identity rotation. As north pole has cartessian coordinates $(0,0,\pi)$, according to first paragraph it must be understood as a rotation around z axis by a rotation angle $\pi$, that is not the identity rotation.

Could be due to this misunderstood, I do not reach my true target, understand why rotations has a fundamental group of order $2$.

答: The north pole is a clockwise rotation with angle $\pi$ and the south pole is a counterclockwise rotation with angle $\pi$. But these are equal (since the angle is $\pi$; otherwise, they would be two distinct rotations). And, while you go from the north pole to the south pole through the center of the sphere, you pass throuh that center, which corresponds to the identity rotation.

Title: Topology of 3D rotations


问:
Limit of periodic function at infinity
Tried this problem but I am out of ideas.

Let $f: \Bbb R \to \Bbb R$ periodic function with period $T>0$. Prove, using the definition of limits at infinity, that $f$ cannot be infinite while $x \to \infty$
or else $\lim_{x\to\infty}f(x) = \infty$

答: Suppose that $\lim\limits_{x\rightarrow \infty}f(x)=\infty$, i.e. $\forall E>0~ \exists \Delta(E)$ such that $\forall x$ with $|x|> \Delta$ holds $|f(x)|>E$. For $E>|f(0)|$ and $n\in \mathbb{N}$ such that $nT> \Delta(E)$ we have $|f(nT)|=|f(0)|<E$, which contradicts with our assumption.

Title: Limit of periodic function at infinity


问:
Do $\lim_{x \downarrow 0} f(x)$ and $\lim_{x \uparrow 1} f(x)$ exist if $f$ is concave over $[0, 1]$?
Let $f: [0, 1] \rightarrow \mathbb{R}$ be a concave function. We know that $f$ is continuous over $(0, 1)$. Is it true that both $\lim_{x \downarrow 0} f(x)$ and $\lim_{x \uparrow 1} f(x)$ exist? Normal graphs of a concave function suggest these properties. But I have no idea how to prove them. Thanks.

答:
If $f$ is concave on $[0, 1]$ then
$$ \tag{*}
f(b) \ge \frac{c-b}{c-a} f(a) + \frac{b-a}{c-a} f(c)
$$
for $0 \le a < b < c \le 1$. This can be rewritten as
$$
\frac{c-b}{c-a} \bigl(f(b) - f(a)\bigr) \ge \frac{b-a}{c-a}\bigl(f(c) - f(b)\bigr) \, .
$$
In particular we have the following implication:
$$
\tag{**}
\bigl( 0 \le a < b < c \le 1 \text{ and } f(a) > f(b) \bigr)\implies f(b) > f(c) \, .
$$

Now consider two cases:

$f$ is increasing on $(0, 1)$. $(*)$ with $0 = a < b=1/2 < c < 1$ gives an upper bound for $f(c)$, so that $\lim_{x \to 1-} f(x)$ exists as a finite value.

Otherwise there is $0 < x_1 < x_2  < 1$ with $f(x_1) > f(x_2)$. Then $(**)$ implies that $f$ is decreasing on $[x_2, 1)$. $(*)$ with $a=0 < b < c=1$ gives a lower bound for $f(b)$, so that $\lim_{x \to 1-} f(x)$ exists as a finite value in this case as well.

A similar argument works for $\lim_{x \to 0+} f(x)$.

It follows from the concavity condition $(*)$ that
$$
\lim_{x \to 0+} f(x) \ge f(0) \\
\lim_{x \to 1-} f(x) \ge f(1)
$$
(i.e. $f$ is lower semi-continuous at the boundary points) but strict inequality can hold.


问:
$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $ Did I do it right? How to do the last step?
I need to find:

$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $$

So I calculate:

$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}|  = \lim_{n \to +\infty} |\frac{\sum_{k=1}^{2n} (-1)^k \sqrt{k}}{\sqrt{n}} | $$

Denominator grows to $\infty$, so I can use stolz theorem:

$$\lim_{n \to +\infty} |\frac{\sum_{k=1}^{2n} (-1)^k \sqrt{k}}{\sqrt{n}} |  = \lim_{n \to +\infty} |\frac{\sqrt{2n}}{\sqrt{n} - \sqrt{n-1}} | \implies[\frac{\sqrt{2}}{0} ] $$

So I have to transform that a bit:

$$\lim_{n \to +\infty} |\frac{\sqrt{2n}(\sqrt{n} + \sqrt{n-1})}{n - n + 1} | = \lim_{n \to +\infty} |\sqrt{2}n + \sqrt{2n^2-2n}| \implies \infty - \infty$$

Another transformation:

$$\lim_{n \to +\infty} |\frac{2n^2 - 2n^2 + 2n}{\sqrt{2}n - \sqrt{2n^2-2n}}| = \lim_{n \to +\infty} |\frac{2n}{n(\sqrt{2} - \sqrt{2-\frac{2}{n}})}| = \lim_{n \to +\infty} |\frac{2}{\sqrt{2} - \sqrt{2-\frac{2}{n}}}| \implies[\frac{2}{0} ]$$

... and I feel like it doesn't lead nowhere. Can somebody tell what do I do wrong?

Body:
$$\begin{split}
\sum_{k=1}^{2n} (-1)^k \sqrt{k} &= \sum_{k=1}^{n} \left((-1)^{2k-1} \sqrt{2k-1} + (-1)^{2k} \sqrt{2k}\right)\\
&= \sum_{k=1}^{n} \left( \sqrt{2k}-\sqrt{2k-1}\right)\\
&= \sum_{k=1}^{n} \frac{1}{\sqrt{2k}+\sqrt{2k-1}}
\end{split}$$
Thus $$\sum_{k=1}^n\frac{1}{2\sqrt{2k}}\leq \sum_{k=1}^{2n} (-1)^k \sqrt{k}\leq \sum_{k=1}^n\frac{1}{2\sqrt{2k-1}}$$
We can use integrals to estimate the bounds
$$\frac 1 {2\sqrt 2}\int_1^{n+1}\frac{dx}{\sqrt x}\leq \sum_{k=1}^{2n} (-1)^k \sqrt{k}\leq \frac 1 2+ \int_2^{n+1}\frac{dx}{2\sqrt{2x-1}}$$
to see that $$\sum_{k=1}^{2n} (-1)^k \sqrt{k}\sim \sqrt{\frac n {2}}$$
Thus
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}|=\frac 1 {\sqrt 2}$$


答:
The decomposition into partial fractions supposes   these are proper rational fractions, where the degree of the numerator is less than the degree of the irreducible factor in the denominator. When the given fraction has only simple poles, as is the case here, this means the numerator is a constant, and actually, there exists a formula for a simple pole $\alpha$ of the rational function $\frac{P(x)}{Q(x)}$: the coefficient $A_\alpha$ for the pole $\alpha$ is given by
$$A_\alpha=\frac{P(\alpha)}{Q'(\alpha)},$$
which leads to the decomposition
$$\frac x{x^2+3x-4}=\frac x{(x-1)(x+4)}=\frac 15\frac1{x-1}+\frac 45\frac1{x+4}$$
and the integral
$$\frac15\bigl(\ln|x-1|+4\ln|x+4|\bigr)+C=\frac15\ln\bigl(|x-1|(x+4)^4\bigr)+C.$$


问:
Convolution of $1/(1+x^2)$ and $\exp(-x^2/(4t))$
Is there a closed form formula for the convolution of $1/(1+x^2)$ and $\exp(-x^2/(4t))$, where $t>0$, i.e. the integral
$$\int_{-\infty}^\infty \frac{1}{1+u^2}\exp\left(-\frac{(x-u)^2}{4t}\right)\,du?$$
I doubt it, but someone may be able to use special functions.</p>

答:
The integral may be evaluated using the convolution theorem.  The FT of this convolution is the product of the individual FTs; we then take the inverse FT.

The individual functions are $f(x)=1/(1+x^2)$ and $g(x)=e^{-x^2/(4 t)}$.  The FT of the above convolution is then

$$\hat{f}(k) \hat{g}(k) = 2 \sqrt{\pi^3 t} e^{-t k^2} e^{-|k|}$$

The integral is then the inverse FT of the above:

$$\begin{align}\int_{-\infty}^{\infty} \frac{du}{1+u^2} e^{-(u-x)^2/(4 t)} &= \sqrt{\pi t} \int_{-\infty}^{\infty} dk \, e^{-t k^2} e^{-|k|} e^{-i k x}\\ &=\sqrt{\pi t} \int_{-\infty}^{0} dk \, e^{-t k^2} e^{k} e^{-i k x} + \sqrt{\pi t} \int_{0}^{\infty} dk \, e^{-t k^2} e^{-k} e^{i k x}\\ &=2 \sqrt{\pi t} \Re{\left [\int_0^{\infty} dk\, e^{-t k^2} e^{(1-i x)k}  \right ]}\\ &= \pi e^{-(x^2-1)/(4 t)} \Re{\left [e^{-i x/(2 t)} \text{erfc}\left(\frac{-1+i x}{2 \sqrt{t}} \right ) \right]} \end{align}$$


Body: $$\int{\frac{xdx}{x^2 + 3x -4}}=\frac{1}{2}(\int{\frac{(2x+3)dx}{x^2 + 3x -4}}-3\int{\frac{dx}{x^2 + 3x -4}})=\frac{1}{2}(\int{\frac{(2x+3)dx}{x^2 + 3x -4}}+\frac{3}{5}(\int{\frac{dx}{x +4}}-\int{\frac{dx}{x -1}}))=\frac{1}{2}\ln{|x^2+3x-4|}+\frac{3}{5}\ln{|x+4|} -\frac{3}{5}\ln{|x-1|}+C=\frac{1}{2}\ln{|x+4|}+\frac{1}{2}\ln{|x-1|}+\frac{3}{10}\ln{|x+4|} -\frac{3}{10}\ln{|x-1|}+C=\frac{4}{5}\ln|x+4|+\frac{1}{5}\ln|x-1|+C$$


答: Let $m=f'(x_0)>0$.  Suppose by way of contradiction that $f(x)>f(x_0)$ did not hold in some interval $(x_0, x_0+\delta)$.  Then there would be some sequence $x_1, x_2, \ldots$ such that $x_i>x_0$, $\lim_{i\rightarrow \infty}x_i=x_0$, and $f(x_i)<f(x_0)$.  But then this contradicts $$\lim_{x\rightarrow x_0+}\frac{f(x)-f(x_0)}{x-x_0}=m>0$$</p>

Title: $\epsilon$-$\delta$ proof

Body:
I found an interesting problem in my textbook, it asks to prove the following statement:

If $f'(x_{0}) >0$, then there is a $\delta >0$ such that $f(x)<f(x_{0})$ if $x_{0}- \delta < x <x_{0}$, and $f(x) > f(x_{0})$ if $x_{0} <x < x_{0} + \delta$.

How would one prove this from the definitions?


问:
How does a probability density function of a random variable is induced due to Radon-Nikodyn derivative?
If someone can please clarify the following for me:
Given a probability space $(\omega,\Sigma,P)$ and a continous random variable $X:w \rightarrow \mathcal{R}$.
I have read the statement of the Radon-Nikodyn Theorem, but I still can't see how does one relate it to the existance of a probability density function. Can someone please elaborate on this?
Thanks

答:
Let $\mathbb P_X$ the measure on $\mathbb R$ defined by $$\mathbb P_X(B):=\mathbb P\{X\in B\},$$
for all Borel set $B\subset \mathbb R$.

Definition Let $m$ denote the Lebesgue measure on $\mathbb R$. Then, $X$ is continuous if $\mathbb P_X$ is absolute continuous with respect to $m$.

Therefore, by Radon-Nikodym, there is $f:\mathbb R\to \mathbb R$ s.t. $$\frac{\mathrm d \mathbb P_X}{\mathrm d m}=f,$$

i.e. $$\mathbb P_X(B)=\int_B f(x)m(\mathrm d x).$$
Therefore, $X$ has a density $f$.


答:
Your mistake is in your numerator:

Using Stolz-Cesaro you should get
$$\frac{\sqrt{2n}\color{blue}{-\sqrt{2n-1}}}{\sqrt{n} - \sqrt{n-1}}$$

Now, all goes well and converges to $\frac 1{\sqrt 2}$.

Title: $ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $ Did I do it right? How to do the last step?

Body:
I need to find:

$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $$

So I calculate:

$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}|  = \lim_{n \to +\infty} |\frac{\sum_{k=1}^{2n} (-1)^k \sqrt{k}}{\sqrt{n}} | $$

Denominator grows to $\infty$, so I can use stolz theorem:

$$\lim_{n \to +\infty} |\frac{\sum_{k=1}^{2n} (-1)^k \sqrt{k}}{\sqrt{n}} |  = \lim_{n \to +\infty} |\frac{\sqrt{2n}}{\sqrt{n} - \sqrt{n-1}} | \implies[\frac{\sqrt{2}}{0} ] $$

So I have to transform that a bit:

$$\lim_{n \to +\infty} |\frac{\sqrt{2n}(\sqrt{n} + \sqrt{n-1})}{n - n + 1} | = \lim_{n \to +\infty} |\sqrt{2}n + \sqrt{2n^2-2n}| \implies \infty - \infty$$

Another transformation:

$$\lim_{n \to +\infty} |\frac{2n^2 - 2n^2 + 2n}{\sqrt{2}n - \sqrt{2n^2-2n}}| = \lim_{n \to +\infty} |\frac{2n}{n(\sqrt{2} - \sqrt{2-\frac{2}{n}})}| = \lim_{n \to +\infty} |\frac{2}{\sqrt{2} - \sqrt{2-\frac{2}{n}}}| \implies[\frac{2}{0} ]$$

... and I feel like it doesn't lead nowhere. Can somebody tell what do I do wrong?


问:
factory normal distribution
The number of workers in a factory distributes $N(100,7^2)$.

A factory is considered big when it has more than $110$ workers.
Considered small when it has less than $80$ workers and is considered mediocre with $80-110$ workers. I need to calculate the probability for each of this cases.

What I did was,

$P(B>110)=1-P(B<=110)=1-0.9235$
$P(B<80)=P((B-100)/7<-20/7)=1-0.99786$
$P(80<B<110)=0.0765-0.00214=0.07436$

I'm still trying to grasp the whole normal distribution thing and am pretty sure that I did somethings wrong or used the table wrong with the negetive values, would love to get some feedback to wether what I did was correct and wether these answers are applicable final answers or am I supposed to do something in addition.

thank you.

Title: factory normal distribution

Body:
Only the last one is wrong

$$\mathbb{P}[80\leq X\leq 110]=\Phi\Bigg[\frac{110-100}{7}\Bigg]-\Phi\Bigg[\frac{80-100}{7}\Bigg]=0.9234-0.0021\approx 92.13\%$$

The sum of the three probabilities has to be 1:

$$7.66\%+0.21\%+92.13\%=100\%$$


问:
A question about linear, bounded map in $L^p$ spaces and $C[0,1]$
Let $l^{\infty}$ be the space of all complex bounded sequences with the supremum norm.
Let $a_n\in l^{\infty}$
Define $T: l^1\to C[0,1]$ such that:

$\forall b_n\in l^1 , x\in [0,1]$,   $$T(b_n)=\sum_{n=1}^{\infty} a_nb_nx^n.$$

Show that $T$ is well-defined, linear continuos and $||T||=||a_n||_{\infty}$ where $C[0,1]$ with the maximum norm.

Well defined: I think it means to show that for $b_n\in   l^1$ the defined sum is in $C[0, 1].$ Given that $b_n\in l^1 , a_n\in l^{\infty}$ thus $a_nb_n\in l^1$ and $||a_nb_n||_1\leq ||a_n||_\infty ||b_n||_1$ and this is a finite value since $x\in [0, 1]$ so the sum converges and thus is continuos. I'm not sure about it.

T is linear:
Let $b_n , c_n \in l^1$ and $d, e \in \mathbb{C}$ then
\begin{align}
T(cb_n+dc_n)(x)&= \sum_n a_n(cb_n+dc_n) x^n\\
&= \sum_n(ca_nb_n+da_nc_n)x^n\\
&= \sum_n ca_nb +\sum_n da_nc_n\\
&= c\sum_n a_nb_n+d\sum_n a_nc_n\\
&= cT(b_n)(x)+dT(c_n)(x)
\end{align}

$T$ is continuous:

We can show that $T$ is bounded so since $T$ is linear, by theorem we get that $T$ is continuous.

According to the definition we have to show that there is $M>0$ such that
$$||T(b_n)||\leq M ||b_n||.$$

To this end, \begin{align}
||T(b_n)(x)||_{C[0,1]}&=||\sum_n a_nb_nx^n||_{C[0,1]}\\
&=\max |\sum_n a_nb_nx^n|\\
&\leq \max \sum_n |a_nb_nx^n|.
\end{align}
I did not know how to use the information that $a_n\in l^\infty$  and $b_n\in l^1$ to finish this method.

The last part can be shown by the equivalent definitions
$||T||=sup_{||_n||=1} ||T(b_n)||=sup_{b_n\neq0 \in l^1} \frac{||T(b_n)||}{||b_n||}$.

I had a problem with applying the definition that I know, in this question.
I would be glad if you can help me in understanding this.

答:
Well defined: we know that $\sum\limits_{n=1}^{\infty}a_nb_n$ converges absolutly. Since for every $x$ in $[0,1]$ we have $|a_nb_nx^n| \leq |a_nb_n|$, then Weierstrass M-test gives us uniformly convergance of series $\sum\limits_{n=1}^{\infty}a_nb_nx^n$ on $[0,1]$. Since $u_n(x)=a_nb_nx^n$ is continuous, then $\sum\limits_{n=1}^{\infty}a_nb_nx^n$ is continuous on $[0,1]$

Boundness: $|T(b_n)(x)|=|\sum\limits_{n=1}^{\infty}a_nb_nx^n|\leq \sum\limits_{n=1}^{\infty}|a_nb_nx^n| \leq \sum\limits_{n=1}^{\infty}|a_nb_n| \leq \|a_n\|_{\infty}\|b_n\|_1$. So, $T$ is bounded (and then continuos) and $\|T\|\leq \|a_n\|_{\infty}.$

Norm: let $b=(0,0,...,b_n=1,0,...) \in l^1$ with $\|b\|_1=1$. Then $\max\limits_{[0,1]}|T(b)(x)|=\max\limits_{[0,1]}|a_nx^n|=|a_n|\|b\|_1$, that is why $\|T\| \geq |a_n|$ for any $n$. Taking supremum over all positive integer $n$ we have $\|T\| \geq \|a_n\|_{\infty}$ compliting the proof.


答:
If $f(x_0)>0$ then for all sufficiently small $\epsilon>0$ there is $\delta>0$, whit $x<x_0$, such that
\begin{align}
0<|x-x_0|<\delta
 \implies
&
\left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0) \right|<\epsilon
\\
-\delta <x-x_0<\delta
\implies
&
0<f'(x_0) -\epsilon <\frac{f(x)-f(x_0)}{x-x_0} <+f'(x_0) +\epsilon
\\
x_0-\delta <x<x_0+\delta
\implies
&
0<f'(x_0)(x-x_0)-\epsilon (x-x_0)+f(x_0)< f(x)
\end{align}</p>

Title: $\epsilon$-$\delta$ proof


问:
Is it possible to have a graph with chromatic number 3 that does not have the subgraph K3?
I am asked to either give an example for explain why this graph cannot exist and I'm a bit stuck. While I can't seem to think of an example of a 3-colorable graph without the complete graph K3 as a subgraph, I also don't know how to prove that every graph that has chromatic number 3 has the subgraph K3. Any help would be greatly appreciated!</p>

答: How about the cycle graph on $5$ vertices? In fact any cycle graph $G$ on an odd number $n \ge 5$ is 3-colorable (start at one node colored black, alternate black-white-black-white... and finish with red). And the chromatic number of such a cycle graph is $3$ exactly since clearly it can't be colored with $1$ color (there are adjacent vertices) nor can it be colored with $2$ colors, since starting anywhere with say black would then need to alternate black with white and run into impossibility when trying to color the final vertex in the loop. But if the odd $n$ is greater than $3$ then $G$ does not have a subgraph $K_3.$


答:
You are almost there. Just a small point, you are right that to show that $T$ is well-defined, you need to show that $T(b_n)\in C[0, 1].$ The proof, however, you have given above only argues that $T(b_n)(x)<\infty$ for each $x\in [0, 1].$ This is because you have only argued that the sum converges for each $x.$ But it does not give you continuity. Recall that pointwise limits of continuous functions may fail to be continuous. To summarize, you proof is incomplete. It might be a helpful idea to try the Weiestrass-M test to obtain uniform convergence and hence conclude that the limiting function is indeed continuous.

Your proof of linearity is all right.

Finally to prove that $T$ is bounded. You have already obtained
\begin{equation}
||T(b_n)||_{C[0, 1]}\le \max_{x\in [0, 1]}\sum_{n}| a_nb_nx^n|.
\end{equation}
To finish, just observe that for $|a_nb_nx^n|\le |a_nb_n|$ for all $x\in [0, 1].$ Therefore,
$$||T(b_n)||_{C[0, 1]}\le \sum_{n}| a_nb_n|\le ||a_n||_{\infty}\sum_{n}|b_n|=||a_n||_{\infty}||b_n||_1.$$

Title: A question about linear, bounded map in $L^p$ spaces and $C[0,1]$

Body:
Let $l^{\infty}$ be the space of all complex bounded sequences with the supremum norm.
Let $a_n\in l^{\infty}$
Define $T: l^1\to C[0,1]$ such that:

$\forall b_n\in l^1 , x\in [0,1]$,   $$T(b_n)=\sum_{n=1}^{\infty} a_nb_nx^n.$$

Show that $T$ is well-defined, linear continuos and $||T||=||a_n||_{\infty}$ where $C[0,1]$ with the maximum norm.

Well defined: I think it means to show that for $b_n\in   l^1$ the defined sum is in $C[0, 1].$ Given that $b_n\in l^1 , a_n\in l^{\infty}$ thus $a_nb_n\in l^1$ and $||a_nb_n||_1\leq ||a_n||_\infty ||b_n||_1$ and this is a finite value since $x\in [0, 1]$ so the sum converges and thus is continuos. I'm not sure about it.

T is linear:
Let $b_n , c_n \in l^1$ and $d, e \in \mathbb{C}$ then
\begin{align}
T(cb_n+dc_n)(x)&= \sum_n a_n(cb_n+dc_n) x^n\\
&= \sum_n(ca_nb_n+da_nc_n)x^n\\
&= \sum_n ca_nb +\sum_n da_nc_n\\
&= c\sum_n a_nb_n+d\sum_n a_nc_n\\
&= cT(b_n)(x)+dT(c_n)(x)
\end{align}

$T$ is continuous:

We can show that $T$ is bounded so since $T$ is linear, by theorem we get that $T$ is continuous.

According to the definition we have to show that there is $M>0$ such that
$$||T(b_n)||\leq M ||b_n||.$$

To this end, \begin{align}
||T(b_n)(x)||_{C[0,1]}&=||\sum_n a_nb_nx^n||_{C[0,1]}\\
&=\max |\sum_n a_nb_nx^n|\\
&\leq \max \sum_n |a_nb_nx^n|.
\end{align}
I did not know how to use the information that $a_n\in l^\infty$  and $b_n\in l^1$ to finish this method.

The last part can be shown by the equivalent definitions
$||T||=sup_{||_n||=1} ||T(b_n)||=sup_{b_n\neq0 \in l^1} \frac{||T(b_n)||}{||b_n||}$.

I had a problem with applying the definition that I know, in this question.
I would be glad if you can help me in understanding this.


答: Does this figure help you understand the integrals?

Title: factory normal distribution

Body:
The number of workers in a factory distributes $N(100,7^2)$.

A factory is considered big when it has more than $110$ workers.
Considered small when it has less than $80$ workers and is considered mediocre with $80-110$ workers. I need to calculate the probability for each of this cases.

What I did was,

$P(B>110)=1-P(B<=110)=1-0.9235$
$P(B<80)=P((B-100)/7<-20/7)=1-0.99786$
$P(80<B<110)=0.0765-0.00214=0.07436$

I'm still trying to grasp the whole normal distribution thing and am pretty sure that I did somethings wrong or used the table wrong with the negetive values, would love to get some feedback to wether what I did was correct and wether these answers are applicable final answers or am I supposed to do something in addition.

thank you.


问:
How to find the confidence interval for a prediction using a multiple linear regression model?
I have a dataset of 100 observations and 5 numerical variables ($x_1, x_2, ..., x_{5}$) which I used to generate a linear model which to predict some value $y$. I plugged in some values into the model and got a result for $\hat{y}$.

I would like to calculate a 95% confidence interval for this value.

Using Python I get the following model summary:

```<code>Adj. R-squared: 0.892
F-statistic:    27.63
Prop(F-stat):   2.45e-18

        coef   std err   t      P>|t|   [0.025    0.975]
const  -12.71  3.713    -3.424  0.001   -20.156  -5.268
x1       0.29  0.255     1.134  0.262   -0.222   -0.800
x2       0.17  0.031     5.283  0.000    0.103    0.228
x3       0.44  0.089     4.970  0.000    0.262    0.617
x4      -0.47  0.194    -2.414  0.019   -0.856   -0.079
x5       0.36  0.226     1.587  0.118   -0.094    0.811
</code>```

I am aware that in order to calculate a 95% confidence interval for a simple linear regression the formula is $\hat{\beta} \pm t_{0.95, f}\: \cdot \: \hat{\sigma}_{B_0}$ where $f$ is the degree of freedom and $\hat{\sigma}_{B_0}$ is the std err. But how do I calculate this for multiple linear regression?

答:
Quantile regression has a different function you are minimizing:

$(\tau - 1) \int_{-\infty}^u (y-u)f(y)dy + \tau \int_{u}^\infty (y-u)f(y)dy $

, where $\tau$ is the quantile you are looking for.

Using the leibniz integral rule, you can find the derivative of it as:

$F(u)-\tau$

You can solve this exactly, but this is quite a long derivation.
Now, with the data you have and the simple derivative, you can also solve it iteratively via a constantly updating your output based on this derivative (also known as gradient boosting). Scikit-learn has a gradient boosting regression model where you can specify the quantile and find the 95%-quantile. Here is a link.


答:
As you have noted, the linearity of T is clear. Then, boundedness (continuity) is proved as you have suspected. Here are the details:
\begin{align*} \|T(b)\|_{C[0,1]}&=\sup_{x\in[0,1]}\left|\sum_{n=1}^{\infty}a_{n}b_{n}x^{n}\right| \\ &\leq \sum_{n=1}^{\infty}|a_{n}||b_{n}|\leq\|a\|_{\infty}\|b\|_{1}\end{align*}
where $a=(a_{n})_{n=1}^{\infty}\in\ell_{\infty}$ and $b=(b_{n})_{n=1}^{\infty}\in\ell_{1}$. This is because $|a_{n}|\leq\|a\|_{\infty}$ for all $n$ and because $b$ is absolutely summable by assumption. Well-definedness also follows as noted in the first (and second) answer.


问:
If every linear form $f: E \to \mathbb R$ that vanishes on $F \subset E$ is zero, then $E = F$.
I am workin in Galdi's Introduction to the Mathematical Theory of the Navier-Stokes Equations and somewhere in a proof we want to show that, for $E$ a Banach space and $F$ a closed subspace of $E$, $F = E$. To this end, he use the fact that if every linear functional $f \in E'$ that vanishes on $F$ is zero, then $F = E$.

I can add the definitions of $E$ and $F$ if it is needed but he presents this fact as a quite general result. I tried to find a proof in the literature but I couldn't find it anywhere. Does anyone have any idea where I could get the proof or how to show it?

答:
This is a direct application of the Hahn-Banach Theorem. Suppose that $F\subsetneq E$. Take $e\in E\setminus F$. Since $F$ is a subspace, it is linearly independent from $e$. So you can define, on $\operatorname{span}\{F\cup\{e\}\}$,
$$
f(\lambda e+\sum_jc_j g_j)=\lambda,\ \ g_1,\ldots,g_n\in F.
$$
We need to show that $f$ is bounded:
\begin{align}
\|f\|&=\sup\big\{\tfrac{|f(\lambda e+ g)|}{\|\lambda e + g\|}:\ \lambda\in\mathbb C,\ g\in F\}\\[0.3cm]
&=\sup\big\{\tfrac{|f( e- g)|}{\|\lambda e - g\|}:\ \lambda\in\mathbb C,\ g\in F\}\\[0.3cm]
&=\sup\big\{\tfrac{1}{\|\lambda e - g\|}:\ \lambda\in\mathbb C,\ g\in F\}\\[0.3cm]
&=\frac1{\inf\big\{{\|\lambda e - g\|}:\ \lambda\in\mathbb C,\ g\in F\}}\\[0.3cm]
&=\frac1{\operatorname{dist}(e,F)}.
\end{align}
The fact that $F$ is closed guarantees that $\operatorname{dist}(e,F)>0$, so $f$ is bounded. Now Hahn-Banach gives you an extension $f$ to all $E$, bounded, with $f(e)=1$ and $f|_F=0$.


问:
Prove that $\lVert \exp(A) - \exp(B) \rVert \leq \lVert A-B \rVert e^{\max\{\lVert A \rVert, \lVert B \rVert \}}$
, where A and B are square Matrices, exp(A) is the matrix exponential

I can see where the $\lVert A-B \rVert$ comes from, but can't figure out how to get the $ e^{max\{\lVert A \rVert, \lVert B \rVert\}} $. The definition for the matrix exponential we're using is the following:
$\exp(A)= \sum \frac{A^k}{k!}$.

答:
Partial answer

Define

$$F(t) = \exp((1-t)A + tB)$$ for $t\in [0,1]$. According to the Mean Value Theorem

$$\Vert \exp(A) - \exp(B) \Vert = \Vert F(1) - F(0) \Vert \le \sup_{t \in [0,1]} \Vert F^\prime(t) \Vert \vert 1 - 0 \vert =\sup_{t \in [0,1]} \Vert F^\prime(t) \Vert$$

As $$F^\prime(t) = \exp((1-t)A + tB) \cdot (B-A)$$ we're left to prove that

$$\sup_{t \in [0,1]} \Vert \exp((1-t)A + tB) \Vert \le e^{\max(\Vert A \Vert,\Vert B \Vert)}$$ which is clear if $A,B$ commutes as in this case
$$\exp((1-t)A + tB) = \exp((1-t)A) \exp(tB).$$

If not, I don't see how to conclude...


问:
How to prove $\int_{0}^{\infty}\frac{\tan^{a} (x) \;dx}{x^2+b^2}=\frac{\pi}{2b}\frac{\tanh^{a} (b)}{\cos\frac{a\pi}{2}}$
I'm having trouble to prove the following:
$$
\int_{0}^{\infty}\frac{\tan^{a}\left(x\right)}
{x^{2} + b^{2}}\,\mathrm{d}x =
\frac{\pi}{2b}\,\frac{\tanh^{a}\left(b\right)}{\cos\left(\pi a/2\right)}\quad
\mbox{where}\ \left\vert a\right\vert < 1\
\mbox{and}\ b > 0
$$
The main difficulty is that the sign of the $\tan\left(x\right)$  alternates between positive and negative and so we need $\left(-1\right)$ to raise to power of noninteger number $a$.

I have no idea how to proceed.

答:
if we try letting $x=iy\Rightarrow dx=idy$ and $(0,\infty)\rightarrow(0,\infty)$. We then know that:
$$\tan(iy)=i\tanh(y)$$
$$\frac{1}{(iy)^2+b^2}=\frac{1}{b^2-y^2}$$
so our integral becomes:
$$\int_0^\infty\frac{i^{a+1}\tanh^a(y)}{b^2-y^2}dy$$
now you can do factor: $b^2-y^2=(b+y)(b-y)$ and also notice that:
$$i^{a+1}=(e^{i\pi/2})^{a+1}=\cos\left(\frac{(a+1)\pi}{2}\right)+i\sin\left(\frac{(a+1)\pi}{2}\right)$$


问:
Movie Earnings and Relative Ticket Sizes
Given that movie A made 308 million and a 1 by 3 cm ticket represented A and movie B made 609 million, assuming that prices have been adjusted for inflation and ticket B is the same shape as ticket A, what would the size of ticket B be? </p>

<p>I said if the area of A was 3 cm which represented 308 million than ticket B would just need to be about 1.97 $cm^2$ larger i.e. have an area of about 5.91. I got this because since its a graphical representation then ticket B should be proportional to ticket A. Not really sure if it's right so if anyone would check it out and let me know, thanks! </p>

答: As you say, the ratio of areas should be $\frac  {609}{308}\approx 1.977$.  Note that this ratio has no units: both ticket areas have the same unit:  cm$^2$ so the ratio has none.  To get the dimensions for ticket B, you need to multiply each dimension of ticket A by the square root of this, $\sqrt {1.977}\approx 1.406$ so the ticket size is $1.406 \times 4.218$ cm.</p>


问:
General solution for an integral with Hermite polynomials?
I'm currently studying quantum mechanics and thus dealing with Hermite polynomials.

By looking on Wikipedia I found this identity for Hermite polynomials:

$$\int\limits_{-\infty}^{\infty} H_m(x)H_n(x)e^{-x^2} \ dx=\sqrt{\pi}2^nn!\delta_{mn}$$

While doing a problem, after a little bit of calculations, I came to the following integral:
$$\int\limits_{-\infty}^{\infty} H_m(x)\cdot x\cdot H_n(x)e^{-x^2} \ dx$$

By just looking at this integral, I figured that it has to be zero when $m-n=\pm2k, k=1,2,3...$ (cause the expression would be uneven and thus the integral would be zero).

I thought I could make the same argument by saying the integral is nonzero whene $m-n=\pm(2k+1)$. While this is true for $k=0$ (so $m-n=\pm1$), I turns out that this seems to be not true for $k>0$. For example I calculated:
$$\int\limits_{-\infty}^{\infty} H_0(x)\cdot x\cdot H_3(x)e^{-x^2} \ dx=\int\limits_{-\infty}^{\infty} e^{-x^2}x(8x^3-12x)dx=0$$

I tried to understand why this is the case by using the recursion formula of the Hermite polynomials and plugging it into the integral but didn't succeed...

I wanted to ask if there is another already existing general identity for my integral (like the one on Wikipedia). If not it would be awesome if someone could help me out here and explain why the argumentation breaks down for $k>0$... Thanks!

答:
The answer's already been accepted, but expanding on metamorphy's comment, an explicit formula for the integral of interest is

$$\int_{-\infty}^\infty \exp{(-x^2)} \ x H_m(x)H_n(x) \ dx = \sqrt{\pi} \ 2^n n! \sum_{j=0}^n \binom{m}{n-j}\frac{2^{-j}}{j!} \delta_{1,m-n+2j} $$

This is derivable from the orthogonality relationship and the explicit linearization formula
$$   H_m(x)H_n(x) = 2^n n!\sum_{j=0}^n \binom{m}{n-j}\frac{2^{-j}}{j!} H_{m-n+2j}(x) $$ since $x = \tfrac{1}{2}H_1(x).$


问:
Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$
I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.

The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$

So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so that'd be a go-to. Also, since the degree of th enumerator is one less than the denominator we can maybe treat this as $du/u$ which would imply a natural log, though we don't quite know that yet.

So we decompose: I know that $x^2+3x-4$ can be factored as $(x-1)(x+4)$.

That gets me

$$\int{\frac{x}{(x-1)(x+4)}dx}$$
so I can do this:
$$\frac{Ax}{x-1}+\frac{B}{x+4} = \frac{x}{(x-1)(x+4)}$$
Which implies
$$A(x+4)+B(x-1) = x$$
and since the roots are at $x=1$ and $x=-4$, I can set it up like this:
$$5A = 1;-5B=1 \text{  and } A = \frac{1}{5}, B=-\frac{1}{5}$$

Leading to:

$$\frac{x}{5(x-1)}-\frac{1}{5(x+4)}$$
Which I can set up the integral like so:
$$\int{\frac{x}{5(x-1)}-\frac{1}{5(x+4)}dx}=\int{\frac{x}{5(x-1)}dx-\int{\frac{1}{5(x+4)}dx}}$$
I can now integrate by addition here  $$\int{\frac{x}{x-1}}dx=\int{\frac{x-1+1}{x-1}}dx=x+\int{\frac{1}{x-1}}dx=x-\ln(x-1)$$
And doing the same thing for the second term and bringing back my $\frac{1}{5}$
$$\frac{1}{5}(x-\ln(x-1)+\ln(x+4))$$

I suspect there is a further simplification I could do. On a problem like this I also saw it integrated as an arctangent, but that seemed needlessly complex? In any case I was curious if I did this correctly.

答:
Since the denominator has two linear factors, you can set up your partial fraction decomposition as

$$\frac{x}{(x+4)(x-1)} = \frac{A}{x+4} + \frac{B}{x-1}$$

This implies

$$ x = A(x-1) + B(x+4) $$

So $A+B=1$ and $4B-A=0$ $\Rightarrow$ $(A,B) = (\frac{4}{5},\frac{1}{5})$.

This means that

$$\int \frac{x}{(x-4)(x+1)} dx = \frac{1}{5} \int \frac{4}{x+4} + \frac{1}{x-1}dx $$

Then you can continue from here.

This is my first response, so I would very much appreciate any feedback on formatting or other site etiquette I may be unaware of. I hope this helps!


问:
Find all the solutions of $x^2 \equiv 1 \mbox{ mod }365$.

Find all the solutions of $x^2 \equiv 1 \mbox{ mod }365$.

We know that $365=5\cdot 73$. So if I could find the solutions of $x^2 \equiv 1 \mbox{ mod }5$ and $x^2 \equiv 1 \mbox{ mod }73$, using CRT I could find the solutions of the given equation.

I can solve $x^2 \equiv 1 \mbox{ mod }5$ by hand, but I'm sure that there is an easier way to solve $x^2 \equiv 1 \mbox{ mod }73$.

Would appreciate your help:)

答:
I'm sure that there is an easier way to solve $x^2 \equiv 1 \mbox{ mod }73$.

Would appreciate your help:)

$x^2\equiv1\bmod73\iff x\equiv\pm1\bmod73$


答: For any prime $p$, you can show that $x^2\equiv1\bmod p\iff x\equiv \pm1\bmod p$. To this end, notice that $$x^2\equiv 1\bmod p\iff \exists k\in\mathbb Z: x^2=k\cdot p+1\iff (x+1)(x-1)=k\cdot p$$ Euclid's Lemma implies now that either $p\mid x+1\iff x\equiv -1\bmod p$ or $p\mid x-1\iff x\equiv 1\bmod p$.

Body:
Find all the solutions of $x^2 \equiv 1 \mbox{ mod }365$.

We know that $365=5\cdot 73$. So if I could find the solutions of $x^2 \equiv 1 \mbox{ mod }5$ and $x^2 \equiv 1 \mbox{ mod }73$, using CRT I could find the solutions of the given equation.

I can solve $x^2 \equiv 1 \mbox{ mod }5$ by hand, but I'm sure that there is an easier way to solve $x^2 \equiv 1 \mbox{ mod }73$.

Would appreciate your help:)


问:
Cauchy sequence in $\mathcal{L}^1$
For $\Omega = \mathbb{N}$ with counting measure is $\mathcal{L}^1$ complete.
Let $(f_k)_{k \in \mathbb{N}}$ be Cauchy sequence in $\mathcal{L}^1$ such that  every $f_k$ is a sequence $(f^n_k)_{n \in \mathbb{N}}$.

Show that a sequence $f=(f^n)_{n \in \mathbb{N}}$ exists such that $f_k$ converges pointwise to $f$.

答:
Attention: Since the functions $f_k$ are functions defined on $\mathbb{N}$, it is clearer to write $f_k(n)$ instead of $f_k^n$.

Let us prove the result.

Since $(f_k)_{k \in \mathbb{N}}$ be Cauchy sequence in $\mathcal{L}^1(\mathbb{N},\#)$ (where $\#$ is the counting measure), we have that, for every $\varepsilon >0$, there is $M \in \mathbb{N}$, such that, if $r,s >M$ then
$$ \sum_{i \in \mathbb{N}} |f_r(i) -f_s(i)| = \int |f_r -f_s| d\# < \varepsilon$$

So, for all $j \in \mathbb{N}$, for every $\varepsilon >0$, there is $M \in \mathbb{N}$, such that, if $r,s >M$
$$ |f_r(j) -f_s(j)| \leqslant \sum_{i \in \mathbb{N}} |f_r(i) -f_s(i)| = \int |f_r -f_s| d\# < \varepsilon$$

So, for all $j \in \mathbb{N}$, $(f_k(j))_{k \in \mathbb{N}}$ is a Cauchy sequence (in $\mathbb{R}$ or $\mathbb{C}$). So,  for all $j \in \mathbb{N}$, there is $a_j$ such that  $(f_k(j))_{k \in \mathbb{N}}$ converges to $a_j$.

Define $f$ from $\mathbb{N}$ to $\mathbb{R}$ (or $\mathbb{C}$), by $f(j)=a_j$, for each $j \in \mathbb{N}$.

It is clear that $f_k$ converges pointwise to $f$.

Title: Cauchy sequence in $\mathcal{L}^1$


问:
Finding $b$ such that $y=mx+c$ is tangential to $y=b^x$.
I was thinking about the following problem:

Find $b$ such that $y=mx+c$ is tangential to $y=b^x$.

My work so far:

To satisfy the condition we need to find $x$ such that $\frac d{dx}\left[b^x\right]=\frac d{dx}[mx+c]$. Solving the derivatives, we get $\ln b\cdot b^x=m\Rightarrow b^x=\frac m{\ln b}\Rightarrow x=\log_b\left(\frac m{\ln b}\right)=\frac{\ln \left(\frac m{\ln b}\right)}{\ln b}=\frac{\ln m-\ln\ln b}{\ln b}$.

At the same time, we have $b^x=mx+c$. Using the substitution $u=mx+c$ and noting $mx=u-c\Rightarrow x=\frac1mu-\frac cm$, we have:

$$b^{\frac um-\frac cm}=u$$

$$b^{-\frac cm}=u\left(\frac1b\right)^\frac um$$
$$ue^{\ln\left(\frac1b\right)\frac um}=ue^{-\frac{\ln b}mu}=b^{-\frac cm}$$
$$-\frac{\ln b}mue^{\ln\left(\frac1b\right)\frac um}=ue^{-\frac{\ln b}mu}=-b^{-\frac cm}\frac{\ln b}m$$
Let $z=-\frac{\ln b}mu$
$$ze^z=-b^{-\frac cm}\frac{\ln b}m$$
$$z=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$-\frac{\ln b}mu=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$-\frac{\ln b}mu=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$u=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$mx+c=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$mx=-\frac m{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$x=-\frac 1{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac cm$$
Putting the two equations together, we have:
$$\frac{\ln m-\ln\ln b}{\ln b}=-\frac 1{\ln b}W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac cm$$
$$\ln m-\ln\ln b=-W\left(-b^{-\frac cm}\frac{\ln b}m\right)-\frac{c\ln b}m$$
$$\ln m-\ln\ln b+\frac{c\ln b}m=-W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$\ln\ln b-\ln m-\frac{c\ln b}m=W\left(-b^{-\frac cm}\frac{\ln b}m\right)$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)e^{\ln\ln b-\ln m-\frac{c\ln b}m}=-b^{-\frac cm}\frac{\ln b}m$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)e^{\ln\ln b-\ln m-\frac cm\ln b}=-b^{-\frac cm}\frac{\ln b}m$$
$$\left(\ln\ln b-\ln m-\frac{c\ln b}m\right)\frac{\ln b}mb^{-\frac cm}=-b^{-\frac cm}\frac{\ln b}m$$
$$\ln\ln b-\ln m-\frac{c\ln b}m=-1$$
$$\ln\ln b-\frac{c\ln b}m=\ln m-1$$
$$e^{\ln\ln b-\ln b\frac cm}=e^{\ln m-1}$$
$$\ln b\cdot b^{-\frac cm}=\frac me$$

I was not able to get any further than this. The equation is probably not solvable through elementary means but I would be interested in a closed solution using special functions (such as the Lambert W function) as well as a derivation.

答:
$y=b^x;\;y'=b^x\log b;\;b>0$ $

Derivative must be equal to the slope of $t:y=mx+c$

$$b^x\log b=m\to b^x=\frac{m}{\log b}\to x^*=\frac{\log \left(\frac{m}{\log (b)}\right)}{\log (b)}$$

Plug this value in the equation of the tangent. We get
$$y=m\,\frac{\log \left(\frac{m}{\log (b)}\right)}{\log (b)}+c$$
This value must be equal to the value we get substituting $x^*$ in the given equation
$$y=b^{x^*}\to y=\frac{m}{\log b}$$
The two $y$-values must be equal so we get the equation
$$m\,\frac{\log \left(\frac{m}{\log (b)}\right)}{\log (b)}+c=\frac{m}{\log b}$$
$$c \log b-\log (\log b)+m \log m=m$$

Set $\log b=u$

$$cu-\log u=m-m\log m$$

$$u -\frac{1}{c}\log u =\frac{m-m\log m}{c}$$

$$\log b=-\frac{m W\left(-\frac{c}{e}\right)}{c}$$
$$b=\exp\left(-\frac{m W\left(-\frac{c}{e}\right)}{c}\right)$$
where $W(z)$ is Lambert function.

Edit

Example.

If we have $y=2x-1$  we get
$$b=\exp\left(-2 W\left(-\frac{1}{e}\right)\right)\approx 1.7453$$
This is shown in the graph below

$$
...
$$


问:
Basic Geometric Series Question-Stuck
I'm studying Calc 2 and I have a basic series question.

A geometric series $\sum _{n=0}^{\infty }\:Ar^n$ is convergent if |r|<1 and the sum equals $\frac{a}{1-r}$ if the series is convergent.

Question: $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$

and $\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$ totals $\frac{5}{1-\frac{1}{\pi }}=\frac{5\pi \:}{\pi \:-1}$.

Therefore, $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$ = $\frac{5\pi \:}{\pi \:-1}\:-\frac{5}{\pi }$.

I know this is wrong per my textbook, but after an hour working on this problem I cannot figure out my error.

答:
I know this is wrong per my textbook, but after an hour working on this problem I cannot figure out my error.

$-\dfrac5{\pi^0}=-5$, because $\pi^0=1$.

Title: Basic Geometric Series Question-Stuck


答:
You did almost everything right. You took a sum from 1 to $\infty$ and added a $n=0$ term, and then subtracted it (your $\frac{-5}{\pi}$ term).

But you wrote it wrong, because
$$
\frac{5}{\pi^0}  = \frac{5}{1} = 5.
$$

Nice work otherwise!

There's another way to deal with a series starting at $1$. You write
\begin{align}
S
&= \sum_{n=1}^\infty \frac5{\pi^n} \\
\end{align}
and then you let $k = n-1$ in that expression, or $n = k + 1$ (which si the same thing), and rewrite the sum in terms of $k$. As $n$ goes from $1$ to $\infty$, $k$ goes from $0$ to $\infty$. So you get
\begin{align}
S
&= \sum_{n=1}^\infty \frac5{\pi^n} \\
&= \sum_{k=0}^\infty \frac5{\pi^{k+1}} \\
\end{align}
Now that's not actually in geometric-series form, but a little algebra -- $\pi^{k+1} = \pi \cdot \pi^k$, is enough to get you there:
\begin{align}
S
&= \sum_{n=1}^\infty \frac5{\pi^n} \\
&= \sum_{k=0}^\infty \frac5{\pi^{k+1}} \\
&= \sum_{k=0}^\infty \frac5{\pi\cdot\pi^{k+1}} \\
&= \sum_{k=0}^\infty \frac5{\pi}\cdot\frac{1}{\pi^{k}} \\
&= \frac{5}{\pi}\sum_{k=0}^\infty \frac{1}{\pi^{k}} \\
&= \frac{5}{\pi}\sum_{k=0}^\infty \left(\frac{1}{\pi}\right)^{k} \\
\end{align}
and that's a geometric series with $A = 1$ and $r = \frac1\pi$.

Title: Basic Geometric Series Question-Stuck

Body:
I'm studying Calc 2 and I have a basic series question.

A geometric series $\sum _{n=0}^{\infty }\:Ar^n$ is convergent if |r|<1 and the sum equals $\frac{a}{1-r}$ if the series is convergent.

Question: $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$

and $\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$ totals $\frac{5}{1-\frac{1}{\pi }}=\frac{5\pi \:}{\pi \:-1}$.

Therefore, $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$ = $\frac{5\pi \:}{\pi \:-1}\:-\frac{5}{\pi }$.

I know this is wrong per my textbook, but after an hour working on this problem I cannot figure out my error.


问:
Prove that running time T(n)=$(n + a)^b$ is $O(n^b)$ with a and b > 0
Image of the question

I'm studying asymptotic notation, but i'm stuck at this question, can anyone help me?

答:
By definition of $O(n^b)$ we need to show that it exists a constant $C>0$ such that $(n+a)^b \leq Cn^b$. (There should be absolute values, but assuming $a\geq0$ you don't need them; if $a<0$ the problem is trivial with $C=1$)

Now, can $f$ defined as $f(x) = \frac{(x+a)^b}{x^b}$ be unbounded as a function from $[1,+\infty)$ to $\mathbb{R}$?


问:
Green's theorem and conversion to polar coordinates
While subject to the force:

$$F(x,y) = y^3i + (x^3 + 3xy^2)j$$
a particle travels counterclockwise once around circle of radius 3. Find work done using Green's theorem.

So $P = y^3$, $Q=x^3+3xy^2$, and $dP/dy =3y^2$ and $dQ/dx = 3x^2 + 3y^2$

So where do I go from here? Why do I have to convert to polar coordinates from here? How do I do this?

答:
By Green's theorem
$$\int_{C}\vec F\cdot \,d\vec r=\int\int_{A}\bigg(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\biggr)\,dA$$
where $A$ is the area that the circle of radius $3$ encloses. I.e. $A=\{(x,y)\in \mathbb{R^2}\,:\, x^2+y^2\leq 9\}$. Substituting $\frac{\partial Q}{\partial x},\frac{\partial P}{\partial y}$ the second integrals equals to
$$\iint_{\{(x,y):\,x^2+y^2\leq 9\}}3 x^2\,dx\,dy $$

Now the easiest way to solve this is to use polar coordinates. Set $x=r\cos \theta$ and $y=r\sin \theta$. In polar coordinates the integral becomes
\begin{align}
\int_{0}^{2\pi}\int_{0}^{3}3\cdot r^3\cdot \cos ^2\theta\,dr\,d\theta&=\frac{81}{4}\cdot \int_{0}^{2\pi}\cos^2\theta\,d\theta\\
&=\frac{81}{4}\int_{0}^{2\pi}\frac{\cos 2\theta+1}{2}\,d\theta\\
&=\frac{81\pi}{4}+\int_{0}^{2\pi}\frac{\cos 2\theta}{2}\,d\theta\\
&=\frac{81\pi}{4}+\frac{\sin 2\theta}{4}\biggl|_{0}^{2\pi}\\
&=\frac{81\pi}{4}
\end{align}
where in the second equality we used the inequality $\cos 2\theta = 2\cos^2\theta-1$.


问:
Fourier transformation with the similarity theorem
I'm trying to calculate a simple fourier transformation but I got confused at a part and still don't know why .

$$f(t) = e^{−at^2}$$ knowing that  $$ F(w) = F({e^{−t^2}}) = \sqrt{\pi} . e^{-w^2/4}     $$

We can calulate the Fourier transformation of f(t) . From the fourier table the result should be :
$ \sqrt{\pi/a} . e^{-w^2/4a}    (I)$

But when calculating the function with the similarity theorem $  (1/|a|) . F(w/a) $
I get : $(\sqrt{\pi}/a)  . e^{-w^2/4a^2}  $ which is not the same as ( I ) what I'm doing wrong here ?

isn't $ 1/|a| F(w/a) = (1/a) .  e^{-(w^2/a^2)/4} = 1/|a| . e^{-w^2/4a^2} $    ?

答:
It looks like you are using $a^2$ instead of $a$, i.e., you're computing $F(e^{-(at)^2})$ instead of $F(e^{-at^2})$.

It is not too hard to see what the result should be using a change of variable $s = \sqrt{a}\,t$ in the integral defining the FT (note your normalization of where the factors of $\pi$ or $\sqrt{\pi}$ go in the definition of the FT may be different; I was lazy, so I used this convention—I didn't bother to check if this definition of the FT works on Gaussians like it does in your post, but that doesn't matter, the important thing is that the change of variable approach works no matter what your normalizations in the definition of the FT are):

\begin{align*}
F(e^{-at^2})(\omega) &= \int e^{-at^2}e^{i\omega t}\,dt\\
&= \int e^{-s^2}e^{i\omega(s/\sqrt{a})}\,d(\frac{s}{\sqrt{a}})\\
&= \frac{1}{\sqrt{a}}F(\omega/\sqrt{a})\\
&= \sqrt{\frac{\pi}{a}}\,e^{-(\omega/\sqrt{a})^2/4}\\
&= \sqrt{\frac{\pi}{a}}\,e^{-\omega^2/4a}.
\end{align*}

If we plug in $a^2$ instead of $a$, we see we would get at the end
$$
F(e^{-a^2t^2})(\omega) =\sqrt{\frac{\pi}{a^2}}\,e^{-\omega^2/4a^2} = \frac{\sqrt\pi}{a}\,e^{-\omega^2/4a^2},
$$
which is what you were getting.


问:
Solutions in primes of $\,\frac{p+1}2=q=\frac{r+1}4$
Why the only solutions in primes of
$$\,\frac{p+1}2=q=\frac{r+1}4$$
seem to be $\,(p,q,r)=(3,2,7)\,$ and $\,(p,q,r)=(5,3,11)$?

Said in other terms: "if $\,r\,$ is a prime, different from $\,7\,$ and $\,11$, and $\,r-1=2p\,$ is a semiprime, then $\,r+1=4q\,$ must have at least $\,4\,$ factors $\,[2, 2, s, t]$".

Many thanks.

答:
Note that the conditions imply that $(p,q,r)=(2q-1,q,4q-1)$. If $q>3$ then either $q=3n+1$ or $q=3n-1$ for some $n$.

If $q=3n+1$ then $$r=4q-1=12n+4-1=12n+3$$ is divisible by $3$.

If $q=3n-1$ then $$p=2q-1=6n-2-1=6n-3$$ is divisible by $3$.


答:
To prove it, show:
$$
\lim_{n \to \infty} \frac{(n+a)^b}{n^b} = 1
$$

Title: Prove that running time T(n)=$(n + a)^b$ is $O(n^b)$ with a and b > 0

Body:
Image of the question

I'm studying asymptotic notation, but i'm stuck at this question, can anyone help me?


问:
Find a invertible matrix $Q$ such that $AQ$ = $B$
Hi I have calculate this matrices:

$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$

$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$

And I'm trying to find a invertible matrix $Q$ such that $AQ$ = $B$, but I'm stuck.

Can you help me?

I have not seen determinats yet

答:
From linear independence of the rows we see that $A$ is invertible and so $Q = A^{-1}B$. You can use adjugate matrix and determinant to compute $A^{-1}$

$$
A^{-1} = \frac{{\rm adj} A}{\det A} = \frac{\begin{pmatrix}
1/3 & 1/3 & 0 \\
-1/3 & 1/3 & 1/3 \\
0 & -1 & 0
\end{pmatrix}}{\frac{1}{3}} =
\begin{pmatrix}
1 & 1 & 0 \\
-1 & 1 & 1 \\
0 & -3 & 0
\end{pmatrix}.
$$

Alternatively, you can find $A^{-1}$ using Gaussian elimination. Finally, we get

$$
Q = A^{-1} B  = \begin{pmatrix}
1 & 0 & 1 \\
0 & 2 & 1 \\
1 & 1 & 0
\end{pmatrix}.
$$


问:
Calculation of angles in regular polyhedrons
For any regular polyhedron (tetrahedron, cube, octahedron, dodecahedron, icosahedron, ...), how do I calculate an angle from the center of the regular polyhedron to the center of each face? Obviously, the angle decreases with the increasing number of faces. What is the formula, when the number of faces n is the variable?

For cube, n= 6, the angle is 90 degrees:

答:
What you're describing is the supplementary angle of the dihedral angle of the solid. This latter angle $\theta$ is the interior angle between two connected faces, with

$$\sin\frac\theta2 = \frac{\cos\frac\pi q}{\sin\frac\pi p}$$

for a solid with Schläfli symbol $\{p,q\}$. See the full chart in the Wikipedia article on Platonic solids.

Therefore, taking the supplement, you're looking for
$$2\pi - \theta.$$


问:
Showing uniqueness of Riemann's Integral
I am given the definition: Le $f$ be defined on $[a,b]$. we say that $f$ is Riemann Integrable on $[a,b]$ if there is a number $L$ with the following property: for every $\epsilon>0$, there is a $\delta > 0$ such that $\left\|P\right\|< \delta$ implies $| \sigma -L| < \epsilon$ where $\sigma$ is the Riemann Sum of $f$ over the partition $P$ of $[a,b]$. In this case, we say that $L$ is the Riemann Integral of $f$ over $[a,b]$, and write $\int_{a}^{b} f(x) dx=L$

I am then asked to show why $L$ is a unique limit. It does make sence if not unique but how to show is well above me.

答: suppose there is another limit different form $L$ call it $P$ then $\exists \epsilon : |L-P|>2\epsilon$ now by definition $2\epsilon>|\sigma-L|+|\sigma-P|\geq|L-P|>2\epsilon$</p>


答:
You can start by restating the following inequality.

Lemma:

$\forall n \in \mathbb{N}$, we have:
$$ \| A^n -B^n \| \le n \|A-B\| c^{n-1}$$
where $c= \max( \| A \|, \| B\| )$.

Demonstration (Outline)

It's just a direct consequence of the submultiplicity of $\| \|$. For example, for $n=3$ , $$ A^3-B^3 = (A-B)A^2+B(A-B)A+B^2(A-B)$$
$\square$

Then you can see that:
$$ \| \exp(A) -\exp(B) \| \le \sum_{n \ge 1} \| A^n-B^n \| \dfrac{1}{n!}$$

The rest is simple $\square$

Remark: A similar proof by MVT like given by mathecounterexample can also be carried out, some final arguments must ofcourse be completed by using the submultiplicity.

Title: Prove that $\lVert \exp(A) - \exp(B) \rVert \leq \lVert A-B \rVert e^{\max\{\lVert A \rVert, \lVert B \rVert \}}$

Body:
, where A and B are square Matrices, exp(A) is the matrix exponential

I can see where the $\lVert A-B \rVert$ comes from, but can't figure out how to get the $ e^{max\{\lVert A \rVert, \lVert B \rVert\}} $. The definition for the matrix exponential we're using is the following:
$\exp(A)= \sum \frac{A^k}{k!}$.


问:
Find constant such that inequality is true
I have to find $c>0$, such that following inequality is true:
$$-3x^2+2|x||y|-y^2\leq -c(x^2+y^2)$$
for all $x,y \in \mathbb{R}$.
Is there any way to find $c$ except trying random values? What if I want to find the best constant?

答:
The proposed inequality is equivalent to
\begin{align*}
3x^{2} - 2|x||y| + y^{2} \geq c(x^{2} +y^{2}) \Longleftrightarrow (3-c)|x|^{2} - 2|x||y| + (1-c)|y|^{2} \geq0
\end{align*}

Which is equivalent to $3 - c > 0$ and $\Delta \leq 0$. More precisely, one has
\begin{align*}
4y^{2} - 4(3-c)(1-c)y^{2} \leq 0 & \Longleftrightarrow 4y^{2} - 4(3 - 4c + c^{2})y^{2} \leq 0\\\\
& \Longleftrightarrow (4 - 12 + 16c - 4c^{2})y^{2} \leq 0\\\\
& \Longleftrightarrow 4c^{2} - 16c + 8 \geq 0\\\\
& \Longleftrightarrow c^{2} - 4c + 2 \geq 0\\\\
& \Longleftrightarrow (c-2)^{2} - 2 \geq 0\\\\
& \Longleftrightarrow c\in(-\infty,2-\sqrt{2}]\cup[2+\sqrt{2},+\infty)
\end{align*}

Since $0 < c < 3$, we conclude that $c\in(0,2-\sqrt{2}]$.

Hopefully this helps!


问:
Is the fact that $g$ is defined from $\mathbb{R}$ to $\mathbb{R}$ enough for me to claim the following?
The questions was:

Given $f, g$ from $\mathbb{R}$ to $\mathbb{R}$, proof or disproof that if
$f * g$ (or $f(g(x))$) is $*$on $\mathbb{R}$, then $f$ is on $\mathbb{R}$.

I wrote that the claim is true, citing that because $g$ is a function from $\mathbb{R}$ to $\mathbb{R}$, then if $f$ wasn't on $\mathbb{R}$, $f * g$ would be undefined.

I am not sure whether this proof is valid or not.

$*$on is defined as:
function $f:B\to A$ is on group $A$, iff for a given $a\in A$, there is $b\in B$ such that $f(b) = a$

答: Consider the functions $g:\mathbb{R} \to \mathbb{R}$ and $f:\mathbb{R} \to \mathbb{R}$ You are asked if $f * g$ is on $\mathbb{R}$ if $f$ is onto $\mathbb{R}$.  So you are taking $g$ which is by definition on $\mathbb{R}$ and going to $\mathbb{R}$.  Does this seem to be evident by definition?  It does to me.


问:
How to calculate how long each side of an regular polygon should be to fit on a grid of x*x size?
I found this formula (see link) that's supposed to calculate the length of each side of my polygon. But I end up with an negative size, my polygon has 16 sides and an radius of 24 cm. So how do I calculate how long each side should actually be?

https://i.stack.imgur.com/1G2gI.png

答:
You've most likely used a calculator to compute
$$
\sin(\frac{180}{16})
$$
and gotten $-0.967$, but that's because the calculator is using radians, not degrees. The thing you really wanted to compute was
$$
\sin(\frac{\pi}{16}) \approx 0.19509032201
$$
which, when multiplied by 48 (i.e., twice the radius) gives you a side-length of approximately
$
9.36433545677
$ centimeters.

(You could also switch your calculator to "degrees" mode to get the same answer.)
I hope this helps.


问:
Can there be any representation of a finite group which is reducible but indecomposable?

Can there be any representation of a finite group which is reducible but indecomposable?

I know that it is true that irreducibility implies indecomposability. But the converse is not necessarily true. For instance we could take $\Bbb Z$ and the representation $\varphi_n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ which is not irreducible but indecomposable. Is there any analogous example for the representation of finite groups? Any help in this regard will be highly appreciated.

Thanks in advance.

答:
There are analogues in finite characteristic -- the same as your example for $n=1$ if the group is cyclic of order $m$, and $m \bmod p=0$.
In characteristic zero, irreducibility for finite group representations is equivalent to indecomposability. This is due to Maschke's theorem, see for example
Understanding Maschke's Theorem


问:
How to compare the angles of pixels of an image in different noises?
I am working on a project that needs the angles of pixels of images. I have found the way to do it in matlab for a clean image. Now, I need to add different amounts of additive white Gaussian noise to the image and I need to see whether the code works well and I also need to find out how the angles change when the amount of additive noise is decreased or increased. I used hist() function in matlab to see the angle vlaues but it is very difficult to compare in this way. For example in the case of peppers test image:

The following is the histogram of angles that I obtained using matlab:

I really appreciate it if someone could please help me to find a better way to
visually represent the angles values in order to find out whether the code works fine for different amount of noise. These are part of values for the angles of edge pixels of peppers:

答:
Your question is still a bit unclear, but I am going to try to fill in the gaps and make a suggestion. First, here is what I believe you are trying to ask:

You have performed an edge detection on an image, and have a set of pixels identified as 'edge' pixels. For each of those pixels, you have a gradient vector, which makes some angle $\theta$ from horizontal. Now you want to test the robustness of your algorithm to additive noise in the image by running the same calculation on the image with noise added to it, and comparing the value of the angle you got for each pixel. Is this a correct interpretation of your question?

If so, here is a suggestion. Assign each pixel in the entire image the angle $-180\lt\theta\leq 180$ calculated for the edge it is part of. For pixels not part of an edge, assign them a fake angle outside of your range of actual angles, like 200 degrees, so that you can identify them easily on a plot. Let $\theta_n^{(1)}$ and $\theta_n^{(2)}$ be the angle of the $n^{th}$ pixel in the first and second images, respectively. The integer $n$ runs through every pixel of your image (say, the flattened array). Then make a scatter plot of $\theta^{(2)}$ vs  $\theta^{(1)}$. Make sure that each $n$ refers to the same pixel in each image.

If your algorithm were completely insensitive to noise and you got exactly the same angles for every pixel in the two images, the points in the scatter plot would fall on a diagonal line running through the origin, with unit slope. The extent to which the points deviate from this line will give you a visualization of how different your answers are for the two calculations.


答:
Maybe your teacher want you to solve the equations one by one for
each column of $B$. For example let the first column of $Q$ be $q_{1}=\begin{pmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{pmatrix}$ Then you have the equations $Aq_{1}=b_{1}$, that is
\begin{align*}
x_{1}+\frac{1}{3}x_{3} & =\frac{4}{3}\\
-\frac{x_{3}}{3} & =-\frac{1}{3}\\
x_{1}+x_{2}+\frac{2}{3}x_{3} & =\frac{5}{3}
\end{align*}
From the second line $x_{3}=1$, from the first line $x_{1}=1$ and
from the third line $x_{2}=0$. Hence the first column of $Q$ is
$$
q_{1}=\begin{pmatrix}1\\
0\\
1
\end{pmatrix}.
$$
You can find the other columns simply, in a similar way.

Title: Find a invertible matrix $Q$ such that $AQ$ = $B$

Body:
Hi I have calculate this matrices:

$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$

$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$

And I'm trying to find a invertible matrix $Q$ such that $AQ$ = $B$, but I'm stuck.

Can you help me?

I have not seen determinats yet


答:
Suppose that we have $$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$

$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$

and we want $Q$ such that $AQ =B$. Let

$$ Q = \begin{pmatrix} a & b &c \\  d & e & f \\ g & h & i \end{pmatrix}$$

Then using matrix multiplication you'd see that

$$ AQ =  \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}   \begin{pmatrix} a & b &c \\  d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix} a + \frac{g}{3}  & b + \frac{h}{3} & c + \frac{i}{3} \\ -\frac{g}{3} & \frac{-h}{3} & \frac{-i}{3} \\ a + d + \frac{2g}{3} & b + e + \frac{2h}{3} & c + f + \frac{2i}{3} \end{pmatrix} $$

now set it equal to $B$

$$ \begin{pmatrix} a + \frac{g}{3}  & b + \frac{h}{3} & c + \frac{i}{3} \\ -\frac{g}{3} & \frac{-h}{3} & \frac{-i}{3} \\ a + d + \frac{2g}{3} & b + e + \frac{2h}{3} & c + f + \frac{2i}{3} \end{pmatrix} =  \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$

Multiply by $3$ on both sides and we get

$$ \begin{pmatrix} 3a + g  & 3b + h & 3c + i \\ -g & -h & -i \\ 3a + 3d + 2g & 3b + 3e 2h & 3c + 3f + 2i \end{pmatrix} =  \begin{pmatrix} 4 & 1 & 3 \\ -1 & -1 & 0 \\ 5 & 8 & 6\end{pmatrix}$$

Then you directly get $g,h,i$

$$ -g = -1 \implies g = 1 \\ -h = -1 \implies h = 1 \\ -i = 0 \implies i=0$$

Now substitute those into the top row

$$ 3a +1 =  4 \implies a =1 \\ 3b + 1 = 1 \implies b = 0 \\ 3c = 3 \implies c =1 $$

then continue and get the last three


问:
A Measure For The Space of Probability Density Functions
Consider the space of all joint probability density functions of two variables. I want to know what the measure is of the portion of this space that is filled by uncorrelated joint pdfs relative to the size of the entire space. In other words, what is the fraction of all possible joint pdfs that are uncorrelated?

The motivations for the question is a desire to quantify how much it matters if I take into account correlations or not, for a generic joint pdf.

Any ideas for directions to look would be greatly appreciated.

答:
I think that you're attacking this from the wrong angle.

First, your idea is fraught with technical difficulties. You didn't mention what kind of space you envision this space of PDFs to be. You could start with just the set, and simply define a $\sigma$-algebra on its powerset. But there's a myriad of ways to do that, which one do you pick?

Or you could find a topology, and try to create the analog of the Borel $\sigma$-algebra on your set. That means picking a topology first. Now, if you look at all distributions (CDFs), I think the Lévy-metric might give you a topology which corresponds to convergence in measure. For PDFs, however, things are harder, because not all distributions have PDFs.

Say you have picked a suitable topology, and used that to define a $\sigma$-algebra. How are you going to define a measure? Doing this involves some kind of idea about which sets are supposed to have large measures, and which sets small ones. In other words, you need to have an idea about which PDFs are likely, and which are unlikely. Even picking a kind of "uniform" measure won't solve this - it'll then be the metric (or topology) you picked which actually defines this.

I think you have to start with asking yourself which PDFs you want to consider. It seems that what you're actually interested in is estimating (for a specific situation?) how "likely" it is for two random variables to be correlated. For that question to be meaningful, however, you have to define the circumstances in much more detail than you have done here.


问:
Is a matrix row equivalent to its transpose?
If we know that a matrix (A) nxn is invertible wich implies that also its transpose, can we demonstrate that A is row equivalent to its transpose?</p>

答: Yes! Using Gauss' algorithm you can show that every invertible matrix is row-equivalent to the identity matrix. That means that both $A$ and $A^T$ are  row-equivalent to the identity matrix thus, according to the transitive property of equivalence, $A$ and $A^T$ are row-equivalent.


问:
Relationship between singular values and eigenvalues of a square matrix A?
What is the relationship between the singular values and the eigenvalues of $$AA^T$$ if $A$ is a square matrix (if that even matters)?

So what I did:

$$AA^T = USV^TVSU^T = USSU^T = US^2U^T$$

I read on google that the singular values of $A$ are the square root of the eigenvalues of $AA^T$ but I don't really know how to derive this from the equation above (or at all)?

Any hints how I can proof this?

答:
That's the relationship exactly. If you take the eigen decomposition of $AA^T$ just rewrite it.

$$ AA^T = U \Sigma^{2} U^T = U \Lambda U^T $$

where $\Lambda$ is the matrix of eigenvalues for $AA^T$. If $\Sigma^{2} = \Lambda$ then

$$ \sigma_{i} = \sqrt{\lambda_{i}} $$

Note that $AA^T$ is positive semidefinite so the eigenvalues are non-negative. If you multiply a diagonal matrix with another diagonal matrix it just multiplies each element by the other one so this squares them.


问:
How limit of multivariable function with scalar inputs are related to vector inputs
I was watching lectures by professor Herbert gross(from YouTube) on multivariable calculus and he gradually move from introduction of vectors to n-tuples and then multivariable calculus .

And what I understand that there are some properties  of vectors that are followed by vector spaces but not all properties . And  how a multivariable function takes   n-tuple as input and output a scalar function .

And limit in both cases i.e when input of a multivariable scalar function is a vector and when input of same function is n-tuple ,in both cases definition of limit behaves exactly similar.

But how can we prove using vector space properties only that limits definition would be same in both cases ?

答:
We can define multivariable functions, and we can define vector functions (meaning vectors as inputs, not as outputs).

A multivariable function in $n$ variables is a function defined on a subset of $X_1\times\dots\times X_n$, where $X_n$ are arbitrary sets.

A vector function is a function defined on a subset of a vector space.

Now we can notice a certain overlap between these concepts. Let $X_1=\dots=X_n=\mathbb R$. Let $U\subseteq X_1\times\dots\times X_n=\mathbb R^n$. Then a function defined on $U$ is both a multivariable function, since it's defined on a subset of a Cartesian product, and a vector function, since it's defined on a subset of the vector space $\mathbb R^n$. So in this case, vector functions and multivariable functions are one and the same thing. Whatever we prove for vector functions, we automatically prove for multivariable functions with inputs from a field, since such multivariable functions are vector functions.

That said, in analysis it is really for the best to think of them as vector functions, not multivariable functions, since it's the structure of the vector space which allows us to do analysis in the first place. The fact that the functions we consider often happen to be multivariable ones is a nice bit of trivia, but really has no bearing on the theory at all. For instance, the vector space of functions of the form $a\sin x+b\cos x$ is a perfectly fine vector space on which we can do analysis, but it's not a Cartesian product, so functions defined on this space aren't multivariable ones.


问:
Is $\limsup_{t} \frac{a_t}{ h(t+1) - h(t) }$ finite when $0 \leq \sum_{i=t}^\infty a_i \leq h(t) = t^{-\alpha}$?
Consider a monotone decreasing sequence $\{a_i\}_{i=1}^\infty$, where $a_i  \geq 0$ for all $i$. This sequence is assumed to satisfy the bound on its residual series
$$
\sum_{i=t}^\infty a_i \leq h(t) = t^{-\alpha} \;\;\;\forall t\in\mathbb{N}
\;,
$$
where $\alpha \geq 1$. From the above bound, it is clear that $\{a_i\}$ is summable since we have $\sum_{i=1}^\infty a_i \leq 1 < \infty$.

My objective is to determine whether
$$\limsup_{i\rightarrow \infty} \frac{a_i}{-\Delta h_i} < \infty \,,$$
where $\Delta h_i = h(i+1) - h(i)$.

Thus far, I have only been able to prove the weaker property that
$$ \liminf_{i\rightarrow\infty}  \frac{a_i}{-\Delta h_i} \leq 1 \;,$$
which is done by showing that $a_i \leq -\Delta h_i$
is true for infinitely many $i$ (see below for proof).

I have found that the answer may be related to the Stolz-Cesaro Theorem or its converse, but I have been unable to make the exact requirements of the theorem line up with what I need. I have also found that this may be connected to the following post.

Any help on this would be appreciated. I am also curious about the more general case when we have some convex function $h:(0,\infty)\rightarrow (0,\infty)$ satisfying $\lim_{i\rightarrow\infty} h(i) = 0$.

Proof that $a_i \leq -\Delta h_i$ for infinitely many $i$: Assume the converse for all but finitely many $i$. Then there exists a $T\in\mathbb{N}$ such that for all $t \geq T$,
$$
\sum_{i=t}^\infty a_i > - \sum_{i=t}^\infty \Delta h_i = h(t)
\;,
$$
which is a contradiction. Hence we must have that $a_i \leq -\Delta h_i$ holds infinitely often.

答:
I think the proof gets easier if you replace all the discrete stuff by continuous stuff (eg integrals instead of sums and $h'$ instead of $h(t+1)-h(t)$). I think it is not hard to do some bounds why the continuous problem is equivalent.

If $\frac{a(T)}{h'(T)}=c$, then $a(t)\ge c h'(t)\ \forall t\le T$ because of monotonicity of $a$ and $a(t)\ge 0\ \forall t> T$

For $\alpha=1$ one can show this way that $\limsup_{t\to\infty}\frac{a(t)}{h'(t)}\leq 4$ by solving this optimization problem: https://www.wolframalpha.com/input/?i=min+1%2Ft-%28T-t%29%28T%29%5E%28-2%29*c+from+t%3D0+to+%E2%88%9E (one should optimize $t$ from $0$ to $T$, but it is equiavelnt to optimize from $0$ to $\infty$ which is easier for WolframAlpha) coming from the integral condition $h(t)-\int_t^{\infty}a(s)ds\ge 0$. (You see that the minimum would be negative for $c>4$).

Probably similar argumetns are posssible for $\alpha>1$.


问:
I have to find a matrix$A_{4 \times 4}$ with eigenvalues $2$ and $4$ and that both $E(2, A)$ and $E(4, A)$ are $1$ dimensional
I know that matrix with eigenvalues $2$ and $4$ can be made from just putting $2$ and $4$ on the diagonal. I have tried many different matrices that have the eigenvalue of $2$ and $4$ but I can not find one that satisfies the rule that that both $E(2, A)$ and $E(4, A)$ are $1$ dimensional. I have come up with some that satisfy either $E2$ or $E4$ but I can not come up with one that satisfies both of them.

答:
If a matrix has all eigenvalues distinct, it is diagonalisable and all eigenspaces are one-dimensional. Recall too that the eigenvalues of a diagonal matrix are simply the diagonal entries.

Thus a matrix with eigenvalues $2,4$, both with one-dimensional eigenspaces, is the diagonal matrix with non-zero entries $1,2,3,4$.


问:
Probability of coin toss there are 68% chance that % heads in the range 50% plus or minus SD?
A coin is tossed $2500$ times. There is about a $68\%$ chance that the percent of heads is in the range $50\%$ plus or minus ($0.5$, or $1$, or $1.5$, or $2$, or $2.5$)?

$\text{P(of coin tossed is 1/2)}=0.50  \pm  \text{SD}$

$n=2500$

$μ=E[X]=x_1∗p_1+x_2∗p_2+x_3∗p_3+...+ x_n∗p_n$
$μ=1*(1/2) + (0*(1/2))=1/2$

The standard error is:

$EX=\sqrt{E[(X−μ)^2]}=P$

$=\sqrt{(x_1−μ)^2∗p_1+(x_2−μ)^2∗p_2+(x_3−μ)^2∗p_3+\cdots+(x_n−μ)^2∗p_n}$
$= \sqrt{(1-1/2)^2*(1/2) + (0-1/2)^2*(1/2)}$
$= \sqrt{2/8}$

$P= 0.5$

$σ=\sqrt{(p*(1 - p))/n}$

$σ=\sqrt{(0.5*(1-0.5))/2500}= \pm 1% = 0.01$

答:
Hint:  If you let $x_1=1$ for heads and $x_2=0$ for tails and plug into the equations what do you get?  What are the probabilities?

Added:  Your added work is correct except that you have expressed $\sigma$ as a percentage without indicating it.  The absolute standard deviation is $\sqrt {2500 \cdot 0.5 \cdot (1-0.5)}=25$.  You have written $\sqrt {0.5 \cdot (1-0.5)/2500}$ which is $0.01=1\%$


问:
Prove that $\angle ABC <$ $ 84^\circ$.
Angles A, B and C meet the circle at P, Q and R respectively.

I made the picture below, but what would the solution be?

答:
$\begin{array}{} \text{Euler's relation} & IO=\sqrt{R(R-2r)} & IO=r=1 & R=1+\sqrt{2} \end{array}$

$\begin{array}{} sin(\frac{ π }{6})=\frac{1}{2} & AI=2 & AI^2=1+y_{A}^2 & A=(1,y_{A}=\sqrt{3}) \end{array}$

$\begin{array}{} {O}=λ ∩ ω & O=(h,k) \end{array}$

$\left\{ \begin{array}{} x^2+y^2=1 \\ (x-1)^2+(y-\sqrt{3})^2=(1+\sqrt{2})^2  \end{array}   \right.$

$\begin{array}{} h=\frac{1-\sqrt{2}+\sqrt{3+6\sqrt{2}}}{4} & k=\frac{-3+3\sqrt{2}+\sqrt{3+\sqrt{2}}}{4\sqrt{3}}  \end{array}$

$\begin{array}{} {C}=ψ∩t & \left\{ \begin{array}{} x=1 \\ (x-h)^2+(y-k)^2=(1+\sqrt{2})^2  \end{array}  \right. \end{array}$

$\begin{array}{} y_{C}=k+\sqrt{2h-h^2+2\sqrt{2}+2} & y_{C}=\frac{3\sqrt{3}+3\sqrt{6}+\sqrt{9+18\sqrt{2}}}{6}  \end{array}$

$\begin{array}{} AC=\sqrt{3}+y_{C} & AC=\frac{3\sqrt{3}+\sqrt{6}+\sqrt{1+2\sqrt{2}}}{2}  \end{array}$

$\begin{array}{} ϕ= \angle(AOB)& θ=\frac{ϕ}{2} & AO=OC=1+\sqrt{2} \end{array}$

$\begin{array}{} AC^2=AO^2+OC^2-2·AO·OC·cos(ϕ) & cos(ϕ)=1-\left( \frac{AC}{OC}  \right) ^2  \end{array}$

$cos(ϕ)=\frac{-7+4\sqrt{2}-5\sqrt{3+6\sqrt{2}}+3\sqrt{6+12\sqrt{2}}}{4}$

$\begin{array}{} θ=\angle(ABC) & θ=\frac{ϕ}{2}  \end{array}$

$θ=\frac{1}{2}arccos\left( \frac{-7}{4}+\sqrt{2}-\frac{5}{4}\sqrt{3+6\sqrt{2}}+\frac{3}{4}\sqrt{6-12\sqrt{2}}  \right)·\frac{180}{π}=83.905711$

$\begin{array}{} \text{checking} & θ<\frac{7π}{15} & ϕ=2θ<\frac{14π}{15} \end{array}$

$\begin{array}{} cos(ϕ)>cos(\frac{14π}{15}) & cos(ϕ)>cos(\frac{π}{15}) & cos(\frac{π}{15}>-cos(ϕ)) \end{array}$

$\frac{1}{8}\left( -1+\sqrt{5}+\sqrt{6(5+\sqrt{5}}  \right) >\frac{-1}{4}\left( -7+4\sqrt{2}-5\sqrt{3+6\sqrt{2}}+3\sqrt{6-12\sqrt{2}}  \right)$

how to prove it exactly?

$\begin{array}{} \text{True (numerical verification)} & \text{difference} & 6.896×10^{-4} \end{array}$

($λ$) Incircle of a $triangleABC$, (locus of point $O$) with center $I=(0,0)$ and radius $r=1$, $T=(1,0)$, ($t$) vertical line $x=1$ (locus of points $A$ and $C$), [$\frac{π}{3}$ angle construction] ($μ$) circle with center $I$ and radius=2, $A=(1,\sqrt{3})$ intersection $μ$ and $t$, $AI=2$, $D=(-2,0)$ intersection $μ$ and $x-axis$, $AD$, $\angle(IAT)=\frac{π}{6}$, $sin(\frac{π}{6})=\frac{1}{2}$, $\angle(DAI)=\frac{π}{6}$, ($ω$) circle with center $A$ and $R=1+\sqrt{2}$, (Euler's relation), $O=(h,k)$ intersection $λ$ and $ω$, ($ψ$) circle by $A$ with center in $O$ (locus points $A$, $B$ and $C$), $B$ intersection $AD$ and $ψ$, $C$ intersection $ψ$ and $t$, $BC$

Title: Prove that $\angle ABC <$ $ 84^\circ$.


问:
Find the probability density function of the random variable y=y(x)=x^2 , x with known probability density function.
Find the probability density function of the random variable y=y(x)=x^2, x with known probability density function,

$f(x) =
     \begin{cases}
       0 &\quad\text{if} \; \infty \lt x \lt -1 \\
       \frac{1}{2}  &\quad\text{if} \; -1 \le x \lt 0 \\
       \frac{1}{2}e^{-x} &\quad\text{if} \; 0 \le x \lt \infty \\
     \end{cases}$

I want to use the theorem that says $g(y)=\lvert \frac{dx}{dy} \rvert f(x)$ but since y=x^2 is not bijective I don't know how to proceed. I know that I have to work with $x=-\sqrt{y}$ separately from $x=\sqrt{y}$ but don't know in wich way. Any other method to solve this is fine. Thanks

答: Try finding the CDF of $Y$ and differentiating. Note that for $y \ge 0$ we have $P(Y \le y) = P(-\sqrt{y} \le X \le \sqrt{y})$.


答:
For another class of examples, consider the matrix $D=\begin{bmatrix}2&0\\0&4\end{bmatrix}$ and consider any invertible matrix $P$.

The matrix $PDP^{-1}$ then has eigenvalues $2$ and $4$ as well, each of which occurring with dimension $1$.

Title: I have to find a matrix$A_{4 \times 4}$ with eigenvalues $2$ and $4$ and that both $E(2, A)$ and $E(4, A)$ are $1$ dimensional

Body: I know that matrix with eigenvalues $2$ and $4$ can be made from just putting $2$ and $4$ on the diagonal. I have tried many different matrices that have the eigenvalue of $2$ and $4$ but I can not find one that satisfies the rule that that both $E(2, A)$ and $E(4, A)$ are $1$ dimensional. I have come up with some that satisfy either $E2$ or $E4$ but I can not come up with one that satisfies both of them.


问:
Prove that there are no natural numbers to satisfy the equation.
Prove that there are no natural numbers to satisfy the equation:

$a^2+b^2=3(c^2+d^2)$.

I tried to factorize it:
$a^2 + b^2 - 3 (c^2 + d^2)=0$  but I am not getting anywhere.
I can't think of anything else!!

答: Hint: do you know the criterion for a number to be the sum of two squares?</p>


问:
Definition of Arithmetic Mean
I am currently in school and we are learning about Arithmetic Sequences. In one of the sections in our learning material, the arithmetic mean is defined as the term of the arithmetic sequence that is between two given terms. For me, at least, I know this is wrong. In all the reliable sources I found about the definition of arithmetic mean, its definition is the sum of the numbers divided by the count of the numbers. Here, an example is given that the arithmetic mean(s) of 2, 4, 6, 8, 10 are 4, 6, and 8 (which should be 6). Our material also mentions about inserting arithmetic means between two numbers like "Insert 4 arithmetic means between 7 and 47," wherein the answer would be 15, 23, 31, and 39 because it will form an arithmetic sequence. As far as I know I don't know anywhere where this definition is taught. I think this is stupid. So my questions are,

Is this really accepted? or where is it accepted?

What should I do about it? We have been taught this 2 months ago but no one bothered to ask the teacher about it that time

答: Why should the $8$ (which is arithmetic mean of $6$ and $10$) be instead $6$ as you say in the parentheses of your example starting "Here, an example"? It looks like in that sequence of your cited example that each term except first and last is indeed the arithmetic mean of the terms before and after it. The inserted sequence doesn't seem to be related to any use of arithmetic means of more than two terms.

Title: Definition of Arithmetic Mean


问:
If a function is discontinuous, why doesn't it contradict the Fundamental Theorem of Calculus
So for example, suppose $g(x) = \int_o^x F(t) dt$ where $$F(t) =
\begin{cases}
t & 0 \leq t \leq 1 \\
t - 1 & 1 < t \leq 2
\end{cases}
$$

The function $g$ is not differentiable at 1. I am curious to why it doesn't contradict the Fundamental Theorem of Calculus. Some thoughts and assistance

答: Because the Fundamental Theorem of Calculus states that if $f$ is continuous, then$$x\mapsto\int_a^xf(t)\,\mathrm dt$$is differentiable. And your function $F$ is not continuous.


问:
how do I prove that the matrix certain numbers has a determinant that is divisible by the same common divisor of said numbers?
Recently, I have been given the following task, which is practically unsolvable for me: The numbers 2014, 1484, 3710 and 6996 are all divisible by 106. Without calculating it, show that the determinant of \begin{pmatrix}2&0&1&4\\ \:\:1&4&8&4\\ \:\:3&7&1&0\\ \:\:6&9&9&6\end{pmatrix} is also divisible by 106. I simply don't know where to begin, I have tried reducing the matrix, finding a similar matrix, finding some 2 matrices that are divisible by 106 and if multiplied with each other yield the given matrix, but everything as failed so far. Can anyone give me a hint, what I could use to make this prove?

答:
If your matrix is $M$, note that
$$ M \pmatrix{1000\cr 100\cr 10\cr 1} = \pmatrix{2014\cr 1484\cr 3710\cr 6996}$$
The fact that each term on the right is divisible by $106$ tells you that your matrix is singular modulo certain primes...


答:
Splicing the answer from here. Number the columns $C_1,C_2,C_3,C_4$ from left to right. The operation
$$C_4\leftarrow 1000C_1+100C_2+10C_3+C_4$$
does not change the determinant, but leaves the last column having numbers all divisible by $106$ (per the givens in the question). If now we divide the last column by $106$, the determinant is divided by the same amount, and we still have an integer matrix with integer determinant. Thus the original matrix has a determinant divisible by $106$.

Title: how do I prove that the matrix certain numbers has a determinant that is divisible by the same common divisor of said numbers?

Body: Recently, I have been given the following task, which is practically unsolvable for me: The numbers 2014, 1484, 3710 and 6996 are all divisible by 106. Without calculating it, show that the determinant of \begin{pmatrix}2&0&1&4\\ \:\:1&4&8&4\\ \:\:3&7&1&0\\ \:\:6&9&9&6\end{pmatrix} is also divisible by 106. I simply don't know where to begin, I have tried reducing the matrix, finding a similar matrix, finding some 2 matrices that are divisible by 106 and if multiplied with each other yield the given matrix, but everything as failed so far. Can anyone give me a hint, what I could use to make this prove?


答:
The matrix
$$
\pmatrix{2&0&1&4\\1&4&8&4\\3&7&1&0\\6&9&9&6} \pmatrix{1&0&0&1000\\0&1&0&100\\0&0&1&10\\0&0&0&1} =
\pmatrix{2&0&1&2014\\1&4&8&1484\\3&7&1&3710\\6&9&9&6996}
$$
has the same determinant. Dividing the determinant of this matrix by $106$ gives us the determinant of the matrix
$$
\pmatrix{2&0&1&2014/106\\1&4&8&1484/106\\3&7&1&3710/106\\6&9&9&6996/106}.
$$
This is a matrix with integer entries, so this determinant is an integer.

Equivalently: in terms of column operations, the operations
$$
C_4 \to 1000 C_1 + 100 C_2 + 10 C_3 + C_4\\
C_4 \to \frac 1{106}C_4
$$
gives us a matrix whose determinant is an integer $k$. It follows that the determinant of the original matrix is equal to $106k$.


答:
You see for all $t$,by monotonicity of $a$,  we have:
$$ t^{\alpha+1} a_{2t}  \le t^{\alpha} \sum_{n \ge t+1} a_n \le 1$$
Hence, $$ \limsup t^{\alpha+1} a_{t} \le 2^{\alpha+1}$$
Q.E.D

Title: Is $\limsup_{t} \frac{a_t}{ h(t+1) - h(t) }$ finite when $0 \leq \sum_{i=t}^\infty a_i \leq h(t) = t^{-\alpha}$?

Body:
Consider a monotone decreasing sequence $\{a_i\}_{i=1}^\infty$, where $a_i  \geq 0$ for all $i$. This sequence is assumed to satisfy the bound on its residual series
$$
\sum_{i=t}^\infty a_i \leq h(t) = t^{-\alpha} \;\;\;\forall t\in\mathbb{N}
\;,
$$
where $\alpha \geq 1$. From the above bound, it is clear that $\{a_i\}$ is summable since we have $\sum_{i=1}^\infty a_i \leq 1 < \infty$.

My objective is to determine whether
$$\limsup_{i\rightarrow \infty} \frac{a_i}{-\Delta h_i} < \infty \,,$$
where $\Delta h_i = h(i+1) - h(i)$.

Thus far, I have only been able to prove the weaker property that
$$ \liminf_{i\rightarrow\infty}  \frac{a_i}{-\Delta h_i} \leq 1 \;,$$
which is done by showing that $a_i \leq -\Delta h_i$
is true for infinitely many $i$ (see below for proof).

I have found that the answer may be related to the Stolz-Cesaro Theorem or its converse, but I have been unable to make the exact requirements of the theorem line up with what I need. I have also found that this may be connected to the following post.

Any help on this would be appreciated. I am also curious about the more general case when we have some convex function $h:(0,\infty)\rightarrow (0,\infty)$ satisfying $\lim_{i\rightarrow\infty} h(i) = 0$.

Proof that $a_i \leq -\Delta h_i$ for infinitely many $i$: Assume the converse for all but finitely many $i$. Then there exists a $T\in\mathbb{N}$ such that for all $t \geq T$,
$$
\sum_{i=t}^\infty a_i > - \sum_{i=t}^\infty \Delta h_i = h(t)
\;,
$$
which is a contradiction. Hence we must have that $a_i \leq -\Delta h_i$ holds infinitely often.


问:
$\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2^n -1}$ is not true for every $n > 1, n \in \mathbb{N}$, right?
So I found an exercise where it is required to prove that for every $n > 1, n \in \mathbb{N}$ the inequality is true:

$$\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2^n -1} < n$$

I found that disturbing since I think that for n > 5 that is simply not true (the left hand side is wrong). Can somebody ensure me in my thinking or tell me if I'm wrong?

答: The inequalities are true. For each $i$, $$\frac 1 {2^{k+1}} \leq \frac 1 i \leq \frac 1 {2^k}$$ for some $k$. Summing over $i$ the way it's done here, you'll get your two inequalities.


问:
Proving a given set of vectors on one side of a hyperplane is a basis
Let $v_1, \ldots, v_n$ be $n$ vectors in $\mathbb{R}^n$.  Assume that there exists a hyperplane $H$ through the origin such that $v_1, \ldots, v_n$ are strictly on one side of $H$.  Also assume that the angle between any two vectors $v_i$ and $v_j$ is obtuse.  Then show that $\{v_1, \ldots, v_n\}$ is a basis of $\mathbb{R}^n$.

I'm at a loss for how to proceed.  Since they explicitly mention angle I'm assuming I might need to use the dot product or something, but that's about as far as I got.

答:
Suppose the vectors are all in the half-space $ax > 0$, and that some linear combination of them is zero, $c_1 v_1 + \dots + c_n v_n = 0$.

Then, $a(c_1v_1 + \dots + c_nv_n) = c_1(av_1) + c_2(av_2) + \dots + c_n(av_n) = 0$, so if any $c_i$ is non-zero, there must be coefficients of both positive and negative signs. Now we can relabel so that $c_1, c_2, \dots, c_j$ are the positive coefficients and $c_{k}, c_{k+1},\dots,c_n$ are the negative coefficients.

Now, both of the vectors $v = c_1v_1 + c_2v_2 + \dots + c_jv_j$ and $w = (-c_{k})v_{k} + (-c_{k+1})v_{k+1} + \dots + (-c_n)v_n$ have positive coefficients, and $v = w$, so what happens if we try to compute $\|v\|^2 = v \cdot w$?


问:
If $f_n$ converges uniformly to $f$ and $|f_n| \leq g$, then $|f| \leq g$
Suppose $g$ and $f_n$ are defined on $(0,\infty)$, $f_n$ converges uniformly to $f$ on every compact subset of $(0,\infty)$, and $|f_n| \leq g$.  I want to prove that $|f(x)| \leq g(x)$.

This seems so obviously true because $f_n$ can get as close to $f$ as you want but I don't know how to formally show this.

答: All you need is that they converge pointwise, and that $\{x: |x| \le y\}$ is a closed set for all $y$.


问:
Looking for math audio resources online
It is pretty difficult to find advanced math lectures online, though, with some effort, one can find a few (by "advanced" I mean anything beyond 1st or 2nd year of college, and preferably aimed at mathematicians, rather than engineers or scientists).  The difficulty of this search becomes orders of magnitude greater if one is looking for audio-only lectures (that can be easily readily played in an mp3 player).  Of course, much of this has to do with the fact that it is difficult to communicate math without visuals, although it is definitely possible.

Does anyone know of any such resources (preferably free) online?

Thanks!

PS: one could, with considerable additional effort, extract the audio from, say, math video lectures on YouTube or iTunes, but I'm not too keen on this option, and not only because of the work involved.

答: I wrote a book called Math Fundamentals for Audio which you can find here: https://www.areditions.com/gaston-bird-math-fundamentals-for-audio-das027.html</p>

Title: Looking for math audio resources online


答: Alternatively, suppose for a contradiction that there exists $x\in(0,\infty)$ such that $f(x)>g(x)$. Let $\epsilon=f(x)-g(x)>0$. Then, use uniform convergence to get the contradiction that $f_{n}(x)>g(x)$ for large enough $n$.

Title: If $f_n$ converges uniformly to $f$ and $|f_n| \leq g$, then $|f| \leq g$

Body:
Suppose $g$ and $f_n$ are defined on $(0,\infty)$, $f_n$ converges uniformly to $f$ on every compact subset of $(0,\infty)$, and $|f_n| \leq g$.  I want to prove that $|f(x)| \leq g(x)$.

This seems so obviously true because $f_n$ can get as close to $f$ as you want but I don't know how to formally show this.


问:
How to use the Method of Undetermined Coefficients to solve Non-Homogeneous ODEs
I have a very general question about the Method of Undetermined coefficients when the non-homogeneous part is a polynomial (say $n$th degree). In this case, I know that if you try:
$$ y_p(x) = A_nx^n+A_{n-1}x^{n-1}+...+A_1x+A_0 $$

then you will be guaranteed to get the correct answer. However, when I look through solutions I find that my professor is often simplifying her choice by choosing $A_i=0$ for some $i<n$.

Is there a way to know exactly when you can do this and when you have to assume that none of the $A_i$ are zero?

If it helps to have a specific example, one I just saw was:
$$ x^2y''+xy'+y=x^4 $$
This is a standard Euler equation which can be solved from guessing $y=x^r$. Doing so gives, $r=\pm i$ and so our homogeneous solution is:
$$ y_h(x) = c_1 \cos(\ln x)+c_2\sin(\ln x) $$
All of this I totally understand and can replicate. However, when she solves the non-homogeneous part she only chooses:
$$ y_p(x) =Ax^4 $$
This is what I'm confused about. Is there something special about this problem that allows you to assume the particular solution is only a function of $x^4$?

Conversely, for the equation:
$$ y''-y'-2y=x $$
we can similarly just solve the characteristic equation to get $r=-1,2$. So our homogeneous solution is:
$$ y_h(x) = c_1e^{-x}+c_2e^{2x} $$
Now in this example, however, she chooses:
$$ y_p(x) = Ax+B $$
and after solving it turns out both are non-zero. What are the differences - is i.t because the first is an Euler equation? If so, why does that change things? Or are there no differences and I should just always assume $A_i$ are nonzero?

Any help is much appreciated!

答:
The rule for constant-coefficient equations (e.g. your $y'' - 2 y' + y = x$) is that if the right side is of the form $p(x) e^{ax}$ where $p$ is a polynomial and $a$ is not a root of the characteristic polynomial, there will be a particular solution of the form $q(x) e^{ax}$ where $q$ is a polynomial
of the same degree as $p$; if $a$ is a root of the characteristic polynomial of multiplicity $m$,
then there will be a particular solution of the form $x^m q(x) e^{ax}$ where $q$ is a polynomial of the same degree as $p$.

Euler equations can be transformed to constant-coefficient equations by the transformation $x = e^t$.  If the right side of the Euler equation is a constant times a power of $x$ (or more generally a power of $x$ times a polynomial in $\log(x)$, you can apply the paragraph above to the transformed equation.


答:
$$x^2y''+xy'+y=x^4$$
For the Euler equation it's a bit particular because all the derivatives ( also $y$)  are on the form $Cx^4$. So that $y_p=Ax^4$ will do the job. (If we suppose of course that the solution to the homogeneous Euler equation has not the form : $y_h=C_1x^4+...$. In this case the guess $y_p=Ax^4$ will not work. )

And both differential equations are different. Euler equation has variable coefficients. When the second DE has constants coefficients.

$$y''-y'-2y=x$$
Here all the coefficents are constants. And $y_p=Ax$ will not work. The derivatives are not all of the form $Cx$. You have constants too. $y'_p$ is a constant not on the form $Cx$.

Title: How to use the Method of Undetermined Coefficients to solve Non-Homogeneous ODEs

Body:
I have a very general question about the Method of Undetermined coefficients when the non-homogeneous part is a polynomial (say $n$th degree). In this case, I know that if you try:
$$ y_p(x) = A_nx^n+A_{n-1}x^{n-1}+...+A_1x+A_0 $$

then you will be guaranteed to get the correct answer. However, when I look through solutions I find that my professor is often simplifying her choice by choosing $A_i=0$ for some $i<n$.

Is there a way to know exactly when you can do this and when you have to assume that none of the $A_i$ are zero?

If it helps to have a specific example, one I just saw was:
$$ x^2y''+xy'+y=x^4 $$
This is a standard Euler equation which can be solved from guessing $y=x^r$. Doing so gives, $r=\pm i$ and so our homogeneous solution is:
$$ y_h(x) = c_1 \cos(\ln x)+c_2\sin(\ln x) $$
All of this I totally understand and can replicate. However, when she solves the non-homogeneous part she only chooses:
$$ y_p(x) =Ax^4 $$
This is what I'm confused about. Is there something special about this problem that allows you to assume the particular solution is only a function of $x^4$?

Conversely, for the equation:
$$ y''-y'-2y=x $$
we can similarly just solve the characteristic equation to get $r=-1,2$. So our homogeneous solution is:
$$ y_h(x) = c_1e^{-x}+c_2e^{2x} $$
Now in this example, however, she chooses:
$$ y_p(x) = Ax+B $$
and after solving it turns out both are non-zero. What are the differences - is i.t because the first is an Euler equation? If so, why does that change things? Or are there no differences and I should just always assume $A_i$ are nonzero?

Any help is much appreciated!


问:
How many distinct multiplications?
I have a multiset $\left\{ {2,3,3,3,4,4,5,5,6,7} \right\}$. How many distinct multiplications of three numbers chosen from the multiset?

e.g. $2×3×3$; $2×3×4$; $4×4×3$;... not $2×3×3$; $3×3×2$;... and $4×3×3$; $6×2×3$; ... (equal multiplication) are considered as the same.

答: Now that we have clarification from Piseth as to what was intended, it seems to me that there's no really nice way to do the problem. I'd say either bung it on a computer and let it count the number of different products, or use the methods suggested by svenkatr or Yuval to get a number from which you can subtract the duplications by hand. I don't think there are that many duplications: 2,6,7 and 3,4,7; 2,5,6 and 3,4,5; 2,4,6 and 3,4,4; 2,3,6 and 3,3,4; have I missed any?</p>

Title: How many distinct multiplications?


答: Take $a\in(0,+\infty)$. Then $\{a\}$ is compact, and therefore $(f_n)_{n\in\Bbb N}$ converges uniformly to $f$ there, which simply means that $\lim_{n\to\infty}f_n(a)=f(a)$. And therefore $\lim_{n\to\infty}\bigl|f_n(a)\bigr|=\bigl|f(a)\bigr|$. So, since each $\bigl|f_n(a)\bigr|$ is smaller than or equal to $g(a)$, $\bigl|f(a)\bigr|\leqslant g(a)$.