问:
Writing a symmetric matrix in the form $BDB'$ where $D$ is a diagonal matrix and $B'$ is transpose of $B$.
I have been given a $3×3$ symmetric matrix and I am required to write this in the form $BDB'$ where $D$ is a diagonal  matrix  and $B'$ is transpose of $B$.

From my knowledge of eigen vectors and diagonalization of matrices, I know that a diagonalizable matrix $A$ can be expressed as $A= PDP^-1$ where $P$ is the invertible matrix formed by arranging the eigen vectors of $A$ as column vectors and $D$ is the diagonal matrix formed by taking the eigen values of $A$ as diagonal elements.

But the above question asks me to have transpose matrix in place of inverse matrix. How can I get that? Please suggest.

答:
(Real) symmetric matrices are automatically diagonalisable with real eigenvalues. This means it should be possible to find a basis of eigenvectors, which we do by smooshing together bases for the eigenspaces (e.g. if you have a two-dimensional eigenspace, then you simply pick two linearly independent eigenvectors from that eigenspace when forming your basis of eigenvectors).

Moreover, symmetric matrices have orthogonal eigenspaces, meaning that eigenvectors for distinct eigenvectors will automatically be orthogonal. If we take orthonormal bases for each eigenspace (if the eigenspace is $1$-dimensional, then just take a unit-length eigenvector), then the basis we get from smooshing these vectors together is an orthonormal basis for $\Bbb{R}^3$.

Now, if you diagonalise with respect to an orthonormal basis, the change-of-basis matrices will be orthogonal, meaning that the inverse of the change-of-basis matrix will be the same thing as the transpose of the change-of-basis matrix.

So, basically, you just need to perform the usual process of diagonalisation, just ensuring that your basis of eigenvectors is orthonormal.


问:
Construct a multiplication table for $\mathbb{Z}_2[i]$.
Construct a multiplication table for $$\mathbb{Z}_2[i]=\{a+bi|a,b \in \mathbb{Z}_2\}$$.

I have figured this table out for the most part. I know that each side has $0,1,i,1+i$, but $(1+i)(i)=i-1$, and what would that become in $\mathbb{Z}_2[i]$? I have the same question for $(1+i)(1+i)$, I don't know how to reduce these in terms of the mod.

答:
As others also mentioned , we can generate the field ${Z_2[x]}$ with irreducible polynomial ${x^2+x+1}$ ,and the elements are ${\{0,1,x,x+1\}}$  , here ${x^2=x+1}$,and  ${-1=1}$ , ${0=(1+i)^2 = i+i}$

Here ${(x+1)^{-1}=x}$, cayley table may looks like..

ContentLicense: CC BY-SA 4.0


问:
Prove that this function define a norm on $\mathbb{R}^2$.
Let $\| \;\|:\mathbb{R}^2\rightarrow\mathbb{R}$ be a function such that
$ \| (x,y)\|=\sqrt{|x|^2+|y|^2}$ for all $(x,y)\in\mathbb{R}^2$.

I need to show that $\|\;\|$ defines a norm. I would like a hint in order to prove that $\parallel \; \parallel$ satisfies the triangle inequality:

$ \sqrt{|x_1+x_2|^2+|y_1+y_2|^2}\leq \sqrt{|x_1|^2+|y_1|^2}+\sqrt{|x_2|^2+|y_2|^2}$

Thanks!

答:
You can do it directly, for $\mathbb{R}^2$; it becomes really complicated in higher dimension, where the Cauchy-Schwarz inequality is the best approach.

Square your inequality: you can because both terms are non negative. You can also remove the absolute values, because $(-a)^2=a^2$. Thus your inequality is equivalent to
$$
(x_1+x_2)^2+(y_1+y_2)^2\leq
x_1^2+y_1^2+x_2^2+y_2^2
+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}.
$$
Expand the terms and you get, after rearranging terms,
$$
2(x_1x_2+y_1y_2)\le+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}
$$
that's obviously satisfied when $x_1x_2+y_1y_2<0$. So we proceed under the additional assumption that $x_1x_2+y_1y_2\ge0$, so the inequality is equivalent to the one we obtain by squaring (first remove the common factor $2$):
$$
(x_1x_2+y_1y_2)^2\le(x_1^2+y_1^2)(x_2^2+y_2^2)
$$
Expand the terms
$$
x_1^2x_2^2+2x_1x_2y_1y_2+y_1^2y_2^2\le
x_1^2x_2^2+x_1^2y_2^2+x_2^2y_1^2+y_1^2y_2^2
$$
and transport all terms to the right hand side
$$
0\le x_1^2y_2^2-2x_1x_2y_1y_2+x_2^2y_1^2
$$
which can be written
$$
0\le (x_1y_2-x_2y_1)^2
$$
that's certainly true.

ContentLicense: CC BY-SA 3.0


问:
Combinations of three mutually perpendicular edges of a cube.
I need help in determining the possible combinations of mutually perpendicular edges of a cube, taken three at a time. Two examples are shown in the image attached. enter image description here</p>

答:
A cube has $3 \cdot 4$ edges, so each edge can be chosen from one set of four.  That means there are $4 \cdot 4 \cdot 4 = 4^3 = 64$ unique combinations.

This assumes that each set of edges is considered as an unordered set, that it does not matter which of the three edges corresponds to $x$, $y$, or $z$.  Each set has $3 \cdot 2 \cdot 1 = 6$ such combinations, so if the order does matter, there are $64 \cdot 6 = 384$ combinations total.

If we consider each set of unordered edges, but exclude those configurations that are just rotated versions of existing sets, we are left with just six unique edge configurations:

(This has been verified with a program that examines all 64 unique combinations, dropping those configurations that are just rotated versions of others, using 90° rotations around each axis.)

Here is a Python 3 program, generate.py, you can use to generate and/or verify the sets:

```<code>#!/usr/bin/python3
# SPDX-License-Identifier: CC0-1.0
# -*- coding: utf-8 -*-

# Edges are numbered thus:
#   0: (0,0,0)-(0,0,1)  z
#   1: (0,0,0)-(0,1,0)  y
#   2: (0,0,0)-(1,0,0)  x
#   3: (0,0,1)-(0,1,1)  y
#   4: (0,0,1)-(1,0,1)  x
#   5: (0,1,0)-(0,1,1)  z
#   6: (0,1,0)-(1,1,0)  x
#   7: (0,1,1)-(1,1,1)  x
#   8: (1,0,0)-(1,0,1)  z
#   9: (1,0,0)-(1,1,0)  y
#  10: (1,0,1)-(1,1,1)  y
#  11: (1,1,0)-(1,1,1)  z

class Mapping(tuple):

    def __new__(cls, mapping=None):
        d = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ]
        if isinstance(mapping, dict):
            for k,v in mapping.items():
                d[k] = v
        elif mapping is not None:
            raise TypeError("Unsupported mapping type (%s)" % type(mapping))
        return tuple.__new__(cls, d)

    def __init__(self, *args, **kwargs):
        return

    def apply(self, mapping):
        d = list(self)
        if isinstance(mapping, dict):
            for k,v in mapping.items():
                d[k] = self[v]
        elif isinstance(mapping, (list, tuple)):
            if len(mapping) != 12:
                raise ValueError("Mapping needs 12 elements, not %d" % len(mapping))
            for k in range(0, 12):
                d[k] = self[mapping[k]]
        else:
            raise TypeError("Unsupported mapping type (%s)" % type(mapping))
        return tuple.__new__(self.__class__, d)

class Triplet(tuple):

    def __new__(cls, *args):
        if len(args) == 1:
            if isinstance(args[0], int):
                args = [ args[0] // 256, (args[0] // 16) & 15, args[0] & 15 ]
            elif isinstance(args[0], (list, tuple)):
                args = args[0]
            else:
                raise TypeError("Invalid triplet definition (%s)" % type(args[0]))
        if len(args) != 3:
            raise ValueError("Triplet needs exactly three arguments, not %d" % len(args))
        return tuple.__new__(cls, sorted(args))

    def __init__(self, *args, **kwargs):
        return

    @property
    def key(self):
        return 256*self[0] + 16*self[1] + self[2]

    def apply(self, mapping):
        return tuple.__new__(self.__class__, sorted([mapping[self[0]], mapping[self[1]], mapping[self[2]]]))

def SVG(*triplets):
    w = 8 + 80*len(triplets)
    out = [ '<?xml version="1.0" encoding="UTF-8" standalone="no"?>\n',
            '<svg version="1.1" viewBox="0 0 %d 136">\n' % w,
            '<rect x="0" y="0" width="%d" height="136" fill="#ffffff" stroke="none"/>\n' % w ]

    def line(x0, x1, y1, x2, y2, t, selected):
        if selected:
            return '<path d="M %d,%d %d,%d" fill="none" stroke="#cc0000" stroke-width="2" stroke-linecap="round"/>\n' % (x0+x1, y1, x0+x2, y2)
        else:
            return '<path d="M %d,%d %d,%d" fill="none" stroke="#000000" stroke-width="%s" stroke-linecap="round"/>\n' % (x0+x1, y1, x0+x2, y2, t)

    for i in range(0, len(triplets)):
        x = 80*i
        out.append('<path d="M %d,80 %d,128 %d,104 %d,56 Z" fill="#eff7ff" stroke="none"/>\n' % (x+8, x+32, x+80, x+56))
        out.append('<path d="M %d,32 %d,80 %d,56 %d,8 Z" fill="#dfefff" stroke="none"/>\n' % (x+8, x+8, x+56, x+56))
        out.append('<path d="M %d,56 %d,104 %d,56 %d,8 Z" fill="#bfdfff" stroke="none"/>\n' % (x+56, x+80, x+80, x+56))
        t = Triplet(triplets[i])
        out.append(line(x, 56,8, 56,56, 1, 0 in t))
        if (2 in t):
            out.append('<path d="M %d,80 %d.6,71.2 M %d.8,69.6 %d,56" fill="none" stroke="#cc0000" stroke-width="2" stroke-linecap="round"/>\n' % (x+8, x+25, x+28, x+56))
        else:
            out.append('<path d="M %d,80 %d,71 M %d.4,69.8 %d,56" fill="none" stroke="#000000" stroke-width="1" stroke-linecap="round"/>\n' % (x+8, x+26, x+28, x+56))
        if (1 in t):
            out.append('<path d="M %d,56 %d.2,64.4 M %d.4,66.8 %d,104" stroke="#cc0000" stroke-width="2" stroke-linecap="round"/>\n' % (x+56, x+60, x+61, x+80))
        else:
            out.append('<path d="M %d,56 %d,64 M %d.6,67.2 %d,104" fill="none" stroke="#000000" stroke-width="1" stroke-linecap="round"/>\n' % (x+56, x+60, x+61, x+80))
        out.append(line(x,  8,80,  32,128, 2, 9 in t))
        out.append(line(x, 80,104, 32,128, 2, 6 in t))
        out.append(line(x,  8,80,   8,32,  2, 8 in t))
        out.append(line(x, 80,104, 80,56,  2, 5 in t))
        out.append(line(x,  8,32,  56,8,   2, 4 in t))
        out.append(line(x, 56,8,   80,56,  2, 3 in t))
        out.append(line(x, 32,128, 32,80,  1, 11 in t))
        out.append(line(x,  8,32,  32,80,  1, 10 in t))
        out.append(line(x, 32,80,  80,56,  1, 7 in t))

    out.append('</svg>\n')
    return ''.join(out)

class Known:

    def __init__(self, mappings=[]):
        self.known = set()
        self.mappings = mappings

    def __str__(self):
        return "(%d, %d, %d)" % self

    def unique(self, triplet):
        for combination in range(0, 2**len(self.mappings)):
            t = Triplet(triplet)
            for m in range(0, len(self.mappings)):
                if combination & (1 << m):
                    t = t.apply(self.mappings[m])
            k = t.key
            if k in self.known:
                return False
        self.known.add(Triplet(triplet).key)
        return True

    def prefer(self, triplet, scores={}):
        w = [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ]
        for k,s in scores.items():
            w[k] = s

        best = Triplet(triplet)
        bestscore = 0

        for combination in range(0, 2**len(self.mappings)):
            t = Triplet(triplet)
            for m in range(0, len(self.mappings)):
                if combination & (1 << m):
                    t = t.apply(self.mappings[m])
            score = w[t[0]] + w[t[1]] + w[t[2]]
            if bestscore < score:
                bestscore = score
                best = t

        return best

if __name__ == '__main__':
    from sys import stdout, stderr

    # Edges parallel to axis
    x_edges = ( 2, 4, 6, 7 )
    y_edges = ( 1, 3, 9, 10 )
    z_edges = ( 0, 5, 8, 11 )

    # Each operation is described as a dictionary of old:new edge mappings.
    rotate_x = Mapping({ 0:3,3:5,5:1,1:0, 8:10,10:11,11:9,9:8, 2:4,4:7,7:6,6:2 })
    rotate_y = Mapping({ 1:9,9:10,10:3,3:1, 5:6,6:11,11:7,7:5, 2:8,8:4,4:0,0:2 })
    rotate_z = Mapping({ 2:9,9:6,6:1,1:2, 8:11,11:5,5:0,0:8, 4:10,10:7,7:3,3:4 })
    mirror_x = Mapping({ 0:8,8:0, 3:10,10:3, 5:11,11:5, 1:9,9:1 })
    mirror_y = Mapping({ 8:11,11:8, 2:6,6:2, 4:7,7:4, 0:5,5:0 })
    mirror_z = Mapping({ 2:4,4:2, 1:3,3:1, 9:10,10:9, 6:7,7:6 })

    # To include mirroring or not?
    mirrors = False

    # All known operations:
    if mirrors:
        known = Known([ rotate_x, rotate_x.apply(rotate_x),
                        rotate_y, rotate_y.apply(rotate_y),
                        rotate_z, rotate_z.apply(rotate_z),
                        mirror_x, mirror_y, mirror_z ])
    else:
        known = Known([ rotate_x, rotate_x.apply(rotate_x),
                        rotate_y, rotate_y.apply(rotate_y),
                        rotate_z, rotate_z.apply(rotate_z) ])

    # Unique triplets:
    unique = []

    # Try all 4*4*4 = 64 edge combinations
    for x in x_edges:
        for y in y_edges:
            for z in z_edges:
                t = Triplet(x, y, z)
                if known.unique(t):
                    unique.append(known.prefer(t, { 2:10, 1:10,  # Prefer edges 2 and 1,
                                                    0:8,         # Then edge 0,
                                                    8:6, 5:6,    # Then edges 8 and 5,
                                                    9:4, 6:4,    # Then edges 9, 6, 3, 4,
                                                    3:4, 4:4,    #
                                                    7:1, 10:1, 11:1 }))
    stderr.write("Found %d unique edge triplets:\n" % len(unique))
    for t in unique:
        stderr.write("\t%s\n" % str(t))

    stdout.write(SVG(*unique))
</code>```

Run it using e.g. python3 generate.py > nomirrors.svg and it will examine all unique edge triplets, where one edge is parallel to the $x$ axis, one to the $y$ axis, and one to the $z$ axis, excluding rotationally symmetric configurations.  The result is an SVG file (that you can open in for example your browser) describing the above six configurations.

Modify the line mirrors = False to mirrors = True, and then run e.g. python3 generate.py > mirrors.svg, and the result is an SVG file describing the four unique configurations, when both mirrored and rotated configurations are excluded.

Mapping is a class which describes how the edges are permuted when an operation is applied to the triplet.  There are six rotations – 90° around the $x$ axis, 180° around the $x$ axis, 90° around the $y$ axis, 180° around the $y$ axis, 90° around the $z$ axis, and 180° around the $z$ axis – so $2^6 = 64$ possible rotation combinations; and three reflections (along the $x$, $y$, or $z$ axis) for $2^3 = 8$ possible reflection combinations.

Triplet is a tuple subclass, describing each triplet.  It has a key property, which encodes the triplet to a single integer.

Known is a class that contains the mappings through which we recognize duplicates (as the mappings member).  Its known member is an unordered set of known triplet keys.  The unique(triplet) method returns True if the specified triplet is a new one, adding it to the known set; and False if the specified triplet is already known.  The prefer(triplet, scoreboard) is a helper function, which "scores" the permutations of the triplet using the specified scoreboard dictionary (with unspecified edges getting a score/weight of 1), returning the permutation of the triplet that yields the highest score.  This is used to select the "nicest" view for each unique edge set.

The SVG method returns a string containing an SVG file describing each of the specified triplets, using the same format/visual layout as the image of the six configurations above.

Here is how the program enumerates the edges 0 through 11:

Note that $000$ refers to origin, $100$ to $+x$, $010$ to $+y$, $001$ to $+z$, and so on; these identify the vertices of the cube.

Body:
A cube has $3 \cdot 4$ edges, so each edge can be chosen from one set of four.  That means there are $4 \cdot 4 \cdot 4 = 4^3 = 64$ unique combinations.

This assumes that each set of edges is considered as an unordered set, that it does not matter which of the three edges corresponds to $x$, $y$, or $z$.  Each set has $3 \cdot 2 \cdot 1 = 6$ such combinations, so if the order does matter, there are $64 \cdot 6 = 384$ combinations total.

If we consider each set of unordered edges, but exclude those configurations that are just rotated versions of existing sets, we are left with just six unique edge configurations:

(This has been verified with a program that examines all 64 unique combinations, dropping those configurations that are just rotated versions of others, using 90° rotations around each axis.)

Here is a Python 3 program, generate.py, you can use to generate and/or verify the sets:

#!/usr/bin/python3
# SPDX-License-Identifier: CC0-1.0
# -*- coding: utf-8 -*-

# Edges are numbered thus:
#   0: (0,0,0)-(0,0,1)  z
#   1: (0,0,0)-(0,1,0)  y
#   2: (0,0,0)-(1,0,0)  x
#   3: (0,0,1)-(0,1,1)  y
#   4: (0,0,1)-(1,0,1)  x
#   5: (0,1,0)-(0,1,1)  z
#   6: (0,1,0)-(1,1,0)  x
#   7: (0,1,1)-(1,1,1)  x
#   8: (1,0,0)-(1,0,1)  z
#   9: (1,0,0)-(1,1,0)  y
#  10: (1,0,1)-(1,1,1)  y
#  11: (1,1,0)-(1,1,1)  z

class Mapping(tuple):

    def __new__(cls, mapping=None):
        d = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ]
        if isinstance(mapping, dict):
            for k,v in mapping.items():
                d[k] = v
        elif mapping is not None:
            raise TypeError("Unsupported mapping type (%s)" % type(mapping))
        return tuple.__new__(cls, d)

    def __init__(self, *args, **kwargs):
        return

    def apply(self, mapping):
        d = list(self)
        if isinstance(mapping, dict):
            for k,v in mapping.items():
                d[k] = self[v]
        elif isinstance(mapping, (list, tuple)):
            if len(mapping) != 12:
                raise ValueError("Mapping needs 12 elements, not %d" % len(mapping))
            for k in range(0, 12):
                d[k] = self[mapping[k]]
        else:
            raise TypeError("Unsupported mapping type (%s)" % type(mapping))
        return tuple.__new__(self.__class__, d)

class Triplet(tuple):

    def __new__(cls, *args):
        if len(args) == 1:
            if isinstance(args[0], int):
                args = [ args[0] // 256, (args[0] // 16) & 15, args[0] & 15 ]
            elif isinstance(args[0], (list, tuple)):
                args = args[0]
            else:
                raise TypeError("Invalid triplet definition (%s)" % type(args[0]))
        if len(args) != 3:
            raise ValueError("Triplet needs exactly three arguments, not %d" % len(args))
        return tuple.__new__(cls, sorted(args))

    def __init__(self, *args, **kwargs):
        return

    @property
    def key(self):
        return 256*self[0] + 16*self[1] + self[2]

    def apply(self, mapping):
        return tuple.__new__(self.__class__, sorted([mapping[self[0]], mapping[self[1]], mapping[self[2]]]))

def SVG(*triplets):
    w = 8 + 80*len(triplets)
    out = [ '<?xml version="1.0" encoding="UTF-8" standalone="no"?>\n',
            '<svg version="1.1" viewBox="0 0 %d 136">\n' % w,
            '<rect x="0" y="0" width="%d" height="136" fill="#ffffff" stroke="none"/>\n' % w ]

    def line(x0, x1, y1, x2, y2, t, selected):
        if selected:
            return '<path d="M %d,%d %d,%d" fill="none" stroke="#cc0000" stroke-width="2" stroke-linecap="round"/>\n' % (x0+x1, y1, x0+x2, y2)
        else:
            return '<path d="M %d,%d %d,%d" fill="none" stroke="#000000" stroke-width="%s" stroke-linecap="round"/>\n' % (x0+x1, y1, x0+x2, y2, t)

    for i in range(0, len(triplets)):
        x = 80*i
        out.append('<path d="M %d,80 %d,128 %d,104 %d,56 Z" fill="#eff7ff" stroke="none"/>\n' % (x+8, x+32, x+80, x+56))
        out.append('<path d="M %d,32 %d,80 %d,56 %d,8 Z" fill="#dfefff" stroke="none"/>\n' % (x+8, x+8, x+56, x+56))
        out.append('<path d="M %d,56 %d,104 %d,56 %d,8 Z" fill="#bfdfff" stroke="none"/>\n' % (x+56, x+80, x+80, x+56))
        t = Triplet(triplets[i])
        out.append(line(x, 56,8, 56,56, 1, 0 in t))
        if (2 in t):
            out.append('<path d="M %d,80 %d.6,71.2 M %d.8,69.6 %d,56" fill="none" stroke="#cc0000" stroke-width="2" stroke-linecap="round"/>\n' % (x+8, x+25, x+28, x+56))
        else:
            out.append('<path d="M %d,80 %d,71 M %d.4,69.8 %d,56" fill="none" stroke="#000000" stroke-width="1" stroke-linecap="round"/>\n' % (x+8, x+26, x+28, x+56))
        if (1 in t):
            out.append('<path d="M %d,56 %d.2,64.4 M %d.4,66.8 %d,104" stroke="#cc0000" stroke-width="2" stroke-linecap="round"/>\n' % (x+56, x+60, x+61, x+80))
        else:
            out.append('<path d="M %d,56 %d,64 M %d.6,67.2 %d,104" fill="none" stroke="#000000" stroke-width="1" stroke-linecap="round"/>\n' % (x+56, x+60, x+61, x+80))
        out.append(line(x,  8,80,  32,128, 2, 9 in t))
        out.append(line(x, 80,104, 32,128, 2, 6 in t))
        out.append(line(x,  8,80,   8,32,  2, 8 in t))
        out.append(line(x, 80,104, 80,56,  2, 5 in t))
        out.append(line(x,  8,32,  56,8,   2, 4 in t))
        out.append(line(x, 56,8,   80,56,  2, 3 in t))
        out.append(line(x, 32,128, 32,80,  1, 11 in t))
        out.append(line(x,  8,32,  32,80,  1, 10 in t))
        out.append(line(x, 32,80,  80,56,  1, 7 in t))

    out.append('</svg>\n')
    return ''.join(out)

class Known:

    def __init__(self, mappings=[]):
        self.known = set()
        self.mappings = mappings

    def __str__(self):
        return "(%d, %d, %d)" % self

    def unique(self, triplet):
        for combination in range(0, 2**len(self.mappings)):
            t = Triplet(triplet)
            for m in range(0, len(self.mappings)):
                if combination & (1 << m):
                    t = t.apply(self.mappings[m])
            k = t.key
            if k in self.known:
                return False
        self.known.add(Triplet(triplet).key)
        return True

    def prefer(self, triplet, scores={}):
        w = [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ]
        for k,s in scores.items():
            w[k] = s

        best = Triplet(triplet)
        bestscore = 0

        for combination in range(0, 2**len(self.mappings)):
            t = Triplet(triplet)
            for m in range(0, len(self.mappings)):
                if combination & (1 << m):
                    t = t.apply(self.mappings[m])
            score = w[t[0]] + w[t[1]] + w[t[2]]
            if bestscore < score:
                bestscore = score
                best = t

        return best

if __name__ == '__main__':
    from sys import stdout, stderr

    # Edges parallel to axis
    x_edges = ( 2, 4, 6, 7 )
    y_edges = ( 1, 3, 9, 10 )
    z_edges = ( 0, 5, 8, 11 )

    # Each operation is described as a dictionary of old:new edge mappings.
    rotate_x = Mapping({ 0:3,3:5,5:1,1:0, 8:10,10:11,11:9,9:8, 2:4,4:7,7:6,6:2 })
    rotate_y = Mapping({ 1:9,9:10,10:3,3:1, 5:6,6:11,11:7,7:5, 2:8,8:4,4:0,0:2 })
    rotate_z = Mapping({ 2:9,9:6,6:1,1:2, 8:11,11:5,5:0,0:8, 4:10,10:7,7:3,3:4 })
    mirror_x = Mapping({ 0:8,8:0, 3:10,10:3, 5:11,11:5, 1:9,9:1 })
    mirror_y = Mapping({ 8:11,11:8, 2:6,6:2, 4:7,7:4, 0:5,5:0 })
    mirror_z = Mapping({ 2:4,4:2, 1:3,3:1, 9:10,10:9, 6:7,7:6 })

    # To include mirroring or not?
    mirrors = False

    # All known operations:
    if mirrors:
        known = Known([ rotate_x, rotate_x.apply(rotate_x),
                        rotate_y, rotate_y.apply(rotate_y),
                        rotate_z, rotate_z.apply(rotate_z),
                        mirror_x, mirror_y, mirror_z ])
    else:
        known = Known([ rotate_x, rotate_x.apply(rotate_x),
                        rotate_y, rotate_y.apply(rotate_y),
                        rotate_z, rotate_z.apply(rotate_z) ])

    # Unique triplets:
    unique = []

    # Try all 4*4*4 = 64 edge combinations
    for x in x_edges:
        for y in y_edges:
            for z in z_edges:
                t = Triplet(x, y, z)
                if known.unique(t):
                    unique.append(known.prefer(t, { 2:10, 1:10,  # Prefer edges 2 and 1,
                                                    0:8,         # Then edge 0,
                                                    8:6, 5:6,    # Then edges 8 and 5,
                                                    9:4, 6:4,    # Then edges 9, 6, 3, 4,
                                                    3:4, 4:4,    #
                                                    7:1, 10:1, 11:1 }))
    stderr.write("Found %d unique edge triplets:\n" % len(unique))
    for t in unique:
        stderr.write("\t%s\n" % str(t))

    stdout.write(SVG(*unique))

Run it using e.g. python3 generate.py > nomirrors.svg and it will examine all unique edge triplets, where one edge is parallel to the $x$ axis, one to the $y$ axis, and one to the $z$ axis, excluding rotationally symmetric configurations.  The result is an SVG file (that you can open in for example your browser) describing the above six configurations.

Modify the line mirrors = False to mirrors = True, and then run e.g. python3 generate.py > mirrors.svg, and the result is an SVG file describing the four unique configurations, when both mirrored and rotated configurations are excluded.

Mapping is a class which describes how the edges are permuted when an operation is applied to the triplet.  There are six rotations – 90° around the $x$ axis, 180° around the $x$ axis, 90° around the $y$ axis, 180° around the $y$ axis, 90° around the $z$ axis, and 180° around the $z$ axis – so $2^6 = 64$ possible rotation combinations; and three reflections (along the $x$, $y$, or $z$ axis) for $2^3 = 8$ possible reflection combinations.

Triplet is a tuple subclass, describing each triplet.  It has a key property, which encodes the triplet to a single integer.

Known is a class that contains the mappings through which we recognize duplicates (as the mappings member).  Its known member is an unordered set of known triplet keys.  The unique(triplet) method returns True if the specified triplet is a new one, adding it to the known set; and False if the specified triplet is already known.  The prefer(triplet, scoreboard) is a helper function, which "scores" the permutations of the triplet using the specified scoreboard dictionary (with unspecified edges getting a score/weight of 1), returning the permutation of the triplet that yields the highest score.  This is used to select the "nicest" view for each unique edge set.

The SVG method returns a string containing an SVG file describing each of the specified triplets, using the same format/visual layout as the image of the six configurations above.

Here is how the program enumerates the edges 0 through 11:

Note that $000$ refers to origin, $100$ to $+x$, $010$ to $+y$, $001$ to $+z$, and so on; these identify the vertices of the cube.


答:
Pick $1$ edge of the $\color{red}{12}$ possible.

Your next choice will be from $\color{red} 8$ of the remaining $11$. Pick one.

There will only be $\color{red} 4$ possible choices for the third edge.

So the number of choices will be $12 \cdot 8 \cdot 4 = 384$

Title: Combinations of three mutually perpendicular edges of a cube.

Body: I need help in determining the possible combinations of mutually perpendicular edges of a cube, taken three at a time. Two examples are shown in the image attached. enter image description here</p>


问:
Closure of union of two sets
Show that

$$\overline{A\cup B} = \overline A\cup\overline B$$

$\overline A=A\cup A'$ where $A'$ are the limit points

My attempt:

Since $\overline A$ is closed and $\overline B$, the union of two closed sets is closed.

Hence $\overline A\cup \overline B$ can be rewritten as $\overline{A\cup B}$

My Second attempt:

$\overline{A\cup B}=(A\cup B)\cup (A\cup B)'=(A\cup B)\cup (A'\cup B')=A\cup A' \cup B \cup B' = \overline{A}\cup \overline{B}$

答:
Note that closures preserve set inclusion, i.e. if $A \subset B$ then $\overline A \subset \overline B$.

We want to show $\overline{A\cup B} = \overline{A} \cup \overline{B}$. $A \subset A\cup B$ and $B \subset A\cup B$ so $\overline{A} \subset \overline{A\cup B}$ and $\overline{B} \subset \overline{A\cup B}$. So, $\overline{A} \cup \overline{B} \subset \overline{A\cup B}$. On the other hand, $A \subset \overline{A}, B\subset \overline{B}$, so $A\cup B \subset \overline{A} \cup \overline{B}$. The latter is closed, so $\overline{A\cup B} \subset \overline{A} \cup \overline{B}$.

We conclude $\overline{A\cup B} = \overline{A} \cup \overline{B}$. $\ \  \blacksquare$

Title: Closure of union of two sets


问:
$K$-theory of formal power series.
I was wondering whether there is a calculation of algebraic $K$-groups of the formal power series $\mathbb{F}_p[[x]]$?

答:
Here are some thoughts which became a bit too long for a comment.

Since $\mathbb{F}_p[[X]]$ is local, we have $K_0(\mathbb{F}_p[[X]]) \simeq \mathbb Z$ and $K_1(\mathbb{F}_p[[X]]) \simeq (\mathbb{F}_p[[X]])_{\mathsf ab}^\times = \mathbb{F}_p[[X]]^\times$. This is proven for arbitrary, non necessarily commutative local rings in in Rosenberg's book.

Example 4.4.3 of Weibel's K-book, Chapter III cites a result of Drinfeld which says that $K_{-1}(A) = 0$ for all henselian local rings, which $\mathbb{F}_p[[X]]$ is an example of.

I don't know about the higher $K$-groups, they are famously hard to compute and usually involve some homotopical algebra, which I am not that familiar with.

One thing to note is that $\mathbb{F}_p[[X]]$ is regular noetherian, hence its $K$-theory coincides with Weibel's homotopy $K$-theory,

$$
K_\bullet(\mathbb{F}_p[[X]]) = KH_\bullet(\mathbb{F}_p[[X]]).
$$

This is Corollary 12.3.1 of the $K$-book, Chapter IV. The advantage of the latter theory is that it is matricially stable, polynomially homotopy invariant, excisive, and commutes with filtering colimits.

Edit: for example, since $\mathbb{F}_p[[X]]$ sits in an exact sequence of non unital rings

$$
(X) \to \mathbb{F}_p[[X]] \to \mathbb{F}_p[[X]]/(X) \simeq \mathbb{F}_p,
$$

we have a long exact sequence

$$
\ldots \to KH_\bullet((X)) \to KH_\bullet\mathbb{F}_p[[X]] \to KH_\bullet(\mathbb{F}_p) \to KH_{\bullet-1}((X)) \to \ldots.
$$

A result due to Quillen characterizes the positive $K$-theory of finite fields (which by a previous argument, coincides with $KH$): for each $n \geq 1$ we have

$$
K_n(\mathbb F_p) = \begin{cases}\mathbb F_p^\times &\text{$n$ odd}\\ 0&\text{otherwise}\end{cases}
$$

On the other hand, the negative $KH$-theory of a commutative artinian ring vanishes (e.g. by Example 12.5.1, Chapter IV of the $K$-book), so $K_n(\mathbb F_p) = 0$ for negative $n$.

I don't know how much information this gives; for what it's worth we see that for a vast amount of indices the $K$-theory of $\mathbb F_p[[X]]$ is the same as the $KH$-theory of its maximal ideal $(X)$ viewed as a rng.

It would be interesting to know any results on the polynomial homotopy type of these algebras.

Edit': upon further inspection of the aformentioned Chapter IV, Example 12.5.2 shows that $KH_n(R) = 0$ for $n < 0$ and $R$ commutative, noetherian and $1$-dimensional.

Hence $K_n(\mathbb F_p[[X]]) = 0$ for negative $n$, and by excision, the same holds for the negative groups of $KH_\bullet((X))$.

Edit'': to sum up, so far we have
$$
K_n(\mathbb F_p[[X]]) = K_n(\mathbb{F}_p) = \delta_{n,0}\mathbb Z
$$

for all $n \leq 0$, $K_1(\mathbb F_p[[X]]) = \mathbb F_p[[X]]^\times$ and exact sequences

$$
0 \to KH_{2n-1}((X)) \to K_{2n-1}(\mathbb F_p[[X]]) \to \mathbb F_p^\times \to K_{2n-2}((X)) \to KH_{2n-2}(\mathbb F_p[[X]]) \to 0
$$

for each $n > 1$.

Title: $K$-theory of formal power series.


问:
The twisted cubic is an affine variety.

Let $Y=\{(t,t^2,t^3)\mid t\in k\}$ be the twisted cubic curve. I'm trying to prove this curve is a variety, i.e., it's irreducible and affine algebraic set.

The easier part is to prove the twisted cubic curve is an affine algebraic set $(Y=Z(x^2-y,x^3-z))$.

I don't know how to prove that $Y$ is irreducible, I'm trying to prove that $(x^2-y,x^3-z)$ is prime, I think if I do this I proved what I want, but I found this hard to prove.

I need help to finish this question.

Thanks a lot.

答:
Expand a little on Martin's answer.

Define $f:X \to Y$ as $$f(x) = (x, x^2, x^3)$$
Define $g:Y \to X$ as $$g(x,y,z) = x$$

Then $f^{-1} = g$, $f$ is an isomorphism between $A^1$ and Y.

Now by p.g.29 of Shafarecich's Basic algebraic Geometry 1, there exists an isomorphism between coordinate rings of $A^1$ and $ Y$, which means $$k[x] \cong k[x,y,z]/((x^2-y),x^3-z)$$

Since k[x] is an integral domain, $k[x,y,z]/((x^2-y),x^3-z))$ is an integral domain, which means that $((x^2-y),x^3-z))$ is irreducible.

Title: The twisted cubic is an affine variety.


问:
Prove $a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right| < 1$ without use of log properties
$a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right|  < 1 $

Hint $u_{2n}$ = $u_{n}^2$

I have totally no idea how to prove this, this looks obvious but i found out proof is really hard...
I am doing a real analysis course and there's a lot of proving and I stuck there.
Any advices? Practice makes perfect?

答:
Let $u_n = a^n$ with $|a|< 1$

Let $v_n=|u_n|$

$v_{n+1} = a v_n < v_n$ so $(v_n)$ is decreasing

$0 \le v_n$ so $(v_n)$ is minored

Since $(v_n)$ is minored and decreasing, it converges.

Now let $v_\infty=\lim\limits_{n\to \infty}v_n$

You have $v_{n+1} = a v_n$

By making $n\to \infty$ on both sides (which you can do because $x\mapsto ax$ is continuous) you get:

$v_\infty = a v_\infty$

$0 = (a-1) v_\infty$

But $|a|<1$ so $a-1\not= 0$ so $v_\infty=0$

That is $|u_n|\underset{n\to\infty}{\longrightarrow}0$

So we can conclude that $u_n\underset{n\to\infty}{\longrightarrow}0$


问:
Can someone please explain how the answer is 48.
Three couples are to be seated on one side of a table for lunch. One of the couples, Edward and Bella, are quarreling and would like to be seated apart. How many arrangements are possible if each of the other two couples would like to be seated together?

Can someone please explain how the answer is 48.

答:
First seat Edward & Bella. There are two of them and exactly three places (relative to the other couples who sit together.

$$ ? == ? == ?$$

So there are $3 \cdot 2 = 6$ ways of seating Bella & Edward before we seat the other two couples.

Now seat the other two couples. There are two places for the couples, and each couple can switch places with each other, hence there are
$6 \cdot 2 \cdot (2 \cdot 2) = 48 $ places.


问:
Computing knot/link groups
The knot group of a knot $K$ is the fundamental group of $\mathbb R^3 \smallsetminus K$; that is, the set of possibly self-crossing closed paths (starting and ending at any single point in space) which you can take through real-space which are not equivalent to each other if you forbid passing through the locus of the knot; and where you can compose the paths in the natural way. For instance, the knot group of the un-knot is just $\mathbb Z$, corresponding to the number of times that a path winds about a cycle in space.

We can consider similar questions for links ("knots" having more than one closed loop) as for knots of a single component: given a link, we can consider the group of closed paths in space which avoids crossing either of the two components.

I have a friend who has a tattoo of a knot (more precisely, a link of two components) in the shape of a maple leaf; we were curious what properties it had. A presentation of the link is given below. How would one obtain a presentation of the link group? Does the link group of this particular link correspond to any relatively nice group?

答:
I plugged this into SnapPy and got the following presentation:

Generators: a,b,c,d

Relators: abCBACADadcDAdabCBADadCDAdadcDAdabcBADadCDAdacabcBADadCBCbcDAdabCBACdcDadCDcabcBADadCBcbcDAd, abCBCbDAdacACADadBcbcDAd, abCBACdcDAdCDcDadCDADadcDAdabCBADadCDAdadcDAdCdcDadCDcabcBADadCBcbcDAd

(That is, four generators, three relators.)  The link complement has volume approximately 33.331392948.  Since the link is alternating and has no flypes, the diagram shown is minimal - thus the crossing number is 16.  One more thing - the diagram has an involution about the central vertical line.  You could mod out by this involution to get a smaller orbifold - there might be something known about the quotient...

Title: Computing knot/link groups


答:
Well, even I am surprised that the answer given is $48$,

because simply seating couples together and switching husband/wife in $3!*2^3 = 48$ ways !

To keep the quarrelsome couple apart, first seat the two harmonious couples in $2!*2^2 = 8$ ways, then push in two chairs for the quarrelsome couple at two of the uparrows in the diagram below

$\uparrow \fbox{HW}\uparrow \fbox{WH}\uparrow$

so the final answer is  $8\times 3\times 2 = 48$

which is the same if they weren't quarrelsome !

Body:
Three couples are to be seated on one side of a table for lunch. One of the couples, Edward and Bella, are quarreling and would like to be seated apart. How many arrangements are possible if each of the other two couples would like to be seated together?

Can someone please explain how the answer is 48.


问:
Does $\lim_{a\to 0}\int_1^\infty f\left(\sqrt{a^2+x^2}\right)dx=\int_1^\infty f\left(x\right)dx$ for smooth, rapidly decaying $f$?
Let $f:[1,\infty)\to\mathbb R$ be infinitely smooth and suppose $f$ and its derivatives decay faster than any power of $x$ as $x\to\infty$. In other words, $f\in\mathscr S([1,\infty)),$ where $\mathscr S$ denotes Schwartz space.

Then does $\lim_{a\to 0}\int_1^\infty f\left(\sqrt{a^2+x^2}\right)dx=\int_1^\infty f\left(x\right)dx$?

I would expect yes, and that a dominated convergence argument should apply somehow. I am tempted to use identities like $x<\sqrt{a^2+x^2}<a+x,$ except I can't assume $f$ is increasing/decreasing—it could even be oscillatory at infinity.

This is a "real" problem, by the way. I have been assuming this fact in my research, but am not sure how to prove it (or even if it is true).

答: Since $f$ is rapidly decaying $x^{4} f(x)$ is bounded. If $|f(x)| \leq \frac M {x^{4}}$ the $|f(\sqrt {a^{2}+x^{2}})|$ is bounded by $\frac M {x^{2}}$ which is integrable in $(1,\infty)$.


问:
Absolute maximum and minimum on the closed disk
Find the absolute maximum and minimum of $f(x,y)=exp(x^2-y^2)$ on the set $R^2$ with $x^2+y^2 \leq1$.

I have applied Lagrange Multipliers method, but found that exponent function should be equal to zero if $f$ function has minimum. Any help will be strongly appreciated.

答:
A geometrical interpretation of this problem can also be helpful.  Since we are looking for level curves for the function, $ \ f(x,y) \ = \ e^{x^2 \ - \ y^2} \ = \ c \ \ , $ we could work with the curve equation  $ \  x^2 \ - \ y^2  \ = \ \ln c \ \ . $  This is the (square) "horizontal" hyperbola centered on the origin, with its major axis on the $ \ x-$ axis and with asymptotes $ \ y\  = \ \pm x \ $ .

Its vertices then lie at $ \ ( \ \pm \sqrt{\ln c} \ , \ 0 ) \ \ . $  The largest possible value of $ \ \ln c \ $ for which the hyperbola will make contact with the boundary of the disk, $ \ x^2 \ + \ y^2 \ = \ 1 \ , $ is thus $  \ \ln c \ = \ 1 \ \ , $ so the absolute maximum value of our function occurs for $ \ f(\pm 1 \ , \ 0) \ =  \ e \ \ . $

As we reduce $ \ \ln c \ $ , the vertices close in on the origin, reducing the hyperbola to its "degenerate form" of the two asymptotes.  The two vertices have merged with the center of the hyperbola at $ \ (0 \ , \ 0 ) \ $ for $ \ \ln c \ = \ 0 \ \ \Rightarrow \ \ c \ = \ 1 \ . $

If we then continue to reduce $ \ \ln c \ $ by making it negative, the curve equation becomes

$$ \  x^2 \ - \ y^2  \ = \ \ln c \ \ \ \rightarrow \ \ \  y^2 \ - \ x^2  \ = \ -| \ \ln c \ |  \ \ , $$

which is a (square) "vertical" hyperbola with the same center and asymptotes, but now with its major axis on the $ \ y-$ axis.  The hyperbola will be just tangent to the boundary for $ \ -| \ \ln c \ | \ = \ 1 \ \ \Rightarrow \ \ \ln c \ = \ -1 \ \ . $  Hence, the absolute minimum value of the function occurs for the vertices  $ \ f(0 \ , \ \pm 1) \ =  \ e^{-1} \ \ . $

In the neighborhood of the origin, as $ \ \ln c \ $ increases from values slightly less than zero to those slightly greater than zero, the two points on the $ \ y-$ axis draw closer together, merge at the origin for $ \ c \ = \ 1 \ $ , and then separate along the $ \ x-$ axis.  Thus, the critical point at the origin behaves as a "saddle point".


问:
Continuity of a solution of a differential equation with respect to a time-dependent input?
I am interested in a question of the following form, and I'm asking because the question seems so simple to pose (if not to answer) that I figure there must be some discipline interested in answering it.  I am interested for some particular cases but I would like to know where it would fit into the study of differential equations more generally.

The question can be written like this: given $\epsilon > 0$, we want to find a $\delta > 0$...

$$
\lvert\lvert \phi(t)-\psi(t) \rvert\rvert < \delta
$$
Such that we can say for the initial value problem $x'(t) = f(x(t),g(t)),\ x(t_0)=x_0$:
$$
\lvert\lvert x(t)_{g(t) = \phi(t)}-x(t)_{g(t)=\psi(t)}\rvert \rvert<\epsilon
$$
My case has all sorts of nice properties one might expect ($f \in C^\infty$, non-negativity of $x$, smoothness and boundedness of $g$, etc), but I figured that someone might recognize it more immediately if I posed it in greater generality.

It seems to have superficial similarities with structural stability (my problem came out of similar motivations to what structural stability intends to address), but in that case the input is not time-based, and with control theory, but in that case they're interested in (as far as I can tell) different kinds of questions.

答:
If one assumes that $f$ is nice enough to be jointly Lipschitz continuous in both arguments,
$$
\|f(x_1,g_1)-f(x_2,g_2)\|\le L\|x_1-x_2\|+M\|g_1-g_2\|,
$$
then that directly results in a differential inequality that can be handled by the Grönwall lemma
$$
\|\dot x_1(t)-\dot x_2(t)\|\le L\|x_1(t)-x_2(t)\|+M\|ϕ(t)-ψ(t)\|
$$
which gives a bound
$$
\|x_1(t)-x_2(t)\|\le \|x_1(0)-x_2(0)\|e^{Lt}+M\int_0^t e^{L(t-s)}\|ϕ(s)-ψ(s)\|\,ds
$$
so that under the further assumptions
$$
\|x_1(t)-x_2(t)\|\le Mδ\frac{e^{Lt}-1}{L}.
$$
For a finite time interval $t\in [0,T]$ this bound can be made smaller than $ϵ$ by selecting $δ$ accordingly.


问:
Open set on an arbitrary metric on any finite set
I know that for the standard metric on R, every singleton is closed. However, I have a claim that on a finite set for any arbitrary metric every singleton is open, how can I prove this?</p>

答: If $F$ is finite, and $x \in F$, then the set $D_x:=\{d(x,y) : y \in F, y \neq x\}$ is also a finite set and does not contain $0$ (as $d(x,y)=0$ iff $x=y$ and $y$ is never equal to $x$) so $r=\min D_x$ is well-defined and $>0$ (set it to $1$ if $D_x$ is empty, which happens for $F=\{x\}$). Then for all $y \neq x \in F$, $d(x,y) \ge r$, so $y \notin B(x,r) \cap F$. It follows that $B(x,r) \cap F = \{x\}$ and so $\{x\}$ is relatively open in $F$.


问:
Conditional expectation of one random variable
Let $X$ be a continuous random variable whose probability density function is
$$
f(x)= \left\{ \begin{array}{lcc}
             \alpha^2xe^{-\alpha x} &   if  & x > 0 \\
             \\ 0&  otherwise
             \end{array}
   \right.
$$
With $\alpha >0$. Calculate $E(X|X<1)$.

I know that by definition, I have
$$
E(X|X<1)=\int_{0}^{\infty}x \cdot P(X|X<1) dx = \int_{0}^{\infty}x \cdot \frac{P(X=x_k \land X<1 )}{P(X<1)}dx
$$
Is ok if i change the limits of the integral and put this?
$$
E(X|X<1)=\int_{0}^{1}x\cdot \frac{P(X<1 )}{P(X<1)}dx=\frac{1}{2}
$$
It confuses me a bit that the condition is on the same random variable. Thanks for the help

答:
The correct expression is
$$\operatorname{E}[X \mid X < 1] = \frac{\int_{x=0}^1 x f_X(x) \, dx}{\int_{x=0}^1 f_X(x) \, dx}.$$  You cannot write $Pr[X = x_k \wedge X < 1]$ because first of all, you did not define $x_k$, and second, it would need to depend on the variable of integration $x$.  Then even if you did write $\Pr[X = x \wedge X < 1]$, this is problematic because $X$ has a density, not a probability mass function.  Fourth, even if you ignore all of the above, the events $X = x$ and $X < 1$ are not independent.


问:
How to conclude that $\sum_{n>e^t} f_t(n)<e^{-t}?$
Given a Poisson process $N_t$. If we set
$$f_t(n):=P(N_t=n)=\frac{e^{-t}t^n}{n!}.$$
We know $\sum_{n\ge 0} f_t(n)=1$.

My question is that how to conclude that
$$\sum_{n>et} f_t(n)<e^{-t}?$$

This is equivalent as to show that
$$\sum_{n>et}\frac{t^n}{n!}<1$$

Note that $n!\approx\sqrt{2\pi n}n^ne^{-n}$, then
$$\sum_{n>et}\frac{t^n}{n!}\le \sum_{n>et}\frac{t^ne^n}{\sqrt{2\pi n} n^n}$$

答:
The mean value form of the remainder $\sum_{n>m} \frac{t^n}{n!}$ of the $m$th-degree Taylor approximation to $e^t$ is $\frac{1}{(m+1)!} \xi^{m+1}$ where $\xi$ is between $0$ and $t$.

When $m \approx et$ (hand-waving a bit here, you can be more rigorous with floor/ceiling functions), we have $\xi^{m+1} \le t^{et + 1}$ and $$\frac{1}{(m+1)!} \overset{*}{\le} \left(\frac{e}{m+1}\right)^{m+1} \le \frac{e^{et+1}}{(et)^{et+1}} = \frac{1}{ t^{et+1}}$$

where the starred inequality is due to $e^k = \sum_{n \ge 0} \frac{k^n}{n!} \ge \frac{k^k}{k!}$.


问:
$\mathbb{Q}$ is not a Baire space?
The condition of 'X is a Baire space': (cite from Topology by Munkres)

Given any countable collection $\{ A_n \}$ of closed sets of X each of which has empty interior in X, their union $\bigcup A_n$ also has empty interior in X.

So for rational numbers $\mathbb{Q}$ with the usual topology of $\mathbb{R}$:

(1) countable collection $\{ Q_i \}$ : one can index all rational numbers with  positive integers, so let $Q_i = \{q_i\}$ be
one-point set.

(2) $Q_i$ is closed, since $\mathbb{R}- Q_i= (-\infty, q_i) \cup (q_i, \infty)$ is open in $\mathbb{R}$.

(3) $Q_i$ has empty interior, since $Q_i$ contains no open set of $\mathbb{R}$.

(4) $\mathbb{Q}=\bigcup Q_i$ has empty interior, since $\mathbb{Q}$ contains no open set of $\mathbb{R}$.

Thus, $Q_i$ is a Baire space. What's wrong here?

答:
You have to consider the topology of $\Bbb Q$ in its own right: it has the order topology, or equivalently the metric topology induced by $d(x,y)= \max(y-x,x-y)$ for $x,y \in \Bbb Q$. There is no need to refer to the topology on $\Bbb R$ to define it.

Indeed $\{q\}$ is closed (as in any ordered or metric space), and your proof explicitly shows this for the order topology: $\Bbb Q\setminus \{q\} = \{r\in \Bbb Q\mid r < q\} \cup \{r \in \Bbb Q\mid r > q\}$ and both right hand sets are subbasic order-open sets.

$\{q\}$ has empty interior because if $q$ were an interior point there must be rationals $r_1, r_2$ such that $r_1 < q < r_2$ and $\langle r_1,r_2\rangle \subseteq \{q\}$, as $q$ is not the max or min and so has a local base of open intervals $\langle r_1,r_2\rangle = \{r \in \Bbb Q: r_1 < r < r_2\}$ in the order topology. But inbetween any two rationals there is a third ($\Bbb Q$ is order dense) so any $r_3$ with $r_1 < r_3 < q$ contradicts the inclusion $\langle r_1,r_2\rangle \subseteq \{q\}$, contradiction. Or simply note that all open sets of $\Bbb Q$ are infinite.

And the interior of $\Bbb Q$ in $\Bbb Q$ is $\Bbb Q$ itself, of course. Any space is open in itself. We don't need to refer to points outside $\Bbb Q$; ignore them.

Title: $\mathbb{Q}$ is not a Baire space?


问:
How to find all possible curvatures for any smooth curve in a donut?
The question is to find all possible curvature for any smooth curve in a donut. I parametrize the donut as
$$((3+\sin v)\sin u, (3+\sin v)\cos u, \cos v)$$

My thought is, since if we keep shrinking a circle, curvature goes to infinity, we only need to find the minimum curvature of any normal section. But I have trouble finding that...

答:
If instead of unit normal curvature we have a different normal curvature of the donut ( produced by tightening a string around the tube meridional section for instance ) then its parameterization would be:

$$ [ (3+ \frac{\sin v}{kn})\sin u, (3+ \frac{\sin v}{kn}  )\cos u, \frac{\cos v}{kn}]$$

Yes, when $ kn \to \infty $ we are  left with a thin circle around the middle of torus.

It is then no more a 2D  surface embedded in 3-space, but just a line in 3-space.


问:
Prove that for every natural number $n$, there exists a natural number $m$ such that $\frac{1}{m}-\frac{1}{m+1} < \frac{1}{n}$
I am not sure how to proceed with my proof. I can easily explain in words why this is so (just by maximizing the values of the LHS and RHS then noting the left will always decrease faster than the RHS due to denominator getting smaller), but I do not think this is mathematically valid and doesn't necessarily prove anything.

This is how I did it.

$\frac{1}{m} - \frac{1}{m+1} < \frac{1}{n}$

$=\frac{1}{m(m+1)} < \frac{1}{n}$

Now, I am sadly stuck and have no way to proceed. I am thinking to expand this inequality like $\frac{1}{m(m+1)}  < \text{something} < \frac{1}{n}$

or

$\text{something}  < \frac{1}{m(m+1)}  < \frac{1}{n}$, but I am not sure how to go about this or if this is even the right approach.

答: Even more directly, if $m=n$, then $\frac 1m - \frac {1}{m+1} = \frac 1n - \frac{1}{n+1} \lt \frac 1n$ because $\frac {1}{n+1} \gt 0$.


问:
Algebraic Closure is Unique up to Isomorphism in Dummit Foote
I know that there are many proofs of this, but I am wondering if someone could help me understand what is "swept under the rug" in Dummit and Foote. I couldn't find anything about that on this website.

I am currently self-studying field theory for the first time from Dummit and Foote, and just read the proof demonstrating the existence of an algebraic closure for all fields. The procedure that it takes is proving via Zorn's lemma that for any field $F$, there exists an algebraically closed field $K$ containing $F$. Then the subfield $\overline{F} \subset K$ of elements of $K$ algebraic over $F$ is shown to be an algebraic closure for $F$. After this proof, the authors note that it is easy to see that $\overline{F}$ is unique up to isomorphism by Zorn's lemma and from results about splitting fields.

I have tried for the last hour to see why this is true, but think I am missing something. I assume the fact that splitting fields are unique up to isomorphism is used, but I can't fit it in anywhere. Could someone explain why this is the case? Thanhs!

答:
Let's first state clearly the question you have.

Let $F$ be a field. An "algebraic closure" of $F$ is a field extension $C/F$ such that

$C$ is algebraic over $F$;

$C$ is algebraically closed.

Claim:

If $C_1/F, C_2/F$ are two algebraic closures of $F$, then there exists an $F$-isomorphism $\iota: C_1 \simeq C_2$.

Remark: This is not a universal property, as we don't require (and it is usually impossible) that the isomorphism $\tau$ be unique.

Proof: We first use Zorn's lemma to show that there is an $F$-embedding (i.e. a homomorphism of fields which is identity on $F$) from $C_1$ to $C_2$.

Let $A$ be the set of pairs $(K, \iota)$, where:

$K/F$ is a subextension of $C_1/F$;

$\iota: K \rightarrow C_2$ is an $F$-embedding.

The set $A$ is non-empty, as it contains the element $(F, id)$ where $id$ stands for the canonical embedding of $F$ into $C_2$.

A partial order $\leq$ is defined on $A$ as follows. We say that $(K, \iota) \leq (K', \iota')$ if $K$ is a subextension of $K'$ and $\iota'|_K = \iota$.

Now if we have a chain $(K_i, \iota_i)_{i \in I}$ in $A$, we build an element $(K, \iota)$ with $K = \bigcup_{i\in I}K_i$ and $\iota: K \rightarrow C_2$ defined by $\iota(x) = \iota_i(x)$ for any $i\in I$ and any $x\in K_i$, which is well defined because of the chain condition.

This is then an upper bound of the chain.

Therefore, our set $A$ satisfies the condition in Zorn's lemma, and consequently $A$ contains at least one maximal element, say $(K, \iota)$.

Suppose that $K \neq C_1$. Then there exists $x \in C_1 \backslash K$. We let $K'$ be the field $K(x)$.

Since $C_1$ is algebraic over $F$ and hence algebraic over $K$, we may take the minimal polynomial $f$ of $x$ over $K$. Now $f$ has at least one root $y$ in $C_2$, and from results of rupture fields (see below *), we see that $\iota$ extends to a homomorphism $\iota': K' \rightarrow C_2$ such that $\iota'|_K = \iota$ and $\iota(x) = y$.

This means that we have $(K, \iota)$ is strictly smaller than $(K', \iota')$, which contradicts the maximality of $(K, \iota)$.

We have thus proved that $K = C_1$ and hence the existence of an $F$-embedding $\iota: C_1\simeq C_2$.

It only remains to see that $\iota$ is surjective.

For any $y\in C_2$, let $g\in F[X]$ be the minimal polynomial of $y$ over $F$ (or any polynomial in $F[X]$ such that $g(y) = 0$).

Since $C_1$ is algebraically closed, $g$ decomposes in $C_1$ as a product of linear factors: $g = \prod_{i = 1}^d(X - x_i)$.

Taking images under $\iota$, we get an identity $g = \prod_{i = 1}^d(X - \iota(x_i))$ in $C_2[X]$.

Putting $X = y$ gives $0 = g(y) = \prod_{i = 1}^d(y - \iota(x_i))$ as an identity in $C_2$.

But $C_2$ is a field, hence $y$ must be equal to one of the $\iota(x_i)$.

Thus we have shown that $\iota$ is surjective.

(*) In fact, we have $K' = K(x) \simeq K[X]/(f)$, so we may first define a ring homomorphism $K[X]\rightarrow C_2$ sending $X$ to $y$. Since $f(y) = 0$, the ring homomorphism passes to quotient and gives $\iota'$.


问:
Doubt in handling double sums of the type: $\sum_{n,k\in\mathbb{N}_1}\frac{1}{n^2k^2(n+k)}$
I'm having a little trouble calculating the double series given by the proposition below:

$$\int_0^1\frac{\left[\text{Li}_2(x)\right]^2}{x}\mathrm dx=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2(n+k)}=2\zeta(2)\zeta(3)-3\zeta(5).\tag{1}$$

In fact, this integral has already been evaluated here and here, but i have no intention of evaluating it by integration or by using harmonic series. My interest is to attack the double sum in $(1)$ only using the manipulation of sums. Having clarified my interest in this topic, i naively proceeded as follows:
$$\begin{align*}\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2(n+k)}&=\sum_{n,k\in\mathbb{N}_1}\frac{1}{n^2k^2(n+k)}\\
&=\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n=k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n>k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n<k}}}\frac{1}{n^2k^2(n+k)}\\
&=\frac{\zeta(5)}{2}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n>k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n<k}}}\frac{1}{n^2k^2(n+k)}\tag{a}\end{align*}$$

Here's my difficulty with that approach! How can i solve the last two double sums in $(\text a)$ with the indexes $ n <k $ and $ n> k $ ?. Studying other posts (links at the end of the question) related to my question, i found here another possible approach using the fact that: $ \frac{1}{n^2k^2}=\frac{1}{(n+k)^2n^2}+\frac{1}{(n+k)^2k^2}+\frac{2}{(n+k)^3k}+\frac{2}{(n+k)^3n}$, then multiplying by $\frac{1}{n+k}$ on both sides and taking the double sum, we get:
$$\sum_{n,k\in\mathbb{N}_1}\frac{1}{n^2k^2(n+k)}=2\sum_{n,k\in\mathbb{N}_1}\frac{1}{(n+k)^3n^2}+4\sum_{n,k\in\mathbb{N}_1}\frac{1}{(n+k)^4n}.\tag{b} $$

According to this article, $(\text b)$ is equal to $T(2,2,1)=2T(2,0,3)+4T(1,0,4)$, where $T(a,b,c)$ is Tornheim’s double series defined by $T(a,b,c)=\sum_{n,m=1}\frac{1}{n^am^b(n+m)^c}$. In this second approach, the expression in $(\text b)$ seems to me a little less aggressive, but if i apply the concepts of the mentioned article, i would feel a little difficult to understand, since it starts with already established formulas. In short, two questions hang in my mind: How to work $(\text a)$ and $(\text b)$ without using integrals or harmonic numbers?

Related links:

A. $\displaystyle \sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$

B. $\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$

C. $\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$

Body:
\begin{align*}
\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2(n+k)}&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2}\int_0^1 x^{n+k-1}\,\text{d}x\\
&=\int_0^1 \left(\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right)\left(\sum_{n=1}^\infty\frac{x^{k-1}}{k^2}\right)\,\text{d}x\\
&=\int_0^1 \left(\text{Li}_2(x)\right)\left(\sum_{n=1}^\infty\frac{x^{k-1}}{k^2}\right)\,\text{d}x\\
&=\sum_{k=1}^\infty\frac1{k^2}\int_0^1 x^{k-1}\text{Li}_2(x)\,\text{d}x\\
&\overset{\text{IBP}}{=}\sum_{k=1}^\infty\frac1{k^2}\left(\frac{\zeta(2)}{k}-\frac{H_k}{k^2}\right)\\
&=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\frac{H_k}{k^4}\\
&=\zeta(2)\zeta(3)-[3\zeta(5)-\zeta(2)\zeta(3)]\\
&=2\zeta(2)\zeta(3)-3\zeta(5).\\
\end{align*}


问:
Show that if $m^∗(E) < ∞$ and there exist intervals $I_1, . . . , I_n$ such that $m∗E(∆(∪_{i=1}^{n}I_i))< ∞$, then each of the interval are finite
Show that if $m^∗(E) < ∞$ and there exist intervals $I_1, . . . , I_n$ such that $m^∗(E(∆(∪_{i=1}^{n}I_i)))< ∞$, then each of the interval $I_i$
is finite.

I have been asked to prove this and I am thinking along the lines of Littlewood's First Principle, but I haven't made much progress yet. Any help would be appreciated

答:
There exists an open set $U$ such that $E \subseteq U$ and $m(U) <m^{*}(E)+1$. We can write $U$ as  countable disjoint union of open intervals $(a_i,b_i)$. Now
$\bigcup_{i=1}^{n} (a_i,b_i)) \subseteq  (E\Delta \bigcup_{i=1}^{n} (a_i,b_i)) \cup E$ so $m(\bigcup_{i=1}^{n} (a_i,b_i))<\infty$. This implies that each of the $I_i$'s is a finite interval.


问:
Does $[0,+ \infty)$ counts for a closed interval in just real number system?
This is different from my previous question which asked about the situation in extended real number system.

So, we are discussing the situation when $X=\mathbb{R}$. Without the elements $\{-\infty, +\infty \}$, we could just define $[0,+\infty)$ formally to be $\{t\in X: t\geq 0 \}$. Since for any $x,y \in X, x<y$, we have $[x,y] \neq [0,+ \infty)$ because there exists $|y|+1 \in [0,+\infty)$ with $|y|+1$ not in $[x,y]$. Hence $[0,+\infty)$ is not a closed interval.

Well, is there any flaw in my understanding?

For reference, please see Topology by Munkres, page 84, chapter 2, section 14 the order topology:

page 85:

答:
Munkres is quite clear here: $[0,+\infty)$ is not a closed interval as it does not have an upper limit (the $b$ in the interval notation) in $\Bbb R$ but is a closed ray.

Both closed intervals and closed rays are closed sets in the order topology, as their name already suggests.

This is in $\Bbb R$

In the extended reals $+\infty$ is a point, so $[0,+\infty)$ is then a half-open interval, but it still contains the same points. It's no longer closed, as $+\infty$ is in its closure but not in it.


问:
Is there a simpler approach to these system of equations?
I recently came across the following system of equations:

$$x + y + z = 1 \\
x^2 + y^2 + z^2 = 2  \\
x^3 + y ^3 + z^3 = 3$$

And I have two questions:

One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set :

$$x + y + z = 1 \\ x^2 + y^2 + z^2 = 2 \\ ...\\x^n + y^n + z^n = n$$

Two, is there a simpler approach for the prior solution set than substitution? As of now, I'm not getting anywhere with this method. I end up getting into a long-winded series of substitution and isolation that yields something like $z = z$ or $1 = 1$.

Sorry if it's a repeat question; I couldn't exactly find an accurate way to search for the equations.

答:
As lab bhattacharjee has mentioned, you can view $x,y,z$ as roots of a degree 3 polynomial, and thereby obtain:

$$x+y+z=1$$
$$xy+yz+zx=-\frac{1}{2}$$
$$xyz=\frac{1}{6}$$

Vieta's Formulas
 relates the above expressions to the coefficients of polynomials, from which we can solve for $x,y,z$

In response to your other question:

Note that the first three relations:
$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$ completely define $x,y,z$ (up to symmetry), as these relations define a degree $3$ polynomial, which clearly has $3$ (not necessarily distinct) roots, which must be $x,y,z$ respeectively.

Thus $x^4+y^4+z^4$, for instance, is already defined as $\frac{31}{6}$, and not $4$, so the general question has no solutions.

The general identities $x_1^k+x_2^k+\dots+x_m^k=n_k$ for some polynomial with degree $m$ and roots $x_1,\dots,x_m$ and their relation to the coefficients of the polynomial are actually well-known as Newton's Identities, so you can actually easily derive the relations that lab bhattacharjee mentioned in his posts.


问:
Intuition behind $\exists x \in S, \forall y \in \emptyset, P(x)$
I am trying to prove the statement in the form:

$\exists x \in S, \forall y \in \emptyset, P(x)$

where $P(x): x > y$ and $S$ is an arbitrary non-empty set.

My proof looks something like:

Let $x =$ some number in set S. Let $y \in \emptyset$. My intuition is that since we cannot check whether $x > y$ since $y$ is an empty set, the statement is vacuously true. My questions is, how do I write this "intuition" in a rigorous way?

答: You could restate the claim as $\exists x\in S\neg\exists y\in\emptyset\neg P(x)$, which is trivial for each $x\in S$.


问:
Positive $a$, $b$, $c$ with $a^2+b^2+c^2=3$, show $(2-a)(2-b)(2-c) \geq \frac{25}{27}$

Let $a;b;c>0$ and $a^2+b^2+c^2=3$

Prove: $$(2-a)(2-b)(2-c) \geq \frac{25}{27}$$

I tried to use uvw method, but then  $$
(2−a)^2(2−b)≥25/27$$ with $$2a^2+b^2=3$$ is hard to finish by "paper and pencil"

Can anyone prove that problem without using Lagrange multipliers method?

Thank you.

答:
A hint. Suppose $c = \min \{a,\,b,\,c\}$ then $0 < c \leqslant 1.$

(1) Prove
$$(2-a)(2-b) \geqslant \frac{c^2+1}{2}.$$
(2) Find the minimum value of the expression
$$\frac{(2-c)(c^2+1)}{2}, \quad 0 < c \leqslant 1.$$


答:
This reduces to: $\exists x \colon y \in \{\} \implies P(x)$.

But $y \notin \{\}$.

The truth table of material implication/conditionals ('IF') definitionally maps False to True.

QED.

Title: Intuition behind $\exists x \in S, \forall y \in \emptyset, P(x)$

Body:
I am trying to prove the statement in the form:

$\exists x \in S, \forall y \in \emptyset, P(x)$

where $P(x): x > y$ and $S$ is an arbitrary non-empty set.

My proof looks something like:

Let $x =$ some number in set S. Let $y \in \emptyset$. My intuition is that since we cannot check whether $x > y$ since $y$ is an empty set, the statement is vacuously true. My questions is, how do I write this "intuition" in a rigorous way?


问:
BV[a,b]$\cap$C[a,b]$\neq$AC[a,b]
Take an $f\in$BV[0,1]$\cap$C[0,1] e.g. the Cantor function.
I take the Lebesgue Stiltigies measure of $f$:

$$
\mu_f((a,b])=f(b)-f(a).
$$

Now I have a finite positive measure, so I can do the Radon-Nikodim decomposition:

$$
\mu_f=\mu_1+\mu_2
$$

(I denote | $\ $| the Lebesgue measure)

With $\mu_1\perp |\ \ |$ and $\mu_2<<|\ \ |$. So $\mu_2= h $ d$|\ \ |$

I have that $\mu_f(\{x\})=0$ for all $x\in[0,1]$ because $f\in$C[0,1].

Can I say that $\mu_1=0$?

In this case, why have I that $f$ is not AC[0,1]? $f(x)=\mu_2([0,x])=\int_0^xh $ d$|\ \ |$ for all $x\in[0,1]$.

Thank you in advance! I think that the point is that $\mu_1\neq 0$ but I don't understand how a measure without 'atoms' could have the singular part not $0$.

答: A basic property of the Cantor function $f$ is $f'(x)=0$ a.e. (w.r.t Lebesgue measure). If $\mu_1=0$ then $\mu_f << ||$ which is equivalent to the fact that $f$ is an absoluetly continuous function.  If $f$ is absolutely continuous then $f(b)-f(a)=\int_a^{b}f'(t)dt=0$, a contradiction.

Title: BV[a,b]$\cap$C[a,b]$\neq$AC[a,b]


答:
Without loss of generality, let $a\geqslant b\geqslant c\geqslant0$. The target inequality is equivalent to: $$\left(\frac{1+{a}^{3}}{1+a{b}^{2}}-1\right)+\left(\frac{1+{b}^{3}}{1+b{c}^{2}}-1\right)+\left(\frac{1+{c}^{3}}{1+c{a}^{2}}-1\right)\geqslant0$$
That is
\begin{align*}
a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1+bc^2}+c\frac{c^2-a^2}{1+ca^2}\geqslant0\Leftrightarrow\\
a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1+bc^2}+c\left(\frac{c^2-b^2}{1+ca^2}+\frac{b^2-a^2}{1+ca^2}\right)\geqslant0\Leftrightarrow\\
(a^2-b^2)\left(\frac{a}{1+ab^2}-\frac{c}{1+ca^2}\right)+(b^2-c^2)\left(\frac{b}{1+bc^2}-\frac{c}{1+ca^2}\right)\geqslant0
\end{align*}
Think of the numerators of $\frac{a}{1+ab^2}-\frac{c}{1+ca^2}$ and $\frac{b}{1+bc^2}-\frac{c}{1+ca^2}$ are $$(a-c)+ac(a^2-b^2)\geqslant0~\mbox{and}~(b-c)+bc(a^2-c^2)\geqslant0$$Hence we complete the proof.</p>

Title: Prove ${\frac {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac {1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $

Body:
Let $a,b,c \ge0$, prove the on equality: $${\frac
 {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{  \frac
 {1+{c}^{3}}{1+c{a}^{2}}}\ge 3  $$

I tried: $$LHS = \sum\frac 1{1+ab^2}+\sum \frac {a^4}{a+a^2b^2} \ge\frac 9{3+\sum ab^2} + \frac {(a^2+b^2+c^2)^2}{\sum a+ \sum (ab)^2}\ge...$$
but it seem like useless.

Look like better inequality is $ \left( 1+{a}^{3} \right)  \left( 1+{b}^{3} \right)  \left( 1+{c}^{3}
 \right) \ge \left( 1+a{b}^{2} \right)  \left( 1+b{c}^{2} \right)
 \left( 1+c{a}^{2} \right)$ (but still can't prove it)


问:
If we know that $P^2=P$, show that $(I+P)^{-1}=I-\frac{1}{2}P$
the problem is as follows. given that P is a matrix with the property $P^2=P$ show that $(I+P)^{-1}=I-\frac{1}{2}P$.
I tried squaring both quantities and using the fact that I can interchange the exponents on the RHS but it did not yield anything helpful.

Body: Since $(I+P)(I-\tfrac12P)=I+\tfrac12P-\tfrac12P^2=I$, $(I+P)^{-1}=I-\tfrac12P$.


答:
$(I+P)(I-\frac{1}{2}P) = I - \frac{1}{2}P + P - \frac{1}{2}P^2 = I - \frac{1}{2}P + P - \frac{1}{2}P = I$.

Thus $I-\frac{1}{2}P$ is an inverse of $I+P$. But the inverse is uniquely determined and therefore $(I+P)^{-1} = I-\frac{1}{2}P$.

Title: If we know that $P^2=P$, show that $(I+P)^{-1}=I-\frac{1}{2}P$

Body:
the problem is as follows. given that P is a matrix with the property $P^2=P$ show that $(I+P)^{-1}=I-\frac{1}{2}P$.
I tried squaring both quantities and using the fact that I can interchange the exponents on the RHS but it did not yield anything helpful.


答:
If you have a candidate matrix that you'd like to show is the inverse of $(I+P)$, simply multiply it by $(I+P)$ and verify that you get the identity matrix $I$.

Try it yourself, then click to reveal the calculation:

! $$
\begin{align*}
(I - \tfrac{1}{2}P) (I + P)
&= I + P - \tfrac{1}{2}P - \tfrac{1}{2}\color{red}{P^2} \\
&= I + P - \tfrac{1}{2}P - \tfrac{1}{2}\color{red}{P} = I
\end{align*}
$$


问:
Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$
Let $ABC$ be a triangle. Prove that: $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) \ge \sqrt{3} $$
My attempt:
$$P=\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) = \frac{1}{2}\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)$$
$$\alpha = \cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$$
$$\Leftrightarrow \alpha \ge3.\sqrt[3]{\alpha} \Leftrightarrow \alpha^2\ge27\Leftrightarrow \alpha\ge3\sqrt{3}$$
$$\Rightarrow P=\frac{1}{2}\alpha\ge\frac{3\sqrt{3}}{2}$$
Hmm where I was wrong ?

答:
Let $p$ is the semi-perimeter, $R$ is circumradius and $r$ is radius.

We have
$$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C} = \frac{2R}{a} + \frac{2R}{b} + \frac{2R}{c} = \frac{4R+r}{2p}+\frac{p}{2r},$$
and
$$\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2} = \sum \frac{r}{b+c-a} = \frac{4R+r}{p}.$$
The inequality become
$$p \geqslant 2\sqrt3 r.$$
Which is true by know inequality $$p \geqslant 3\sqrt 3 r.$$


问:
Deriving the expected value of the normal distribution via a substitution
I am trying to compute the expected value, E$[x]$, of a random variable $X\sim\mathcal{N}(\mu,\sigma^2)$. The density function of the normal distribution is
$$f_X(x)=\frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)^2\right), \ \ \ -\infty<x<\infty.$$

I am attempting to use the following substitution to help find the expected value:
$$y=\frac{x-\mu}{\sigma\sqrt{2}}\implies dx=\sigma\sqrt{2} dy \tag{1}.$$

The expected value is computed as
\begin{align}
\text{E}[x]&=\int_{-\infty}^{\infty} x f_X(x) \ dx \\
&=\frac{1}{\sigma \sqrt{2\pi}}\int_{-\infty}^{\infty} x\exp\left(-\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)^2\right) \ dx \\
&=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}(\mu+\sigma\sqrt{2} y)\exp(-y^2)\ dy \ \ \ \ \text{(using substitution $(1)$)} \\
&=\frac{\mu}{\sqrt{\pi}}\int_{-\infty}^{\infty} \exp(-y^2) \ dy \ + \ \sigma\sqrt{\frac{2}{\pi}}\int_{-\infty}^{\infty} y\exp(-y^2) \ dy \\
&=\sigma\sqrt{\frac{2}{\pi}}\left(\left[-\frac{1}{2}\exp(-y^2)\right]_{-\infty}^{\infty}+\frac{1}{2}\int_{-\infty}^{\infty}\frac{\exp(-y^2)}{y} \ dy\right).
\end{align}
I am unsure what this simplifies to (specifically how to deal with the final integral). I've noticed that the integrand is odd (does the integral simply cancel?).

答:
Your substitution is terrible but it must work as well! the natural substitution is

$$Y=\frac{X-\mu}{\sigma}$$

Using this (in this case you are standardizing your Gaussian) the result is very easy.

Considering valid your procedure, reading your last but one passage, the sum is the following

$$\frac{\mu}{\sqrt{\pi}}\cdot \sqrt{\pi}+0=\mu$$

this because the first integral is the Gaussian integral (in the link you can find the easy proof too) and the second is the integral of an odd function over a symmetric domain around zero.


问:
A nonsensical argument
I am reading An Introduction to Formal Logic by Peter Smith. I was checking if the following argument ( Question no-8, Exercise-1 ) is valid or not:

All the slithy toves did gyre and gimble in the wabe. Some mome raths are slithy toves. Hence some mome raths did gyre and gimble in the wabe.

I concluded that it is a valid argument based on its inference pattern. But in the answer sheet, the author has written following:

This has the look of an inference of the shape:
All Ss are G. Some Ms are S. So, some Ms are G.

And a genuine inference of that form will of course be valid. However, our Jabberwockian
example is not a genuine inference, since the premisses and conclusion are nonsense, and make
no contentful claims. And if it isn’t a genuine inference, where we infer a contentful claim
from two other contentful claims, it can’t be a valid genuine inference in particular!

Only assuming that the words (slithy toves, gyre, gimble, wabe and mome raths) are meaningless makes this argument deductively invalid.The problem is that English is not hard and fast language, it cannot be that all meaningful words are written in dictionary. So, shouldn't this argument be valid?

答:
I wonder whether the author is trying to make the point that certain argument forms do not apply when one of the objects is vacuous.

However, in this case it is an $AII$ of the first figure, where this restriction does not apply.

(See https://proofwiki.org/wiki/Definition:Figure_of_Categorical_Syllogism for a link to where such things are discussed.)

It is stated that "some mome raths are slithy toves". Hence the existence of mome raths is asserted. From that assertion it follows that slithy toves exist, and it is further asserted that all such slithy toves also in fact gyre and gimble.

Therefore it is completely valid that:

given the fact that there exist some mome raths which are slithy toves, and

given that all slithy toves gyre and gimble

it is the case that some mome raths gyre and gimble.

It is important that we are not scared off by arguments making deductions about entities that may or may not exist, because in mathematics it is often the case that we don't yet know whether such entities exist -- but if they do exist, then we can deduce things about them which, in turn, may allow us to deduce whether or not such entities can exist or can be proven to exist.

Further considerations, which may or may not shed further light:

The validity of an argument is independent of the truth of its premises. An argument with false premises may well be valid, but you may deduce only that the conclusion is true if the premises are true.

Hence the difference between a valid argument and a theorem.

I have a battered old copy of E.J. Lemmon's "Beginning Logic" on my shelf which I used to teach myself this stuff. It contains an entire section on the syllogism, which is analysed in some breadth, if not depth.

Title: A nonsensical argument


问:
31st IMO 1990 shortlist p1
Question :  Is there a positive integer which can be written as the sum of 1990 consecutive positive integers and which can be written as a sum of two or more consecutive positive integers in just 1990 ways? (IMO 1990 shortlisted)

I know that $N=m+(m+1)+...+(m+1989)=5×199×(2m+1989)$, but it's hard to me to solve ' written as the sum of 1990 consecutive positive integers'

Please help me.

答:
If we sum up $a, a + 1, \dots, a + k$ with $a, k > 0$, then we get $(k + 1)a + \frac{k(k + 1)}2 = \frac 1 2(k + 1)(2a + k)$. Note that $k + 1 < 2a + k$ and they have different parities.

Conversely, given a number $N$ and a factorization $2N = uv$ with $1 < u < v$ and $u, v$ having different parities, we may solve out $k = u - 1$ and $a = \frac12(v - u + 1)$.

Therefore "the number of ways that $N$ can be written as a sum of two or more consecutive positive integers" is equal to "the number of ways that $2N$ can be written as a product $uv$ with $1 < u < v$ and $u\neq v \mod 2$".

Let's now consider the latter. We write $2N = 2^e t$ with $t$ odd. It is clear that, if we omit the condition $1 < u < v$, then the number of ways to write $2N$ as $uv$ with $u \neq v \mod 2$ is simply two times the number of divisors of $t$: every divisor $d$ of $t$ contributes to two ways, namely $(u, v) = (2^e d, \frac t d)$ or $(\frac t d, 2^e d)$.

Taking into account the condition $u < v$, we divide the result by $2$ and get just the number of divisors of $t$. Removing the case $u = 1$, we finally get the following:

If $N = 2^et$ with $t$ odd, then the number of ways that $N$ can be written as a sum of two or more consecutive positive integers is equal to $\tau(t) - 1$, where $\tau(t)$ stands for the number of divisors of $t$.

This being equal to $1990$, we see that $t$ has $1991$ divisors. Since $1991 = 11\times 181$, there are just two cases:

$t = p^{10}q^{180}$ with $p, q$ primes.

$t = p^{1990}$ with $p$ prime.

Now we also require that $N = 5 \times 199 \times (2m + 1989)$ for some integer $m$. This forces $N$ to be odd, and hence $e = 0$ and $N = t$.

Since both $5$ and $199$ divide $N$, the second case above is already impossible. Thus we are in the first case, and $p, q$ are $5$ and $199$, in some order.

This leads to two solutions:

$5 \times 199 \times (2m + 1989) = 5^{10}199^{180}$, or

$5 \times 199 \times (2m + 1989) = 199^{10}5^{180}$.

Both are obviously valid.

Hence the answer is "yes, there are in fact two such integers".

Title: 31st IMO 1990 shortlist p1


问:
Better way to prove $\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}a+\sqrt[3]{2}b+c|$
Question:- Let $a,b$ and $c$ be integers, not all equal to $0$. Show that

$$\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}a+\sqrt[3]{2}b+c|$$

This problem was proposed in a canadian journal. The presented solution is very hard to grasp. Is there a better way to prove this(may be using calculus)?

答:
Essentially the same proof, without the error of assuming $a,b,c>0$ and motivating how we come up with the first note:

The algebraic integer $\alpha_0=\sqrt[3]4a+\sqrt[3]2b+c$ has conjugates $\alpha_1=\zeta^2\sqrt[3]4a+\zeta\sqrt[3]2b+c$ and $\alpha_2=\zeta^4\sqrt[3]4a+\zeta^2\sqrt[3]2b+c$, $\zeta$ a primitive cube root of unity.

Since $\alpha_0\neq 0$ for all $(a,b,c)\in\mathbb{Z}^3-\{(0,0,0)\}$, the complex conjugates $\alpha_1$ and $\alpha_2$ are nonzero, hence the real numbers $\alpha_0$ and its number-theoretic norm $$
N_{\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q}}(\alpha_0)=\alpha_0\alpha_1\alpha_2=4a^3+2b^3+c^3-6abc
$$
have the same sign.  Since $\alpha_0\neq 0$, we have $\lvert N_{\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q}}(\alpha_0)\rvert\geq 1$.  So it remains to prove
$$
\alpha_1\alpha_2\leq 4a^2+3b^2+2c^2.
$$
Note that $\alpha_1\alpha_2=2\sqrt[3]2a^2+\dots$ is automatically positive, so the $-2\sqrt[3]2a^2-\dots\leq 4a^2+3b^2+2c^2$ part of the given proof is never required.


问:
Why is $\int_0^t g(s)dB_s \sim \mathcal{N}\left(0, \int_0^t g^2(s) ds\right)$
I've found in different references that $\int_0^t g(s)dB_s$ is a Wiener integral, where $g(s)$ is a bounded, continuous function and $dB_s$ is the standard Brownian Motion. Also, that $\int_0^t g(s)dB_s \sim \mathcal{N}(0, \int_0^t g^2(s) ds)$. Why is this?

My attempt so far, is to approximate it as a series of normal random variables. Since the increments $B_{t_{i+1}}-B_{t_i}$ have a distribution $\mathcal{N}(0, t_{i+1}-t_i)$ and they are pairwise independent. Then, thinking it as a Riemann integral $\sum_{i=0}^{\infty} g_i(s) (B_{t_{i+1}}-B_{t_i}) \sim \mathcal{N}(0, \sum_{i=0}^{\infty} g_i^2(s)(t_{i+1}-t_i))$ for some $s\in [t_i, t_{i+1}]$. My question is, is it posible to take the limit $\lim_{t_{i+1} \to t_i} \sum_{i=0}^{\infty} g_i^2(s) (t_{i+1}-t_i) = \int_0^t g^2(s)ds$ in this case? Why?

答:
We can "derive" the fact that Ito Integral $I(t)$ is normally distributed with mean zero and variance $\int_{s=0}^{s=t}g(s)^2ds$ using two fundamental properties of Ito Integrals:

(i) The martingale property

(ii) Ito Isometry

Sketch of why $\mathbb{E}[I(t)]=0$ and why $I(t) \sim N$:

The integral $I(t):=\int_{s=0}^{s=t}g(s)dB_s$ is defined as:

$$I(t):=\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)\left(B_{t_{i+1}}-B_{t_i}\right)$$

The limit above is in probability. Notice that the integrator $B(t)$ is (by definition of Ito Integral) forward-looking, and thus at any time $t_i$, the difference $\Delta(B_{t_i}):=B_{t_{i+1}}-B_{t_i}$ is independent from the integrand $g(t_i)$. Therefore:

$$\mathbb{E}[I(t)]=\mathbb{E}\left[\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)\Delta(B_{t_i})\right]=\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)\mathbb{E}\left[\Delta(B_{t_i})\right]=0$$

Because each $\Delta(B_{t_i})$ is normally distributed with mean zero (by definition of Brownian motion), the sum of all the $\Delta(B_{t_i})$ terms is also normally distributed, so indeed $I(t)$ is normally distributed.

Sketch of Ito Isometry: because $\mathbb{E}[I(t)]=0$, clearly the variance of $I(t)$ must equal to $\mathbb{E}[I(t)^2]$. Now:

$$\mathbb{E}\left[I(t)^2\right]=\mathbb{E}\left[\left(\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)\Delta(B_{t_i})\right)^2\right]=\mathbb{E}\left[\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)^2\Delta(B_{t_i})^2\right]$$

The above is true because all the cross-terms of the type $g(t_i)g(t_j)\Delta(B_{t_i})\Delta(B_{t_j})$ have expectation equal to zero whenever $j\neq i$.

The expectation $\mathbb{E}[\Delta(B_{t_i})^2]=t_{i+1}-t_i$ (by definition of Brownian motion), and so one can see that:

$$\mathbb{E}\left[\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)^2\Delta(B_{t_i})^2\right]\rightarrow\lim_{n \to \infty}\sum_{i=0}^{i=n-1}g(t_i)^2\Delta(t_i)\rightarrow \int_{s=0}^{s=t}g(s)ds$$


问:
Piecewise equivalence of trigonometric inverses
In this website, it is a theorem that:

$$ 2 \tan^{-1}(x)=  \begin{cases} \sin^{-1} \frac{2x}{1+x^2}, x \in [-1,1] \cr
 \pi-\sin^{-1} \frac{2x}{1+x^2}, x >1 \cr -\pi - \sin^{-1} \frac{2x}{1+x^2}, x<1  \end{cases}$$

Why is there such a piecewise definition? I considered proving equivalence of tan inverse and sine inverse and I was only able to achieve the first definition.

My proof:

We begin with $ \sin^{-1} \frac{2x}{1+x^2}$ , substitute $ x = \tan \phi$ , the previous result simplfies as $ \sin^{-1} ( 2 \tan \phi \cos^2 \phi) = \sin^{-1} ( 2 \sin \phi \cos \phi) = 2 \phi = 2\tan^{-1} (x)$.  Where is my mistake?

答:
The arcsine function, $\arcsin : [-1,1] \to [-\frac \pi 2, \frac \pi 2]$ ($\color{red}{\mathit{note\ the \  codomain}}$ , and the function is often written as $\sin^{-1}$ but I prefer $\arcsin$) is defined as follows : if $a \in [-1,1]$ we can find a unique $x \in [-\frac \pi 2, \frac \pi 2]$ such that $\sin x  =a$. We let $x = \arcsin a$.

Since there may be many $x$ such that $\sin x = a$, it is important to specify the codomain of the arcsine, so that we need not worry about multiple values.

Similarly, $\arctan : \mathbb R \to (-\frac \pi 2, \frac \pi 2)$ is defined as : for $a \in \mathbb R$ there is a unique $-\frac \pi 2 < x <\frac \pi 2$ with $\tan x =a$. We let $x = \arctan a$.

Now, we must prove the given statement. First of all, note that $-1\leq \frac{2x}{1+x^2} \leq 1$ for all values of $x$, so certainly $\arcsin \frac{2x}{1+x^2}$ is defined.

Now let $x \in \mathbb R$ and $\phi = \arctan x$ so that $x = \tan \phi$. Note that $-\frac \pi 2 < \phi < \frac \pi 2$. The argument you make shows that $$
\frac{2x}{1+x^2} = \sin 2 \phi
$$

and therefore, $$
\arcsin\left(\frac{2x}{1+x^2}\right) = \arcsin(\sin 2 \phi)
$$

Now, we would like to simplify the RHS. For this, we cannot simply cancel out the two functions, because of codomain issues.

For example, let $x = 2$. Then the LHS is $\arcsin \frac 45 = 53.13^\circ$. On the RHS : we have $\phi = \arctan 2 = 63.43^\circ$ so $2 \phi = 126.86^\circ$. So clearly, $LHS \neq 2 \phi$ always, which is what one would have expected if one could cancel the $\sin$ and $\arcsin$.

In fact, we have $\arcsin(\sin 2 \phi) = \theta$ where $\theta$ is the unique angle in $[-\frac \pi 2, \frac \pi 2]$ such that $\sin 2 \phi = \sin \theta$. Then we can write $\arcsin(\frac{2x}{1+x^2}) = \theta$. Now, we need the dependence between $2 \phi$ and $\theta$.

For this, note that the points of problem are those where the codomain breaks. For example at $\phi = \frac \pi 4$, we have $2 \phi = \frac \pi 2$ which is still in the $\arcsin$ codomain. Take $\phi$ a little higher and that is lost. So the relation between $\phi$ and $\theta$ is expected to change at the point $\frac \pi 4$, and similarly at $\phi = \frac {-\pi}4$ as well, for the lower end of the codomain breaks then.

So we can break into three cases depending upon these break points :

If $2 \phi \in [-\frac \pi 2 , \frac \pi 2]$ then of course $\theta = 2 \phi$.

If $2\phi \in (\frac \pi 2 , \pi]$ then $\sin(\pi - x) = \sin x$ gives $\sin(\pi - 2\phi) = \sin 2 \phi$ so $\theta = \pi - 2\phi$ (which lies in the codomain).

If $2 \phi \in [-\pi , -\frac \pi 2)$ then $\sin(-\pi-x) = -\sin(\pi + x) = \sin x$ so $\sin(-\pi - 2\phi) = \sin(2 \phi)$, and hence $\theta = -\pi - 2 \phi$.

Now, we need to express $\theta$ in terms of $x$, not $\phi$, and the break needs to be in terms of $x$. From our usual trigonometry relations, we know that $x>1, x<1,x \in [-1,1]$ if and only if $2 \phi > \frac \pi 2 , 2\phi < - \frac \pi 2, 2\phi \in [-\frac \pi 2, \frac \pi 2]$ respectively.

Therefore, we get :

If $-1 \leq x \leq 1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) = 2 \phi$.

If $x>1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) =\pi- 2 \phi$.

If $x<1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) =-\pi- 2 \phi$

Bring $2\phi$ to one side in each of these equations makes it the subject of the equation, and you have the definition given by the website. So the piecewise nature of the definition reflects the breaks in the arcsin codomain which can only be compensated on a case-by-case basis.


问:
Continuous Image of $S^{n-1}$ in $\mathbb{R}$
Let $f:S^{n-1}\to \mathbb{R}$ be a nonconstant continuous function.

$S^{n-1}$ is Compact & Connected$\implies$ Its image is of the form $[a,b]$ in $\mathbb{R}$.

For any $c \in (a,b)$: $\#\{f^{-1}(c)\}>1$.

If not, then $f:S^{n-1}\setminus \{f^{-1}(c)\}\to \mathbb{R}$ is continuous and since $S^{n-1}\setminus \{f^{-1}(c)\}$ is connected it image i.e $[a,c)\cup(c,b]$ is connected. Contradiction.

There exist $x \in S^{n-1}$ such that $f(x)=f(-x)$

Consider $g(x)=f(x)-f(-x) \implies g$ is an odd function. If $g(x)=0$ then we are done. If not, as $g(S^{n-1})$ is connected, either $g(x)>0$ or $g(x)<0$ for all $x \in S^{n-1}$. Contradicting the fact that $g$ is odd.

Geometrically I can see that:

In $S^1$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ has atleast two points.

In $S^2$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ should atleast contain a shape of homeomorphic to a circle i.e $S^1$ (Is this true?)

My Question: Can we tell that in $S^{n-1}(n\ge3)$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ contains a set homeomorphic to $S^{n-2}$?

答:
My Question: Can we tell that in $S^{n-1}(n\ge3)$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ contains a set homeomorphic to $S^{n-2}$?

No.

Consider any closed subset $A\subseteq S^{n-1}$ such that $A$ disconnects the sphere into two nonempty open components $U_1,U_2$. Since $S^{n-1}$ is perfectly normal then we can construct $f:S^{n-1}\to [-1,1]$ such that $f^{-1}(0)=A$ and $f^{-1}(-1)$ is a singleton in $U_1$ while $f^{-1}(1)$ is a singleton in $U_2$.

All we need to have now is a subset $A\subseteq S^{n-1}$ that is closed, disconnects the sphere but it does not contain a copy of $S^{n-2}$. For $n=3$ this can be simply done by taking the (closed) topologist's sine curve around $S^{2}$. This then can be lifted to higher dimensions.


Title: Positive $a$, $b$, $c$ with $a^2+b^2+c^2=3$, show $(2-a)(2-b)(2-c) \geq \frac{25}{27}$

Body:
Let $a;b;c>0$ and $a^2+b^2+c^2=3$

Prove: $$(2-a)(2-b)(2-c) \geq \frac{25}{27}$$

I tried to use uvw method, but then  $$
(2−a)^2(2−b)≥25/27$$ with $$2a^2+b^2=3$$ is hard to finish by "paper and pencil"

Can anyone prove that problem without using Lagrange multipliers method?

Thank you.

Body:
$WLOG$ $\min\left \{ a,b,c \right \}=c$ , then $0<c\le 1$.

Claim 1: $(2-a)(2-b)\ge \frac{c^2+1}{2}$.

Claim 1 $\Leftrightarrow (2-a)(2-b)\ge \frac{4-a^2-b^2}{2}$

$\Leftrightarrow (a+b-2)^2\ge 0$ , which is obvious.

Claim 2: $\frac{(2-c)(c^2+1)}{2}\ge \frac{25}{27}$

Let $f(x)=(2-x)(x^2+1)=-x^3+2x^2-x+2$.

Then $f'(x)=-3x^2+4x-1=-(3x-1)(x-1)$.

So $f(c)\ge f(\frac{1}{3})=\frac{50}{27}\Rightarrow \frac{(2-c)(c^2+1)}{2}\ge \frac{25}{27}$.

Hence, we are done.


问:
Words of length $10$ in alphabet $\{a,b,c\}$ such that the letter $a$ is always doubled

Compute the number of words of length $10$ in alphabet $\{a,b,c\}$ such that letter $a$ is always doubled (for example "$aabcbcbcaa$" is allowed but "$abcbcaabcc$" is forbidden).

I am looking for a quick/efficient way to resolve this problem. I thought of fixing "$aa$" in the beginning then draw a tree of the next possibilities but this tree will end up to be a whole forest. Can you help me ?

答:
Say there are $i$ occurrence of 'aa' that you "tape" together.  The you have a total of $10-i$ slots, of which $i$ has to be 'aa' and the other $10-2i$ are either 'b' or 'c'.  That gives the total
$$
\sum_{i=0}^5\sum_{j=0}^{10-2i}\binom{10-i}i \binom{10-2i}j.
$$


问:
Prove $f(x) = x^2 + 3$ is continuous at $x=3$
Prove that $f(x) = x^2 + 3$ is continuous at $x=3$.

I have tried using $\delta = \sqrt{\epsilon + 9} - 3$.

I tried to split $|x^2-9| = |(x-3)(x+3)|$ and tried to make $x+3$ in terms of $\delta$.

But I get $\delta^2 + 6\delta$.

I don't really get what I'm doing wrong or right so I need some help with finding the right $\delta$.

Also, I cannot really understand the goal of the proof, I read on elsewhere that I need to make $\delta$ as small as possible?

答:
You need to solve the equation

$$0<|x-3|<\delta\implies|(x^2+3)-(3^2+3)|<\epsilon$$ for $\delta$.

As the function $x^2+3$ is monotonic around $x=3$, you can solve the equation

$$|x^2-9|=\epsilon$$ and use any $\delta$ such that $(3-\delta,3+\delta)$ is wholly contained between the two solutions, $x=\sqrt{9\pm\epsilon}.$ (Other possibilities are $x=-\sqrt{9\pm\epsilon}$, but this does not straddle $x=3$.)

The largest value of $\delta$ is thus

$$\min(\sqrt{9+\epsilon}-3,3-\sqrt{9-\epsilon})$$
and this is always $\sqrt{9+\epsilon}-3$.

Check:

$$-\delta<x-3<\delta\implies3-\delta<x<3+\delta\implies x^2<(6-\sqrt{9+\epsilon})^2,(\sqrt{9+\epsilon})^2.$$

But

$$(6-\sqrt{9+\epsilon})^2=36-12\sqrt{9+\epsilon}+9+\epsilon\ge 9-\epsilon$$

because

$$36+2\epsilon\ge 12\sqrt{9+\epsilon}$$
$$1296+144\epsilon+4\epsilon^2\ge 1296+144\epsilon$$

and $$\epsilon^2>0.$$


答:
You can break this up into cases based on how many aa pairs there are, which can range from $0$ to $5$. Consider, for example, the case that you have $3$ aa pairs. Then you have four other letters that are all either b or c, so there are $2^4$ ways to choose the subsequence formed by those letters. Then thinking of each aa pair as a single letter, you have $7$ letters, and you need to choose $3$ of them to be an aa pair. This gives $2^4{7\choose 3}$ words with $3$ aa pairs.

In general, the number of words of this form with $i$ aa pairs is $2^{10-2i}{10-i\choose i}$. Then add up all terms of this form as $i$ goes from $0$ to $5$.

Title: Words of length $10$ in alphabet $\{a,b,c\}$ such that the letter $a$ is always doubled

Body:
Compute the number of words of length $10$ in alphabet $\{a,b,c\}$ such that letter $a$ is always doubled (for example "$aabcbcbcaa$" is allowed but "$abcbcaabcc$" is forbidden).

I am looking for a quick/efficient way to resolve this problem. I thought of fixing "$aa$" in the beginning then draw a tree of the next possibilities but this tree will end up to be a whole forest. Can you help me ?


问:
Ring of integers of p-adic field
My tutor recently mentioned the following result:

"For any complete discretely valued extension $K$ of $\mathbb{Q}_p$ with perfect residue field $k$, the ring of integers $\mathcal{O}_K$ can be written as a quotient of $W(k)[[T]]$ by a principal ideal generated by an Eisenstein polynomial"

I have tried looking at various books on local fields but this has not helped. Many hours of Internet search have not helped either. Could someone please tell me where I could find a proof of this fact?

答:
When $k$ is a finite field (ie. $K/\Bbb{Q}_p$ is algebraic) then $W(k)$ is a fancy way to say $\Bbb{Z}_p[\zeta_{q-1}]$ where $q=|k|$, the ring of integers of the largest unramified extension $\subset K$.

$\zeta_{q-1}\in K$ by Hensel lemma and $$O_K=\Bbb{Z}_p[\zeta_{q-1}][[\pi_K]]=\Bbb{Z}_p[\zeta_{q-1},\pi_K]\cong \Bbb{Z}_p[\zeta_{q-1}][T]/(f(T))$$ where $\pi_K$ is a generator of the maximal ideal of $O_K$ and $f(T)\in \Bbb{Z}_p[\zeta_{q-1}][T]$ is its minimal polynomial, which is Eisenstein, because $\pi_K^e = pu$ for some $e$ and $u\in O_K^\times$,

from which $\deg(f)= [K:\Bbb{Q}_p[\zeta_{q-1}]]=e$, which  follows from $$O_K=\{ \sum_{j\ge 0} c_j \pi_K^j, c_j\in 0\cup \{\zeta_{q-1}^l\}\} = \sum_{j=0}^{e-1}\pi_K^j  \Bbb{Z}_p[\zeta_{q-1}]$$

When $k$ is a more general residue field then the idea is the same: by Hensel lemma the unramified polynomials have roots in $K$ so there is a largest unramified extension $L$ such that $O_L/(p)=k$ and $O_K=O_L[[\pi_K]]\cong O_L[[T]]/(f(T))$.

Title: Ring of integers of p-adic field


问:
Question about asymptotic distribution of order statistics.
Let $U_i$'s be i.i.d. Uniform$(0,1)$ random variables, and let $X_n = min\{U_1, \dots , U_n \}, Y_n = max\{ U_1 , \dots , U_n  \}, Z_n = Y_n - X_n $.
Then find asymptotic distribution of $n(1-Z_n)$.

I understand how to find asymptotic distribution of $X_n , Y_n$. But this problem makes me confused. Also I tried to use this method($X_{(r)} \sim Beta(r,n-r+1)$), but it failed.

Please help!

I know $Z_n=Y_n-X_n \sim Beta(n-1,2)$.
So, cdf of $Z_n$ is

$P(Z_n \leq t) = n(n-1)\int_{0}^{t} x^{n-2}(1-x)dx$

$P(1-Z_n \leq t) = 1-n(n-1) \int_{0}^{1-t}x^{n-2}(1-x)dx$

$P(n(1-Z_n) \leq t) = 1-n(n-1)\int_{0}^{1-t/n}x^{n-2}(1-x)dx$

I don't know what to do after this.

答:
starting from your Range density that is a $Z_n\sim Beta(n-1;2)$ transforming your

$$Y_n=n(1-Z_n)$$

with usual technics:

$$f_Y(y)=f_X(g^{-1}(y))|\frac{d}{dy}g^{-1}(y)|$$

you get

$$f_{Y_n}(y)=\frac{n-1}{n}\cdot y\cdot \left(1-\frac{y}{n}\right)^{n-2}$$

whose limit results to me

$$\lim\limits_{n \to \infty}f_{Y_n}(y)=ye^{-y}$$

which is the density of a $Gamma(2;1)$


问:
Divisibility of the difference between two very large numbers
I am learning from chapter 1 of the algorithm textbook known as DPV.

There is a question 1.11 that asks:

Is $4^{1536} - 9^{4824}$ divisible by 35?

The whole chapter is on modular arithmetic, modular exponentiation, Euclid GCD (as far I got so far) so I broke down the first term to

$2^{3072}$ and second term to

$3^{2(2048)} * 3^{2(2048)} * 3^{3(512)} * 3^{2(128)} * 3^{3(64)} * 3^{3(16)} * 3^{3(8)}$ thinking it would lead to some sort of aha moment but the only conclusions I can draw is that the first term's factors are all divisible by 2, the second term's are all divisible by 3 and 35 has factors 5 and 7 so neither is divisible by 35 on its own.  How can I apply any of the modular arithmetic or Euclid's GCD to this problem to solve it?

答:
Note that if $4^{1536}\equiv 9^{4824} \bmod 35$ then $35$ divides $4^{1536}- 9^{4824}$, otherwise not.

There are various tools that could help you decide this, and if you have heard of Euler's theorem and the Chinese remainder theorem these could certainly help, but with just the idea of modular exponentiation you can investigate the powers of $4$ and $9$ $\bmod 35$ and reach a conclusion.

Successive powers of $4\bmod 35$ look like:
$4^1 \equiv \color{blue}{4},
4^2\equiv \color{blue}{16},
4^3\equiv 64\equiv \color{blue}{29},
4^4\equiv 116\equiv \color{blue}{11},
4^5\equiv 44\equiv \color{blue}{9},
4^6\equiv 36\equiv \color{red}{1}$

Successive powers of $9\bmod 35$ look like:
$9^1 \equiv \color{blue}{9},
9^2\equiv 81\equiv \color{blue}{11},
9^3\equiv 99\equiv \color{blue}{29}
9^4\equiv 261\equiv \color{blue}{16},
9^5\equiv 144\equiv \color{blue}{4},
9^6\equiv 36\equiv \color{red}{1}$

(If you know about modular inverse, you can also see from the powers of $4$ that $4^{-1}\equiv 9 \bmod 35$ which is why the two power value sequences are mirror images of each other.)

And then casting out multiples of $6$ in the exponent allows us to finish the calculations without difficulty; for example $4^{23} = 4^{3\cdot\color{red}{6}+\color{blue}{5}}\equiv \color{red}{1}^3\cdot 4^{\color{blue}{5}}\equiv \color{red}{1}\cdot \color{blue}{9}\equiv 9\bmod 35$.


答:
The definition of a logical sentence requires it to only use terms of some predefined language (in this case, English) and possibly some free variables. Even if English is naturally a flexible language, when you talk logic, you must treat it as a formal language, and therefore you're not allow to freely add new terms.

The expressions that you're given look like they are logical sentences, they seem to have the correct grammar, but since they use terms that don't exist in the accepted language, they are not. And you cannot apply logic to expressions that are not logical sentences.

You could however define a new language, let's call it an expanded English that does include these terms and then it will be a logical sentence in that language, and the reasoning will be valid. The thing is, it should be done beforehand, and if it is not stated, it is presumed that the language you're working in is normal English.

Title: A nonsensical argument

Body:
I am reading An Introduction to Formal Logic by Peter Smith. I was checking if the following argument ( Question no-8, Exercise-1 ) is valid or not:

All the slithy toves did gyre and gimble in the wabe. Some mome raths are slithy toves. Hence some mome raths did gyre and gimble in the wabe.

I concluded that it is a valid argument based on its inference pattern. But in the answer sheet, the author has written following:

This has the look of an inference of the shape:
All Ss are G. Some Ms are S. So, some Ms are G.

And a genuine inference of that form will of course be valid. However, our Jabberwockian
example is not a genuine inference, since the premisses and conclusion are nonsense, and make
no contentful claims. And if it isn’t a genuine inference, where we infer a contentful claim
from two other contentful claims, it can’t be a valid genuine inference in particular!

Only assuming that the words (slithy toves, gyre, gimble, wabe and mome raths) are meaningless makes this argument deductively invalid.The problem is that English is not hard and fast language, it cannot be that all meaningful words are written in dictionary. So, shouldn't this argument be valid?