问:
Simplex Method row operations help?
before programming an algorithm which implements the simplex method, I thought I'd solve an issue before the actual programming work begins.

For some reason, I can NEVER get the correct answer. I've understood the method, but the problem is with the row operations - where you try to get a column to have all 0 values except for the pivot element which has a value of '1'.

To do this, I play around with the rows by doing R1-R2, R2+5R1, etc. I always manage to get the pivot column to be 1 and the rest 0's, however my answers never match the correct ones. Ive narrowed it down to a problem with the row operations - are there any rules related to this, or can I just play around with the rows as much as I like? Also, can I mix between older tableaux and the current one?

Thanks

答:
I'll assume you are using a nice full table that includes a P column and P row, where P is the objective function.

The critical thing is that to read a final answer from your tableau, you must always:

Ensure the tableau describes the system in a (pivoted) RREF

Ensure the right hand sides are all non-negative

Ensure the bottom row (the objective row) is all non-negative

Ensure that P is a basic variable

This means that at the end your table describes a linear system in which some of the variables are free and some (including P) are solved in terms of those.  The first • ensures you can read the answers off easily.  The second ensures that you can set the free variables to 0 without breaking the law.  The third ensures that doing this is good for the objective, since those variables were just costing you profit anyways.  The fourth is very important: you don't want to set profit = 0!

Here are some common mistakes I see:

Deciding which element becomes a pivot has some small amount of freedom in it, but you can choose a "wrong" pivot.  This will break the second • point, or best case scenario, fail to make progress on the third • point.

Make sure that you only use the chosen pivot as your second row in the row operations.  Otherwise you are likely to break the first • point (look in your tables and see if you still have the same number of 0-0-1-0 type columns after your row op shenanigans as before).

Technically you could use previous tableaux rows to alter things, but I think this is very likely to break the first • point.

One somewhat rare mistake I see with very creative row ops (that are legal in solving linear systems) is that people manage to swap P out of the basis.  Sometimes this means they have a "2" as the pivot for P, and so the last row means 3x + 5y + 2P = 26.  x and y are free and hurt profit, so set x=y=0, but then you have 2P=26 instead of P=26, and so maybe P=13 is the right answer (or maybe something is horribly wrong).  Sometimes people end up with P a free variable.  Then you are forced to set P=0!  That is very rarely the most profitable move.

For example: Consider a simple company with two products and two resources.  Both products make them a dollar each.  The first product uses 1 resource of each type.  The second product uses 2 of the first resource and none of the second.  The company decides to make X units of the first product and Y units of the second.  It has 100 units of both resources.

The profit is P = X + Y, or written linear style, (-1)X + (-1)Y + 0U + 1P = 0.

The first resource constraint is 1X + 2Y + U = 100, where U is the unused portion of the first resource.

The second resource constraint is 1X + 0Y + V = 100, where V is the unused portion of the second resource.

To be feasible, we must have X ≥ 0, Y ≥ 0, U ≥ 0, and V ≥ 0

To be optimal, we want P to be a maximum

The initial tableau is:

```<code> X  Y  U  V  P  RHS
 1  2  1  0  0  100
 1  0  0  1  0  100
-1 -1  0  0  1    0
</code>```

Currently our "basic" solution is feasible but not optimal: X=Y=0 are free, and U=100, V=100, and P=0 have pivots.  Each row is a linear equation.

Both X and Y are profitable, so we can choose either to pivot (into the basis).  Let's choose X since it is first alphabetically.  We can choose either row as the pivot row, so let's choose the first as it is "owned" by U and U is first alphabetically.  So we want R1 column 1 to be a "1", R2 column 1 to be a "0", R3 column 1 to be a "0".

Let's do it wrong(ly).  For my first row operation, I will choose R3 + R2, instead of the correct R3+R1.  I guess I'll also do the correct R2−R1, and the oh-so-easy "leave R1 alone".

Our next tableau is:

```<code> X  Y  U  V  P  RHS
 1  2  1  0  0  100
 0 -2 -1  1  0    0    (wrong)
 0 -1  0  1  1    0
</code>```

First column looks good.  What's the big deal?  Look at the V column.  It used to be a nice RREF looking 0-1-0.  Now it is all messy, 0-1-1.  Is V a free variable now?  If so who are the pivots?  X has a pivot.  P has a pivot.  Nobody else. Uh oh.  First • is broken.

Ok, so backtrack:

```<code> X  Y  U  V  P  RHS
 1  2  1  0  0  100
 1  0  0  1  0  100
-1 -1  0  0  1    0
</code>```

Now do the right row ops: R1 stays the same.  R2 becomes R2−R1.  R3 becomes R3+R1.  We get:

```<code> X  Y  U  V  P  RHS
 1  2  1  0  0  100
 0 -2 -1  1  0    0
 0  1  1  0  1  100
</code>```

Now what do we see?  Let's look at the basic solution:  X, V, P have pivots, so we can solve for them X=100, V=0, P=100.  Y,U are free, so we set them to 0.  How much unused resources? U=0 of the first resource, and V=0 of the second.  So efficient.  How much do we produce?  X=100 of the first product, and Y=0 of the second.  So easy.  How much money do we make?  P=100.  Sounds good.

Can we do better?  No.  Y+U+P=100.  The bigger we make Y, the less room we have for profit.  The bigger we make U, the less room we have for profit.  The only reasonable thing is to set Y=U=0.  Yay.

Title: Simplex Method row operations help?

Body:
I'll assume you are using a nice full table that includes a P column and P row, where P is the objective function.

The critical thing is that to read a final answer from your tableau, you must always:

Ensure the tableau describes the system in a (pivoted) RREF

Ensure the right hand sides are all non-negative

Ensure the bottom row (the objective row) is all non-negative

Ensure that P is a basic variable

This means that at the end your table describes a linear system in which some of the variables are free and some (including P) are solved in terms of those.  The first • ensures you can read the answers off easily.  The second ensures that you can set the free variables to 0 without breaking the law.  The third ensures that doing this is good for the objective, since those variables were just costing you profit anyways.  The fourth is very important: you don't want to set profit = 0!

Here are some common mistakes I see:

Deciding which element becomes a pivot has some small amount of freedom in it, but you can choose a "wrong" pivot.  This will break the second • point, or best case scenario, fail to make progress on the third • point.

Make sure that you only use the chosen pivot as your second row in the row operations.  Otherwise you are likely to break the first • point (look in your tables and see if you still have the same number of 0-0-1-0 type columns after your row op shenanigans as before).

Technically you could use previous tableaux rows to alter things, but I think this is very likely to break the first • point.

One somewhat rare mistake I see with very creative row ops (that are legal in solving linear systems) is that people manage to swap P out of the basis.  Sometimes this means they have a "2" as the pivot for P, and so the last row means 3x + 5y + 2P = 26.  x and y are free and hurt profit, so set x=y=0, but then you have 2P=26 instead of P=26, and so maybe P=13 is the right answer (or maybe something is horribly wrong).  Sometimes people end up with P a free variable.  Then you are forced to set P=0!  That is very rarely the most profitable move.

For example: Consider a simple company with two products and two resources.  Both products make them a dollar each.  The first product uses 1 resource of each type.  The second product uses 2 of the first resource and none of the second.  The company decides to make X units of the first product and Y units of the second.  It has 100 units of both resources.

The profit is P = X + Y, or written linear style, (-1)X + (-1)Y + 0U + 1P = 0.

The first resource constraint is 1X + 2Y + U = 100, where U is the unused portion of the first resource.

The second resource constraint is 1X + 0Y + V = 100, where V is the unused portion of the second resource.

To be feasible, we must have X ≥ 0, Y ≥ 0, U ≥ 0, and V ≥ 0

To be optimal, we want P to be a maximum

The initial tableau is:

 X  Y  U  V  P  RHS
 1  2  1  0  0  100
 1  0  0  1  0  100
-1 -1  0  0  1    0

Currently our "basic" solution is feasible but not optimal: X=Y=0 are free, and U=100, V=100, and P=0 have pivots.  Each row is a linear equation.

Both X and Y are profitable, so we can choose either to pivot (into the basis).  Let's choose X since it is first alphabetically.  We can choose either row as the pivot row, so let's choose the first as it is "owned" by U and U is first alphabetically.  So we want R1 column 1 to be a "1", R2 column 1 to be a "0", R3 column 1 to be a "0".

Let's do it wrong(ly).  For my first row operation, I will choose R3 + R2, instead of the correct R3+R1.  I guess I'll also do the correct R2−R1, and the oh-so-easy "leave R1 alone".

Our next tableau is:

 X  Y  U  V  P  RHS
 1  2  1  0  0  100
 0 -2 -1  1  0    0    (wrong)
 0 -1  0  1  1    0

First column looks good.  What's the big deal?  Look at the V column.  It used to be a nice RREF looking 0-1-0.  Now it is all messy, 0-1-1.  Is V a free variable now?  If so who are the pivots?  X has a pivot.  P has a pivot.  Nobody else. Uh oh.  First • is broken.

Ok, so backtrack:

 X  Y  U  V  P  RHS
 1  2  1  0  0  100
 1  0  0  1  0  100
-1 -1  0  0  1    0

Now do the right row ops: R1 stays the same.  R2 becomes R2−R1.  R3 becomes R3+R1.  We get:

 X  Y  U  V  P  RHS
 1  2  1  0  0  100
 0 -2 -1  1  0    0
 0  1  1  0  1  100

Now what do we see?  Let's look at the basic solution:  X, V, P have pivots, so we can solve for them X=100, V=0, P=100.  Y,U are free, so we set them to 0.  How much unused resources? U=0 of the first resource, and V=0 of the second.  So efficient.  How much do we produce?  X=100 of the first product, and Y=0 of the second.  So easy.  How much money do we make?  P=100.  Sounds good.

Can we do better?  No.  Y+U+P=100.  The bigger we make Y, the less room we have for profit.  The bigger we make U, the less room we have for profit.  The only reasonable thing is to set Y=U=0.  Yay.


问:
Oscillation and Hölder continuity
I am studying a proof of a theorem. And I have the following situation in the proof:

Consider $\Omega$ is a bounded open set of $\mathbb R^n$ and $u: \Omega \to \mathbb R$ is a function satisfying:

$$\operatorname{osc}_{B(x_0,R)} u \leq (1-\delta) \operatorname{osc}_{B(x_0,4R)},$$ for all ${B(x_0,R)} \subset \Omega$ for some $0<\delta <1$ ($\delta$ is independent of the open ball).

The book says: Iterating this inequality we have that $u$ is Hölder continuous.

Someone can help me understand the proof in the part of the "Iteration".

Thank you!

答:
I prefer to write the inequality as
$$\operatorname{osc}_{B(x_0,4^{-1}R)} u \leq (1-\delta) \operatorname{osc}_{B(x_0,R)} \tag1$$
Iteration means that applying (1) with $B(x_0,R)$ replaced by $B(x_0,4^{-1}R)$, then replaced by $B(x_0,4^{-2}R)$, etc., and then chaining these inequalities together. Thus,
$$\operatorname{osc}_{B(x_0,4^{-2}R)} u \leq (1-\delta) \operatorname{osc}_{B(x_0,4^{-1}R)} u \leq (1-\delta)^2 \operatorname{osc}_{B(x_0,R)} \tag2$$
and in general,
$$\operatorname{osc}_{B(x_0,4^{-k}R)} u \leq (1-\delta)^k \operatorname{osc}_{B(x_0,R)} \tag3$$

Inequality (3) can be written in the form of the Hölder condition. Indeed, let $\alpha>0$ be the number such that $(1-\delta)=4^{-\alpha}$. Given $x$ near $x_0$, let $k$ be the integer such that $4^{-k-1}\le |x-x_0|< 4^{-k}R $. By (3),
$$|u(x)-u(x_0)|\le (1-\delta)^k \operatorname{osc}_{B(x_0,R)} \le C |x-x_0|^\alpha \tag4$$
where $C=4^\alpha \operatorname{osc}_{B(x_0,R)}$ is independent of $x$. Thus, $u$ is  Hölder continuous.

ContentLicense: CC BY-SA 3.0

Title: Oscillation and Hölder continuity


问:
Let $V$ be a vector space and $u,v∈V$. Prove {$u,v$} is linearly independent if and only if {$u+v,u-v$} is linearly independent.
Let $V$ be a vector space and $u,v∈V$. Prove {$u,v$} is linearly independent if and only if{$u+v,u-v$} is linearly independent.

consider $(c+d)u+(c-d)v=0$.

we obtain $cu+du+cv-dv=c(u+v)+d(u-v)=0$ but because $u+v,u-v$ are linearly independent $c=d=0$. it follows that $c-d=0$ and $c+d=0$. so $u,v$ are linearly independent.

Now for the other directions..

consider $c(u+v)+d(u-v)=0$.

we obtain $cu+cv+du-dv=(c+d)u+(c-d)v=0$ but because $u,v$ are linearly independent $c+d=0$ and $c-d=0$. therefore, $c=d=0$ and $u+v,u-v$ are linearly independent. q.e.d.

Is this  a valid proof? For the first direction, is starting with $(c+d)u+(c-d)v=0$ allowed?

答:
It's not ideal to start with $(c+d)u+(c-d)v=0$. There's an implicit assumption here that any linear combination $au + bv$ can be expressed in the form $(c + d)u + (c - d)v$. Or, to put it another way, the map $(c, d) \mapsto (c + d, c - d)$ is surjective. If I, as a reader, wanted proof that $2u + 3v \neq 0$, I would then have to figure out which values of $c$ and $d$ would make $(c + d)u + (c - d)v = 2u + 3v$, or just trust that it could be done.

Now, the implicit assumption turns out to be totally correct, but that's beside the point. The proof should either prove this assumption explicitly, or should tackle this direction a different way.

Inverting the map $(c, d) \mapsto (c + d, c - d)$ actually gives us a good way forward here. If we have $(c + d, c - d) = (a, b)$, then
\begin{align*}
c + d &= a \\
c - d &= b.
\end{align*}
Adding the equations, we get $2c = a + b$, and subtracting, we get $2d = a - b$. So, the inverse map is
$$(a, b) \mapsto \left(\frac{a + b}{2}, \frac{a - b}{2}\right).$$
So, if my powers of intuition are correct, then
$$au + bv = \frac{a + b}{2}(u + v) + \frac{a - b}{2}(u - v).$$
Try using the above fact to finish your proof.

ContentLicense: CC BY-SA 4.0


问:
If $B=5I_2-3A+A^2$ can I show this matrix is invertible?
Let $D=\begin{pmatrix} 2 & 0\\ 0 & 7 \end{pmatrix}$, then $A=PDP^{-1}$ and $B=5I-3A+A^2$. Show that B is diagonalizable.

I have tried algebraically, trying to get $B^{-1}$, but get nothing.

Body:
Since $A=PDP^{-1}$, $P^{-1}AP=D$,

so $P^{-1}BP=P^{-1}(5I-3A+A^2)P=P^{-1}(5I)P-3P^{-1}(A)P+P^{-1}(A^2)P$

$=5I-3D+D^2$, which is diagonal.


问:
An example of function in $\textit{L} ^p([0,1])$ but not in $\textit{L} ^{\infty}([0,1])$.
I would like to show that $\bigcap_{p\ge1}\textit{L}^p([0,1])\setminus \textit{L}^{\infty}([0,1])\not=\emptyset$, so does somebody know a function that is in this set?

答:
A function in that set is $f(x)=\log(x)$.

$\forall p\ge 1$ there exists a neighborhood $U(0^+)$ s.t. $|f(x)|^p\le \frac{1}{\sqrt{x}}$ in it, so
$$\int_{(0,1)}|\log(x)|^pdx\le \int_{U(0^+)}\frac{1}{\sqrt{x}}dx+\int_{(0,1)\setminus U(0^+)}|\log(x)|^pdx < +\infty$$
and we can conclude that $f\in \textit{L}^p([0,1])$ for all $p\ge 1$.

This function is not in $\textit{L}^{\infty}([0,1])$, because it satisfies:
$$\forall M\in\mathbb{R} \mbox{ I have that }|\{x\in[0,1]:|f(x)|>M\}|>0$$
in fact, given $M>0$ the set described above is the interval $(0,e^{-M})$ so it has positive measure.


问:
Determine the range of $p$ respectively when the series converges or diverges

Determine the range of $p$ respectively, when $$\sum_{n=1}^\infty \left (\dfrac{\sin n}{n^p + \sin n} + \dfrac{\cos ^2 n}{n^{2p}} \right)$$ converges or diverges.

That is, when does the series converge? And when does it diverge?

I have no idea about the problem. It seems so hard to determine the range of $p$, since the denominator of the items is complicated. Can anyone help?

Body:
Split partial sums. For $0 < p < 1$, work with

$$\frac{\cos^2 n}{n^{2p}}= \frac{1}{2n^{2p}}+ \frac{\cos 2n}{2n^{2p}},$$
$$\frac{\sin n}{n^p + \sin n} = 1 - \frac{1}{1+ \frac{\sin n}{n^p}} = \frac{\sin n}{n^p} - \frac{\sin ^2n}{n^{2p}} + \mathcal{O}(n^{-3p})$$


问:
Why is $n=\frac{2p^2+2pq+2pr+q^2+2qr+r^2}{p+q+r}$, where $n$ is $\text{prime}$, of form such that $p\pm a,p\pm b,$ are $ \text{prime}, 1<a<b<n$
Why is $n= \left\lfloor \frac{2p^2+2pq+2pr+q^2+2qr+r^2}{p+q+r} \right\rfloor$, where $n$ is $\text{prime}$, of form such that $p\pm a,p\pm b,$ are $ \text{prime}, 1<a<b<n$?

Consider this:

Let $p,q,r$ be $\mathbb{prime}$ where $r<q<p$.

Let $x\pm y$ denote $(x+y\quad\mathbb{and}\quad x-y)$

I conjecture that if the following conditions are true;
$$\mathbb{isprime}( n),\quad n = \left\lfloor \frac{2 p^2+2 p q+2 p r+q^2+2 q r+r^2}{p+q+r} \right\rfloor$$ Then $n$ is a prime of such form that $n \pm a$, and $n \pm b$ are prime numbers where $a$ and $b$ are distinct positive integers with $a < b < n$.$\quad$
(A137669(n) where $n>2$)

However, I have no clue as to why this is the case.

Example:

$x = 29, y = 17,z = 11, a = 12, b = 18$

$$71 = \left\lfloor \frac{2 \cdot 29^2+2 \cdot 29 \cdot 17+2 \cdot 29 \cdot 11+17^2+2 \cdot 17 \cdot 11+11^2}{29+17+11} \right\rfloor \,\text{is prime}$$

$$71\pm 12 = \{83,59\} = \{71+12,71-12\},\,\text{are both prime}$$

$$71\pm 18 = \{89,53\} = \{71+18,71-18\},\,\text{are both prime}$$

Series representation:

$$
\sum_{n = -\infty}^{\infty}=\left\lfloor\left (
\begin{cases}
1, & n = 1 \\
y+z, & n = 0 \\
{y+z\over(-y-z)^n}, & n\geq2
\end{cases}
\right )
x^n \right\rfloor = \left\lfloor \frac{2 p^2+2 p q+2 p r+q^2+2 q r+r^2}{p+q+r} \right\rfloor
$$

If anyone can explain why this is the case, it would be very much appreciated.

Body:
Actually, you can represent any odd integer $n$ this way. With your original letters, the fraction is $$ \frac{p^2 + (p+q+r)^2}{p+q+r}.   $$
You can take $$  p = \frac{n-1}{2}, \; \; q+r = \lceil \sqrt n \rceil  $$

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

   n           p         q + r
   3           1           2
   5           2           3
   7           3           3
   9           4           3

  11           5           4
  13           6           4
  15           7           4
  17           8           5
  19           9           5

  21          10           5
  23          11           5
  25          12           5
  27          13           6
  29          14           6

  31          15           6
  33          16           6
  35          17           6
  37          18           7
  39          19           7

  41          20           7
  43          21           7
  45          22           7
  47          23           7
  49          24           7

  51          25           8
  53          26           8
  55          27           8
  57          28           8
  59          29           8

  61          30           8
  63          31           8
  65          32           9
  67          33           9
  69          34           9

  71          35           9
  73          36           9
  75          37           9
  77          38           9
  79          39           9

  81          40           9
  83          41          10
  85          42          10
  87          43          10
  89          44          10

  91          45          10
  93          46          10
  95          47          10
  97          48          10
  99          49          10

 101          50          11
 103          51          11
 105          52          11
 107          53          11
 109          54          11

 111          55          11
 113          56          11
 115          57          11
 117          58          11
 119          59          11

 121          60          11
 123          61          12
 125          62          12
 127          63          12
 129          64          12

 131          65          12
 133          66          12
 135          67          12
 137          68          12
 139          69          12

 141          70          12
 143          71          12
 145          72          13
 147          73          13
 149          74          13

 151          75          13
 153          76          13
 155          77          13
 157          78          13
 159          79          13

 161          80          13
 163          81          13
 165          82          13
 167          83          13
 169          84          13

 171          85          14
 173          86          14
 175          87          14
 177          88          14
 179          89          14

 181          90          14
 183          91          14
 185          92          14
 187          93          14
 189          94          14

 191          95          14
 193          96          14
 195          97          14
 197          98          15
 199          99          15

 201         100          15
 203         101          15
 205         102          15
 207         103          15
 209         104          15

 211         105          15
 213         106          15
 215         107          15
 217         108          15
 219         109          15

 221         110          15
 223         111          15
 225         112          15
 227         113          16
 229         114          16

 231         115          16
 233         116          16
 235         117          16
 237         118          16
 239         119          16

 241         120          16
 243         121          16
 245         122          16
 247         123          16
 249         124          16

 251         125          16
 253         126          16
 255         127          16
 257         128          17
 259         129          17

 261         130          17
 263         131          17
 265         132          17
 267         133          17
 269         134          17

 271         135          17
 273         136          17
 275         137          17
 277         138          17
 279         139          17

 281         140          17
 283         141          17
 285         142          17
 287         143          17
 289         144          17

 291         145          18
 293         146          18
 295         147          18
 297         148          18
 299         149          18

 301         150          18
 303         151          18
 305         152          18
 307         153          18
 309         154          18

 311         155          18
 313         156          18
 315         157          18
 317         158          18
 319         159          18

 321         160          18
 323         161          18
 325         162          19
 327         163          19
 329         164          19

 331         165          19
 333         166          19
 335         167          19
 337         168          19
 339         169          19

 341         170          19
 343         171          19
 345         172          19
 347         173          19
 349         174          19

 351         175          19
 353         176          19
 355         177          19
 357         178          19
 359         179          19

 361         180          19
 363         181          20
 365         182          20
 367         183          20
 369         184          20

 371         185          20
 373         186          20
 375         187          20
 377         188          20
 379         189          20

 381         190          20
 383         191          20
 385         192          20
 387         193          20
 389         194          20

 391         195          20
 393         196          20
 395         197          20
 397         198          20
 399         199          20

 401         200          21
 403         201          21

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


问:
Prove that $(\log(n))^k=O(n)$ for any $k$
I'm having problem proving that $(\log(n))^k=O(n)$. I tried googling it but can't find an answer.

答:
Without lost of generality let me consider $\ln$ in place of $\log$. Having $\frac{(\ln n)}{n}^k=\left( \frac{\ln n}{n^{1/k}} \right)^k$ is enough to consider

$$\lim\limits_{n \to \infty}\frac{\ln n}{n^{1/k}} =\lim\limits_{n \to \infty}\frac{\frac{1}{n}}{\frac{1}{k}n^{\frac{1}{k}-1}}=\lim\limits_{n \to \infty}\frac{k}{n^{1/k}} = 0$$


问:
The applicable cases for the sandwich theorem
I've only seen the definition for the sandwich theorem in the form $f(x) \leq g(x)\leq h(x)$. But I've  seen multiple questions where one or even both of the inequality signs are just $\lt$.

These cases feel intuitively possible, but why aren't they included in the official statement for the theorem? Are they incorrect?

答: If $f(x) < g(x)$ then $f(x) \le g(x)$ as well, so you can still use the sandwich theorem.


问:
Divide by a number without dividing.
Can anyone come up with a way to divide any given x by any given y without actually dividing?

For example to add any given x to any given y without adding you would just do:

$x-(-y)$

And to subtract any given x from any given y (that is, y-x) you could do:

$y+xe^{iπ}$

*edit: well since (i) is ($\sqrt{-1}$) and that is technically subtracting this one might not work perfectly but for the sake of the riddle and for the sake of example, I'm using that equation :)

How can you divide without dividing? Can anyone come up with equations that work for all $x$ and $y$ values? (For all intents and purposes we will leave out dividing by zero issues and what-not... don't worry about that...

答:
Take the logarithm that maps multiplication/division into addition/subtraction:

$$\frac{x}{y}=e^{\log{x/y}}=e^{\log x- \log y}.$$

$x,y >0$.

Also, see my answer for multiplying natural numbers here:
Advocating base 12 number system

Title: Divide by a number without dividing.


问:
How do you find the sum of this infinite series?
The actual series that I want to solve:
$$
\sum_{n=2}^{\infty}\frac{n 2^{n-1} + \frac{n(n-1)}{1\cdot 2}2^{n-2}+\frac{n(n-1)(n-2)}{1\cdot 2\cdot 3}2^{n-3}+\cdots+n 2^1+1}{3^n}
$$

答:
HINT

Use Binomial Theorem to write the numerator as $3^n-2^n$ and then simply use the sum of geometric series $$\sum_{n=0}^{\infty}ar^n=\frac{1}{1-r}, \ |r|<1$$


问:
Finding roots at right half-plane
Let $P(z)\in\mathbb{C}[z]$ be a polynomial with complex coefficients.

(1) Prove that if all zeroes of $P$ are in the right plane $Re z\geq 0$, then the same is true for $P'(z)$.

(2) Deduce that $x^5+2x^4-3x^3-4x^2-5x-6$ has at least one root in the left half-plane $Rez\leq 0$.

Is there any hint on how to approach this question? I was thinking to use the Mobius transformation to convert right half plane to unit circle and apply the $\int_C\frac{f'(w)}{f(w)}dw$ but I do not know how to proceed afterwards.

Title: Finding roots at right half-plane

Body: $\Re(\sum_j \frac1{z-a_j})<0$ for $\Re(z)< 0$


问:
Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
Prove using contour integration that $\displaystyle \int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$

I am at a loss at how to start this problem and which contour to pick. I have been trying to get the sector with angle $2\pi/3$ to work with a bump around the pole at $e^{i2\pi/3}$ and the origin, but I am getting 5 or 6 different integrals and it is not really getting me anywhere.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\large\tt Alternative:\ A\  "\Re eal"\ Integration,}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{x} \over x^{3} - 1}
\,\dd x} =
\left.\bracks{\nu^{1}}\int_{0}^{\infty}{1 - x^{\nu} \over
1 - x^{3}}\,\dd x\,
\right\vert_{\,\Re\pars{\nu}\ \in\ \pars{-1,2}}
\\[5mm] = &\
{1 \over 3}\bracks{\nu^{1}}
\int_{0}^{\infty}{x^{-2/3} - x^{\nu/3 - 2/3}\,\,\, \over 1 - x}
\,\dd x
\\[5mm] = &\
{1 \over 3}\bracks{\nu^{1}}
\int_{0}^{\infty}{x^{\color{red}{1/3} - 1}\ -\
x^{\color{red}{\nu/3 + 1/3} - 1}\,\,\, \over 1 - x}\,\dd x
\end{align}
However,$\ds{{1 \over 1 - x} = \sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{1 + k}\expo{\ic\pi k}}\,
{\pars{-x}^{k} \over k!}}$

Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{x} \over x^{3} - 1}
\,\dd x}
\\ = &\
-\,{1 \over 3}\bracks{\nu^{1}}
\\ &
\overbrace{%
\Gamma\pars{\!%
\color{red}{{\nu \over 3} + {1 \over 3}}\!}
\!\Gamma\pars{\!1 -
\bracks{\color{red}{{\nu \over 3} + {1 \over 3}}}\!}
\!\exp\pars{\!%
-\ic\pi\bracks{\color{red}{{\nu \over 3} + {1 \over 3}}}\!}}
^{\substack{\ds{Ramanujan's} \\[0.5mm] \ds{Master}\\[0.5mm] \ds{Theorem}}}
\\[5mm] = &\
-\,{1 \over 3}\bracks{\nu^{1}}
\bracks{{\pi \over \sin\pars{\pi\nu/3 + \pi/3}}\,
\exp\pars{-\ic\pi\bracks{{\nu \over 3} + {1 \over 3}}}}
\\[5mm] = &\
-\,{\pi \over 3}\bracks{\nu^{1}}
\cot\pars{{\pi\nu \over 3} + {\pi \over 3}} =
\left.-\,{\pi \over 3}
\totald{\bracks{\cot\pars{{\pi\nu \over 3} + {\pi \over 3}}}}{\nu}\,\right\vert_{\ \nu\ =\ 0}
\\[5mm] = &\
-\,{\pi \over 3}\bracks{-\,{\pi \over 3}
\csc^{2}\pars{\pi \over 3}} =
\bbx{4\pi^{2} \over 27} \approx 1.4622 \\ &
\end{align}


问:
Application closed in Zariski topology
Consider the integral homomorphism $f: A \rightarrow  B$.
How do I prove that the map $f^*:Spec(B) \rightarrow Spec(A)$ given by $f^*(Q) = f^{-1}(Q)$ is a closed application.
Note:
1) I believe we need to show that $f^*(V(I)) = V(f^{-1}(I))$. I have already shown that $f^*(V(I))$ is contained in $V(f^{-1}(I))$ but I can not show the other inclusion.
2) $f^{-1}$ is the inverse image of $f$.
3) $I$ is any ideal of the ring $B$.</p>

答:
The massive hint that you need to solve your problem is the following result

Theorem 5.10 Atiyah - Macdonald:
    Let $f : A \to B$ be an integral morphism. For every prime ideal $\mathfrak{p} \subseteq A$ there is a prime ideal $\mathfrak{q} \subseteq B$ such that $A \cap \mathfrak{q} = \mathfrak{p}$.

Title: Application closed in Zariski topology


问:
How are these linear algebra equations equal?
From equations 3.13, 3.14, and 3.17 in Gaussian Processes

3.13 and 3.14

$$
\begin{aligned}
\nabla\psi(f) &= \nabla\log p(y|f) - K^{-1}f \\
\nabla \nabla\psi(f) &= \nabla \nabla \log p(y|f) - K^{-1} = -W - K^{-1}
\end{aligned}
$$

So we have defined the derivative and the second derivative of $\psi$. The next page then goes on to explain using newtons method to find where $\nabla\psi = 0$ which makes sense, but I dont get the derivation of the iterative update in equation 3.18...

3.18

$$
\begin{aligned}
f^{new} &= f - (\nabla \nabla\psi)^{-1}\nabla\psi \\
&= f + (K^{-1} + W)^{-1} (\nabla \log p(y|f) - K^{-1}f) \\
&= (K^{-1} + W)^{-1} (Wf + \nabla \log p(y|f))
\end{aligned}
$$

I can't follow the last step in this equation at all. I cannot see why the $f$ disappears in the front and somehow gets put into the last term and then $Wf$ pops up. Can anyone follow this and explain the last step of the derivation?

答:
Take
$$f + (K^{-1} + W)^{-1} (\nabla \log p(y|f) - K^{-1}f),$$
and add and subtract $Wf$ inside the parentheses, i.e.
$$f + (K^{-1} + W)^{-1} (\nabla \log p(y|f) - K^{-1}f - Wf + Wf).$$
Factor $K^{-1}f - Wf = (K^{-1} + W)f$, and partially expand the parentheses:
$$f + (K^{-1} + W)^{-1} (\nabla \log p(y|f) + Wf) - (K^{-1} + W)^{-1}(K^{-1} - W)f.$$
This clearly simplifies to
$$(K^{-1} + W)^{-1} (Wf + \nabla \log p(y|f)),$$
as required.


问:
Any good undergrad analysis research problems?
I'm looking for some ideas for longer and more research type problems to do in real analysis. The coursework we get is not very demanding and it would be nice to have some longer and harder projects to do on the side. At the moment i have background in basic real analysis, (limits and sequences, differentiation and continuity, integration, sequences of functions and series), linear algebra, and basic set theory and construction and the axioms of real/rational/natural numbers, but as i'm studying and intend to study more all the time, any suggestions about more advanced topics are also highly appreciated and would give something to work towards and do later on. Something that would require a bit more time and effort and looking into, something a bit more researchier than proving specific limits and differentiation rules.</p>

答:
How about the problem of finding a function which is differentiable, but whose derivative is not continuous?

You could look it up, and there's a standard example ($f(x)=\begin{cases}x^2\sin(1/x),x\gt0\\0,x\le0\end{cases}$, I believe).

But in addition to studying it, maybe you could look for other examples.


问:
Nested Interval Theorem: Another part
Let [a1, b1] ⊇ [a2, b2] ⊇ · · · be a sequence of intervals. If lim(bi − ai) → 0, then ∩[an, bn] is a singleton set.

I have proved the first part which says that the infinite intersection of such subsets is not empty and is a closed interval. But how do I prove this part? I tried solving it by assuming that the intersection contains two unequal numbers x and y. And then by generating real numbers between them I was trying to reach some kind of contradiction but was not able to proceed.
Can someone help me with this proof?

答: If you have proved that the intersection is a closed interval, then if you assume there exist $x$ and $y$ distinct in the intersection you have the whole interval $[x, y] $ in it as well. This is impossible since it has positive measure.

Title: Nested Interval Theorem: Another part


问:
Finding Power Law constants
I need to find the constant α and β of the power law formular (d=α*D^(β)).

For that I have the folowing graph:

I'm reading an article that says: Knowing that log10(d)= mlog10(D)+b. A power-law is then d=αD^(β) where α=10^(b) and β= m.

For this case, if I set α=10^(0.684), converting b from meters to km, and β= 0.054177, I get a strange result that do not match with the graph.

I don't know what am I doing wrong. Any insights?

The data for the blue line and points are (the data for diameter and depth are in meters):

```<code>crater Diameter Depth
    1   9703    1071
    2   11121   1668
    3   11198   1113
    4   12667   1221
    5   12786   1843
    6   13386   2055
    7   13462   903
    8   13821   961
    9   13870   963
    10  14104   1513
    11  14126   1813
    12  14830   1800
    13  16405   1983
    14  18915   1318
    15  19753   2184
    16  21074   1160
    17  22140   2162
    18  24052   1669
    19  24134   1716
    20  28690   1655
    21  29107   2591
    22  33187   3335
    23  33561   2468
    24  34412   2360
    25  34825   2585
    26  43102   2368
    27  49474   3855
</code>```

Title: Finding Power Law constants

Body:
In order to be in the same conditions, I started defining
$$x_i=\frac{\text{Diameter}_i} {1000} \qquad \text{and} \qquad y_i=\frac{\text{depth}_i } {1000}$$

Then, the log-log regression  written as
$$\log_{10}(y)=a + b \log_{10} x$$ leads to
$$a=-1.29342\qquad \text{and} \qquad b=0.62046 \qquad (R^2=0.553897)$$ which for the model
$$y =\alpha\, x^\beta$$ gives as estimates
$$\alpha_0=10^{-1.29342}=0.0509\qquad \text{and} \qquad \beta_0=0.6205$$ Starting with these values, the nonlinear regression gives
$$\alpha=0.255262\qquad \text{and} \qquad \beta=0.652586\qquad (R^2=0.953088)$$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval}
   \\
 \alpha & 0.255262 & 0.084156 & \{0.081572,0.428952\} \\
 \beta  & 0.652586 & 0.100473 & \{0.445219,0.859952\} \\
\end{array}$$

Back to the original data
$$\left(
\begin{array}{cccc}
\text{crater} & \text{Diameter} & \text{Depth} & \text{predicted} \\
 1 & 9703 & 1071 & 1125 \\
 2 & 11121 & 1668 & 1229 \\
 3 & 11198 & 1113 & 1235 \\
 4 & 12667 & 1221 & 1338 \\
 5 & 12786 & 1843 & 1347 \\
 6 & 13386 & 2055 & 1387 \\
 7 & 13462 & 903 & 1393 \\
 8 & 13821 & 961 & 1417 \\
 9 & 13870 & 963 & 1420 \\
 10 & 14104 & 1513 & 1436 \\
 11 & 14126 & 1813 & 1437 \\
 12 & 14830 & 1800 & 1483 \\
 13 & 16405 & 1983 & 1584 \\
 14 & 18915 & 1318 & 1739 \\
 15 & 19753 & 2184 & 1789 \\
 16 & 21074 & 1160 & 1866 \\
 17 & 22140 & 2162 & 1927 \\
 18 & 24052 & 1669 & 2034 \\
 19 & 24134 & 1716 & 2038 \\
 20 & 28690 & 1655 & 2282 \\
 21 & 29107 & 2591 & 2303 \\
 22 & 33187 & 3335 & 2509 \\
 23 & 33561 & 2468 & 2528 \\
 24 & 34412 & 2360 & 2569 \\
 25 & 34825 & 2585 & 2589 \\
 26 & 43102 & 2368 & 2976 \\
 27 & 49474 & 3855 & 3256
\end{array}
\right)$$


问:
What can I say about a map multiplication?
If I know the following:

If I have the following homomorphism $f: \mathbb Z_{p^a} \rightarrow \Bbb Z_{p^b}$ defined by multiplication by $n/d$ where $n = p^b$ and $d = \gcd(p^a, p^b).$

And I know the following:

if $a\geq b$ then $f$ is onto.

if $a \leq b$ then $f$ is 1-1.

How can I use this piece of information to conclude something about this function: multiplication by $rn/d$ for $r= p^t x$ and $0 \leq r < d$? when is this map 1-1 and when it is onto?

Could anyone help me in that please?

Is this map trivially onto?

答:
Let us handle the natural generalisation of the problem you present. Consider $m, n \in \mathbb{N}^{\times}$ together with the notation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for any $r \in \mathbb{Z}$. Let $d\colon=(m; n)$ denote the greatest common divisor of the two numbers and also consider $m'\colon=\frac{m}{d}, n'=\frac{n}{d}$ (the fractions exist since $d \neq 0$).

Let us fix $k \in \mathbb{N}^{\times} \cap n'\mathbb{Z}$ and let $f \in \mathrm{End}_{\mathbf{Gr}}(\mathbb{Z})$ be the additive group endomorphism given by $f(r)=kr$ for arbitrary $r \in \mathbb{Z}$. It is obvious that $f[\mathbb{Z}]=f\left[\langle m \rangle\right]=\left\langle f(m)\right\rangle \leqslant n\mathbb{Z}$, since $f(m)=mk \in mn'\mathbb{Z}=m'dn'\mathbb{Z}=m'n\mathbb{Z} \leqslant n\mathbb{Z}$, which means that $f$ induces a quotient morphism $g \in \mathrm{Hom}_{\mathbf{Gr}}\left(\mathbb{Z}_m, \mathbb{Z}_n\right)$, described by $g\left(\overline{r}\right)=\widehat{f(r)}=\widehat{kr}$, where $\overline{\bullet}$ denotes classes modulo $m$ respectively $\widehat{\bullet}$ refers to classes modulo $n$.

By virtue of the general properties of quotient morphisms, we have $\mathrm{Ker}g=f^{-1}\left[n\mathbb{Z}\right]/m\mathbb{Z}$. It is easy to see that $f^{-1}\left[n\mathbb{Z}\right]=\frac{n}{(k; n)}\mathbb{Z}$, whence $\mathrm{Ker}g=\frac{n}{(k; n)}\mathbb{Z}/m\mathbb{Z} \approx \mathbb{Z}_{\frac{m(k;n)}{n}} \ (\mathbf{Gr})$. In particular, $g$ is injective if and only if $\frac{m(k; n)}{n}=1$, relation equivalently expressed as $m(k; n)=n \Leftrightarrow m \mid n \wedge (k; n)=\frac{n}{m}$.

As to the image, we have that $\mathrm{Im}g=\left(\mathrm{Im}f+n\mathbb{Z}\right)/n\mathbb{Z}=\left(k\mathbb{Z}+n\mathbb{Z}\right)/n\mathbb{Z}=(k; n)\mathbb{Z}/n\mathbb{Z}=\frac{n}{(k; n)}\mathbb{Z} \ (\mathbf{Gr})$. Therefore, $g$ is surjective if and only if $(k; n)=1$, which entails $n'=1$ and thus $n \mid m$. The surjectivity of $g$ is thus attained if and only if $n \mid m$ and $(k; n)=1$.

P.S. On the existence and properties of quotient morphisms in the category (context) of groups. I won't fully go into details by proving the statements below (we could do this at a later time, as I confess I currently don't have too much time to spare in order to be active on this forum).

Proposition 1. Let $f \colon G \to G'$ and $g \colon G \to G''$ two group morphisms such that $g$ is surjective and $\mathrm{Ker}g \leqslant \mathrm{Ker}f$. Then there exists a unique group morphism $h \colon G'' \to G'$ such that $f=h \circ g$ (in other words, the diagram below is commutative). We refer to $h$ as the quotient of $f$ with respect to $g$. We have the following descriptions:
$$\begin{align}
\mathrm{Ker}h&=g\left[\mathrm{Ker}f\right]\\
\mathrm{Im}h&=\mathrm{Im}f.
\end{align}$$

Proposition 2 (particular case of the previous). For arbitrary group $G$ and normal subgroup $H \trianglelefteq G$ we adopt the notation $\sigma^G_H \colon G \to G/H$ to refer to the canonical surjection associated to the quotient by $H$. Let $f \colon G \to G'$ be a group morphism together with two normal subgroups $H \trianglelefteq G$ and $H' \trianglelefteq G'$ such that $f[H] \leqslant H'$ (which is equivalent to $H \leqslant f^{-1}\left[H'\right]$). Then there exists a unique morphism $g \colon G/H \to G'/H'$ such that $\sigma^{G'}_{H'} \circ f=g \circ \sigma^G_H$ (in other words, the diagram below is again commutative). We have the descriptions:
$$\begin{align}
\mathrm{Ker}g&=f^{-1}\left[H'\right]/H\\
\mathrm{Im}g&=\sigma^{G'}_{H'}\left[\mathrm{Im}f\right]=\left(H'\mathrm{Im}f\right)/H'.
\end{align}$$

Proposition 2 is obtained from proposition 1 by considering $\sigma^{G'}_{H'} \circ f$ in the role of the original $f$ and $\sigma^G_H$ in the role of the original $g$.


答: Assume two different points x and y. Both of them are of the form  $a(n)<X<b(n) , a(n)<y<b(n)$. So $0<|x-y|<|a(n)-b(n)|< €$. Difference of two constant number is less than any positive number €>0. Deduce that $X=y$

Title: Nested Interval Theorem: Another part

Body:
Let [a1, b1] ⊇ [a2, b2] ⊇ · · · be a sequence of intervals. If lim(bi − ai) → 0, then ∩[an, bn] is a singleton set.

I have proved the first part which says that the infinite intersection of such subsets is not empty and is a closed interval. But how do I prove this part? I tried solving it by assuming that the intersection contains two unequal numbers x and y. And then by generating real numbers between them I was trying to reach some kind of contradiction but was not able to proceed.
Can someone help me with this proof?


问:
A question in Functional Analysis.
I want to prove the following statement:

$\Omega$ is a compact set. Let $A_N $ be an $N$ dimensional subspace of $L_\infty(\Omega)$. $\{g_i\}_{i=1}^N$ be any orthonormal basis of $A_N\subset L_2(\Omega)$. Prove that for any $f=\sum_{i=1}^Nc_ig_i(x)$ , ($c_1,...,c_N\in \mathbb{R}$): $$\sup_{f\in A_N,f\neq0}\frac{|f(x)|}{||f||_2}=\sup_{(c_1,...,c_N)\in\mathbb{R}^N\\\sum_{i=1}^Nc_i^2=1}\big|\sum_{i=1}^Nc_ig_i(x)\big|.$$

My understanding is that we transfer the characterization of the norm on $f$ to the coefficients ${c}_i$ by considering a unit sphere: $\sum_{i=1}^N c_i^2=1$. But I have no idea to show this identity.

Any suggestions would be welcome! Thank you!

答:
Show that the left-hand side can be rewritten as $$\sup_{f \in A_N, \|f\|_2=1} |f(x)|.$$

The supremum over functions $f \in A_N$ can be instead over functions of the form $\sum_{i=1}^N c_i g_i(x)$.

Show that then the $\|f\|_2=1$ constraint then becomes $\|\sum_{i=1}^n c_i g_i\|_2^2 = \sum_{i=1}^N c_i^2 = 1$.

Title: A question in Functional Analysis.


问:
understanding $\mathbb{R}$/$\mathbb{Z}$
I am having trouble understanding the factor group, $\mathbb{R}$/$\mathbb{Z}$, or maybe i'm not.  Here's what I am thinking.

Okay, so i have a group $G=(\mathbb{R},+)$, and I have a subgroup $N=(\mathbb{Z},+)$.  Then I form $G/N$.  So this thing identifies any real number $x$ with the integers that are exactly 1 unit step away.  So if $x=\frac{3}{4}$, then $[x]=({...,\frac{-5}{4},\frac{-1}{4},\frac{3}{4},\frac{7}{4},...})$ and i can do this for any real number.  So therefore, my cosets are unit intervals $[0,1)+k$, for integers $k$.  Herstein calls this thing a circle and I was not sure why, but here's my intuition.  The unit interval is essentially closed and since every real number plus an integer identifies with itself, these "circles" keep piling up on top of each other as if its one closed interval.  Since it's closed it is a circle.  Does that make sense?

Now how do I extend this intuition to this?

$G'=[(a,b)|a,b\in{\mathbb{R}}], N'=[(a,b)|a,b\in{\mathbb{Z}}].$  What is $G'/N'$?  How is this a torus?  I can't get an intuitive picture in my head...

EDIT:  Actually, are the cosets just simply $[x]=[x\in{\mathbb{R}}|x+k,k\in{\mathbb{Z}}]?$

答:
Just try to add some intuitive answers:

The essential intuition that $\mathbb{R}/\mathbb{Z}$ is isomorphic to unit circle in the complex plane is that we can map a real number $x$ to the radian so we get $e^{2\pi i x}$, and so if two number differ by integers they map to the same radian (we circle back).

The same intuition carries over to the isomorphism that $\mathbb{R}^2/\mathbb{Z^2}$ is isomorphic to the torus: We just need to tweak how to generate the coordinates: It turns out that any coordinate on a torus can be indexed by two radians. See this video at 0:33, there is one horizontal circle, and one vertical circle, so you can think of the first real number as a radian that maps you to some position in the horizontal circle, and the second number maps you to some position in the vertical circle. Again, differing by integers does not matter any more.

Title: understanding $\mathbb{R}$/$\mathbb{Z}$


问:
What is the difference between the variable 2X and the two identically distributed variable X+X?
Not sure if I phrase the question quite right, but lets says there is a random variable ${X\sim N(\mu,\sigma^2)}$, what would be the difference between a multiple of the variable and the sum of the same amount of the variable? Examples would be appreciated, since our textbook doesn't elaborate. As context, I was wondering why ${Var(X+X) \neq Var(2X)}$, at least according to my textbook.

答:
$X+X$ and $2X$ are exactly the same.

What you probably are asking is how the above quantity differs from $X + Y$, where $Y$ is an independent random variable with the same distribution as $X$.

To understand the difference, it is easier to consider a simpler scenario, where $X \sim \text{Bernoulli}(1/2)$ (i.e. it takes values $1$ and $0$ with equal probability $1/2$).

On one hand, $2X$ only takes two values: $0$ and $2$ (with equal probability $1/2$).

On the other hand, $X + Y$ takes values $0, 1, 2$ with probabilites $1/4, 1/2, 1/4$ respectively.

The difference is that in $2X$, you only have one source of randomness, and you are doubling whatever you get. In $X + Y$, you have two separate sources of randomness, so there are opportunities for things to "cancel" in the sum. This is a rough explanation of the intuition for why $\text{Var}(2X) > \text{Var}(X+Y)$.


问:
A simple proof about $e^x$?
Do you guys think this is correct? I am trying to prove that there is no single-term polynomial function (oxymoron, I know) $f(x)$ which is always (or at least as x approaches infinity) greater than $e^x$ (I will try to expand this to any polynomial function later). Let $f(x)$ be such a function with the lowest possible degree. Then that means that the slope of $f$ will have to be greater than the slope of $e^x$ as x approaches infinity. This means that $f'(x)>e^x$ as x approaches infinity. However, we already assumed that $f(x)$ is the lowest degree which is greater than $e^x$, but $f'(x)$ is one degree lower than $f$. Therefore, this is a contradiction. What do you guys think? Is this valid? Thanks!</p>

答:
For any polynomial $p(x)$, we have $\displaystyle\lim_{x\to\infty}\dfrac{p(x)}{e^x}=0$. This can be shown, for example, by using L'Hospital's Rule, and in other ways.

Your argument uses the derivative in a different way. The argument is somewhat informal, but it can be made formal. One could use induction. Let us prove by induction on degree that $\lim_{x\to\infty }\frac{p(x)}{e^x}=0$.

We need to prove the degree $1$ base case. For the induction step, suppose the result holds for polynomials of degree $k$. We show it holds for polynomials of degree $k+1$. After a certain point $B$, by the induction hypothesis, we have $|p'(x)|\lt 1$. So $|p(B)+x)|\lt |p(B)|+x$. Thus the ratio $\frac{p(B+x)}{e^{B+x}}$ approaches $0$ as $x\to\infty$.

Remark: If we are allowed to use the power series expansion of $e^x$, there is an easy argument. Let $p(x)$ have degree $n$. For positive $x$, the power series expansion of $e^x$ shows us that $e^x\gt \frac{x^{n+1}}{(n+1)!}$. Now we can show easily that
$$\lim_{x\to\infty}\frac{p(x)}{\frac{x^{n+1}}{(n+1)!}}=0.$$

Title: A simple proof about $e^x$?


问:
Convergence Rate and Newton's Method?
I have a few nonlinear functions that I am using Newton's method to solve and am mainly interested in computing the convergence rate (or lack there of) each:

$x^2 = 0$

$x^3 = 0$

$x + x^3 = 0$

$x + x^4 = 0$

Taking the first equation for example:
$$x_{k+1} = x_k - \frac{x_k^2}{2x_k} = \frac{x_k}{2}$$
Now my textbook simply states $x_k$ converges to zero but only with a linear rate. Apparently this is obvious, but I don't see why. It also goes on to state:
$$\lim_{x \to 0}\frac{|\nabla^2f(x)|}{|\nabla f(x)|} = \infty$$
which means a quadratic convergence is unbounded.

Can someone explain the test for convergence rate of Newton's method?

From my understanding it is the following:

if $|x^{k+1} - x_*| \leq \gamma |x^{k} - x_*|$ then $\{x^k\}$ is said to converge to $x_*$ linearly

if $|x^{k+1} - x_*| \leq c|x^{k} - x_*|^P$ then $\{x^k\}$ is said to converge to $x_*$ with rate at least P

Questions:

Are $\gamma$, $c$, and $P$ simply arbitrary constants?

Why is it obvious that equation 1 converges linearly? What about the others?

This response to a similar question seems to compute the multiplicity of the roots of each equation but divergence is never mentioned

答:
Let $c$ be a fixed point of a function $g$, i.e. $g(c) = c$. I will assume the definition that the iteration sequence $x_k = g^{k}(x_0)$ has order of convergence $p \geq 1$ if
$$\lim_{k \to \infty}\frac{|x_{k + 1} - c|}{|x_{k} - c|^p} = C$$
for some $C \neq 0$. If $p = 1$ it is required that $C < 1$. Of course definitions vary from place to place, so adjust accordingly.
In your example,
$$\lim_{k \to \infty}\frac{|x_{k + 1} - 0|}{|x_{k} - 0|} = \frac{1}{2} < 1$$
so the convergence is linear.

Here is a rather simple test for convergence rate:

Suppose you already know $x_k$ converges to $c$ and you want to find the order of convergence. Let $p$ be the smallest positive integer for which $g^{(p)}(c) \neq 0$. Then by the Lagrange remainder theorem,
$$x_{k + 1} = g(x_k) = g(c) + \frac{g^{(p)}(\xi_k)}{p!}(x_k - c)^p = c + \frac{g^{(p)}(\xi_k)}{p!}(x_k - c)^p.$$
where $\xi_k$ is between $c$ and $x_k$. By the squeeze theorem, $\lim_{k \to \infty}\xi_k = c$. Hence assuming $g^{(p)}$ is continuous at $c$,
$$\lim_{k \to \infty}\frac{|x_{k + 1} - c|}{|x_{k} - c|^p} = \frac{|g^{(p)}(c)|}{p!}$$
so the convergence is of order $p$, except that if $p = 1$ it is also necessary that $|g'(c)| < 1$.

Whether or not the iteration converges, and what fixed point it converges to depends on the function $g$ and the starting point $x_0$.

Edit:
To find a solution $c$ to $f(c) = 0$, Newton's method constructs a function $$g(x) = x - \frac{f(x)}{f'(x)}$$
and tries to find a fixed point of $g$ via the iteration $x_k = g^{k}(x_0)$ since $g(c) = c \iff f(c) = 0$. So
$$g'(x) = \frac{f(x)f''(x)}{f'(x)^2},$$
$$g''(x) = \dots$$
etc. You can verify that if $c$ is a root of $f$ of multiplicity $m$, that is, $$f(x) = (x - c)^{m}h(x)$$
for some function $h$ with $h(c) \neq 0$, then $g'(c) = 1 - \frac{1}{m}$ and thus the convergence is only linear when $m > 1$.

Title: Convergence Rate and Newton's Method?


问:
Continuous Random Variables with Joint PDF problem
Suppose two random variables X and Y have joint pdf $f(x,y) = 3x^2e^{-y} $ when 0 < x < 1, y > 0. f(x,y) = 0 otherwise.

a. Find the marginal pdf of X

b. Find the conditional density of Y given X and evaluate P(Y > 1|X = 0.5).

c. Show that Cov(X,Y) = 0. Are X and Y independent?

d. Compute E[2X(3X + Y)].

I've compute the marginal pdf of X to be 3x^2 when 0 < x < 1. Then that means the conditional density of Y given X is $\frac{f(x,y)}{f_x(x)} = \frac{3x^2e^{-y}}{(3x^2)}$. I'm having trouble calculating P(Y > 1|X = 0.5). Would it be correct to calculate 1 - P(Y <= 1|X = 0.5)? And would I just integrate $\frac{f(x,y)}{f_x(x)}$ from 0 to 1?

For part C I know that Cov(X,Y) = E(XY) - E(X)E(Y). Would E(XY) = $\int^ \infty_0\int^1_0(xy*f(x,y))dxdy$? Have I chosen my bounds correctly?

答:
You have $f_{\small X,Y}(x,y)= 3 x^2\mathrm e^{-y}\mathbf 1_{x\in(0;1)}\mathbf 1_{y\in(0;\infty)}$.

So indeed:

(a) $f_{\small X}(x)=3x^2\mathbf 1_{x\in(0;1)}\require{cancel}\cancelto{1}{~\int_0^1\mathrm e^{-y}\mathrm d y~}$

(b) $f_{\small Y\mid X}(y\mid x)=e^{-y}\mathbf 1_{y\in(0;\infty)}\mathbf 1_{x\in(0;1)}$ and hence:$$\begin{align}\mathsf P(Y\gt 1\mid X=0.5) ~&=~\int_1^\infty \mathrm e^{-y}\,\mathrm d y\\[1ex]&=~1-\int_0^1\mathrm e ^{-y}\,\mathrm d y\end{align}$$

(c) $\mathsf E(XY)=\int_0^1\int_0^\infty xy\cdot 3x^2\mathrm e^{-y}\,\mathrm d y\,\mathrm d x$ and hence $$\begin{align}\mathsf E(XY)&=\int_0^1 x\cdot 3x^2\,\mathrm d x\cdot\int_0^\infty y\cdot \mathrm e^{-y}\,\mathrm d y\\[2ex]&=\text{hmm...}\end{align}$$

Well done.

(d) So you are on track to evaluate $\mathsf E(2X(3X+Y))$.


问:
Computing the Variance of The Number of Descents of a Permutation
I am trying to solve the following problem:

Let $\sigma: [n] \longrightarrow [n]$ be a random permutation in $S_n$, the set of all permutations, and let $d(\sigma)$ be the number of descents of $\sigma$ defined by the cardinality of the set $\{ i \in [n-1] \;| \;\sigma(i+1) < \sigma(i)\}$. Give a closed form expression for the variance of $d(\sigma)$.

My approach thus far: Let $X$ be the random variable denoting the number of descents, and let us express it as the sum of indiator random variables. That is, $X = X_1 + X_2 + ... + X_{n-1}$. We have that
\begin{align*}
\mathbb{V}(X) &= \mathbb{E}(X^2) - \mathbb{E}(X)^2 \\
&= \mathbb{E}(XX) - \mathbb{E}(X)^2 \\
&= \sum_{i=1, j=1}^{n-1} \mathbb{E}(X_iX_j) - \left( \sum_{i=1}^{n-1}\mathbb{E} (X_i)\sum_{j=1}^{n-1} \mathbb{E}(X_j)\right) \\
&= \sum_{i=1, j=1}^{n-1} \mathbb{E}(X_i X_j) - \mathbb{E}(X_i)\mathbb{E}(X_j)
\end{align*}
It looks like we obtain a sum of covariances for muliple pairs of $X_i$'s. Nonetheless, this is where I got stuck because I am not sure how to proceed from here. It would be easier if the $X_i$'s were independent and identically distributed, but it is not the case. I would truly appreciate any help on computing this variance!

答:
For each $i\in[n-1]$ let $X_i=1 \iff \sigma(i+1)>\sigma(i)$ and $X_i=0$ otherwise. Put $$X=X_1 + \dots + X_{n-1}$$
As you said, $X$ counts the number of descents in your random permutation. We need to compute $V(X)$ which we'll express as $$V(X)=\sum_{k=1}^{n-1}V(X_k)+2\sum_{i<j}\text{cov}(X_i,X_j)$$ Notice  $V(X_1)=V(X_i)$ and $\text{cov}(X_1,X_2)=\text{cov}(X_i,X_{i+1})$. Moreover $\text{cov}(X_1,X_3)=\text{cov}(X_i,X_j)$ for $i,j\in[n-1]$ such that $j-i>1$. Since there are $n-1$ terms in the first sum and $n-2$ out of ${n-1 \choose 2}$ terms of the form $\text{cov}(X_i,X_{i+1})$ in the second sum, we can further simply $V(X)$ into the following expression: $$V(X)=(n-1)V(X_1)+2(n-2)\text{cov}(X_1,X_2)+(n-2)(n-3)\text{cov}(X_1,X_3)$$ With a little counting we see that $$P(X_1=1)=\frac{{n \choose 2}(n-2)!}{n!}=\frac{1}{2} \\ P(X_1=1,X_2=1)=\frac{{n \choose 3}(n-3)!}{n!}=\frac{1}{6} \\ P(X_1=1,X_3=1)=\frac{{n \choose 2}{n-2 \choose 2}(n-4)!}{n!}=\frac{1}{4}$$ Using these values we obtain $V(X_1)=\frac{1}{4},\text{cov}(X_1,X_2)=-\frac{1}{12},$ and $\text{cov}(X_1,X_3)=0$. Putting everything together, we finally get $V(X)=\frac{n+1}{12}$.


问:
Find equations of the ellipses given conditions on the directrices, foci, and vertices
The ellipses have their centers at the origin and their major axes on the $x$-axis. Find the equation:

with distance between directrices $27$, and between foci $3$;

with a focus at $(-\sqrt{13},0)$ and a vertex at $(0,2)$.

答:
The equation of an ellipse with the center at the origin and the  major axes on the x-axis is $$\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$$

where $2a,2b$ are the major & minor axes respectively.

We know the coordinate of the foci are $(\pm ae,0)$ and the equation of directrices are $x=\pm\frac ae$ where $e$ is the Eccentricity   $e=\frac{\sqrt{a^2-b^2}}b$

So the distance between the foci is $2ae\implies 2ae=3$ and

the distance between the directrices is $\frac{2a}e\implies \frac{2a}e=27$

Solve for $a,e$

$2.$ As the coordinate of the foci are $(\pm ae,0),  ae=\sqrt{13}$ as $ae>0$

The vertices are $(\pm a,0)$ and $(0,\pm b)\implies b=2$ as $b>0$

As $b^2=a^2(1-e^2)\iff a^2=b^2+(ae)^2=2^2+13=17$


问:
Quadratic and arithmetic mean inequality
I want to prove by induction that

$$\left(\tfrac{1}{n}\sum_{k=1}^{n} x_k\right)^2\leq\tfrac{1}{n}\sum_{k=1}^{n} x_k^2$$

But I have no idea how. For n=1 it is trivial that
$$\left(\tfrac{1}{1}\sum_{k=1}^{1} x_k\right)^2\leq\tfrac{1}{1}\sum_{k=1}^{1} x_k^2=x_k^2$$

But if i want to conclude n+1 from n I get

$$\left(\tfrac{1}{n+1}\sum_{k=1}^{n+1} x_k\right)^2\leq\tfrac{1}{n+1}\sum_{k=1}^{n+1} x_k^2$$

And now I cannot seperate $\tfrac{1}{n+1}$ in order to substitute the rest with the induction hypothesis.

Maybe one needs to use a different version of induction? Thanks for any help.

答:
With a little rewriting, the Induction hypothesis is
$$\color{blue}{\left(\sum_{k=1}^n x_k\right)^2 \leqslant n \sum_{k=1}^n x_k^2}$$

For the inductive step, note
$$\left(\sum_{k=1}^n x_k +x_{n+1} \right)^2 = \color{blue}{\left(\sum_{k=1}^n x_k \right)^2} + 2x_{n+1}\left(\sum_{k=1}^n x_k \right)+x_{n+1}^2  \color{blue}{ \leqslant n \sum_{k=1}^n x_k^2} + x_{n+1}^2 + 2x_{n+1}\left(\sum_{k=1}^n x_k \right)$$

Hence it remains to show
$$2x_{n+1}\left(\sum_{k=1}^n x_k \right) \leqslant \sum_{k=1}^n x_k^2+nx_{n+1}^2 \iff \sum_{k=1}^n \left(x_{n+1}-x_k \right)^2 \geqslant 0$$


问:
How to understand 【140 is almost 100 times less than 13000.】
How to understand these two sentences?

(1)【140 is almost 100 times less than 13000.】

(2)【7 is nearly 2 times greater than 3.】

In my opinion,
if the first sentences is correct: 140 (* nearly 100 times) < 13000 ,
by analogy, the second sentence should be 【7 is nearly 1/2 times greater than 3】: 7 (* almost 1/2) > 3.
Am I correct?

答:
No, because the word "greater" is the opposite of the word "less".

The number that is $2$ times less than $3$ is $3/2$.
The number that is $2$ times greater than $3$ is $6$, which is almost $7.$

The number that is $1/2$ times greater than $3$ is $3/2$ again, which is not near $7$ at all. So I would never say $7$ is about $1/2$ times greater than $3.$

By the way, sentence $(2)$ is badly worded because "nearly" implies that $7$ is not quite as large as $6,$ which is incorrect.
The sentence should say something like,
"$7$ is more than $2$ times greater than $3.$"

Actually, the wording of the sentences in the question seems altogether a little strange to me. The phrase "$100$ times smaller" is slightly better than "$100$ times less", and "smaller by a factor of nearly $100$" is better than either of those. In any case I think we are venturing into ordinary English rather than mathematical writing here.


问:
What is $\mathbb{Q}(\pi^2) \cap \mathbb{Q}(\pi^2-\pi)??$
I'm understanding transcendental extension and I've found this question: Find $\mathbb{Q}(\pi^2) \cap \mathbb{Q}(\pi^2-\pi)$.

I guess the answer is $\mathbb{Q}$, but I still didn't get the proof. If $\alpha \in \mathbb{Q}(\pi^2) \cap \mathbb{Q}(\pi^2-\pi)$ then there exists $p,q,r,s \in \mathbb{Q}[x]$ s.t

$$\alpha = \frac {p(\pi^2)}{q(\pi^2)} = \frac{r(\pi^2-\pi)}{s(\pi^2-\pi)}$$.

It could help: If we call $\alpha = \phi(\pi)$, since $\alpha \in \mathbb{Q}(\pi^2)$ we have $\phi(\pi) = \phi(-\pi)$ and $\alpha \in \mathbb{Q}(\pi^2- \pi)$ we have $\phi(\pi) =  \phi(1-\pi)$.

Do you have any hint about that??

答: Your helpful observation is on point. First though, for psychological reasons, lets think of rational functions of $x$ rather than of $\pi$. There is no harm here as $\mathbb{Q}(x)\cong \mathbb{Q}(\pi)$. Hence, we have a rational function $\phi(x)$ with the property that $\phi(-x) = \phi(x)$ and $\phi(1-x) = \phi(x)$. Consequently, $\phi(x) = \phi(1-x) = \phi(x-1)$, and $\phi$ is a periodic function. However, the only rational periodic functions are the constant functions: Suppose $\phi = p(x)/q(x)$ is nonconstant, $p(x)$, $q(x)$ relatively prime polynomials. Without loss of generality, we may suppose the degree of $p$ is positive (otherwise replace $\phi$ with $1/\phi$). Then $\phi$ has a positive finite number of zeros, and is not periodic.


问:
How to calculate the dot product of two functions?
Hopefully someone can help me out with this.
I have a question by my prof that goes like:

Prove that the following function is orthogonal to each other (i.e., they form a basis in the space of square integrable functions):

$$\sin(kt)\text{ and }\cos(mt)$$

($k\neq m$ natural numbers, $-\pi \leq  t \leq  \pi$.) and he gave me a hint, which is a definition: Dot product

$$\langle f, g\rangle  = \int_{-\pi}^{\pi} f(x)g(x) dx$$

and if the dot product is $0$, then the functions are orthogonal.

Could anyone please explain me this definition and give me an example on how I am supposed to use this?

答:
Two vectors $x,y \in \mathbb{R}^n$ are called orthogonal iff there is a $90^\circ$ angle between them, which happens iff $\left<x,y\right> = 0$. Generalizing this to some space $S$ where elements are functions, we define an inner product on two functions $f,g \in S$ as
$$
\left<f,g\right> = \int_{-\pi}^\pi f(x)g(x)dx
$$
and your question asks to prove that $\sin(kt),\cos(mt)$ are orthogonal. To do that, you must show that $\left<\sin(kt),\cos(mt)\right> = 0$ under your assumptions. Can you finish?


问:
Number of ordered pairs (a,b) of this equality.
To find the number of ordered pairs (a,b) for which the equality $a(\cos x-1)+b^2=cos(ax+b^2)-1$ holds true for all $x\in R$

I subbed the value of x as zero an obtained $b^2=cos(b^2)-1$ which gives me the value of b as zero,Now I have:

$a(\cos x-1)=cos(ax)-1$

By observation like last time I get a=1, Is (1,0) the only possible pair?

Is there a more rigorous method to solve this question? I guessed my way through it and it's bugging me.

答:
Note that $a = 0$ doesn't work. For $a \neq 0$, the left hand side has minimum period $2\pi$, while the right hand side has minimum period $2\pi / a$. Hence, $a = 1$ is the only possibility.

You could also look in terms of the ranges of the respective functions. The left hand side has range $[-2a, 0]$ if $ a \ge 0$, or range $[0, -2a]$ if $a < 0$. The right hand side has range $[-2, 0]$. This can only match the left hand side if $a = 1$.


问:
Help with proving a sequence is convergent
Prove that the sequence $\{\ln(1+\frac{1}{n})^n\}_{n=1}^\infty$ converges. I don't know how to approach this problem without the use of limits. Any insight would be helpful.

答: If you know that $\left(1+\frac{1}{n}\right)^n$ goes to $e$, as $n$ goes to infinity, you only have to use the continuity of the $\ln$ function.


答: You can prove this using the Monotone Convergence theorem. $\ln(x)$ is monotone increasing (to show this just look at its derivative). Using the binomial theorem one can prove that $(1+\frac{1}{n})^n $ is bounded (by $e$). In fact, $(1+\frac{1}{n})^n $ this is also monotone. Here is a post discussing this: Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing. This completes the proof, since $\ln(1+\frac{1}{n})^n$ is monotone increasing and bounded above by $\ln(e)=1$

Title: Help with proving a sequence is convergent

Body: Prove that the sequence $\{\ln(1+\frac{1}{n})^n\}_{n=1}^\infty$ converges. I don't know how to approach this problem without the use of limits. Any insight would be helpful.


问:
Geometrically explained, why do Linear Transformations Take a circle to an ellipse
Short Version:
How can it be geometrically shown that non-singular 2D linear transformations take circles to ellipses?

(Also, its probably important to state I'd prefer an explanation that doesn't use SVD, as I don't really understand it yet...although I see it everywhere)

Long Version:
Let's use the definition of an ellipse as being a circle stretched in two perpendicular directions. The two directions its stretched in will correspond to the two axes of the ellipse.

We begin by defining the circle as the endpoints of all possible 2D unit vectors. The ellipse (or at least I"m TOLD it's an ellipse) is the shape drawn by the endpoints of all these vectors after they have all been transformed in the same way (aka multiplied by the same nonsingular matrix $A$).

For linear transformations represented by diagonal matrices, it's easy to see. We're just stretching the circle in the X and Y directions.

For linear transformations represented by symmetric matrices...its a little harder, but I can see the transformation because the eigenvectors of the symmetric matrix are perpendicular, and if we change to a basis where those eigenvectors are the basis vectors, the transformation can be represented by a diagonal matrix (as for WHY symmetric matrices can be decomposed this way I don't yet really understand - but for the purpose of this question I'm just accepting that they can; I'm accepting that the eigenvectors of symmetric matrices are perpendicular to one another).

So, just like diagonal matrices, symmetric matrices also correspond to stretching a unit circle in perpendicular directions - but unless the symmetric matrix is diagonal, these are perpendicular directions different from the X and Y directions.

Buuut...what about for nonsymmetric matrices?

Thanks!

EDIT:

I've now learned of the polar decomposition of any real matrix, and that provides a beautiful explanation for why any real matrix takes a circle to an ellipse!

$A=QS$, where $Q$ is an orthogonal matrix (rotation) and $S$ is a symmetric matrix (stretching in the direction of the eigenvectors).

The symmetric matrix will definitely correspond to making an ellipse (since it scales in orthogonal directions, although perhaps not our regular $x$ and $y$ directions) and all the orthonormal matrix will do is rotate this ellipse.

However, all the explanations I've seen so far that PROVE that polar decompositions of real matrices are always possible use an algebraic explanation instead of a geometric one...so they aren't really what I'm looking for.

Thanks again!

答:
The answers on this thread are quite insightful but I am attempting here a rather very geometric answer, as the OP demanded so. For this I am going to use another interesting geometric interpretation of linear transformation (which is easier to imagine)

An alternate geometric interpretation of (dimension-preserving) linear transformation is that it's a transformation such that any line in original space is always transformed into a line and origin is not shifted.

Now imagine a freshly made pizza base/crust. Even more, let's make grids on it.

Our job is now to transform it's shape such that all the grid lines (or any possible line) remain lines. So you see at best what we can do is *stretch it with equal forces on opposite sides. You are free to choose where to stretch and Of course we can simply rotate it also after or before all the stretching.

It's not difficult to imagine that we can only get an ellipse (or bigger circle if we stretched from all sides with equal force).

Other interesting points:
(1): The direction of stretch is eigen vector and extent to which it is stretched is eigen value;
(2): Each rectangle from the gridlines is now parallelogram.


答:
$lim_{n\rightarrow\infty}\ln(1+1/n)^n=\ln[lim_{n\rightarrow\infty}(1+1/n)^n]=\ln_e e =1.$

As explained in other answers the limit property that applied here follows from the continuity of $\ln x.$


答:
From $0 \le b^2 = \cos(b^2)-1 \le 1-1$ we see that $b^2 = 0$. Hence $b=0$.

Now we have $a(\cos x-1) = \cos (ax)-1$.

Substituting $x=2\pi$, we have $0=\cos (2\pi a)-1$. $\cos (2\pi a)=1$ implies that $a \in \mathbb Z$.

Substituting $x=\pi$, we have $-2a = \cos (a\pi)-1 = (-1)^a-1 = 0$ or $-2$. Hence $a=0$ or $1$.

Substituting both values of $a$ into our original equation, we have

$$0 = \cos 0 - 1$$

$$\cos x - 1 = \cos x - 1$$

which are both valid for all $x \in \mathbb R$.

Title: Number of ordered pairs (a,b) of this equality.

Body:
To find the number of ordered pairs (a,b) for which the equality $a(\cos x-1)+b^2=cos(ax+b^2)-1$ holds true for all $x\in R$

I subbed the value of x as zero an obtained $b^2=cos(b^2)-1$ which gives me the value of b as zero,Now I have:

$a(\cos x-1)=cos(ax)-1$

By observation like last time I get a=1, Is (1,0) the only possible pair?

Is there a more rigorous method to solve this question? I guessed my way through it and it's bugging me.


问:
If $p^k m^2$ is an odd perfect number, then what is the optimal constant $C$ such that $\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$?
The topic of odd perfect numbers likely needs no introduction.

The question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, then what is the optimal constant $C$ such that $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}?$$

(Note that the special prime $p$ satisfies $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.)

MY OWN PROOF FOR $C = 1$

It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$.  This implies that
$$\frac{\sigma(m^2)}{p^k} \neq m^2 - p^k$$
since $\sigma(m^2)/p^k$ is always odd.

Now, suppose to the contrary that
$$\sigma(m^2) > p^k m^2 - p^{2k}.$$
$$p^{2k} + \sigma(m^2) > p^k m^2$$
$$2p^{2k} + 2\sigma(m^2) > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$
$$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 2\bigg)$$
$$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 2} \leq \frac{2p^{2k}}{p^k - 1}$$
where we have used the lower bound $\sigma(p^k) \geq p^k + 1$.
But from the paper Dris (2012), we have the lower bound
$$3p^k \leq \sigma(m^2)$$
while we also have the upper bound
$$\frac{2p^{2k}}{p^k - 1} = 2(p^k + 1) + \frac{2}{p^k - 1}.$$

This implies that
$$3p^k < 2(p^k + 1) + \frac{2}{p^k - 1}$$
$$p^k - 2 < \frac{2}{p^k - 1}$$
$$(p^k - 1)(p^k - 2) < 2.$$
This last inequality is a contradiction, as $p^k \geq 5$.

We therefore conclude that the inequality
$$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$$
holds when $C=1$.

Does the inequality still hold when the constant $C > 1$?  If so, then what is the optimal value for $C$?

答:
Let $p^k m^2$ be an odd perfect number with special prime $p$.

Suppose to the contrary that
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{C}$$
and that $C=2$.

(Note that
$$\frac{\sigma(m^2)}{p^k} \neq \frac{m^2 - p^k}{2}$$
since the LHS is odd while the RHS is even, as $m^2 - p^k \equiv 0 \pmod 4$.)

Following in the footsteps of the proof in the OP, we have
$$2\sigma(m^2) > p^k m^2 - p^{2k}$$
$$4\sigma(m^2) + 2p^{2k} > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$
$$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 4\bigg)$$
(Note that, since $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$, then $p^k \geq 5$, so that $\sigma(p^k) \geq p^k + 1 \geq 6$.  This shows that the factor $\bigg(\sigma(p^k) - 4\bigg)$ is always positive.)
This implies that
$$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 4} \leq \frac{2p^{2k}}{p^k - 3}$$
where we have used the lower bound $\sigma(p^k) \geq p^k + 1$.
But the upper bound can be rewritten as
$$\frac{2p^{2k}}{p^k - 3} = \frac{2p^{2k} - 18}{p^k - 3} + \frac{18}{p^k - 3} = 2(p^k + 3) + \frac{18}{p^k - 3}.$$
However, as before, we have the lower bound $3p^k \leq \sigma(m^2)$.  This implies that
$$3p^k < 2(p^k + 3) + \frac{18}{p^k - 3}$$
which means that
$$(p^k - 6)(p^k - 3) < 18,$$
an inequality that is satisfied when $p=5$ and $k=1$.

We infer that we need to use a larger lower bound for $\sigma(m^2)/p^k$.  We use the one by Dris and Luca (2016):
$$7p^k \leq \sigma(m^2) < 2(p^k + 3) + \frac{18}{p^k - 3}$$
$$(5p^k - 6)(p^k - 3) < 18.$$
This is a contradiction as $p^k \geq 5$ implies that
$$(5p^k - 6)(p^k - 3) \geq 38.$$

We therefore have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$, then
$$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{2}.$$


问:
Let $Y$ be a Geometric random variable, find the expected value of $e^{r(Y-1)}$
Let $Y$ be a Geometric random variable with $p\leq \frac{1}{2}$. I know that $f(y) = (1-p)^y p$ for $y=0,1,2,...$Therefore,

$$E(e^{r(Y-1)}) = \sum_{y=0}^{\infty} e^{r(y-1)}(1-p)^yp$$

I want to calculate that expected value to prove that if $E(e^{r(Y-1)})=1$ then $r=\log(\frac{p}{1-p})$.

Going backwards from what I need to prove I get that
$$e^r\left(\frac{1-p}{p}\right) = 1$$

Therefore I can say that $E(e^{r(Y-1)})$ should be $e^r\left(\frac{1-p}{p}\right)$ but I'm not aware of how to get there analytically. I would appreciate some advice in this one.

Body:
Use the sum of infinite GP.
$$E(e^{r(1-y)})=\sum_{y=0}^{\infty} e^{r(y-1)} (1-p)^y p=\frac{pe^{-r}}{1-e^{r(1-p)}} =1 \implies pe^{-r}=1-e^{r}(1-p).$$
$$\implies p=\frac{1-e^{-r}}{1-e^{-2r}}=\frac{1}{1+e^{-r}} \implies e^{r}=\frac{p}{1-p}\implies r =\ln\frac{p}{1-p} $$


问:
Solve:$x^x=5$ and $x^{(1/x)}=1/5$
The first equation comes from standard log equation:
$\log(x)= \frac{1}{x}$

When base of log is $5$ then it will be reduced to
$x^x=5$

The second equation comes from standard log equation:
$\log(x)=-x$

When base of log is $5$ then it will be reduced to
$x^{1/x}= \frac{1}{5}$

答:
As @Leonhard Euler answered, the only analytical solutions involve Lambert function.

If, for any reason, you cannot use it, you will need some numerical method (Newton being the simplest) but, as usual, you will need to provide a starting guess.

For the first one, consider that you look for the zero('s) of function
$$f(x)=x^x-5$$ It is varying very fast. So, assuming $x>0$, consider instead the much smoother function
$$g(x)=x\log(x)-\log(5)$$ Its derivatives are
$$g'(x)=\log(x)+1 \qquad \text{and} \qquad g''(x)=\frac 1 x \quad >0 \quad \forall x >0$$
The first derivative cancels at
$$x_*=\frac 1e \implies g(x_*)=-\frac{1}{e}-\log (5) <0$$ and the second derivative test confirms that $x_*$ correspond to a minimum.

Around this point, build a Taylor series to have as an estimate
$$g(x)=g(x_*)+\frac 12 g''(x_*)(x-x_*)^2  +O\left((x-x_*)^3\right)$$ Ignoring the higher order terms, solving the abova equation
$$x_0=x_*\pm \sqrt{-2\frac{g(x_*)}{g''(x_*)}}$$
$$x_0=\frac{1}{e}\pm\frac{\sqrt{2 (1+e \log (5))}}{e}$$ The negative solution must be dicarded because we assumed $x>0$ and our guess is then
$$x_0=\frac{1}{e}+\frac{\sqrt{2 (1+e \log (5))}}{e}\approx 1.57404$$ while the exact solution is $2.12937$. Now, using Newton, the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1.5740423 \\
 1 & 2.1899953 \\
 2 & 2.1298472 \\
 3 & 2.1293725
\end{array}
\right)$$

For the second one, in the same spirit
$$f(x)=x^{\frac 1x}-\frac 15$$
$$g(x)=\log(x)+x\log(5)$$
$$g'(x)=\frac{1}{x}+\log (5) \qquad \text{and} \qquad g''(x)=-\frac{1}{x^2}\quad <0 \quad \forall x >0$$
The first derivative cancels at
$$x_*=-\frac{1}{\log (5)} \implies g(x_*)=-\frac{1}{e}-\log (5) <0$$ and the second derivative test confirms that $x_*$ correspond to a minimum.
Just continue the same process.

Title: Solve:$x^x=5$ and $x^{(1/x)}=1/5$


答:
I like this proof a lot. I think we can try to tighten it up a bit, though.

The main thing I would like you to justify is that "the slope of $f$ will have to be greater than the slope of $e^x$ as $x$ approaches infinity." In other words, there exists some $N\in \mathbb{R}$ such that $f'(x) \geq e^x$ for all $x\geq N$. Can you prove this?

Hint: Try supposing $f(x)\geq e^x$ for $x\geq M$, but $e^x \geq f'(x)$ for all $x\geq N$ and get a contradiction through integration.

Title: A simple proof about $e^x$?

Body: Do you guys think this is correct? I am trying to prove that there is no single-term polynomial function (oxymoron, I know) $f(x)$ which is always (or at least as x approaches infinity) greater than $e^x$ (I will try to expand this to any polynomial function later). Let $f(x)$ be such a function with the lowest possible degree. Then that means that the slope of $f$ will have to be greater than the slope of $e^x$ as x approaches infinity. This means that $f'(x)>e^x$ as x approaches infinity. However, we already assumed that $f(x)$ is the lowest degree which is greater than $e^x$, but $f'(x)$ is one degree lower than $f$. Therefore, this is a contradiction. What do you guys think? Is this valid? Thanks!</p>


问:
How to find the value of this product of tensors?
I was asked to find the value of $A_{lm}x^lx^m$ under the condition that the tensor $A_{lm}$ is skew-symmetric. I´ve tried using coordinate transformation for $A_{lm}$, $x^l$ and $x^m$, then applying the defintion of the skew-symmetric tensor to the result, but I cannot progress further this. Is there something that I´m missing?

答:
The expression is zero, because it is both symmetric and skew-symmetric in the indices $\ell$ and $m$. Namely, one can write $$A_{\ell m}x^\ell x^m \stackrel{(1)}{=} A_{m\ell}x^mx^\ell \stackrel{(2)}{=} -A_{\ell m}x^mx^\ell \stackrel{(3)}{=} -A_{\ell m}x^\ell x^m,$$and the conclusion follows. In (1), one renames the dummy indices $m \leftrightarrow \ell$. In (2), we use skew-symmetry of $A$. In (3), we just switch the order $x^mx^\ell = x^\ell x^m$ (because multiplication of real numbers is commutative).

For some weird reason people may have a really hard time understanding (1). It is for the same reason one can write things such as $$\sum_{i=1}^n b_i = \sum_{j=1}^n b_j = \sum_{k=1}^n b_k, \qquad \int_0^1 f(x)\,{\rm d}x = \int_0^1f(t)\,{\rm d}t  = \int_0^1 f(s)\,{\rm d}s, \qquad \mbox{etc.}$$


答:
(Last updated on March 2, 2021 - 5:04 PM Manila time)

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Suppose to the contrary that
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{C}$$
and that $C=3$.

Following in the footsteps of this answer, we have
$$3\sigma(m^2) > p^k m^2 - p^{2k}$$
$$6\sigma(m^2) + 2p^{2k} > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$
$$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 6\bigg)$$
Assuming $\sigma(p^k) \neq 6$ (that is, assuming it is not the case that $p=5$ and $k=1$ holds), then
$$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 6} \leq \frac{2p^{2k}}{p^k - 5} = \frac{2p^{2k} - 50}{p^k - 5} + \frac{50}{p^k - 5} = 2(p^k + 5) + \frac{50}{p^k - 5}.$$

But according to Broughan et al. (2013), we have the lower bound $315p^k \leq \sigma(m^2)$.

This implies that
$$315p^k < 2(p^k + 5) + \frac{50}{p^k - 5}$$
$$(313p^k - 10)(p^k - 5) < 50$$
which is clearly a contradiction if $p \neq 5$ or $k \neq 1$.

We therefore have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$ satisfying $\sigma(p^k) \neq 6$, then
$$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{3}.$$

SECOND ATTEMPT

Using mathlove's approach in his/her answer below, together with Broughan et al. (2013)'s result, we have the lower bound $\frac{\sigma(m^2)}{p^k} \geq 315$.
As in mathlove's answer, let the quantity $F$ be given by
$$F = \frac{\sigma(p^k)}{2} - \frac{p^{2k}}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{315} = \frac{p^{k+1} - 1}{2(p-1)} - \frac{p^k}{315} := f_1(k).$$
Since $f_1(k)$ is increasing, we get
$$F \geq f_1(k) \geq f_1(1) = \frac{p+1}{2} - \frac{p}{315} \geq \frac{188}{63} \approx 2.98412698412698$$
whence we have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$, then
$$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{\frac{188}{63}}.$$

THIRD ATTEMPT

Using mathlove's approach in his/her answer below, together with mathlove's answer to a closely related question, we have the lower bound $\frac{\sigma(m^2)}{p^k} \geq 3^3 \times 5^3 = 3375$.
As in mathlove's answer, let the quantity $F$ be given by
$$F = \frac{\sigma(p^k)}{2} - \frac{p^{2k}}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{3375} = \frac{p^{k+1} - 1}{2(p-1)} - \frac{p^k}{3375} := f_2(k).$$
Since $f_2(k)$ is increasing, we get
$$F \geq f_2(k) \geq f_2(1) = \frac{p+1}{2} - \frac{p}{3375} \geq \frac{2024}{675} = 2.99\overline{851}$$
whence we have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$, then
$$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{\frac{2024}{675}}.$$

Title: If $p^k m^2$ is an odd perfect number, then what is the optimal constant $C$ such that $\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$?

Body:
The topic of odd perfect numbers likely needs no introduction.

The question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, then what is the optimal constant $C$ such that $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}?$$

(Note that the special prime $p$ satisfies $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.)

MY OWN PROOF FOR $C = 1$

It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$.  This implies that
$$\frac{\sigma(m^2)}{p^k} \neq m^2 - p^k$$
since $\sigma(m^2)/p^k$ is always odd.

Now, suppose to the contrary that
$$\sigma(m^2) > p^k m^2 - p^{2k}.$$
$$p^{2k} + \sigma(m^2) > p^k m^2$$
$$2p^{2k} + 2\sigma(m^2) > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$
$$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 2\bigg)$$
$$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 2} \leq \frac{2p^{2k}}{p^k - 1}$$
where we have used the lower bound $\sigma(p^k) \geq p^k + 1$.
But from the paper Dris (2012), we have the lower bound
$$3p^k \leq \sigma(m^2)$$
while we also have the upper bound
$$\frac{2p^{2k}}{p^k - 1} = 2(p^k + 1) + \frac{2}{p^k - 1}.$$

This implies that
$$3p^k < 2(p^k + 1) + \frac{2}{p^k - 1}$$
$$p^k - 2 < \frac{2}{p^k - 1}$$
$$(p^k - 1)(p^k - 2) < 2.$$
This last inequality is a contradiction, as $p^k \geq 5$.

We therefore conclude that the inequality
$$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$$
holds when $C=1$.

Does the inequality still hold when the constant $C > 1$?  If so, then what is the optimal value for $C$?


问:
A question in Functional Analysis.
I want to prove the following statement:

$\Omega$ is a compact set. Let $A_N $ be an $N$ dimensional subspace of $L_\infty(\Omega)$. $\{g_i\}_{i=1}^N$ be any orthonormal basis of $A_N\subset L_2(\Omega)$. Prove that for any $f=\sum_{i=1}^Nc_ig_i(x)$ , ($c_1,...,c_N\in \mathbb{R}$): $$\sup_{f\in A_N,f\neq0}\frac{|f(x)|}{||f||_2}=\sup_{(c_1,...,c_N)\in\mathbb{R}^N\\\sum_{i=1}^Nc_i^2=1}\big|\sum_{i=1}^Nc_ig_i(x)\big|.$$

My understanding is that we transfer the characterization of the norm on $f$ to the coefficients ${c}_i$ by considering a unit sphere: $\sum_{i=1}^N c_i^2=1$. But I have no idea to show this identity.

Any suggestions would be welcome! Thank you!

答: The equation simply says that $\sup \{\frac {\{|\sum c_ig_i(x)|} {|\sqrt {\sum |c_i|^{2}}}\}=\sup \{|\sum c_ig_i(x)|: \sum |c_i|^{2}=1\}$ where the supremum  on the left is taken over all choices of $c_1,c_2....,c_N$. [I have used the  orthonormlaity of $g_i$'s to get $\|f\|_1^{2}=\sum |c_i|^{2}$]. Now this identity has noting to do with $L^{2}$ functions. For any real numbers $a_1,a_2,..,a_N$ we have $\sup \{\frac {\{|\sum c_ia_i|} {|\sqrt {\sum |c_i|^{2}}}\}=\sup \{|\sum c_ia_i|: \sum |c_i|^{2}=1\}$ and this is elementary. Can you prove this? [I will post a detailed answer if you cannot see why this is true].

Title: A question in Functional Analysis.


问:
Exercise Baby Rudin 4.8
Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb{R}^1$. Prove that $f$ is bounded on $E.$

I am struggling to show that $E$ is compact (I know how to resolve the problem after that). I was originally going to say that it is a subset of a 1-cell since it's bounded but I know that $(0,1)$ isn't compact when $[0,1]$ is. Is it possible to make an argument about limit points?

答:
Why does $E$ need to be compact? You only know that it is bounded, it need not be closed. You can solve it using the Bolzano-Weierstrass Theorem like this.

Suppose otherwise: without loss of generality, let $f$ be not bounded above. Then for any $n\in\mathbb{N}$, there exists $x_n\in E$ such that $f(x_n)>n$. Because $(x_n)_{n\in\mathbb{N}}$ is a sequence in a bounded subset of $\mathbb{R}$, it has a convergent, and thus Cauchy, subsequence $(x_{n_k})_{k\in\mathbb{N}}$.

Since $f$ is uniformly continuous, we can fix some $\delta>0$ such that for all $x,y\in E$,
$$|x-y|<\delta\implies |f(x)-f(y)|<1.$$
As $(x_{n_k})_{k\in\mathbb{N}}$ is Cauchy, there exists some $N\in\mathbb{N}$ such that for all $l,m>N$, $|x_{n_k}-x_{n_l}|<\delta$. Then for all $l,m>N$, $|f(x_{n_k})-f(x_{n_l})|<1$.

This contradicts the unboundedness of $(f(x_{n_k}))_{k\in\mathbb{N}}$ (Why?) and therefore, $f$ is bounded.

Title: Exercise Baby Rudin 4.8


答:
Alternatively, if $r = \log \frac{p}{1-p}$, then $$r(Y-1) = (Y-1) \log \frac{p}{1-p} = \log \left(\frac{p}{1-p}\right)^{Y-1},$$ hence $$e^{r(Y-1)} = \left(\frac{p}{1-p}\right)^{Y-1}.$$  It follows that $$\begin{align}
\operatorname{E}[e^{r(Y-1)}] &= \operatorname{E}\left[\left(\frac{p}{1-p}\right)^{Y-1}\right] \\
&= \sum_{y=0}^\infty \left(\frac{p}{1-p}\right)^{y-1} (1-p)^y p \\
&= \sum_{y=0}^\infty p^y (1-p) \\
&= (1-p) \sum_{y=0}^\infty p^y \\
&= (1-p) \cdot \frac{1}{1-p} \\
&= 1. \end{align}$$

Or you can evaluate directly:

$$\begin{align}
\operatorname{E}[e^{r(Y-1)}]
&= \sum_{y=0}^\infty e^{r(y-1)} (1-p)^y p \\
&= \sum_{y=0}^\infty e^{ry} (1-p)^y e^{-r} p \\
&= e^{-r} p \sum_{y=0}^\infty (e^r (1-p))^y \\
&= e^{-r} p \frac{1}{1 - e^r (1-p)}.
\end{align}$$

Then we solve for $r$ such that this expression equals $1$; i.e.,
$$e^{-r} p = 1 - e^r (1-p).$$  Letting $u = e^r$, we get the quadratic
$$(1-p) u^2 - u + p = 0,$$ hence $$u \in \{1, p/(1-p)\}.$$  The first root gives $r = 0$ which we discard; the second root gives $r = \log \frac{p}{1-p}$ as claimed.

Title: Let $Y$ be a Geometric random variable, find the expected value of $e^{r(Y-1)}$

Body:
Let $Y$ be a Geometric random variable with $p\leq \frac{1}{2}$. I know that $f(y) = (1-p)^y p$ for $y=0,1,2,...$Therefore,

$$E(e^{r(Y-1)}) = \sum_{y=0}^{\infty} e^{r(y-1)}(1-p)^yp$$

I want to calculate that expected value to prove that if $E(e^{r(Y-1)})=1$ then $r=\log(\frac{p}{1-p})$.

Going backwards from what I need to prove I get that
$$e^r\left(\frac{1-p}{p}\right) = 1$$

Therefore I can say that $E(e^{r(Y-1)})$ should be $e^r\left(\frac{1-p}{p}\right)$ but I'm not aware of how to get there analytically. I would appreciate some advice in this one.


问:
Finding a one sided limit algebraically (not plugging in numbers)
I'm looking for a way to determine a one sided limit algebraically, such as

$$\color{blue}{f(x) = \frac {|x|}{x} , x \neq 0}$$

I know that you can find the limit by plugging in numbers or graphing it, but there must be a way to find it without using either of those as a crutch.

答:
Recall that $$|a| = \begin{cases} a, & \mbox{if }  a \ge 0  \\ -a,  & \mbox{if } a < 0. \end{cases} $$

Using this definition you should be able to use normal limit techniques ($\epsilon-\delta$ or what have you)

Notice, of course, that your limit does not exist as $x$ approaches zero.


问:
Probability chain rule: Does $p(x_{1},...,x_{n}) = p(x_{1},...,x_{n-1}|x_{n})p(x_{n})$?
Typically the chain rule is written $p(x_{1},...,x_{n}) = p(x_{1}|x_{2},...,x_{n})p(x_{2}|x_{3},...,x_{n})...p(x_{n-1}|x_{n})p(x_{n})$.  While it seems logical that I should be able to write it the way noted in the title, I didn't really know how to deduce/induce this formally.

答: That was derived by induction. You have $p(x_1,\ldots,x_n)=p(x_1\,|\,x_2,\ldots,x_n)p(x_2,\ldots,x_n)$ and you keep on doing that, you will obtain the chain rule. As for the title, it is similar to this, just by definition of conditional probability.


答:
If $f = \sin(kt)$ and $g = \cos(kt)$, then

$$
\left<f,g \right> = \left<\sin(kt),\cos(kt)\right> = \int_{-\pi}^{\pi}\sin(kt)\cos(kt) dt.
$$

But if $h(t) = \sin(kt)\cos(kt)$, then

$$
h(-t) = \sin(-kt)\cos(kt) = - \sin(kt)\cos(kt) = -h(t).
$$
So, $h$ is odd. Since $\left[-\pi,\pi \right]$ is a symmetrical interval around $0$,

$$
\int_{-\pi}^{\pi}\sin(kt)\cos(kt) dt = 0.
$$
You can calculate this integral aswell by substitution, calling $kt = u$ and $s = \cos(u)$.

Title: How to calculate the dot product of two functions?

Body:
Hopefully someone can help me out with this.
I have a question by my prof that goes like:

Prove that the following function is orthogonal to each other (i.e., they form a basis in the space of square integrable functions):

$$\sin(kt)\text{ and }\cos(mt)$$

($k\neq m$ natural numbers, $-\pi \leq  t \leq  \pi$.) and he gave me a hint, which is a definition: Dot product

$$\langle f, g\rangle  = \int_{-\pi}^{\pi} f(x)g(x) dx$$

and if the dot product is $0$, then the functions are orthogonal.

Could anyone please explain me this definition and give me an example on how I am supposed to use this?


问:
Variance of the MLE for a geometric distribution
I have derived the MLE for $p$ in the geometric distribution, $p=1/\overline X$.  I now need to find its asymptotic variance.  Trying to take the variance of $1/\overline X$ directly seems intractable.  I do have the following facts about the variance:

$$ \operatorname{Var}(\hat \theta_\text{MLE}) = -E[\ell (\theta\mid\vec X)] \approx [-\ell''(\hat \theta_\text{MLE}\mid\vec X)]^{-1} $$

But trying to find the given expected value seems no easier.  Since I’m looking for the asymptotic variance maybe the last approximation above is fine.

$$ \ell'' = \sum \frac{\partial^2}{\partial \theta^2}\ln f(X_i \mid \theta) $$

$$ = \sum \frac{\partial^2}{\partial \theta^2} \ln [(1-\theta)^{X_i-1} \theta] = \sum \frac{\partial^2}{\partial\theta^2}[(X_i-1)\ln(1-\theta)+\ln\theta] $$

$$ = \sum \frac{\partial}{\partial \theta} \left[\frac{-(X_i-1)}{1-\theta} + \frac{1}{\theta}\right]  = \sum \frac{X_i-1}{(1-\theta)^2}-\frac{1}{\theta^2}$$

But after all this I don’t think it’s going anywhere productive.  At least I don’t see how I would find the sum as $n\rightarrow \infty$.

I also have that

$$ \sqrt{nI(\theta_0)}(\hat\theta-\theta_0) $$

tends to a standard normal distribution, but I have no sense of what it tells me or whether it tells me anything useful for this problem. Besides that, computing the information seems to require computing a sum that is equally mysterious as the one above.

答:
You are almost there.

Asymptotic variance of $\hat{\theta}_\text{MLE}$ is

$$\frac{1}{nI(\theta)}=\frac{1}{-n\mathbb{E}\Big\{\frac{\partial^2}{\partial \theta^2}\log f(x\mid\theta)\Big\}}$$

where

$$\frac{\partial^2}{\partial \theta^2}\log f(x\mid\theta)=-\frac{1}{\theta^2}-\frac{X-1}{(1-\theta)^2}$$

And knowing that $\mathbb{E}[X]=\frac{1}{\theta}$ you immediately get

$$-n\mathbb{E}\Bigg\{\frac{\partial^2}{\partial \theta^2}\log f(x \mid \theta)\Bigg\} = \frac{n}{\theta^2(1-\theta)}$$

Thus, asymptotically,

$$\mathbb{V}[\hat{\theta}]= \frac{\theta^2(1-\theta)} {n} $$


问:
How does $\int\limits_0^\infty{{x^{v-1}}{e^{-\frac{\beta}{x}-\gamma x}}dx}=2{({\frac{\beta}{\gamma}})^{\frac{v}{2}}}{K_v}({2\sqrt{\beta\gamma}})$?
I have been reading Gradshteyn and Ryzhik's Table of Integrals, Series, and Products
7th Edition and see the 9th integral in chapter 3 section 3.471.

$$\int\limits_0^\infty{{x^{v-1}}{e^{-\frac{\beta}{x}-\gamma x}}dx}=2{({\frac{\beta}{\gamma}})^{\frac{v}{2}}}{K_v}({2\sqrt{\beta\gamma}})$$

Where $\operatorname{Real} \left( \beta  \right) > 0,\operatorname{Real} \left( \gamma  \right) > 0$

The book does not provide any detail explanations on why the Bessel function just pop up in the answer so are there anyway to solve this integral explicitly step by step ?

Thank you for your enthusiasm !

答:
Recall the definition of $K$ (second equation after the the graphs https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:_I%CE%B1,_K%CE%B1
\begin{eqnarray*}
K_{\alpha}(x)= \int_0^{\infty} e^{x \cosh t } \cosh( \alpha t) dt.
\end{eqnarray*}
Now try the substitution
\begin{eqnarray*}
x= \sqrt{ \frac{\beta}{\gamma} } e^t \\
\frac{dx}{x} =  dt
\end{eqnarray*}

and your integral becomes
\begin{eqnarray*}
\int_{- \infty}^{\infty} \left( \sqrt{ \frac{\beta}{\gamma} } e^t\right)^{\nu} e^{- \sqrt{ \beta \gamma} (e^t+e^{-t})} dt
\end{eqnarray*}
Now split the interval integral at zero and sub $e^t+e^{-t}=2 \cosh(t)$
\begin{eqnarray*}
=  \left( \sqrt{ \frac{\beta}{\gamma} } \right)^{\nu} \int_{0}^{\infty} \left( e^{ \nu t} + e^{-\nu t} \right) e^{- 2\sqrt{ \beta \gamma}  \cosh (t) } dt
\end{eqnarray*}
and the result follows.


问:
How do I find the roots of $x\sin^2(x) - \frac1x$ on the open interval $(0, 2\pi)$?
Sorry for the poor formatting, this is my first post. I was asked to find the number of critical points of a function, the derivative of which is $x\sin^2(x) - 1/x$, on the open interval $(0, 2\pi)$. Thus, I'm trying to find how many values of $x$ for which $x\sin^2(x) - 1/x = 0$ is true. I used Desmos and saw that the answer was $4$, but I want to know how to solve the problem analytically. I've tried substituting $(1-\cos^2(x))$ for $\sin^2(x)$ and manipulating the function a number of other ways but I still can't figure out how to solve it. This has been giving me a headache for over an hour; help would be highly appreciated.

答:
This is an highly transcendal equation and then it will not show any analytical solution. So, numerical methods are required and, as a consequence, you need to generate a good starting value for each root.

I think that is would be better to look for the zero's of
$$f(x)=x^2\sin^2(x) - 1$$

If you plot  it, you will notice a root close to $\frac \pi 3$, two roots on each side of $\pi$ and last root close to $2\pi$.

For the first root, make a Taylor expansion
$$f(x)=\left(\frac{\pi ^2}{12}-1\right)+\frac{1}{18} \left(9 \pi +\sqrt{3} \pi ^2\right)
   \left(x-\frac{\pi }{3}\right)+O\left(\left(x-\frac{\pi }{3}\right)^2\right)$$ Neglecting the higher order terms, solve for $x$
$$x_1=\frac{\pi }{3}+\frac{36-3 \pi ^2}{18 \pi +2 \sqrt{3} \pi ^2}\approx 1.11763$$ while the exact solution is $1.11416$.

For the second and third roots
$$f(x)=-1+\pi ^2 (x-\pi )^2+O\left((x-\pi )^3\right)$$ which gives
$$x_2=\pi -\frac 1 \pi\approx 2.82328 \qquad\qquad x_3=\pi +\frac 1 \pi\approx 3.45990$$ while the exact solutions are $2.77260$ and $3.43683$.

For the last root
$$f(x)=-1+4 \pi ^2 (x-2 \pi )^2+O\left((x-2 \pi )^3\right)$$ which gives
$$x_4=2\pi-\frac 1 {2\pi} \approx 6.12403$$ while the exact solution is $6.11902$.

Using these estimates (easily derived), to polish the roots for any accuracy, Nexton method will work like a charm.


问:
Is there a Continuous Multinomial Distribution??
In Multinomial Distribution, we have
 \begin{align}
f(x_1,\ldots,x_k;n,p_1,\ldots,p_k) & {} = \Pr(X_1 = x_1\mbox{ and }\dots\mbox{ and }X_k = x_k) \\  \\
& {} = \begin{cases} { \displaystyle {n! \over x_1!\cdots x_k!}p_1^{x_1}\cdots p_k^{x_k}}, \quad &
\mbox{when } \sum_{i=1}^k x_i=n \\  \\
0 & \mbox{otherwise,} \end{cases}
\end{align}

where $x_i$ is an integer. Besides, we should note that x_i have a constant sum, and the sum of p_k equals to 1.0 (another constant sum).

But now, I need a Continuous Multinomial Distribution, where $x_i$ doesn't need to be an integer, and the sum of $x_i$ still equals $n$.
I cannot find such a distribution, could any one help me?

p.s. I found a related question in this site. Someone says that Dirichlet Distribution can be helpful. However, the alpha parameters in Dirichlet do not have a constant sum, which is not perfect for my problem.

Thanks very much!

答: I am not aware of the distribution you were looking for, but there is a "continuous" categorical distribution (special case of multinomial distribution with $n=1$) recently popped up in the machine learning community.


问:
Calculating the Accuracy of a Taylor Approximation for $\cos(x)$
I recently have been going over Power Series and Taylor Series (and their theorems) but ran into the following multiple choice problem.  I am not allowed to use a calculator so I am unsure as to how we find the interval of accuracy.

On which interval does the polynomial $y=-\cfrac{3\pi}{2}+ x - \cfrac{(2x-3\pi)^3}{48}$ best approximates the function $y=\cos (x)$.

By playing with the given function I did find that it can be re-written as $y=\bigg(x-\cfrac{3\pi}{2}\bigg) - \cfrac{8(x-\frac{3\pi}{2})^3}{48}$ and this leads me to the fact that this is a Taylor Series for $\sin (x)$ centered at $\cfrac{3\pi}{2}$.

By graphing it I find the solution is roughly the interval [3,6], but I am not sure how to do this without calculator.  Any help is appreciated. Thank you!

答:
The key point is that you properly identified that the given expression if the expansion of $\sin(x)$ around $x=\frac 32\pi$. So the next term would be
$$\frac{1}{120} \left(x-\frac{3 \pi}{2}\right)^5$$ If you use the range $\pi \leq x \leq 2\pi$, the value of this correction is, in absolute value, $$\frac{\pi ^5}{3840}\approx 0.08$$ which, compared to $1$ is quite small. So, the approximation is more than decent for this interval.

What you could also check is where happen the maximum and minimum values of the approximation. they are are
$$x_\pm=\frac{3 \pi}{2}\pm \sqrt 2$$ and $\sqrt 2$ is not "so far" from $\frac \pi 2$. At these points, the function value is $\pm \frac{2 \sqrt{2}}{3}$ which is not "very far" from $1$.

All of the above confirms the range $\pi \leq x \leq 2\pi$.


问:
Question about the 0 eigenvalue
Suppose I have a linear transformation T: V to V, and I know that $Tx=0, x\neq0$ (not injective). Then can I say that $0$ is an eigenvalue of T because $Tx=0x$? I am thinking yes because since $x\neq0$ and $0x = 0$ and I can induce an x. Am I correct or missing something?

答: Yes you are right! And null space is zero eigenspace of that matrix.</p>

Title: Question about the 0 eigenvalue