问:
Group law of elliptic curve under transformation
For example, elliptic curve $E:y^2＝x^3-x$ can be transformed
to move the identity element to the origin, $(z,w)=(0,0)$, we do the change of coordinates $z=-x/y$, $w=-1/y$. The equation of this curve then becomes
$w=z^3+zw^2\quad$

My question is
; Does E':$w=z^3+zw^2 $have group structure, just like ' take two points from $E'$,and line through those points intersects at $E'$ , and connect the point with $（0,0）∈（z,w）$, then the line intersects with $E'$ again, and this is the sum of first two points' ?

The graph of $w=z^3+zw^2\quad$seems completely different in affine plane from usual $y^2＝x^3-x$, so I'm suspicious. But maybe we can say the group structure above holds because they are both same Weierstrass from in projective plane, so is isomorphic to Picard group of $E$.

答:
We divide by $-y^3=(-y)^3$ both sides in $y^2 =x^3-x$.

Then we get
$(-1/y)=(-x/y)^3 \color{red}{-} (-x/y)(-1/y)^2$, so the new curve has the affine equation $w=z^3-zw^2$.

For the homogenized equation,
$$
\begin{aligned}
&E'\ &:\ WT^2&=Z^3-ZW^2\ ,\qquad\text{associated to}\\
&E'_{\text{affine}}\ &:\ w&=z^3-zw^2\ ,
\end{aligned}
$$
obtained using the correspondence
$w=W/T$, $z=Z/T$,
so that an affine point $(z,w)$ goes to $[Z:W:T]=[z:w:1]$.

There are three points "at infinity" on $E'$ (w.r.t. the chosen affine part), the ones with $T=0$: They are $\Omega_a=[a:1:0]$ with $a$ among $-1$, $0$, $1$.
They correspond to the three points $(a,0)$ on $E$.

We will work in the sequel over $\Bbb Q$.

The line $Z=0$ intersects the curve $E'$ in three points, $O:=[0:0:1]=(0,0)$ with multiplicity one and $\Omega$ with multiplicity two.
This should be considered when writing formulas. Below, i will consider only the "generic case", which is the case when all constructed points involved are $\ne \Omega$.

We will describe the operations which makes a posteriori $O\in E'(\Bbb Q)$ to be the neutral element of the group $(E'(\Bbb Q),+)$.

Consider two $\Bbb Q$-rational affine points, $P_1=(z_1,w_1)$, $P_2=(z_2,w_2)$.
The equation of the line $P_1P_2$ is then of the shape
$$
(L)\ :\ w = mz+n\ ,
$$
with $m=\displaystyle\frac{w_1-w_2}{z_1-z_2}$, which makes sense for power series purposes around $O=(0,0)$ since $w$ has order $3$, $z$ order $1$ around zero, and unfortunately we will also need
$n=\displaystyle\frac{z_1w_2-z_2w_1}{z_1-z_2}$. In case the two points conicide, $P_1=P_2$, the line has the "same" equation, but $m,n$ are obtained by formally differentiating $(E')$, to get
$$dw =3z^2\; dz-w^2\; dz-2zw\; dw\ .$$
Formally replacing $(z,w)$ by $(z_1,w_1)$, and $dz, dw$ by $(z-z_1)$ and respectively $(w-w_1)$ we obtain the first order Taylor polynomial approximation around $P_1$. This gives $(1+2z_1w_1)\; dw= (3z_1^2-w_1^2)\; dz$. So $\displaystyle m=\frac
{3z_1^2-w_1^2}{1+2z_1w_1}$, which is also fine around $O$. Then set $n$ again so that $w_1=mz_1+n$.

The system with two equations, $(E')$ and $(L)$, in the two unknowns has three solutions, Bezout, we obtain an equation of degree three in $z$ by the linear substitution of $w$ from $(L)$ in $(E')$, thus getting:
$$
0 = z^3-z(mz+n)^2-(mz+n)\ .
$$

The coefficient in $z^3$ is $1-m^2$.

The coefficient in $z^2$ is $-2mn$.

So the third point of intersection $P_3'=(z_3',w_3')$, counted with multiplicities, has the $z$-component delivered by Vieta:
$$z_3'=-z_1-z_2+\frac{2mn}{1-m^2}\ .$$

The $w$ component of $P_3'$ is then
$$w_3'=mz_3'+n\ .$$

Now we consider the line $OR_3'$, which intersects $E'$ in a third point $R_3$. This point is the point we need.
Now note that together with a point $(z,w)$ on $(E')$ the point $(-z,-w)$ is also a point on $(E')$. With this observation we can directly accept the formula for $R_3=(z_3,w_3):=(-z_3',-w_3')$:
$$
\begin{aligned}
z_3 &=z_1+z_2-\frac{2mn}{1-m^2}\ ,\\
w_3 &=mz_3-n\ .
\end{aligned}
$$

To have some power series experiment, i will use sage, and some point with coordinates in the power series ring $R=\Bbb Q[[ q ]]$. If this feels alien while regarding  the target in the OP, please ignore the following. I insert this since some power series application was suggested in the comments.

Investigations on this path have their origin in

Arithmetic of Elliptic Curves, John T. Tate, §3, Expansion near Zero, The Formal Group.

With a completely different starting point and target, rather for my purposes, there is still some open project in my bottom drawer related to the following experiment, which uses power series coming from the modular parametrization...

I will use sage in the following.

The addition and multiplication of "blind tuples", corresponding to affine points on the given curve, is implemented as follows.

```<code>def add(P1, P2):
    z1, w1 = P1
    z2, w2 = P2
    if z1 != z2:
        m = (w1 - w2)/(z1 - z2)
        # and n = (z1*w2 - z2*w1)/(z1 - z2)
    else:
        m = (3*z1^2 - w1^2)/(1 + 2*z1*w1)
    n = w1 - m*z1    # in both cases

    z3 = z1 + z2  - 2*m*n/(1 - m^2)
    w3 = m*z3 - n
    return (z3, w3)

def mul(n, P):
    if n not in ZZ: return None
    if n == 0: return (0,0)
    if n == 1: return P
    if n < 0:
        z, w = mul(-n, P)
        return (-z, -w)
    if n % 2:
        return add(P, mul((n-1), P))
    # else n is an even positive integer and we compute
    Q = mul(ZZ(n/2), P)
    return add(Q, Q)
</code>```

Now we can start experimenting.

We may start with $z=q$, and ask for the corresponding lift of $w$ in the power series ring in $q$. Since in our special case $w$ is the solution of the equation of second degree $qw^2+w-q^3=0$, we already have a formula for a starting point.
We will compute some multiples of this point.

```<code>R.<q> = PowerSeriesRing(QQ, default_prec=40)
w = 1/2/q*(-1 + sqrt(1+4*q^4))
P = (q, w)

for n in [1..5]:
    Q = mul(n, P)
    print(f"{n} * P has the components:")
    print('\t', Q[0] + O(q^20))
    print('\t', Q[1] + O(q^20))
</code>```

This gives:

```<code>1 * P has the components:
     q + O(q^20)
     q^3 - q^7 + 2*q^11 - 5*q^15 + 14*q^19 + O(q^20)
2 * P has the components:
     2*q + 12*q^5 + 44*q^9 + 184*q^13 + 716*q^17 + O(q^20)
     8*q^3 + 16*q^7 + 112*q^11 + 288*q^15 + 1712*q^19 + O(q^20)
3 * P has the components:
     3*q + 96*q^5 + 2432*q^9 + 62976*q^13 + 1628160*q^17 + O(q^20)
     27*q^3 + 405*q^7 + 13014*q^11 + 314793*q^15 + 8325594*q^19 + O(q^20)
4 * P has the components:
     4*q + 408*q^5 + 34136*q^9 + 2904688*q^13 + 247097240*q^17 + O(q^20)
     64*q^3 + 3200*q^7 + 326528*q^11 + 26521856*q^15 + 2285282688*q^19 + O(q^20)
5 * P has the components:
     5*q + 1248*q^5 + 257920*q^9 + 54156800*q^13 + 11371264000*q^17 + O(q^20)
     125*q^3 + 15475*q^7 + 3862810*q^11 + 777959167*q^15 + 165141476790*q^19 + O(q^20)
</code>```

We can check that the above points are on the curve. For instance for the last point...

```<code>sage: z5, w5 = mul(5, P)
sage: w5 == z5^3 - z5*w5^2
True

sage: z5 + O(q^20)
5*q + 1248*q^5 + 257920*q^9 + 54156800*q^13 + 11371264000*q^17 + O(q^20)
sage: w5 + O(q^20)
125*q^3 + 15475*q^7 + 3862810*q^11 + 777959167*q^15 + 165141476790*q^19 + O(q^20)
</code>```

The experiment using power series related to the modular parametrization are of the shape

```<code>E = EllipticCurve(QQ, [-1, 0])
X, Y = E.modular_parametrization().power_series(40)
R = X.parent()    # power series ring in q truncated with precision 40
R.inject_variables()

print("Modular parametrization of the Fermat curve y² = x³ - x:")
print(f"X = {X}")
print(f"Y = {Y}")
print("\nWe associate Z = -X/Y and W = -1/Y:")
Z, W = -X/Y, -1/Y
print(f"Z = {Z}")
print(f"W = {W}")

print("\n\nThen using the point P = (Z, W):")
P = (Z, W)
for n in [1..5]:
    Q = mul(n, P)
    print(f"{n} * P has the components:")
    print('\t', Q[0] + O(q^20))
    print('\t', Q[1] + O(q^20))
</code>```

This gives:

```<code>Modular parametrization of the Fermat curve y² = x³ - x:
X = q^-2 + q^2 + q^6 - q^18 - q^22 + q^30 + 2*q^34 + O(q^38)
Y = -q^-3 - q - 2*q^5 - q^9 + 2*q^17 + 2*q^21 + q^25 - q^29 - 4*q^33 + O(q^37)

We associate Z = -X/Y and W = -1/Y:
Z = q - q^9 + 2*q^17 - 3*q^25 + 4*q^33 + O(q^41)
W = q^3 - q^7 - q^11 + 2*q^15 + q^19 - 2*q^23 - 2*q^27 + 2*q^31 + 4*q^35 - 4*q^39 + O(q^43)

Then using the point P = (Z, W):
1 * P has the components:
     q - q^9 + 2*q^17 + O(q^20)
     q^3 - q^7 - q^11 + 2*q^15 + q^19 + O(q^20)
2 * P has the components:
     2*q + 12*q^5 + 42*q^9 + 124*q^13 + 324*q^17 + O(q^20)
     8*q^3 + 16*q^7 + 88*q^11 + 176*q^15 + 552*q^19 + O(q^20)
3 * P has the components:
     3*q + 96*q^5 + 2429*q^9 + 62496*q^13 + 1606278*q^17 + O(q^20)
     27*q^3 + 405*q^7 + 12933*q^11 + 311958*q^15 + 8182683*q^19 + O(q^20)
4 * P has the components:
     4*q + 408*q^5 + 34132*q^9 + 2902648*q^13 + 246790024*q^17 + O(q^20)
     64*q^3 + 3200*q^7 + 326336*q^11 + 26499456*q^15 + 2281691456*q^19 + O(q^20)
5 * P has the components:
     5*q + 1248*q^5 + 257915*q^9 + 54150560*q^13 + 11368942730*q^17 + O(q^20)
     125*q^3 + 15475*q^7 + 3862435*q^11 + 777850842*q^15 + 165098987005*q^19 + O(q^20)
</code>```

Body:
We divide by $-y^3=(-y)^3$ both sides in $y^2 =x^3-x$.

Then we get
$(-1/y)=(-x/y)^3 \color{red}{-} (-x/y)(-1/y)^2$, so the new curve has the affine equation $w=z^3-zw^2$.

For the homogenized equation,
$$
\begin{aligned}
&E'\ &:\ WT^2&=Z^3-ZW^2\ ,\qquad\text{associated to}\\
&E'_{\text{affine}}\ &:\ w&=z^3-zw^2\ ,
\end{aligned}
$$
obtained using the correspondence
$w=W/T$, $z=Z/T$,
so that an affine point $(z,w)$ goes to $[Z:W:T]=[z:w:1]$.

There are three points "at infinity" on $E'$ (w.r.t. the chosen affine part), the ones with $T=0$: They are $\Omega_a=[a:1:0]$ with $a$ among $-1$, $0$, $1$.
They correspond to the three points $(a,0)$ on $E$.

We will work in the sequel over $\Bbb Q$.

The line $Z=0$ intersects the curve $E'$ in three points, $O:=[0:0:1]=(0,0)$ with multiplicity one and $\Omega$ with multiplicity two.
This should be considered when writing formulas. Below, i will consider only the "generic case", which is the case when all constructed points involved are $\ne \Omega$.

We will describe the operations which makes a posteriori $O\in E'(\Bbb Q)$ to be the neutral element of the group $(E'(\Bbb Q),+)$.

Consider two $\Bbb Q$-rational affine points, $P_1=(z_1,w_1)$, $P_2=(z_2,w_2)$.
The equation of the line $P_1P_2$ is then of the shape
$$
(L)\ :\ w = mz+n\ ,
$$
with $m=\displaystyle\frac{w_1-w_2}{z_1-z_2}$, which makes sense for power series purposes around $O=(0,0)$ since $w$ has order $3$, $z$ order $1$ around zero, and unfortunately we will also need
$n=\displaystyle\frac{z_1w_2-z_2w_1}{z_1-z_2}$. In case the two points conicide, $P_1=P_2$, the line has the "same" equation, but $m,n$ are obtained by formally differentiating $(E')$, to get
$$dw =3z^2\; dz-w^2\; dz-2zw\; dw\ .$$
Formally replacing $(z,w)$ by $(z_1,w_1)$, and $dz, dw$ by $(z-z_1)$ and respectively $(w-w_1)$ we obtain the first order Taylor polynomial approximation around $P_1$. This gives $(1+2z_1w_1)\; dw= (3z_1^2-w_1^2)\; dz$. So $\displaystyle m=\frac
{3z_1^2-w_1^2}{1+2z_1w_1}$, which is also fine around $O$. Then set $n$ again so that $w_1=mz_1+n$.

The system with two equations, $(E')$ and $(L)$, in the two unknowns has three solutions, Bezout, we obtain an equation of degree three in $z$ by the linear substitution of $w$ from $(L)$ in $(E')$, thus getting:
$$
0 = z^3-z(mz+n)^2-(mz+n)\ .
$$

The coefficient in $z^3$ is $1-m^2$.

The coefficient in $z^2$ is $-2mn$.

So the third point of intersection $P_3'=(z_3',w_3')$, counted with multiplicities, has the $z$-component delivered by Vieta:
$$z_3'=-z_1-z_2+\frac{2mn}{1-m^2}\ .$$

The $w$ component of $P_3'$ is then
$$w_3'=mz_3'+n\ .$$

Now we consider the line $OR_3'$, which intersects $E'$ in a third point $R_3$. This point is the point we need.
Now note that together with a point $(z,w)$ on $(E')$ the point $(-z,-w)$ is also a point on $(E')$. With this observation we can directly accept the formula for $R_3=(z_3,w_3):=(-z_3',-w_3')$:
$$
\begin{aligned}
z_3 &=z_1+z_2-\frac{2mn}{1-m^2}\ ,\\
w_3 &=mz_3-n\ .
\end{aligned}
$$

To have some power series experiment, i will use sage, and some point with coordinates in the power series ring $R=\Bbb Q[[ q ]]$. If this feels alien while regarding  the target in the OP, please ignore the following. I insert this since some power series application was suggested in the comments.

Investigations on this path have their origin in

Arithmetic of Elliptic Curves, John T. Tate, §3, Expansion near Zero, The Formal Group.

With a completely different starting point and target, rather for my purposes, there is still some open project in my bottom drawer related to the following experiment, which uses power series coming from the modular parametrization...

I will use sage in the following.

The addition and multiplication of "blind tuples", corresponding to affine points on the given curve, is implemented as follows.

def add(P1, P2):
    z1, w1 = P1
    z2, w2 = P2
    if z1 != z2:
        m = (w1 - w2)/(z1 - z2)
        # and n = (z1*w2 - z2*w1)/(z1 - z2)
    else:
        m = (3*z1^2 - w1^2)/(1 + 2*z1*w1)
    n = w1 - m*z1    # in both cases

    z3 = z1 + z2  - 2*m*n/(1 - m^2)
    w3 = m*z3 - n
    return (z3, w3)

def mul(n, P):
    if n not in ZZ: return None
    if n == 0: return (0,0)
    if n == 1: return P
    if n < 0:
        z, w = mul(-n, P)
        return (-z, -w)
    if n % 2:
        return add(P, mul((n-1), P))
    # else n is an even positive integer and we compute
    Q = mul(ZZ(n/2), P)
    return add(Q, Q)

Now we can start experimenting.

We may start with $z=q$, and ask for the corresponding lift of $w$ in the power series ring in $q$. Since in our special case $w$ is the solution of the equation of second degree $qw^2+w-q^3=0$, we already have a formula for a starting point.
We will compute some multiples of this point.

R.<q> = PowerSeriesRing(QQ, default_prec=40)
w = 1/2/q*(-1 + sqrt(1+4*q^4))
P = (q, w)

for n in [1..5]:
    Q = mul(n, P)
    print(f"{n} * P has the components:")
    print('\t', Q[0] + O(q^20))
    print('\t', Q[1] + O(q^20))

This gives:

1 * P has the components:
     q + O(q^20)
     q^3 - q^7 + 2*q^11 - 5*q^15 + 14*q^19 + O(q^20)
2 * P has the components:
     2*q + 12*q^5 + 44*q^9 + 184*q^13 + 716*q^17 + O(q^20)
     8*q^3 + 16*q^7 + 112*q^11 + 288*q^15 + 1712*q^19 + O(q^20)
3 * P has the components:
     3*q + 96*q^5 + 2432*q^9 + 62976*q^13 + 1628160*q^17 + O(q^20)
     27*q^3 + 405*q^7 + 13014*q^11 + 314793*q^15 + 8325594*q^19 + O(q^20)
4 * P has the components:
     4*q + 408*q^5 + 34136*q^9 + 2904688*q^13 + 247097240*q^17 + O(q^20)
     64*q^3 + 3200*q^7 + 326528*q^11 + 26521856*q^15 + 2285282688*q^19 + O(q^20)
5 * P has the components:
     5*q + 1248*q^5 + 257920*q^9 + 54156800*q^13 + 11371264000*q^17 + O(q^20)
     125*q^3 + 15475*q^7 + 3862810*q^11 + 777959167*q^15 + 165141476790*q^19 + O(q^20)

We can check that the above points are on the curve. For instance for the last point...

sage: z5, w5 = mul(5, P)
sage: w5 == z5^3 - z5*w5^2
True

sage: z5 + O(q^20)
5*q + 1248*q^5 + 257920*q^9 + 54156800*q^13 + 11371264000*q^17 + O(q^20)
sage: w5 + O(q^20)
125*q^3 + 15475*q^7 + 3862810*q^11 + 777959167*q^15 + 165141476790*q^19 + O(q^20)

The experiment using power series related to the modular parametrization are of the shape

E = EllipticCurve(QQ, [-1, 0])
X, Y = E.modular_parametrization().power_series(40)
R = X.parent()    # power series ring in q truncated with precision 40
R.inject_variables()

print("Modular parametrization of the Fermat curve y² = x³ - x:")
print(f"X = {X}")
print(f"Y = {Y}")
print("\nWe associate Z = -X/Y and W = -1/Y:")
Z, W = -X/Y, -1/Y
print(f"Z = {Z}")
print(f"W = {W}")

print("\n\nThen using the point P = (Z, W):")
P = (Z, W)
for n in [1..5]:
    Q = mul(n, P)
    print(f"{n} * P has the components:")
    print('\t', Q[0] + O(q^20))
    print('\t', Q[1] + O(q^20))

This gives:

Modular parametrization of the Fermat curve y² = x³ - x:
X = q^-2 + q^2 + q^6 - q^18 - q^22 + q^30 + 2*q^34 + O(q^38)
Y = -q^-3 - q - 2*q^5 - q^9 + 2*q^17 + 2*q^21 + q^25 - q^29 - 4*q^33 + O(q^37)

We associate Z = -X/Y and W = -1/Y:
Z = q - q^9 + 2*q^17 - 3*q^25 + 4*q^33 + O(q^41)
W = q^3 - q^7 - q^11 + 2*q^15 + q^19 - 2*q^23 - 2*q^27 + 2*q^31 + 4*q^35 - 4*q^39 + O(q^43)

Then using the point P = (Z, W):
1 * P has the components:
     q - q^9 + 2*q^17 + O(q^20)
     q^3 - q^7 - q^11 + 2*q^15 + q^19 + O(q^20)
2 * P has the components:
     2*q + 12*q^5 + 42*q^9 + 124*q^13 + 324*q^17 + O(q^20)
     8*q^3 + 16*q^7 + 88*q^11 + 176*q^15 + 552*q^19 + O(q^20)
3 * P has the components:
     3*q + 96*q^5 + 2429*q^9 + 62496*q^13 + 1606278*q^17 + O(q^20)
     27*q^3 + 405*q^7 + 12933*q^11 + 311958*q^15 + 8182683*q^19 + O(q^20)
4 * P has the components:
     4*q + 408*q^5 + 34132*q^9 + 2902648*q^13 + 246790024*q^17 + O(q^20)
     64*q^3 + 3200*q^7 + 326336*q^11 + 26499456*q^15 + 2281691456*q^19 + O(q^20)
5 * P has the components:
     5*q + 1248*q^5 + 257915*q^9 + 54150560*q^13 + 11368942730*q^17 + O(q^20)
     125*q^3 + 15475*q^7 + 3862435*q^11 + 777850842*q^15 + 165098987005*q^19 + O(q^20)


问:
Number of types in simple theories

Let $T$ be the theory of random graphs. I want to count the number of 1-types in this theory, i.e $|S_1(T)|.$

My claim is this: $|S_1(T)|=1$, because every two points have the same types by extension axiom. (I cannot prove my claim precisely!)

We know the theory of random graph is simple. So there is a following natural question.

If $T$ is an arbitrary simple theory, what can we say about the number of $n$-types. In particular what can we say about the size of $S_1(T)$?

I have no idea about question 2.

答:
You are right that the random graph has precisely one $1$-type. This follows directly from the fact that the random graph has quantifier elimination.

We can generally not say something about the cardinality of $S_1(T)$, even for stable $T$. For example, take the theory of $(\mathbb{Z}, +, 0, 1)$, which is stable. In this theory congruence modulo $n$, which I will write as $\equiv_n$, is definable for any $n \in \mathbb{N}$. Let $p_1, p_2, \ldots$ enumerate all the primes. Then for any $I \subseteq \mathbb{N}$ we define:
$$
\Sigma_I(x) = \{ x \equiv_{p_i} 0 : i \in I \} \cup \{x \not \equiv_{p_i} 0 : i \not \in I\}.
$$
For any such $I$ we have that $\Sigma_I(x)$ is finitely satisfiable: any finite part of $\Sigma_I(x)$ is realised by the product of the primes that appear in that finite part. Clearly, for $I \neq I'$ we have that $\Sigma_I(x)$ and $\Sigma_{I'}(x)$ are incompatible types. So we find at least $2^{\aleph_0}$ many $1$-types. That is as bad as it can get generally, because for any theory $T$ there are at most $2^{|T|}$ many $1$-types.

ContentLicense: CC BY-SA 4.0

Title: Number of types in simple theories


问:
Riemann Zeta Function at $z=1+i \cdot t$
First, we define the Riemann Zeta functon as a infinite sum:

$$\zeta(z)=\sum_{n=1}^{\infty}\frac{1}{n^z}, \ Re(z)>1$$

This is an analytic function in this domain. We can not extend this function with this summation because we have

$$\left|\frac{1}{n^z}\right|\leq \left|\frac{1}{n^{Re(z)}}\right|$$

and then by comparision the entire sum converges when $Re(z)>1$.

We also know that we can extend $\zeta$ via holomorphic continuation to a meromorphic function with a simple pole in $z=1$.

My question is, what about the values of the extended $\zeta$ function in $1+it$, where $t\in\mathbb{R}$? I mean, if you take the limit of the summation by the right way, you have this

$$\lim_{z\rightarrow (1+it)^+}\sum_{n=1}^{\infty}\frac{1}{n^z}$$

and this converges to $\zeta(1+it)$ because is an analytic continuiation, hence continuous, right? But the sum goes to infinity, or am I wrong?

Thank you very much!

答:
You seem to think that because $\lim_{\sigma\to1^+}\sum_n n^{-(\sigma+it)}$ exists (which is true by the reasons you mention), it should be equal to $\sum_n n^{-(1+it)}$. But that's not true unfortunately. It's the whole shebang about when we can interchange limits and sums. As a simpler analogous example:
$$
f(x) := \frac{1}{1-x} = \sum_{n\ge 0}x^n
$$
for $|x|<1$. Here $f$ is continuous at $x=-1$ with $f(-1)=\frac12$, but the series is $1-1+1-1+\cdots$ and diverges. It's just how it is.

Note also that the series diverges without blowing up to infinity, just as the $\zeta$ funciton series at $1+it$.


问:
Continuity of $f(x)=[x]+ \sqrt{x-[x]}$
Consider the function $f(x)=[x]+ \sqrt{x-[x]}, \, x\in \Bbb R$ ; where "$[ \space ]$" denotes the greatest integer function. It is obvious that if $b$ is an integer, then </p>

<p>$$\lim_{x\to b-} f(x)=\lim_{x\to b-}[x] + \sqrt { \lim_{x\to b-}x - \lim_{x\to b-}[x] } =b-1+\sqrt {b-(b-1)}=b-1+1=b$$ </p>

<p>$$\lim_{x\to b+} f(x)=\lim_{x\to b+}[x] + \sqrt { \lim_{x\to b+}x - \lim_{x\to b+}[x] } =b+\sqrt {b-b}=b,$$ and </p>

<p>$$f(b)=[b]+ \sqrt{b-[b]}=b+\sqrt {b-b}=b,$$ hence $f(x)$ is continuous for all integer values of $x$.  Is $f(x)$ continuous for all non-integer values of $x$ also, i.e. is it true that $f(x)$ is continuous for all $ x \in \Bbb R$ ? </p>

答:
Write $f(x)$ as

$$ f(x) = x + g(x) \quad \text{where} \quad g(x) = \sqrt{x - [x]} - (x - [x]). $$

It is clear that $g(x) = \sqrt{x} - x$ on $[0, 1)$ and $g(x+1) = g(x)$.

From this observation it is easy to prove that $g$ is continuous on $[0, 1]$ (the point $x = 1$ only matters), and then this continuity extends to $\Bbb{R}$ by periodicity of $g$. Therefore $g$ is continuous and hence the same is true for $f$.

ContentLicense: CC BY-SA 3.0

Title: Continuity of $f(x)=[x]+ \sqrt{x-[x]}$


问:
Unsure on Riemann Sum to Definite Integral
I've taught definite integrals for high school twice, but everytime I run into this particular Riemann Sum to Definite Integral question that I can't seem to solve.

The question is:

If $n$ is a positive integer, then $\lim\limits_{n\to \infty} \frac{4}{n} \bigg[ \bigg(2+ \frac{4}{n}\bigg)+ \bigg(2+ \frac{5}{n}\bigg) + \bigg(2+ \frac{6}{n}\bigg) + \ldots \bigg(2+ \frac{n}{n}\bigg) \bigg]$ can be expressed as what definite integral?

I have attempted a solution by letting $\Delta x= \frac{4}{n}=(b-a)/n$ and based on the function inside, it seems like $f(x)=x$.  Trying to find the bounds of integration is where my solution is dependent on my assumption since I usually assume $a=2$ based on how the sum is written.  This does not seem right.

After attempting it once again, I ended up with something like  $\lim\limits_{n\to \infty} \sum\limits_{k=1}^n  \bigg[ \bigg(2+ \frac{k(1+\frac{3}{k})}{n}\bigg) \bigg] \frac{4}{n}$
Thanks!

答: $\int_0^{1}4(2+x)dx=\lim \frac  1 n \sum\limits_{i=1}^{n} 4(2+\frac i n)$. This is same as the given limit since the limit of $\frac  1 n  4(2+\frac i n)$ is $0$ for $i\leq 3$. So the answer is $10$.


问:
How many possible seat arrangements are there?

Four boys Bobby, Benson Benny, Benjamin and three girls Grace, Gloria, Georgia are to be seated in a row according to the following rules:

a boy will not sit next to another boy and a girl will not sit next to another girl.

Benjamin must not sit next to Grace. How many possible seat arrangements are there?

My solution:

$$3!4! - 6 = 138$$
The $3!4!$ is the answer if the rule is only #$1$. I subtracted $6$ because it’s supposedly the number of possibilities Benjamin and Grace sitting next to each other.

I’m sure there are a lot of flaws in my solutions since in the answer key, the answer is $72$.

I have just recently dived into these type of problems so please give me constructive criticism for the many flaws and preferably some detailed explanations. Thank you.

答:
Let name them $B_1,B_2,B_3,B_4,G_1,G_2,G_3$ ,respectively.

Let arrange girls such that $-G_1-G_2-G_3-$ , there are $4$ place to place boys and we do not want $G_1,B_4$ together.

So , there are $3$ possible choices for left of $G_1$ and $2$ choices for the right of it (Because we cannot place $B_4$ right or left of $G_1$ so we have $3 \times 2$ boys to place next to her).

Moreover , when you place boys like that , there are  $3 \times 2 \times 2 \times1 = 12$ arrangement for boys and there are $3!$ for girls .

So $12 \times 3! =72$


问:
Adding powers of $i$
I've been struggling with figuring out how to add powers of $i$.

For example, the result of $i^3 + i^4 + i^5$ is $1$. But how do I get the result of $i^3 + i^4 + ... + i^{50}$? Writing it all down would be pretty mundane...

It has to do something with division by 4, since the "power cycle" of $i$ repeats every fourth power.

Thank you for any clues.

Title: Adding powers of $i$

LastEditorDisplayName: anonymous

Body:
We have $$i^{3} + i^{4} + i^{5} = 1 = i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} = i^{3} + i^{4} + \cdots +i^{4n+1}$$

Now $i^{50}=1 \times -1$, therefore we have the sum is $0$.

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
using finitely generated in proof.
I am trying to show that:Let $R$ be commutative with unity. if $N$ be a finitely generated $R-$module, then the following conditions are equivalent:

(a) $N$ is a sum of simple modules.

(b) $N$ is a direct sum of simple modules.

(c) Given any submodule $M \subset N,$ there exists a unique submodule $M' \subset N$ such that $N = M \oplus M'. $

For $(a) \implies (b),$ I said the following:

Assume $N$ is a sum of simple modules, we want to show that $N$ is a direct sum of simple modules i.e. $N = \oplus_{i=1}^n N_i$ where $N_i$ are simple modules. So to prove this, it is enough to show that $N_i \cap N_j = \phi.$ Assume towards contradiction that $M$ is a submodule of $N_i \cap N_j,$ then $M$ is a submodule of both $N_i, N_j$ but this is not possible because $N_i, N_j$ are simple which means they have no nontrivial submodule.

For $(b) \implies (c),$ I said the following:

Suppose that $N = \oplus_{i=1}^n N_i$ where $N_i$ are simple modules and suppose that $M$ is a submodule $N,$

$(1)$ if $M$ is simple then $M$ will be one of those $N_i.$ Let $M = N_1$ but then I do not know how to complete. Any help will be appreciated in that case.

$(2)$ if $M$ is not simple, also I do not know how to handle that case and I will be beyond grateful for any help.

It is clear to me that I have not used the main assumption that $N$ is a finitely generated $R-$module but I do not know how to use it.

For $(c) \implies (a),$ I said the following:

Given any submodule $M \subset N,$ there exists a unique submodule $M' \subset N$ such that $N = M \oplus M'$ and so $N$ is clearly a sum of modules but how to show that they are simple?

Also, any help will be greatly appreciated!

答:
I think your hypotheses lack some necessary precision. Is $R$ commutative? unital? Noetherian, left or right? Is $N$ a left $R$-module? I think you intended $R$ to be commutative but what I wrote below doesn't require it.

$a) \Rightarrow b)$ seems incorrect or at least incomplete. It is possible for a vector space to be a sum of three vector subspaces which aren't in direct sum, even though they are pairwise in direct sum. For instance, $K^2=K(0,1) + K(1,0) + K(1,1)$, and the lines have pairwise intersection zero, but the overall sum isn't direct.

I'd prove it this way: let $N=\sum_i{N_i}$ where $N_i$ is simple. Because $N$ is finitely generated, we can assume that the $N_i$ are finitely many, and we can thus name them $N_1,\ldots,N_m$. For each $i$, call $N_i$ new if its intersection with $\sum_{j < i}{N_j}$ is zero. As $N_i$ is simple, if $N_i$ is not new then $N_i \subset \sum_{j < i}{N_j}$. It is then easy enough to check that $N$ is the direct sum of the new $N_i$.

Uniqueness in $b) \Rightarrow c)$ seems false. For instance, in $K^2$, we have $K^2=K(0,1)\oplus K(1,0)=K(0,1)\oplus K(1,1)$. But let's see if we can show existence: write a finite sum $N=\bigoplus_{i \in I}{N_i}$ where the $N_i$ are simple. Let $M$ be a submodule of $N$.

Let $J \subset I$ be a maximal subset such that $\sum_{j \in J}{N_j}$ is in direct sum with $M$. Let's show that $N=M \oplus \sum_{j \in J}{N_j}$. Clearly, it's enough to show that $P = M \oplus \sum_{j \in J}{N_j}$ contains every $N_i$. If $i \in J$, that's clear. If not, we know that $P$ isn't in direct sum with $N_i$ (by maximality of $J$), that is, $P \cap N_i$ is nonzero. As $N_i$ is simple, it follows $N_i \subset P$.

Again, how about existence in $c)$ implying $a)$?
I'll simply prove that every submodule of $N$ satisfies existence in $c)$ as well. Indeed, let $M_1 \subset M$ be submodules, let $M' \subset N$ be such that $M \oplus M'=N$. Let then $M''$ be such that $M' \oplus (M'+M_1)=N$. Let $\pi:N \rightarrow M$ be the projection with kernel $M'$. It's easy to check that $M = \pi(M'') \oplus M_1$.

Note also that every submodule of $N$ is the image of $N$ under some projection so is finitely generated. In particular, every sequence of submodules $(M_n)_n$ such that $M_n \subset M_{n+1}$ is stationary.

Let $(M_n)$ be a sequence of submodules such that $M_n \supset M_{n+1}$. Let $M_n=N_n \oplus M_{n+1}$. Then $\left(\bigoplus_{p \leq n}{N_p}\right)_n$ is a nondecreasing sequence of submodules of $N$ so is stationary, which implies that for $n$ large enough $N_n=0$, ie $M_n=M_{n+1}$ so $M_n$ is stationary.

From this, it follows that the set of nonzero submodules of $N$ has minimal elements (the simple submodules). Let $S$ be their sum, we can write $N=S \oplus S_1$ for some submodule $S_1$. Assume $S_1$ is nonzero: since every non-increasing/nondecreasing sequence of submodules of $S_1$ is stationary, $S_1$ contains a simple submodule $P$. But $P \subset S$ by definition of $S$, so $P \subset S\cap S_1 = \{0\}$, a contradiction. So $N=S$ is sum of simple modules.

Title: using finitely generated in proof.


问:
Convergence of the sequence $\frac{1}{nx}$ in $\mathbb{B}(0,1)$
Let $X=\mathbb{B}(0,1)$ denote the space of bounded functions from $(0, 1) \rightarrow \mathbb{R}$ with the supremum-metric. Does the sequence $(f_n)_{n=0}^\infty,f_n(x)=\frac{1}{nx}$ converge in $X$?

So we know that $X$ is complete and we can show that $f_n$ is Cauchy: Given $\epsilon >0$, we need $N\in\mathbb{N}$ such that $\forall m,n \in\mathbb{N}$ with $m,n>N$ , we have $sup_{x\in(0,1)}|\frac{1}{nx}-\frac{1}{mx}|<\epsilon \iff sup_{x\in(0,1)}(\frac{1}{|x|}|\frac{1}{n}-\frac{1}{m}|)<\epsilon\iff \frac{1}{|x|}sup_{x\in(0,1)}(|\frac{1}{n}-\frac{1}{m}|)=\frac{1}{|x|}(|\frac{1}{n}-\frac{1}{m}|)<\epsilon\iff (|\frac{1}{n}-\frac{1}{m}|)<\epsilon|x|$. Picking for simplicity $m=n+1$, we have $(|\frac{1}{n}-\frac{1}{n+1}|)=|\frac{1}{n(n+1)}|=\frac{1}{n(n+1)}<\frac{1}{n^2}<\epsilon|x|\iff n>\sqrt{\frac{1}{\epsilon|x|}}$ By similar calculations, we can compute the point-wise limit to be $0$. Is what I've done correct? Does there exist a point-wise limit that the function converges to?

答:
Your calcualtions are not valid. You cannot pull out $\frac 1 {|x|}$ from the supremum over $x$.

For every fixed $x$ $\lim f_n(x)=0$ but the convergence in the supremum metric fails. This is because $\sup_x |f_n(x)-0|\geq |f_n(\frac1 n)| =1$.


问:
Local rings of affine space in etale topology.
This is probably a very straightforward question. What does the etale local rings of $\mathbb{A}_{\mathbb{F}_q}^n$ look like? (In other words the strict henselization of local rings at closed points)

答:
The strict henselization of a DVR is (up to isomorphism) the smallest DVR extending the valuation where Hensel lemma holds and with separably closed residue field.

The strict henselization of $\Bbb{F}_q[T]_{(T)}$ is $\overline{k(T)}\cap k[[T]]$ where $k=\overline{\Bbb{F}_q}$:

If $a(T^p)\in k[[T^p]]$ is algebraic over $k(T)$ then $\sum_{n=0}^N c_n(T)a(T^p)^n=0$. Write $c_n(T)=\sum_{m=0}^{p-1} T^m c_{n,m}(T^p)$ with $c_{n,m}(T^p)\in k(T^p)\subset k((T^p))$ then
$\sum_{m=0}^{p-1} T^m \sum_{n=0}^N c_{n,m}(T^p)a(T^p)^n=0$ implies that each $\sum_{n=0}^N c_{n,m}(T^p)a(T^p)^n=0$. Whence the minimal polynomial of $a(T^p)$ is in $k(T^p)[Y]$ from which $\overline{k(T)}\cap k[[T]]$ is separable over $k(T)$.

I doubt there is any simple characterization of the elements of $\overline{k(T)}\cap k[[T]]$.

For $f(T)\in \Bbb{F}_q[T]$ irreducible, $f(T)=(T-\alpha)g(T)\in k[T]$, the strict henselization of $\Bbb{F}_q[T]_{(f(T))}$ is the same as the strict henselization of $\Bbb{F}_q[T]_{(f(T))}[\alpha,g(T)^{-1}]\cong \Bbb{F}_q(\alpha)[T]_{(T)}$ which is again $\overline{k(T)}\cap k[[T]]$.

For the local ring $\Bbb{F}_q[T]_{\{0\}}=\Bbb{F}_q(T)$ at the generic point, its strict henselization is $\Bbb{F}_q(T)^{sep}$.


问:
Show that if $a,b,c \in \mathbb{R}$ are such that for all $\lambda \in \mathbb{R}$, $a\lambda^²+b\lambda+c\geqslant0$, then $b^2-4ac \leqslant0$.

Show that if $a,b,c \in \mathbb{R}$ are such that for all $\lambda \in \mathbb{R}$, $a\lambda^²+b\lambda+c\geqslant0$, then $b^2-4ac \leqslant0$.

I'm reading the book Introduction to Topology by Gamelin and Greene and found this problem there. I'm curious why is this on a topology book?

The reasoning I see here is that since $\lambda \in \mathbb{R} \implies b^2-4ac \geqslant 0$ and this would imply that $b^2-4ac=0$ so $\lambda = \frac{-b}{2a}$? I'm wondering if there is some deeper meaning here from topology point of view?

答:
Looking at the book, this is part 1 of a longer exercise. You are meant to use part 1 to prove part 2, which is known as the Cauchy-Schwarz inequality.

The Cauchy-Schwarz inequality is important in a book about topology, because it can be used to show that Euclidean space is an example of a metric space, an important type of topological space. (The particular chapter you are reading deals with metric spaces.)

As for why this is true: by considering the behaviour as $\lambda \to \infty$, we see that we must have $a\ge 0$. Therefore, after multiplying through by $4a$, we get
$$
4a^2\lambda^2+4ab\lambda+4ac\ge0\,.
$$
Now, completing the square:
$$
(2a\lambda + b)^2  - b^2 + 4ac \ge 0\,,
$$
and so
$$
(2a\lambda+b)^2 \ge b^2 - 4ac
$$
for all $\lambda$.

Now we have to split into two cases. Firstly, if $a=0$, then $b$ must also be $0$, or $a\lambda^2+b\lambda+c=b\lambda+c$ must necessarily take negative values. Otherwise, $2a\lambda + b$ is a linear function with non-zero gradient and therefore intersects the $x$ axis for some value of $\lambda$. It follows that for some $\lambda$ we have $(2a\lambda + b)^2 = 0$ and therefore that $b^2-4ac \le 0$.


答:
"I subtracted $6$ because it was supposedly the number of possibilities of Benjamin and Grace sitting next to each other"

This number was too small.  $6$ is just the number of places that you could seat Benjamin and Grace in the row... it being B G b g b g b, or b G B g b g b or b g B G b g b or ... or b g b g b G B where the uppercase B stands for where Benjamin sits and G where Grace sits.  The lowercase b's and g's still need to be determined which person it was specifically who was sitting there before we have uniquely determined an overall possibility.

Once having chosen which of the six possibilities for how Ben and Grace sit, we then from left-to-right decide which other boys sit in each of the seats designated for boys and which other girls sit in the seats designated for girls.  This can be done in $3!$ and $2!$ ways respectively for a total of $6\cdot 3!\cdot 2!=72$ overall outcomes for how Ben and Grace can sit.

Our grand total is then

$$4!\cdot 3! - 6\cdot 3!\cdot 2! = 144-72=72$$

Body:
Four boys Bobby, Benson Benny, Benjamin and three girls Grace, Gloria, Georgia are to be seated in a row according to the following rules:

a boy will not sit next to another boy and a girl will not sit next to another girl.

Benjamin must not sit next to Grace. How many possible seat arrangements are there?

My solution:

$$3!4! - 6 = 138$$
The $3!4!$ is the answer if the rule is only #$1$. I subtracted $6$ because it’s supposedly the number of possibilities Benjamin and Grace sitting next to each other.

I’m sure there are a lot of flaws in my solutions since in the answer key, the answer is $72$.

I have just recently dived into these type of problems so please give me constructive criticism for the many flaws and preferably some detailed explanations. Thank you.


问:
For $a \in \mathbb R$, find polynomials $P$ such that $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$
Find all polynomials $P(x)$ with real coefficients such that:-    $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$

This is $a$ creating some big problems for me. I don't know what to do.

I am not able to figure out anything because of that $a$. The best I can figure out is that I will find the roots of $P(x)$ because $a$ is not computable as there is no way of finding out the zeros of $P(x-3)$. If I would have been able to find the roots of $(x^2-ax+18)$ then I would have been able to figure out what to do.

If there would have been no $a$ I would have found of the roots of $P(x)$ like for example $\alpha, \beta$ then I would have written out $P(x)$ in the form of $(x-\alpha)(x-\beta)Q(x)$ for $Q(x)$ being any polynomial. Then I would have tried to calculate the answer.

Any help would be appreciated

Body:
We have

$$ (x²-ax+18)P(x) = (x²+3x)P(x-3) \ \ \dots  eq.(1)$$

Put $x=-3$,
$$ \rightarrow (27+3a)P(-3) = (9-9)P(-6) $$ $ \ \  \ \ i.e. (27+3a)=0$  $ or $ $ P(-3)=0$

$CASE\ 1:$ If $(27+3a)=0$ $gives \
    a=-9$

Then we have: $ from \ \ Equation \ \ \ (1)$

$$ P(x)(x+6) \ \ = \ \ xP(x-3)$$

Put $x=0$ $\ \  \rightarrow P(0) \ \ = \ \ 0 $

Put $x=3$ $\ \  \rightarrow P(3) \ \ = \ \ 0 $

Put $x=6$ $\ \  \rightarrow P(6) \ \ = \ \ 0 $

$This\  goes \ on\  and \ on\  till \ infinity,\  implying \ P(x) \ has\ infinite\ roots$

$$ \rightarrow P(x) \ \ = \ \ 0$$

$CASE\  2:$ If $P(-6)\ \ = \ \ 0$

Put $x=-6$ $\ \  \rightarrow P(-9) \ \ = \ \ 0 $

Put $x=-9$ $\ \  \rightarrow P(-12) \ \ = \ \ 0 $

$$\ \ \ \ \ \ \ \ \    .... \dots$$

Thus both the cases lead to the same conclusion that
$$ P(x) \ \ \ \ = \ \ \ \ 0 $$


问:
why $Log(e^{z})=z$?
$Log(z)=log(\left | z \right |)+iArg(z)$ where $\left \{ z \in \mathbb{C}:-\pi <Imz\leq \pi \right \}$

Now  $Log(e^{z})=log(e^x)+iArg(e^{x+iy})=x+iArg(e^{x+iy})$
my question is why $Arg(e^{x+iy})=y$?
So it can be  $Log(e^{z})=z$

答:
Notice that : $$ \mathrm{e}^{x+\mathrm{i}y}=r\,\mathrm{e}^{\mathrm{i}\theta} $$

were $ r $ here is $ \mathrm{e}^{x}>0 $, and $ \theta $ is none other than $ y\in\mathbb{R} $.

Therefore : $ \arg\left(\mathrm{e}^{x+\mathrm{i}y}\right)\equiv\theta \equiv y\ \left(\mathrm{mod}\ 2\pi\right) $.

Title: why $Log(e^{z})=z$?


问:
Does there exist a compact submanifold of $\mathbb{R}^3$ whose fundamental group is $\mathbb{Z}^3$?

Does there exist a compact submanifold of $\mathbb{R}^3$ whose fundamental group is $\mathbb{Z}^3$ ?

The question in the title is a generalization of the question that really interests me:

Does there exist a connected finite set of unit cubes of a Cartesian lattice whose fundamental group is $\mathbb{Z}^3$ ?

It is clear that, for example, $\mathbb{Z}$ and $\mathbb{Z}^2$ are simply realized as a solid torus and a thickened torus, respectively (it is clear that, up to homeomorphism, they are assembled from a finite number of closed unit cubes).

It is clear that it would be sufficient to embed the CW-complex corresponding to the presentation $$\langle a, b, c \mid [a, b] = 1, [b, c] = 1, [a, c] = 1\rangle$$ but (as far as I can see) it doesn't embed in $\mathbb{R}^3$.

答:
I am surprised to see your question has no answer. There exists no such submanifold of $\Bbb R^3$. There are simpler proofs than the below but perhaps less illuminating.

You may find the following theorem in Hempel's book on 3-manifolds.

If $M$ is a 3-manifold with fundamental group $\Bbb Z^3$, then $M$ is homotopy equivalent to $T^3$, and homeomorphic to $T^3 \# S$ for some homotopy 3-sphere $S$.

Since Perelman, we know that a homotopy 3-sphere is a 3-sphere, so in fact your purported submanifold is homeomorphic to $T^3$.

But not only is $T^3$ not a submanifold of $\Bbb R^3$, in fact, there is no compact submanifold of $\Bbb R^3$ without boundary: this follows from the invariance of domain theorem.

(A shorter proof indicates that your space is aspherical using the "sphere theorem", and then that it must be homotopy equivalent to $K(\Bbb Z^3, 1) \simeq \Bbb Z^3$, and then that such a space has non-trivial third homology, which is only the case for compact 3-manifolds without boundary. Then run the same invariance of domain argument.)

Note that neither argument assumes that your 3-manifold is compact. You prove this.


答:
One way to look at this is that there are $4$ possible places to seat Benjamin as per first condition. Now if we cannot seat him next to Grace, that leaves only $2$ possible places for him out of $4$. As there are equal number of seating arrangements for Benjamin being in each of these $4$ places,

Number of seating arrangements meeting both conditions $ = \displaystyle \frac{2}{4} \cdot 4! \cdot 3! = 72$

Title: How many possible seat arrangements are there?


问:
Multipliciousness within an inner product space.

Question:

Let $V$ be an inner product space and $v,w\in V$. Prove that $\lvert\langle v,w\rangle\rvert=\lVert v\rVert \lVert w\rVert$ if and only if one of the vectors $v$ or $w$ is a multiple of the other.

Attempt:

Assume the identity holds and $y\neq 0$. Let
$$ a=\frac{\langle x,y\rangle }{\lVert y \rVert^2}, $$
and let
$$ z=x-ay. $$
Now, note that $y$ and $z$ are orthogonal because if $z=x-ay$, then by the definition of $a$ we have
$$a=\frac{\langle x,y\rangle }{\lVert y \rVert^2}=\frac{\langle z+ay, y\rangle }{\langle y,y\rangle }=\frac{\langle z,y\rangle }{\langle y,y\rangle }+a\frac{\langle y,y\rangle }{\langle y,y\rangle }=\frac{\langle z,y\rangle }{\langle y,y\rangle }+a,$$
which means that
$$ \frac{\langle z,y\rangle }{\langle y,y\rangle }=0~\text{or}~ \langle z,y\rangle =0,$$
namely $y$ and $z$ are orthogonal. Furthermore, since $\lvert \langle x, y \rangle \rvert= \lVert x \rVert \lVert y \rVert$, then
$$ \frac{\lvert \langle x, y \rangle \rvert}{\lVert y \rVert} =\lVert x \rVert ~\overset{1/\lVert y \rVert}{\implies}~\frac{\lvert \langle x, y \rangle \rvert}{\lVert y \rVert^2} =\frac{\lVert x \rVert}{\lVert y \rVert} =\lvert a \rvert.$$

Lemma: Let $V$ be an inner product space, and suppose that $x$ and $y$ are orthogonal vectors in $V$. Then $\lVert x+y \rVert^2 = \lVert x \rVert^2 + \lVert y \rVert^2$.

Now, we know that
$$ \lVert x \rVert^2 = \lVert ay+z \rVert^2,$$
but, by the lemma, this means that
$$ \lVert ay+z \rVert^2 = \lVert ay \rVert^2 + \lVert z \rVert^2~~\overset{\text{not sure where to go}}{\dots}$$

Note:

I'm trying to follow along with Friedberg's description on pages 337 and 338 in his Linear Algebra:

EDIT$^1$:

OK, so I think I've got an idea. To complete the proof I need to notice that $\frac{\lVert x \rVert}{\lVert y \rVert} =\lvert a \rvert$, which means that $\lVert x \rVert = \lvert a \rvert\lVert y \rVert$ and more importantly that $\lVert x \rVert^2 = \lvert a \rvert^2\lVert y \rVert^2,$ but this is exactly the $\lVert ay\rVert^2$ from the application of the lemma as $\lVert ay\rVert^2=(\lVert ay\rVert)^2=(\lvert a\rvert\lVert y\rVert)^2=\lvert a\rvert^2\lVert y\rVert^2$. This will give the result.

CONCLUSION:

Here is my end proof:

EDIT$^2$:

I found here in E. B. Vinberg's A Course in Linear Algebra a different flavor of this proof (if anybody happens to care):

答:
Notice that:
$$ \begin{align*}
\lVert x \rVert^2 &= \lVert ay+z \rVert^2 \\
\lVert x \rVert^2 &= \lVert ay \rVert^2 + \lVert z \rVert^2 \\
\lVert x \rVert^2 &= (|a|\lVert y \rVert)^2 + \lVert z \rVert^2 \\
\lVert x \rVert^2 &= \left( \dfrac{\lVert x\rVert}{\lVert y\rVert}\lVert y \rVert \right)^2 + \lVert z \rVert^2 \\
\lVert x \rVert^2 &= \lVert x \rVert^2 + \lVert z \rVert^2 \\
0 &= \lVert z \rVert^2 \\
\lVert z \rVert &= 0 \\
z &= 0 \qquad \text{ (the zero vector)}
\end{align*} $$

Thus, we have $0=z=x-ay \iff x=ay$ so that $x$ and $y$ are scalar multiples of each other, as desired.

It remains to prove the converse. Suppose $x=ay$ for some scalar $a$. Then we have:
$$\lvert\langle x,y\rangle\rvert
=\lvert\langle ay,y\rangle\rvert
=\lvert a\langle y,y\rangle\rvert
=\lvert a \left( \lVert y\rVert \lVert y\rVert \right) \rvert
=\lvert a \rvert \left( \lVert y\rVert \lVert y\rVert \right)
=\lVert ay\rVert \lVert y\rVert
=\lVert x\rVert \lVert y\rVert$$
as desired.


答:
A geometrical point of view

$a\lambda² + b\lambda +c = 0$ is the equation of a parabola with a vertical axis. The $y$ coordinate of its vertex is $\frac{4ac-b^2}{4a}$. Furthermore, when $a>0$, the parabola points upward, whereas when $a<0$ it points downward.

So when $a>0$, the expression $a\lambda² + b\lambda +c ≥ 0$ means that the whole parabola is above the $x$-axis, thus meaning that the vertex is also above (or on, when $b²=4ac$) the $x$-axis and thus $4ac-b² ≥ 0$, i.e. $b²-4ac ≤ 0$.

When $a<0$, instead, the expression $a\lambda² + b\lambda +c ≥ 0$ cannot hold for each $\lambda \in \mathbb{R}$, but we can state an analogous result:

If $a<0$ and if $a\lambda² + b\lambda +c ≤ 0 \quad \forall \, \lambda \in \mathbb{R}$ (which means that the parabola is pointing upward and is all below the $x$-axis), then $b²-4ac ≤ 0$ (the vertex as well must be below the $x$-axis).

Body:
Show that if $a,b,c \in \mathbb{R}$ are such that for all $\lambda \in \mathbb{R}$, $a\lambda^²+b\lambda+c\geqslant0$, then $b^2-4ac \leqslant0$.

I'm reading the book Introduction to Topology by Gamelin and Greene and found this problem there. I'm curious why is this on a topology book?

The reasoning I see here is that since $\lambda \in \mathbb{R} \implies b^2-4ac \geqslant 0$ and this would imply that $b^2-4ac=0$ so $\lambda = \frac{-b}{2a}$? I'm wondering if there is some deeper meaning here from topology point of view?


问:
why the ending symbol dx is also changed into a function of d theta and then multiplied by the integral after substitution?
This is a question about trig substitution used in integrals.

Because it is difficult to solve an integral when there is radical in it, we use a trig function of theta to substitute x from the original integral.

The only thing I don't understand is why the ending symbol dx, which indicates that we are integrating the function with respect to the "x" variable, is also changed into a function of d theta and then multiplied by the integral after substitution.

I think after substitution, we can just rewrite dx into d theta.

答:
$dx$ and $d\theta$ are not the same thing.  For example, if $x = \sin \theta$, then $\frac{dx}{d\theta} = \cos \theta $ and so $dx = \cos {\theta}  d\theta$.

This is important because if that were not true then just about any substitution would become trivial and integration would be as easy as differentiation.


问:
On Similar Matrices' Transformations
Say we have $A,B,C,D\in GL_n(\mathbb{C})$ such that $A$ and $B$ are conjugates,$C$ and $D$ are conjugates, can we find a matrix $P$ in $GL_n(\mathbb{C})$ such that $A=PBP^{-1}$ and $C=PDP^{-1}$?

答:
No. The desired property implies that $AC$ is similar to $BD$, but this is not a consequence of the similarities between $A$ and $B$ and between $C$ and $D$. E.g. consider
$$
A=B=C=\pmatrix{1\\ &0},\ D=\pmatrix{0\\ &1}.
$$
Then $A$ is similar to $B$ and $C$ is similar to $D$, but $AC\ne0$ is not similar to $BD=0$.

Title: On Similar Matrices' Transformations


答: Why you'd find that problem in such a book has already been addressed, so I'll just give an approach to it. Since $a\lambda^2+b\lambda+c$ is asymptotic for sufficiently large $|\lambda|$ to $b\lambda$ if $a=0$ or $a\lambda^2$ otherwise, $a>0$. Then$$a\lambda^2+b\lambda+c=a(\lambda+b/2a)^2-(b^2-4ac)/(4a)$$is non-negative on $\Bbb R$ iff it's non-negative at its minimum (obtained with $\lambda=-b/2a$), i.e. iff $(b^2-4ac)/(4a)\le0$.

Title: Show that if $a,b,c \in \mathbb{R}$ are such that for all $\lambda \in \mathbb{R}$, $a\lambda^²+b\lambda+c\geqslant0$, then $b^2-4ac \leqslant0$.


答: Let $$f(\lambda) = a\lambda^2 + b\lambda + c$$We know that if $\Delta = b^2 - 4ac\gt0$ then there are two distinct roots $\lambda_1 \lt \lambda_2$. Furthermore, when $\lambda_1 \lt\lambda \lt \lambda_2$ we have $\text{sign}\{f(\lambda)\} = -\text{sign}\{a\}$ and when $\lambda\gt \lambda_2$ or $\lambda\lt \lambda_1$ we have $\text{sign}\{f(\lambda)\} = \text{sign}\{a\}$. This means $\Delta>0$ always leads to a sign change in $f(\lambda)$ and $\forall \lambda \in \mathbb{R}, \ f(\lambda)\ge0$ can't holds. The only option is $\Delta \le 0$ which leads to no sign change.


问:
How to check if a power series formula is correct?
Given the next power series convergent within the interval $(-3,-1)$:

$$\sum_{n=0}^\infty (-1)^{n+1}\frac{2^{2n+1}(x+2)^{2n}}{7^{2n}}$$

I need to express it as a formula, so using the Geometric series:

First I reestructure the series to make it similar to the geometric series

$\sum_{n=0}^\infty \frac{(-1)^{n+1}2^{2n+1}}{7^{2n}}(x+2)^{2n}$

$(-2)\sum_{n=0}^\infty \frac{(-1)^{n}2^{2n}}{7^{2n}}(x+2)^{2n}$

$(-2)\sum_{n=0}^\infty (-1)^{n}\biggl(\frac{2^2}{7^2}(x+2)^2\biggr)^n$

$(-2)\sum_{n=0}^\infty \biggl((-1)\frac{2^2}{7^2}(x+2)^2\biggr)^n$

According to the geometric series $\sum_{n=0}^\infty z^n = \frac{1}{1-z}$ so $z=-\frac{4}{49}(x+2)^2$

That leaves us with:

$$S(x)=\frac{-2}{1+\frac{4}{49}(x+2)^2} $$

Now my question is: How do I check this formula is correct?

答: One way is to calculate derivatives of your power series and your solution and see that they match.  Another is to evaluate both at specific values of x.</p>


问:
Conjecture about a point inside an equilateral triangle divided by integer angles

In the equilateral triangle below, $CD, AD$, and $BD$ concur at point $D$. The measures of the angles ($a, b, c, d, e,$ and $f$) are also defined as given in the figure below.
Conjecture:
There aren't any $a, b, c, d, e,$ and $f$ such that all are
different integers without using the combinations of the numbers: $6, 54, 42, 18, 48$, and $12$. (to clarify, all six of these numbers can not be used together, but they can be used together in groups)
How can I prove this conjecture mathematically, and what is so special about these six numbers?

My thoughts and work on the problem

I am intentionally omitting the degree signs for convenience.

Using, the Trigonometric Ceva's Theorem, I obtained the following expression:

$$\frac{\sin(a)}{\sin(b)}\cdot\frac{\sin(c)}{\sin(d)}\cdot\frac{\sin(e)}{\sin(f)}= 1$$

which is equivalent to:
$$\begin{aligned} \sin(a) \sin(c) \sin(e) = \sin(b) \sin(d) \sin(f)
 \end{aligned} \label{a}\tag{1}$$

We also know that,

$$a+b=c+d=e+f=60$$

To convert the triple sine product to summations, I used the following "trick". Since,
$$\sin(a+c+e)=\sin(b+d+f)$$

and also due to $Eq. (1)$, it is true that:
$$\sin(-a+c+e)+\sin(a-c+e)+\sin(a+c-e)$$

$$=\sin(-b+d+f)+\sin(b-d+f)+\sin(b+d-f)$$

Let's substitute $b=60-a, d=60-c$, and $f=60-e$ and let $x=(c+e-a), y=(a+e-c),$ and $z=(a+c-e)$ to simplify things a bit:

$$\sin(x) +\sin(y) +\sin(z)=\sin(60-x)+\sin(60-y) +\sin(60-z)$$

I expanded the $(60-x)$'s using compound angle formulae:

$$\sqrt{3}/2 (\cos(x) + \cos(y) + \cos(z)) - 3/2 (\sin(x) + \sin(y) + \sin(z)) = 0$$

which simplifies to:

$$\sin(30-x) +\sin(30-y) +\sin(30-z)=0$$

but I feel lost in the algebra and don't know what to do next. I want to get to a point where I can easily plug numbers and show that no other numbers satisfy this equation, other than the combinations of the six numbers given at the beginning.

The trigonometric proof of one of the cases may aid to formulate a proof, so I decided to include it as well:
For $a=6, b=54, c=42, d=18, e=48, f=12$,  $Eq. (1)$ is satisfied, so let's prove it.

Proof.
$$ \sin(12°) \sin(18°) \sin(54°)\stackrel{?}{=} \sin(6°)\sin(42°)\sin(48°)$$
Now we use the "trick":
$$ \sin(60°)+ \sin(48°)- \sin(24°)\stackrel{?}{=} \sin(84°)+\sin(12°)+\sin(0°)$$
$$\sin(60°)=\frac{\sqrt{3}}{2} \stackrel{?}{=} \sin(12°)+ \sin(24°) - (\sin(48°) -\sin(84°))$$
$$ = 2 \sin(18°) \cos(6°) + 2 \cos(66°) \sin(18°)$$
$$ = 2 \sin(18°) (\cos(6°) + \cos(66°)$$
$$ = 2 \sin(18°) (2\cos(36°)\cos(30°) )$$
$$1 \stackrel{?}{=} 4 \sin(18°) \cos(36°)$$
$$\frac{1}{2} =  \cos(36°) \cos(72°)$$
which is true as it is a well-known identity.

Here is the figure of this triangle (rotated):

Given only 6 and 12 degrees (the red angles) the rest of the angles can be found. For the sake of brevity, I will omit the synthetic proofs. For the interested reader, see this link, as it also shows the beauty of synthetic solutions in these types of problems from the perspective of a geometrician. The combinations of $6, 54, 42, 18, 48$, and $12$ often come up in synthetic geometry problems: I even managed to find one in SE.

@KorayUlusan helped me develop a code to test the conjecture by brute force in Python. Turns out that the conjecture is correct, but I'm looking for a mathematical explanation. Here is the code:

```<code>import math

epsilon = 2.22044604925e-14  # for integer overflow

for a in range(60):
    b = 60 - a  # because a+b=60
    for c in range(60):
        d = 60 - c
        for e in range(60):
            f = 60 - e
            if a != b and b != c and c != d and d != e and e != f:
                try:
                    rightHandSide = (
                        math.sin(math.radians(a))
                        * math.sin(math.radians(c))
                        * math.sin(math.radians(e))
                    ) / (
                        math.sin(math.radians(b))
                        * math.sin(math.radians(d))
                        * math.sin(math.radians(f))
                    )

                    if abs(1 - rightHandSide) < epsilon:
                        print(f"a is {a}, b is {b}, c is {c}, d is {d}, e is {e}, f is {f}")

                except ZeroDivisionError:
                    pass
</code>```

The Python output for the code (of course, most of these results are the same triangles, just rotated):

```<code>a is 6,  b is 54, c is 42, d is 18, e is 48, f is 12
a is 6,  b is 54, c is 48, d is 12, e is 42, f is 18
a is 12, b is 48, c is 18, d is 42, e is 54, f is 6
a is 12, b is 48, c is 54, d is 6,  e is 18, f is 42
a is 18, b is 42, c is 12, d is 48, e is 54, f is 6
a is 18, b is 42, c is 54, d is 6,  e is 12, f is 48
a is 42, b is 18, c is 6,  d is 54, e is 48, f is 12
a is 42, b is 18, c is 48, d is 12, e is 6,  f is 54
a is 48, b is 12, c is 6,  d is 54, e is 42, f is 18
a is 48, b is 12, c is 42, d is 18, e is 6,  f is 54
a is 54, b is 6,  c is 12, d is 48, e is 18, f is 42
a is 54, b is 6,  c is 18, d is 42, e is 12, f is 48

</code>```

Thanks in advance for any sort of help or contribution.

答:
Not an answer but an extended comment.

First: you have already proved the conjecture. You may think that the exhaustive search is not a nice tool, but you checked all $59\times 59\times 59$ possibilities and proved that there are no other solutions.

Second: there is no mystics about the integer numbers. They appear to be integer because some ancient human being decided that it is nice to divide the circle in 360 parts which were named "degrees". The obtained solutions would remain integer if he/she decided to divide it instead in 60 parts but cancel to be integer if he/she would prefer 100 parts.

Third: the problem can be however formulated as a really challenging one.

The following formulation of the problem is due to @Blue

Namely: find all triples $\{p,q,r\}$ of non-zero rational multiples of $\pi$: $$-\frac\pi6 <p,q,r<\frac\pi6$$
such that
$$
\sin\left(\frac{\pi}{6}+p\right)
\sin\left(\frac{\pi}{6}+q\right)
\sin\left(\frac{\pi}{6}+r\right)
=
\sin\left(\frac{\pi}{6}-p\right)
\sin\left(\frac{\pi}{6}-q\right)
\sin\left(\frac{\pi}{6}-r\right)\tag1
$$
which neatly reduces to:
$$
\cot p\cot q+\cot q\cot r+\cot r\cot p=−3.\tag2
$$

I have checked all triplets with denominator up to $n=120$ and the result is: there are no others except for
$$
\frac\pi{30}\{2,3,-4\},
$$
which you have found. Is the triplet the only one?


问:
Riemann Zeta Function at $z=1+i \cdot t$
First, we define the Riemann Zeta functon as a infinite sum:

$$\zeta(z)=\sum_{n=1}^{\infty}\frac{1}{n^z}, \ Re(z)>1$$

This is an analytic function in this domain. We can not extend this function with this summation because we have

$$\left|\frac{1}{n^z}\right|\leq \left|\frac{1}{n^{Re(z)}}\right|$$

and then by comparision the entire sum converges when $Re(z)>1$.

We also know that we can extend $\zeta$ via holomorphic continuation to a meromorphic function with a simple pole in $z=1$.

My question is, what about the values of the extended $\zeta$ function in $1+it$, where $t\in\mathbb{R}$? I mean, if you take the limit of the summation by the right way, you have this

$$\lim_{z\rightarrow (1+it)^+}\sum_{n=1}^{\infty}\frac{1}{n^z}$$

and this converges to $\zeta(1+it)$ because is an analytic continuiation, hence continuous, right? But the sum goes to infinity, or am I wrong?

Thank you very much!

答:
For $s\ne 1$
$$\lim_{N\to \infty} \sum_{n= 1}^N n^{-s}=\lim_{N\to \infty}\left( \int_1^{N+1} x^{-s}dx+\sum_{n= 1}^N (n^{-s}-\int_n^{n+1} x^{-s}dx)\right)$$
$$ = \lim_{N\to \infty}\left(\frac{1-(N+1)^{1-s}}{s-1}+ \sum_{n= 1}^N \int_n^{n+1}\int_n^x st^{-s-1}dtdx\right)$$

$\lim_{N\to \infty}\sum_{n= 1}^N \int_n^{n+1}\int_n^x st^{-s-1}dtdx$ converges for $\Re(s) > 0$

$\lim_{N\to \infty}\frac{1-(N+1)^{1-s}}{s-1}$ converges iff $\Re(s) > 1$, for $\Re(s)=1,s\ne 1$ it oscillates staying bounded.

Whence $\lim_{N\to \infty} \sum_{n= 1}^N n^{-s}$ converges iff $\Re(s) > 1$. The same holds for the real and imaginary part.

The analytic continuation to $\Re(s)>0,s\ne 1$ is $\frac1{s-1}+\sum_{n= 1}^\infty \int_n^{n+1}\int_n^x st^{-s-1}dtdx$.


答:
One more way: Let $a>0$

$f(x)=ax^2+bx+c$, if $f(x) \ge 0, \forall x \in Re$, then $min(f(x))\ge 0$
As $f'(x)=0 \implies x_0=-b/(2a), f''(x_0)=2a>0$ so min is at $x=x_0$, therefore $f(x_0) \ge 0 \implies c-b^2/(4a)\ge0 \implies b^2-4ac\le 0.$


问:
How does this line work in a problem about restricted compositions of $n$?
I'm trying to follow an example problem of calculating how many $k$-part compositions of $n$ are there, with the restriction that each part is at most 5.

During the calculation there is this mess:

$$\begin{eqnarray*}
[x^{n}]x^{k}(1-x^{5})^{k}(1-x)^{-k} & = & [x^{n-k}]\sum_{i\geq0}{k \choose i}\left(-x^{5}\right)^{i}\sum_{j\geq0}{k+j-1 \choose j}x^{j}\\
 & = & [x^{n-k}]\sum_{i\geq0}\sum_{j\geq0}{k \choose i}(-1)^{i}{k+j-1 \choose j}x^{5i+j}\\
 & = & \sum_{i\geq0,5i+j=n-k}\sum_{j\geq0}{k \choose i}(-1)^{i}{k+j-1 \choose j}\\
 & = & \sum_{i=0}^{\left\lfloor \frac{n-k}{5}\right\rfloor }{k \choose i}(-1)^{i}{k+n-k-5i-1 \choose n-k-5i}
\end{eqnarray*}$$

I don't exactly understand how the second $\sum$ was gotten rid of. How would one mathematically work with having two conditions under a sum sign, as in the next-to-last line? Where did the floor come from?

答:
First note that $k$ is fixed throughout. The original summation is over all $i,j\ge 0$. However, we want the coefficient of $x^{n-k}$, which is $5i+j$, so the only combinations of $i$ and $j$ that interest us are those satisfying $5i+j=n-k$. This completely determines $j$ as a function of $i$ and vice versa. In particular, given $i$, $j=n-k-5i$. Thus the inner summation $\sum_{j\ge 0}$ in the penultimate line of the computation is feather-bedding: it serves no purpose whatsoever, since the outer summation already determines the unique $j$ that goes with each possible $i$, and the double summation is exactly equivalent to

$$\sum_{i\ge 0,5i+j-n-k}\binom{k}i(-1)^i\binom{k+j-1}j\;.$$

Now substitute $j=n-k-5i$ to get

$$\sum_{i\ge 0}\binom{k}i(-1)^i\binom{n-5i-1}{n-k-5i}$$

and realize that $\binom{k}i=0$ if $i>k$, so that $i$ really just runs from $0$ through $\left\lfloor\frac{n-k}5\right\rfloor$.


问:
Reduction formula for $\int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $
I'm struggling to go any further. Anyone have any hints?

$$ I _ n = \int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $$

I used integration by parts where I differentiated $ x ^ n $, but it resulted in an $ \arcsin x $ term which didn't get me anywhere.

$$ u = x ^ n $$
$$ v = \arcsin x $$
$$ \mathrm d v = \frac { \mathrm d x } { \sqrt { 1 - x ^ 2 } } $$
$$ I _ n = \int u \ \mathrm d v = u v - \int v \ \mathrm d u = x ^ n \arcsin x - ( n - 1 ) \int x ^ { n - 1 } \arcsin x \ \mathrm d x $$

答:
Integrate by parts as follows
\begin{align}
 \int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } d x
&= -\int \frac { x ^{n-1} } { (1 - x ^ 2 )^{\frac{n-1}2}} d \left(\frac{(1-x^2)^{\frac n2}}n\right)\\
&=-\frac1n \frac{x^{n-1}}{\sqrt { 1 - x ^ 2 } }+ \frac{n-1}n\int \frac { x ^ {n-2} } { \sqrt { 1 - x ^ 2 } } d x \\
\end{align}
hence the reduction formula

$$I_n = -\frac1n \frac{x^{n-1}}{\sqrt { 1 - x ^ 2 } }+ \frac{n-1}n I_{n-2}$$


问:
Nonzero ideals in a discrete valuation ring
We say that $K$ is a discrete valuation field if there exists a surjective map $v:K\to\mathbb{Z}\cup\{\infty\}$ such that

$v(xy)=v(x)+v(y)$

$v(x+y)=\inf\{v(x),v(y)\}$

$v(x)=\infty$ if and only if $x=0$

We then define the discrete valuation ring to be $A=\{x\in K\mid v(x)\geq 0\}$.

I am trying to show that if $A$ is a discrete valuation ring, and $m=(\pi)$ with $v(\pi)=1$, then all ideals in $A$ are of the form $(\pi^n)$ for some $n$. Now one way to do this is to show that $A$ is noetherian and to invoke Krull's intersection theorem, but neither have been discussed in class, so I'm wondering: is there a way to show this directly?

答:
Hint:

This results from the fact that if $x\in A$ is such that $v(x)=n>0$, then we can write
$$x=u\pi^n,\quad\text{where }u \text{ is a unit in }A.$$


问:
Make a matrix symmetric and positive semi-definite if it already approximately is
I have a matrix that ought to be symmetric and positive semi-definite (with real entries). However, due to small rounding errors that can accumulate over time, that might not be strictly true. For example, the smallest eigenvalue might be $-10^{-10}$ even though it ought to be $0$ if we had been using exact calculations. This isn't particularly important overall, except that some computations will fail altogether with even this small negative number.

Is there a simple expression that will fix this? As an example of what I'm looking for:

If I were asking "how do I make a real number non-negative if it already nearly is?" I'd like the answer "take the absolute value". This leaves non-negative numbers precisely unchanged, and only has a small effect on small strictly negative numbers.

If I were asking "how do I make a matrix $M$ symmetric" I'd like the answer "use $M + M^t$". Actually that is one of the things I want so that will be my first step!

So what I'm asking, analogously, is: "how do I make the matrix positive semidefinite?"

答:
One possible method is to take the operator
$$
|M| = (M^T M)^{1/2}.
$$
This is a positive semidefinite matrix whose eigenvalues are the singular values of $M$. For example, if $M$ is a symmetric matrix then $|M|$ is a symmetric matrix matrix whose eigenvalues are the absolute value of the eigenvalues of $M$. Indeed, if $M$ is symmetric then let $M = U D U^T$ be the spectral decomposition of $M$ where $U$ is an orthogonal matrix and $D$ is diagonal. Then
$$
|M| = ((UDU^T)^T (UDU^T))^{1/2} = (UDU^T U DU^T)^{1/2} = U(D^2)^{1/2}U^T = U |D| U^T.
$$
It also follows then that if $M$ is already positive semidefinite then $|M| = M$.

If you want this to work in general for matrices acting on $\mathbb{C}^n$ then you can also take $|M| = (M^* M)^{1/2}$ where $~^*$ indicates the conjugate transpose.


问:
What's the expected value of Y for this?
Boston and Toronto play a series of basketball games.  Suppose game results are independent,both teams are equally good, and there are no ties, so each game is like a fair coin flip.  The teamskeep playing until one team has won four games.  Let $Y$ be the total number of games the teams play(so the possible outcomes of $Y$ are $4, 5, 6,$ and $7$.)  What is the expected value of $Y$?  (Hints: $Y$ isn’t binomial.  You might try a counting method.)

答:
First of all, $Y \in  \{4,5,6,7\}$ as suggested in the text.

Now we calculate the probability that the first team wins for each possible number of matches and then, because the problem is simmetric, we will need to multiply by $2$ .

The  last match has to be won by the first team while the other $3$ matches can be in any order between the previous games so you can use the binomial distribution here, you should obtain that for $Y=k$ such that $4 \leq k \leq 7$ you have

$$p_k = {k-1 \choose3}\frac{1}{2^{k-1}} \cdot \frac{1}{2} $$

Now you shouldn't have problems to calculate the expected value given by

$$E[Y]=\sum_{k=4}^7 k\cdot 2p_k = \sum_{k=4}^7 k{k-1 \choose3}\frac{1}{2^{k-1}}  $$


问:
What's wrong in my approach in evaluating $\cos^{-1}(\cos\ 10)$
The Question

$\large\cos^{-1}(\cos 10)$

My Approach

$\large\cos^{-1}(\cos\ (10-2\pi))$ As $2\pi$ is the period of $\cos x.$ Now our angle is-

Now we know that for $\large\cos^{-1}(\cos x) = x$, $\large x\ \epsilon\ [0,\pi]$ but currently our $\large x$ is in the 3rd quadrant, we need to get it into 1st or 2nd quadrant. Therefore let us subtract $\large \pi$ from our current angle i.e $\large 10-2\pi$

$\large\cos^{-1}(\cos\ -((10-2\pi)-\pi))$ Now you must be wondering why did I put this minus sign, w.k.t $\large \cos x$ is -ve in third quadrant, by subtracting $\large \pi$ we brought it into 1st quadrant and $\large \cos x$ is +ve in 1st quad, so to compensate that I have put a minus sign.

Now its within range so we can use $\large\cos^{-1}(\cos x) = x$.

So Final answer $\large -((10-2\pi)-\pi)$ = $\large 3\pi - 10$

But this answer is wrong and the correct answer is $\large 4\pi - 10$.

They have used the property $\large \cos\  (\pi - \theta) = \cos\  (\pi + \theta)$ But why my approach is wrong?

Kindly help.

答:
The angle $\theta$ that you're looking for corresponds to the point (on the unit circle) directly above the black dot in your second drawing, which corresponds to the angle $10-2\pi$. The average of these two angles is $\pi$, so from

$${1\over2}(\theta+(10-2\pi))=\pi$$

we find $\theta=4\pi-10$.


问:
Commutation of symmetric and skew-symmetric part of orthogonal matrix
Can we claim that for an orthogonal matrix $A$ (satysfying $AA^T=I$),
its symmetric and skew-symetric parts are always commuting?

Symmetric part is calculated as $S=\frac{1}{2}(A+A^T)$ and skew symmetric part as $K=\frac{1}{2}(A-A^T)$.

It's easy to show the claim for dimension $2$.

Also for dimension $3$ we can see commutation for orthogonal matrices with determinant equal to 1.

For a rotation matrix $R$ it's evident from the Rodrigues formula,

when $R=\color{blue}{I+(1-\cos\theta) V^2 }+ \color{green}{\sin\theta V }$ for some skew-symmetric matrix $V $.

Here commutation  follows because   symmetric part   and skew-symmetric part are expressed as polynomials of the same matrix $V $.

Is it possible to find similar explanation for higher dimensions of orthogonal matrices?

答:
Presumably the characteristic of the underlying field is not $2$.

The orthogonality condition implies that $(S+K)(S-K)=4AA^T=4A^TA=(S-K)(S+K)$. Hence $S$ commutes with $K$.

Alternatively, note that the inverse of a matrix $A$ is a polynomial in $A$. This is a consequence of Cayley-Hamilton theorem. It follows that when $A$ is orthogonal, $A^T$ is a polynomial in $A$. In turn, $S$ and $K$ are polynomials in $A$ as well. Hence they commute.


问:
Homework Help. Probability Density Functions.
$X$ is $N(10,1)$. Find $f(x|(x-10)^2 < 4)$

This is a homework question. I can only figure out that X is normally distributed with mean 10 and variance 1.

Can you please explain what is meant to be found here.

Thanks.

答:
It is best to rewrite $\{x\;|\;(x-10)^2<4\}$ as  $\{x\;|\;-2<x-10<2\}$. In terms of the normalized quantity $z=(x-\mu)/\sigma$  this is  $\{z\;|\;-2<z<2\}$.

I don't know what $f$ is doing in this problem. If what you really meant was  $P\{x\;|\;(x-10)^2<4\}$, then a table of probabilities for $z$-scores is all you need.


问:
For $a \in \mathbb R$, find polynomials $P$ such that $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$
Find all polynomials $P(x)$ with real coefficients such that:-    $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$

This is $a$ creating some big problems for me. I don't know what to do.

I am not able to figure out anything because of that $a$. The best I can figure out is that I will find the roots of $P(x)$ because $a$ is not computable as there is no way of finding out the zeros of $P(x-3)$. If I would have been able to find the roots of $(x^2-ax+18)$ then I would have been able to figure out what to do.

If there would have been no $a$ I would have found of the roots of $P(x)$ like for example $\alpha, \beta$ then I would have written out $P(x)$ in the form of $(x-\alpha)(x-\beta)Q(x)$ for $Q(x)$ being any polynomial. Then I would have tried to calculate the answer.

Any help would be appreciated

Body:
the function is not a polynomial
$$ (x^2-ax+18)P(x) = (x^2+3x)P(x-3) $$

Put $x=-3$,
$$ (27+3a)P(-3) = (9-9)P(-6) $$ $ (27+3a) \cdot P(-3)=0$, $27+3a=0$, then $a= -9$
$$ (x²-(-9)x+18)P(x) = (x²+3x)P(x-3) $$
$$(x^2+9x+18)P(x) = (x^2+3x)P(x-3) $$
$$(x+3)(x+6)P(x) = x(x+3)P(x-3) $$
$$(x+6)P(x) = xP(x-3)$$
$$(x+6)P'(x)+ xP(x) = xP'(x)+P(x-3)$$
$$(x+6)P'(x)+ xP(x) = xP'(x)+ \frac{(x+6)}{x} P(x)$$
$$6P'(x) = \frac{(x+6)}{x} P(x)-xP(x) $$
$$6P'(x) = ( 1+ \frac{6}{x}-x)P(x) $$
$$ \frac{P'(x)}{P(x)} = \frac{1}{6}+\frac{1}{x}-\frac{x}{6} $$
$$ \int \frac{P'(x)}{P(x)} = \int ( \frac{1}{6}+\frac{1}{x}-\frac{x}{6} ) $$
$$\log(P(x))+c = \frac{x}{6}-\frac{x^2}{12}+\log(x) $$
$$\log(P(x)) = \frac{x}{6}-\frac{x^2}{12}+\log(x)-c $$
$$ P(x) = \exp ( \frac{x}{6}-\frac{x^2}{12}+\log(x)-c ) $$
$$P(x) = x \cdot \exp(  \frac{x}{6}-\frac{x^2}{12}-c) $$


问:
Find quaternion rotation
Given units quaternions $p$ and $q$, how do you find the quaternion $r$ such as:

$$
rpr^{-1} = q
$$

In other terms, how to find the rotation transforming $p$ to $q$?

答: The rotation applies on vectors of $\mathbb{R}^3$, not on quaternions.

Title: Find quaternion rotation


问:
Euler's number inherently in $\mathbf{x^a=a^x}$ and $\mathbf{x.a=a^x}$?
I was solving some equations graphically, when I came across an identity of Euler's number in two cases:

Case 1: $\mathbf{a.x = a^x}$

Ever wondered how for two numbers $\mathbf{\left\{{ a,x,y} \right \}\in \mathbb{R_{> 0}}
 :\space a.x = a^x= y}$?

For example natural numbers: $\mathbf{2.2=2^2=4}$

What is the general formula for $\mathbf{x}$, given that $\mathbf{x.a=a^x}$?

It has multiple solutions. One is always $\mathbf{x=1}$ And interestingly, if
$\mathbf{x_1=x_2: \space a=e}$. In this case, actually: $\mathbf{x_1=x_2=1}$. So as $\mathbf{a\to e\implies x_1=x_2\to1}$.

$\mathbf{e}$ "came out of nowhere". I am sure the analytical solution would introduce something like Lambert W function which would explain the presence of $\mathbf{e}$.

Case 2: $\mathbf{x^a = a^x}$

Or second case when: $\mathbf{\left\{{ a,x,y} \right \}\in \mathbb{R_{> 0}}
 :\space x^a = a^x= y}$?

For example $\mathbf{2^2=2^2=4}$ or how $\mathbf{3^{2.478...}\approx2.478...^3\approx15.216}...$

What is the general formula for $\mathbf{x}$, given that $\mathbf{x^a=a^x}$?

It has multiple solutions. And interestingly, as $\mathbf{x\to e\implies x_1\to x_2}$.

=======================================

What are the two formulas for $x$ and how to obtain them?

Solution attempt

How to solve for $x$?

I tried this:

$\mathbf{log_ax^a = log_aa^x\\a.log_ax=x\\a.x^{-1}log_ax=1\\a.log_ax^{x^{-1}}=1}\\$

Lambert W function comes into mind. But how?

答:
This answer deals with the solution to the first equation:

$$x^a=a^x$$

$$a\ln x=x\ln a$$

$$\frac{\ln x}{x}=\frac{\ln a}{a}$$

Let, $x=t^{-1}$

$$t\ln t=-\frac{\ln a}{a}$$

$$W\left(\ln te^{\ln t}\right)=W\left(-\frac{\ln a}{a}\right)$$

$$\ln t=W\left(-\frac{\ln a}{a}\right)$$

$$\ln x=-W\left(-\frac{\ln a}{a}\right)$$

$$x=e^{-W\left(-\frac{\ln a}{a}\right)}.$$


答: If your matrix $M$ is "nearly" positive semi-definite then the smallest eigenvalue $\lambda_1$ is just below zero. Hence, If $I$ is the identity matrix, $M+\epsilon I$ should be positive semi-definite with smallest eigenvalue $\lambda_1+\epsilon\geq 0$ (I guess finding the smallest epsilon as to not perturb your matrix too much is the hard job).

Title: Make a matrix symmetric and positive semi-definite if it already approximately is

Body:
I have a matrix that ought to be symmetric and positive semi-definite (with real entries). However, due to small rounding errors that can accumulate over time, that might not be strictly true. For example, the smallest eigenvalue might be $-10^{-10}$ even though it ought to be $0$ if we had been using exact calculations. This isn't particularly important overall, except that some computations will fail altogether with even this small negative number.

Is there a simple expression that will fix this? As an example of what I'm looking for:

If I were asking "how do I make a real number non-negative if it already nearly is?" I'd like the answer "take the absolute value". This leaves non-negative numbers precisely unchanged, and only has a small effect on small strictly negative numbers.

If I were asking "how do I make a matrix $M$ symmetric" I'd like the answer "use $M + M^t$". Actually that is one of the things I want so that will be my first step!

So what I'm asking, analogously, is: "how do I make the matrix positive semidefinite?"


问:
standard error for ratio of two poisson random variables
I'm collecting data for two Poisson random variables, that are completely independent of each other. My current analysis method yields 95% (or 90% or other) confidence intervals for the two $\lambda_1$, $\lambda_2$ for the two Poissons.

My question: how can I come up with a confidence interval for the ratio of the two lambdas?

答:
If $\lambda_1$ belongs to $[a_1,b_1]$ with probability at least $p_1$ and $\lambda_2$ belongs to $[a_2,b_2]$ with probability at least $p_2$ for some $0<a_1<b_1$ and $0<a_2<b_2$, then, by independence, $\lambda_1/\lambda_2$ belongs to $[a_1/b_2,b_1/a_1]$ with probability at least $p_1p_2$.

For example, $p_1=p_2=90\%$ yields $p_1p_2=81\%$.


问:
Geodesic flow and Killing vector field
Let $\gamma$ be a geodesic, $\gamma'$ its tangent vector, $X$ a Killing vector field and $X_\gamma$ the restriction of $X$ to the curve $\gamma$. Let $g$ be the metric on the manifold considered.

Prove that $g(\gamma',X_\gamma)$ is constant along $\gamma$.

The definition I have for a Killing field $X$ is that it satisfied $\mathcal{L}_Xg=0$ where $\mathcal{L}$ denotes the Lie derivative.

I thought about using the induced covariant derivative on the curve $\gamma$, but all I have is a definition and no real explanation on how to put this into pratice.

答: Hint: You will first have to understand how the fact that $X$ is a Killing field is expressed in terms of the covariant derivative $\nabla X$. This is a standard result, it can be directly derived from exapanding the equation $0=(\mathcal L_Xg)(Y,Z)$. Having done that, you can express $\tfrac{d}{dt}g(\gamma'(\gamma(t)),X(\gamma(t)))$ in terms of covariant derivatives (along the curve) and use the equations satisfied by $\gamma'$ and by $X$ to conclude that this vanishes.

Title: Geodesic flow and Killing vector field


问:
Set theory in practice - indexed sets

Hi, I have a problem as above. And below is my first few attempts to solve a, and b.

a) $B_1 \cap \bigcup C_k = \{1, 2, 3\}$. My understanding is that the $\bigcup C_k$ in a is the set of elements that belong to at least one $C_k$ so as k is in $[1,3]$, my answer was that.

b) = $\{1, 2, 3, 4, 5, 12, 13, 14, 15\}$

I have to admit that the above is all my guesses. Can someone explain the indexed sets to me in a simpler and easy-to-apply way? All theoretical definitions make me confused. It would be best if you take one problem as an example. I would really appreciate any help for this advanced course. I have to explain my steps so if you can give me answers to check, that would be great, too.

答:
You can translate $\cup$ to "or" and $\cap$ to "and" when it comes the the conditions defining the sets. So a) is for example "the people that are both (born in January) and (born Sunday, Monday or Tuesday)". Or in natural words, "the people born in Januray on a Sunday, Monday or Tuesday".

See if you can do the others now.

Your guesses right now ($\{1,2,3\}$ and $\{1,2,3,4,5,12,13,14,15\}$) are wrong. In fact they're nonsensical because the sets contain people, not numbers. What's true is e.g.
$$
B_1\cap\left(\bigcup_{1\le k \le 3}C_k\right)
= B_1\cap\left(C_1\cup C_2\cup C_3\right)
$$
So $x\in B_1\cap\left(\bigcup_{1\le k \le 3}C_k\right)$ iff $x$ is in $B_1$ and $x$ in (at least) one of $C_1$, $C_2$, $C_3$. This translates to what I wrote above.

Edit:

In (d), (e), (f), we also have the $'$ meaning complement, which translates to "not". So (d) is "the people that are not (born on Sun, Mon, Tue, Wed, or Thu)" = "the people that are born Fri or Sat".

There are also the general rules $\left(\bigcup_i A_i\right)' = \left(\bigcap_i A_i'\right)$ and $\left(\bigcap_i A_i\right)' = \left(\bigcup_i A_i'\right)$, which might be nice to know, but not necessary for the problem.

Help for b and e:

As I hinted in the comments (your first b was correct), b turns out to be "(something impossible) or (birthday 12th-15th)". If something's impossible it's the empty set, and in gereral $\emptyset\cap A=\emptyset$ and $\emptyset\cup A=A$. Use that.

Your e is wrong. The literal translation would be "(not born in Jan) or (not born in Feb) or ... or (not born in May)". Or equivalently with my rule above: "not (born in Jan and Feb and ... and May)". Can you figure it out from there?

Title: Set theory in practice - indexed sets


问:
Meaning of = in antisymmetric relation
We have a set of students.

$$
\{\text{Bob}, \text{John}, \text{Tom} \}
$$

Scores of a exam were:

$$
\begin{align}
\text{Bob} & : 5 \\
\text{John} & : 10 \\
\text{Tom} & : 10 \\
\end{align}
$$

In the real life, we can sort by scores and say students are ordered.

$$\text{Bob} \le \text{John}$$
$$\text{Bob} \le \text{Tom}$$
$$\text{John} \le \text{Tom}$$
$$\text{Tom} \le \text{John}$$

But order relations are said to be antisymmetric.

$$
xRy \land yRx \Rightarrow x = y
$$

In the example,

$$
\text{John} \le \text{Tom }\land \text{Tom} \le \text{John} \Rightarrow \text{John} = \text{Tom}
$$

So, John and Tom must be equal, if the example is an ordered set. But it is only the scores that matches. If you compare the elements, they are different and John $\ne$ Tom.

Is it allowed to compare scores and say $\text{John} = \text{Tom}$? Can't we say that the example is an ordered set?

Adding some comment to make my question clear:

As nachosemu said, I am doing

$$
x R y \Leftrightarrow score(x) \leq score(y), \,\,\, x,y\in \{\text{Bob}, \text{John}, \text{Tom} \}
$$

For this example, should I read antisymmetric relation as comparing elements:

$$
xRy \land yRx \Rightarrow x = y
$$

or as $=$ in $\lt, =, \gt$:

$$
xRy \land yRx \Rightarrow score(x) = score(y)
$$

答:
The terminology "ordered set" does not really have a fixed meaning. This seems to be a perfectly reasonable confusion in the original question. It is especially confusing as the terminology "partially ordered set" does almost always mean the same thing. That is, a set together with a relation which is reflexive, antisymmetric and transitive.

There is no general agreement about what exactly an "order" is, and even more confusingly the word "total" can mean different things in combination with other words. For example, the difference between partial orders and total orders is not the same as the difference between partial functions and total functions even though all four kinds of thing are binary relations.

As other answers have already made clear, a given set can have lots of different binary relations. When modelling the world, more than one relation may capture something useful.

Title: Meaning of = in antisymmetric relation


问:
Mixed finite element methods and time-dependence
I've looked all over stack exchange and can see the solution to this so here's hopping somebody knows. I'm Trying to solve an time-dependent parabolic equation using mixed finite element method and backwards Euler. So I have two coupled equations:

${\partial u \over \partial x} + \alpha \omega = \gamma{\partial \omega \over \partial t} $

${\partial \omega \over \partial x} + \beta u = 0$

Where $u$ and $\omega$ are the parameters to be solved for and $\alpha$, $\beta$ and $\gamma$ are constant (not depending on either $u$ or $\omega$). We can reduce the system down to a matrix equation of the form:

$M_1  \psi^t = M_2  \psi^{t-1}$

where $\psi = (u\:\:\omega)^\text{T}$ and  solve like normal. However this is not stably for every $\Delta t$ (time stepping) - only a small range of $\Delta t$ give the correct result i.e if $t_{min} <\Delta t < t_{max} $ .
$t_{max}/t_{min}$ obviously this is varying with $\Delta x$, $\alpha$, $\beta$ and $\gamma$.

So my question is:

1) I've double checked my algorithm and believe it to be correct so why is backwards Euler not always stable for some values of $\Delta t$? How does backwards Euler deal with mixed finite elements? Is it still stable?! What am I missing?

In mixed finite elements there are two conditions that must be met:
1) the inf-sup condition
2) the invertibility on the kernel condition

I believe both are still intact - however these are derived for a stationary state (no time dependence) so maybe they are different for time dependence?

I'm really struggling on this so any help is really appreciated. I've been on this for nearly 4 days and I must be missing something. As far as I know the Courant-friedrichs-levy condition doesn't apply to backwards Euler.  Also it should be noted that setting $\gamma = 0$ (non-time dependent) gives the correct solutions - so the code is working-ish. Many thanks guys!

答:
Why are you writing this as two equations?

You could just substitute the second one into the first, and get
$$
\frac{\partial{\omega}}{\partial t} = \frac{\alpha}{\gamma}\omega -\frac{\beta}{\gamma} \frac{\partial^2 \omega}{\partial^2 x}
$$

This is a standard diffusion equation with source term. There is a constrain for the equation to be stable:
$$
-\frac{\beta}{\gamma} > 0
$$
otherwise, this is anti-diffusion, which is a ill-posed problem. Such a ill-posed problem would always blow up, even if you're using an unconditionally stable method like backward-Euler. It blows up because the equation blows up.

Further, watch your boundary conditions. If you want to have a stationary solution, then you cannot use insulated boundaries, i.e. $\frac{\partial \omega}{\partial x} = 0$ on both sides.


问:
Reducing a second order ODE to a hypergeometric equation
I want to solve the following second order ODE with non-constant coefficients: $$y^{\prime\prime} + (ax+b) y^{\prime} + (cx+d)y = 0,\tag{1}$$
where $a,b,c,d\in\mathbb{R}$. From what I understand, its solution involves confluent hypergeometric functions of the first and second kind; e.g., see this post. In particular, it appears that finding the solution of (1) involves finding the solution of the degenerate hypergeometric equation
$$xy^{\prime\prime} +\left(\frac{1}{2} -x\right)y^{\prime} -ey=0,\tag{2}$$
where $e\in\mathbb{R}$ is some appropriately determined constant (not to be confused with Euler's number). Alas, I could not quite figure out the procedure for finding the general solution from that post.

Not only would I like to understand the solution of (1), but I would like to know how to "reduce" the problem from solving (1) to solving (2). (The Wikipedia page on the confluent hypergeometric function lays out the steps for this reduction when we have a non-constant affine coefficient on the $y^{\prime\prime}$ term in (1), but not when the coefficient is a constant.) I have skimmed over the book by Abramowitz and Stegun but didn't find it particularly helpful, although I'm sure that my question is embedded in their development. Other references would be delightful.

答:
The equation you posted is much more complex than the one in the linked post.
$$y^{\prime\prime} + (ax+b)\,y^{\prime} + (cx+d)\,y = 0$$ The idea is first to find a transform such that the multiplying factor of $y$ does not depend on $x$.

So, let $y=z\,e^{f(x)}$. Cheating a little, I found that
$$f(x)= -\frac{a }{2}x^2+ \left(\frac{c}{a}-b\right)x$$ does the job.

Replacing, we end with
$$z''-(ax+\alpha)z'+\beta z=0\tag 1$$ where
$$\alpha=b-\frac{2 c}{a} \qquad \text{and} \qquad \beta=\frac{-a^3+a^2 d-a b c+c^2}{a^2}$$ As you can see here changing variable, we have Hermite differential equation.

So the general solution of $(1)$ is a linear combination of a confluent hypergeometric function of the first kind and an Hermite polynomial.


问:
Multipliciousness within an inner product space.

Question:

Let $V$ be an inner product space and $v,w\in V$. Prove that $\lvert\langle v,w\rangle\rvert=\lVert v\rVert \lVert w\rVert$ if and only if one of the vectors $v$ or $w$ is a multiple of the other.

Attempt:

Assume the identity holds and $y\neq 0$. Let
$$ a=\frac{\langle x,y\rangle }{\lVert y \rVert^2}, $$
and let
$$ z=x-ay. $$
Now, note that $y$ and $z$ are orthogonal because if $z=x-ay$, then by the definition of $a$ we have
$$a=\frac{\langle x,y\rangle }{\lVert y \rVert^2}=\frac{\langle z+ay, y\rangle }{\langle y,y\rangle }=\frac{\langle z,y\rangle }{\langle y,y\rangle }+a\frac{\langle y,y\rangle }{\langle y,y\rangle }=\frac{\langle z,y\rangle }{\langle y,y\rangle }+a,$$
which means that
$$ \frac{\langle z,y\rangle }{\langle y,y\rangle }=0~\text{or}~ \langle z,y\rangle =0,$$
namely $y$ and $z$ are orthogonal. Furthermore, since $\lvert \langle x, y \rangle \rvert= \lVert x \rVert \lVert y \rVert$, then
$$ \frac{\lvert \langle x, y \rangle \rvert}{\lVert y \rVert} =\lVert x \rVert ~\overset{1/\lVert y \rVert}{\implies}~\frac{\lvert \langle x, y \rangle \rvert}{\lVert y \rVert^2} =\frac{\lVert x \rVert}{\lVert y \rVert} =\lvert a \rvert.$$

Lemma: Let $V$ be an inner product space, and suppose that $x$ and $y$ are orthogonal vectors in $V$. Then $\lVert x+y \rVert^2 = \lVert x \rVert^2 + \lVert y \rVert^2$.

Now, we know that
$$ \lVert x \rVert^2 = \lVert ay+z \rVert^2,$$
but, by the lemma, this means that
$$ \lVert ay+z \rVert^2 = \lVert ay \rVert^2 + \lVert z \rVert^2~~\overset{\text{not sure where to go}}{\dots}$$

Note:

I'm trying to follow along with Friedberg's description on pages 337 and 338 in his Linear Algebra:

EDIT$^1$:

OK, so I think I've got an idea. To complete the proof I need to notice that $\frac{\lVert x \rVert}{\lVert y \rVert} =\lvert a \rvert$, which means that $\lVert x \rVert = \lvert a \rvert\lVert y \rVert$ and more importantly that $\lVert x \rVert^2 = \lvert a \rvert^2\lVert y \rVert^2,$ but this is exactly the $\lVert ay\rVert^2$ from the application of the lemma as $\lVert ay\rVert^2=(\lVert ay\rVert)^2=(\lvert a\rvert\lVert y\rVert)^2=\lvert a\rvert^2\lVert y\rVert^2$. This will give the result.

CONCLUSION:

Here is my end proof:

EDIT$^2$:

I found here in E. B. Vinberg's A Course in Linear Algebra a different flavor of this proof (if anybody happens to care):

答:
You basically want to show that
$$\left\langle\frac{v}{\|v\|},\frac{w}{\|w\|}\right\rangle=\frac{\langle v,w\rangle}{\|v\|\cdot\|w\|}=1$$
if and only if $v$ is a multiple of $w$, which will be the case if and only if $\frac{v}{\|v\|}=\frac{w}{\|w\|}$. Hence, let us assume without loss of generality that $\|w\|=\|v\|=1$ and prove that $\langle v,w\rangle = 1$ if and only if $w=v$. Indeed, "$\Leftarrow$" is easy since $v=w$ implies $\langle v,w\rangle = \langle v,v\rangle=\|v\|^2 = 1$. On the other hand, let $V=\mathrm{Span}(v) \perp U$ and write $w=\lambda v + u$. Then,
$$1 = \langle v,w\rangle = \langle v, \lambda v + u\rangle = \lambda + \langle v,u\rangle = \lambda.$$
Therefore, $w=v+u$. Then,
$$1 = \langle w,w\rangle = \langle v+u,v+u\rangle = \langle v,v\rangle + \langle u,u\rangle + 2\langle v,u\rangle = \langle v,v\rangle + \langle u,u\rangle  = 1 + \|u\|$$
implies $\|u\|=0$, so $u=0$ and $v=w$.</p>


问:
Please if someone help me to find the winning Probability
you are playing a chess match against a skilled opponent .for a given game ,there is a 75% chance the game will end in a draw.of the game that do not end in a draw,there is a 2/3 chance you will win,and a 1/3 chance you will lose.the first player to win 2 games in a row is declared the winner of the chess match.what is the probability you will win the match?</p>

答:
It is easy to see that the probabilities that you win, lose, and draw one game are

$\frac16,\; \frac1{12},\;$  and $\;\frac34$ respectively

Let there be $3\,$ states, s (start=level), a (ahead by one ) and b (behind by one)

We seek the probability that you ultimately win starting from s

Note that from any stage you can draw and return to s with $Pr = 3/4$.

Tracking your movement from state to state from s to victory,

$s = 3s/4 + a/6 + b/12\quad$   [I]

Similarly, $a = 3s/4 + 1/6 +b/12\quad$ [II]

(Note that the second term , a win from a completes a win for you)

$b =  3s/4 + a/6\quad$ [III]

solving these $3$ linear equations gives us $P(\,you\; win) = s = \frac{26}{33}$


问:
Show that if $j:[0,t]\to [0,t]$ is continuous on $[0,t]$ and then there exists $x^{**}\in [0,t]$ such that $j(x^{**})=x^{**}$
Suppose $h:[0,1] \to \mathbb{R}$ is continuous on $[0,1]$ and $\{h(x)\space|\space x\in [0,1]\} \subseteq [0,1]$ then there exists $x^* \in [0,1]$ such that $h(x^*)= x^*$.

Let $t>0$. Using only the following results,

Let $X, Y \subseteq \mathbb{R.}\space$For $f:X \to \mathbb{R}, g:Y\to\mathbb{R}\space$with $\{f(x)\space|\space x\in X\}\subseteq Y.\space$If $f$ is continuous at $c\in X$ and $g$ is continuous at $f(c)$ then $g\circ f:X\to\mathbb{R}\space$is continuous on X.

Let $a>0.\space$Then $\mathbb{R}\to\mathbb{R}\space$given by $f(x)=ax$ is continuous on $\mathbb{R}.$

show that if $j:[0,t]\to\mathbb{R}$ is continuous on $[0,t]$ and  $\{j(x)\space|\space x\in [0,t]\}\subseteq [0,t]$ then there exists $x^{**}\in [0,t]$ such that $j(x^{**})=x^{**}$

答:
Consider the curve $f:[0,t]\to[0,1]$ given by $f(x)=\frac{x}{t}$. Note by 2. we have $f(x)=\frac{x}{t}=\frac{1}{t}.x$ is continuous since $\frac{1}{t}>0.$

Then by 1. we have $h\circ f$ is continuous on $[0,t]$

The composition of functions $t\circ h\circ f(x)$ where $t:\mathbb{R}\to\mathbb{R}$ given by $t(x)=tx$ maps the range from $[0,1]\to[0,t].$ And we know this is continuous since this is the composition of the continous functions and $t$ is continuous by 2.

Thus define $j:[0,t]\to[0,t]$ by $t\circ h\circ f:[0,t]\to[0,t].$

Now for $x^{*}\in [0,1]$ such that $h(x^{*})=x^{*},$ choose $x^{**}=tx^{*}\in [0,t],$ which gives us:

$$j(x^{**})=j(tx^{*})=t\circ h\circ f(tx^{*})=t\circ h(x^{*})=t(x^{*})=tx^{*}=x^{**}$$
as required.


问:
A countable family in $\mathcal P (\mathbb N)$ such that every set in $\mathbb N$ contains some element of the family.
Does there exists a countable family of infinite sets $\{A_n:n\in\mathbb N\}\subset\mathcal P(\mathbb N)$ satisfying the following property:
$$\text{For every infinite set }I\in\mathcal P(\mathbb N),\text{ there is }n\in\mathbb N\text{ such that }A_n\subset I\text{ ?}$$

If we require the family to have cardinality $\mathfrak c$, then the question is trivial, but for the countable case I'm stuck.

答:
No. Let $\{A_n\}$ be any sequence of infinite sets. For each $n$ choose an element $x_n$ of $A_n$ that's  different from all the elements $x_k$ chosen so far
and greater than $2n$.   Then the complement of the set of all $x_k$ has none of the $A_n$ as a subset. It is infinite because  it is the complement of an increasing sequence that omits at least half the numbers less than $2n$ for every $n$.


问:
Decomposing Hilbert space into orthogonal complements
I'm revising some elementary theory about Hilbert spaces and there is a question that's been bugging me. A common theorem which you find in many books is the following: given a closed subspace $W$ of a Hilbert space $H$, we can decompose $H$ into the direct sum of $W$ and its orthogonal complement: $H = W \oplus{{W^\perp}}$.

Now as far as I can tell $W^\perp$ will be a closed subspace regardless of whether $W$ is closed or a subspace. Using the projection theorem, we can then write $H = W^\perp \oplus{({W^\perp})^\perp}$. So the only 'advantage' we gain by requiring that $W$ is closed subspace is that $W=({W^\perp})^\perp$. Is this correct?

答: No, that's not correct. If $W$ is not closed, $H = W \oplus{{W^\perp}}$ is not necessarily true: just take a dense subspace ($W=C[0,1]$, when $H=L^2[0,1]$, say), and you'll have $W^\perp=\{0\}$, and $H \neq W \oplus{{W^\perp}}=W$.


问:
Inequality for an integrable real valued function with a compactly supported Fourier transform
Let $f$ be an integrable function on $\mathbb{R}$ where  $\operatorname{support}(\widehat{f}) \subseteq [-\gamma, \gamma]$ for some $ 0 < \gamma < 1$

Prove that $\lvert f(x) - f(0)\rvert  \leq c \gamma \lvert x\rvert \sup\limits_{ y \in \mathbb{R}}\left\{(1+|y|)\lvert f(y)\rvert\right\}$ for some absolute constant $c$.

答: Write Fourier inversion formula for $f(x)$ and $f(0)$. You get an expression of $f(x)-f(0)$ as an integral on the compact set $[-\gamma, \gamma]$. Then you just have to bound all the terms in the integral (to bound $1-e^{iyx}$, you may want to express it as an integral).</p>


问:
curl of what yields $(0,s^{-1},0)$ in cylindrical coordinates?
In cylindrical coordinates $(s,\theta,z)$, what function $\mathbf{A}$ has the property $$\nabla\times \mathbf{A} = (0, \frac{1}{s} , 0) $$

I know generally that $$\nabla\times \mathbf{A} = \left(\frac{1}{s}\frac{\partial A_3}{\partial \theta}-\frac{\partial A_2}{\partial z}, \frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial s} , \frac{1}{s} \left( \frac{\partial}{\partial z}(s\cdot A_2) - \frac{\partial A_1}{\partial \theta}\right)\right) $$

Is there some better way of solving for $\mathbf{A}$ other than a lot of ugly equations?

答:
There is a path integral formula to construct the inverse of the exterior derivative operator for closed differential forms, for curl in three dimensions:
$$
\newcommand{\A}{\mathbf{A}}\newcommand{\ps}{\boldsymbol{\psi}} \newcommand{\p}{\mathbf{p}} \mathscr{R}(\ps) = -(\p - \p_0)\times \int^1_0 \ps\big(\p_0 + t(\p- \p_0)\big)t\,dt,\tag{1}
$$
where $\p = (x,y,z)$ and $\ps$ has to be divergence free. We have that $\mathscr{R}(\ps) $ satisfies:
$$
\nabla \times \mathscr{R}(\ps) = \ps.
$$

The vector field you gave, $(0,1/s,0) = (0,\dfrac{1}{\sqrt{x^2+y^2}},0)$, is not divergence free, there does not exist any vector potential $\A$ satisfying:
$$
\nabla \times \A = (0,\frac{1}{s},0).
$$
However, we can just modify a little bit to make your question a well-posed problem, we can move $1/s$ to the $z$-component, and it is divergence free:
$$
\nabla \cdot \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right) = 0,
$$
and
$$
\nabla \times \A = (0,0,\frac{1}{s})
$$
has a solution (not unique though).

Let $\ps$ be the above vector, and in (1) we take $\p_0$ as the origin we have:
\begin{align}
\A = &\mathscr{R}(\ps) = -\p\times \int^1_0 \ps(t\p)t\,dt
\\
&= -(x,y,z)\times \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right)
\\
&=  \left(-\frac{y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}},0\right).\tag{2}
\end{align}

We can check:
\begin{align}
\nabla \times\A &= \left(0,0, \frac{\partial }{\partial x}\Big(\frac{x}{\sqrt{x^2+y^2}}\Big) + \frac{\partial }{\partial y}\Big(\frac{y}{\sqrt{x^2+y^2}}\Big)\right)
\\
&= \left(0,0, \frac{\sqrt{x^2+y^2} - x\frac{x}{\sqrt{x^2+y^2}}}{x^2+y^2} + \frac{\sqrt{x^2+y^2} - y\frac{y}{\sqrt{x^2+y^2}}}{x^2+y^2}\right)
\\
& = \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right).
\end{align}

Or you can rewrite (2) in cylindrical coordinates:
$$
\A = (-\sin\theta, \cos\theta,0),
$$
and use the curl formula you gave.


问:
Question about Bézout's Lemma
I'm somewhat confused about the following: Let $\gcd(a,b)=1$ and $\gcd(a,c)=1$ so that by Bézout there are integers $m_1,n_1,m_2,n_2$ s.t. $$m_1a-n_1b=1,\quad m_2a-n_2c=1 \, .$$

Obviously we can make the simultaneous replacements for instance $m_1 \rightarrow m_1 + k_1 b$ and $n_1 \rightarrow n_1 + k_1 a$ that don't change the result for any integer $k_1$. Similarly this can be done for the second equation. This implies there are infinitely many such integer pairs. Now fixing $m_1$ and $n_1$, apparently we can always choose $m_2=m_1$. Why?

答:
The asserted property is false.

Take a counterexample: $a=2$, $b=3$, $c=9$. Then $\gcd(a,b)=\gcd(a,c)=1$. We can express the $\gcd$ of $a$ and $b$ as:
$$1 = 2a -b = 2(2) - 3.$$
The asserted claim is that we can express $1=\gcd(a,c)$ in the form
$$1= 2a-n_2c$$
for some integer $n_2$. But in this case, that would mean solving $1 = 4-9c$, and this equation has no solution: you would need $9c_2=3$.

What is true is that you can find expressions for both $\gcd(a,b)$ and $\gcd(a,c)$ where the coefficient of $a$ is the same (you can’t just decide what it will be). To verify this, note that since $\gcd(a,b)=\gcd(a,c)=1$, then $\gcd(a,bc)=1$. Therefore, there exist integers $m$ and $n$ such that
$$ma + nbc = 1.$$
From this, you get equations $ma + (nc)b = 1$ and $ma + (nb)c = 1$. You can also replace $bc$ with $\mathrm{lcm}(b,c)$.

Title: Question about Bézout's Lemma


问:
Rolle-like problem
Let $f:[a,b]\to \mathbb{R}$ a function that is  continuous on $[a,b]$ and differentiable on $(a,b)$. Prove that there exists $c\in (a,b)$ such that
$$f'(c)\cdot \int\limits_a^c f(x) ~dx=-f^2(c).$$
This kind of problems usually reduces to find an appropriate function and to apply Rolle theorem for this function, in order to get the required equality. The obvious/easy choices did not get me there.

答:
You claim seems incorrect.
take $$f(x)=1$$
then
$$f(c)=0$$

Title: Rolle-like problem


答:
This is not true, as the example $a=0,b=1$ and $f(x)=x$ shows.

But if we have in addition that $f(b)=0$, then define

$g(x):= f(x)\int\limits_a^x f(t) ~dt$. Then $g(a)=g(b)=0$ and

$$g'(x)=f'(x)\int\limits_a^x f(t) ~dt+f(x)^2.$$

Title: Rolle-like problem

Body:
Let $f:[a,b]\to \mathbb{R}$ a function that is  continuous on $[a,b]$ and differentiable on $(a,b)$. Prove that there exists $c\in (a,b)$ such that
$$f'(c)\cdot \int\limits_a^c f(x) ~dx=-f^2(c).$$
This kind of problems usually reduces to find an appropriate function and to apply Rolle theorem for this function, in order to get the required equality. The obvious/easy choices did not get me there.


问:
Proving a series of matrices to be positive
Assume that $A \geq 0$, $(I-A)$ is invertible. I have shown that if we have a sequence $$ S_n = I + A + A^2 + \cdots $$ that $$(I-A)S_n = I - S^{n+1} = I$$
as $n \rightarrow \infty$.

How can I now show that $S_{n \rightarrow \infty} \geq 0$? "Positive" here means that all entries of the matrix are  $\geq 0$.

答:
Let $S = \lim_{n \to \infty} S_n$, where $S_n = I + A + \cdots + A^{n-1}$. Because $S_n$ converges, each entry of $S_n$ converges to the corresponding entry of $S$. That is, for each $i,j,$ $\lim_{n \to \infty} S_n(i,j) = S(i,j)$.

Now, if a sequence $x_n$ satisfies $x_n \to x$ with $x_n \geq 0$ for all $n$, then it must hold that $x \geq 0$. Conclude that each entry of $S$ is non-negative, which is to say that $S \geq 0$.


答:
It might be easier to take $F'(x)=f(x)$ so:
$$F''(c)\left[F(c)-F(a)\right]=-\left[F'(c)\right]^2$$
now can such a point exist?

Title: Rolle-like problem