问:
Approximating a double sum by a double integral
Related to this question, I'm interested in bounding from above the following sum
$$
S:=\sum_{x=0}^\infty \sum_{y=0}^\infty (x+y)^m e^{-\frac{x^2}{2i} - \frac{y^2}{2j}},
$$
which I hope to do by relating it to the integral
$$
I:=\int_0^\infty \int_0^\infty (x+y)^m e^{-\frac{x^2}{2i} - \frac{y^2}{2j}} dx\,dy.
$$

Answers to the previous questions confirmed my expectation that $I = O\left(\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)\sqrt{ij}\right)$, the intuition for which is probably that the function behaves approximately like a gaussian around its maximum at $(x_0,y_0) = \left(i \sqrt{\frac{m}{i+j}},j \sqrt{\frac{m}{i+j}} \right)$, where the function takes the value $\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)$.

However I've been unable to show that the difference $|I-S|$ is significantly smaller than this bound. For simple one dimensional integrals, for example with a unique maximum, it's not too hard to bound this difference in terms of the maximum by considering appropriate telescoping sums. However, a naive analogue of this argument doesn't seem to work in two dimensions, and trying to apply this argument to each `slice' of the integral led to some pretty horrendous calculations. I also looked into using the Euler-Maclaurin formula but it's a bit out of my area of expertise.

I suspect that there should be a relatively standard way to approximate $|I-S|$, and I also wouldn't be surprised if someone more proficient in computing can get a CAS to provide a proof. The former would be more useful, just so that I have a tool for approaching similar questions.

So, very explicitly, I would like to know if
$$
|I-S| = o\left(\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)\sqrt{ij}\right),
$$
where even big-O would be sufficient for the application I have in mind, and I wouldn't be surprised if the difference is even bounded by a multiple of the maximum of the function. I'm interested in the asymptotics for $i$ and $j$ tending to infinity, $m$ can be fixed or also a function of $i$ and $j$. For the application I have in mind it would probably be sufficient to have such a result for $i = (1+o(1))j$ and $m = o(i)$.

答:
```<code>%94 // FullSimplify

(1/(m (1 + m)))2^(
 m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
   Gamma[1 + m/
      2] (j^(1 + m/2)
        m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
      i^(m/2) (1 + m) (-i^2 +
         i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
           2, -(1/2), -(j/i)]))/(j (-2 + m))))
</code>```

This shows a professional grade solution. It is meant to be exact and not an approximation of some kind. It sets the requirements ot the question properly first and then integrate for all values that appear in the integrant and integral correctly. To type this in MathML is a really tedious task. So this is Mathematica, Wolfram Language code and can for some extent be verified on Wolfram Alpha. Have a try on the free Maple app too.

The names of the functions appear in the reference question partly and are not reduced to somehow more elementary functions in this calculation. This respects Mathematica great internal convention in favour to some universitary or college simplicity.

This is still a function of {i,j,m} and very lengthy. Only two terms have this $\sqrt{i j}$. Others are dependent on m in the exponent.

This can be calculated to numbers. For example for $j=10=i$ and $m=1,2,3$:
{
{79.2665},
{514.159},
{3963.33},
{34849.6},
{340846.},
{3.6449610^6},
{4.2090510^7},
{5.1989410^8},
{6.8177210^9},
{9.43489*10^10}
}
That looks pretty divergent for bigger ${i,j}$.

For example for $i=10$ and $j$ free and $m=1,2,3,...0$ this looks this way:

$
 5 \sqrt{j} \sqrt{2 \pi} + j \sqrt{5 \pi},$

$20 j + \sqrt{5/2} \sqrt{j} (10 + j) \pi,$

$ \sqrt{j} (100 \sqrt{2} + 30 \sqrt{5} \sqrt{j} + 15 \sqrt{2} j +
 2 \sqrt{5} j^(3/2)) \sqrt{\pi},$
$ (80 j (10 + j) + 3 \sqrt{5/2} \sqrt{j} (10 + j)^2 \pi,$
$ \sqrt{j} (4000 \sqrt{2} + 1500 \sqrt{5} \sqrt{j} + 1000 \sqrt{2} j +
 200 \sqrt{5} j^(3/2) + 75 Sqrt{2} j^2 +
 8 \sqrt{5} j^(5/2)) \sqrt{\pi}, ...$

These are polynomials in j and $\sqrt{j}$.

If there would be another condition for the solution from the Mathematica engine for the input it would show that in the output or on the message cue. But there is none. OK. I am lazy because this is a tedious expression. I added an answer to the reference question too to earn more seriosity for my answer. This is not an expansion of any form and high mathematical intuition to take and offer that solution path for this highly complex problem and the restrictions on it.

$I$ is a function independent on $x$ and $y$. $I(i,j,m)$!

S rises strongly. If the summation is trunkated for fixed $m$ than the function may be plotted. The higher the terms that are taken into the sum the higher the function value. The plot is taken over $l$ and $k$.

I is a type of convolution integral. It acts in information theory and related as s low pass filter.

The binomial polynom is dominating for lower summation parts. The summands to not converge fast enough. This is what concerns one factor: Convolution. This requires that the other function is finite on the reals. That is not the case for $(x+y)^m$ unless it is limited to an finite interval. That is not the case in the question. Therefore the infinite summation fails. $S$ is returned unevaluated by Mathematica and other CAS.

The cognitive dissonance is the the summation is over discretized $x$ and $y$ with out and exhaustion. Integers are not dense in the reals. Approximation of convolution integrals require the finiteness of the convoluted function. This binomial is unlimited.

I selected $m=1$.

```<code>p1 = Plot3D[
  With[{m = 1},
   1/(m (1 + m)) 2^(
    m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
      Gamma[1 + m/
         2] (j^(1 + m/2)
           m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
         i^(m/2) (1 + m) (-i^2 +
            i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
              2, -(1/2), -(j/i)]))/(j (-2 + m))))], {i, 1, 20}, {j, 1,
    20}]

p2 = Plot3D[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(x = 0\), \(100\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(y = 0\), \(100\)]
\*SuperscriptBox[\((x + y)\), \(1\)] Exp[\(-1\)\ \((
\*SuperscriptBox[\((
\*FractionBox[\(x\), \(k\)])\), \(2\)] +
\*SuperscriptBox[\((
\*FractionBox[\(y\), \(l\)])\), \(2\)])\)]\)\), {k, 1, 20}, {l, 1,
   20}]
</code>```

Apparently not very much alike. I seems to have something of a Puiseux_series.

```<code>1/(m (1 + m)) 2^(
   m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
     Gamma[1 + m/
        2] (j^(1 + m/2)
          m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
        i^(m/2) (1 + m) (-i^2 +
           i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
             2, -(1/2), -(j/i)]))/(j (-2 + m)))) /.
  m -> 1 // FullSimplify
</code>```

$(\sqrt{i}*j + \sqrt{i*j}*\sqrt{i + j} -
   i*\sqrt{j}*(-1 + \sqrt{\frac{(i + j)}{i}}))*\sqrt{\frac{\pi}{2}}$

On the diagonal $i=j$ or $k=l$ for $m=1$

```<code>Plot[{\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(x = 0\), \(100\)]\(
\*SuperscriptBox[\((2  x)\), \(1\)] Exp[\(-1\)\ 2
\*SuperscriptBox[\((
\*FractionBox[\(x\), \(k\)])\), \(2\)]]\)\), (k^(
     3/2) - (-1 + Sqrt[2]) k^(3/2) +
     Sqrt[2] Sqrt[k] Sqrt[k^2]) Sqrt[\[Pi]/2]}, {k, 1, 1000},
 PlotRange -> Full, AxesLabel -> Automatic]
</code>```

The sum S exits for all values, since exponential is stronger than polynomial and polynomial is kept fix while the argument of the exponential drops to zero with inverse square order. Blue is S and I is goldish. The infinite summation takes endless on my machine. The FullSimplify might get complex and needs extra care.

It is in general very complicated or delicate to approximate convolution integrals. The Gaussian function is problematic in representation and integration. Such profile function play an important roles in all measurements of modern technologies. So they are studied with great care. Since we know this function we just have to handle it with care. Mathematica has extra feature for dealing with the representation of very large and very small numbers. This both are needed unless the solution is symbolic.

There some complexity function around on this community. Mixing polynomials with exponentials is always very hard and complicated. Help might get easier if the wordings would be presented.

There might be impulses for a different approach by interpretation as a Kampe de Feriet Function. The given sum is related to the answer of this question: evaluation-of-sum-x-0-infty-e-x2. Mathematica can do this:

```<code>Sum[E^(-r^2), {r, 0, Infinity}]
1/2 (1 + EllipticTheta[3, 0, 1/E])
</code>```

If we do this:

```<code>Sum[D[D[E^(-r^2), r], r], {r, 0, Infinity}]
Sum[-2/E^r^2 + (4*r^2)/E^r^2, {r, 0, Infinity}]
</code>```

Nice for it is this Jacobi Theta Functions. So S is the second derivative of the EllipticTheta with first argument 3 with respect to the second argument evaluated at 0 for the second argument and 1/e for the third up to $2^3$ as a factor following the formula for the corresponding derivative on EllipticThetaPrime4. Now we have a representation for $i=j=1$ and $m=1$. This is a very simple cases of $S$.

We can plug our $e{-(\frac{x^2}{i^2}+\frac{y^2}{j^2})}$ into this too. That is only representation. But for $m>1$ there is still no help. The $i$ and $j$ make the exponent still elliptical. The $m$ add circularity.

So the next step is hard again. Find relationships between EllipticTheta and Hypergeometric2F1. There are some for derivatives but this needs identities. Then generalize. Candidates are Elliptic hypergeometric series. Perhaps there is need for a revision of the hypothesis.

Body:
%94 // FullSimplify

(1/(m (1 + m)))2^(
 m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
   Gamma[1 + m/
      2] (j^(1 + m/2)
        m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
      i^(m/2) (1 + m) (-i^2 +
         i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
           2, -(1/2), -(j/i)]))/(j (-2 + m))))

This shows a professional grade solution. It is meant to be exact and not an approximation of some kind. It sets the requirements ot the question properly first and then integrate for all values that appear in the integrant and integral correctly. To type this in MathML is a really tedious task. So this is Mathematica, Wolfram Language code and can for some extent be verified on Wolfram Alpha. Have a try on the free Maple app too.

The names of the functions appear in the reference question partly and are not reduced to somehow more elementary functions in this calculation. This respects Mathematica great internal convention in favour to some universitary or college simplicity.

This is still a function of {i,j,m} and very lengthy. Only two terms have this $\sqrt{i j}$. Others are dependent on m in the exponent.

This can be calculated to numbers. For example for $j=10=i$ and $m=1,2,3$:
{
{79.2665},
{514.159},
{3963.33},
{34849.6},
{340846.},
{3.6449610^6},
{4.2090510^7},
{5.1989410^8},
{6.8177210^9},
{9.43489*10^10}
}
That looks pretty divergent for bigger ${i,j}$.

For example for $i=10$ and $j$ free and $m=1,2,3,...0$ this looks this way:

$
 5 \sqrt{j} \sqrt{2 \pi} + j \sqrt{5 \pi},$

$20 j + \sqrt{5/2} \sqrt{j} (10 + j) \pi,$

$ \sqrt{j} (100 \sqrt{2} + 30 \sqrt{5} \sqrt{j} + 15 \sqrt{2} j +
 2 \sqrt{5} j^(3/2)) \sqrt{\pi},$
$ (80 j (10 + j) + 3 \sqrt{5/2} \sqrt{j} (10 + j)^2 \pi,$
$ \sqrt{j} (4000 \sqrt{2} + 1500 \sqrt{5} \sqrt{j} + 1000 \sqrt{2} j +
 200 \sqrt{5} j^(3/2) + 75 Sqrt{2} j^2 +
 8 \sqrt{5} j^(5/2)) \sqrt{\pi}, ...$

These are polynomials in j and $\sqrt{j}$.

If there would be another condition for the solution from the Mathematica engine for the input it would show that in the output or on the message cue. But there is none. OK. I am lazy because this is a tedious expression. I added an answer to the reference question too to earn more seriosity for my answer. This is not an expansion of any form and high mathematical intuition to take and offer that solution path for this highly complex problem and the restrictions on it.

$I$ is a function independent on $x$ and $y$. $I(i,j,m)$!

S rises strongly. If the summation is trunkated for fixed $m$ than the function may be plotted. The higher the terms that are taken into the sum the higher the function value. The plot is taken over $l$ and $k$.

I is a type of convolution integral. It acts in information theory and related as s low pass filter.

The binomial polynom is dominating for lower summation parts. The summands to not converge fast enough. This is what concerns one factor: Convolution. This requires that the other function is finite on the reals. That is not the case for $(x+y)^m$ unless it is limited to an finite interval. That is not the case in the question. Therefore the infinite summation fails. $S$ is returned unevaluated by Mathematica and other CAS.

The cognitive dissonance is the the summation is over discretized $x$ and $y$ with out and exhaustion. Integers are not dense in the reals. Approximation of convolution integrals require the finiteness of the convoluted function. This binomial is unlimited.

I selected $m=1$.

p1 = Plot3D[
  With[{m = 1},
   1/(m (1 + m)) 2^(
    m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
      Gamma[1 + m/
         2] (j^(1 + m/2)
           m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
         i^(m/2) (1 + m) (-i^2 +
            i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
              2, -(1/2), -(j/i)]))/(j (-2 + m))))], {i, 1, 20}, {j, 1,
    20}]

p2 = Plot3D[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(x = 0\), \(100\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(y = 0\), \(100\)]
\*SuperscriptBox[\((x + y)\), \(1\)] Exp[\(-1\)\ \((
\*SuperscriptBox[\((
\*FractionBox[\(x\), \(k\)])\), \(2\)] +
\*SuperscriptBox[\((
\*FractionBox[\(y\), \(l\)])\), \(2\)])\)]\)\), {k, 1, 20}, {l, 1,
   20}]

Apparently not very much alike. I seems to have something of a Puiseux_series.

1/(m (1 + m)) 2^(
   m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
     Gamma[1 + m/
        2] (j^(1 + m/2)
          m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
        i^(m/2) (1 + m) (-i^2 +
           i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
             2, -(1/2), -(j/i)]))/(j (-2 + m)))) /.
  m -> 1 // FullSimplify

$(\sqrt{i}*j + \sqrt{i*j}*\sqrt{i + j} -
   i*\sqrt{j}*(-1 + \sqrt{\frac{(i + j)}{i}}))*\sqrt{\frac{\pi}{2}}$

On the diagonal $i=j$ or $k=l$ for $m=1$

Plot[{\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(x = 0\), \(100\)]\(
\*SuperscriptBox[\((2  x)\), \(1\)] Exp[\(-1\)\ 2
\*SuperscriptBox[\((
\*FractionBox[\(x\), \(k\)])\), \(2\)]]\)\), (k^(
     3/2) - (-1 + Sqrt[2]) k^(3/2) +
     Sqrt[2] Sqrt[k] Sqrt[k^2]) Sqrt[\[Pi]/2]}, {k, 1, 1000},
 PlotRange -> Full, AxesLabel -> Automatic]

The sum S exits for all values, since exponential is stronger than polynomial and polynomial is kept fix while the argument of the exponential drops to zero with inverse square order. Blue is S and I is goldish. The infinite summation takes endless on my machine. The FullSimplify might get complex and needs extra care.

It is in general very complicated or delicate to approximate convolution integrals. The Gaussian function is problematic in representation and integration. Such profile function play an important roles in all measurements of modern technologies. So they are studied with great care. Since we know this function we just have to handle it with care. Mathematica has extra feature for dealing with the representation of very large and very small numbers. This both are needed unless the solution is symbolic.

There some complexity function around on this community. Mixing polynomials with exponentials is always very hard and complicated. Help might get easier if the wordings would be presented.

There might be impulses for a different approach by interpretation as a Kampe de Feriet Function. The given sum is related to the answer of this question: evaluation-of-sum-x-0-infty-e-x2. Mathematica can do this:

Sum[E^(-r^2), {r, 0, Infinity}]
1/2 (1 + EllipticTheta[3, 0, 1/E])

If we do this:

Sum[D[D[E^(-r^2), r], r], {r, 0, Infinity}]
Sum[-2/E^r^2 + (4*r^2)/E^r^2, {r, 0, Infinity}]

Nice for it is this Jacobi Theta Functions. So S is the second derivative of the EllipticTheta with first argument 3 with respect to the second argument evaluated at 0 for the second argument and 1/e for the third up to $2^3$ as a factor following the formula for the corresponding derivative on EllipticThetaPrime4. Now we have a representation for $i=j=1$ and $m=1$. This is a very simple cases of $S$.

We can plug our $e{-(\frac{x^2}{i^2}+\frac{y^2}{j^2})}$ into this too. That is only representation. But for $m>1$ there is still no help. The $i$ and $j$ make the exponent still elliptical. The $m$ add circularity.

So the next step is hard again. Find relationships between EllipticTheta and Hypergeometric2F1. There are some for derivatives but this needs identities. Then generalize. Candidates are Elliptic hypergeometric series. Perhaps there is need for a revision of the hypothesis.


问:
How do I evaluate this summation/series?
$$\sum_{i=0}^{\infty} (1+r)^{-i} = \ ?$$

I have so far:
$$\sum_{i=0}^{\infty} (1+r)^{-i} = 1+ \frac{1}{1+r}+\frac{1}{(1+r)^2}+\ ... \ \frac{1}{(1+r)^\infty}$$

But I think it's supposed to be:
$$\sum_{i=0}^{\infty} (1+r)^{-i} = \frac{1+r}{r} $$

Edit: I think the key is that I'm failing to understand how the denominators converge to r

答:
Use this "lemma":

For $x$ with $|x|<1$ we have $$\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}$$

Using this lemma, your series is in fact equal to $$\frac{1}{1-(1+r)^{-1}}=\frac{1}{1-\frac{1}{1+r}}=\frac{1}{\frac{r}{r+1}}=\frac{r+1}{r}$$

Now, here are $2$ proofs of the lemma:

$\text{Proof 1:}$

Let $$S=\sum_{i=o}^{\infty}x^i\text{ and observe that } x\cdot S=\sum_{i=1}^{\infty}x^i=\sum_{i=0}^{\infty}x^i-1=S-1$$ so $$x\cdot S=S-1\text{ which gives us }S=\frac{1}{1-x}$$

$\text{Proof 2:}$

Observe that $$\sum_{i=o}^{\infty}x^i=\lim_{n\to\infty}\sum_{i=0}^{n}x^i=\lim_{n\to\infty}\frac{x^{n+1}-1}{x-1}=\frac{1}{1-x}$$

ContentLicense: CC BY-SA 4.0


问:
Help with a Groebner basis
I need to calculate a Grobner basis of
$$I=(u-x^2,v-y^2,w-xy)$$
with lexicographic order $x>y>u>v>w$.

I am trying to proceed with Buchberger's algorithm, but i probably didn't quite understand how to apply the multivariate division algorithm.

Let $f_1:=u-x^2$, $f_2:=v-y^2$, $f_3:=w-xy$.

Denote by $g_i$ the leading term of $f_i$ with respect to the given
ordering $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; g_1:=x^2$, $g_2:=y^2$, $g_3:=xy$.

Denote by $a_{ij}$ the least common multiple of $g_i$ and $g_j$.$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;a_{1,2}=x^2y^2$, $a_{1,3}=x^2y$, $a_{2,3}=xy^2$

Choose two polynomials in G and let $S_{ij} = (a_{ij} / g_i) f_i −
   (a_{ij} / g_j) f_j. \;\;\;$I chose $S_{1,2}=y^2(u-x^2)-x^2(v-y^2)$

But now, from multivariate division algorithm of $S_{ij}$ by $f_1,f_2,f_3$
always gives $r=0$.  What am I doing wrong?

答:
We calculate $S_{1,3}=\frac{x^2y}{-x^2}f_1-\frac{x^2y}{-xy}f_2=xw-yu$ . Now by the lexicographic order the leading term of $S_{1,3}$ is given by $xw$ but this is not divisible by the leading terms of any of the three given polynomials. Now we initiate the multivariate division algorithm with the following data:

1)$r^{(0)}=0$, $p^{(0)}=S_{1,3}$

Now the leading term of $p$ is given by $xw$ and this is not divisible by any of the leading terms of the $f_i$ so following the algorithm we have

2)$p^{(1)}=p^{(0)}-lt(p^{(0)})=-yu$ and $r^{(1)}=r^{(0)}+lt(p^{(0)})=xw$

Now we notice again that the leading term of $p^{(1)}$ is not divisible by any of the leading terms of the $f_i$ so doing another step in the algorithm we obtain:

3)$p^{(2)}=p^{(1)}-lt(p^{(1)})=-yu+yu=0$ and $r^{(2)}=r^{(1)}+lt(p^{(1)})=xw-yu=S_{1,3}$

Now our $p^{(2)}=0$ hence we stop the algorithm and add the rest $r=S_{1,3}$ to the divisors and start from scratch.

Note that in the multivariate division divisibility only depends on the leading terms. I hope this clarifies some of your problems!
The full Gröbner basis is given by $[x^2 - u, xy - w, xv - yw, xw - yu, y^2 - v, uv - w^2]$. Just adding this to check the results of your further calculations :)

Lg Mo

Title: Help with a Groebner basis


问:
How to prove that $B_{d_{H}}(x, \varepsilon) = B_{d}(x, \varepsilon) \cap H$.
Let (X, d) be a metric space and H$\subset$X. ¿How can I prove the question above?

I’ve tried proving that the first set is included in the second one, but I’m not sure if I can say that $\{h\in H / d(x,h) < \varepsilon\} \subset \{a \in X / d(x,a) < \varepsilon \} \cap H$ just by stating that every $h \in H$ is also in X since H is a subset of X.

Also, I don’t know how to demonstrate the second inclusion to show that both sets are equal.

答:
If I'm not mistaken, $d_H$ is simply the subspace metric on $H$ right?

If that's correct then your initial logic for the first containment is correct. Since $h \in B_{d_H}(x,\epsilon)$, we have that $h \in H$ and $d(x,h) < \epsilon$. But this exactly means that $h$ is an element of both $H$ AND $B_d(x,\epsilon)$.

The other way, simply note that if $h$ is an element of the set $B_d(x,\epsilon) \bigcap H$, means exactly that $d(x,h)< \epsilon$  (because it is in $B_d(x,\epsilon)$) AND that $h \in H$. But this is exactly the definition of the set $B_{d_H}(x,\epsilon)$. Hence we have inclusion both ways.

Another way to think about it is to consider the definition of the subspace metric. It's the restriction of the metric on $X$ to $H$. I.e. $d_H = d \bigcap \big((H\times H)\times \mathbb{R}\big)$. This looks confusing, but it just shows that the metric $d_H$ is essentially the exact same as the metric $d$, so long as you are working 'inside' $H$. That's to say that by definition, $d_H(x,y) = d(x,y)$ if both $x,y \in H$. Then intuitively, taking the intersection of $B_d(x,\epsilon)$ with $H$ allows you to 'work inside' H, by restricting the set of values to be lie H.


问:
Solve the recurrence $\ T(n) = \sqrt{n}\cdot T(\sqrt{n}) + n\cdot \log(n) $
Solve the following recurrence:

$$\ T(n) = \sqrt{n}\cdot T(\sqrt{n}) + n\cdot \log(n)$$

I tried to just unroll it to see if I can understand, but it is too complicated .

$\ T(n) = n^{\frac{1}{2}} \cdot T(n^{\frac{1}{2}}) +n\cdot \log(n)\\
= n^{\frac{1}{2}} \cdot \left( n^{\frac{1}{4}} \cdot T(n^{\frac{1}{4}} ) + \frac{n}{2} \cdot \log(\frac{n}{2}) \right) + n \cdot \log(n) \\
= n^{\frac{1}{2}} \left( n^{\frac{1}{4}} \cdot \left( n^{\frac{1}{8}} \cdot T(n^{\frac{1}{8}}) +  \frac{n}{4}\cdot \log(\frac{n}{4})  \right) + \frac{n}{2} \cdot \log(\frac{n}{2}) \right) + n\cdot \log(n) $

Not really clear how to proceed here? I also tried recurrence tree, but couldn't see a clear pattern. I've found a similar question here but I could not understand the answer

答:
I find this answer much more simpler and efficient

This is my attempt:

$T(n) = \sqrt{n} T(\sqrt{n}) + n \log n$

First a change of variable $n = 2^k$ leading to:

$T(2^k)  = 2^{k/2} T(2^{k/2}) + k 2^k$

Now a change of function: $T(2^k) = S(k)$, leading to:

$S(k) = 2^{k/2} S(k/2) + k2^k$

Unfortunately we cannot apply Master theorem on this because of the non-constant multiplier $2^{k/2}$.

Instead we have to expand this using repeated substitution:

$S(k) = 2^{k/2 + k/4 + \ldots + k/2^d}T(k/2^d) + k/2^d 2^{k/2^d} + \ldots + k2^k$

where $d$ is the depth of this tree. It is decided by the base case. You haven't mentioned a base case but assuming it's a constant, we can say $k/2^d = B$ and T(B) = C, whee $B,C$ are constants:

$S(k) = 2^{k/2 + k/4 + \ldots + k/2^d} C + k/2^d 2^{k/2^d} + \ldots + k2^k$.

The first term is 2 powered to a decaying GP series. The second term is a AGP series.

The GP series converges to $2$ (thus the $2^2C$ can be ignored for asymptotic bounds). The AGP series is equal to $(k^2)^k + 2^{k/4 - 2} k + 2^{k/2 - 1} k$.

We can now move back to the original variable $n$:
$\Rightarrow ((\log n)^2)^{\log n} + 2^{(\log n)/4 - 2} \log n + 2^{(\log n)/2 - 1} \log n$

$\Rightarrow  ((\log n)^{2\log n} + 2^{(\log n)/4 - \log 4} \log n + 2^{(\log \sqrt{n}) - \log 2} \log n$

$\Rightarrow  ((\log n)^{2\log n} + 2^{(\log n^{1/4}/4} \log n + 2^{\log \sqrt{n}/2 } \log n$

$\Rightarrow  ((\log n)^{2\log n} + n^{1/4} \log n /4 + \sqrt{n} \log n/2 $


问:
Convergence proof of sequence
I want to show that the sequence $a(n):= \frac{n!}{n^n}$ converges.

Therefore I've shown that a(n+1)<a(n):

$\frac{a(n+1)}{a(n)}=\frac{n+1)!}{n+1)^(n+1)}\cdot\frac{n^n}{n!}=\frac{n^n}{(n+1)^n}<1$

and because this is smaller than one the sequence is monotonously decreasing.

Next I have to show that the sequence is limited from below, which overall would mean that the sequence converges.

Indeed I don't really know how to go on here.

Can someone give me a hint please?

答:
You have shown that $a_n \leq 1$, but it is also true that $a_n >0$. So, you have that $0 < a_n \leq 1$. Since the sequence is bounded and monotonous, it is convergent. Using Stirling's formula, you also see that
$$
\lim \frac{n!}{n^n} = \lim \frac{\sqrt{2 \pi n}}{e^n} = 0.
$$

Title: Convergence proof of sequence


问:
Divergence of a matrix vector product.
Let take a square constant matrix $A\in \mathbb{R}^{d\times d}$, and a vector field $v:\mathbb{R}^d \to \mathbb{R}^d$. Is there a way to write the divergence of their product in terms of the divergence of $v$?

That is can we write

$$ \text{div}(Av) $$ as something involving $A$ and $\text{div}(v)$?

答:
Using index notation, divergence is interpreted as the trace of the Jacobian matrix of a vector i.e.

$$ \operatorname{div} \mathbf{v} = \partial_i v^i$$

with Einstein summation convention. Therefore we can correctly interpret the divergence of the product as follows

$$\partial_i(A_j^i v^j) = (\partial_i A_j^i)v^j + A_j^i (\partial_i v^j)$$

The object on the left is the dot product of $\mathbf{v}$ with a vector whose elements are the divergences of the column vectors of $\mathbf{A}$. The object on the right is the "matrix dot product" (element wise multiplication then adding) of $\mathbf{A}$ with the Jacobian matrix of $\mathbf{v}$.

Title: Divergence of a matrix vector product.


问:
Are the area of a circle inscribed in a square and the area of the "spandrels" (the four corners that remain) commensurable?
And how would you demonstrate that most simply?

See the beginning of my blog post for a little more:
http://seekecho.blogspot.fr/2013/02/different-ilks.html

答:
So you want to show that the area of the circle, i.e. $\pi$, is not commensurable with the area of “square minus circle”, i.e. $4-\pi$, right? Two things are commensurable if their fraction is a rational number. Now look at it this way:

$$\frac{4-\pi}\pi = \frac4\pi-1 = r\not\in\mathbb Q$$

Suppose $r$ were a rational number, then all the following would be rational numbers as well:

\begin{align*}
\frac4\pi &= r + 1 \\
\frac1\pi &= \frac{r+1}4 \\
\pi &= \frac4{r+1}
\end{align*}

But as you know that $\pi$ is irrational, you know that $r$ cannot be rational.

Given two incommensurable numbers $a$ and $b$, the set $\{pa+qb\;\vert\;p,q\in\mathbb Q\}$ can be interpreted as a vectorspace over $\mathbb Q$ with dimension two. You can check all the vector space axioms to see that this is true. The switch from “circle and square” to “circle and spandrel” is simply a change in basis vectors, but does not make the vectors linearily dependent. So as a general conclusion, two numbers $p_1a+q_1b$ and $p_2a+q_2b$ will still be incommensurable unless they are linearily dependent in the vector space sense, i.e. unless

$$\begin{vmatrix} p_1 & p_2 \\ q_1 & q_2 \end{vmatrix} =
p_1q_2 - p_2q_1 = 0$$

ContentLicense: CC BY-SA 3.0


问:
Prove that $X_n \sim N(0,\frac{1}{n})$ converges in probability to 0
Problem

Prove that $X_n \sim N(0,\frac{1}{n})$ converges in probability to 0.

Discussion

There a problems similar to this question, but I want to ask specifically about the convergence of an integral that arises in the solution.

Attempt

Let $\epsilon>0$. To show $X_n \overset{p}{\to} 0$, we must show $P(|X_n| > \epsilon) \to 0$.

$$
\begin{align}
P(|X_n|>\epsilon)
&= P(X_n>\epsilon) + P(X_n < -\epsilon)\\
&= P(Z>\epsilon\sqrt{n}) + P(X_n<-\epsilon\sqrt{n})
\end{align}
$$
where $Z$ is the standard normal distribution. Thus we want to show
$$P(|X_n|>\epsilon)=\int_{\epsilon\sqrt{n}}^\infty\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\text{dx}
+
\int_{-\infty}^{-\epsilon\sqrt{n}}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\text{dx}\to 0$$

I understanding intuitively that these integrals should converge to zero, but how can I show rigorously that they go to zero?

答: Your integrals contain spurious instances of $\sqrt n$. You are evaluating $P(|Z|>\epsilon\sqrt n)$ where $Z$ is standard normal, so the integrand should not involve $\sqrt n$. Anyhow, you can prove $P(|Z|>\epsilon\sqrt n)\to0$ rigorously using Markov's inequality.


问:
Relation Between Dependent Type Theory And Categories
I have been trying to understand the relationship between a dependent type theory and a corresponding locally Cartesian closed category $\mathbb C,$ as described in

R. A. G. Seely, Locally cartesian closed categories and type theory, Math. Proc. Camb. Phil. Soc. (1984) 95

But I am having some difficulty. As I understand it, the objects of $\mathbb C$ are types without free variables, and the arrows of $\mathbb C$ are members of independent function types.

If $\mathbb U$ denotes a universe of types, then am I correct in thinking that a type family $B: A \rightarrow \mathbb U$ corresponds to a (projection) function $\pi _1 : \Sigma _{x:A} B(x) \rightarrow A$ ? It seems that the above reference states that this type family $B$ also corresponds a member of the slice category $\mathbb{C} / A.$ I am trying to understand this. I guess that a type family $B$ corresponds to an object of $\mathbb{C} / A$ because $B$ corresponds to an arrow $\pi _1 : \Sigma _{x:A} B(x) \rightarrow A$ of $\mathbb C,$ that goes into $A$ (and the objects of the slice category $\mathbb{C} / A$ correspond to arrows of $\mathbb C,$ that go into $A$).

I would appreciate it if somebody could tell me if I am on the right track with this thinking.

The other thing I am very confused about is that the dependent type theory described in the homotopy type theory book I read had a hierarchy of universes $\mathbb U _0 : \mathbb U _1 ,$ and $\mathbb U _1 : \mathbb U _2 , \dots$ etc. I am confused about how these universes interplay with the correspondance between the type theory and the locally Cartesian closed category described above ? I presume there is no object of $\mathbb C$ which actually corresponds to a type universe in the theory, but it would be great if somebody could explain more about how type universes enter into the correspondence (if they do).

答:
As I understand it, the objects of ℂ are types without free variables, and the arrows of ℂ are members of independent function types.

This is right, to first approximation.  Technically it's better to take the objects of $\mathbb{C}$ to be contexts, but in a theory with $\Sigma$-types satisfying a uniqueness/eta rule every context is isomorphic to a single closed type, so the first-approximation answer gives an equivalent category.

A type universe is itself a type, hence an object of this category, and so a type family $B:A\to \mathbb{U}$ is at its most basic level just a morphism in the category.  There is also the projection function $\pi_1 : (\sum_{x:A} B(x)) \to A$ as you say, which is an object of the slice category $\mathbb{C}/A$, and the correspondence between them is that $\pi_1$ is the pullback along $B$ of the universal projection $\pi_1 : (\sum_{X:\mathbb{U}} X) \to \mathbb{U}$.  So yes, you are basically on the right track here too.

The deal with hierarchies of universes is just that in a consistent theory it can't be that every type is an element of the same universe $\mathbb{U}$, or put categorically it can't be that every object of $\mathbb{C}/A$ is classified by some map $B:A\to \mathbb{U}$ for the same $\mathbb{U}$.  Each type universe in the hierarchy yields an object of the category that classifies some, but not all, of the objects of each slice category.


问:
The Sobolev Space $H^{1/2}(\partial \Omega)$ as the Quotient Space $H^1(\Omega)/\ker(\text{tr})$
In both questions Reference request: norm of the image of a bounded linear operator and
The Sobolev Space  $H^{1/2}$, the Sobolev space
$$H^{1/2}(\partial Ω) = \{ u ∈ L^2(\partial Ω) \;|\; ∃ \tilde u ∈ H^1(Ω)\colon u = \text{tr}(\tilde u) \}$$
is introduced together with the norm
$$\| u \|_{H^{1/2}(\partial Ω)} = \inf \{ \| \tilde u \|_{H^1(Ω)} \;|\; \text{tr}(\tilde u) = u\}.$$
On the one hand, this space is instantly recognisable as the range of the trace operator, $\text{tr}$, but on the other hand, I have read that this space, $H^{1/2}(\partial\Omega)$, is constructed as the quotient of $H^1(\Omega)$ by $\ker(\text{tr})$.

The Mathonline page Quotient Normed Linear Spaces defines the quotient normed linear space, for a given normed linear space $(X,\|\cdot\|_X)$ and linear subspace $M\subseteq X$, as $X / M := \{ x + M : x \in X \}$ where $x + M := \{ x + m : m \in M \}$. A seminorm is defined on this quotient space by $\| x + M \|_{X / M} = \inf \{ \| x + m \|_X : m \in M \}$, which is a genuine norm when $M\subseteq X$ is closed. I understand that in our case $M=\ker(\text{tr})$, which is closed since $\text{tr}$ is a bounded linear operator. Question: why is it that in our case there is no representative $m$ of $M$ in the definition of $\| \cdot \|_{H^{1/2}(\partial Ω)}$?

Question: If $H^{1/2}(\partial Ω)$ is instantly recognisable as the range of $\text{tr}$, how does the need arise to make use of a quotient construction? How is the identification of $\text{ran}(\text{tr})$ with the quotient $H^1(\Omega)/\ker(\text{tr})$ a natural one to make?

答:
I expanded my original answer to the question you linked, hopefully the updated version should clear up the confusion regarding point 1.

As for point 2, the space was first investigated due to its role as a trace of $H^1$-functions, where the quotient construction is a natural abstract characterization of the space. The other characterization was not directly evident, but a result of that investigation, which was a major driving force behind the development of interpolation theory (some historical background can be found in the book "Tartar, L. (2007). An Introduction to Sobolev Spaces and Interpolation Spaces.").


答:
$$E[X_n]=0$$

$$\lim_{n \to \infty}\mathrm{Var}[X_n]=0$$

This is a necessary and sufficient condition for

$$X_n\xrightarrow{L^2}0$$

Convergence in $L^2$ implies convergence in probability

Body:
Problem

Prove that $X_n \sim N(0,\frac{1}{n})$ converges in probability to 0.

Discussion

There a problems similar to this question, but I want to ask specifically about the convergence of an integral that arises in the solution.

Attempt

Let $\epsilon>0$. To show $X_n \overset{p}{\to} 0$, we must show $P(|X_n| > \epsilon) \to 0$.

$$
\begin{align}
P(|X_n|>\epsilon)
&= P(X_n>\epsilon) + P(X_n < -\epsilon)\\
&= P(Z>\epsilon\sqrt{n}) + P(X_n<-\epsilon\sqrt{n})
\end{align}
$$
where $Z$ is the standard normal distribution. Thus we want to show
$$P(|X_n|>\epsilon)=\int_{\epsilon\sqrt{n}}^\infty\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\text{dx}
+
\int_{-\infty}^{-\epsilon\sqrt{n}}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\text{dx}\to 0$$

I understanding intuitively that these integrals should converge to zero, but how can I show rigorously that they go to zero?


问:
Finding The Shortest Distance Between Two 3D Line Segments
I have two Line Segments, represented by a 3D point at their beginning/end points.

Line:

```<code>class Line
{
    public string Name { get; set; }
    public Point3D Start { get; set; } = new Point3D();
    public Point3D End { get; set; } = new Point3D();
}
</code>```

The 3D points are just 3 doubles for coordinates X,Y and Z.

3DPoint:

```<code>class Point3D
{
    public double X { get; set; }
    public double Y { get; set; }
    public double Z { get; set; }
}
</code>```

The Question:

Can I find the distance between two 'Lines' and the endpoints of that distance 'Line'. Here is an Image to Better Illustrate What I am trying to Achieve I know that this question seems programmatically oriented. But the true issue is a math question at heart. Thank you in advance for your understanding.

What I have:

Currently, I can successfully get the distance between the two lines with this code (Adapted From Here Using the Segment To Segment Section):

```<code>    public double lineNearLine(Line l1, Line l2)
    {
        Vector3D uS = new Vector3D { X = l1.Start.X, Y = l1.Start.Y, Z = l1.Start.Z };
        Vector3D uE = new Vector3D { X = l1.End.X, Y = l1.End.Y, Z = l1.End.Z };
        Vector3D vS = new Vector3D { X = l2.Start.X, Y = l2.Start.Y, Z = l2.Start.Z };
        Vector3D vE = new Vector3D { X = l2.End.X, Y = l2.End.Y, Z = l2.End.Z };
        Vector3D w1 = new Vector3D { X = l1.Start.X, Y = l1.Start.Y, Z = l1.Start.Z };
        Vector3D w2 = new Vector3D { X = l2.Start.X, Y = l2.Start.Y, Z = l2.Start.Z };
        Vector3D u = uE - uS;
        Vector3D v = vE - vS;
        Vector3D w = w1 - w2;
        double a = Vector3D.DotProduct(u, u);
        double b = Vector3D.DotProduct(u, v);
        double c = Vector3D.DotProduct(v, v);
        double d = Vector3D.DotProduct(u, w);
        double e = Vector3D.DotProduct(v, w);
        double D = a * c - b * b;
        double sc, sN, sD = D;
        double tc, tN, tD = D;
        if (D < 0.01)
        {
            sN = 0;
            sD = 1;
            tN = e;
            tD = c;
        }
        else
        {
            sN = (b * e - c * d);
            tN = (a * e - b * d);
            if (sN < 0)
            {
                sN = 0;
                tN = e;
                tD = c;
            }
            else if (sN > sD)
            {
                sN = sD;
                tN = e + b;
                tD = c;
            }
        }
        if (tN < 0)
        {
            tN = 0;
            if (-d < 0)
            {
                sN = 0;
            }
            else if (-d > a)
            {
                sN = sD;
            }
            else
            {
                sN = -d;
                sD = a;
            }
        }
        else if (tN > tD)
        {
            tN = tD;
            if ((-d + b) < 0)
            {
                sN = 0;
            }
            else if ((-d + b) > a)
            {
                sN = sD;
            }
            else
            {
                sN = (-d + b);
                sD = a;
            }
        }
        if (Math.Abs(sN) < 0.01)
        {
            sc = 0;
        }
        else
        {
            sc = sN / sD;
        }
        if (Math.Abs(tN) < 0.01)
        {
            tc = 0;
        }
        else
        {
            tc = tN / tD;
        }
        Vector3D dP = w + (sc * u) - (tc * v);
        double distance1 = Math.Sqrt(Vector3D.DotProduct(dP, dP));
        return distance1;
    }
</code>```

What I Need:

Is there any way to determine the endpoints of the displacement vector 'dP' from the code above? If not, can anyone suggest a better method for finding minimum distance and the endpoints of that distance?

Thank you for Reading, and Thanks in advance for any suggestions!

For those interested in the solution, I have posted a code complete version on my question HERE

答:
I'm not sure about the correctness of the webpage linked by @amd (I don't have enough points to add a comment below it).

The author claims: 'However, in these cases, the minimum always occurs on the boundary of G, ...'.

Imagine the line segments defined by $ P_1=(-1,1,0)$, $ P_2=(1,-1,0) $ and $ Q_1=(-1,-1,1) $ and $ Q_2=(1,1,1) $. The shortest distance is not at the boundary of G. It is at s=t=0.5.

I think the correct theory is provided by this PDF: https://www.geometrictools.com/Documentation/DistanceLine3Line3.pdf

There is even an implementation linked inside the PDF: https://www.geometrictools.com/GTE/Mathematics/DistSegmentSegment.h


问:
What does infinity as an upper/lower limit mean in a definite integral? Eg, $\int_0^\infty e^{-x} dx$
I was finding
$$\int_0^\infty e^{-x} dx$$
So the integral becomes
$$\left.(-e)^{-x}\right\rvert_0^\infty$$

Now here's the confusing thing for me.  What exactly does infinity mean in upper/lower limit?

How do I actually calculate definite integrals when one of the limit is infinity?

答: Look at the area here, and imagine extending to the right "forever":


问:
Integral of $\int \frac{\sin(3x)}{\sin(5x)} \, dx$
How to find integral of

$$\int \frac{\sin(3x)}{\sin(5x)}dx$$

I wrote $\sin(3x)=\sin(8x-5x)$ but it generated $\frac{\sin(8x) \cos(5x)}{\sin(5x)}$.

How should I proceed?

答:
Express

\begin{align}
\sin(3x)& =2\sin x  \left(\cos(2x)- \cos\frac{2\pi}{3}\right) \\
\sin(5x)& =4\sin x  \left(\cos(2x)- \cos\frac{2\pi}{5}\right) \left(\cos(2x)- \cos \frac{4\pi}{5}\right)
\end{align}

Then, decompose the integrand as follows
\begin{align}\frac{\sin (3x)}{\sin(5x)}
=&\frac{\cos(2x)- \cos\frac{2\pi}{3}} {2\left(\cos(2x)- \cos\frac{2\pi}{5}\right) \left(\cos(2x)- \cos \frac{4\pi}{5}\right)}\\
=&\frac{\frac1{\sqrt5}\cos\frac{2\pi}{5} }{\cos(2x)- \cos\frac{4\pi}{5} }
-\frac{ \frac1{\sqrt5} \cos\frac{4\pi}{5}}{\cos(2x)- \cos\frac{2\pi}{5} }
\\
=&\frac{\frac1{2\sqrt5}\cos\frac{2\pi}{5} }{\sin(\frac{2\pi}5+x)\sin(\frac{2\pi}5-x) }-\frac{\frac1{2\sqrt5} \cos\frac{4\pi}{5} }{\sin(\frac{\pi}5+x)\sin(\frac{\pi}5-x) }\\
=&\frac15\sin\frac\pi5\left(\cot(\frac{2\pi}5+x)+\cot(\frac{2\pi}5-x) \right)\\
&\hspace{1cm}+\frac15\sin\frac{2\pi}5\left(\cot(\frac{\pi}5+x)+\cot(\frac{\pi}5-x) \right)
\end{align}
Apply $(\ln \sin t)’= \cot t$ to integrate
$$\int \frac{\sin (3x)}{\sin(5x)}dx= \frac15\sin\frac{\pi}5\>\ln \frac{\sin(\frac{2\pi}5+x) }{\sin(\frac{2\pi}5-x) } +\frac15\sin\frac{2\pi}5\>\ln \frac{\sin(\frac{\pi}5+x) }{\sin(\frac\pi5-x) }+C
$$


问:
Construct a bijection $h:\{0,1\}^{\mathbb{N}} \to \{3,4\}^{\mathbb{P}}$
I have to construct a bijection $h:\{0,1\}^{\mathbb{N}} \to \{3,4\}^{\mathbb{P}}$ where $\mathbb{P}=\{n \in \mathbb{N} \mid \exists k \in \mathbb{N} \ n=2k\}$. I came up with something like this: $$f:\mathbb{N} \to \{0,1\}\\g:\mathbb{P} \to \{3,4\}\\h(f)=g\\g(n)=f(n)+3$$Now I have to proof that $h(f)$ is a bijection. In order to do that, I need to proof that it is injective and surjective. The problem is I'm not sure how.

答:
I think you are having a mental block of functions being "points" as space and conceiving of a function that maps the set functions to another set of functions.  I had the same block when I was a student.

If $f\in \{0,1\}^{\mathbb N}$ then that means $f$ is function $f:\mathbb N \to \{0,1\}$ so that for any $n\in \mathbb N$ then $f(n) =\begin{cases}0\\1\end{cases}$.

(  There are, of course, an infinite number such function elements of $\{0,1\}^{\mathbb N}$.  In fact, if we define $a_n := f(n)$ then $\{f(1),f(2),f(3),.....\} = \{a_1,a_2,a_3\}$ could be any infinite sequence of $0,1$ terms and we can define the elements of $\{0,1\}^{\mathbb N}$ by considering any such sequence $\{b_k\}$ so that $h(n) := b_n$ and therefore $h\in \{0,1\}^{\mathbb N}$.)

And the same is true for $\{3,4\}^{\mathbb P}$.  If $h\in \{3,4\}^{\mathbb P}$ then $g:\mathbb P\to \{3,4\}$ and for any $m=2k\in \in \mathbb P$ then $h(m)=\begin{cases}3\\4\end{cases}$.

Now it should be intuitive that this are equivalent.  For every $f \in \{0,1\}^{\mathbb N}$ so that $f: n\mapsto \begin{cases}0\\1\end{cases}$ we can just make a $g:\mathbb P \to \{3,4\}$ via $g: 2n \mapsto \begin{cases}3&\text{if }f(n)=0\\4&\text{if }f(n)=1\end{cases}$

So.... lets do that:  Consider $\tau:\{0,1\}^{\mathbb N}\to \{3,4\}^{\mathbb P}$ via $\tau{f} = g$ where $g: \mathbb P\to \{3,4\}$ via $g(2n)=\begin{cases}3&\text{if }f(n)=0\\4&\text{if }f(n)=1\end{cases}$.

Now we have to prove

$\tau$ is a well-defined function.

$\tau$ is onto

$\tau$ is one to one

Proving 1: for any $f\in \{0,1\}^{\mathbb N}$ will be a well defined function the maps any $k\in \mathbb N$ to a distinct value of either $0$ or $1$.  For ever $p\in \mathbb P$ there is a unique $m\in \mathbb N$ so that $p = 2m$.  For that distinct $m$ there is a distinct value of $0$ or $1$ and for that distinct value we can assign $p \mapsto 3$ if $f(m)=0$ or $p\mapsto 4$ if $f(m)=1$.  As there is no ambiguity in this mapping and each element $p$ is mapped to a distinct element of $\{3,4\}$ this mapping is a function mapping $\mathbb P \to \{3,4\}$ and is an element of $\{3,4\}^{\mathbb P}$.  So $\tau$ is well defined.

Proving 2:  If $g:\mathbb P \to \{3,4\}$ we can define $f:\mathbb N \to \{0,1\}$ via $f(n) = \begin{cases}0&\text{if }g(2n)=3\\1&\text{if }g(2n)=4\end{cases}$.  Then $\tau(f)(p)=\begin{cases}3&\text{if }f(\frac p2)=0\\4&\text{if }f(\frac p2)=1\end{cases}=\begin{cases}3&\text{if }g(2\cdot\frac p2)=3\\4&\text{if }g(\frac 2\cdot \frac p2)=4\end{cases}=g(n)$.  So $\tau(f)= g$.

Proving 3:

Suppose $f,h \in \{0,1\}^{\mathbb N}$ but $f\ne h$.  That means there is a $k\in \mathbb N$ where $f(k) \ne h(k)$.  If $f(k) = 0$ then $h(k) = 1$ and therefore $\tau(f)(2k) = 3$ while $\tau(h)(2k) = 4$.  And if $f(k) =1$ then $h(k) = 0$ and therefore $\tau(f)(2k) = 4$ while $\tau(h)(2k) =3$.

In either case if $f \ne h$ then $\tau(f) \ne \tau(h)$.

So $\tau$ is one-to-one.


问:
equalities between conditional expectation
How can I make the following equation correct by changing the order of $X,Y$ and $E$ :
$$E[X]=E[X|E[Y]]$$
and how can I prove it ?

答:
$\mathbb E[Y]$ is a constant, hence
\begin{align*}
\mathbb E[X|E[Y]] &= \mathbb E[X|\sigma(E[Y])]\\
&=\mathbb E[X|\{\emptyset,\Omega \}]\\
&=\mathbb E[X]
\end{align*}

The last line is because any $\{\emptyset,\Omega \}$ measurable random variable is a constant so that $\mathbb E[X|\{\emptyset,\Omega \}]$ is the constant $a$ that satisfies $\mathbb E[Xb]=\mathbb E[ab]$ for all constant $b$, now whenever $b\neq 0$, then we directly get $\mathbb E[X]=a$ by linearity of expectation.


问:
Prove questions similar to the binomial theorem for complex numbers

1

How should I prove that $$\forall n \in \mathbb{Z}, n\geq0, (1+i)^{n}+(1-i)^{n}=2\sum_{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor}\binom{n}{2k}(-1)^{k}$$

From what I can see here, it looks really similar to the binomial theorem, but alternates between negative and positive numbers. Is it possible for me to invoke De Moivre's Theorem in this case?

2

Also, a similar question that I wanted to prove was $$\forall n \in \mathbb{Z}, n\geq0, (1+i)^{n}+(1-i)^{n}=0\leftrightarrow n\equiv2(mod \text{ 4})$$

This seems to be a very tricky if and only if statement, but I made a very key observation that $(1+i)^{4}=(1-i)^{4}=-4$. I'm a bit lost as to how I can apply it though. Especially to both the forward and backward implications.

答:
Hint for 1:

Expand the binomials $(1+i)^n$ and $(1-i)^n$ and add them, noting that $i^2=(-i)^2=-1$.

Hint for 2:

It does help here to use de Moivre's theorem,

because $1+i=\sqrt2e^{i \pi/4}$ and $1+i=\sqrt2e^{- i \pi/4}$,

so $(1+i)^n+(1-i)^n=\sqrt2^n(e^{in\pi/4}+e^{-in\pi/4})=\sqrt2^n2\cos(n\pi/4).$


答:
Naively, you can do this, and it gets the right answer:
$$\int_0^\infty e^{-x}dx = \left[-e^{-x}\right]_0^\infty=e^0-e^{-\infty}=1-0=1$$

More formally, you can define
$$\int_0^\infty e^{-x}dx \,\,\,\underset{\text{def}}{=}\,\,\,\lim_{t\to\infty} \int_0^t e^{-x}dx = \lim_{t\to\infty} \left[-e^{-x}\right]_0^t=\lim_{t\to\infty}(e^0-e^{-t})=1-\lim_{t\to\infty}e^{-t}=1$$

Title: What does infinity as an upper/lower limit mean in a definite integral? Eg, $\int_0^\infty e^{-x} dx$

Body:
I was finding
$$\int_0^\infty e^{-x} dx$$
So the integral becomes
$$\left.(-e)^{-x}\right\rvert_0^\infty$$

Now here's the confusing thing for me.  What exactly does infinity mean in upper/lower limit?

How do I actually calculate definite integrals when one of the limit is infinity?


答:
You can use the tail bound for a standard normal random variable:
$$
\int_x^\infty \exp(-y^2/2)\,dy \le x^{-1}\exp(-x^2/2),
$$
which can be proved changing variables $y = x + z$ and using $\exp(-z^2/2) \le 1$.

Title: Prove that $X_n \sim N(0,\frac{1}{n})$ converges in probability to 0


问:
How to prove the existence of odd numbers $a$ and $b$ in $2m\equiv a^{20}+b^{11}\pmod{2^n}$

Show that for any natural numbers $m,n$, there exist odd numbers $a,b$ such that
  $$2m\equiv a^{20}+b^{11}\pmod{2^n}$$

Thank you everyone.

答:
We proceed by induction on $n$.

When $n=1$, we have $1^{20}+1^{11} \equiv 0 \equiv 2m \pmod{2^1}$.

Suppose that the statement holds for $n=k$, so $\exists a, b \in \mathbb{Z}$, $a, b$ odd, such that $2m \equiv a^{20}+b^{11} \pmod{2^k}$. Thus $a^{20}+b^{11} \equiv 2m, 2m+2^k \pmod{2^{k+1}}$. If $a^{20}+b^{11} \equiv 2m \pmod{2^{k+1}}$, we are done. Otherwise we have $a^{20}+b^{11} \equiv 2m+2^k \pmod{2^{k+1}}$. Then $a^{20}+(b+2^k)^{11} \equiv a^{20}+b^{11}+11b^{10}2^k \equiv (a^{20}+b^{11})+2^k \equiv 2m \pmod{2^{k+1}}$.  We are thus done by induction.

Therefore for any $m, n \in \mathbb{Z}^+$, there exist odd $a, b \in \mathbb{Z}$ such that $2m \equiv a^{20}+b^{11} \pmod{2^n}$.


问:
A difficult problem concerning functional equations: $ f ( x ) f ( y ) = x ^ a f \left( \frac y 2 \right) + y ^ b f \left( \frac x 2 \right) $
While studying  functional equation, I came across the following problem.

Problem. find all functions $ f : \mathbb R \to \mathbb R $ such that for some $ a , b \in \mathbb R $,
$$ f ( x ) f ( y ) = x ^ a f \left( \frac y 2 \right) + y ^ b f \left( \frac x 2 \right) $$
holds for all reals $ x $ and $ y $.

The solution presented by the author is as follows.

Solution. If $ a = b $, we can denote
$$ \frac { f ( x ) } { x ^ a } = g ( x ) \text , $$
so that
$$ g ( x ) g ( y ) = g ( x + y ) \text , \quad \forall x , y \in \mathbb R \text , $$
and here use the new function $ \require {color} \color {red} h ( x ) = g ( e ^ x ) $ to get $ h ( x + y ) = h ( x ) + h ( y ) $. This is the classic Cauchy equation with the solution $ h ( x ) = k x $ for all real $ x $ and any real $ k $.

I am not able to understand that how we can obtain
$$ h ( x + y ) = h ( x ) + h ( y ) $$
by using the new function $ h ( x ) = g ( e ^ x ) $.

My try:

We have $ h ( x ) = g ( e ^ x ) $ which is same as saying $ h ( \ln x ) = g ( x ) $. On substituting $ h ( \ln x ) = g ( x ) $ in
$$ g ( x ) g ( y ) = g ( x ) + g ( y ) $$
we get
$$ h ( \ln x ) h ( \ln y ) = h ( \ln x ) + h ( \ln y ) \text . $$
But I am really not able proceed any further.

A detailed explanation of what's actually going on and how to proceed further would be highly helpful. As it is mentioned in the problem "... for some $ a , b \in \mathbb R $", could you please explain that why and how $ a = b $ in the solution?

Also another elegant solution for the very problem  would be highly appreciated.

答:
As I've explained in a comment, the question is not well-posed. To make this go away, one can add the requirement that $ a $ and $ b $ are positive integers. Alternatively, on can consider $ f : \mathbb R ^ + \to \mathbb R $ and require that the functional equation
$$ f ( x ) f ( y ) = x ^ a f \left( \frac y 2 \right) + y ^ b f \left( \frac x 2 \right) \tag 0 \label 0 $$
holds for all positive reals $ x $ and $ y $. Some (seemingly more complicated) alternatives are possible, too.

Assuming we've taken care of the ambiguity of the problem, one can see that in case $ a \ne b $, the only solution is the constant zero function. To see this, use \eqref{0} and see that
$$ x ^ a f \left( \frac y 2 \right) + y ^ b f \left( \frac x 2 \right) = f ( x ) f ( y ) = f ( y ) f ( x ) = y ^ a f \left( \frac x 2 \right) + x ^ b f \left( \frac y 2 \right) \text , $$
which in particular shows that
$$ f ( x ) = \frac { f ( 1 ) } { 2 ^ a - 2 ^ b } \left( ( 2 x ) ^ a - ( 2 x ) ^ b \right) \text . \tag 1 \label 1 $$
\eqref{1} shows that $ f \left( \frac 1 2 \right) = 0 $, and then letting $ x = y = 1 $ in \eqref{0} we get $ f ( 1 ) = 0 $, and thus by \eqref{1}, $ f $ is the constant zero function. It's trivial to check that this indeed gives a solution.

For the case $ a \ne b $, you can follow the given solution. But you need to note that as $ g ( x ) $ is defined to be equal to $ \frac { f ( x ) } { x ^ a } $, this is only possible when $ x ^ a \ne 0 $, and thus $ 0 $ cannot be assumed to be in the domain of $ g $. Also, as $ h ( x ) $ is defined to be equal to $ g ( e ^ x ) $, determining $ h $ only determines the restriction of $ g $ to the set of positive reals (since $ e ^ x > 0 $). If we've chosen the alternative in which we have $ f : \mathbb R ^ + \to \mathbb R $, we're indeed done at this point. But if we've chosen the alternative in which $ f : \mathbb R \to \mathbb R $ but $ a $ and $ b $ couldn't be just any real numbers, we need to continue further, and find out what $ f ( x ) $ can be for negative $ x $ and also for $ x = 0 $. That shouldn't be much difficult, assuming you've made your peace with the given solution in the original post.


问:
Prove this endomorphism cannot have real eigenvalues?
$f$ is an endomorphism of an $\Bbb R$-vector space $E$ with dimension $n\geq1$. $f$ satisfies $f^2+f+$Id$_E = 0$.

How can I go about showing that $f$ can't have real eigenvalues?

答:
Hint

The minimal polynomial $q(x)$ of $f$ is the smallest anulator polynomial in the sense that if $p(f)=0$ then $q(x)\mid p(x)$. Moreover, $\lambda $ is an eigenvalue of $f$  if and only if $q(\lambda )=0$.


答:
Well, if $x$ is an eigenvector of $f$ with $f(x) = \lambda x$.

Then $0 = 0(x) = (f^2+f+id)(x) =f^2(x) + f(x) + x = \lambda^2 x + \lambda x + x = (\lambda^2+\lambda + 1)x$.

If $x\ne 0$, then $\lambda^2+\lambda+x=0$. But this polynomial has negative discriminant $D=1^2-4\cdot 1\cdot 1 = -3$ and so no real-valued solution.

Title: Prove this endomorphism cannot have real eigenvalues?

Body:
$f$ is an endomorphism of an $\Bbb R$-vector space $E$ with dimension $n\geq1$. $f$ satisfies $f^2+f+$Id$_E = 0$.

How can I go about showing that $f$ can't have real eigenvalues?


答: Any eigenvalue $\lambda$ would have to satisfy $\lambda^2+\lambda+1=0\implies \lambda=(1\pm\sqrt3i)/2$.


问:
Find the integer value of a, b such that $\lim_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.
Find the integer value of $a$, $b$ such that  $\lim\limits_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.

My work:

I multiplied with $\frac{ax+\sqrt{x}+b}{ax+\sqrt{x}+b}$, and obtained:

$=\lim\limits_{x \rightarrow 4} \frac{(ax+b)^2-x}{(x-4)(ax+b+\sqrt{x})}$

$=\lim\limits_{x \rightarrow 4} \frac{(a^2x^2+2abx+b^2-x)}{(x-4)(ax+b+\sqrt{x})}$

I factored $(x-4)$ from the numerator:

$(x-4)(a^2x+2ab-1+4a^2)+(b^2+8ab-4+16a^2)$

We have the first equation

$(b^2+8ab-4+16a^2)=0$

$=\lim\limits_{x \rightarrow 4} \frac{(x-4)(a^2x+2ab-1+4a^2)}{(x-4)(ax+b+\sqrt{x})}$
$=\frac{4a^2+2ab-1+4a^2}{4a+b+2}=\frac{3}{4}$

So, the second equation is
$16a^2+8ab-4+16a^2=12a+3b+6$

$32a^2-12a+8ab-3b-10=0$

But I can't find the integer value of $a$, $b$.

答:
Substitute $\sqrt x=y$, so that

$$\lim_{x\to4}\frac{ax-\sqrt x+b}{x-4}=\lim_{y\to2}\frac{ay^2-y+b}{y^2-4}=\frac34$$

We have

$$\frac{ay^2-y+b}{y^2-4}=a-\frac{y-4a-b}{(y-2)(y+2)}$$

For the limit to exist, $y=2$ must be a removable singularity, so $y-2$ must divide $y+4a+b$. We have

$$\frac{y-4a-b}{y-2}=1-\frac{4a+b-2}{y-2}$$

and the remainder should be such that $4a+b-2=0$. So,

$$\lim_{y\to2}\frac{ay^2-y+b}{y^2-4}=\lim_{y\to2}\left(a-\frac1{y+2}-\frac{4a+b-2}{(y+2)(y-2)}\right)=a-\frac14=\frac34$$

Then $a=1$ and $b=-2$.


问:
In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

What I Tried: Here is the figure :-

I could conclude that $ABCD$ is cyclic, but I really could not use that property in a useful way till now. I thought about Ptolemy's Theorem but I am not sure since I don't know enough lengths.

Instead, Pythagoras Theorem gives the required values, assuming $AD = AB = x$ . But I am still stuck as that is not enough for finding the side-lengths of the quadrilateral, only then I can find it's area. The red angles came to be $45^\circ$, and I am stuck right here. I can't seem to use something with the cyclic quadrilaterals.

Can anyone help me? Thank You.

答:
Take $\angle BDC=\theta $, so $DE = xcos(\theta+45), xsin(\theta+45)=5, EC = x\sqrt{2}cos\theta-xcos(\theta+45), DC = x\sqrt{2}cos\theta, \\
BC = x\sqrt{2}sin\theta$

$\Delta ADB+\Delta BDC = \Delta ADE+ \square AECB$

$\Delta ADB+\Delta BDC = \dfrac{x^2}{2} + x^2sin\theta cos\theta = \dfrac{x^2}{2}(sin\theta + cos\theta )^2 $

$\Delta ADE+ \square AECB = x^2sin\theta cos\theta - \dfrac{x^2}{\sqrt{2}}sin\theta cos(\theta +45)+  \dfrac{5}{\sqrt{2}}xcos\theta  $

Equating both we get $x = \dfrac{5\sqrt{2}}{(sin\theta + cos\theta)}$

Using this in $\dfrac{x^2}{2}(sin\theta + cos\theta )^2 $  we get area $=25$


问:
How to compute $\int_0^\infty \frac{\tanh\left(\pi x\right)}{x\left(1+x^2\right)} \, \mathrm{d}x$?
How do I compute the following integral?

$$\int_0^\infty \frac{\tanh\left(\pi x\right)}{x\left(1+x^2\right)} \, \mathrm{d}x$$

I tried some basic substitutions but they only make it more complicated. WolframAlpha says the answer is $2$ but I have no clue how to get there.

答:
Let $N$ be a positive integer, and consider the contour integral of

$$ f(z) = \frac{\tanh(\pi z)}{z(z^2+1)} $$

along the boundary of the rectangle with the corners $\pm N$ and $\pm N+ iN$. Noting that $\tanh(\pi z)$ has a simple zero at $ki$ and a simple pole at $z_k := \bigl(k+\frac{1}{2}\bigr)i$ for each $ k \in \mathbb{Z}$, the function $f$ has simple poles only at $z_k$'s. (The poles at $0$ and $\pm i$ are cancelled out by the zeros of $f$.) So by the residue theorem,

\begin{align*}
\int_{-N}^{N} f(x) \, \mathrm{d}x
&= 2\pi i \sum_{k=0}^{N-1} \mathop{\mathrm{Res}}_{z=z_k} f(z) - \int_{\Gamma_N} f(z) \, \mathrm{d}z,
\end{align*}

where $\Gamma_N$ is the piecewise linear path from $N$ to $N+iN$ to $-N+iN$ to $-N$. Now it is not hard to show that the integral of $f$ along $\Gamma_N$ vanishes as $N\to\infty$, and so, letting $N\to\infty$ yields

\begin{align*}
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x
&= 2\pi i \sum_{k=0}^{\infty} \mathop{\mathrm{Res}}_{z=z_k} f(z) \\
&= i \sum_{k=0}^{\infty} \frac{2}{z_k (z_k + i)(z_k - i)} \\
&= i \sum_{k=0}^{\infty} \left( - \frac{1}{z_{k-1}} + \frac{2}{z_k} - \frac{1}{z_{k+1}} \right) \\
&= \frac{i}{z_0} - \frac{i}{z_{-1}} \\
&= 4.
\end{align*}

Therefore the answer is $\frac{1}{2} \cdot 4 = 2$.


答:
Assume $x$ is an eigenvector of $f$ with eigenvalue $\lambda$ then evaluating your equation at $x$ gives $f(f(x))+f(x)+id(x)=0$, but since $x$ is an eigenvector $f(x)=\lambda x$ and $f(f(x))=f(\lambda x)=\lambda f(x)=\lambda^2 x$ where we used the linearity of $f$, so inserting into the first equation yields $$\lambda^2 x+\lambda x+x=0$$ which by the vector space axioms of scalar multiplication is equivalent to $$(\lambda^2+\lambda+1)x=0$$ Now since all eigenvector are assumed nonzero we have that $$\lambda^2+\lambda+1=0$$
But this equation has no real solutions so our linear map cannot have real eigenvalues.


问:
If the number of positive integer $n$ for which $n \leq 1991$ and $6$ is a factor of $n^2 + 3n + 2$ is given by $K$, Find the value of $\sqrt{K - 32}$
I tried by factorising $n^2 + 3n + 2$ into $(n + 1)(n + 2)$ and tried by
assuming $ n = 2k + 1 $ and $ n = 2k $ for odd and even respectively and puting them in place of n

For odd : $(2k + 2)(2k+ 3)$ $\Rightarrow$ $2k + 3$ should be divisible by 3

For even :
$(2k + 1)(2k + 2)$ $\Rightarrow$ $2k + 1$ should be divisible by 3

But I can't proceed any further... Can anyone explain how to go through this question?

答:
Let $f(n)=(n+1)(n+2)$. Each six consecutive integers from $n=1$ till $n=1991$ contains two numbers $f(n)$ which have not $6$ as a factor and these are $$3,6,9,12,\cdots,1989$$ They formed an evident arithmetic progression so there are $1989:3=663$ numbers involved that are not multiples of $6$. It follows that one has $1991-663=1328$ numbers having $6$ as a factor.

Thus $K=1328$ and the asked number is then $\sqrt{1296}=36$.


Title: Find the integer value of a, b such that $\lim_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.

Body:
Find the integer value of $a$, $b$ such that  $\lim\limits_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.

My work:

I multiplied with $\frac{ax+\sqrt{x}+b}{ax+\sqrt{x}+b}$, and obtained:

$=\lim\limits_{x \rightarrow 4} \frac{(ax+b)^2-x}{(x-4)(ax+b+\sqrt{x})}$

$=\lim\limits_{x \rightarrow 4} \frac{(a^2x^2+2abx+b^2-x)}{(x-4)(ax+b+\sqrt{x})}$

I factored $(x-4)$ from the numerator:

$(x-4)(a^2x+2ab-1+4a^2)+(b^2+8ab-4+16a^2)$

We have the first equation

$(b^2+8ab-4+16a^2)=0$

$=\lim\limits_{x \rightarrow 4} \frac{(x-4)(a^2x+2ab-1+4a^2)}{(x-4)(ax+b+\sqrt{x})}$
$=\frac{4a^2+2ab-1+4a^2}{4a+b+2}=\frac{3}{4}$

So, the second equation is
$16a^2+8ab-4+16a^2=12a+3b+6$

$32a^2-12a+8ab-3b-10=0$

But I can't find the integer value of $a$, $b$.

Body:
Note

$$\frac{ax-\sqrt{x}+b}{x-4}
= a - \frac{1}{\sqrt x+2} + \frac{4a+b-2}{x-4}
$$
Then,  $\lim_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ requires
$$a-\frac1{\sqrt4+2}= \frac34,\>\>\>\>\>4a+b-2=0$$
which yields $a=1$ and $b=-2$.


问:
Transformation of an equation with a fraction in the numerator of a fraction
Can someone explain the steps here -- how does it get the second part where it is equals to $20$? I was confused because of the fraction $1/2$ in the numerator.

I am studying some examples but I don't understand this one.

答:
If you have
$$
\frac{(1/2)x}{21}+\frac{(1/2)x}{24}=10
$$
then you can take a factor of $\frac{1}{2}$ out the front of the left-hand side
$$
\frac{1}{2}\left(\frac{x}{21}+\frac{x}{24}\right)=10
$$
and then multiply both sides of the equation by $2$
$$
\frac{x}{21}+\frac{x}{24}=20
$$</p>


问:
Solving a complex equation with complex conjugate
Short and sweet, the problem to solve is:

$$(\overline{z}-i)^4 = \cos{\frac{4\pi}{3}} +i\sin{\frac{4\pi}{3}}$$

My process of solving was as follows:

First, take the fourth root using the polar coordinate representation of complex numbers as follows: $$\overline{z}-i = r^\frac{1}{4}e^{\frac{\phi+2k\pi}{4} i} $$

Note that there are four solutions for $k=0,1,2,3$, but for the sake of concision, I will only do the case of $k=0$. The rest can be calculated with similar methods.

$$\overline{z}-i = \frac{1}{2} +i\frac{\sqrt{3}}{2}$$

Then, move $i$ to the right side and get

$$\overline{z} = \frac{1}{2} + \large i(\frac{\sqrt{3}}{2}+1) $$

Lastly, since we have $z$ conjugated on the left side, take the conjugate of the right side and get:

$$z = \frac{1}{2} - \large i(\frac{\sqrt{3}}{2}+1) $$

Can someone please verify if this is correct? If not, what is the correct method to do this?

Once again, I am aware that this equation has four possible solutions in the complex plane.

Body:
$$\bar z-i=e^{(2n\pi i+i4\pi/3)/4}=e^{i\pi(3n+2)/6}$$ where $n=0,1,2,3$

$$\implies \bar z=i+\cos\dfrac{\pi(3n+2)}6+i\sin\dfrac{\pi(3n+2)}6$$

$$\implies z=\bar{\bar z}=-i+\cos\dfrac{\pi(3n+2)}6-i\sin\dfrac{\pi(3n+2)}6$$


问:
Image of a totally geodesic surface under immersion equals image of exponential map?
I was wondering the following. Suppose we have some ambient manifold $B$, and we have a totally geodesic surface (meaning second fundamental form vanishes) $M \subset B$.

I know from a proposition that due to $M$ being totally geodesic, the geodesics of $M$ are also geodesics in $B$.

Now also suppose that $B$ is a complete Riemannian manifold. I want to know the following: Suppose we have an immersion $F: M \rightarrow B$. If we let $p \in M$ be a point, is it true in general that $F(M) = \exp_p (T_p M)$? If so, how can we prove this? Or is it trivial?

答:
Say $(M',g)$ is a complete riemannian manifold.

Suppose $M \subset M'$ is a totally geodesic submanifold. This means that if $p \in M$ and $v \in T_pM$, then the geodesic $\gamma_{p,v} : t \mapsto \exp_p(tv)$  lies in $M$, where $\exp_p$ denotes the exponential map of $M'$ at $p$.

This shows that if $p \in M$, $\exp_p(T_pM) \subset M$. If $M$ is connected, then every point in $M$ can be connected to $p$ by a geodesic in $M$, that is, $M \subset \exp_p(T_pM)$, and the the equality follows.


问:
Maximum point of $ 1- \sqrt{x^2+y^2} $
Consider the function $ 1- \sqrt{x^2+y^2} $.

The point (0,0) clearly is the maximum point of the function, and it is also an inner point because $ f $ is defines for any (x,y). But the partial derivative according to $ x $ is not 0 at (0,0). How's that possible?

答: The problem is that the map $(x,y) \mapsto \sqrt{x^2+y^2}$ is not differentiable at $(0,0)$. Like the real map $x \mapsto \vert x \vert$ is not differentiable at $0$.

Title: Maximum point of $ 1- \sqrt{x^2+y^2} $


答:
This type of integral is called an 'improper integral'. I have always assumed that the term 'improper' refers to the fact that an improper integral is not really an integral, but rather the limit of one:
$$
\int_a^\infty f(x) \, dx := \lim_{n \to \infty} \int_a^n f(x) \, dx \, .
$$
Thus, '$\int_{0}^{\infty}e^{-x}$' is just a shorthand for $\lim_{n \to \infty} \int_{0}^{n}e^{-x}$, in the same way that the 'infinite sum'
$$
\sum_{n=1}^{\infty} \frac{1}{n}
$$
is really just a shorthand for
$$
\lim_{N \to \infty}\sum_{n=1}^{N} \frac{1}{n} \, .
$$
And just like series can be either convergent or divergent, the same is true of improper integrals. The example you gave is convergent:
\begin{align}
\int_{0}^{\infty}e^{-x} &= \lim_{n \to \infty} \int_{0}^{n}e^{-x} \\
&=  \lim_{n \to \infty} \left[-e^{-x}\right]_{0}^{n} \\
&= \lim_{n \to \infty} (-e^{-n}+1) \\
&= 1 \, .
\end{align}
Other improper integrals are not so well-behaved. Consider
$$
\int_{1}^{\infty}\frac{1}{x} \, dx \, .
$$
There's another class of improper integrals that are a little different. This is when the bounds of the integral are finite, but the function itself is unbounded. Try computing
$$
\int_{0}^{1}\frac{1}{\sqrt{x}} \, dx \,
$$
using the fact that
$$
\int_{a}^{b} f(x) \, dx := \lim_{n \to a^+} \int_{n}^{b} f(x) \, dx
$$
if $f(x)$ is undefined at $x=a$.

Informally, the improper integral
$$
\int_{a}^{\infty} f(x) \, dx
$$
represents the area under the curve between $x=a$ and $x=\infty$. I feel a little uncomfortable making such a statement, since it is hard to see what 'area' means in the infinite case. However, this gives us a great intuitive picture of what improper integrals are all about, something which David G. Stork's answer (+1) alludes to. Here is another exercise. How would you define
$$
\int_{-\infty}^{\infty} x \, dx \, ?
$$
There isn't a correct answer to this question, since definitions can't be 'true' or 'false', but notice that different definitions yield different values. If you are interested in the typical way such an improper integral would be defined, feel free to ask.


问:
Tensor product of maps between Hilbert spaces
Let $H_1,H_2, K_1,K_2$ be Hilbert spaces. Let $S_1: H_1 \to K_1$ and $S_2: H_2 \to K_2$ be bounded linear maps. Is it true that there exists a bounded linear map
$$S_1 \otimes S_2: H_1 \otimes H_2 \to K_1 \otimes K_2$$ such that $(S_1 \otimes S_2) (\xi_1 \otimes \xi_2) = (S_1 \xi_1) \otimes (S_2 \xi_2)$? Here the tensor products are not algebraic tensor products! Rather, they are completions of the algebraic tensor product.

I know this result is true when $H_1 = K_1$ and $H_2 = K_2$, but is this true as well?

答:
The approach  based on decomposing $S_1$ and $S_2$ as  linear combinations of unitary operators  does prove  that $S_1\otimes S_2$ is
bounded but, as far as I can remember, does not directly give the sharp estimate
$$
  \|S_1\otimes S_2\|\leq   \|S_1\|\|S_2\|,
  $$
since the coefficients in the aforementioned linear combinations have little relationship with the norm of the
operators involved.

Here is a different approach,  completely avoiding  unitaries, and which gives the desired norm estimate.

Lemma 1.  If $T:H_1\to H_2$ is a bounded linear operator between  Hilbert spaces $H_1$ and $H_2$, and given $x_1,x_2,\ldots ,x_n\in H_1$, let $A$ and $B$ be the $n\times n$
matrices defined  by
$$
  a_{ij} = \langle T(x_i),T(x_j)\rangle ,
  \quad \text{and} \quad
  b_{ij} = \langle x_i,x_j\rangle .
  $$
Then $A\leq  \|T\|^2 B$.

Proof.  Given scalars  $\lambda _1,\lambda _2,\ldots ,\lambda _n$, we have
$$
  \sum_{i,j}\lambda _i\bar\lambda _ja_{ij} =
  \sum_{i,j}\lambda _i\bar\lambda _j\langle T(x_i),T(x_j)\rangle  =
  \sum_{i,j}\langle T(\lambda _ix_i),T(\lambda _jx_j)\rangle  = $$$$ =
  \left\langle \sum_iT(\lambda _ix_i),\sum_jT(\lambda _jx_j)\right\rangle  =
  \left\|T\left(\sum_i \lambda _ix_i\right)\right\|^2 \leq  $$$$ \leq
  \|T\|^2\left\|\sum_i \lambda _ix_i\right\|^2 =
  \|T\|^2  \sum_{i,j}\langle \lambda _ix_i,\lambda _jx_j\rangle  =  $$$$ =
  \|T\|^2   \sum_{i,j}\lambda _i\bar\lambda _j\langle x_i,x_j\rangle  =
  \|T\|^2    \sum_{i,j}\lambda _i\bar\lambda _jb_{ij}.
  $$
QED

Lemma 2.  If $T:H_1\to H_2$ is a bounded linear operator between  Hilbert spaces $H_1$ and $H_2$, and $K$ is another Hilbert
space, then there is a unique  linear operator
$$
  T\otimes I:H_1\otimes K\to H_2\otimes K,
  $$
such that
$$
  T(x\otimes y) = T(x)\otimes y, \quad \forall x\in  H_1, \quad \forall y\in  K.
  $$
Moreover $\|T\otimes I\|\leq \|T\|$.

Proof.  By replacing $T$ with $T/\|T\|$, we may assume that $\|T\|=1$.
The crucial step in the proof is then showing the inequality
$$
  \left\| \sum_i T(x_i)\otimes y_i \right\| \leq    \left\| \sum_i x_i\otimes y_i \right\|,
  $$
whenever $x_1,x_2,\ldots ,x_n\in H_1$, and $y_1,y_2,\ldots ,y_n\in K$.

In this case,
observe that the matrices $A$ and $B$ defined in Lemma (1)  satisfy $A\leq  B$, whence $B-A\geq 0$,
so we may find an $n\times n$ scalar matrix $C$ such that
$B-A=CC^*$.  Therefore, for every $i$ and $j$ one has that
$$
  \langle x_i, x_j\rangle  = \langle T(x_i), T(x_j)\rangle  +  \sum_k c_{ik}\bar c_{j k}.
  $$
Multiplying by   $\langle y_i, y_j\rangle $ and summing on $i$ and $j$, we get
$$
  \sum_{i, j}\langle x_i, x_j\rangle   \langle y_i, y_j\rangle  = \sum_{i, j}\langle T(x_i), T(x_j)\rangle \langle y_i, y_j\rangle  +  \sum_{i, j}\sum_k c_{ik}\bar c_{jk}\langle y_i, y_j\rangle ,
  $$
which means that
$$
  \left\|\sum_i x_i\otimes y_i \right\|^2 = \left\|\sum_i T(x_i)\otimes y_i \right\|^2 +  \sum_{i, j}\sum_k c_{ik}\bar c_{j k}\langle y_i, y_j\rangle ,
  $$
so it suffices to show that the last sum above is nonnegative.  But this is easy because
$$
  \sum_{i, j}\sum_k c_{ik}\bar c_{j k}\langle y_i, y_j\rangle  =
  \sum_k \left\langle \sum_i c_{ik}y_i,\sum_jc_{j k}y_j\right\rangle  =
  \sum_k \left\|\sum_i c_{ik}y_i\right\|^2 \geq 0.
  $$
QED

Adopting the notation used by the OP, we may now easily prove  that $S_1\otimes S_2$ is bounded, and indeed that
$\|S_1\otimes S_2\|\leq \|S_1\|\|S_2\|$, as follows
$$
  \|S_1\otimes S_2\| =   \|(S_1\otimes I)(I\otimes S_2)\| \leq  \|S_1\otimes I\|\|I\otimes S_2\| \leq   \|S_1\|\|S_2\|.
  $$


问:
Problem with a Arithmetico-Geometric Series
Good afternoon to everyone, I have the following question:

What does the arithmetico-geometric series:

$$S = \sum^{\infty}_{n=1} ne^{-nrt}$$

Converge to?

($r > 1$, $t > 1$)

I tried to break it up in the series of the even number and the series of the odd numbers multipliers:
\begin{align*}
S &= e^{-rt} + 2e^{-2rt} + ... \\
\\
S &= S_1 + S_2 \\
\\
S_1 &= (3-2)e^{-rt} + (3-0)e^{-3rt} + (3+2)e^{-5rt} + ... \\
\\
    &= \frac{3e^{-rt}}{1-e^{-2rt}} -2e^{-rt} +\frac{2e^{-5rt}}{1-2e^{-2rt}} \\
\\
S_2 &= 2e^{-2rt} + 4e^{-4rt}+... \\
\\
S_2 &= \frac{2e^{-2rt}}{1-2e^{-2rt}}
\end{align*}

Hence,

$$ S =\frac{2(e^{-5rt}+e^{-2rt})}{1-2e^{-2rt}} + \frac{3e^{-rt}}{1-e^{-rt}}-2e^{-rt} $$

But honestly, I'm not sure at all... Plus I bet $1\$$ that I made some computational mistake...

Ok, now I see it. I've never integrated / differentiated a series before...

\begin{equation*}
S = \sum^{\infty}_{n=1} ne^{nrt} \\
\end{equation*}

Let $e^{-rt}=x$.

\begin{align*}
S &= \sum^{\infty}_{n=1} nx^n\\
\\
  &= x\sum^{\infty}_{n=1} nx^{n-1} \\
\\
  &= x\sum^{\infty}_{n=1} \frac{\partial S_1}{\partial x}
\end{align*}

It follows that it must be:

\begin{align*}
 S_1 &= \sum^{\infty}_{n=1} x^n
     &= \frac{x}{1-x}
\end{align*}

Hence,

\begin{equation*}
\frac{\partial S_1}{\partial x} = \frac{1}{(1-x)^2}
\end{equation*}

Therefore,

\begin{align*}
S &= \frac{x}{(1-x)^2} \\
\\
  &= \frac{e^{-rt}}{(1-e^{-rt})^2}
\end{align*}

答:
You can see that
$$
\sum_{n=1}^\infty nx^n = \sum_{n=1}^{\infty}\sum_{k=1}^n x^n = \sum_{k=1}^\infty\sum_{n=k}^\infty x^n = \sum_{k=1}^\infty \frac{x^k}{1-x} = \frac{x}{(1-x)^2}.
$$
Now it is just a matter of replacing $x$ with $e^{-rt}$.


问:
Minimum number of relatively isolated nodes in a graph
Let G be a simple graph with $2n$ nodes, with each node having $m$ edges ($m$, $n$ are both unfixed integers).
Prove that there are at least $$k=\left\lceil \frac{2n-1}{m}\right \rceil$$
nodes that are not connected to each other (there is no edge between any of the two nodes).

I found this in a discrete math book  and I have no idea how to solve it. I would seriously appreciate some help.

答:
Hint: Use Brook's theorem.

Solution:

If $G$ is complete, then $k=1$ and the statement is trivial. Otherwise, by Brook's theorem the vertices of $G$ can be colored by $m$ colors such that each color class forms an independent set. One of these color classes must have size at least $\lceil2n/m\rceil \geq k$, as required.


问:
Companion Matrix of first degree monomial $x$
What is the Companion matrix of $x$? Is it $[0]$ or $[1]$?

答: Using the definition that the characteristic polynomial of the companion matrix is equal to the original polynomial, you have $C(p) = [0]$ if $p(x)=x$.


问:
Showing $(0,0,0)$ is not a saddle point of $f(x,y,z)=x^2+y^2-z^3$
Given the function
$$f(x,y,z)=x^2+y^2-z^3$$
I found out $(0,0,0)$ is its critical point. The Hessian at this point is equal to zero so it's inconclusive but it's very easy tho show that $(0,0,0)$ is not the maximum or the minimum of $f$.

However, according to Wolfram Alpha, $f$ has no saddle points. So, how am I supposed to show this?

答: Wolfram Alpha is wrong. Consider $f(0,0,z)$ to see that $(0,0,0)$ is a saddle point.


问:
Can first order logic be defined on the domain of existing and non-existing objects?
Within the First Order Logic we can make statements like this:

$\forall x : [[A(x) \wedge B(x)] \lor [\lnot D(x)]]$

Which, in plain language, means that for all $x$ the specified condition is true. If we think of $x$ as "objects", then we can say that for any "existing" object the considered condition returns true. This also means, that for any "imaginary" object (which can also be "existing" or "not-existing") for which the condition return false, we can say that this object does not exist. In other words, if object exists, the condition returns true, and if condition returns false, then the object does not exist.

Now I wonder if we can use the language of the First Order Logic to formulate the following statement (for some given "condition"): If, for an "imaginary" object, the condition returns true,  then this object is guarantied to exist. This also would mean that, if an object does not exist, the condition should always return false.

To make it more clear, we can formulate it in terms of sets.

In the first example (that I have expressed using FOL), the defined set (let's call it $C$) contains the set of real / existing objects (let's call it $R$): $R \subseteq C$. As a consequence, all the existing / real objects satisfy the condition (belong to the set $C$) and, if condition is not satisfied, then an object does not exist (does not belong to $R$).

In the second example, I want to use the language of FOL to describe a situation in which the set defined by the condition ($C$) is a subset of the real objects ($R$): $C \subseteq R$. As I have already mentioned, it would mean that any object from the set $C$ is guarantied to exist (belong to the set $R$) and if an object does not exist than it also does not belong to $C$.

The existential operator ($\exists$) unfortunately does not solve my problem since it states that overlap between the set $R$ and set $C$ is not equal to zero (which is not exactly the same as what I want to say).

答:
In first-order logic proper, this can be handled by tweaking the language a bit.

We add to the language we're considering a new distinguished unary predicate symbol $Real$. We think of this as describing the collection of objects which exist.

As usual, existential and universal quantifiers range over the whole domain, but we can bound existential quantifiers to $Real$ if we so choose as "$\exists x(Real(x)\wedge ...)$ and "$\forall x\in Real(...)$," respectively. So for example $$\mbox{Everything which exists has property $P$}$$ would be expressed as $$\forall x(Real(x)\rightarrow P(x))$$ and $$\mbox{Everything which has property $P$ exists}$$ would be expressed as $$\forall x(P(x)\rightarrow Real(x)).$$

The point is that all your distinction between "existent"/"nonexistent" objects is really doing is partitioning a broad universe (of "all objects" in some sense) into two pieces, and we can forget the fancy language of existence and just talk about such a partition.

Alternatively, we could replace first-order logic with a close relative more suited, at least philosophically, to the idea of nonexistent objects. Free logic is the standard candidate in this context. But per the above, we can translate appropriately between free logic and standard first-order logic; the distinction is one of form rather than essential expressivity.


问:
How do I find the lengths of the two legs of a right triangle if one leg is half the size of the other and I'm given the hypotenuse?
I've tried squaring the hypotenuse and then dividing up that number into 3 parts. I then give 2 of those parts to the bigger leg and one to the smaller leg. Afterward, I find the square root of each part of the hypotenuse squared. Whenever I do this, I end up with two numbers, but neither of them are half of the other.

(P.S. I'm sorry if this question is confusing to read.) (I'm not sure if I put down the right tag either. This is my first time on this website.)

答:
The proportion of the sides should be  $(1,2, \sqrt5)$. Even if not required after seeing $\sqrt 5$ was curious to find how this right triangle whose sides can be  related to the Golden Ratio $ \phi = \dfrac{\sqrt5 +1}{2}$ using Pythagorean triples... even if not asked for strictly in the question.

In the Pythagorean triplet let

$$(m^2+n^2)^2=(2mn)^2+(m^2-n^2)^2,\; \text{when} \;2mn > m^2-n^2\tag1 $$

$$2mn= 2(m^2-n^2),\; m^2-mn-n^2 =0 \tag 2 $$

Solving the quadratic,

$$ m = n \dfrac{1\pm \sqrt5}{2} = n \phi  \tag3$$

We should discard negative sign of a line segment.

Plugging  into first part of(1)

$$ n^2 \left((\phi^2+1),(\phi^2-1),2\phi \right ) \tag4$$

By the Golden Ratio property Equation (2) has the form $\; \phi^2= \phi+1$ we have

$$ n^2 \left((\phi+2), \phi, 2\phi \right ) \tag5$$

and wlog the proportion of three sides turn out related to $\phi$ as expected.

$$ \left(\phi+2, \phi, 2\phi \right ) \tag6$$

which are essentially in the proportion

$$ (\sqrt 5, 1,2). \tag7 $$


问:
Series of characteristic polynomials
Consider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $-1$, e.g.
$$
M_4= \begin{pmatrix}
               2 & -1 & 0 & 0 \\
               -1 & 2 & -1 & 0\\
               0 & -1 & 2 & -1\\
               0 & 0 & -1 & 2\\
              \end{pmatrix}
$$

I've found out that the characteristic polynomials are
$$
\begin{cases}
P_1(x)=2-x\\
P_2(x)=(2-x)^2-1\\
P_n(x) = (2-x)P_{n-1}(x)-P_{n-2}(x)
\end{cases}
$$

Or with a variable change
$$
\begin{cases}
Q_1(y)=y\\
Q_2(y)=y^2-1\\
Q_n(y) = y Q_{n-1}(y)-Q_{n-2}(y)
\end{cases}
$$

Looking at the first 8 $P_n$

I see that all eigenvalues are real (as for any symmetric matrix), they are between 0 and 4.

How can I prove that all eigenvalues are between 0 and 4?

Are these polynomials known (have a name)?

How can I prove that the polynomial are sandwitched between
$$
\frac{1}{x}+\frac{1}{4-x}\quad\text{and}\quad
-\frac{1}{x}-\frac{1}{4-x}
$$

答:
From a very attenuated literature search, I see that the Lucas polynomials of the second kind obey the recursion
$$ L_{k+1}(x) = x \ L_k(x) - L_{k-1}(x), \quad  L_1(x) = 1 \ , L_2(x)=x$$
This matches the recursion for $Q_n,$ with an index shift of 1.  The one reference I've seen (I don't have access to journal papers from home) states that this polynomial set obeys the relation

$$L_k(2 \cos(\theta) ) = 2 \cos( k \ \theta) $$

This functional relationship would explain the location of the zeros, and likely the envelope property as well.  I suggest that the proposer start his research with 'Lucas polynomials of the second kind.'

Title: Series of characteristic polynomials


问:
Prove that if $\{ f_n\}$ converges to $f$ pointwise on $(0,1)$, then $\{ f_n \}$ converges to $f$ uniformly on $(0,1)$
Let $\{ f_n \}$ be sequence of differentiable functions on the open interval $(0,1)$ such that for any $x \in (0,1)$ and any positive integers $n$

$$|f'_n(x)| \le 1.$$
Prove that if $\{ f_n\}$ converges to $f$ pointwise on $(0,1)$, then $\{ f_n \}$ converges to $f$ uniformly on $(0,1)$.

How can I solve this?

答:
By the MVT, each $f_n$ is 1-Lipschitz on $(0,1).$ Therefore

(i)... $\lim_{x\to 0^+}f_n(x)$ and $\lim_{x\to 1^-} f_n(x)$ exist, so each $f_n$ can be extended to a continuous function on the domain $[0,1].$

(ii)... And the extension is also $1$-Lipschitz. So we can apply the methods in the link in a comment to the Q by @MartinR.

Proofs:

(i)... By contradiction, suppose $g:(0,1)\to\Bbb R$ is $1$-Lipschitz but $\lim_{x\to 0^+} g(x)$ does not exist. Then there exists $r\in (0,1)$ such that for any $y\in (0,1)$ there exist $x'$ and $x''$ in $(0,y)$ with $|g(x')-g(x'')|>r.$ But if $y=r/2,$ this implies $|g(x')-g(x'')|>r>2|x'-x''|,$ contrary to $g$ being $1$-Lipschitz...  Similarly for $\lim_{x\to 1^-} f_n(x),$

(ii)... The continuous extension of $f_n$ to $[0,1]$ is $1$-Lipschitz because if $0\le x<y\le 1$ then   any $x',y'$ in $(x,y)$ belong to $(0,1)$ so $|f_n(x')-f_n(y')|\le |x'-y'|$ so $$|f_n(x)-f_n(y)|=\lim_{x'\to x^+\land y'\to y^-}|f_n(x')-f_n(y')|\le \lim_{x'\to x^+\land y'\to y^-} |x'-y'|=|x-y|.$$


问:
Computing the reduction of a quotient of a quartic polynomial defining an extension over $\mathbb{Q}_3$
Consider the field $L = \mathbb{Q}_3(\alpha)$ where the minimal polynomial of $\alpha$ over $\mathbb{Q}_3$ is $f = x^4 - 3x^2 + 18$. Over $L$, the polynomial $f$ splits into
$$
f = (x-\alpha)(x+\alpha)(x-\alpha')(x+\alpha')
$$
where $\alpha' = \frac{(2\alpha^2-3)u}{\alpha}$ and $u = \sqrt{\frac{2}{-7}}$ (cf. this post, credits go to user KCd).

Question What is $\alpha/\alpha'$ modulo $\pi_L$ (a uniformizer in $L$)?

I know that result is only unique up to sign but I hope that does not cause any further difficulties.

I computed via Newton polygons that the valuations of $\alpha$ and $\alpha'$ are both $1/2$ (here, I assume $v(3) = 1$), so the reduction of $\alpha/\alpha'$ is non-zero.

I tried to apply a technique similar to an answer from this post:

I considered the polynomial
$$
\frac{f(\alpha x)}{\alpha^4} = x^4 - \frac{3}{\alpha^2} x^2 + \frac{18}{\alpha^4}
$$
whose roots are $\pm 1$ and $\pm \frac{\alpha'}{\alpha}$.

This means that the polynomial
$$
\frac{x^4 - \frac{3}{\alpha^2} x^2 + \frac{18}{\alpha^4}}{x^2-1} = x^2 + \left(1-\frac{3}{\alpha^2}\right)
$$
has the roots $\pm \frac{\alpha'}{\alpha}$.

Now I would like to compute the reduction of this polynomial. However, I am not sure that the reduction of $3/\alpha^2$ must be in this case. This seems to be a similar problem but it seems to be as easy an in the last mentioned post since we only have the second power of $\alpha$ (instead of the fourth power of $\alpha$) here.

答:
Let $\beta=\frac{\alpha^2}{3}$. It follows that $\beta^2-\beta+2=0$, hence $\beta=\frac{1 \pm \sqrt{-7}}{2}$.

It follows that $(\frac{\alpha}{\alpha’})^2=\frac{-3+\sqrt{-7}}{4}$, if we choose $\beta=1+\sqrt{-7}$.

Now, note that in $\mathbb{F}_9$, $-7$ is actually a fourth power, as $(-7)^{8/4}=49=1$.

Thus the reduction of $\frac{\alpha}{\alpha’}$ in the residue field is $\frac{(-7)^{1/4}}{2}=-(-7)^{1/4}$.

If we want a more implicit description, we can check that by changing signs and considering the inverse ratio, we find that $\alpha/\alpha’$ can be any generator of $\mathbb{F}_9^{\times}$.


问:
Find the integer value of a, b such that $\lim_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.
Find the integer value of $a$, $b$ such that  $\lim\limits_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.

My work:

I multiplied with $\frac{ax+\sqrt{x}+b}{ax+\sqrt{x}+b}$, and obtained:

$=\lim\limits_{x \rightarrow 4} \frac{(ax+b)^2-x}{(x-4)(ax+b+\sqrt{x})}$

$=\lim\limits_{x \rightarrow 4} \frac{(a^2x^2+2abx+b^2-x)}{(x-4)(ax+b+\sqrt{x})}$

I factored $(x-4)$ from the numerator:

$(x-4)(a^2x+2ab-1+4a^2)+(b^2+8ab-4+16a^2)$

We have the first equation

$(b^2+8ab-4+16a^2)=0$

$=\lim\limits_{x \rightarrow 4} \frac{(x-4)(a^2x+2ab-1+4a^2)}{(x-4)(ax+b+\sqrt{x})}$
$=\frac{4a^2+2ab-1+4a^2}{4a+b+2}=\frac{3}{4}$

So, the second equation is
$16a^2+8ab-4+16a^2=12a+3b+6$

$32a^2-12a+8ab-3b-10=0$

But I can't find the integer value of $a$, $b$.

答:
You were almost there!
$$
\begin{cases}
b^2+8ab-4+16a^2=0\\
32a^2-12a+8ab-3b-10=0\\
\end{cases}
$$
gives two solutions$^*$

$a= -\frac{1}{4},b= -1$ discarded because there are fractions

$a= 1, b = -2$, the solution you were looking for.

$^*$

from the first equation $(b+4a)^2=4\to b+4a=\pm 2\to b=-4a\pm 2$

let $b=-4a+2$ and plug in the second equation

$32a^2-12a+8a(-4a+2)-3(-4a+2)-10=0\to 16 (a-1)=0\to a=1$

and $b = -2$

the other solution $b=-4a-2$ leads to the fractions.


问:
Compute $\oint_{|z|=4} \frac{e^{1 /(z-1)}}{z-2} d z$
Here is an integral.

$\oint_{|z|=4} \frac{e^{1 /(z-1)}}{z-2} d z$

Obviously, the function is not analytic in $D(0,4)$, thus, we should apply residue theorem

$\oint_{c} f(z) d z=2 \pi i \sum_{j=1}^{n} b_{1 j}$

where $b_{1j}$ is the first coefficient in Laurent series at the point of singularity.

So,

at the point $z_1 = 2$ we have a pole of the first order.

$Res_{z_1} = e^1 $

at the point $z_2 = 1$ we have an essential singularity. Compute Laurent series of the function in the point $z=1$. WolframAlpha:

$\frac{e^{1 /(z-1)}}{z-2} = \sqrt[x-1]{e}\left(-1-(z-1)-(z-1)^{2}-(z-1)^{3}+O\left((z-1)^{4}\right)\right)$

Thus, $Res_{z_2} =  -1$

And we get:

$\oint_{c} f(z) d z=2 \pi i (e - 1)$

It does not make sense, for a correct answer is $2i$

What is wrong? Tell me, please.

答: It is clear that$$\operatorname{res}_{z=2}\left(\frac{e^{1/(z-1)}}{z-2}\right)=e^{1/(2-1)}=e.$$On the other hand, near $1$ we have$$e^{1/(z-1)}=1+\frac1{z-1}+\frac1{2!(z-1)^2}+\frac1{3!(z-1)^3}+\cdots\tag1$$and$$\frac1{z-2}=-\frac1{1-(z-1)}=-1-(z-1)-(z-1)^2-(z-1)^3+\cdots\tag2$$If we compute the Cauchy product of the series $(1)$ and $(2)$, then the coefficient of $(z-1)^{-1}$ shall be$$-1-\frac1{2!}-\frac1{3!}-\cdots=1-e.$$In other words,$$\operatorname{res}_{z=1}\left(\frac{e^{1/(z-1)}}{z-2}\right)=1-e$$and therefore$$\oint_{|z|=4}\frac{e^{1/(z-1)}}{z-2}\,\mathrm dz=2\pi i\bigl(e+(1-e)\bigr)=2\pi i.$$


问:
What does infinity as an upper/lower limit mean in a definite integral? Eg, $\int_0^\infty e^{-x} dx$
I was finding
$$\int_0^\infty e^{-x} dx$$
So the integral becomes
$$\left.(-e)^{-x}\right\rvert_0^\infty$$

Now here's the confusing thing for me.  What exactly does infinity mean in upper/lower limit?

How do I actually calculate definite integrals when one of the limit is infinity?

答:
Just like

$$\sum_{k=0}^\infty t_k$$ is a shorthand for

$$\lim_{n\to\infty}\sum_{k=0}^n t_k\ ,$$

$$\int_0^\infty f(x)\,dx$$ is short for

$$\lim_{u\to\infty}\int_0^u f(x)\,dx.$$


问:
How to compute $\int_0^\infty \frac{\tanh\left(\pi x\right)}{x\left(1+x^2\right)} \, \mathrm{d}x$?
How do I compute the following integral?

$$\int_0^\infty \frac{\tanh\left(\pi x\right)}{x\left(1+x^2\right)} \, \mathrm{d}x$$

I tried some basic substitutions but they only make it more complicated. WolframAlpha says the answer is $2$ but I have no clue how to get there.

答:
Another thought is to try the following substitution:
$$I(t)=\int_0^\infty\frac{\tanh(tx)}{x(x^2+1)}dx$$
$u=-x\Rightarrow du=-dx$ and so:
$$I(t)=\int_0^{-\infty}\frac{\tanh(-tu)}{-u(u^2+1)}(-du)=\int_{-\infty}^0\frac{\tanh(tx)}{x(x^2+1)}dx$$
Which means that we can write:
$$I(t)=\frac12\int_{-\infty}^\infty\frac{\tanh(tx)}{x(x^2+1)}dx$$
$$I'(t)=\frac12\int_{-\infty}^\infty\frac{\tanh(tx)}{x^2+1}$$
$$I''(t)=\frac12\int_{-\infty}^\infty\frac{x\tanh(tx)}{x^2+1}dx$$
$$I'''(t)=\frac12\int_{-\infty}^\infty\frac{x^2\tanh(tx)}{x^2+1}=\frac12\int_{-\infty}^\infty\tanh(tx)dx-\frac12\int_{-\infty}^\infty\frac{\tanh(tx)}{x^2+1}dx$$

$$I'''(t)=-I'(t)$$
$$I'''+I'=0$$
now we just need to solve this ode:
$$I=Ae^{\lambda t},I'=A\lambda e^{\lambda t},I'''=A\lambda^3e^{\lambda t}$$
$$A\lambda e^{\lambda t}+A\lambda^3e^{\lambda t}=0$$
$$\lambda+\lambda^3=0,\,\lambda(\lambda^2+1)=0$$
$$\lambda=0,\pm i$$
$$I=A+Be^{it}+Ce^{-it}\Rightarrow I=c_1+c_2\cos(t)+c_3\sin(t)$$
now we just need to work out what these constants are. This part is proving difficult, but it seems obvious that $I'(t)=0$ everywhere since it is an odd function. we also know that $I(0)=0$ and that $I''(t)$ is going to be divergent everywhere other than $I''(0)=0$. This does not seem to fit well so maybe something is wrong but we can say:
$$I'(t)=0\Rightarrow I(t)=C$$
but $I(0)=0$ and so $I(t)=0$, which is clearly untrue. I've spent a while looking at this working out why it is wrong and I believe it is because $I''$ is not convergent.

This means that one of the main issues here is:
$$\lim_{t\to 0}\int_{-\infty}^\infty\frac{x\tanh(tx)}{x^2+1}dx$$


答:
The Chebyshev polynomials of the second kind satisfy the recurrence relation
$$
\begin{cases}
U_0(y) = 1 \\
U_1(y)=x\\
U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y)
\end{cases}
$$
so that $Q_n(y) = U_n(y/2)$ and $P_n(x) = U_n(1-x/2)$.

The zeros of $U_n$ are
$$
 y_k = \cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n
$$
in the range $(-1, 1)$, so that
$$
 x_k = 2 - 2\cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n
$$
are the zeros of $P_n$ in the range $(0, 4)$.

Also for $|x| < 1$
$$
U_n(x) = \frac{\sin((n+1)\arccos(x))}{\sqrt{1-x^2}}
$$
which implies
$$
|U_n(x)| \le \frac{1}{\sqrt{1-x^2}}
$$
and therefore
$$
| P_n(x)| \le \frac{2}{\sqrt{x(4-x)}} \le \frac 1x + \frac{1}{4-x}
$$
for $0 < x < 4$, the last estimate follows from the inequality between harmonic and geometric mean.

Title: Series of characteristic polynomials

Body:
Consider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $-1$, e.g.
$$
M_4= \begin{pmatrix}
               2 & -1 & 0 & 0 \\
               -1 & 2 & -1 & 0\\
               0 & -1 & 2 & -1\\
               0 & 0 & -1 & 2\\
              \end{pmatrix}
$$

I've found out that the characteristic polynomials are
$$
\begin{cases}
P_1(x)=2-x\\
P_2(x)=(2-x)^2-1\\
P_n(x) = (2-x)P_{n-1}(x)-P_{n-2}(x)
\end{cases}
$$

Or with a variable change
$$
\begin{cases}
Q_1(y)=y\\
Q_2(y)=y^2-1\\
Q_n(y) = y Q_{n-1}(y)-Q_{n-2}(y)
\end{cases}
$$

Looking at the first 8 $P_n$

I see that all eigenvalues are real (as for any symmetric matrix), they are between 0 and 4.

How can I prove that all eigenvalues are between 0 and 4?

Are these polynomials known (have a name)?

How can I prove that the polynomial are sandwitched between
$$
\frac{1}{x}+\frac{1}{4-x}\quad\text{and}\quad
-\frac{1}{x}-\frac{1}{4-x}
$$


问:
Rationalizing Denominators with Radicals
How do you rationalize this fraction:

$$\frac{1}{\sqrt{n+1}}$$

答:
well... you should agree that $\frac{\sqrt{n+1}}{\sqrt{n+1}}=1$, given $n\neq-1$; thus
$$\frac{1}{\sqrt{n+1}}=\frac{1}{\sqrt{n+1}}\frac{\sqrt{n+1}}{\sqrt{n+1}}=\frac{\sqrt{n+1}}{\sqrt{n+1}\sqrt{n+1}}=\frac{\sqrt{n+1}}{n+1}\,.$$
Is this what you wanted?</p>

OwnerDisplayName: Joseph


问:
$[0,\omega_1]$ is not first countable.
I was looking for examples of spaces that are not first countable. On wikipedia they give the example of $[0,\omega_1]$. I think I understand that $\omega_1$ is  a limit point of $[0,\omega_1)$ since if $(\alpha,\beta)$ is an open containing $\omega_1$ it contains an ordinal smaller than $\omega_1$ and thus this ordinal is in $[0,\omega_1)$. Now why is there no sequence of $[0,\omega_1)$ converging to $\omega_1$. It is not entirely clear to me how to prove this. Also I don't really understand why this implies that $\omega_1$ doesn't have a countable basis.

答:
If there exist a sequence $x_n\in [0,\omega_1)$ converging to $\omega_1$, then $\omega_1$ would be countable, which is not.

To see this you can use the model where ordinals are defined as

$0=\emptyset$

$1=\{\emptyset\}=\{0\}$

$2=\{\emptyset,\{\emptyset\}\}=\{0,1\}$

...

$\alpha=\{\beta: \beta<\alpha\}$ the set of ordinals smaller than $\alpha$

Then, if $x_n<\omega_1$ and $x_n\to \omega_1$ then $\omega_1=\cup_n x_n$ is coutable union of countable set (because $\omega_1$ is the first uncountable) and so it would be countable.

Title: $[0,\omega_1]$ is not first countable.


问:
Distributions for vector-valued functions (Question from Navier-Stokes equation)
I am currently studuying the Navier-Stokes equation, more precisely, The Leray's Theorem (1933) that gives existence of weak solutions. In order to study it, our teacher gave us a problem to solve and in it there is a question about the Leray's Projector :

$$ P: u \in L^2(R^d) \mapsto P(u) \in K$$

where $ K = \{ u \in L^2(R^d) : \operatorname{div}(u) = 0 \text{ in } \mathcal{D}'(R^d)\}$

My problem is that, even if I think I have understood the Distribution Theory for scalar valued-functions, I don't know of we make the similar generalization from vector-valued functions. Can someone try to explain me that ? And how we then define the Divergence and other differential Operators ?

Thank you very much.

答:
A $\mathbb{R}^d$-valued distribution $u$ can be treated as $d$ scalar-valued distributions, $u_1, \ldots, u_d \in \mathcal{D}'(\mathbb{R}^d)$.

The divergence is then simply $\operatorname{div}u = \nabla\cdot u = \partial_1 u_1 + \cdots + \partial_d u_d \in \mathcal{D}'$.


问:
What is the dimension of the image of T?
I have a question
Let $T : \mathbb{R}^{28} →  \mathbb{R}^{31}$ be a linear transformation whose kernel has dimension 9. What is the dimension of the image of $T$?

答:
We use the dimension formula https://en.m.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces so we get $$\dim(\ker(T))+\dim(Im(T))=\dim(R^{28})=28$$
So plugging in your assumption that $\dim(\ker(T))=9$ we simply solve the equation to obtain the dimension of the image.


问:
I need help with Question 13, this is regarding the Loan Amortization Formula

I have been working on 13, it is especially frustrating since I am so close to being done, but if there is anyone who could explain how to derive this, it would mean so much! I have retraced my work using Questions 4-6 and 11-12, but still no luck on getting the variable substitutions to work right!

答:
First of all you have this equation:

$$B\cdot (1+i)^N=m\cdot \frac{(1+i)^N-1}{i} \qquad (*)$$

Explanation: Assume you borrow $B$ monetary units with an interest rate of $i\cdot 100$ percent per month. Then after N month the loan ($B$) has an value of $B\cdot (1+i)^N$. This is the future value of $B$. On the right hand side you have N monthly repayments of the loan. Every payment has a an amount of $m$. Every payment which is made at month $t$ is compounded so that it represents the future value as well. So you have an geometric series on the right hand side. Let $q=1+i$ then the right hand side is

$$m\cdot \sum_{x=0}^{N-1}q^x=m\cdot \left(q^0+q^1+q^2+\ldots +q^{N-1} \right)$$

The closed term for a partial sum of a geometric series is
$$\sum_{x=0}^{N-1} q^x= \frac{q^N-1}{q-1}$$

This is what you have with the definition of q: $\frac{(1+i)^N-1}{(1+i)-1}=\frac{(1+i)^N-1}{i}$

Hopefully the equation $(*)$ ist clear now. Now we can start to solve the equation for $m$. Multiplying both sides by $i$.

$$B\cdot (1+i)^N\cdot i=m\cdot \left((1+i)^N-1\right) $$

Dividing the equation by $\left((1+i)^N-1\right)$

$$B\cdot \frac{(1+i)^N\cdot i}{(1+i)^N-1}=m $$

Next we expand the fraction on the right by $(1+i)^{-N}$. Every summand has to be multiplied by that factor. Here the exponent rules are helpful. For instance, $a^b\cdot a^{-b}=a^{b-b}=a^0=1$

$$B\cdot \frac{(1+i)^N\cdot (1+i)^{-N}\cdot i}{\underbrace{(1+i)^N\cdot (1+i)^{-N}}_{=1}-1\cdot (1+i)^{-N}}=m $$

$$\boxed{B\cdot \frac{ i}{1- (1+i)^{-N}}=m} $$


问:
Finite algorithm for guessing positive integer
Question: $A$ and $B$ play a game. $B$ chooses a positive integer in mind. $A$ can ask questions of the form "Is your number $x$?" for any $x$ that $A$ wants. Does $A$ have a finite algorithm to correctly guess $B$'s number?

Answer 1: Yes. $A$ sequentially asks the question for $1,2,3,\dots$, so no matter which number $y$ that $B$ chooses, $A$'s algorithm will finish in $y$ steps.

Answer 2: No. No matter what $A$ asks, $B$'s strategy is to always answer no. So $A$'s algorithm cannot terminate.

Obviously, one of these answers must be wrong. But which one?

答:
Answer 1 is correct, and Answer 2 is incorrect.

According to the specification of this problem, $B$ is to choose a number and stick to that. So, $A$ will eventually hit on that number using the $1,2,3,...$ strategy.

Of course, $B$ can lie and say "no" even if $A$ guesses the number $B$ has in mind. But that does not mean that $A$ did not pick correctly. The question is not whether $A$ has a strategy that will, after some finite guesses, tell $A$ what the number is. The question is simply:

Does A have a finite algorithm to correctly guess B's number?

So that has nothing to do with $A$ knowledge, but simply about the fact whether $A$ spoke out the number that $B$ picked.

Here, by the way, is another interesting scenario. Suppose that $A$'s strategy is predictable, as the $1,2,3,..$ strategy is. With that, $B$ can anticipate what $A$ is going to ask and always pick a new number that $A$ is not going to ask, and hasn't asked before either.  In this scenario, $B$ will never have to lie and can always confidently say "No" forever. However, now we are describing a game that is different from the original (that is, if $B$ does this, then $B$ is cheating). For the original game, where $B$ chooses a number once, and then has to stick with it, there is a finite winning strategy for $A$

In sum, as long as we stick to the rules of the game, then the answer to the question:

Does A have a finite algorithm to correctly guess B's number?

is yes. And the reasoning given by Answer 1 is exactly correct.