问:
Can any positive real be approximated as $2^m/3^n$ with $(m,n)$ large enough?
Conjecture.

There exist positive integers $(m,n)$ large enough, such that for any positive real number $r$ and a given error $\epsilon$ :
$$
   \left| r - \frac{2^m}{3^n} \right| < \epsilon
$$
There is numerical evidence for this conjecture. I have tried $r = \sqrt{2}$ and $\epsilon = 10^{-3}$.

Below is a little Delphi Pascal program with accompanying output.

But .. can somebody prove the conjecture?

```
program apart;<br>
procedure test(r : double; eps : double);
var
  a : double;
  m,n : integer;
begin
  a := 1;
  m := 0; n := 0;
  while true do
  begin
    if a < r then
    begin
      m := m + 1;
      a := a * 2;
    end else begin
      n := n + 1;
      a := a / 3;
    end;
    if abs(r-a) < eps then Break;
  end;
  Writeln(r,' = 2^',m,'/3^',n,' =',a);
end;<br>
begin
  test(sqrt(2),1.E-3);
end.
```

Output:

```
 1.41421356237310E+0000 = 2^243/3^153 = 1.41493657935359E+0000
```

UPDATE.

The answer by lhf does look like a very concise proof. But for me - as an retired physicist by education - it's a bit beyond comprehension.

Furthermore, it leaves a few issues untouched. One might ask for example whether there are estimates for $m$ and $n$ when $r$ and $\epsilon$ are given.

Note. The question can also be formulated as:
Can any positive real be approximated as $3^m/2^n$ with $(m,n)$ large enough? Which is the same as allowing negative integers with the original formulation. In this form, it shows some resemblance to the (in)famous Collatz problem.

EDIT.

As suggested by the answers, an approach with logarithms could be more effective:

```
program anders;<br>
procedure proef(r : double; eps : double);
var
  a,l2,l3,lr : double;
  m,n : integer;
begin
  l2 := ln(2); l3 := ln(3);
  lr := ln(r); a := 0;
  m := 0; n := 0;
  while true do
  begin
    a := m*l2 - n*l3 - lr;
    if abs(a) < eps then Break;
    if a < 0 then m := m + 1 else n := n + 1;
  end;
  Writeln(r,' = 2^',m,'/3^',n,' =',exp(a)*r);
end;<br>
begin
  proef(sqrt(2),1.E-3);
  proef(sqrt(2),1.E-9);
end.
```

Output:

```
 1.41421356237310E+0000 = 2^243/3^153 = 1.41493657935356E+0000
 1.41421356237310E+0000 = 2^911485507/3^575083326 = 1.41421356125035E+0000
```

The first line in the output is almost identical to the result obtained previously .

The last line in the output shows that the latter approach indeed is more effective.
The error plays the same role in both approaches. Oh well, almost.
Let's take a look at the places where the 'Break's are. First program:
$$
\left| r - \frac{2^m}{3^n} \right| < \epsilon
$$
Second program:
$$
-\epsilon < m\ln(2) - n\ln(3) - \ln(r) < +\epsilon \\
\ln(1-\epsilon) < \ln\left(\frac{2^m/3^n}{r}\right) < \ln(1+\epsilon) \\
-\epsilon < \frac{2^m/3^n}{r} - 1 < +\epsilon \\
\left| r - \frac{2^m}{3^n} \right| < \epsilon.r
$$
So $\epsilon$ in the first program is an absolute error,
while $\epsilon$ in the second program is a relative error.

Continuing story at:

Can the Stern-Brocot tree be employed for better convergence of $2^m/3^n$?

答:
Yes, there are always solutions $(m, n)$ for any positive real $r$ and $\epsilon$ for
$$\left| r - \frac{2^m}{3^n} \right| < \epsilon$$
And there's a much more efficient way to find those solutions than stepping through $m$ and $n$ values one by one.

We have
$$r \approx 2^m/3^n$$
Taking logarithms,
$$\log r \approx m\log 2 - n\log 3$$
$$\log r/\log 2\approx m - n\log 3 / \log 2$$
i.e.,
$$\log_2 r\approx m - n\log_2 3$$

[Incidentally,
$$1 = \frac m{\log_2r}-\frac n{\log_3r}$$
which is a line in the $(m,n)$ plane with $m$ intercept $\log_2r$ and $n$ intercept $-\log_3r$. We want to find when that line passes close to integer $(m, n)$ lattice points].

We can find rational approximations to both those base 2 logarithms to any desired precision. However, to satisfy that equation with integer $m$ and $n$, the denominators of our approximations must be commensurate.

Let
$$\log_2 r = f \approx s/t$$
and
$$\log_2 3 \approx p/q$$
with the fractions being in lowest terms, i.e., $\gcd(s,t)=gcd(p,q)=1$.

Then
$$\frac st = m - n \frac pq$$
$$sq = (qm - pn)t$$
Thus $t|sq$. But $s$ & $t$ are coprime, hence $t|q$.

Let $q=tk$.
$$f \approx \frac st = \frac{sk}{tk}=\frac{sk}{q}=\frac dq$$
for some integer $d$.

So, for a given approximation $\frac pq$ to $\log_2 3$, the best rational approximations to $f$ with commensurate denominators are
$\frac{d_0}q$ and $\frac{d_1}q$, where $d_0=\lfloor fq\rfloor$ and $d_1=\lceil fq\rceil$. That is,
$$\frac{d_0}q \le f \le \frac{d_1}q$$
If $f$ is rational (eg, when $r=\sqrt 2$), then $d_0$ and $d_1$ may be equal.

So for a given $p$ & $q$ we just need to find integers $m$ & $n$ that solve our revised equation
$$\frac dq = m - n \frac pq$$
$$d=qm-pn$$
for both $d_0$ and $d_1$. There are solutions for any integer $d$ because $p$ & $q$ are coprime. And those solutions can be found using the extended Euclidean algorithm.

But we also need to find suitable $p$ & $q$. That can be done using the convergents of the continued fraction expansion of $\log_2 3$. The standard algorithm for computing a continued fraction is closely related to the extended Euclidean algorithm, and as that Wikipedia article explains (in Theorem 3), if the $n$th convergent of a continued fraction is $\frac{h_n}{k_n}$ then
$$k_nh_{n-1} - k_{n-1}h_n = (-1)^n$$
which enables us to find $m$ and $n$ without doing a separate Euclidean algorithm calculation.

The continued fraction convergent $\frac hk$ of a number $x$ gives the best rational approximations to $x$ for any denominator $\le k$. The error is
$$\left|x - \frac hk\right| \le \frac 1{k^2}$$
and it can often be much better. In contrast, the error for an approximation $\frac hk$ with a "random" denominator (i.e., not a continued fraction convergent) is generally around $\frac 1{2k}$.

Unfortunately, because of the need for commensurate denominators in our approximations to the two logarithms, we don't get the full $\frac 1{k^2}$ goodness. But we do generally get better than $\frac 1{k}$.

So to find solutions with better error than a given $\epsilon$, we just need to look at the convergents to $\log_2 3$ with denominators in the neighbourhood of $\frac 1\epsilon$.

Here is some Sage / Python code that performs that task. Sage is a collection of mathematical libraries built on top of the popular Python programming language. It has arbitrary precision arithmetic, and facilities for performing symbolic algebra, but I've (mostly) avoided Sage features in this code (apart from the arbitrary precision arithmetic), to make it easier to port to other languages, if desired; I've also avoided most "Pythonisms", apart from Python's ability to return multiple values from a function.

```<code># Numeric precision. Standard IEEE 754 binary64
# numbers (aka doubles) have 53 bits of precision.
bits = 53

# Limit the length of the continued fraction
depth = 20

def dio(q, p, x, y, d):
    """ Given q, p, x, y: q*x - p*y == 1,
        find the smallest m, n > 0:
        q*m - p*n == d
    """
    m = x * d
    n = y * d
    u = min(m // p, n // q)
    m -= u * p
    n -= u * q
    assert q*m - p*n == d
    return m, n

log2 = log(2).n(bits)
log3 = log(3).n(bits)
def func(m, n):
    """ Calculate 2**m / 3**n """
    # The naive form is too slow for large args,
    # and chews up a lot of RAM because it uses
    # arbitrary precision integer arithmetic.
    # So we use logs instead.
    #return (2**m / 3**n).n(bits)
    return exp(m * log2 - n * log3).n(bits)

def cont_frac(f, depth):
    """ Build lists of the convergents of
        the continued fraction of f
    """
    f = f.n(bits)
    num = [0, 1]
    den = [1, 0]
    for i in range(depth):
        a = floor(f)
        n = num[-2] + a * num[-1]
        d = den[-2] + a * den[-1]
        #print(a, n, d)
        num.append(n)
        den.append(d)
        f -= a
        if f < 1e-10:
            break
        f = 1 / f
    return num, den

num, den = cont_frac(log(3, 2), depth)

@interact
def main(r=sqrt(2), epsilon=1/1000):
    print("r:", r.n(bits))
    f = log(r, 2)
    s = 1
    digits = 2
    for i in range(3, depth+2):
        s = -s
        p = num[i]
        q = den[i]
        x = num[i-1] * s
        y = den[i-1] * s
        assert q*x - p*y == 1
        fq = f * q
        d0 = floor(fq)
        d1 = ceil(fq)
        print(f"\n{i}: {p} / {q}, {d0} {d1}")
        dseq = [d0]
        if d0 < d1:
            dseq = [d0, d1]
        else:
            dseq = [d0]
        for d in dseq:
            m, n = dio(q, p, x, y, d)
            v = func(m, n)
            eps = abs(r - v).n(bits)
            if eps > 0:
                digits = 1 - floor(log(eps, 10))
            print(f"m: {m}, n: {n}")
            print(f"v: {v:.{digits}f}, eps: {eps:.3e}")
            if eps < epsilon:
                return
</code>```

Here's the output of that program, searching for solutions with $\epsilon=10^{-6}$:

```<code>r: 1.41421356237310

3: 2 / 1, 0 1
m: 0, n: 0
v: 1.00, eps: 4.142e-1
m: 1, n: 0
v: 2.00, eps: 5.858e-1

4: 3 / 2, 1 1
m: 2, n: 1
v: 1.333, eps: 8.088e-2

5: 8 / 5, 2 3
m: 2, n: 1
v: 1.333, eps: 8.088e-2
m: 7, n: 4
v: 1.58, eps: 1.660e-1

6: 19 / 12, 6 6
m: 10, n: 6
v: 1.4047, eps: 9.550e-3

7: 65 / 41, 20 21
m: 10, n: 6
v: 1.4047, eps: 9.550e-3
m: 56, n: 35
v: 1.440, eps: 2.603e-2

8: 84 / 53, 26 27
m: 10, n: 6
v: 1.4047, eps: 9.550e-3
m: 75, n: 47
v: 1.4209, eps: 6.645e-3

9: 485 / 306, 153 153
m: 243, n: 153
v: 1.41494, eps: 7.230e-4

10: 1054 / 665, 332 333
m: 812, n: 512
v: 1.41343, eps: 7.844e-4
m: 243, n: 153
v: 1.41494, eps: 7.230e-4

11: 24727 / 15601, 7800 7801
m: 12891, n: 8133
v: 1.414196, eps: 1.800e-5
m: 11837, n: 7468
v: 1.414257, eps: 4.373e-5

12: 50508 / 31867, 15933 15934
m: 12891, n: 8133
v: 1.414196, eps: 1.800e-5
m: 37618, n: 23734
v: 1.4142213, eps: 7.728e-6

13: 125743 / 79335, 39667 39668
m: 88126, n: 55601
v: 1.4142110, eps: 2.546e-6
m: 37618, n: 23734
v: 1.4142213, eps: 7.728e-6

14: 176251 / 111202, 55601 55601
m: 88126, n: 55601
v: 1.4142110, eps: 2.546e-6

15: 301994 / 190537, 95268 95269
m: 88126, n: 55601
v: 1.4142110, eps: 2.546e-6
m: 213869, n: 134936
v: 1.4142162, eps: 2.637e-6

16: 16785921 / 10590737, 5295368 5295369
m: 8241964, n: 5200100
v: 1.414213479, eps: 8.295e-8
</code>```

And here is a live version that you can play with on the SageMath server. My code isn't stored on the server, it's encoded in the URL.

If you get weird behaviour with small $\epsilon$, try increasing the number of the bits global variable (at the top of the file). The default setting of 53 should be ok for $\epsilon > 10^{-8}$ or so. You may also need to increase the depth of the continued fraction.

FWIW, $\log_2 3$ is rather important in the mathematical music theory of equally-tempered scales. The standard 12 tone scale uses the convergent $19/12$.

Body:
Yes, there are always solutions $(m, n)$ for any positive real $r$ and $\epsilon$ for
$$\left| r - \frac{2^m}{3^n} \right| < \epsilon$$
And there's a much more efficient way to find those solutions than stepping through $m$ and $n$ values one by one.

We have
$$r \approx 2^m/3^n$$
Taking logarithms,
$$\log r \approx m\log 2 - n\log 3$$
$$\log r/\log 2\approx m - n\log 3 / \log 2$$
i.e.,
$$\log_2 r\approx m - n\log_2 3$$

[Incidentally,
$$1 = \frac m{\log_2r}-\frac n{\log_3r}$$
which is a line in the $(m,n)$ plane with $m$ intercept $\log_2r$ and $n$ intercept $-\log_3r$. We want to find when that line passes close to integer $(m, n)$ lattice points].

We can find rational approximations to both those base 2 logarithms to any desired precision. However, to satisfy that equation with integer $m$ and $n$, the denominators of our approximations must be commensurate.

Let
$$\log_2 r = f \approx s/t$$
and
$$\log_2 3 \approx p/q$$
with the fractions being in lowest terms, i.e., $\gcd(s,t)=gcd(p,q)=1$.

Then
$$\frac st = m - n \frac pq$$
$$sq = (qm - pn)t$$
Thus $t|sq$. But $s$ & $t$ are coprime, hence $t|q$.

Let $q=tk$.
$$f \approx \frac st = \frac{sk}{tk}=\frac{sk}{q}=\frac dq$$
for some integer $d$.

So, for a given approximation $\frac pq$ to $\log_2 3$, the best rational approximations to $f$ with commensurate denominators are
$\frac{d_0}q$ and $\frac{d_1}q$, where $d_0=\lfloor fq\rfloor$ and $d_1=\lceil fq\rceil$. That is,
$$\frac{d_0}q \le f \le \frac{d_1}q$$
If $f$ is rational (eg, when $r=\sqrt 2$), then $d_0$ and $d_1$ may be equal.

So for a given $p$ & $q$ we just need to find integers $m$ & $n$ that solve our revised equation
$$\frac dq = m - n \frac pq$$
$$d=qm-pn$$
for both $d_0$ and $d_1$. There are solutions for any integer $d$ because $p$ & $q$ are coprime. And those solutions can be found using the extended Euclidean algorithm.

But we also need to find suitable $p$ & $q$. That can be done using the convergents of the continued fraction expansion of $\log_2 3$. The standard algorithm for computing a continued fraction is closely related to the extended Euclidean algorithm, and as that Wikipedia article explains (in Theorem 3), if the $n$th convergent of a continued fraction is $\frac{h_n}{k_n}$ then
$$k_nh_{n-1} - k_{n-1}h_n = (-1)^n$$
which enables us to find $m$ and $n$ without doing a separate Euclidean algorithm calculation.

The continued fraction convergent $\frac hk$ of a number $x$ gives the best rational approximations to $x$ for any denominator $\le k$. The error is
$$\left|x - \frac hk\right| \le \frac 1{k^2}$$
and it can often be much better. In contrast, the error for an approximation $\frac hk$ with a "random" denominator (i.e., not a continued fraction convergent) is generally around $\frac 1{2k}$.

Unfortunately, because of the need for commensurate denominators in our approximations to the two logarithms, we don't get the full $\frac 1{k^2}$ goodness. But we do generally get better than $\frac 1{k}$.

So to find solutions with better error than a given $\epsilon$, we just need to look at the convergents to $\log_2 3$ with denominators in the neighbourhood of $\frac 1\epsilon$.

Here is some Sage / Python code that performs that task. Sage is a collection of mathematical libraries built on top of the popular Python programming language. It has arbitrary precision arithmetic, and facilities for performing symbolic algebra, but I've (mostly) avoided Sage features in this code (apart from the arbitrary precision arithmetic), to make it easier to port to other languages, if desired; I've also avoided most "Pythonisms", apart from Python's ability to return multiple values from a function.

# Numeric precision. Standard IEEE 754 binary64
# numbers (aka doubles) have 53 bits of precision.
bits = 53

# Limit the length of the continued fraction
depth = 20

def dio(q, p, x, y, d):
    """ Given q, p, x, y: q*x - p*y == 1,
        find the smallest m, n > 0:
        q*m - p*n == d
    """
    m = x * d
    n = y * d
    u = min(m // p, n // q)
    m -= u * p
    n -= u * q
    assert q*m - p*n == d
    return m, n

log2 = log(2).n(bits)
log3 = log(3).n(bits)
def func(m, n):
    """ Calculate 2**m / 3**n """
    # The naive form is too slow for large args,
    # and chews up a lot of RAM because it uses
    # arbitrary precision integer arithmetic.
    # So we use logs instead.
    #return (2**m / 3**n).n(bits)
    return exp(m * log2 - n * log3).n(bits)

def cont_frac(f, depth):
    """ Build lists of the convergents of
        the continued fraction of f
    """
    f = f.n(bits)
    num = [0, 1]
    den = [1, 0]
    for i in range(depth):
        a = floor(f)
        n = num[-2] + a * num[-1]
        d = den[-2] + a * den[-1]
        #print(a, n, d)
        num.append(n)
        den.append(d)
        f -= a
        if f < 1e-10:
            break
        f = 1 / f
    return num, den

num, den = cont_frac(log(3, 2), depth)

@interact
def main(r=sqrt(2), epsilon=1/1000):
    print("r:", r.n(bits))
    f = log(r, 2)
    s = 1
    digits = 2
    for i in range(3, depth+2):
        s = -s
        p = num[i]
        q = den[i]
        x = num[i-1] * s
        y = den[i-1] * s
        assert q*x - p*y == 1
        fq = f * q
        d0 = floor(fq)
        d1 = ceil(fq)
        print(f"\n{i}: {p} / {q}, {d0} {d1}")
        dseq = [d0]
        if d0 < d1:
            dseq = [d0, d1]
        else:
            dseq = [d0]
        for d in dseq:
            m, n = dio(q, p, x, y, d)
            v = func(m, n)
            eps = abs(r - v).n(bits)
            if eps > 0:
                digits = 1 - floor(log(eps, 10))
            print(f"m: {m}, n: {n}")
            print(f"v: {v:.{digits}f}, eps: {eps:.3e}")
            if eps < epsilon:
                return

Here's the output of that program, searching for solutions with $\epsilon=10^{-6}$:

r: 1.41421356237310

3: 2 / 1, 0 1
m: 0, n: 0
v: 1.00, eps: 4.142e-1
m: 1, n: 0
v: 2.00, eps: 5.858e-1

4: 3 / 2, 1 1
m: 2, n: 1
v: 1.333, eps: 8.088e-2

5: 8 / 5, 2 3
m: 2, n: 1
v: 1.333, eps: 8.088e-2
m: 7, n: 4
v: 1.58, eps: 1.660e-1

6: 19 / 12, 6 6
m: 10, n: 6
v: 1.4047, eps: 9.550e-3

7: 65 / 41, 20 21
m: 10, n: 6
v: 1.4047, eps: 9.550e-3
m: 56, n: 35
v: 1.440, eps: 2.603e-2

8: 84 / 53, 26 27
m: 10, n: 6
v: 1.4047, eps: 9.550e-3
m: 75, n: 47
v: 1.4209, eps: 6.645e-3

9: 485 / 306, 153 153
m: 243, n: 153
v: 1.41494, eps: 7.230e-4

10: 1054 / 665, 332 333
m: 812, n: 512
v: 1.41343, eps: 7.844e-4
m: 243, n: 153
v: 1.41494, eps: 7.230e-4

11: 24727 / 15601, 7800 7801
m: 12891, n: 8133
v: 1.414196, eps: 1.800e-5
m: 11837, n: 7468
v: 1.414257, eps: 4.373e-5

12: 50508 / 31867, 15933 15934
m: 12891, n: 8133
v: 1.414196, eps: 1.800e-5
m: 37618, n: 23734
v: 1.4142213, eps: 7.728e-6

13: 125743 / 79335, 39667 39668
m: 88126, n: 55601
v: 1.4142110, eps: 2.546e-6
m: 37618, n: 23734
v: 1.4142213, eps: 7.728e-6

14: 176251 / 111202, 55601 55601
m: 88126, n: 55601
v: 1.4142110, eps: 2.546e-6

15: 301994 / 190537, 95268 95269
m: 88126, n: 55601
v: 1.4142110, eps: 2.546e-6
m: 213869, n: 134936
v: 1.4142162, eps: 2.637e-6

16: 16785921 / 10590737, 5295368 5295369
m: 8241964, n: 5200100
v: 1.414213479, eps: 8.295e-8

And here is a <a href="https://sagecell.sagemath.org/?z=eJx1VU2P2zYQvetXDLwXSSt7JTtpAWNdJC0WRYE0h6S37SKgJcomIlESSXltGP7vnaG-qHTjgyQO33CG7z3Sd_C5LbkSKdSKp0KLSq7gq2EyYyqDv56enuDX9-9gLyRTl1_eeXcg23LPlQaffWeQVe2-4DqAIztxeL9BoNFQ5c5qng3tcNLD7E-iFAbMkUPB5cEcCUujtJJGyJZnkCuWGkz0Ml7j_A7WsYffOWSi8psI6gjOEVwiyIKtB_hbLBbwpzhxCdPsFprwDEuowwvsdpBEFkm_XMjMVtQlK7B1A2UEEn6DeDtimrC0uZJys6GIfZfY0BnCPooAuIyjFkelkH4JDw_UiKR3E_R5yx0CQqj7xH7Y2CHTmivzVl3FTauk7dHziuqwxhr48tfBSvrEbEDRTR_dTFFiLG9l6lOqw9QfrEjbghkO6xDLPcAmxHrD9u7gH6RGMqQT8kqVIDSYqgJdVK8UgIKpAwd86KhPQKtAeuSvGtoaGLZhSNMvH_-GPU9Zqzmg4PjSA15hgwrdNHkEhDT8wBXOCXMsuRHpqkd_reCVUzrtTyNQG86yfrYnx3d2MhHgsMfPNYoSguVvieTbT4csyxZZ8Bu5z8_RXGQ-h7bfW1FkUAjd2bu37IkjG9KGRvO87WZKymdOylGzfNYtniyMPccRJC82kHEy2HMSQdwFSAGBJIBi8sB9t0lrI1qyqCrl58EYpCVw5efl-gXuERJ2o76ELYMILOUg7MhB3NUKJfIZWonO3bR2W65YXXOZ-XKKYvYQdbA5WZ6NQ4F8wCMkfJk4J49-e8XZdycNjy-Km7uCYl2SCI_E8IWoST97FCJYB4OOnveBLEZaWKlLhsdU7XSjjE8oXmtRVHKXPCRxHPeMdnteqO0iAjUIFYzaURFFRWyEbrikE00cugtv_ZZkm76l-7WjG6GXehzWvWJiEqDpJXJC5wGFQqFkU_plwP44Md4y7s04EU018vFKsluJJ0M1jr4Jsc1FMYt2dOWLf-VV3LZwrW-o2rW5RXDN4hs-ktvCWUNzqvecxS-uJbDiI64_N8SERfIcV_JC858hJxQJkJEANDmH25t_98Yfywx1IhLGm3Q2hb7BSbbXvkJKT_PLx9kW4Wb_MGPDg1kSzO-YJl8hHi-BOJgvNDBcIrsl8irxLV1SXdAJJ0_b1bWrcMtv1uUYpOdqw3_M67t8HM7C_3vtzt5_MaAeNQ==&lang=sage&interacts=eJyLjgUAARUAuQ==" rel="noreferrer">live version that you can play with on the SageMath server. My code isn't stored on the server, it's encoded in the URL.

If you get weird behaviour with small $\epsilon$, try increasing the number of the bits global variable (at the top of the file). The default setting of 53 should be ok for $\epsilon > 10^{-8}$ or so. You may also need to increase the depth of the continued fraction.

FWIW, $\log_2 3$ is rather important in the mathematical music theory of equally-tempered scales. The standard 12 tone scale uses the convergent $19/12$.


问:
What exactly is a function or mapping? Is it an object?
What exactly is a function or mapping? Is it an object?  I'm not talking about the domain or the codmain which I know are Sets and therefore objects. The mapping itself, what is it? From what I understand everything is a set, but a function is made of two sets plus something else. What is that something else?  As far as I understand in ZFC everything is a Set.

From Abstract Algebra  by Dummit and Foote

"A group is an ordered pair (G, *) where G is a set and * is a binary operation on G satisfying the following axioms:  ..."

"A binary operation * on a set G is a function $* : G \to G$."

What is confusing is that I just finished learning ZFC Set Theory and it shows that an ordered-pair is the Set:

(x,y) = { {x}, {x, y}}

When the y is not set a but a function now I'm confused of how "(G, *)" can be an ordered pair when the * is not simply a Set.

答: No, it's also a set. A function with domain $A$ and codomain $B$ is a subset $f$ of $A\times B$ such that, for each $a\in A$, there is one and only one $b\in B$ such that $(a,b)\in f$.

ContentLicense: CC BY-SA 4.0


问:
Fixed Point Iteration Methods - Convergence
We wish to solve the equation $f_a(x) = 0$, where $a ∈ R$ is a parameter and $f_a(x) = (1−a)x+ax^3$. We will study the fixed point iteration $x^{k+1} = φ(x^k)$, where the function $φ(x) = ax(1 − x^2)$.

(i) Show that the fixed point method is consistent and that φ : [0, 1] → [0, 1], provided $0\leqslant a \leqslant \frac{3\sqrt{2}}{2}$.

(ii) Find the positive values of $a$ such that the fixed point iteration converges to the root $α_1 = 0$ for any initial guess $x^0 ∈ [0, 1]$.

(iii) Find a condition on $a$ under which a second zero $α_2 > 0$ exists in the interval [0, 1] and determine for which $a$ the fixed point iteration can approximate $α_2$.

(iv) For what value of $a$ can the fixed point iteration approximate $α_2$ with second order convergence?

Solved part (i): Used the consistency condition $f(\alpha)=0 \iff \phi(\alpha)=\alpha$ which gives LHS: $x=0$, $x=\pm\sqrt\frac{a-1}{a}$. These values both give $\phi(x)=x$.
Solving the RHS gives: $a-a\alpha^2 -1=0$, hence $\alpha=\pm\sqrt\frac{a-1}{a}$ as required.
Then for the next part I calculated the maximum of $\phi(x)$ by finding $\phi'(x)=0$ which gives $x=\pm\sqrt\frac{1}{3}$. Since we are concerned with [0,1], we take $x=+\sqrt\frac{1}{3}$, and find $\phi(\sqrt\frac{1}{3})$ which gives $x=2a\frac{\sqrt3}{9}$. Setting $0\leqslant 2a\frac{\sqrt3}{9} \leqslant 1$ and rearranging achieves the result $0\leqslant a \leqslant \frac{3\sqrt{2}}{2}$.

Attempt at part (ii) Using the local linear convergence theorem, we will try to find the values of a for which the method converges. We first check whether $\phi(x)$ is continuously differentiable or not.

$\phi(x) = ax(1-x^2)$

$\phi'(x) = a - 3ax^2$

Hence we can see that $\phi(x)$ is in fact continuously differentiable.

Then we check that $\alpha_1$ is, in fact, a fixed point of $\phi$:
$\phi(\alpha_1 = 0) = a(0)(1-0) = 0 = \alpha_1$.
Then by the local linear convergence theorem, the fixed point iteration converges if $|\phi'(\alpha_1)| < 1$, for all initial guesses $x^0 \in [\alpha_1-\delta, \alpha_1+\delta]$.
Therefore we have,
$|\phi'(\alpha_1)| = |a-3a(0)^2| = |a|$.
Hence, we have convergence for $|a|<1$.

The following part is the bit that I am unsure about (where you apply the final part of theorem with delta):

I just said, let $\delta = 1$, then for $|a|<1$ the method converges to $\alpha_1 = 0$, for all initial guesses $x^0 \in [-1,1]$

Since we are considering strictly positive values of a:
For $0<a<1$, the fixed point iteration method converges to $\alpha_1 = 0$, for all initial guesses $x^0 \in [0,1]$.

Attempt at part (iii): I am getting $1<a\leqslant a+1$ i.e. $a\in(1,a+1]$ which seems wrong, then I am not sure how to apply part (ii)?

答:
In (ii), you should have learned that you are guaranteed local convergence if $|\varphi'(r)|<1$ and you have global convergence if $|\varphi'(x)|<1$ for all $x$ in the domain. So this is just a matter of messing with some explicit inequalities now.

For (iii), as you say, the other possible zeros are $\pm \sqrt{\frac{a-1}{a}}$. These are real if $a \geq 1$ and the positive one is strictly positive if $a>1$. Is the positive one always in $(0,1]$ for $a>1$, or not? Once you know under what conditions it is in $(0,1]$, you can repeat what you did in (ii). (IMO the prompt is a bit vague, because it says "for which $a$ you can approximate $\alpha_2$?" and it isn't clear whether that means "for some $x^0 \in [0,1]$" or "for all $x^0 \in [0,1]$". If it's the former, then you can just check local convergence instead of global convergence.)

For (iv) you should have learned that the condition for second order convergence is $\varphi'(r)=0$, which is just an algebraic equation you can solve for $a$ since you know the root is $\sqrt{\frac{a-1}{a}}$.


问:
A generalization of Kirkman's schoolgirl problem
A friend of mine asked me this question. "I have $3n$ elements, and I want to know which is the maximum number of triplets $(a,b,c)$ so that no two triplets have more than one element in common".

The first thing which came into my mind was the Kirkman's schoolgirl problem: with 15 elements there are 35 different triplets (which may be arranged in 7 groups of five triplets each, but this is not important for the problem my friend asked me). A bit of research led me to Steiner triples, and I understand that there is always a maximal solution. But is there an algorithm to find one?

答:
I see no reason to assume that the number of elements is divisible by $3$; so I'll work with $m$ elements, where $m$ is your $3n$. This answer has gone through a lot of edits; I now have a complete answer for all $m$.

For $m \equiv 1$ or $3 \bmod 6$, the answer is $(m^2-m)/6$. For $m \equiv 5 \bmod 6$, the answer is $(m^2-m-8)/6$. For $m$ even, the answer is $\lfloor (m^2-2m)/6 \rfloor$, which is $(m^2-2m)/6$ for $m \equiv 0$ or $2 \bmod 6$ and $(m^2-2m-2)/6$ for $m \equiv 4 \bmod 6$.

Upper bounds We have the upper bound $(m^2-m)/6$. The upper bound is because any unordered pair $\{ a,b \}$ can only occur in at most one triple. If $k$ is the number of triples, then every triple contains $3$ pairs, so $3k \leq \binom{m}{2}$ which simplifies to $k \leq (m^2-m)/6$.

For $m \equiv 5 \mod 6$, I can do better. First of all, $(m^2-m)/6$ isn't an integer, so we can improve to $\lfloor (m^2-m)/6 \rfloor = (m^2-m-2)/6$. But, in fact, we can do one better; the true bound is $(m^2-m-8)/6$. Proof: For the sake of contradiction, suppose we could achieve $(m^2-m-2)/6$. Since each triple contains $3$ pairs, there would be $3 (m^2-m-2)/6 = \binom{m}{2} -1$ pairs contained among our triples. In other words, there would be precisely one pair $(x,y)$ not in a triple. There are $m-2$ points other than $x$. By hypothesis, each of these points IS in a triple with $x$, and these triples overlap only at $x$. So these triples look like $(x, z_1, z_2)$, $(x, z_3, z_4)$ etcetera, and we conclude that $m-2$ is even. This contradicts our hypothesis that $m \equiv 5 \bmod 6$. So $(m^2-m-2)/6$ is not achievable and the true bound is $(m^2-m-8)/6$.

Now suppose $m$ is even. For every index $i$ in $\{1,2,\ldots,m\}$, there can only be $m-1$ triples containing $i$. Thus, for every $i$, there is some $f(i)$ such that $(i, f(i))$ is not contained in any triple. That means at least $m/2$ pairs are in no triple. So $3k \leq \binom{m}{2} - (m/2)$ or $k \leq (m^2-2m)/6$.

Lower bounds

If $m \equiv 1$ or $3 \bmod 6$, then $(m^2-m)/6$ is achievable. The configurations which achieve this are called Steiner triple systems; see here or here for constructions.

If $m \equiv 0$, $2 \bmod 6$ then there is a Steiner system of size $m+1$. Deleting all the $m/2$ triples from this that contain $m+1$ achieves $(m^2-2m)/6$ triples, again hitting the optimal bound.

If $m \equiv 5 \bmod 6$, then we can partition $\{1,2,\ldots, m\}$ into $(m-5)(m+4)/6$ triples and one quintuple so that every pair is in exactly one of these sets. See the last slides of these notes. Calling the quintuple $\{a,b,c,d,e\}$, delete it and create the triples $\{ a,b,c \}$ and $\{ c,d,e \}$. This achieves $(m-5)(m+4)/6 + 2 = (m^2-m-8)/6$.

If $m \equiv 4 \mod 6$, then take the above system of size $((m+1)^2 - (m+1)-8)/6$ and delete a point which lies in $m/2-1$ triples. This achieves $(m^2-2m-2)/6 = \lfloor(m^2-2m)/6 \rfloor$.

ContentLicense: CC BY-SA 3.0


问:
What is the meaning of transposing a vector in linear algebra quesitons all the time?
I am currently taking a course in linear algebra and on a lot of questions have this notation and the more I read about linear algebra online the more I see this notation:

$$2(-1, 2)^T\cdot(4,1)^T=4$$

Why to the power of T? I get that it is the transpose and how the transpose works, but why?

答:
Not everybody does it this way, but the idea is that

the vectors we prefer to deal with, and the vectors we've defined dot products for, are column vectors, like $\begin{bmatrix}-1 \\ 2\end{bmatrix}$ and $\begin{bmatrix}4 \\ 1\end{bmatrix}$.

the notation $(-1,2)$ and $(4,1)$ denotes row vectors $\begin{bmatrix}-1 & 2\end{bmatrix}$ and $\begin{bmatrix}4 & 1\end{bmatrix}$.

To get column vectors out of the row vectors, we take the transpose.


答: As you mentioned, "to the power of T" is actually the transpose. Often people write the transpose of a (column) vector instead of the vector since it takes up less vertical space. It is pure notational convenience and can be useful in inline equations.</p>

Title: What is the meaning of transposing a vector in linear algebra quesitons all the time?

Body:
I am currently taking a course in linear algebra and on a lot of questions have this notation and the more I read about linear algebra online the more I see this notation:

$$2(-1, 2)^T\cdot(4,1)^T=4$$

Why to the power of T? I get that it is the transpose and how the transpose works, but why?


问:
A differentiable function with $df = 0$ is locally constant

Let $U$ be an open set and $f : U \subset\mathbb{R}^d \to \mathbb{R}^m$ be a differentiable function with $df=0, \forall x \in U.$

Show that $f$ is locally constant.

So my idea was, given some arbitrary $x_0 \in U$, first taking some ball $B_{\varepsilon}(x_0) \subset U$ which exists since $U$ is open, we can assume $f$ isn't constant there i.e there is $y_o$ s.t $f(x_0) \neq f(y_0)$.

Then defining $\gamma(t) = (1-t)x_0 +t y_0$, we have $f(\gamma(0)) \neq f(\gamma(1))$ meaning they are different in at least one coordiante, so without loss of generality $f^1(\gamma(0)) \neq f^1(\gamma(1))$. and then using lagrange for an $\mathbb{R} \to \mathbb{R}$ function we have $(f^1(\gamma(c)))' \neq 0$.

But on the other hand using chain rule $(f^1(\gamma(c)))' = d(f^1(\gamma(c))) =
\underbrace{df^1(\gamma(c))}_\text{0 since $df=0$}
  \circ d \gamma(c) = 0$

I don't really like my solution since it relies on coordinates, if someone would like to share a different solution it would be great.

答:
Your solution seems correct to me.
I think we could rely directly on mean value ineq., because of convexity of euclidean balls, no matter the norm :
$$ \|f(b) - f(a)\| \leq \sup\| df\||b-a\| = 0$$


问:
Space of functions from [0,1] to IR and from IR to IR
I am convinced that the cardinality of the set S defined as
$\Re^\Re$ is the same as the set Z defined as $[0,1]^\Re$, since we can create an operator $\Phi:S\rightarrow Z$ that sends  each function f to f$\circ$h where h is a bijection from [0,1] to $\Re$, and this operator seems to be bijective.
My question is, if we now look at the natural vector spaces formed by S and Z, can we say the operator $\Phi$ is a linear map? And if it is, does it imply those two vector spaces are isomorphic?

答: Sure:\begin{align}\Phi(f_1+f_2)&=(f_1+f_2)\circ h\\&=f_1\circ h+f_2\circ h\\&=\Phi(f_1)+\Phi(f_2)\end{align}and, by a similar argument, $\Phi(\alpha f)=\alpha\Phi(f)$. And $\Phi^{-1}(f)=f\circ h^{-1}$.


问:
Algebraic Algebras
Let $A$ be an algebra over a field $K$. Consider the following statements.

(1) If $A$ is an integral domain and algebraic over $K$, then $A$ is a field.

(2) If $A$ is a field and is contained in an affine $K$-domain, then $A$is algebraic.

I know these statements to be true as proof's of them were provided to me by an instructor. I was told that I could find examples that show that the hypothesis in both (1) and (2) cannot be omitted in order for each statement to be true. Can anyone please help see that this is the case. Thank you.

this is not a homework assignment.

答:
So, there are four counterexamples you need to find. Already it's nontrivial progress just to state clearly what these are, as follows:

A $K$-algebra $A$ which is an integral domain but not a field. What integral domains do you know how to construct?

A $K$-algebra $A$ which is algebraic over $K$ (I assume this means every element of $A$ satisfies a polynomial with coefficients in $K$) but not an integral domain. Adjoin the root of a polynomial to $K$ and ask yourself when the result is not an integral domain.

A $K$-algebra $A$ which is a field but not algebraic. What fields do you know how to construct? Do you know how to construct fields with transcendental elements?

A $K$-algebra $A$ contained in an affine $K$-domain which is not algebraic. What affine $K$-domains do you know how to construct?

Title: Algebraic Algebras


问:
Are all commutative, associative binary operations isomorphic to addition?
Addition and multiplication are the two classic commutative, associative binary operations on the reals. They satisfy a striking property: they are equivalent up to unary operations. By taking a logarithm of the arguments to addition and exponentiating the result, we get multiplication:

$xy = \exp(\log x + \log y)$

In fact, we can create a new commutative, associative operation from every invertible unary function $f$ in the same way:

$(x,y) \mapsto f^{-1} (f(x) + f(y))$

I actually haven't been able to come up with a commutative, associative operation that doesn't follow this form. Are all commutative, associative operations on the reals isomorphic to addition in a similar way?

EDIT - The answers and comments below give several good counterexamples to the problem as stated. Requiring $f$ to be invertible is too strong a condition. What if we allow this slightly more general functional form

$(x,y) \mapsto g (f(x) + f(y))$

where $f$ maps reals to some field and $g$ maps that field to the reals? This seems to capture the counterexamples. For example:

$(x, y) \mapsto 0$ matches with arbitrary $f$ and $g(z) = 0$

$(x, y) \mapsto x + y - \lfloor{x+y}\rfloor$ matches with $f(x) = x$ and $g(z) = z - \lfloor{z}\rfloor$

I think $(x, y) \mapsto max(x,y)$ matches with $f(x) = (x + a)^n$ and $g(z) = z^{1/n} - a$, where $a, n$ are constants we take to infinity. That choice is inspired by the relationship between the max function and the infinity norm. I get that including limits in $f$ and $g$ is pretty suspect, but even with that hack, it's interesting to me that it's possible.

Modifying the question slightly, can every commutative, associative binary operation be cast in the form $(x,y) \mapsto g (f(x) + f(y))$?

答:
You can consider an operation that does not give the structure of a group. For instance
$$x\circ y\  \colon = \max(x,y)$$
Here we have $x \circ x = x$.  In fact we have a lattice.

But even if the operation would give a commutative group, it may not be isomorphic to $\mathbb{R}$ with addition.  For instance $\mathbb{R}/\mathbb{Z}$ is one such group of the same cardinality as $\mathbb{R}$.

You can check that an abelian group with the same cardinatlity as $\mathbb{R}$ is isomorphic to $\mathbb{R}$ with addition if an only if it is a uniquely-divisible group: for every $x$ and $n\ge 1$ integer, there exist a unique $y$ such that $ny = x$. Since then it is a vector space over $\mathbb{Q}$. However, a concrete isomorphism may not be available without some axiom of choice things. For instance, $\mathbb{R}/\mathbb{Q}$ with addition.


问:
Varieties in which congruences are characterized by subalgebras
For some varieties $V$, for any algebra $\mathbb{A}\in V$, there exists a lattice isomorphism $f:\text{Con}(\mathbb{A})\to \mathbb{B}$ where $\mathbb{B}\leq \text{Sub}(\mathbb{A})$ between the lattice of congruences of $\mathbb{A}$ and a sublattice of subalgebras of $\mathbb{A}$, and the choice of $\mathbb{B}$ is natural in some sense.

For instance, if $V$ is the variety of groups, any congruence corresponds to a normal subgroup, and if $V$ is the variety of rings, congruence corresponds to an ideal.

Can someone point me to literature where such varieties are studied? Names, anything? Proper definition? It doesn't have to be precisely what I've written.

答:
I think that you are looking for the phrase ``ideal determined variety''. Might want to start with this paper:

http://www.mathematik.uni-marburg.de/~gumm/Papers/Ideals%20in%20universal%20algebras.pdf


问:
Arbitrary intersection of open set is not open in metric space
We know that arbitrary intersection of open sets may not be open. The most common example to show this is taking $\mathbb R$ with usual metric and taking $U_n=(\frac{-1}{n},\frac{1}{n})$. Then intersection of $U_n$ is $\{0\}$ which is not open. I am trying to find more examples but in other metric spaces. Do we have some examples of such open sets other then in $\mathbb R^n$ ?

答: All points in a metric space $X$ are closed, so the sets $U_x = X \setminus \{x\}$ are open.  Intersections of these can recover every possible subset of $X$, open or not.


问:
Can infinite unions in some sense be handled like limits?
I'm stuck on a question regarding sets that are covered by an infinite union of sets. If a set $S$ is a subset of an infinite union of sets,

$$S\subset\bigcup_{k=1}^{\infty}{U_k}$$

is it then correct to say that if $x_0\in S$ then $\exists N\in \mathbb{N}$ such that:

$$x_0\in\bigcup_{k=1}^{N}{U_k}$$

similar to what we would say if a limit exists?

答:
Yes, if $x_{0}\in S$ then $x_{0}\in\bigcup_{k=1}^{\infty}{U_{k}}$ and by definition this implies $x_{0}\in U_{m}$ for some $m\in\mathbb{N}$, therefore there exists $N\in\mathbb{N}$ (any $N\geq m$ does the job) such that
$$x_{0}\in\bigcup_{k=1}^{N}{U_{k}}$$


答:
Edit: The new version of your question is, I think, much more general than you intended; the answer is yes, due to the following cheap trick, but it doesn't really have anything to do with addition anymore.

Let $m : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be any commutative operation (we will not need associativity, and operations of the form $g(f(x) + f(y))$ will not be associative in general). We want to find $g, f$ such that $m(x, y) = g(f(x) + f(y))$. You say that $f$ can take values in "some field." We'll take this field to be the function field $K = \mathbb{Q}(f_x, x \in \mathbb{R})$ over $\mathbb{Q}$ on uncountably many variables, one for every real number, and $f : \mathbb{R} \to K$ will be the function $f(x) = f_x$. This guarantees that $f(x) + f(y) = f(z) + f(w)$ iff either $x = z$ and $y = w$ or $x = w$ and $y = z$; in other words, the value of $f(x) + f(y)$ uniquely determines the unordered pair $\{ x, y \}$.

Now we can take $g : K \to \mathbb{R}$ to be any function such that $g(f_x + f_y) = m(x, y)$; since $f_x + f_y$ uniquely determines $\{ x, y \}$ and $m$ is commutative this is well-defined. Of course we have not used any features of $\mathbb{R}$ in this argument, which can be replaced by an arbitrary set.

(This responds to the original question only.)

Your statement about addition and multiplication is not quite right; you only get an isomorphism between $(\mathbb{R}, +)$ and $(\mathbb{R}_{+}, \times)$ (where the subscript denotes taking positive reals). However, it is true that addition and multiplication are locally isomorphic in a suitable sense.

Are all commutative, associative operations on the reals isomorphic to addition in a similar way?

As stated, since you haven't asked for compatibility with any other structure on the reals, you are just asking for commutative monoids with the same cardinality as the reals. There is an enormous diversity of such things, even if we restrict further to abelian groups (which means requiring inverses); as a random example, any countable product $\prod_{i=1}^{\infty} A_i$ of at-most-countable but nontrivial abelian groups has the same cardinality as the reals, for example $\prod_{i=1}^{\infty} \mathbb{Z}$, but such a group is isomorphic to the reals as an abstract group iff each $A_i$ is an at-most-countable-dimensional $\mathbb{Q}$-vector space.

However, we can get a positive result by requiring continuity.

Proposition: Every continuous abelian group structure on $\mathbb{R}$ (with its usual topology) is continuously isomorphic to addition.

This came up recently on MathOverflow. It follows from Gleason-Montgomery-Zippin's solution to Hilbert's fifth problem, which is almost certainly overkill; probably it follows from an easier and known structure theorem for locally compact (Hausdorff) abelian groups but I couldn't quite finish the argument. Maybe it can also be done "by hand."

(Edit: This is overkill, as expected. There is a totally elementary argument that a topological group (not necessarily abelian!) homeomorphic to $\mathbb{R}$ must be isomorphic to $(\mathbb{R}, +)$; see this math.SE question.)

If you further require smoothness then the proof is easier and can probably be done "by hand" more easily; abstractly, $\mathbb{R}$ is the unique $1$-dimensional Lie algebra and hence the unique $1$-dimensional simply connected Lie group, so:

Corollary: Any smooth abelian group structure on $\mathbb{R}$ is smoothly isomorphic to addition.

A related result you may be interested in is the classification of $1$-dimensional formal group laws; over a $\mathbb{Q}$-algebra, and in particular over $\mathbb{R}$, they are all isomorphic to the additive formal group law.

A question I don't know the answer to is whether it suffices to require measurability; that is, I don't know whether a measurable abelian group structure on $\mathbb{R}$ (with the Borel $\sigma$-algebra, say) is measurably isomorphic to addition.

Title: Are all commutative, associative binary operations isomorphic to addition?

Body:
Addition and multiplication are the two classic commutative, associative binary operations on the reals. They satisfy a striking property: they are equivalent up to unary operations. By taking a logarithm of the arguments to addition and exponentiating the result, we get multiplication:

$xy = \exp(\log x + \log y)$

In fact, we can create a new commutative, associative operation from every invertible unary function $f$ in the same way:

$(x,y) \mapsto f^{-1} (f(x) + f(y))$

I actually haven't been able to come up with a commutative, associative operation that doesn't follow this form. Are all commutative, associative operations on the reals isomorphic to addition in a similar way?

EDIT - The answers and comments below give several good counterexamples to the problem as stated. Requiring $f$ to be invertible is too strong a condition. What if we allow this slightly more general functional form

$(x,y) \mapsto g (f(x) + f(y))$

where $f$ maps reals to some field and $g$ maps that field to the reals? This seems to capture the counterexamples. For example:

$(x, y) \mapsto 0$ matches with arbitrary $f$ and $g(z) = 0$

$(x, y) \mapsto x + y - \lfloor{x+y}\rfloor$ matches with $f(x) = x$ and $g(z) = z - \lfloor{z}\rfloor$

I think $(x, y) \mapsto max(x,y)$ matches with $f(x) = (x + a)^n$ and $g(z) = z^{1/n} - a$, where $a, n$ are constants we take to infinity. That choice is inspired by the relationship between the max function and the infinity norm. I get that including limits in $f$ and $g$ is pretty suspect, but even with that hack, it's interesting to me that it's possible.

Modifying the question slightly, can every commutative, associative binary operation be cast in the form $(x,y) \mapsto g (f(x) + f(y))$?


问:
How to get the length or the width of a percentage of a canvas?
I am currently working with the following diagram:

I know the width and height of the original canvas. How can I figure out the width s_w and height s_h of 5% of its overall area?

答:
The area of the smaller rectangle is $s_{w}\cdot s_{h}$, therefore it is $\dfrac{100\cdot s_{w}\cdot s_{h}}{20\cdot 30}\%$ of the total area. If you want this percentage to be $5$ then you need that $\dfrac{100\cdot s_{w}\cdot s_{h}}{20\cdot 30}=5$, so $s_{w}\cdot s_{h}=30$.

Here you have infinitely many possibilities, for example $s_{w}=5$ and $s_{h}=6$, $s_{w}=3$ and $s_{h}=10$, etc. However, if you want that the smallest rectangle is a square then $s_{w}=s_{h}=\sqrt{30}\approx 5.48$.


问:
Area of metric ball on n-sphere
Suppose $S_n = \{x \in \mathbb{R}^{n + 1} : ||x||_2 = 1\}$ is the $n$-sphere. Let $d : S_n^2 \to \mathbb{R}$ be the angle metric on $S_n$, i.e. $d(x, y) = \arccos(x \cdot y)$, where $\cdot$ is the dot product.

What is the area $A(D_n(\theta))$ of the disk $D_n(\theta) = \{x \in S_n : d(x, p) \leq \theta\}$, where $p \in S_n$?

I'm not sure of the terminology here, and whether I'm missing some information. I suspect this has to do with Riemannian manifolds, but I know little about them. Intuitively, I want to define probability densities on $S_n$, those densities being relative to area. At least for $n \leq 3$ the idea seems clear. Clearly $A(D_1(\theta)) = 2$, and $A(D_2(\theta)) = 2 \theta$. Looking at

https://en.wikipedia.org/wiki/Spherical_cap

we have $A(D_3(\theta)) = 2\pi (1 - \cos(\theta))$.

Edit: The same page in fact gives the general formula. It just has to be transformed to use angle rather than height.

答:
Wikipedia's spherical cap article provides the area of (a boundary of) a spherical cap with height $h(\theta)$ and sphere-radius $r$ as

$$A(D_n(\theta)) = (1/2) A_n r^{n - 1} I(\alpha(\theta), (n - 1) / 2, 1 / 2),$$

where

$$A_n = \frac{2 \pi^{n / 2}}{\Gamma(n / 2)},$$
$$\alpha(\theta) = (2rh(\theta) - h(\theta)^2) / r^2,$$
$$h(\theta) = r - r \cos(\theta),$$
$$I(x, a, b) = \frac{B(x, a, b)}{B(a, b)}.$$

Here $A_n$ is the area of the boundary of the unit $n$-ball, $I$ is the regularized incomplete beta function, and $B$ is the (incomplete) beta function. Expanding the definitions,

$$\alpha(\theta) = 2(1 - \cos(\theta)) - (1 - \cos(\theta))^2.$$

This can further be simplified to

$$\alpha(\theta) = \sin(\theta)^2.$$

The formula for $A(D_n(\theta))$ above is valid only when $0 \leq \theta \leq \pi / 2$, corresponding to $0 \leq h(\theta) \leq r$. When $\pi / 2 \leq \theta$, the formula is given by $A_n r^{n - 1} - A(\pi - \theta)$.

Title: Area of metric ball on n-sphere


问:
There are at least 4 global sections on a complete intersection curve
This is Vakil 19.8 C, self-study.

We take a regular complete intersection of a quadric $Q$ and a cubic $K$ in $\mathbb P^3_k$, producing a genus $4$ curve $C$. The problem asks us to show that there are at least 4 sections of $\mathcal O_C(1)$, i.e.

$$h^0(C, \mathcal O_C(1)) \geq 4$$

and says that this translates to showing that $C$ does not lie in a hyperplane. I do not see why either of these statements is true.

First, showing $C$ is not in a hyperplane is trivial it seems. If it were in a degree $d$ hypersurface, the hypersurface would have to look like $f_1Q + f_2K$, where $f_i$ are homogeneous polynomials of the correct degree. This is impossible if $d = 1$.

Second, if I ignore the translated problem and focus on the original problem, the only thing I can think to use is Riemann-Roch and Serre Duality in combination:

$$h^0(\mathcal O_C(1)) = \deg \mathcal O_C(1) - g + 1 + h^1(\mathcal O_C(1)) = 1 - 4 + 1 + h^0(\omega_C \otimes \mathcal O_C(-1))$$

but I do not see how to show that last term is $\geq 6$. We know $h^0(\omega_C) = 4$, for what it's worth.

答:
Since $C$ is a complete intersection, $\deg C = 2\cdot 3 = 6$, which means that actually $\deg \mathcal O_C(1) = 6$ (i.e. $C$ intersects a general hyperplane in $6$ points) giving us $h^0(\mathcal O_C(1)) = 6 - 4 + 1 + h^0(\omega_C \otimes \mathcal O_C(-1))$.

To determine this last term (which needs to be $\geq 1$), note that $\deg \omega_C = 6$ so the tensor product is a line bundle of degree $0$, hence has a section if and only if it is isomorphic to $\mathcal O_C$ which is itself the case if and only if $\mathcal O_C(1) \cong \omega_C$. This follows from iterated adjunction:
$$
\omega_Q \cong \omega_{\mathbb P^3}\otimes \mathcal O_{\mathbb P^3}(Q)|_Q \cong \mathcal O_{\mathbb P^3}(-4)\otimes\mathcal O_{\mathbb P^3}(2)|_Q \cong \mathcal O_{\mathbb P^3}(-2)|_Q,
$$

and then
$$
\omega_C \cong \omega_Q\otimes (\mathcal O_{\mathbb P^3}(K)|_Q)|_C \cong \mathcal O_{\mathbb P^3}(-2)|_Q \otimes (\mathcal O_{\mathbb P^3}(3)|_Q)|_C \cong \mathcal O_{\mathbb P^3}(1)|_C = \mathcal O_C(1),
$$

where I've maybe gone a bit overboard with the restriction notation, but hopefully the idea is clear: everything happening is with line bundles that have been restricted from projective space.


问:
On cases notation
In a formal situation, is it acceptable to write the statement "If $a\geqslant 1$ then $x>y$, if $-1\leqslant a<1$ then $x=y$, and if $a<-1$ then $x<y$." using cases notation as below?
$$x\begin{cases}
>y&\text{if }a\geqslant 1;\\
=y&\text{if }-1\leqslant a<1;\\
<y&\text{if }a<-1.\\
\end{cases}$$
Is anybody aware of a formal resource (books, papers) where such a notation is used?

答:
It is an easy fix. The following is much easier to read:
$$\begin{cases}
x>y&\text{if }a\geqslant 1;\\
x=y&\text{if }-1\leqslant a<1;\\
x<y&\text{if }a<-1.\\
\end{cases}$$

Coming to think about it, the brace is not really needed. So you should just use the align environment

$$\begin{align}
x>y&\quad \text{ if }a\geqslant 1;\\
x=y&\quad \text{ if }-1\leqslant a<1;\\
x<y&\quad \text{ if }a<-1.\\
\end{align}$$

Title: On cases notation


问:
Coefficients of polynomial $f_n(q)=\prod_{k=1}^{n}(1+q^k)$

What is the general formula for the coefficient $c_n(k)$ where
  $$f_n(q)=\prod_{j=1}^{n}(1+q^j)=\sum_{k=0}^{n(n+1)/2}c_n(k)q^k?$$

I came across this problem while researching $q$-analogs. Indeed, the polynomial in question is trivially given by
$$f_n(q)=(-q;q)_n$$
where $(a;q)_n=\prod_{k=0}^{n-1}(1-aq^k)$ is the $q$-Pochhammer symbol. From Wikipedia, I read that
$$f_n(q)=\sum_{j=0}^{n}\left[{{n}\atop{j}}\right]_{q^2}q^j\tag{1}$$
where
$$\left[{{n}\atop{j}}\right]_{q}=\frac{(q;q)_n}{(q;q)_j(q;q)_{n-j}}.$$
While $(1)$ is a cool identity nonetheless, it does not provide me with what I'm looking for, as $\left[{{n}\atop{j}}\right]_{q^2}$ is still dependent on $q$.

I worked out the first few polynomials by hand:
$$\begin{align}
f_1(q)&=1+q\\
f_2(q)&=1+q+q^2+q^3\\
f_3(q)&=1+q+q^2+2q^3+q^4+q^5+q^6\\
f_4(q)&=1+q+q^2+2q^3+2q^4+2q^5+2q^6+2q^7+q^8+q^9+q^{10}\\
f_5(q)&=1+q+q^2+2q^3+2q^4+3q^5+3q^6+3q^7+3q^8+3q^9+3q^{10}+2q^{11}+2q^{12}\\
&+q^{13}+q^{14}+q^{15}.
\end{align}$$

It seems as if, in general, $$c_n(k)=c_n(n(n+1)/2-k).$$

I started trying to find recurrence relations. I saw that
$$c_3(k)=\begin{cases} c_2(k) & 0\le k\le 2 \\ c_2(k)+c_2(k-3) & k=3 \\ c_2(k-3) & 4\le k \le 6 \end{cases}$$
and similarly
$$c_4(k)=\begin{cases} c_3(k) & 0\le k\le 3 \\ c_3(k)+c_3(k-4) & 4\le k\le 6 \\ c_3(k-4) & 7\le k\le 10 \end{cases}$$
as well as
$$c_5(k)=\begin{cases} c_4(k) & 0\le k\le 4 \\ c_4(k)+c_4(k-5) & 5\le k\le 10 \\ c_4(k-5) & 11\le k\le 15  \end{cases}$$
which hints at the general expression
$$c_n(k)=\begin{cases} c_{n-1}(k) & 0\le k\le n-1 \\ c_{n-1}(k)+c_{n-1}(k-n) & n\le k\le n(n-1)/2 \\ c_{n-1}(k-n) & 1+n(n-1)/2\le k\le n(n+1)/2 \end{cases} .\tag{2}$$

This looks promising, but I am not sure if it is correct, because I just found it from basically guessing/recognizing a pattern which is not the most rigorous approach. Could someone verify $(2)$ and/or provide a simpler form of it? I ask this because I want to eventually find the coefficients $C(k)$ in
$$(-q;q)_\infty=\sum_{k\ge0}C(k)q^k$$
by taking $\lim_{n\to\infty}c_n(k)$.

Edit:

It may or may not help to note that, since $f_n(1)=\prod_{k=1}^{n}2=2^n$, we have the interesting identity
$$\sum_{k=0}^{n}c_n(k)=2^n.$$

答:
Another method. Let $S\subseteq\Bbb N$ and $A=(a_n)_{n\in S}$ be a sequence. Then define
$$\Pi(q)=\Pi_{S,A}(q)=\prod_{n\in S}(1-a_nq^n), \qquad |q|<1.$$
Then define $c_n=c_{S,A}(n)$ as the coefficients
$$\Pi(q)=\sum_{n\ge0}c_nq^n,\qquad c_0=1.\tag1$$
Then let $r_n=a_n\chi_S(n)$. Thus we have
$$\Pi(q)=\prod_{n\ge1}(1-r_nq^n).$$
Then
$$\begin{align}
\ln\Pi(q)&=\sum_{n\ge1}\ln(1-r_nq^n)\\
&=-\sum_{n\ge1}\sum_{k\ge1}\frac{r_n^kq^{nk}}{k}.
\end{align}$$
Taking the derivative and multiplying by $q$, we have
$$\begin{align}
\frac{q\Pi'(q)}{\Pi(q)}&=-\sum_{n\ge1}\sum_{k\ge1}nr_n^kq^{nk}\\
&=-\sum_{N\ge1}q^N\sum_{uv=N\\ u,v\ge1}ur_u^v\\
&=-\sum_{n\ge1}q^n\sum_{d|n}dr_d^{n/d}\\
\Rightarrow q\Pi'(q)&=\Pi(q)\sum_{n\ge1}R_nq^n.\tag2
\end{align}$$
We can then use $(1)$ and $(2)$ to multiply power series and compare coefficients to get the recurrence
$$nc_{S,A}(n)=-\sum_{k=1}^{n}c_{S,A}(n-k)\sum_{d|k}\left(\chi_S(d)a_d\right)^{k/d}d.$$
Setting $S=\{1,...,m\}$ and $A=(1)_{n\in S}$, the coefficient $c_{S,A}(n)=\alpha_m(n)$ of $q^n$ in the expansion of $\prod_{i=1}^{m}(1+q^i)$ satisfies the recurrence
$$n\alpha_m(n)=-\sum_{k=1}^{n}\alpha_m(n-k)\sum_{\,\, d|k\\ \, d\le N}d.$$


问:
Prove that $\left(A\cap B\cap C\right)\cup\left(A^{C}\cup B^{C}\cup C^{C}\right)=U$
Prove that $\left(A\cap B\cap C\right)\cup\left(A^{C}\cup B^{C}\cup C^{C}\right)=U$

Since it is an equality be must prove that
$$\left(A\cap B\cap C\right)\cup\left(A^{C}\cup B^{C}\cup C^{C}\right)\subseteq U$$
$$U\subseteq\left(A\cap B\cap C\right)\cup\left(A^{C}\cup B^{C}\cup C^{C}\right)$$

答:
Use De Morgan's Law $S^C\cup T^C=(S\cap T)^C$ twice to get
$$A^C\cup B^C\cup C^C=(A\cap B\cap C)^C.$$
Then clearly the union of $A\cap B\cap C$ with its complement will give the whole universe.


问:
Find x for which for every "a" the equation has solution
$$a^{31x} \equiv a \mod 271$$

I need to find x variable, for which the equation has solution with any a. How can I do this?

Generaly, modular equations have solutions when $GCD(a^{31x}, 271) = 1$, or $GCD(a^{31x}, 271) = d > 1; d|a$

It also looks like I could use Little Fermat's Theorem...

But I can't come up with anything...

Body:
mod $\color{#c00}{27}\!:\, \dfrac{1}{31} \equiv\dfrac{28}4\equiv \color{#c00}7.\ $ mod $\color{#0a0}{10}\!:\, \dfrac{1}1 \equiv \dfrac{1}{31}\equiv \color{#c00}7\!+\!\color{#c00}{27}j\equiv-3\!-\!3j\,$ $\Rightarrow$ $\ 3j\equiv -4\equiv 6\,$ $\Rightarrow$ $\,j\equiv \color{#0b0}2.$

Thus $ $ mod $\,27\cdot 10\!:\ \dfrac{1}{31} \equiv 7\!+\!27j \equiv 7\!+\!27(\color{#0b0}2\!+\!\color{#0b0}{10}k) \equiv 61,\ $ by Chinese Remainder (CRT).


问:
Distribution of 25 book titles between 4 different authors
Can someone please help me? I've been trying to do this for hours:

You are part of a reading group. Each member will read a book from 1 out of 4 different authors(W,X,Y,Z). In how many ways is it possible to select 25 books from theses authors so there are at least 5 books from each author and, at maximum, 6 books from author X?

obs: the titles of the 25 books don't matter, only who the author is

答:
Hint

Suppose that you have the following two problems:

What is the number of positive integer solutions to $x_1 + x_2 + x_3 = n \in \mathbb{Z^+}$, where each of the 3 variables must be $\geq k \in \mathbb{Z^+}$ and $n > 3k$

What is the number of non-negative integer solutions to $x_1 + x_2 + x_3 = (n-3k).$

What is the relationship between the answers to these two problems?


答:
I like to consider elements of the sets in question.

Let $x \in U$. One case is that $x\in A \cap B\cap C$. If this is not true, then $x \in D^C$ for at least one of $D = A, B, C$ . Then, $x\in A^C\cup B^C\cup C^C$. Now, we have proven the second statement.

Body:
Prove that $\left(A\cap B\cap C\right)\cup\left(A^{C}\cup B^{C}\cup C^{C}\right)=U$

Since it is an equality be must prove that
$$\left(A\cap B\cap C\right)\cup\left(A^{C}\cup B^{C}\cup C^{C}\right)\subseteq U$$
$$U\subseteq\left(A\cap B\cap C\right)\cup\left(A^{C}\cup B^{C}\cup C^{C}\right)$$


问:
Simplifying sums that contain factorials and have an index that doesn't start at 0.
I'm trying to simplify the following sum:

$P(a)=\sum_{n=0}^{\infty}\dfrac{x^{n-a}}{(n-a)!}$

I'm tempted to turn this into some sort of exponential function because it closely ressembles the Maclaurin series $e^x=\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$. Below is my attempt at this:

let $N=n-a$ so index now starts at $N=a$

$P(a)=\sum_{N=a}^{\infty}\dfrac{x^N}{N!}$

So now I have succeeded in making a function that is of the same form. Unfortunately, I don't really know what to do about the sum starting at N=a. Rewriting, I get:

$P(a)=\sum_{N=0}^{\infty}\dfrac{x^N}{N!}-\sum_{N=0}^a\dfrac{x^N}{N!}$

Now I definitely can write the first term as an exponential yaaay! But I'm still left with a finite sum that I have to subtract. So I guess what I'm really asking is what do I do about the second term?

Edit: It seems my first equation is incorrect. Here is the context surrounding my question:

Note: The variables were relabeled in my original question and constants factors were pulled out to make things a little easier on the eye.
I'm currently on part b). Subbing the given $P(n|N)$ and $P(N)$ expressions into P(n), I find that a negative factorial is inevitable but maybe I'm missing something.

答:
The partial sum (lower or upper) of the series for $e^x$ is related to the Regularized Incomplete Gamma function
$$
Q\left( {n,z} \right)\quad \left| {\;0 \le n \in Z} \right.\;\;
 = {{\Gamma \left( {n,z} \right)} \over {\Gamma \left( n \right)}}\quad
  = e^{\,\, - z} \sum\limits_{0\, \le \,k\, \le \,n - 1} {{{z^{\,k} } \over {k!\,}}}
$$

Coming to your problem, we have
$$
P(n\;\left| N \right.) = \left( \matrix{  N \cr   n \cr}  \right)
\eta ^{\,n} \left( {1 - \eta } \right)^{\,N - n}
$$
which is clearly the probability of $n$ successes in $N$ trials.

We assume $N$ to follow a Poisson distribution with average $\overline N $
$$
P(N) = e^{\, - \,\overline N } {{\overline N ^{\,N} } \over {N!}}
$$
Now we have
$$
P(n\;\left| N \right.) = {{P\left( {n \wedge N} \right)} \over {P(N)}}
$$
and as rightly hinted in the text, for $P(n)$ we will have
$$
\eqalign{
  & P(n) = \sum\limits_{N = 0}^\infty  {P\left( {n \wedge N} \right)}
  = \sum\limits_{N = 0}^\infty  {P(n\;\left| N \right.)P(N)}  =   \cr
  &  = \sum\limits_{N = 0}^\infty
  {\left( \matrix{  N \cr   n \cr}  \right)\eta ^{\,n} \left( {1 - \eta } \right)^{\,N - n}
 e^{\, - \,\overline N } {{\overline N ^{\,N} } \over {N!}}}  =   \cr
  &  = \eta ^{\,n} \left( {1 - \eta } \right)^{\, - n} e^{\, - \,\overline N }
 \sum\limits_{N = 0}^\infty  {\left( \matrix{  N \cr   n \cr}  \right)
{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N} } \over {N!}}}  =   \cr
  &  = \eta ^{\,n} \left( {1 - \eta } \right)^{\, - n} e^{\, - \,\overline N }
 \sum\limits_{N = n}^\infty  {\left( \matrix{  N \cr   n \cr}  \right)
{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N} } \over {N!}}}  =   \cr
  &  = {{\eta ^{\,n} \left( {1 - \eta } \right)^{\, - n} e^{\, - \,\overline N } } \over {n!}}
\sum\limits_{N = n}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N} }
 \over {\left( {N - n} \right)!}}}  =   \cr
  &  = {{\eta ^{\,n} \left( {1 - \eta } \right)^{\, - n} e^{\, - \,\overline N } } \over {n!}}
\sum\limits_{N - n = 0}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N - n + n} }
 \over {\left( {N - n} \right)!}}}  =   \cr
  &  = {{\eta ^{\,n} \left( {1 - \eta } \right)^{\, - n}
 \left( {\overline N \left( {1 - \eta } \right)} \right)^{\,n} } \over {n!}}e^{\, - \,\overline N }
 \sum\limits_{N - n = 0}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N - n} }
 \over {\left( {N - n} \right)!}}}  =   \cr
  &  = {{\left( {\eta \overline N } \right)^{\,n} } \over {n!}}e^{\, - \,\overline N }
 \sum\limits_{k = 0}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^k } \over {k!}}}  =   \cr
  &  = {{\left( {\eta \overline N } \right)^{\,n} } \over {n!}}e^{\, - \,\overline N }
 e^{\,\overline N \left( {1 - \eta } \right)}  =   \cr
  &  = {{\left( {\eta \overline N } \right)^{\,n} } \over {n!}}
e^{\, - \left( {\eta \,\overline N } \right)}  \cr}
$$

That is, the key passage you got astray is at
$$
\eqalign{
  & \sum\limits_{N = n}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N} }
 \over {\left( {N - n} \right)!}}} \; \Rightarrow
 \sum\limits_{N - n = 0}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N - n + n} }
 \over {\left( {N - n} \right)!}}}  \Rightarrow   \cr
  &  \Rightarrow \left( {\overline N \left( {1 - \eta } \right)} \right)^{\,n}
 \sum\limits_{k = 0}^\infty
 {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,k} } \over {k!}}}  \cr}
$$
But it is good in any case that you learned about the occurring of $Q$ !


问:
What is the connection between sum of tuple products and permutations by cycle?
From my previous question Disjoint sets in a combinatoral sum (continued), let $X = \{x_1, \dots, x_n\}$ be a set and
$$f_X(m) = \sum_{X' \in \binom{X}{m}} \prod_{x \in X'} x$$

Let $S_k = \sum x_i^k$.  We can calculate with inclusion-exclusion:

$$f_X(2) = \frac{1}{2!} (S_1^2 - S_2)$$

$$f_X(3) = \frac{1}{3!} (S_1^3 - 3 S_2 S_1 + 2 S_3) $$

and so on.

In the previous answer I discovered a connection between permutations by cycle type and the coefficients of the expression $f_X(m)$ (table at https://oeis.org/A181897), and came up with a tenuous explanation. Can someone confirm my answer and elucidate if my reasoning is correct? Also what is the connection to integer partitions? I am not that experienced with combinatorics or group theory so an answer that is as elementary as possible (possibly at the expense of generality) is appreciated.
I'm looking for references to this problem which has surely been explored in the past.

答:
This is very classical material, closely related to Newton's identities for the elementary and the power sum symmetric functions; that keyword will bring up a lot of references. I like Chapter 7 of Stanley's Enumerative Combinatorics, Vol. II but be warned that it has a lot of material.

Here is a short but maybe unsatisfying proof. Your functions $f_m$ are more commonly known as the elementary symmetric functions and denoted $e_m$. Your functions $S_k$ are more commonly known as the power sum symmetric functions and denoted $p_k$. The elementary symmetric functions have generating function

$$E(t) = \sum_{k=0}^n e_k t^k = \prod_{i=1}^n (1 + tx_i)$$

whereas the power sum symmetric functions have generating function

$$P(t) = \sum_{k=0}^{\infty} p_k t^k = \sum_{i=1}^n \frac{1}{1 - tx_i}.$$

These generating functions can be related by the logarithmic derivative, which gives

$$\log E(t) = \sum_{i=1}^n \log (1 + tx_i)$$

and hence

$$\frac{d}{dt} \log E(t)  = \sum_{i=1}^n \frac{x_i}{1 + tx_i} = \sum_{k=0}^{\infty} (-1)^k p_{k+1} t^k.$$

Integrating both sides gives

$$\log E(t) = \sum_{k=1}^{\infty} (-1)^{k-1} \frac{p_k}{k} t^k$$

and exponentiating gives

$$E(t) = \exp \left( \sum_{k=1}^{\infty} (-1)^{k-1} \frac{p_k}{k} t^k \right).$$

Expanding this out completely gives an identity relating $e_k$ and $p_k$ in terms of sums over permutations, generalizing the ones you gave, by the permutation form of the exponential formula. It can be stated succinctly as follows: if $\sigma \in S_n$ is a permutation, write $\text{sgn}(\sigma)$ for its signature, and write $p_{\sigma} = \prod_{i=1}^n p_i^{c_i(\sigma)}$ where $c_i(\sigma)$ is the number of $i$-cycles of $\sigma$. Then

$$\boxed{ e_n = \frac{1}{n!} \sum_{\sigma \in S_n} \text{sgn}(\sigma) p_{\sigma} }.$$

Some years ago I tried to work out a purely combinatorial proof of an equivalent version of this identity involving the complete homogeneous symmetric polynomials $h_k$, namely

$$h_n = \frac{1}{n!} \sum_{\sigma \in S_n} p_{\sigma}$$

which is equivalent to the previous one via the substitution $t \mapsto -t$ in the generating function. You can see how far I got in my blog post Newton's sums, necklace congruences, and Zeta functions II; there it's stated somewhat indirectly in terms of walks on graphs (the $x_i$ correspond to the eigenvalues of the graph) and it could be cleaned up to work in terms of the $x_i$ directly. I haven't tried to push through an argument for the $e_k$ version involving inclusion-exclusion but I wouldn't be surprised if it was possible. The $h_k$ version of the identity is, I think, a special case of the Polya enumeration theorem.


问:
logic question inferences
Here are two inferences A and B, one of which is valid, but the other not valid.

Inference A

Every human is mortal.

Socrates is human.

Therefore Socrates is mortal.

Inference B

Every hero likes women.

Taro likes women.

Therefore Taro is a hero.

Which one is correct, and in what sense it is correct ?

Can somebody give the hints to solve it ?

答: Hint: Is everyone who likes women necessarily a hero?</p>

Title: logic question inferences


问:
If $\lim_{h \rightarrow 0}$, then where does $|h|$ spring from?
To avoid typos, please see my screen captures below, and the red underline. The question says $h \rightarrow 0$, thus why $|h|$ in the solution? Mustn't that $|h|$ be $h$?

Spivak, Calculus 2008 4 edn.  His website's errata lists no errata for these pages.

答:
Don't worry. It takes a while and some practice to wrap your head around this stuff. It's easy to get confused by all the different conditions and variables and what depends on what.

Suppose $\lim_{x \to a}f(x) = \ell$

This means that for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $x$, if

$$0 < |x - a| < \delta \text{ then } |f(x) - \ell| < \varepsilon$$

Now, for this same $\delta$, suppose you have a number $h$ such that
$$0 < |h| < \delta$$

We can make this look like the previous condition by noting that $h = (a + h) - a$.
Therefore, for all $h$ if
$$0 < |h| < \delta$$
then
$$0 < |(a + h) - a| < \delta$$

Compare this with the initial $\delta$-requirement placed on $x$, namely $0 < |x - a| < \delta$.

The number $a+h$ satisfies the $x$ requirement from the first limit, so we have $|f(a + h) - \ell| < \varepsilon$.

Putting it all together, if $\lim_{x \to a}f(x) = \ell$ then for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $h$, if

$$0 < |h| < \delta \text { then } |f(a + h) - \ell| < \varepsilon$$

Another way of writing this is $\lim_{h \to 0} f(a+h) = \ell$. If the first limit exists, then so too does the second, and they are equal.

To complete the proof we need to begin with the second limit $\lim_{h \to 0} f(a+h) = \ell$ and use similar arguments to show how it leads to the first limit.

Suppose $\lim_{h \to 0} f(a+h) = \ell$. This tells us that for any $\varepsilon > 0$ there is a $\delta > 0$ such that for all $h$, if
$$0 < |h| < \delta \text { then } |f(a + h) - \ell| < \varepsilon$$

Suppose for this same $\delta$ we have numbers $x$ such that
$$0 < |x - a| < \delta$$

Compare this with the above $\delta$-requirements for $h$. These numbers $x-a$ fulfill the $\delta$-requirements for $h$, therefore

$$|f(a + (x - a)) - \ell| < \varepsilon$$

but this is just
$$|f(x) - \ell| < \varepsilon$$.

Putting it all together, we know that if $\lim_{h \to 0} f(a+h) = \ell$ then for any $\varepsilon > 0$ there is a $\delta > 0$ such that for all $x$, if

$$0 < |x - a| < \delta \text{ then } |f(x) - \ell| < \varepsilon$$

We can of course write this as $\lim_{x \to a}f(x) = \ell$. Again, existence of one limit guarantees existence of the other, and they are equal.


问:
Must a complex power series *fail* to be convergent somewhere on its circle of convergence?
My textbook asserts so, but I can't find its proof of the claim. On the other hand, a lecture slide I'm cross-referencing claims that a power series is allowed to be convergent at ALL points of its circle of convergence.

So who is correct?

答: As @N.S. demonstrated, the power series itself can converge at every point of the circle of convergence. The correct statement is that the function represented by the power series on the interior of the disk of convergence must have a singularity at (at least) one point of the circle. That is, it cannot be analytically continued to give an analytic function on a larger disk. For, if it could, the radius of convergence would have to be larger.


问:
Why is an uncountable set with the discrete topology not separable?
I saw a proof on proofwiki but couldn't understand it. I only know the bare basics of topology (closed, open, limit points, neighbourhoods). Is there a simple proof for this?</p>

答: Assume we have an uncountable discrete space. If you have a discrete topological space, every subset is closed and open, so the closure of a subset is itself. So if you have a dense subset, then it must be the whole space, but the space is uncountable. (I am going off of the definition: a space is separable iff it has a countable dense subset).</p>


问:
Verification of a Proof: Uniqueness of Sequence Limit
I've been reading Abbott's Understanding Analysis with Tao's Analysis I during my university's quarantine period (finally starting freshman year six months late! our class's ten month summer has come to an end. Thanks COVID). I'm still rather early on; an exercise from Abbott's asks the reader to prove the uniqueness of a limit of a sequence, and, after a good bit of searching to quell my lack of confidence in my work, I've noticed that my attempt is significantly different from the "canonical" approach by contradiction + triangle inequality, etc. Is this wrong? Is it necessarily weaker?

Take $a,b$ and a sequence $a_n$ such that $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} a_n = b$. By definition, it follows that, for all $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that, $\forall n > N$, $|a_n - a| < \epsilon$ and $|a_n - b| < \epsilon$. From Theorem 1.2.6,** this is equivalent to saying that $\lim_{n \to \infty}a_n = a$ and $\lim_{n \to \infty}a_n = b$. It quickly follows that $a=b$. $W^5$

Theorem 1.2.6 from Abbott (paraphrased, slightly): $a,b \in \mathbb{R}$ are equal $\iff$ $\forall \epsilon \in \mathbb{R}$, such that $\epsilon > 0$, $|a-b| < \epsilon$.

If anything, I suppose I'm just worried that I'm misinterpreting and/ or misusing 1.2.6; however, the symmetry between that and the convergence definition made it seem obvious. Thanks in advance for your time and feedback.

答:
Hint (1): $\lvert x - y \rvert \le \lvert x - z \rvert + \lvert z - y \rvert $

Hint (2): If $(x_n)$ is a sequence in $\mathbb{R}$ such that $x_n \to x$ then $\forall \epsilon > 0,\exists N \in \mathbb{N}, \forall n \ge N$ we have $\displaystyle \lvert x_n - x \rvert < \frac{\epsilon}{2}$.

These two hints and the application of the theorem you have listed can be used to give the uniqueness of limit of a convergent sequence in $\mathbb{R}$.


问:
Solving for $g(y)$ in the following integral equation
Given the following integral equation where $f, h$ are known:

$$ f(t) = \int_{t-1}^t{g(y)\cdot h(y)\ \textrm{d}y}$$

Is it possible to solve this for $g(t)$, i.e. get an equation of the form $g(t) = \dots$ (supposing $g, h$ are suitably nice)? I have no idea where to start. I acknowledge that an analytical solution may not even exist in the general case, but I remain hopeful!

答:
ok lets suppose we have some function: $z(y)=g(y)h(y)$ then we can say:
$$f(t)=\int_{t-1}^tz(y)dy$$
now suppose that $z$ satisfies the equation $Z'(y)=z(y)$ so we have:
$$f(t)=Z(t)-Z(t-1)$$
now if we differentiate both sides we get:
$$f'(t)=Z'(t)-Z'(t-1)=z(t)-z(t-1)$$
now substituting back in we get:
$$f'(t)=g(t)h(t)-g(t-1)h(t-1)$$
how if we isolate out $g(t)$ we get:
$$g(t)=\frac{f'(t)+g(t-1)h(t-1)}{h(t)}$$
and so:
$$g(t)=\frac{h(t-1)}{h(t)}\color{red}{g(t-1)}+\frac{f'(t)}{h(t)}$$
so whilst we cannot get a pure solution for $g(t)$ we can get this recurrance relation style which I believe you could get a solution from using generating functions depending on what the other functions are. Hope this helps


问:
Probability when doing multiple independent tests with a prior probabilty
A medical diagnostics test is p accurate on people with the disease and q accurate on people without.  Also r fraction of people actually have the disease.  0 <= p,q,r <= 1

If a person tested positive for the disease 4 times what is the probability he actually has the disease expressed in terms of p, q, r?

I'm not sure how accuracies p and q are defined. But I think p = p_true_positive / p_positive and q = q_true_negative / q_negative.

p = p_true_positive / r, q = p_true_negative / (1-r)

Now I was thinking think that the probability that he doesn't have a disease is
(1-p)^4, so that he does is 1 - (1-p)^4, but this seems not to be true.

答:
Use Bayes' rule (and assuming test independence):

$$P(D|++++) = \frac{P(++++|D) P(D)}{P(++++)} \\
= \frac{P(++++|D) P(D)}{P(D) P(++++|D) + P(-D)P(++++|-D)}$$

Here $+$ denotes a positive result on the test, $P(D) = r$ is the prior probability of having the disease, $P(-D) = 1 - P(D) = 1-r$ is the prior probability of not having the disease, and $P(++++|D)$ is the probability of getting four positive tests given the patient has the disease.

Just plug in your numbers.


问:
Contrapositive of: If $a_n > 0$ and $\sum a_n$ converges, then $\sum 1/a_n$ diverges.
I would like to find the contrapositive of: If $a_n > 0$ and $\sum a_n$ converges, then $\sum 1/a_n$ diverges.

My first idea is to rewrite it symbolically: Let $a_n$ be a sequence of real numbers. $\forall i \in \mathbb{N}, a_i > 0 ~ \land \sum a_n$ converges $\implies \sum \frac{1}{a_i}$ diverges.

Then we take the contrapositive: $\exists i \in \mathbb{N}, \sum \frac{1}{a_i}$ converges $\implies a_i \le 0 ~ \lor \sum a_n$ diverges.

However, I think this may be incorrect. A more reasonable contrapositive could be:

$\exists i \in \mathbb{N}, a_i \le 0, \sum \frac{1}{a_i}$ converges $\implies \sum a_n$ diverges.

However, I do not know which is right, or more importantly, why a particular one is right (maybe neither are?)

答:
After the long exchange of comments, I think the best answer we can give is that the contrapositive of:

If, for every $n$, $a_n > 0$, and, if $\sum_n a_n$ converges, then
$\sum_n 1/a_n$ diverges: i.e.,
$$(\forall n (a_n > 0)) \land C(a)) \Rightarrow D(n \mapsto 1/a_n)$$

is:

if $\sum_n 1/a_n$ does not diverge, then, either, for some $n$, $a_n \le 0$ or $\sum_n a_n$ does not converge: i.e.,
$$\lnot D(n \mapsto 1/a_n) \Rightarrow (\exists n (a_n \le 0)) \lor \lnot C(a)$$

where $C(a)$ and $D(a)$ stand for whatever Apostol means by $\sum_n a_n$ converges or $\sum_n a_n$ diverges, respectively.


问:
Meaning of $\lim_{p\to 0^+}$
I would like to know the interpretation of limit expression which calculates the number of non-zero entries in vector $x$ in the expression below:
$$\lVert x\rVert_0 = \#\{i \mid x[i] \neq 0\} = \lim_{p\to 0^+} \left(\sum_{i=1}^{m} |x[i]|^p\right),$$
where $x[i]$ is the $i$th coordinate of $x$ in $\mathbb{R}^m$ and $p \geq 1$</p>

答: $\lim_{p \to 0^{+}}$ indicates that the limit is meant to be taken only from the positive direction; it's a one-sided limit. </p>

Title: Meaning of $\lim_{p\to 0^+}$


问:
Rudin Dedekind Cuts - Define Negative Multiple - Step 8
I am following the construction of the reals using Dedekind Cuts in Rudin's Chapter 1 Appendix.  I would appreciate some review on one of the last steps, Step 8, where he looks at multiplication of the reals.

I am trying to prove property (b), namely that $r^*s^*=(rs)^*$.  Here is a good proof for when $r, s>0$.  But what about when $r>0, s<0$?

A few relevant definitions:  (Note that here, so far, $\mathbb{R}$ is the set of Dedekind cuts and $\alpha, \beta$ are cuts.)

Fix $\alpha \in \mathbb{R}$.  Let $-\alpha$ be the set of all $p\in\mathbb{Q}$ with the following property: There exists $q\in\mathbb{Q}$ with $q>0$ such that $-p-q\notin \alpha$.

$\mathbb{R}^+$ is the set of all $\alpha \in \mathbb{R}$ with $\alpha > 0^*$.  ($0^*$ is the set all negative rational numbers.)  Then if $\alpha, \beta \in \mathbb{R}^+$, we define $\alpha\beta$ to be the set of all $p$ such that $p\leq rs$ for some choice of $r\in\alpha, s\in\beta$ with $r,s>0$.

If $\alpha>0$ and $\beta <0$, $\alpha\beta = -(\alpha(-\beta))$.

We associate with each $r \in \mathbb{Q}$ the set $r^*$ which consists of all $p \in \mathbb{Q}$ such that $p<r$.  These $r^*$ are cuts so the above apply.

Here is what I have.  It relies heavily on manipulating definitions so if someone could check this I would be grateful.  Thank you.

First, I show $r^*s^*\subset (rs)^*$:

Suppose $p\in r^*s^*$.

By (3), $r^*s^*=(-(r^*(-s)^*))$.

Thus, by (1), there exists $q>0$ such that $-p-q\notin r^*(-s)^*$.

By (2), this means that for all $u, v$ with $u\in r^*, v\in (-s)^*$, and $u, v >0$, we have $-p-q>uv$.

By (4), for all such $u$ and $v$, we know $u<r$ and $v<-s$, so this means $-p-q\geq r(-s)$.

Since $q>0$, this means $-p> r(-s)$

This means $p < rs$.

And by (4), this means $p\in (rs)^*$.

Next I show $(rs)^* \subset r^*s^*$:

Suppose $p \in (rs)^*$.  Then by (4), $p<rs$.

Since $r>0, s<0$, we have $rs < 0$, and so $-p>r(-s)$.

Then there exists $q > 0$ such that $-p-q > r(-s)$.

By (4), for all $u \in r^*, v\in (-s)^*$, we have $u<r$ and $v<-s$.  Then for all $u \in r^*$ and $v \in (-s)^*$ with $u, v >0$, we have $-p-q>uv$.

Therefore, by (2), $-p-q\notin r^*(-s)^*$.

Therefore, by (1), $p \in -(r^*(-s)^*)$.

By (3), $-(r^*(-s)^*) = r^*s^*$, and so $p \in r^*s^*$.

EDIT: I have noticed one issue with the above.  Twice I wrote that by (3), $-(r^*(-s)^*) = r^*s^*$.  However, a strict application of the definition of the product of a positive and negative cut would be that $r^*s^* = -(r^*(-(s^*)))$ (the difference is subtle, but there).

However, it is not hard to show that for any rational cut, $-(r^*) = (-r)^*$.

Suppose $p\in (-r)^*$.  Then $p\in\mathbb{Q}$ and $p<-r$.  Then there exists $q\in\mathbb{Q}$ with $q>0$ such that $p+q<-r$.  Then $-p-q>r$, so $-p-q\notin r^*$, and by definition $p\in -(r^*)$.

Next, suppose $p\in -(r^*)$.  Then by definition there exists $q\in\mathbb{Q}$ with $q>0$ such that $-p-q\notin r^*$, i.e. $-p-q\geq r$.  Then $-p>r$, so $p<-r$ and $p \in (-r)^*$.

答: Based on my edit and Brian's comment above, it seems like what is in the post body is correct.</p>


问:
Vector Calculus-Field Lines vs Streamlines
I am an engineer and I have a question related to vector calculus. I find hard to visualise how the field lines of the velocity vector differ from streamlines. I am given that $\bf{V=ui+vj}$.

Also I am given than $\bf{u = \partial ψ/ \partial y}$ and $\bf{v = -\partial ψ/ \partial x.}$ (supposed to be partial derivatives) $\bf{[1]}$

Given the above, it can be shown that for constant $\bf{ψ , \ dψ}$ is equal to 0, hence by using $\bf{dψ=(\partial ψ/ \partial x)dx + (\partial ψ/ \partial y)dy}$ and $\bf{[1],}$
I get that $\bf{dy/dx = v/u. \ [2]}$

What is the physical significance of $\bf{[2]}$? I think it shows that $\bf{ψ}$ corresponds to streamline's equation, but I am not sure "why?".

Also, once I am given the equation of the velocity of the flow, how can I make the corresponding stream line plot?

答:
If I understand your terminology right, the velocity field lines in the steady flow (the lines such that at each point of the line the corresponding velocity vector is tangent to the line) are exactly the same as streamlines (the lines over which a small inertialess particle carried by the flow would travel), so there is no difference except for the name.

As to the second question, what your computation really shows for that particular choice of the velocity field is that
$$
(d\psi)(V)=\frac{\partial\psi}{\partial x}u+\frac{\partial\psi}{\partial y}v=
\frac{\partial\psi}{\partial x}\frac{\partial\psi}{\partial y}+\frac{\partial\psi}{\partial y}\left(-\frac{\partial\psi}{\partial x}\right)=0\,
$$
so $\psi$ is constant along the field lines of $V$, or, if you prefer, along the streamlines, which means that the streamlines are level lines of $\psi$ (lines given by the equation $\psi=const$). Level lines of a known function are usually easier to compute, visualize, and investigate than the field lines of an arbitrary velocity field, so knowing that $\psi$ when it exists really does give you some advantage.


问:
Prove $\prod_{k=0}^{n-1}(a+b\varepsilon^2_k)=a^n+b^n$.
Let $n$ be an odd positive integer and $\varepsilon_0$, $\varepsilon_1$,..., $\varepsilon_{n-1}$ the complex roots of unity of order $n$. Prove that $$\prod_{k=0}^{n-1}(a+b\varepsilon^2_k)=a^n+b^n,$$ for all complex numbers $a$ and $b$.

I believe this problem comes from the 2000 RMO. I'm not too sure where to really start the problem, any hints?

答:
If $b=0,$ the equation holds. Else, let $c=a/b$ and divide by $b^n$ so that we have to show $\prod\limits_{k=0}^{n-1} (c+\epsilon_k^2) = c^n + 1.$ The LHS is simply $\prod\limits_{k=0}^{n-1} (c+\epsilon_k)$ because the $\epsilon_k^2$ form a permutation of the roots of unity as you observed. Now do you see how to finish off with Vieta's Theorem?

Extra hint: What are the roots of $(x-c)^n - 1$?


问:
Kronecker Delta tensor proof
How would I go about proving $\delta_{ij}$ is a rank 2 tensor? Help appreciated

EDIT: Tensor being defined by the following:

$$x'_i = L_{ij} x_j, x_i = L_{ji} x'_j \tag{7.7}$$

A Cartestian Tensor $T_{ij...l}$ of rank $r$ has $r$ indices and transforms under the special orthogonal transformation $(7.7)$ as

$T^{'}_{ij...l}$($x^{'}$) = $L_{ip}$$L_{jq}$...$L_{ls}$$T_{pq...s}$($L^{-1}$$x^{'}$)

答:
The definition of $\delta_{ij}$ is independent of the basis chosen. If $R$ is a rotation matrix, then since $R^\intercal R=I$, we get
$$R_{ip}R_{jq}\delta_{pq}=R_{ip}R_{jp}=\delta_{ij}=\delta_{ij}',$$
which is precisely the transformation rule for a rank $2$ tensor.

Title: Kronecker Delta tensor proof


问:
Find the determinant of this matrix
$k_1,k_2,...,k_n$ are non-negative integers.
Let $M$ be an $n\times n$ matrix with entries:
$a_{i,1} = t^{k_i}, a_{i,j+1} = \frac{da_{i,j}}{dt}$ where $a_{i,j}$ is the element of $i$-th row and $j$-th column.

Prove that there exist $C$ and $r$ such that
$\det(M) = Ct^r$

I was able to reduce it to:

$\det(M) = t^{(k_1+...+k_n) - {n(n-1)\over 2}}$
$\cdot \det\begin{pmatrix}
 1& k_{1} & k_{1}(k_{1}-1)  & ... & {k_1!\over (k_1-n+1)!}\\
 1& k_{2} & k_{2}(k_{2}-1) & \vdots & \vdots\\
 \vdots& \vdots & \vdots &  \ddots \\
 1& k_{n} & k_{3} & ... & {k_n!\over (k_n-n+1)!}
\end{pmatrix}$

I have  already determined $r$, the problem is to calculate  determinant of this thing, any tips ?

答: It seems that taking $C$ to be the above determinant and $r$ to be exponent of $t$ in your expression solves your problem.

Title: Find the determinant of this matrix


问:
Understanding Slope Better
Recently, I have realized how much I have taken for granted in understanding the slope of a line and the slope of a curve in general. With that said, I wanted to clear up my understanding to make sure I am on the right track.

After reading this question: "Understanding the slope of a line as a rate of change", I see now how the slope of the line $y = f(x): = mx + b$ (where $f: \mathbb{R} \to \mathbb{R}$) is the rate of change of $y$ with respect to $x$. However, where I get a little confused is this: since the value $y$ of the function changes by an amount $\Delta y$ for every $\Delta  x$ change in $x$, therefore $y$ changes by an amount $ m =\frac{\Delta y}{\Delta x}$. So, if we were to study the line $y = \frac{2}{3}x$, does that mean the slope of the line can be interpreted as "two units of $y$ every three units of $x$"? With more relatable units, if $y$ is measured in meters and $x$ is measured in seconds, would the rate of change of a particle traveling along this line be read as "two meters every three seconds"?

This seems right, however in general for any differentiable curve $g: \mathbb{R} \to \mathbb{R}$ that may not be a line i've seen for example that $g'(3) = \frac{2}{5} $ (given in meters/seconds) is read as "two-fifths meters per second"; simply because the derivative is the rate of change at a point (or at an instant).

Is my understanding correct at all?

答:
So, if we were to study the line $y = \frac{2}{3}x$, does that mean the rate of change of the line can be interpreted as "two units of  every three units of "? With more relatable units, if  is measured in meters and  is measured in seconds, would the rate of change of a particle traveling along this line would be read as "two meters every three seconds"?

Yes, you are exactly right.  And if it helps you build a more physical intuition, the units of the slope are the units of $y$ divided by the units of $x$.  So the number of hot-dogs Takeru Kobayashi eats in a hot-dog eating competition can be modeled with the line $y = m x$, where $y$ has units of hot dogs, $x$ has units of minutes, and $m \simeq 6.6$ hot dogs / minute.  Very impressive.

As for more general differential curves on any domain: a slope is a slope.  The fact that a derivative exists guarantees that $g(x)$ can be approximated as a line in the neighborhood centered at $x = 3$.  It is possible to overthink this.  Why not eat a hot dog?

Title: Understanding Slope Better


问:
Show $1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$.
Show that $$1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$$
where $|q|<1$ (q can be complex number).

The hint is to convert the left side to a double series.

One can see that the nominators on the left side is double, triple, ... of those on the right side, while denominators on the right side are square of those on the left side. Both sides have 1, which seems redundant.

I'm not sure how to proceed from these observations. Any advice will be helpful. I can try to solve it with some suggestion and if I still can't solve it I will then explain what puzzles me. If anyone gives the answer I guess I will look at only part of it and work from that.

Edits:

I post my answer inspired by the comments and some other thoughts about the question below.

Is there any other ways to solve it?

答:
These series are not equal for all $|q|<1$. Here's an explanation.

First, let
$$A(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n},\qquad |bq|<1$$
and $$B(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2},\qquad |bq|<1.$$
Let
$$F_1(q)=A(-2,-1;-q)=\frac{2q}{1-q}+\frac{4q}{1+q^2}+\frac{8q^3}{1-q^3}+...$$
and $$F_2(q)=B(-1,-1;-q)=\frac{q}{(1-q)^2}+\frac{q^2}{(1+q^2)^2}+\frac{q^3}{(1-q^3)^2}+...$$

You assert that $$1+4F_1(q)=1+8F_2(q).\tag{*}$$ This is incorrect. Below is an explanation.

First of all, define the functions $\phi_n,\sigma_n:\Bbb R\times\Bbb R\to \Bbb R$ to satisfy
$$\begin{align}
A(a,b;q)&=\sum_{n\ge1}\phi_n(a,b)q^n\\
B(a,b;q)&=\sum_{n\ge1}\sigma_n(a,b)q^n.
\end{align}$$
Then we can see that
$$\begin{align}
A(a,b;q)&=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n}\\
&=\frac1b\sum_{n\ge1}a^n\frac{bq^n}{1-bq^n}\\
&=\frac1b\sum_{n\ge1}a^n\sum_{k\ge1}b^kq^{nk}\\
&=\frac1b\sum_{n\ge1}\sum_{k\ge1}a^nb^kq^{nk}\\
&=\frac1b\sum_{m\ge1}q^m\sum_{nk=m}a^nb^k\\
&=\sum_{m\ge1}q^m\sum_{d|n}a^db^{m/d-1}.
\end{align}$$
Thus $$\phi_n(a,b)=\sum_{d|n}a^db^{n/d-1}.$$
Then, preform the index shift $d=n/j$ for $j|n$:
$$\phi_n(a,b)=\sum_{j|n}a^{n/j}b^{j-1}=\frac{a}{b}\sum_{j|n}b^ja^{n/j-1}=\frac{a}{b}\phi_n(b,a).\tag1$$

Then take $\partial_b=\frac{\partial}{\partial b}$ of $A(a/q,b;q)$:
$$\begin{align}
\partial_b A(a/q,b;q)&=\sum_{n\ge1}\partial_b\frac{a^n}{1-bq^n}\\
&=\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2}\\
&=B(a,b;q).
\end{align}$$
Then from $(1)$, we have that $A(a,b;q)=\frac{a}{b}A(b,a;q)$, so that
$$\begin{align}
A(a/q,b;q)&=\frac{a}{bq}A(b,a/q;q)\\
&=a\sum_{n\ge1}\frac{(bq)^{n-1}}{1-aq^{n-1}}\\
&=\frac{a}{1-a}+a\sum_{n\ge1}\frac{(bq)^{n}}{1-aq^{n}}\\
&=\frac{a}{1-a}+bA(a,b;q).
\end{align}$$
Thus,
$$\begin{align}
B(a,b;q)&=\partial_b A(a/q,b;q)\\
&=\partial_b \left\{\frac{a}{1-a}+bA(a,b;q)\right\}\\
&=\partial_b bA(a,b;q)\\
&=A(a,b;q)+b\partial_b A(a,b;q)\\
&=\sum_{n\ge1}q^n\phi_n(a,b)+\sum_{n\ge1}q^nb\partial_b \phi_n(a,b)\\
&=\sum_{n\ge1}q^n\left\{\sum_{d|n}a^db^{n/d-1}+b\sum_{d|n}\left(\tfrac{n}{d}-1\right)a^db^{n/d-2}\right\}\\
&=\sum_{n\ge1}q^n\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1}.
\end{align}$$
Hence $$\sigma_n(a,b)=\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1}.$$

Since $F_1(q)=A(-2,-1;-q)$ and $F_2(q)=B(-1,-1;-q)$, we have
$$F_1(q)=\sum_{n\ge1}q^n\cdot\left((-1)^n\phi_n(-2,-1)\right)=\sum_{n\ge1}p_1(n)q^n,$$
and
$$F_2(q)=\sum_{n\ge1}q^n\cdot\left((-1)^n\sigma_n(-1,-1)\right)=\sum_{n\ge1}p_2(n)q^n.$$
Then your claim $(*)$ is equivalent to $p_1(n)=2p_2(n)$, or
$$\sum_{d|n}(-1)^{d+n/d-1}2^d=2\sum_{d|n}(-1)^{d+n/d-1}\frac{n}{d}.\tag{*'}$$
So $(*)$ is true if and only if $(*')$ is true.

To check this, let
$$f(n)=\sum_{d|n}(-1)^{d+n/d-1}(2^d-2n/d).$$
$(*)$ Is true if and only if $f(n)$ is identical to $0$.

It is quite easy to show that $f(3)=-(2-6)-(2^3-2)=-2$, so $f(n)$ is not identical to $0$ and your claim is false.


问:
Non puntual sequences are not an equicontinous family
Given a sequence $u_{n}$ of (non constant) functions in  $\mathcal{C}(\overline{B},\mathbb{R})$ with $\overline{B}$ the closed unit ball in $\mathbb{R}^{d}$. Then, is it true that if $u_{n}$ does not converge point-wise then it can't be equicontinous?

答: Equicontinuity and boundedness imply a (uniformly) convergent subsequence. But equicontinuity does not imply pointwise convergence: If $a_n$ is a non-convergent sequence of reals, then the functions $f_n(x)=x+a_n$ are equicontinuous but do not converge pointwise.


问:
What is the difference between a partition and a subinterval?
For example the norm of a partition is the widest subinterval</p>

答:
A partition of an interval is a division of the interval into subintervals; the subintervals are the ‘pieces’ of the partition, so to speak, the subintervals of which the partition is composed.

The norm of a partition is not the sidest subinterval of the partition: it is the width of the widest subinterval. Consider, for instance, the partition of $[0,1]$ into subintervals $\left[0,\frac13\right],\left[\frac13,\frac12\right]$, and $\left[\frac12,1\right]$. The partition is this collection of subintervals. The lengths of the subintervals are $\frac13,\frac16$, and $\frac12$, so the norm of the partition is $\frac12$; it is *not the interval $\left[\frac12,1\right]$.


问:
Problem solving earthquake problem magnitude logarithims
I need help solving a simlar type equation to this..... this one was easy though...
An earthquake off the coast of Vancouver Island was measured at 8.9 on the Richter Scale
and an earthquake off the coast of Alaska was measured at 6.5. How many times more intense, to
the nearest whole number, was the earthquake off the coast of Vancouver Island than the one off the coast of Alaska?

my solution for this problem was ((10)^8.9/*10^6.5)
251 times more intense

the is the one I' am having problems with.  A major earthquake of magnitude 8.3 is 120 times as intense as a minor earthquake. Determine
     the magnitude, to the nearest tenth, of the minor earthquake.

I think  I should do 10^8.3 / 10^x =12 and solve fore it?  but it has to be a lograthmatic equation.

Body:
Use subtraction law of exponents:

$$\frac{10^a}{10^b} = 10^{a-b}$$

so that

$$\frac{10^{8.3}}{10^x} = 10^{8.3-x} = 120$$

Therefore

$$8.3 - x = \log_{10}{120} = 2+\log_{10}{1.2} \approx 2.1$$

$$\implies x \approx 6.2$$


问:
How many critical point of $y = 3^{2f(x) + 4x -3}- 2^{-f(x) - 2x+3}$ with given graph of $y = f(x)$
Given $y = f(x)$ with minima is $-3$ at $x= 0$ and maxima is $3$ at $x = -2$ has graph as a conic curve.

How many critical point of $y = g(x) = 3^{\displaystyle{2f(x) + 4x -3}} - 2^{\displaystyle{-f(x) - 2x+3}}$ ?

Firsts step is derivative?

$g'(x) = [2f'(x) + 4]3^{\displaystyle{2f(x) + 4x -3}}.\ln(2f(x) + 4x +3)+ [f'(x) - 2]2^{\displaystyle{-f(x) - 2x+3}}.\ln(-f(x) - 2x +3)$

Then how to find solutions from $g'(x) = 0$ and given information?

I am confused, please help me!!!

Body:
You had the derivatives wrong, recall:

$$ a^{t(x)} = y \Rightarrow y' = a^{t(x)} \cdot t'(x) \cdot \ln(a) ~~ a \in \mathbb{R^+} $$

And so:

$$ g'(x) = 3^{2f(x)+4x-3} ( 2f'(x) + 4) \ln(3) - 2^{-f(x) -2x+3} (-f'(x) -2) \ln(2) \\ g'(x) = 3^{2f(x)+4x-3} (f'(x) + 2) 2\ln(3) + 2^{-f(x) -2x+3} (f'(x)+ 2) \ln(2) \\ (f'(x) +2) \left [ 3^{2f(x)+4x-3} 2 \ln(3) + 2^{-f(x) -2x+3} \ln(2) \right ] = 0$$

Now you can see how many possible values you have, because $$3^{2f(x)+4x-3} 2 \ln(3) + 2^{-f(x) -2x+3} \ln(2) > 0 $$  For all real $x$.


问:
Inverse Fourier Transform of $1/k^2$ in $\mathbb{R}^N $
This comes up in the context of finding the Green's function of Poisson's equation
for $\mathbf{x} \in \mathbb{R}^n $ $$ \nabla^2 G(\mathbf{x}) = \delta(\mathbf{x})$$

Attempt by using Fourier Transforms
$$ \nabla^2 G(\mathbf{x}) = \delta(\mathbf{x}) \Rightarrow -\lVert k\rVert^2 \tilde{G}(\mathbf{k})  = 1$$
$$ G(\mathbf{x}) = -\mathcal{F}^{-1}\left[ \lVert k\rVert^{-2} \right]$$
I reasoned that we could use N-dimensional spherical coordinates and be left with an integral integral over one angle in the plane between $\mathbf{k}$ and $\mathbf{x}$ and a radial integral which would have an element $k^{n-1}\,dk $ where $n$ is the number of dimensions.
$$G\left(\mathbf{x}\right) = \frac{-1}{(2\pi)^n}\int k^{n-1}F(\phi_1,\dots,\phi_{n-3})\sin(\phi_{n-2})\mathop{dk d\phi_1 \dots d\phi_{n-2}d\phi_{n-1}} \frac{e^{ikr\cos \phi_{n-2}}}{k^2} $$
with $u = \cos \phi_{n-2}$
$$G\left(\mathbf{x}\right) = \frac{-1}{(2\pi)^n}\int F(\phi_1,\dots,\phi_{n-3})\mathop{d\phi_1 \dots d\phi_{n-3}d\phi_{n-1} }\int_{-1}^{1} du\int_0^\infty\mathop{dk}  k^{n-3} e^{ikru}$$
Maybe I'm not thinking clearly but the last two integrals don't seem like they are going to converge to anything nice. I don't know if I've made a mistake or I need to take these integrals in a particular order. I know I can show the Green's function is proportional to $|| r||^\alpha $for $n>2$ by using a test function, but the the idea here is to calculate it directly without assuming that much about its form.

答:
Although this can be proven using the fact that the Fourier transform is homogeneous, I think it's quite interesting to see a more constructive proof. Also, let me write this in parts, so that the details do not obscure the overall proof. Also, I will provide 2 different proofs, corresponding to different regularizations, but ultimately give the same answer. The first method is easier intuitively but harder to rigorously regularize, while the 2nd method is easier to rigorously prove, but requires a special trick.

Method 1 (Contour integrals)

Non-rigorous sketch. Notice that by rotating $k$, we see that the inverse Fourier transform is given by
\begin{equation}
\int \frac{1}{k^2} e^{ikx} \frac{dk}{(2\pi)^d}=\int \frac{1}{k^2}e^{ik_1 r} \frac{dk}{(2\pi)^d}, \quad r=|x|
\end{equation}
Let us write $k=(p,q)$ where $p=k_1$ and $q=(k_2,...,k_d) \in \mathbb{R}^{d-1}$. Then we have
$$
\int \frac{1}{p^2+q^2} e^{ipr} \frac{dpdq}{(2\pi)^d}= \int \frac{dq}{(2\pi)^{d-1}}\int \frac{1}{p^2+q^2} e^{ipr} \frac{dp}{2\pi}
$$
(This is a non-rigorous application of the Fubini theorem since the integrand is not absolutely integrable). The inner integral can be computed using contour integrals so that for all $q\ne 0$, we have
$$
\int \frac{1}{p^2+q^2} e^{ipr} \frac{dp}{2\pi}= \frac{1}{2|q|} e^{-|q|r}
$$
Plug this back in and using spherical coordinates, we have
\begin{align}
\int_0^\infty \frac{\rho^{d-2} d\rho}{(2\pi)^{d-1}} \frac{1}{2\rho } e^{-\rho r}\cdot S_{d-2} &= \frac{1}{r^{d-2}}\cdot\frac{1}{2(2\pi)^{d-1}} S_{d-2}\int _0^\infty z^{d-3}e^{-z}dz \\
&= \frac{1}{r^{d-2}}\cdot\frac{1}{2(2\pi)^{d-1}} S_{d-2} \cdot(d-3)! \\
\end{align}
where $S_{d-2}$ is the surface area of the unit sphere in $d-1$-dimensions. Using the Legendre duplication formula, you can check that this is exactly that given in Wiki.

Rigorous Proof. Let me only care about $d>2$ so that $1/k^2$ is locally integrable in $\mathbb{R}^d$. Then the inverse Fourier transform of $1/k^2$ as a tempered distribution is given by
$$
\int \frac{1}{k^2} \check{\varphi}(k) \frac{dk}{(2\pi)^d} = \int \frac{1}{k^2} \frac{dk}{(2\pi)^d}\int\varphi(x)e^{-ikx} \frac{dk}{(2\pi)^d}
$$

where $\check{\varphi}(x)$ is the inverse Fourier transform of $\varphi\in S$ in the Schwartz space.

Now the hard part is choosing the right regularization so that we can apply Fubini's theorem. Let $n\in \mathbb{N}$ be such that $n >(d-2)/2$. Then we see that
$$
\frac{1}{k^2} \left(\frac{\Lambda^2}{k^2+\Lambda^2}\right)^n
$$
is integrable in $\mathbb{R}^d$ and $\to 1/k^2$ in $S'$ as $\Lambda \to \infty$. Since the Fourier transform is continuous on $S'$, we see that we can just compute the Fourier transform of the regularization and take $\Lambda \to \infty$. Now consider
$$
\int \frac{1}{k^2} \left(\frac{\Lambda^2}{k^2+\Lambda^2}\right)^n e^{ikx} \frac{dk}{(2\pi)^d} = \int \frac{1}{k^2} \left(\frac{\Lambda^2}{k^2+\Lambda^2}\right)^n e^{ik_1 r} \frac{dk}{(2\pi)^d}, \quad r=|x|
$$
where we rotated $k$ so that we can replace $kx$ with $k_1r$. Now for simplicity, let $k=(p,q)$ where $p=k_1$ and $q=(k_2,...,k_d)$. Then we see that
$$
\int \frac{1}{k^2} \left(\frac{\Lambda^2}{k^2+\Lambda^2}\right)^n e^{ikx} \frac{dk}{(2\pi)^d} = \int \frac{dq}{(2\pi)^{d-1}} \int \frac{dp}{2\pi} \frac{1}{p^2+q^2} \left( \frac{\Lambda^2}{p^2+q^2+\Lambda^2}\right)^n e^{ipr}, \quad r=|x|
$$
Using contour integrals, we see that
$$
\int \frac{dp}{2\pi} \frac{1}{p^2+q^2} \left( \frac{\Lambda^2}{p^2+q^2+\Lambda^2}\right)^n e^{ipr} = \frac{C_n \Lambda^{2n-2}}{(q^2+\Lambda^2)^{(2n-1)/2}} e^{-r\sqrt{q^2+\Lambda^2}} +\frac{1}{2|q|} e^{-|q|r}
$$
Now the first term $\to 0$ as $\Lambda \to\infty$ and thus by dominated convergence, the statement follows as in the non-rigorous sketch.

Method 2 (Gaussian decay)

Let me only care about $d>2$ so that $1/k^2$ is locally integrable. Notice that $e^{-\epsilon k^2}/k^2 \to 1/k^2$ as tempered distributions. Hence, we only need to compute the Fourier transform of $e^{-\epsilon k^2}/k^2$ and take $\epsilon \to 0$, i.e.,
\begin{align}
\int \frac{e^{ikx}}{k^2} e^{-\epsilon k^2} \frac{dk}{(2\pi)^d} &= \int \frac{dk}{(2\pi)^d} e^{ikx} \int_\epsilon^\infty e^{-tk^2} dt \\
&= \int_\epsilon^\infty dt \int \frac{dk}{(2\pi)^d} e^{-tk^2+ikx}\\
&= \frac{1}{(4\pi)^{d/2}}\int_\epsilon^\infty t^{-d/2} e^{-x^2/4t} dt \\
&= \frac{1}{|x|^{d-2}} \frac{1}{2^2 \pi^{d/2}} \int_0^{x^2/4\epsilon} e^{-y} y^{d/2-2} dy \\
&= \frac{1}{|x|^{d-2}} \frac{1}{2^2 \pi^{d/2}} \Gamma \left(\frac{d}{2}-1 \right) , \quad \epsilon \to 0
\end{align}