问:
If a binary quadratic form primitively represents $n$ and $n^3$, must it be the identity form?
I've been searching numerically, and whenever a positive definite binary quadratic form can primitively represent an integer $n>1$ and $n^3$, it turns out to be equivalent to the identity form in that class group.

Is there some elementary way to see that this is only possible for identity forms?

Here is what initially gave me hope that it could be possible with other forms.

For any binary quadratic form $A x^2 + B xy + C y^2$, if it can represent integers $r$, $s$, $t$ then it can also represent $rst$.

This can be seen as follows:

For any form with coefficients $(A,B,C)$, in the class group its inverse is $(A,-B,C)$.

Both a form and its inverse can represent the same numbers (if $(x,y)$ with the former, use $(x,-y)$ with the later).

If a form can represent $r$ with $(x_1,y_1)$, and $s$ with $(x_2,y_2)$ then the identity form can represent $rs$ with some integers $(u,v)$ due to form composition:

$$
\begin{aligned}
rs &= (A x_1^2 + B x_1 y_1 + C y_1^2)(A x_2^2 + B x_2 y_2 + C y_2^2) \\
   &= (A x_1^2 + B x_1 y_1 + C y_1^2)(A x_2^2 - B x_2 (-y_2) + C (-y_2)^2) \\
   &= I(u,v) \\
\end{aligned}
$$

Futhermore, if the form can represent $t$ with $(x_3,y_3)$, then it can represent $rst$ by some integers $(x_4,y_4)$ due to form composition.

$$rst = I(u,v)(A x_3^2 + B x_3 y_3 + C y_3^2) = (A x_4^2 + B x_4 y_4 + C y_4^2)$$

It is definitely possible to primitively represent cubes, for example

```<code>a) 2 x^2 - xy + 6 y^2 = 3^3  with x=3,y=-1
b) 2 x^2 - xy + 6 y^2 = 8^3  with x=2,y=-9
</code>```

But in example 'a', it is not possible for that form to represent 3. While in example 'b', the form can represent 8, but not primitively (x=2,y=0).

This seems to be the case for any form not equivalent to an identity form.

While for an identity form, I've been able to find a lot of examples that work.

```<code>    x^2 + xy + 12 y^2 = 18     with x=2, y=1, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 18^3   with x=-76, y=7, gcd(x,y)=1

    x^2 + xy + 12 y^2 = 118    with x=2, y=3, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 118^3  with x=964, y=207, gcd(x,y)=1

    x^2 + xy + 12 y^2 = 63     with x=3, y=2, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 63^3   with x=-501, y=2, gcd(x,y)=1
</code>```

Can one prove that if a form can primitively represent $n$ and $n^3$, that is must be equivalent to the identity form?

UPDATE:

A related question On products of ternary quadratic forms $\prod_{i=1}^3 (ax_i^2+by_i^2+cz_i^2) = ax_0^2+by_0^2+cz_0^2$ shows a possible solution for any form with B=0 (all that is left is to analyze if the result is a primitive representation). Such forms are in the same equivalence class as its inverse in the class group, and are called "ambiguous forms".

Below Will Jagy gave an explicit solution for an ambiguous form which is not equivalent to the identity form, showing that the result can be a primitive representation.

The connection to what I tried is that the identity is always an ambiguous form. So that likely refines the question to:

If a form can primitively represents $n$ and $n^3$, is the form necessarily an ambiguous form?

Body:
Well, I recommend sticking with prime $n=p$ and $n^3 = p^3,$  with the addition that $p$ not divide the discriminant.  A good book is David A. Cox, Primes of the form $x^2 + n y^2 $

The example of order 4 was $\langle 5,1,11 \rangle,$ or $5 x^2 + xy + 11 y^2$ The duplicate of this in the class group is $\langle 3,3,19 \rangle.$  Still, as you found, for composite (primitive) values of $\langle 5,1,11 \rangle$ such as $15 \, , \; \; \;$  we may not have a primitive representation of $n^2$ by $\langle 3,3,19 \rangle.$  or by $\langle 1,1,55 \rangle.$ Note that the fourth powers of primes represented by $5,1,11$ such as $5,11,17$ do lead to primitive representation by $1,1,55$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./squareprimitivego
Input three coefficients a b c
3 3 19
Discriminant  -219

          25 = 5^2
         121 = 11^2
         289 = 17^2
         841 = 29^2
        2209 = 47^2
        2809 = 53^2
        3481 = 59^2
        6889 = 83^2
        9025 = 5^2 * 19^2
       10201 = 101^2
       11449 = 107^2
       12769 = 113^2
       15625 = 5^6
       17161 = 131^2
       27889 = 167^2
       32041 = 179^2
       34225 = 5^2 * 37^2
       36481 = 191^2
       38809 = 197^2
       43681 = 11^2 * 19^2
       54289 = 233^2
       57121 = 239^2
       69169 = 263^2
       75625 = 5^4 * 11^2
       78961 = 281^2
       93025 = 5^2 * 61^2

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

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./squareprimitivego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
1 1 55
Discriminant  -219

           1 =  1
         361 = 19^2
         625 = 5^4
        1369 = 37^2
        3025 = 5^2 * 11^2
        3721 = 61^2
        4489 = 67^2
        6241 = 79^2
        7225 = 5^2 * 17^2
        9409 = 97^2
       11881 = 109^2
       14641 = 11^4
       16129 = 127^2
       21025 = 5^2 * 29^2
       32761 = 181^2
       34969 = 11^2 * 17^2
       44521 = 211^2
       49729 = 223^2
       55225 = 5^2 * 47^2
       70225 = 5^2 * 53^2
       80089 = 283^2
       83521 = 17^4
       87025 = 5^2 * 59^2
      101761 = 11^2 * 29^2
      121801 = 349^2
      130321 = 19^4
      134689 = 367^2
      139129 = 373^2
      157609 = 397^2
      172225 = 5^2 * 83^2
      192721 = 439^2

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

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primitive_go
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
5 1 11
Discriminant  219
Maximum number represented?
10000
     5 = 5;     11 = 11;     15 = 3 * 5;     17 = 17;
    29 = 29;     33 = 3 * 11;     47 = 47;     51 = 3 * 17;
    53 = 53;     59 = 59;     83 = 83;     87 = 3 * 29;
    95 = 5 * 19;    101 = 101;    107 = 107;    113 = 113;
   125 = 5^3;    131 = 131;    141 = 3 * 47;    159 = 3 * 53;
   167 = 167;    177 = 3 * 59;    179 = 179;    185 = 5 * 37;
   191 = 191;    197 = 197;    209 = 11 * 19;    233 = 233;
   239 = 239;    249 = 3 * 83;    263 = 263;    275 = 5^2 * 11;
   281 = 281;    285 = 3 * 5 * 19;    303 = 3 * 101;    305 = 5 * 61;
   321 = 3 * 107;    323 = 17 * 19;    335 = 5 * 67;    339 = 3 * 113;
   365 = 5 * 73;    375 = 3 * 5^3;    393 = 3 * 131;    395 = 5 * 79;
   407 = 11 * 37;    425 = 5^2 * 17;    431 = 431;    443 = 443;
   449 = 449;    467 = 467;    485 = 5 * 97;    491 = 491;
   501 = 3 * 167;    521 = 521;    537 = 3 * 179;    545 = 5 * 109;
   551 = 19 * 29;    555 = 3 * 5 * 37;    563 = 563;    569 = 569;
   573 = 3 * 191;    591 = 3 * 197;    599 = 599;    605 = 5 * 11^2;
   617 = 617;    627 = 3 * 11 * 19;    629 = 17 * 37;    635 = 5 * 127;
   647 = 647;    671 = 11 * 61;    677 = 677;    683 = 683;
   699 = 3 * 233;    701 = 701;    717 = 3 * 239;    719 = 719;
   725 = 5^2 * 29;    737 = 11 * 67;    743 = 743;    761 = 761;
   773 = 773;    789 = 3 * 263;    803 = 11 * 73;    825 = 3 * 5^2 * 11;
   843 = 3 * 281;    863 = 863;    869 = 11 * 79;    881 = 881;
   887 = 887;    893 = 19 * 47;    905 = 5 * 181;    915 = 3 * 5 * 61;
   929 = 929;    935 = 5 * 11 * 17;    969 = 3 * 17 * 19;    971 = 971;
   977 = 977;    983 = 983;   1005 = 3 * 5 * 67;   1007 = 19 * 53;
  1037 = 17 * 61;   1055 = 5 * 211;   1061 = 1061;   1067 = 11 * 97;
  1073 = 29 * 37;   1095 = 3 * 5 * 73;   1109 = 1109;   1115 = 5 * 223;
  1121 = 19 * 59;   1139 = 17 * 67;   1151 = 1151;   1163 = 1163;
  1175 = 5^2 * 47;   1181 = 1181;   1185 = 3 * 5 * 79;   1199 = 11 * 109;
  1221 = 3 * 11 * 37;   1241 = 17 * 73;   1275 = 3 * 5^2 * 17;   1283 = 1283;
  1293 = 3 * 431;   1301 = 1301;   1307 = 1307;   1319 = 1319;
  1325 = 5^2 * 53;   1329 = 3 * 443;   1331 = 11^3;   1343 = 17 * 79;
  1347 = 3 * 449;   1361 = 1361;   1367 = 1367;   1373 = 1373;
  1397 = 11 * 127;   1401 = 3 * 467;   1409 = 1409;   1415 = 5 * 283;
  1427 = 1427;   1439 = 1439;   1445 = 5 * 17^2;   1455 = 3 * 5 * 97;
  1473 = 3 * 491;   1475 = 5^2 * 59;   1481 = 1481;   1493 = 1493;
  1499 = 1499;   1511 = 1511;   1523 = 1523;   1553 = 1553;
  1559 = 1559;   1563 = 3 * 521;   1577 = 19 * 83;   1595 = 5 * 11 * 29;
  1601 = 1601;   1613 = 1613;   1619 = 1619;   1635 = 3 * 5 * 109;
  1637 = 1637;   1649 = 17 * 97;   1653 = 3 * 19 * 29;   1689 = 3 * 563;
  1707 = 3 * 569;   1709 = 1709;   1721 = 1721;   1739 = 37 * 47;
  1745 = 5 * 349;   1769 = 29 * 61;   1797 = 3 * 599;   1805 = 5 * 19^2;
  1811 = 1811;   1815 = 3 * 5 * 11^2;   1835 = 5 * 367;   1847 = 1847;
  1851 = 3 * 617;   1853 = 17 * 109;   1865 = 5 * 373;   1877 = 1877;
  1887 = 3 * 17 * 37;   1905 = 3 * 5 * 127;   1913 = 1913;   1919 = 19 * 101;
  1931 = 1931;   1941 = 3 * 647;   1943 = 29 * 67;   1949 = 1949;
  1961 = 37 * 53;   1985 = 5 * 397;   1991 = 11 * 181;   1997 = 1997;
  2013 = 3 * 11 * 61;   2027 = 2027;   2031 = 3 * 677;   2033 = 19 * 107;
  2039 = 2039;   2049 = 3 * 683;   2057 = 11^2 * 17;   2075 = 5^2 * 83;
  2087 = 2087;   2103 = 3 * 701;   2117 = 29 * 73;   2147 = 19 * 113;
  2157 = 3 * 719;   2159 = 17 * 127;   2175 = 3 * 5^2 * 29;   2183 = 37 * 59;
  2195 = 5 * 439;   2207 = 2207;   2211 = 3 * 11 * 67;   2229 = 3 * 743;
  2237 = 2237;   2243 = 2243;   2273 = 2273;   2283 = 3 * 761;
  2285 = 5 * 457;   2291 = 29 * 79;   2297 = 2297;   2315 = 5 * 463;
  2319 = 3 * 773;   2321 = 11 * 211;   2351 = 2351;   2357 = 2357;
  2375 = 5^3 * 19;   2381 = 2381;   2399 = 2399;   2409 = 3 * 11 * 73;
  2423 = 2423;   2435 = 5 * 487;   2453 = 11 * 223;   2465 = 5 * 17 * 29;
  2477 = 2477;   2489 = 19 * 131;   2495 = 5 * 499;   2525 = 5^2 * 101;
  2585 = 5 * 11 * 47;   2589 = 3 * 863;   2607 = 3 * 11 * 79;   2615 = 5 * 523;
  2621 = 2621;   2633 = 2633;   2643 = 3 * 881;   2657 = 2657;
  2661 = 3 * 887;   2675 = 5^2 * 107;   2679 = 3 * 19 * 47;   2687 = 2687;
  2711 = 2711;   2715 = 3 * 5 * 181;   2729 = 2729;   2735 = 5 * 547;
  2741 = 2741;   2753 = 2753;   2787 = 3 * 929;   2789 = 2789;
  2805 = 3 * 5 * 11 * 17;   2813 = 29 * 97;   2819 = 2819;   2825 = 5^2 * 113;
  2837 = 2837;   2861 = 2861;   2867 = 47 * 61;   2903 = 2903;
  2909 = 2909;   2913 = 3 * 971;   2915 = 5 * 11 * 53;   2927 = 2927;
  2931 = 3 * 977;   2949 = 3 * 983;   2963 = 2963;   3021 = 3 * 19 * 53;
  3023 = 3023;   3035 = 5 * 607;   3071 = 37 * 83;   3077 = 17 * 181;
  3083 = 3083;   3095 = 5 * 619;   3111 = 3 * 17 * 61;   3113 = 11 * 283;
  3119 = 3119;   3125 = 5^5;   3149 = 47 * 67;   3161 = 29 * 109;
  3165 = 3 * 5 * 211;   3167 = 3167;   3173 = 19 * 167;   3179 = 11 * 17^2;
  3183 = 3 * 1061;   3191 = 3191;   3201 = 3 * 11 * 97;   3219 = 3 * 29 * 37;
  3233 = 53 * 61;   3245 = 5 * 11 * 59;   3251 = 3251;   3257 = 3257;
  3275 = 5^2 * 131;   3299 = 3299;   3305 = 5 * 661;   3327 = 3 * 1109;
  3329 = 3329;   3345 = 3 * 5 * 223;   3347 = 3347;   3363 = 3 * 19 * 59;
  3365 = 5 * 673;   3371 = 3371;   3389 = 3389;   3401 = 19 * 179;
  3417 = 3 * 17 * 67;   3431 = 47 * 73;   3453 = 3 * 1151;   3461 = 3461;
  3489 = 3 * 1163;   3491 = 3491;   3509 = 11^2 * 29;   3515 = 5 * 19 * 37;
  3525 = 3 * 5^2 * 47;   3533 = 3533;   3543 = 3 * 1181;   3551 = 53 * 67;
  3557 = 3557;   3587 = 17 * 211;   3597 = 3 * 11 * 109;   3599 = 59 * 61;
  3617 = 3617;   3629 = 19 * 191;   3635 = 5 * 727;   3665 = 5 * 733;
  3671 = 3671;   3683 = 29 * 127;   3695 = 5 * 739;   3701 = 3701;
  3713 = 47 * 79;   3723 = 3 * 17 * 73;   3737 = 37 * 101;   3743 = 19 * 197;
  3767 = 3767;   3779 = 3779;   3785 = 5 * 757;   3791 = 17 * 223;
  3803 = 3803;   3839 = 11 * 349;   3849 = 3 * 1283;   3869 = 53 * 73;
  3903 = 3 * 1301;   3911 = 3911;   3921 = 3 * 1307;   3929 = 3929;
  3935 = 5 * 787;   3947 = 3947;   3953 = 59 * 67;   3957 = 3 * 1319;
  3959 = 37 * 107;   3971 = 11 * 19^2;   3975 = 3 * 5^2 * 53;   3989 = 3989;
  3993 = 3 * 11^3;   3995 = 5 * 17 * 47;   4001 = 4001;   4029 = 3 * 17 * 79;
  4037 = 11 * 367;   4049 = 4049;   4055 = 5 * 811;   4073 = 4073;
  4083 = 3 * 1361;   4101 = 3 * 1367;   4103 = 11 * 373;   4119 = 3 * 1373;
  4127 = 4127;   4133 = 4133;   4139 = 4139;   4175 = 5^2 * 167;
  4181 = 37 * 113;   4187 = 53 * 79;   4191 = 3 * 11 * 127;   4205 = 5 * 29^2;
  4217 = 4217;   4227 = 3 * 1409;   4229 = 4229;   4241 = 4241;
  4245 = 3 * 5 * 283;   4265 = 5 * 853;   4281 = 3 * 1427;   4307 = 59 * 73;
  4317 = 3 * 1439;   4335 = 3 * 5 * 17^2;   4337 = 4337;   4349 = 4349;
  4367 = 11 * 397;   4373 = 4373;   4385 = 5 * 877;   4391 = 4391;
  4397 = 4397;   4409 = 4409;   4425 = 3 * 5^2 * 59;   4427 = 19 * 233;
  4443 = 3 * 1481;   4463 = 4463;   4475 = 5^2 * 179;   4479 = 3 * 1493;
  4481 = 4481;   4493 = 4493;   4497 = 3 * 1499;   4505 = 5 * 17 * 53;
  4533 = 3 * 1511;   4541 = 19 * 239;   4547 = 4547;   4559 = 47 * 97;
  4565 = 5 * 11 * 83;   4569 = 3 * 1523;   4625 = 5^3 * 37;   4643 = 4643;
  4659 = 3 * 1553;   4661 = 59 * 79;   4677 = 3 * 1559;   4679 = 4679;
  4685 = 5 * 937;   4703 = 4703;   4731 = 3 * 19 * 83;   4775 = 5^2 * 191;
  4785 = 3 * 5 * 11 * 29;   4787 = 4787;   4803 = 3 * 1601;   4811 = 17 * 283;
  4829 = 11 * 439;   4835 = 5 * 967;   4839 = 3 * 1613;   4847 = 37 * 131;
  4857 = 3 * 1619;   4871 = 4871;   4877 = 4877;   4911 = 3 * 1637;
  4913 = 17^3;   4919 = 4919;   4925 = 5^2 * 197;   4931 = 4931;
  4943 = 4943;   4947 = 3 * 17 * 97;   4985 = 5 * 997;   4997 = 19 * 263;
  5003 = 5003;   5009 = 5009;   5015 = 5 * 17 * 59;   5027 = 11 * 457;
  5051 = 5051;   5063 = 61 * 83;   5081 = 5081;   5093 = 11 * 463;
  5099 = 5099;   5105 = 5 * 1021;   5123 = 47 * 109;   5127 = 3 * 1709;
  5141 = 53 * 97;   5153 = 5153;   5163 = 3 * 1721;   5217 = 3 * 37 * 47;
  5225 = 5^2 * 11 * 19;   5235 = 3 * 5 * 349;   5249 = 29 * 181;   5261 = 5261;
  5273 = 5273;   5303 = 5303;   5307 = 3 * 29 * 61;   5309 = 5309;
  5315 = 5 * 1063;   5339 = 19 * 281;   5351 = 5351;   5357 = 11 * 487;
  5381 = 5381;   5387 = 5387;   5415 = 3 * 5 * 19^2;   5417 = 5417;
  5423 = 11 * 17 * 29;   5433 = 3 * 1811;   5435 = 5 * 1087;   5441 = 5441;
  5465 = 5 * 1093;   5489 = 11 * 499;   5501 = 5501;   5505 = 3 * 5 * 367;
  5519 = 5519;   5531 = 5531;   5541 = 3 * 1847;   5555 = 5 * 11 * 101;
  5559 = 3 * 17 * 109;   5561 = 67 * 83;   5591 = 5591;   5595 = 3 * 5 * 373;
  5631 = 3 * 1877;   5651 = 5651;   5687 = 11^2 * 47;   5711 = 5711;
  5723 = 59 * 97;   5739 = 3 * 1913;   5741 = 5741;   5753 = 11 * 523;
  5757 = 3 * 19 * 101;   5777 = 53 * 109;   5793 = 3 * 1931;   5795 = 5 * 19 * 61;
  5801 = 5801;   5807 = 5807;   5825 = 5^2 * 233;   5829 = 3 * 29 * 67;
  5847 = 3 * 1949;   5855 = 5 * 1171;   5861 = 5861;   5879 = 5879;
  5883 = 3 * 37 * 53;   5885 = 5 * 11 * 107;   5903 = 5903;   5927 = 5927;
  5933 = 17 * 349;   5939 = 5939;   5955 = 3 * 5 * 397;   5969 = 47 * 127;
  5973 = 3 * 11 * 181;   5975 = 5^2 * 239;   5981 = 5981;   5991 = 3 * 1997;
  6017 = 11 * 547;   6029 = 6029;   6059 = 73 * 83;   6081 = 3 * 2027;
  6089 = 6089;   6099 = 3 * 19 * 107;   6101 = 6101;   6117 = 3 * 2039;
  6119 = 29 * 211;   6137 = 17 * 19^2;   6143 = 6143;   6161 = 61 * 101;
  6171 = 3 * 11^2 * 17;   6179 = 37 * 167;   6185 = 5 * 1237;   6215 = 5 * 11 * 113;
  6225 = 3 * 5^2 * 83;   6239 = 17 * 367;   6245 = 5 * 1249;   6257 = 6257;
  6261 = 3 * 2087;   6263 = 6263;   6299 = 6299;   6311 = 6311;
  6317 = 6317;   6323 = 6323;   6329 = 6329;   6341 = 17 * 373;
  6351 = 3 * 29 * 73;   6365 = 5 * 19 * 67;   6395 = 5 * 1279;   6413 = 11^2 * 53;
  6431 = 59 * 109;   6441 = 3 * 19 * 113;   6455 = 5 * 1291;   6467 = 29 * 223;
  6477 = 3 * 17 * 127;   6527 = 61 * 107;   6549 = 3 * 37 * 59;   6557 = 79 * 83;
  6563 = 6563;   6575 = 5^2 * 263;   6581 = 6581;   6585 = 3 * 5 * 439;
  6599 = 6599;   6621 = 3 * 2207;   6623 = 37 * 179;   6653 = 6653;
  6677 = 11 * 607;   6701 = 6701;   6711 = 3 * 2237;   6729 = 3 * 2243;
  6731 = 53 * 127;   6737 = 6737;   6749 = 17 * 397;   6761 = 6761;
  6767 = 67 * 101;   6779 = 6779;   6803 = 6803;   6809 = 11 * 619;
  6815 = 5 * 29 * 47;   6819 = 3 * 2273;   6833 = 6833;   6845 = 5 * 37^2;
  6855 = 3 * 5 * 457;   6857 = 6857;   6869 = 6869;   6873 = 3 * 29 * 79;
  6875 = 5^4 * 11;   6891 = 3 * 2297;   6893 = 61 * 113;   6905 = 5 * 1381;
  6935 = 5 * 19 * 73;   6945 = 3 * 5 * 463;   6963 = 3 * 11 * 211;   6977 = 6977;
  6995 = 5 * 1399;   7001 = 7001;   7013 = 7013;   7019 = 7019;
  7025 = 5^2 * 281;   7053 = 3 * 2351;   7055 = 5 * 17 * 83;   7067 = 37 * 191;
  7071 = 3 * 2357;   7103 = 7103;   7109 = 7109;   7115 = 5 * 1423;
  7121 = 7121;   7125 = 3 * 5^3 * 19;   7139 = 11^2 * 59;   7143 = 3 * 2381;
  7169 = 67 * 107;   7187 = 7187;   7193 = 7193;   7197 = 3 * 2399;
  7205 = 5 * 11 * 131;   7247 = 7247;   7253 = 7253;   7269 = 3 * 2423;
  7271 = 11 * 661;   7283 = 7283;   7289 = 37 * 197;   7295 = 5 * 1459;
  7305 = 3 * 5 * 487;   7307 = 7307;   7331 = 7331;   7359 = 3 * 11 * 223;
  7373 = 73 * 101;   7395 = 3 * 5 * 17 * 29;   7403 = 11 * 673;   7415 = 5 * 1483;
  7431 = 3 * 2477;   7433 = 7433;   7451 = 7451;   7457 = 7457;
  7463 = 17 * 439;   7467 = 3 * 19 * 131;   7485 = 3 * 5 * 499;   7493 = 59 * 127;
  7499 = 7499;   7505 = 5 * 19 * 79;   7529 = 7529;   7541 = 7541;
  7547 = 7547;   7559 = 7559;   7571 = 67 * 113;   7575 = 3 * 5^2 * 101;
  7577 = 7577;   7607 = 7607;   7625 = 5^3 * 61;   7643 = 7643;
  7655 = 5 * 1531;   7685 = 5 * 29 * 53;   7691 = 7691;   7727 = 7727;
  7733 = 11 * 19 * 37;   7745 = 5 * 1549;   7755 = 3 * 5 * 11 * 47;   7769 = 17 * 457;
  7811 = 73 * 107;   7841 = 7841;   7845 = 3 * 5 * 523;   7853 = 7853;
  7863 = 3 * 2621;   7871 = 17 * 463;   7877 = 7877;   7895 = 5 * 1579;
  7899 = 3 * 2633;   7901 = 7901;   7937 = 7937;   7971 = 3 * 2657;
  7979 = 79 * 101;   7985 = 5 * 1597;   7991 = 61 * 131;   7997 = 11 * 727;
  8009 = 8009;   8025 = 3 * 5^2 * 107;   8045 = 5 * 1609;   8051 = 83 * 97;
  8061 = 3 * 2687;   8063 = 11 * 733;   8069 = 8069;   8075 = 5^2 * 17 * 19;
  8081 = 8081;   8093 = 8093;   8117 = 8117;   8123 = 8123;
  8129 = 11 * 739;   8133 = 3 * 2711;   8147 = 8147;   8171 = 8171;
  8187 = 3 * 2729;   8189 = 19 * 431;   8205 = 3 * 5 * 547;   8207 = 29 * 283;
  8219 = 8219;   8223 = 3 * 2741;   8249 = 73 * 113;   8259 = 3 * 2753;
  8279 = 17 * 487;   8291 = 8291;   8315 = 5 * 1663;   8327 = 11 * 757;
  8367 = 3 * 2789;   8369 = 8369;   8375 = 5^3 * 67;   8381 = 17^2 * 29;
  8417 = 19 * 443;   8423 = 8423;   8429 = 8429;   8439 = 3 * 29 * 97;
  8447 = 8447;   8453 = 79 * 107;   8457 = 3 * 2819;   8475 = 3 * 5^2 * 113;
  8483 = 17 * 499;   8501 = 8501;   8507 = 47 * 181;   8511 = 3 * 2837;
  8513 = 8513;   8531 = 19 * 449;   8555 = 5 * 29 * 59;   8583 = 3 * 2861;
  8585 = 5 * 17 * 101;   8597 = 8597;   8601 = 3 * 47 * 61;   8609 = 8609;
  8621 = 37 * 233;   8627 = 8627;   8657 = 11 * 787;   8709 = 3 * 2903;
  8727 = 3 * 2909;   8745 = 3 * 5 * 11 * 53;   8747 = 8747;   8753 = 8753;
  8765 = 5 * 1753;   8777 = 67 * 131;   8781 = 3 * 2927;   8789 = 11 * 17 * 47;
  8807 = 8807;   8819 = 8819;   8843 = 37 * 239;   8861 = 8861;
  8867 = 8867;   8873 = 19 * 467;   8885 = 5 * 1777;   8889 = 3 * 2963;
  8891 = 17 * 523;   8921 = 11 * 811;   8927 = 79 * 113;   8945 = 5 * 1789;
  8951 = 8951;   8969 = 8969;   8999 = 8999;   9005 = 5 * 1801;
  9041 = 9041;   9047 = 83 * 109;   9059 = 9059;   9069 = 3 * 3023;
  9095 = 5 * 17 * 107;   9105 = 3 * 5 * 607;   9125 = 5^3 * 73;   9155 = 5 * 1831;
  9185 = 5 * 11 * 167;   9203 = 9203;   9209 = 9209;   9213 = 3 * 37 * 83;
  9215 = 5 * 19 * 97;   9227 = 9227;   9231 = 3 * 17 * 181;   9249 = 3 * 3083;
  9251 = 11 * 29^2;   9257 = 9257;   9281 = 9281;   9285 = 3 * 5 * 619;
  9293 = 9293;   9299 = 17 * 547;   9305 = 5 * 1861;   9311 = 9311;
  9323 = 9323;   9329 = 19 * 491;   9339 = 3 * 11 * 283;   9357 = 3 * 3119;
  9365 = 5 * 1873;   9375 = 3 * 5^5;   9377 = 9377;   9383 = 11 * 853;
  9395 = 5 * 1879;   9431 = 9431;   9437 = 9437;   9447 = 3 * 47 * 67;
  9461 = 9461;   9473 = 9473;   9479 = 9479;   9483 = 3 * 29 * 109;
  9497 = 9497;   9501 = 3 * 3167;   9519 = 3 * 19 * 167;   9521 = 9521;
  9533 = 9533;   9537 = 3 * 11 * 17^2;   9563 = 73 * 131;   9573 = 3 * 3191;
  9593 = 53 * 181;   9605 = 5 * 17 * 113;   9623 = 9623;   9629 = 9629;
  9647 = 11 * 877;   9665 = 5 * 1933;   9689 = 9689;   9699 = 3 * 53 * 61;
  9719 = 9719;   9731 = 37 * 263;   9735 = 3 * 5 * 11 * 59;   9743 = 9743;
  9749 = 9749;   9753 = 3 * 3251;   9767 = 9767;   9771 = 3 * 3257;
  9797 = 97 * 101;   9803 = 9803;   9825 = 3 * 5^2 * 131;   9833 = 9833;
  9845 = 5 * 11 * 179;   9875 = 5^3 * 79;   9897 = 3 * 3299;   9899 = 19 * 521;
  9911 = 11 * 17 * 53;   9915 = 3 * 5 * 661;   9917 = 47 * 211;   9923 = 9923;
  9935 = 5 * 1987;   9941 = 9941;   9987 = 3 * 3329;

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

The first edition of Cox has a typographical error in the description of Dirichlet composition. It is correct in the second edition:


问:
Three students are selected from a group of 10. What is the probability that George is selected if John is always one of the selected student.
Since John is always selected, that leaves picking 2 students out of 9. So

$$P(G) =\frac{\binom 91}{\binom 92} =\frac 14$$

But this is wrong. What’s the problem here?

答:
To answer your question about what went wrong with your approach:

I think your solution has a numerator of picking one student from $9$, but it should be $1$ student from $8$ since George and John will have been already selected in this scenario.

Then you get $\frac{8\choose1}{9\choose2}$ giving the correct answer of $\frac{8}{36}=\frac29$

ContentLicense: CC BY-SA 4.0


问:
Convergence of $\int_{-\infty}^\infty \frac{1}{1+x^6}dx$
Okay, so I am asked to verify the convergence or divergence of the following improper integrals:

$$\int_{-\infty}^\infty \frac{1}{1+x^6}dx$$

and

$$\int_1^\infty \frac{x}{1-e^x}dx$$

Now, my first attempt was to use comparison criterion with $$\int \frac{1}{x^2}$$

and conclude that both of the improper integrals converge given that they are smaller than the general term $\frac{1}{x^2}$.

Is it the right path? Also, are the antiderivatives of the improper integrals given easy to find?

Thanks.

答:
For the first one you can write:

$$\int_1^\infty \frac{dx}{1+x^6}\le \int_1^{\infty}\frac{dx}{1+x^2} = \frac{\pi}{2} - \frac{\pi}{4} =\frac{\pi}{4}$$

and

$$\int_0^{1} \frac{dx}{1+x^{6}}\le \int_0^{1} dx=1$$

This shows that

$$\int_0^{\infty}\frac{dx}{1+x^6}$$ exists and is finite. You can similarly show $$\int_{-\infty}^{0}\frac{dx}{1+x^6}$$
exists, and adding the two integrals shows the desired integral converges.

ContentLicense: CC BY-SA 3.0


问:
Clarification on Proof that "for $p$ a prime, the elementary abelian group of order $p^2$ has exactly $p+1$ subgroups or order $p$"
I'm reading in Dummit's Abstract Algebra the example (3) p. 156, where they prove the claim "for $p$ a prime, the elementary abelian group of order $p^2$, $E$ (i.e. $Z_{p} \times Z_{p}$, where $Z_{p}$ is the cyclic group of order $p$) has exactly $p+1$ subgroups or order $p$".

They start proving this by saying that, since obviously each nonidentity element of $E$ has order $p$ (since it happens for every element of $Z_{p}$), then they generate a cyclic subgroup of order p. Noting that each of this subgroups has exactly $p-1$ generators (since $p$ is prime) and that by Lagrange's Theorem distinct subgroups of order $p$ intersect trivially, they deduce there are $\frac{p^2-1}{p-1}=p+1$ subgroups of order $p$.

I agree with the fact we have found $p+1$ subgroups of order $p$, but I don't know why they suppose they are all the subgroups of order $p$.

答:
There are $p^2-1$ elements of order $p$.

Each of the non-intersecting subgroups contains $p-1$ of these elements.

Therefore there are precisely $\frac{p^2-1}{p-1}=p+1$ such subgroups.


问:
If every subset of metric space (M,d) is compact, then is M finite?
Given a metric space $(M, d)$ such that any subset E of M is compact.
is M a finite set?

I think it is false but cannot find a counterexample.

答:
No, it is true. notice that every compact set in metric space is closed. So, every subset of $M$ is closed .So, $M$ is closed. Then every subset is open.  Consider the following $$W=\{\{x\}\colon x\in M\}$$ is an open cover for $M$. So, if every element omitted from $W$ the reminder can not cover $M.$
I hope  that helps


问:
Evaluating $\int_{0}^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \mathrm{d}x$

Does the integral $$\int_0^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \text d x \hspace{30pt} (u \le 1) $$ has a closed form? If it has, how do we evaluate it?

I was solving a physics problem which I have asked on physics SE as well (here), and this integral popped out. How do I solve it?

答:
You'll need an elliptic integral for this. (I love them a lot. So much that I put together a whole monograph on them.)

Let the integral be $I(u)$. By symmetry we easily see that $I(u)=I(-u)$, so we may restrict to $u\ge0$. Byrd and Friedman 289.00 then gives
$$I(u)=\frac2{u+1}K\left(\frac{4u}{(u+1)^2}\right)$$
If $|u|<1$ this may be simplified by a descending Gauss transformation to just $2K(u^2)$. Note that I am using the parameter $m$ rather than the elliptic modulus $k=\sqrt m$.


问:
Find all values $x_1$, $x_2$, ..., $x_n$ of a recurrence.
Let $a$ be a real number. Find the real numbers $x_1$, $x_2$, ..., $x_n$ knowing that $$x^2_1+ax_1+(\frac{a-1}{2})^2=x_2$$ $$x^2_2+ax_2+(\frac{a-1}{2})^2=x_3$$ $$...$$ $$x^2_n+ax_n+(\frac{a-1}{2})^2=x_1.$$

It's easy to see that $x_1=x_2$ is a solution as it implies $x_1=x_2=...=x_n$. From here, I split the problem into the two cases: $x_1>x_2$ and $x_1<x_2$, trying to show that both these cases have no solutions. It is intuitively easy to see that $x_1>x_2$ implies $x_1>x_2>...>x_n$ and the other case shows something similar. However, I'm stuck trying to prove these rigorously.

Suppose that $x_{k-1}<x_k$. Therefore $$x_{k-1}^2+ax_{k-1}+(\frac{a-1}{2})^2=x_{k}$$ $$x_{k-1}=\frac{x_{k}-a\pm \sqrt{(a-x_{k})^2-(a-1)^2}}{2}<x_k.$$ Screwing with the above equation along with $(a-x_{k})^2-(a-1)^2\geq0$ hasn't gotten me anywhere so far. Any hints for proving either case?

答: If you sum all equations, you will get $\sum_{k=1}^n x_k^2+(a-1)x_k+(\frac{a-1}{2})^2=\sum_{k=1}^n (x_k+\frac{a-1}{2})^2=0$, from where we get $x_k=\frac{1-a}{2}$ for $k=1,\dots n$.


问:
Relation between a group homomorphism and the composition with itself
Let $G$ be a finite group and $\varphi :G \longrightarrow G$ a group homomorphism. To prove:$\DeclareMathOperator{\Ker}{Ker}\DeclareMathOperator{\Im}{Im}$
$$\Ker(\varphi)=\Ker(\varphi^2) \iff \Im(\varphi)=\Im(\varphi^2)$$

I have some tools but I'm not able to combine them effectively. If $x\in \Ker(\varphi)$, then $\varphi(\varphi(x))=\varphi(e_G)=e_G$, hence $\Ker(\varphi) \subseteq \Ker(\varphi^2)$.
With the first isomorphism theorem follows $\Im(\varphi) \cong \Im(\varphi^2)$.

Any hints?

答: Using the first isomorphism theorem is a good step! It gives $\text{Im}(\varphi)\cong G/\text{Ker}(\varphi)$ and $\text{Im}(\varphi^2)\cong G/\text{Ker}(\varphi^2)$. This should make one direction easy, for the other direction you have already made the first important observation. Good luck!


问:
Show that a point inside an acute triangle is the orthocenter
Point $M$ lies inside an acute triangle $\triangle ABC$ such that the circumscribed circles of $\triangle ABM,\triangle BCM$ and $\triangle ABC$ have equal radii. Show that $M$ is the orthocenter.

I have been trying to solve the problem for an hour using the law of sines because we have just studied it, but I can't come up with anything. What about a solution using vectors?

答:
$\underline{\mathrm{Introduction}}$

As shown in $\mathrm{Fig.\space 1}$, we take an arbitrary point $M$ inside the acute triangle $ABC$ and draw the three cevians $AD$, $BE$, and $CF$ through it to form three subtriangles $MAB$, $MBC$, and $MCA$. After constructing the respective circumcircles of those subtriangles, we assert that two of these circles, namely $MAB$, and $MBC$, have the same radius as that of the triangle $ABC$. Please note that $A_1$, $B_1$, $C_1$, and $O$ are the centers of the mentioned circumcircles.

Our proof consists of two parts. In the first part we show that the radius of the circumcircle of the subtriangle $MCA$ is also equal to that of the triangle $ABC$. In part 2 we prove that, if the radii of all triangles $MAB$, $MBC$, $MCA$, and $ABC$ are equal, then $M$ is indeed the orthocenter of the latter. The proof is lengthy, but not difficult. In fact, this is an extended exercise of angle chasing mostly using the properties of cyclic quadrilaterals and subtended angles at centers and circumferences of circles.

$\underline{\mathrm{Part\space 1}}$

For brevity, let $\measuredangle CAB= \alpha$ and $\measuredangle DAB = \phi$. Also let the radii of the circumcircles $MCA$ and $ABC$ to be equal to $r$ and $R$ respectively. As shown in $\mathrm{Fig.\space 1}$, we draw nine auxiliary line segments $A_1M$, $A_1C$, $B_1C$, $B_1M$, $B_1A$, $C_1A$, $C_1M$, $OA$, and $OC$,

The chord $MB$ of the circle $MAB$ subtends $\measuredangle MAB$ and $\measuredangle MA_1B$ at the circumference and the center of the said circle respectively. Therefore, we have
$$\measuredangle MC_1B = 2\measuredangle MAB = 2\phi.$$

Since each side that forms the quadrilateral $MC_1BA_1$ is a radius of the circumcircle of either $\triangle MAB$ or $\triangle MBC$, we can state that it is a rhombus. Because the opposite angles of a rhombus are equal, we shall write,
$$\measuredangle BA_1M = \measuredangle MC_1B = 2\phi. \tag{1}$$

The chord $BC$ of the circle $ABC$ subtends $\measuredangle CAB$ and $\measuredangle COB$ at the circumference and the center of the said circle respectively. Therefore, we have
$$\measuredangle COB = 2\measuredangle CAB = 2\alpha.$$

Since each side of the quadrilateral $A_1COB$ is a radius of the circumcircle of either $\triangle ABC$ or $\triangle MBC$, it is a rhombus. Because of this, it is evident that,
$$\measuredangle BA_1C = \measuredangle COB = 2\alpha. \tag{2}$$

Now, using (1) and (2), we can express the apex angle $\measuredangle MA_1C$ of the isosceles triangle $MA_1C$ as,
$$\measuredangle MA_1C = \measuredangle BA_1C  - \measuredangle BA_1M  = 2\left(\alpha-\phi\right). \tag{3}$$

The quadrilateral $MA_1CB_1$ is formed by the union of two isosceles triangles $MA_1C$ and $MCB_1$, which are not congruent. Two of its sides, namely $CB_1$ and $MB_1$, are radii of the circle $MCA$ and, therefore, has the length $r$. The other two sides are radii of length $R$, which belong to the circle $MBC$. Therefore $MA_1CB_1$ is a kite.

It is easy to observe that
$$\measuredangle CAM = \measuredangle CAB - \measuredangle DAB = \alpha – \phi.$$

The chord $CM$ of the circle $MAB$ subtends $\measuredangle CAM$ and $\measuredangle CB_1M$ at the circumference and the center of the said circle respectively. Therefore, we have
$$\measuredangle CB_1M = 2\measuredangle CAM = 2\left(\alpha -  \phi\right). \tag{4}$$

We can conclude from (3) and (4) that a pair of opposite angles of the kite $MA_1CB_1$, namely $\measuredangle CB_1M$ and $ \measuredangle MA_1C$, are equal. This means that it is a rhombus as well. Since all four sides of a rhombus are of the same length, we have,
$$r = R. $$

$\underline{\mathrm{Part\space 2}}$

Let $P$ be an arbitrary point on the part of the circumference of the circle $MBC$, which does not contain $M$. Also assume that $\measuredangle AFC = \omega$.

We have already shown in $\mathrm{Part\space 1}$ that $\measuredangle BA_1C$, which is subtended by the arc $CMB$ is equal to $2\alpha$. Therefore, its namesake $\measuredangle BA_1C$ subtended by the arc $BPC$ is equal to $360^o - 2\alpha$. Note that the same arc $BPC$ subtends $\measuredangle CMB$ at $M$ on the arc $CMB$. So, we shall write
$$\measuredangle CMB = \frac{1}{2}\left(360^o - 2\alpha\right) = 180^o – \alpha.$$

$$\therefore\quad \measuredangle FME = 180^o – \alpha\quad\mathrm{as\space well}.$$

Here, we have the situation, where the opposite angles of the quadrilateral $MEAF$, namely $EAF$ and $FME$, are supplementing each other. This means that the quadrilateral $AFME$ is cyclic.

Since an exterior angle of a cyclic quadrilateral is equal to the interior opposite angle, we have
$$\measuredangle CEM = \measuredangle AFM = \omega. \tag{5}$$

It is up to the poster of the question to prove that the quadrilateral $CEMD$ is cyclic too using similar arguments. Start your proof by assuming $\measuredangle BCA = \gamma$ and showing that $\measuredangle AC_1B = \measuredangle  BOA =2\gamma$. Next, establish that the angle subtended by the arc $AQB$, where $Q$ is an arbitrary point on the part of the circumference of the circle $MAB$ that does not contain $M$, is equal to $360^0-2\gamma$. Finally, show that $\measuredangle EMD = \measuredangle BMA = 180^0-\gamma$.

Since $CEMD$ is cyclic, we have from (5),
$$\measuredangle MDC = 180^0  – \measuredangle CEM = 180^0 - \omega. \tag{6}$$

In order to conclude our proof, we need to verify yet another cyclic quadrilateral.
The chord $MB$ of the circle $MBC$ subtends $\measuredangle BCM$ and $\measuredangle BA_1M$ at the circumference and the center of the said circle respectively. Therefore, we have
$$\measuredangle BCM = \frac{1}{2}\measuredangle BA_1M = \phi. $$

Consequently, $\measuredangle DCF$ and $\measuredangle DAF$ of the quadrilateral $AFDC$ are equal making it cyclic.
$$\therefore\quad \measuredangle ADC = \measuredangle AFC = \omega \tag{7}$$

The scenario resulted from the arguments submitted in $\mathrm{Part\space 2}$ is displayed in $\mathrm{Fig.\space 2}$.

Although the two angles $\measuredangle ADC$ and $\measuredangle MDC$ are one and the same, it can be expressed in two different ways according to (6) and (7).  Therefore,
$$\omega = 180^0 - \omega \quad\rightarrow\quad \omega = 90^0.$$

This proves that each cevian is perpendicular to the side opposite to the vertex, from which it is dropped. In other words, the three cevians are the altitudes of $\triangle ABC$. Since altitudes are concurrent at the orthocenter of the triangle, $M$ is the orthocenter of $\triangle ABC$.


问:
Given a relation between $G$ and $g$ prove that $g(x)=x+1$.
I am having several difficulties in solving an exercise that has been put to me.

The exercise says:

If $g: (-1; \infty) \rightarrow \mathbb{R}^+$ is a function that can be antiderivated, let $G$ be the antiderivative of $g$ so that $$G(0)=\frac{1}{2}.$$ Supposing that $$(g(x))^2=2G(x),$$ then prove that $g$ is differentiable and $$\forall x>-1, g(x)=x+1.$$

I needed help trying to unblock the situation because I don't know how to start.

Best regards

答:
If $G$ is the antiderivative (which exists) of $g$ that is to say $G'=g$. We also have that $G(0)=\frac12$ and $g^2(x)=2G(x)$

Firstly, we know that:
$$g(0)=\sqrt{2G(0)}=\sqrt{2(1/2)}=1$$

We can try differentiating the given equation:
$$2g(x)g'(x)=2G'(x)$$
$$2g(x)g'(x)=2g(x)$$
we get this from the first statement we made. Now as long as $g\ne 0$ we can divide through:
$$g'(x)=1$$
$$g(x)=x+C$$
now if we put in our condition for $g(0)$ we obtain:
$$g(x)=x+1$$
Hope this helps :)


问:
Unique prolongation of the solution of a Cauchy's problem to all $\mathbb R$
Given the following
$$(PC_{a,b})\begin{cases}3x^2y''-10xy'+4y=0\\y(2)=a\\y'(2)=b\end{cases}\text{ with }a,b\in\mathbb R$$
(A) We can uniquely extend the solution to all $\mathbb R\iff b=2a$.

(B) we can extend the solution to $\mathbb R$ and there is an infinite number of prolongations of the solution of $(PC_{a,b})$  $\iff b=2a$.

(C) We can extend the solution to $\mathbb R\iff b=a=0$.

(D) The solution is always extendible to $\mathbb R$ without imposing any condition on $a$ and $b$.

$\textbf{My attempt:}$ The equation is defined in $\mathbb R\times\mathbb R$.

I considered the linear operator $\mathcal L$ such that $\mathcal L(y)=3x^2y''-10xy'+4y$ and since the equation is homogeneous, the solutions of $(PC_{a,b})$ correspond to the generators of the space $Ker(\mathcal L)$.

Substituting $\tilde y=x^{\alpha}$, I found the solutions $x^4$ and $x^{1/3}$, so
$$\tilde y(x)=Hx^4+Kx^{1/3}, H,K\in\mathbb R$$
is a solution for $(PC_{a,b})$. I observed that the constant function $y\equiv0$ is a solution for $(PC_{a,b})\iff$$ y(2)=a=y'(2)=b=0$.

Calculating the value of the solution in $0$, we get $\tilde y= 0$, but $y'(0)=4Hx^3+\dfrac{K}{3x^{2/3}}\Big|_{x=0}\to \infty$, so we can prolong $\mathcal C^1$ and then $\mathcal C^2$ the function $\tilde y$ with the function $y\equiv0\iff K=0$.

I don't see other particular existence conditions for the solution $\tilde y$, since $Hx^4+Kx^{1/3}$ is defined in $\mathbb R$.

$\underline{Important:}$ In general, I can't well distinguish the cases in which we have a unique prolongation of the solution or an infinite number of prolongations. What's the main difference between these two situations?

答:
The equation is an ODE only for $x\ne 0$

The equation is of the Euler-Cauchy type. The characteristic polynomial is
$$
0=3m(m-1)-10m+4=3m^2-13m+4=(3m-1)(m-4)
$$
so indeed $x^{1/3}$ and $x^4$ are the basis solutions. Both can be extended to be functions on all of $\Bbb R$.

Note that $x^{1/3}$ has no derivative in $x=0$, so any gluing together of solutions for $x<0$ and $x>0$ can only have the condition of continuity. But then one can glue together arbitrary solutions pairs, with different constants left and right from $x=0$.

To go in direction of uniqueness of the combined solution one needs to add a condition at $x=0$. This can be that both parts be differentiable in $x=0$, with a continuous derivative. This requires the coefficient of $x^{1/3}$ to be zero on both sides. Unfortunately, the coefficient on the right does not uniquely determine the coefficient on the left. This does not change if one adds the continuity of the second derivative. But more one should not demand from a second order equation.

In conclusion, to solve the task you need to know what "can extend" means, as a continuous function, or as a continuously differentiable function, or a twice so function,...?


问:
Question about manipulation of productories

I'm having trouble understanding this manipulation of products.
what steps should i follow to conclude that $\prod_{j=0}^{2m} \sin(z+j*\pi/n)$ is equal to $(-1)^m \prod_{j=0}^{m}\sin(z+j\pi/n)\prod_{j=m+1}^{2m}\sin(z+j\pi/n-\pi)$?

答:
There seem to be two steps within the relevant equality. They first split the product up into $j=0$ to $m$ and $j=m+1$ to $2m$. They leave the former product alone, but for the latter product they then further use $\sin(\theta)=-\sin(\theta-\pi)$ to write $$\prod_{j=m+1}^{2m}\sin(z+j\pi/n)=\prod_{j=m+1}^{2m} (-\sin(z+j\pi/n-\pi))=(-1)^n \prod_{j=m+1}^{2m} \sin(z+j\pi/n-\pi).$$
This yields the desired equality.


问:
Show that $H$ is a normal subgroup of $G$.

Let $N$ be a normal subgroup of $G$. Let $H$ be the set of all elements $h$ of $G$ such
that $hn = nh$ for all $n \in N$. Show that $H$ is a normal subgroup of $G$.

My attempt: I've been trying to do this exercise and so far I've been able to show that $H$ is a normal subgroup of $N$ (the proof is similar to showing that the center is a subgroup), but that $H$ is a normal subgroup of $N$ does not always imply that $H$ be a normal subgroup of $G$. How could I show in this case that $H$ is indeed a normal subgroup of $G$?

答:
\begin{alignat}{1}
H &= \{g\in G\mid gn=ng, \forall n\in N\} \\
&= \bigcap_{n\in N}\{g\in G\mid gn=ng\} \\
&= \bigcap_{n\in N}\{g\in G\mid gng^{-1}=n\} \\
&= \bigcap_{n\in N}G_n \\
\end{alignat}

where $G_n$ is the stabilizer of $n\in N$ under the $G$-action on $N$ by conjugation. Therefore, $H$ is the kernel of such an action, and hence it is a normal subgroup of $G$.


问:
Calabi-Yau conditions for a threefold in the Grassmannian $Gr(2,7)$
I am trying to show a complete intersection $X$ in the grassmannian $G(2,7)$ is a Calabi-Yau in the strict sense. By that I mean $\omega_X\cong\mathcal{O}_X$ and $h^i(\mathcal{O}_X)=0$ for $i=1,2$.

The description of $X$ is given by $X=Gr(2,7)\cap{H}^7$ under the Plücker embedding $i:Gr(2,7)\hookrightarrow\mathbb{P}^{20}$ where $H^7$ is a generic intersection of $7$ hyperplanes in $\mathbb{P}^{20}$.

I showed the part $\omega_X\cong\mathcal{O}_X$ by using adjunction formula and the canonical sheaf $\omega_{G(2,7)}$ of the grassmannian $G(2,7)$.

For the second part I think I know $h^1(\mathcal{O}_X)=h^2(\mathcal{O}_X)$ by Serre duality.

How can I show $h^1(\mathcal{O}_X)=0$?

答: As $X$ is a complete intersection, it follows from the Lefschetz Hyperplane Theorem that $\pi_1(X) \cong \pi_1(\operatorname{Gr}(2, 7)) = 0$, so $H^1(X; \mathbb{C}) = 0$. Since $X$ is Kähler, we have $H^1(X; \mathbb{C}) \cong H^{1,0}_{\bar{\partial}}(X)\oplus H^{0,1}_{\bar{\partial}}(X)$ and hence $H^{0,1}_{\bar{\partial}}(X) = 0$. Finally, by Dolbeault's theorem, we see that $H^1(X, \mathcal{O}_X) \cong H^{0,1}_{\bar{\partial}}(X) = 0$.


问:
Second Order Homogeneous Recurrence Relation Question
I am revising for an exam in a few weeks and I have the following recurrence relation:

```<code>f(1) = 1
f(2) = 2
f(n) = 5f(n/2) - 4f(n/4), n > 2
</code>```

My lecture notes are confusing me and I was hoping to get some help on how to solve this...

I get the characteristic equation:

```<code>a^2 - 5a + 4 = 0
</code>```

EDIT: Not sure if this is right?

Therefore:

p1 = 1 and p2 = 4

I now need to form a general solution for 'f' in terms of k...

I'm not sure how to do this...

I am following the example below:

答:
yes, you got the right characteristic equation.

yes, the roots are 1 and 4

presumably you are doing linear recurrence relations, for which you know how to solve. This problem (and the example given) have an extra things to do which is dealing with the $n = 2^k$ transformation.

So the solution method for the whole problem is: do the transformation first, solve the transformed problem, then transform back.

The general solution of the transformed problem involves the roots 1 and 4 so something like $c_1 4^k + c_2 1^k = c_1 4^k + c_2$. You already know how to solve for $c_1$ and $c_2$ (use a linear system with the base cases (but that's the fun part...what are the base cases -in the transformed system-)).

So now transform back? How do you do that? How do you get a function of $n$ using one that's already in $k$? (you have a relation between $n$ and $k$, use that and simplify).


问:
Prove that limit doesn't exists at all
I have a function $f(x)=\frac{1}{x}+\sin(\frac{1}{x})$ which I have proved that $\lim_{x \to0^+}f(x)=\infty$, and I need to prove that $\lim_{x \to0^+}f'(x) \ne -\infty$.

I have got that $f'(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$

I am trying to prove that:

$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$$

Does not equal to $-\infty$ (or to prove that it doesn't exists at all).

I have tried to assume by contradiction that
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\infty$$

So there exists $\delta_{1}>0$ such that for every $0<x<\delta_{1}$ we say that $$f(x)<\square $$

We choose $\delta=\delta_{1}$, and let $0<x<\delta$ so:

$$f(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\frac{1}{x^2}(1+\cos(\frac{1}{x}))\le -\frac{2}{x^2}<\frac{2}{\delta^2}=k$$

I do not know what to choose instead of $\square$, and if what I did is correct.

Appreciate your help!

I have also tried to think about the continuity of $f$, and saying that it is not bounded so it can't be continuous, so the limit doesn't exists.

Thanks a lot!

答:
With $z:=\dfrac1x$, rewrite as

$$\lim_{z\to\infty}-z^2(1+\cos z).$$

Then it is clear that the function alternates between $0$ and $-2z^2$.

Title: Prove that limit doesn't exists at all


问:
permutation/ arrangement optimization
I would like to obtain the optimized arrangement of integers. Let me give a simple example.

Suppose I have a sequence like 1,2,3,4,5 I would like to re-arrange it according to a given metric function.
Suppose in this case, the metric function is $g(x)$ where $g(x) = \min(|x_{i+1}-x_i|)$. So in our case,  $g(x) = \min(\{2-1, 3-2, 4-3, 5-4\}) = 1$

if we have $x = \{2,5,3,1,4\}\implies g(x) = \min(\{|5-2|, |3-5|, |1-3|, |4-1|\}) = \min(\{3,2,2,3\})=2$

Now Suppose we want to re-arrage the vector such that we maximize this metric function, how can we formulate the problem? Is there a way to solve this without trying all permutations? In my problem, I have a vector of length n and a different metric function which is quite complicated than the one given above.

Let me know if more information is needed

EDIT:

We could think of the problem as $$\underset{x}{\arg\max} ~g(x)~~s.t.~ x_i\in \mathbb N, \mathbf{x} = \{1,...,N\}$$

Can this work? If so how can we proceed?

答:
You can solve the problem via mixed integer nonlinear programming as follows.  Let $m$ be the number of rows and $n$ the number of columns.  For $i\in \{1,\dots,m\}$, $j\in \{1,\dots,n\}$, and $k\in \{1,\dots,mn\}$, let binary decision variable $y_{i,j,k}$ indicate whether $x_{i,j}=k$.  Let $z$ represent the minimum distance over row pairs $(i_1,i_2)$. The problem is to maximize $z$ subject to
\begin{align}
\sum_k y_{i,j,k} &= 1 &&\text{for all $i,j$} \tag1 \\
\sum_{i,j} y_{i,j,k} &= 1 &&\text{for all $k$} \tag2 \\
\sum_k k y_{i,j,k} &= x_{i,j} &&\text{for all $i,j$} \tag3 \\
z &\le \sum_j |x_{i_1,j}-x_{i_2,j}| &&\text{for all $i_1,i_2$ with $i_1<i_2$} \tag4 \\
\end{align}
Constraint $(1)$ assigns exactly one value per cell.
Constraint $(2)$ assigns exactly one cell per value.
Constraint $(3)$ enforces $y_{i,j,k}=1 \iff x_{i,j}=k$.
Constraint $(4)$ enforces the maximin objective; you can replace the right-hand side with any distance formula for row pairs.

For $m=3$ and $n=2$, here is an optimal solution, with $z=|6-5|+|4-2|=3$:
$$
x =
\begin{pmatrix}
1 &3 \\
6 &4 \\
5 &2 \\
\end{pmatrix}
$$

Title: permutation/ arrangement optimization


问:
Basic question on merging functions from different dimensions ($g(f) = g \circ f, f = \mathbb{R}^2 \to \mathbb{R}, g = \mathbb{R}^4 \to \mathbb{R}^2$)
Is it possible to merge 2 functions in that order: $g(f) = g \circ f$ when:

$f(x_1 ; x_2) = -8x_1 + 3x_2$ [$\mathbb{R}^2 \to \mathbb{R}$]

$g(x_1 ; x_2 ; x_3 ; x_4) = (x_1 + x_2 - x_3 - x_4 \ ; \ -2x_1 + 5x_3 - 7x_4)$ [$\mathbb{R}^4 \to \mathbb{R}^2$]

Will the answer be:

$$g(f) = g \circ f = g(-8x_1 + 3x_2) = (-8x_1 + 3x_2 \ ; \ -2(-8x_1 + 3x_2))$$

I in my assumption, while operating on g, I treat everyting as $x_1$ and make $x_2 = 0 , x_3 =0, x_4 =0$
Is that correct?

答: No, this makes no sense. You can do $f\circ g$, but not $g\circ f$. $f(x)\in\Bbb R$ and $g$ needs an input from $\Bbb R^4$, right?


问:
Can anyone see why this lemma is true? Seems very confusing!
Given a function f, we write $\bar{f}(t) = \sup_{u \leq t} f(u).$

Lemma: Let $s_{0},\dotsc,s_{T}$ be real numbers and $h:\mathbb R \to \mathbb R$. Then

$$\sum_{i=0}^{T-1}h(\bar{s}_{i})(s_{i+1}-s_{i})=\displaystyle \sum_{i=0}^{T-1}h(\bar{s}_{i})(\bar{s}_{i+1}-\bar{s}_{i})+h(\bar{s}_{T})(s_{T}-\bar{s}_{T}) \tag{1}
$$

Proof: This followed by properly rearranging summands. Indeed, observe that for a term on the right-hand side there are two possibilities; if $\bar{s}_{i+1}=\bar{s}_{i}$ respectively $s_{T}=\bar{s}_{T}$, it simply vanishes.

Otherwise it equals a sum $h(\bar{s}_{k})(s_{k+1}-s_{k})+.....+h(\bar{s}_{m})(s_{m+1}-s_{m})$ where $  \bar{s}_{k}=\dotsb=\bar{s}_{m}.$ In total, every summand on the left hand side is of (1) is accounted exactly once on the right.

My question is how it is showed that right hand side  equals a sum $h(\bar{s}_{k})(s_{k+1}-s_{k})+\dotsb+h(\bar{s}_{m})(s_{m+1}-s_{m})$? Any idea?

答:
I have given a glance to the paper. Peraphs the tag might be Statistic or Probability theory, but the lemma you have cited seems just a general support to the following proof, so it is not inherent to statistic, is it? I do the following reasoning (but I'm not too sure): starting by a generic term in the r.h.s., say it $\overline s_{i+1}-\overline s_i$,  we have $(1)$ that it may be happened that in $i+1$ the $sup$ is changed (compared to the previous $i$), i.e. $sup$ $s_{i+1}$ is exactly $s_{i+1}$. From this fact we obtain that $\overline s_{i+1}$ can be replaced by $s_{i+1}$. Otherwise $(2)$ it will be the value at some $j<i+1$; in this case we fall in the first possibility, i.e., the term vanishes because $\overline s_{i}=\overline s_{i+1}$ ($\overline s_{i+1}\gt s_{i+1}$, i.e., the supremum falls in an $m+1 \lt i+1$). At this point (case $(1)$) we must consider  $\overline s_i$; the argument is the same: obviously we must start from $m$($=i$), then we repeat until we have the supremum (if the steps end at the second stage then also $sup$ $s_{i}$ equals $s_{i}$, so, in this particular case, we have immediately the result). Finally, in the general case, we can note that proceeding this way  we have that $\overline s_m = \overline s_{m-1} = ... = \overline s_{m-n}$ ($=\overline s_{k}$), and joining the summands (being the $h(\overline s_i)$'s  equal) we obtain

$\qquad$ $s_{m+1}-s_m +s_m-s_{m-1} +s_{m-1}-\;...\;+ \; s_{m-m+1}-s_{m-n} =$

$\qquad$ $s_{m+1}-s_{m-n}=$

$\qquad$ $s_{i+1}-s_k =$

$\qquad$ $\overline s_{i+1} - \overline s_i$.

(p.s. Ask if something is not clear.)


问:
Proving that polynomial f must be of this form if the polynomial is given to be separable

If $f \in K[x] $ is monic irreducible, $\deg (f) \geq  2$, and has all its roots equal (in a splitting
field), then $\text{char }K = p \neq  0$,  and $f = x^{p^n} - a$ for some  $n\geq 1$ and  $a\in K$ .

Characteristic of Field  $F$  is always a prime( Characteristic of Integral Domain is always prime) or 0.

So, the characteristic of the field must be prime. Now, if all roots of $f$ are equal in some splitting field then it will be given in that splitting field by $(x-u)^m $ where $m$ is the degree of $f$ and $u$ is the root that is given equal.

But I don't get it how can I prove f to be equal to  $f = x^{p^n} - a$ with the information given in the question. Can you please give some direction?

Thanks!

答:
First of all, $\text{char}K=0$ means (by convention) that the unique ring homomorphism $\mathbb Z\rightarrow K$ is injective, e.g. when $K=\mathbb Q$, not if the ring is the zero ring. However if $\text{char}K=0$ then all irreducible polynomials are separable, which $f$ isn't. Therefore $\text{char}K\neq0$. Suppose $\text{char} K=p$.

By considering $\gcd(f,f')=1$ and $f$ being irreducible and inseparable, you'd realise $f(x)=g(x^p)$ for some $g\in K[x]$. Now notice $g$ must be irreducible easily, and it has only one root, because the roots of $g$ are simply $p$th powers of the only root of $f$. Therefore, either $g$ is inseparable or $\deg g=1$. Repeat the argument with $g$ in place of $f$ would eventually get you the result.


问:
How to prove that $|\mathcal{F}(\Lambda, \Omega)| = \infty$ for a matrix that induces a non-linear system of equations?
I talked to my advisor about this (twice actually), but I simply do not understand his reasoning unfortunately. Given the following matrix (which is a covariance matrix in the broader context, but I guess that's not important here)
$$
\Sigma = (\sigma_{ij}) = \begin{pmatrix}
\omega_{11} & 0 & 0 \\
0 & \omega_{22} & \omega_{23} + \lambda_{23}\omega_{22} \\
0 & \omega_{23} + \lambda_{23}\omega_{22} & \omega_{33} + 2\omega_{23}\lambda_{23} + \lambda_{23}^2\omega_{22}
\end{pmatrix},
$$
I want to show that there is an infinite amount of choices for the entries of $\Lambda$ and / or $\Omega$  such that the (covariance) matrix stays the same, so $|\mathcal{F}(\Lambda, \Omega)| = \infty$, which denotes the fiber of a function that simply maps $(\Lambda, \Omega)$ onto the (covariance) matrix. The entries of $\Lambda$ and $\Omega$ are unknown coefficients. Now he told me something the lines of treating every entry as an equation with $\omega_{22}$ as a free variable but not the other $\omega_{ij}$, so something like

$\begin{equation}
\sigma_{22} = \omega_{22} \\
\sigma_{23} = \omega_{23} + \lambda_{23}\omega_{22} \Leftrightarrow \omega_{23} = \sigma_{23} - \lambda_{23}\omega_{22}\\
\sigma_{33} = \omega_{33} + 2\omega_{23}\lambda_{23} + \lambda_{23}^2\omega_{22} \Leftrightarrow \omega_{33} = \sigma_{33} - 2\omega_{23}\lambda_{23} - \lambda_{23}^2\omega_{22}
\end{equation}$

I don't understand the point of that. If I treat $\omega_{22}$ as a free variable, wouldn't this always yield a different matrix since $\omega_{22}$ stands alone on the diagonal of $\Sigma$?

答:
I am guessing your advisor meant to say you fix $\omega_{11}=a$ and $\omega_{22}=b$ and then show that the following system has infinitely many solutions:
$$
\begin{cases}
\omega_{23} + \lambda_{23}b = \sigma_{23} \\
\omega_{33} + 2\omega_{23}\lambda_{23} + b^2\lambda_{23}^2 = \sigma_{33}
\end{cases}$$
for given $\sigma_{ij}$ values. In the above two equations, the variables are $\omega_{23}, \lambda_{23}, \omega_{33}$ and so it should have infinitely many solutions because you only have two equations.


问:
Prove that $\int 1/x\, \text{d}x=\ln |x|$
What is the proof that $$\int \frac{1}{x}\, \text{d}x=\ln |x|?$$

Please note that similar questions refer to $\ln x$, and not $\ln |x|$ as I have seen in Calculus textbooks.

Given the formula that $$\frac{\text{d}}{\text{d}x}\ln x=\frac{1}{x},$$ it is obvious that for $x>0$, $\int 1/x\, \text{d}x=\ln x$; however I am unsure how the above definition is proven over all $x\neq 0$.

答:
An "indefinite integral" formula of the form
$$
\int f(x)\,dx=F(x)+C\tag{1}
$$
is understood as $F$ is an antiderivative on some (open) interval $I$, i.e., $F'(x)=f(x)$ for every $x\in I$. When the context is clear, one usually write (1) without mentioning explicitly the interval $I$.

The formula
$$
\int \frac{1}{x}\,dx=\ln|x|+C
$$
is understood valid for any (open) interval $I$ not containing $0$.

The natural logarithm function $\ln x$ is defined only for $x>0$. It does not make sense to prove
$$
\frac{d}{dx}\ln(x)=\frac{1}{x},\quad x<0
$$
since the left hand side is undefined.

However, one can prove that for $x<0$,

$$
\frac{d}{dx}\ln(|x|)=\frac{1}{x}\tag{2}
$$

You can prove (2) using the chain rule if you allow to use the fact $\frac{d}{dx}\ln(x)=\frac{1}{x}$ for $x>0$.

If you want to prove (2) from first principles, then you need to give the definition for the logarithm.


问:
How to find angle in between two right triangles when one side is twice of other?
The problem is as follows:

$\begin{array}{ll}
1.&10^\circ\\
2.&12^\circ\\
3.&15^\circ\\
4.&16^\circ\\
\end{array}$

I'm not sure which sort of construction can be used here to solve this problem?.

I've attempted to draw a perpendicular line from $B$ to segment $AC$. But this did not yielded good results.

Then I've attempted tracing a perpendicular segment to $BC$ intersecting $AC$, from looking on this possibility. I got still stuck.

I don't know how to use the given angles, they suspiciously add up to $3\alpha$.

I still don't know how to use $QC=2HC$. Can someone give me some ideas on what to do?. Should congruence be used here?.

I'm assuming that the intended approach is try to spot triangles such as:

$3-4-5$ or $1-\sqrt{3}-2$ or something along those special right triangles.

I do hope someone could help me on how to solve this problem relying on euclidean geometry can this be done?.

答:
Since they are complementary to angles that subtend $3\alpha$,
$$
\overbrace{\ \angle HPC\ }^{\text{co-}HCP}=\frac\pi2-3\alpha=\overbrace{\ \angle QPB\ }^{\text{co-}PQB}\tag1
$$
Since $\angle BPC=\pi$, we get
$$
\angle QPH=6\alpha\tag2
$$

Extend $\overline{QB}$ by an equal length to $R$; that is, $\overline{BR}=\overline{QB}$. Since $\triangle QCR$ is isosceles,
$$
\angle BCR=\angle BCQ=2\alpha\tag3
$$
Likewise,
$$
\angle BPR=\angle BPQ=\frac\pi2-3\alpha\tag4
$$
Thus, $\angle RPH=\pi$. That is, $R$, $P$, and $H$ are colinear.

Furthermore,
$$
\overline{RC}=\overline{QC}\tag5
$$

Extend $\overline{HC}$ by an equal length to $S$; that is, $\overline{HS}=\overline{HC}$. Thus,
$$
2\overline{HC}=\overline{SC}\tag6
$$
and since $\triangle SRC$ is isosceles, we have
$$
\overline{RS}=\overline{RC}\tag7
$$
Now we use that $\color{#C00}{\overline{QC}=2\overline{HC}}$. $\color{#090}{(7)}$, $\color{#44F}{(5)}$, and $\color{#CA0}{(6)}$ say
$$
\overline{RS}\color{#090}{=}\overline{RC}\color{#44F}{=}\overline{QC}\color{#C00}{=}2\overline{HC}\color{#CA0}{=}\overline{SC}\tag8
$$
Thus, $\triangle RCS$ is equilateral, and therefore,
$$
5\alpha=\angle RCS=\frac\pi3=60^{\large\circ}\tag9
$$
Thus,
$$
\alpha=\frac\pi{15}=12^{\large\circ}\tag{10}
$$


问:
Proof that the cartesian plane is an incidence geometry with only vector definition of a line.
I can get up to showing that it is an abstract geometry, but I cannot figure out how to show that for every two points, there is a unique line. The definition of a line in vector form is given
$L (AB) = \{X \in R \times R~ | ~X=A+t(B-A) ~\text{for some}~~ t \in \mathbb R\}$

$A,B,X$ are points.

I am trying to go about this by supposing A,B are in $L (CD)$ and $L (XY)$. From there I try to get $C=X$ and $D=Y$, but I can't do this. Any help or hint is appreciated.

The Cartesian Plane is the set of Points in $R \times R$ and the set of all lines $L (AB)$.

This is a problem in a textbook but I am not doing this for homework. I just find this subject really interesting so any help is appreciated.

答:
You can't hope to get $C=X$ and $D=Y$, because any two points on a line determine that line. What you have to show is that $L(CD)=L(XY)$. In fact they are both $L(AB)$.

You have $$A=C+s_A(D-C)=X+t_A(Y-X)$$ and $$B=C+s_B(D-C)=X+t_B(Y-X)$$ for some $s_A,s_B,t_A,t_B\in\Bbb R$. Then $B-A=(s_B-s_A)(D-C)$, so

$$D-C=\frac 1{s_B-s_A}(B-A)\;,$$

and

$$C=A-s_A(D-C)=A-\frac{s_A}{s_B-s_A}(B-A)\;.$$

(Here I use the assumption that $A\ne B$.) In similar fashion you can express $D,X$, and $Y$ in terms of $A$ and $B$ to show that $C,D,X,Y\in L(AB)$. Once you have that, a little more work of the same kind will allow you to express every point of $L(CD)$ and $L(XY)$ as a point of $L(AB)$.


问:
Cardinality of the equivalence class of a given braid word?
The symmetric group $S_n$ has presentation $$S_n = \{ s_1,...,s_{n-1} | s_is_{i+1}s_i=s_{i+1}s_is_{i+1}, \text{ } s_i^2=1 , \text{ and } s_is_j = s_js_i \text{ for } |i-j| \geq 2\}$$

If we take away the relation $s_i^2=1$ we get the braid group:

$$B_n = \{ s_1,...,s_{n-1} | s_is_{i+1}s_i=s_{i+1}s_is_{i+1},\ \text{ } s_is_j = s_js_i \text{ for } |i-j| \geq 2\}$$

Thus elements $b \in B_n$ are equivalence classes, with $b \cong c$ if the braid word that represents $b$ differs from the braid word representing $c$ by finite many applications of the relations.

I am trying to understand, given an element $b \in B_n$, what is its cardinality as an equivalence class as an element of $B_n$?

For example, let $b=(1,2,3,1,2,3)$. Then I'm pretty sure, by brute force, I've calculated that there are $7$ different braid words that can represent $b$:

$(1,2,3,1,2,3)$

$(1,2,1,3,2,3)$

$(2,1,2,3,2,3)$

$(2,1,3,2,3,3)$

$(2,3,1,2,3,3)$

$(1,2,1,2,3,2)$

$(2,1,2,2,3,2)$

Does anyone know how I could figure this out in a more sophisitcated way? I know word problems such as this have been well studied. Also, is there a systemic way to determine when two braid words are equivalent? Thank you.

答:
The equivalence classes are all countably infinite.

In fact, from the presentations one reads off a surjective homomorphism $B_n \to S_n$ taking each generator of $B_n$ to the same-named generator of $S_n$. Your equivalence classes are simply the cosets of the kernel of this homomorphism. Since $B_n$ is countably infinite and $S_n$ is finite, and since the cosets all have the same cardinality (a fact true of every group homomorphism), it follows that the cosets are all countably infinite.

By the way, I'll add that the notation of your "brute force calculation" does not make much sense to me; I don't know what those sequences of natural numbers in that calculation are supposed to represent, either as elements of $S_n$ or of $B_n$. Elements of $S_n$ or of $B_n$ can, by definition of a generating set, each be written as a finite word in the generators. For example each element of $B_3$ or of $S_3$ can all be written as a word in the letters $s_1,s_2$; a specific example of such a word is
$$s_1^2s_2^{-1}s_1^{4}s_2^3 = s_1 s_1 s_2^{-1} s_1 s_1 s_1 s_1 s_2 s_2 s_2
$$
Given two such words, to say they represent the same element of $B_3$ means, as you say, that you can get from one word to the other by applying the relations of $B_3$. On top of that, to say those two words represent the same element of $S_3$ means that you can get from one to the other by applying the larger set of relations of $S_3$.


问:
Minimal polynomial of the matrix
Recently I've encountered the following problem

Find minimal polynomial of $A =\begin{pmatrix}
2 & 3 & 4\\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}$ in $\mathbb{R}[x]$

Firstly, we can compute the characteristic polynomial which is obviously $c = (x - 2)^3$, then we know that the minimal polynomial is the one which divides $c$ and has the minimal degree. Straightforward computation yields that $m_A = (x - 2)^2$. And it makes sense, except the fact that minimal polynomial should be irreducible, and here there are two factors. Am I missing anything?

答: The minimal polynomial only has to be irreducible in the scalar setting; $m_A=(x-2)^2$ is perfectly right.

Title: Minimal polynomial of the matrix


问:
Equivalent definitions of $\lim_{x \to c^+}g(x) = -\infty$
In this question, I originally wrote in my answer:

Further edit: I think OP doesn't properly understand what is meant by
$\lim_{x \to c^+}g(x) = -\infty$.

If $g(x)$ is a function, then $\lim_{x \to c^+}g(x) = -\infty$ means
that, for each $\gamma \in \mathbb{R}, \exists \delta > 0 $ such that
$x \in (c,c+\delta) \implies g(x) < \gamma$.

So an equivalent definition of $\lim_{x \to c^+}g(x) \neq -\infty\ $
is:

$\exists \beta \in \mathbb{R}$ such that: for each $x > c, \exists x'\
> $ with $\ c<x'<x$ such that $g(x') > \beta.$

If no such $\beta$ exists, then $\lim_{x \to c^+}g(x) = -\infty\ .$

Note that $g$ need not be continuous at $c$ in either of my
definitions.

The first line is ironic because now it is me asking for a sanity check, so my answer is a little bit of the blind leading the blind, or rather, the slightly confused leading the (probably) more confused. Anyway are my definitions correct?

答:
You definition for
$\displaystyle
\lim_{x\to c+}g(x)=-\infty
$ is okay.

But people usually don't use an expression such as $\displaystyle
\lim_{x\to c+}g(x)\ne-\infty
$,
the meaning of which is ambiguous, and it does not convey much information. One can simply say something like "$\displaystyle
\lim_{x\to c+}g(x)=-\infty
$ is not true", which is the same as saying:

there exists $\beta\in\mathbb{R}$ such that for every $\delta>0$, there exists $x'\in(c,c+\delta)$ with $g(x')\ge \beta$.


问:
Regular Lindelöf space is paracompact
I thought I had a reasonable understanding of basic general topology, but I have been struggling somewhat with the result in the title. After spending a day or so failing to prove it, I looked for a proof in some standard texts. Not everyone uses the same definitions, so for clarity:

(1) $X$ is regular means you can separate points and closed sets with open sets;

(2) $X$ is Lindelöf if every open cover has a countable subcover;

(3) $X$ is paracompact if every open cover has a locally finite refinement.

In Munkres' Topology, 2nd ed, the proof seemed to be largely in Lemma 41.3, although not in a particularly helpful form. With several details filled in (and irrelevant material omitted), it seems to be:

--- start of proof ---

Let $\mathcal{A}$ be any open cover of $X$. Take any $x\in X$ and any $A\in\mathcal{A}$ which contains $x$. Then $X\setminus A$ is closed and $x\notin X\setminus A$. Since $X$ is regular, we can find two disjoint open sets $U,V$ st $x\in U$ and $X\setminus A\subset V$. So the closed set $C=X\setminus V\subset A$. Since $U,V$ are disjoint, we also have $U\subset C$. But $\overline{U}$ is the smallest closed set containing $U$, so we must have $\overline{U}\subset C\subset A$. Hence $U\in\mathcal{B}$, where $\mathcal{B}$ is defined as the collection of all open subsets of $X$ whose closure is contained in an element of $\mathcal{A}$. So $\mathcal{B}$ is a refinement of $\mathcal{A}$ and an open cover of $X$.

Since $X$ is Lindelöf, $\mathcal{B}$ has a countable subcover $\mathcal{C}=\{B_1,B_2,B_3,\dots\}$. Let $D_1=\overline{B_1},D_2=\overline{B_2}\setminus B_1,D_3=\overline{B_3}\setminus(B_1\cup B_2),D_4=\overline{B_4}\setminus(B_1\cup B_2\cup B_3)$ etc. Then $\mathcal{D}=\{D_1,D_2,D_3,\dots\}$ covers $X$ and consists of closed sets. Take any $x\in X$. Let $k$ be the smallest integer such that $x\in B_k$. Each point in $B_k$, which is an open nei of $x$, belongs to at least one of $D_1,\dots,D_k$ but not to any other $D_i$ and hence $B_k$ intersects just finitely many sets in $\mathcal{D}$. In other words, $\mathcal{D}$ is locally finite. Also $D_i\subset\overline{B_i}$ which is a subset of some element of the cover $\mathcal{A}$, so $\mathcal{D}$ is a refinement of $\mathcal{A}$.

Let $\mathcal{E}$ be the collection of all open sets that intersect only finitely many $D_i$. Because $\mathcal{D}$ is locally finite, $\mathcal{E}$ covers $X$. Applying the result in paras 1-2 to $\mathcal{E}$, we can find a refinement $\mathcal{F}$ of $\mathcal{E}$ such that: (i) $\mathcal{F}$ covers $X$; (ii) $\mathcal{F}$ is locally finite; (iii) every set in $\mathcal{F}$ is closed.

For $i=1,2,3,\dots$, let $C_i$ be the union of all sets $F\in\mathcal{F}$ such that $F\cap D_i=\emptyset$. We claim that each $C_i$ is closed. For suppose $x\notin C_i$. Since $\mathcal{F}$ is locally finite we can find an open nei $U$ of $x$ which only intersects finitely many members of $\mathcal{F}$, and hence only intersects finitely many of those whose union is $C_i$. Call those $F_1,\dots,F_k$. Now $F_1,\dots,F_k$ are members of $\mathcal{F}$ and hence closed, so $F_1\cup\dots\cup F_k$ is closed and $U'=U\setminus(F_1\cup\dots\cup F_k)$ is open. So $U'$ is an open set which contains $x$ but does not intersect $C_i$. Hence $x\notin\overline{C_i}$. Hence $C_i$ is closed.

Let $E_i=X\setminus C_i$. So $E_i$ is open and $D_i\subset E_i$ and hence $\{E_1,E_2,E_3,\dots\}$ is an open cover for $X$.

For each $D_i$ take an element $A_i\in\mathcal{A}$ containing $D_i$, and let $G_i=E_i\cap A_i$. Evidently, the collection $\mathcal{G}=\{G_1,G_2,G_3,\dots\}$ is a refinement of $\mathcal{A}$ and consists of open sets. Since $D_i\subset E_i$ and $D_i\subset A_i$, we have $G_i\supset D_i$, and hence $\mathcal{G}$ is an open cover of $X$.

It remains to show that $\mathcal{G}$ is locally finite. Given any $x\in X$, we can find a neighbourhood $U$ of $x$ which intersects only finitely many members of $\mathcal{F}$ (because $\mathcal{F}$ is locally finite). Since $\mathcal{F}$ covers $U$, the sets in $\mathcal{F}$ which intersect $U$ must cover it. So it is enough to show that if $F\in\mathcal{F}$ intersects $U$, then it only intersects finitely many members of $\mathcal{G}$. But $\mathcal{F}$ is a refinement of $\mathcal{E}$ and so by definition $F$ intersects only finitely many members of $\mathcal{D}$. But if $F$ intersects $G_i$, then it intersects $E_i\supset G_i$ and hence $F\not\subset C_i$ and so $F\cap D_i\neq\emptyset$. Hence if $F$ intersects infinitely many members of $\mathcal{G}$, then it intersects infinitely many members of $\mathcal{D}$. Contradiction. So $\mathcal{G}$ is locally finite and we are done. $\Box$

--- end of proof ---

That is all most ingenious, but far more complicated and hard to understand than I expected. My question is simply whether anyone knows of any better proofs. Of course, you can usually shorten proofs by using enough lemmas or theorems. The proof above uses little beyond the basic definitions. So I am interested in better arguments rather than pushing things into lemmas.

答: In this note I wrote up the proof of a classic characterisation of paracompactness for regular spaces (write-up based on my old lecture notes, Engelking and papers by E. Michael). It has the Lindelöf fact as a trivial corollary, as I also note there. The ideas are very similar to the proof you sketched. You’ll recognise many elements.</p>

Title: Regular Lindelöf space is paracompact


问:
Argand's inequality
I'm trying to understand an operation that has been done in the process of prooving the following statement:

If $p$ is a complex poylnomial function thus that $p(z)= \sum_{k=0}^n c_k z^k, n \geq 1$, then we find that every disk $D$ around $a$ contains a $b$ in the interior of D so that $|p(b)|<|p(a)|$.

At the beginning they look at

$$\begin{aligned} p(a+w) &=\sum_{k=0}^{n} c_{k}(a+w)^{k} \\ &=\sum_{k=0}^{n} c_{k} \sum_{i=0}^{k}\left(\begin{array}{c}k \\ i\end{array}\right) a^{k-i} w^{i} =\sum_{i=0}^{n}\left(\sum_{k=i}^{n}\left(\begin{array}{c}k \\ i\end{array}\right) c_{k} a^{k-i}\right) w^{i} \\ &=p(a)+\sum_{i=1}^{n}\left(\sum_{k=i}^{n}\left(\begin{array}{c}k \\ i\end{array}\right) c_{k} a^{k-i}\right) w^{i}=p(a)+\sum_{i=1}^{n} d_{i} w^{i} \end{aligned}$$

Now my question:

Why does this equality hold: $\sum_{k=0}^{n} c_{k} \sum_{i=0}^{k}\left(\begin{array}{c}k \\ i\end{array}\right) a^{k-i} w^{i} =\sum_{i=0}^{n}\left(\sum_{k=i}^{n}\left(\begin{array}{c}k \\ i\end{array}\right) c_{k} a^{k-i}\right) w^{i}$?

答:
Let $I := \{(k,i) \in \{0, \dots, n\}^2: i \leq k\}$. Then it is easy to verify that
$$\sum_{k=0}^n c_k \sum_{i=0}^k { k \choose i} a^{k-i} w^i = \sum_{(k,i) \in I} c_k { k \choose i} a^{k-i} w^i. $$
Using another parameterization for the set $I$ one obtains
$$ \sum_{(k,i) \in I} c_k { k \choose i} a^{k-i} \omega^i = \sum_{i=0}^n \sum_{k=i}^n c_k { k \choose i} a^{k-i} \omega^i, $$
which is the formula on the right hand side in your question.

Title: Argand's inequality


问:
If $f'(x) = -f(x)$ and $f(1)=1$, then what is $f(x)$?
If $f'(x) = -f(x)$ and $f(1)=1$, then $f(x)=$

(a) $1/2e^{-2x+2}$

(b) $e^{-x-1}$

(c) $e^{1-x}$

(d) $e^{-x}$

(e) $-e^{x}$

If you plug in $1$ for $x$ in option c it gives you $e^{1-1}$ or $e^0$ which is $1$ hence what is needed.

答:
Hint: All you need to do in this case is to evaluate $f(1)$: $f_A(1), f_B(1), f_C(1), f_D(1), f_E(1)$.

Show your calculations in each case.

Doing that rules out all possibilities except $(C)$, irregardless of the first condition, though you want to evaluate $f'_C(x)$ to show that $f_C = f'_C$.


问:
Prove that $(x^2+2x+1)$ is continuous by using a $\epsilon$ - $\delta$ argument
I want to prove that the following function is continuous with an Epsilon Delta Argument.
$$ f: \mathbb R \rightarrow \mathbb R \;  (x^2+2x+1)$$

So I started with
$$\left\lvert f(x)-f(y) \right\rvert= \left\lvert (x^2+2x+1) - (y^2+2y+1) \right\rvert = \left\lvert (x^2+2x)-(y^2+2y) \right\rvert$$

$$\le 2\left\lvert \left\lvert \frac{x^2-y^2}{2} \right\rvert + \left\lvert x- y\right\rvert \right\rvert $$
$$\le 2\left\lvert \left\lvert \frac{(x+y)(x-y)}{2} \right\rvert + \delta \right\rvert$$
$$= 2\left\lvert \frac{\left\lvert(x+y)(x-y)\right\rvert}  {2} + \delta \right\rvert $$
$$ \le 2\left\lvert \frac{\left\lvert(x+y)\delta\right\rvert}  {2} + \delta \right\rvert $$
$$= 2\left\lvert \frac{[\left\lvert(x-y)\right\rvert+\left\lvert2y\right\rvert]\delta}  {2} + \delta \right\rvert $$
Now I define $\delta \le1.$
$$= 2\left\lvert \frac{(1+\left\lvert2y\right\rvert)\delta}  {2} + \delta \right\rvert $$
$$= \left\lvert(1+\left\lvert2y\right\rvert)\delta + 2\delta \right\rvert $$
$$= \delta \left\lvert(1+\left\lvert2y\right\rvert) + 2 \right\rvert $$
$$= \delta \left\lvert\left\lvert2y\right\rvert + 3 \right\rvert = \epsilon $$
And so my $\delta$ is $\frac{\epsilon}{\left\lvert\left\lvert2y\right\rvert + 3 \right\rvert}$ and $\delta \le 1$

Can anyone tell my if this is correct or where I went wrong I am studying for exam. :)
Thanks in advance.

答:
As a comparison I did this

$$
\left| x^2 + 2x + 1 - y^2 - 2y - 1 \right| < |x-y|^2 + (2|y| + 2)|x - y| < \epsilon
$$

And I defined $\delta$ as the strictly positive solution of $\delta^2 + (2|y| + 1)\delta = \epsilon$, which you can easily show it exists for every $\epsilon > 0$.


问:
How to do this lagrange multiplier question? (max / min)
I have the following question

Using the method of the Lagrange multipliers, find the maximum and minimum values of the function $$f(x,y,z) = x^2y^2z^2$$
where $(x,y,z)$ is on the sphere $$x^2+y^2+z^2 = r^2$$
Using the result obtained to show that for arbitrary non-negative numbers $a,b$ and $c$,
$$(abc)^\frac{1}{3} \leq \frac{a+b+c}{3}$$

My thoughts are to use the equation $L = f + \lambda g$, where $f$ is the same as above and $g$ is just $x^2+y^2+z^2 - r^2 = 0$

But when I compute this, for example, I get :

$L_{x} = 2xy^2z^2 + 2 \lambda x = 0$ which means that the $x$ cancel ? I think I've made a silly mistake, and was hoping someone could explain how to do this properly.

And if you can, could you guide me on the second part of the question

Thank you in advanced.

答:
$f(x,y,z) = x^2y^2z^2$ and given constraint $g(x,y,z) = x^2+y^2+z^2-r^2 = 0$

Applying Lagrange Multiplier method,

$x^2y^2z^2 = \lambda (x^2+y^2+z^2-r^2)$

$2xy^2z^2 = 2 \lambda x$

$2x^2yz^2 = 2 \lambda y$

$2x^2y^2z = 2 \lambda z$

So if you take the first equation for example, you have $\lambda = y^2z^2, $ if $x \ne 0$

Plugging in the second and third, you get $x^2 = y^2 = z^2$ and given the constraint

$x^2 = y^2 = z^2 = \frac{r^2}{3}$

Similarly check other critical points.

EDIT:
Specifically on the argument sought in the question for non-negative numbers,

Say, $a = x^2, b = y^2, c = z^2$. So they are non-negative and as per given constraint, we have

$a+b+c = r^2$

Maximum value of $\displaystyle abc = \frac{r^6}{27}$ or we say $\displaystyle abc \leq \frac{r^6}{27}$

So, $\displaystyle (abc)^{1/3} \leq \frac{r^2}{3} = \frac{a+b+c}{3}$


问:
For which $z \in \mathbb{C}$ does $f(z) = |z|^2 +2z$ satisfy Cauchy-Riemann equations?
The task is as follows:

$$
\text{Check, for which } z\in\mathbb{C} \text{ function } f(z) = |z|^2 +2z \text{ satisfies the Cauchy-Riemann equations.}
$$
I figured I need to transform $f(z)$ to $f(x,y) = u(x,y) + i \cdot v(x,y)$.

I started by transforming $f(z)$ to $f(x+iy)$ as such:
$$
f(x+iy) = \sqrt{x^2+y^2} + 2(x + iy) = (\sqrt{x^2+y^2}+2x)+(2y)\cdot i
$$
Therefore:
$$
\begin{cases}
u(x,y)=\sqrt{x^2+y^2} + 2x\\
v(x,y) = 2y
\end{cases}
$$

Unfortunately, when checking I started to check the Cauchy-Riemann equations I got the solution "$f(z)$ does not satisfy the Cauchy-Riemann equations for any $z\in \mathbb{C}$" which is wrong.

Here's how I got this erroneous solution:
$$
\frac{\partial u}{\partial x} = \frac{-x}{\sqrt{x^2+y^2}}+2
\\
\frac{\partial v}{\partial y} = 2
\\
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \Leftrightarrow x=0 \wedge y\neq0
$$
$$$$
$$
\frac{\partial u}{\partial y} = \frac{-y}{\sqrt{x^2+y^2}}
\\
-\frac{\partial v}{\partial x} = 0
\\
\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \Leftrightarrow y=0 \wedge x\neq0
$$

And just like that I got the wrong answer. Where did I go wrong? How do I do this task correctly?

答:
You calculated $|z|^2$ wrong.
It should be $x^2+y^2$.

Since $h(z)=2z$ is analytic, you can focus on $g(z)=|z|^2=x^2+y^2$ where $z=x+iy$.

Now $u(x,y)=x^2+y^2$ and $v(x,y)=0$ for $g=u+iv$. You can go on and check the C-R equations.


问:
QR computation only in square matrix A?
I thought the following was true. But the answer is False.
Why so? Could anybody give me some counterexample?

For any matrix A, one can find Q and R such that A = QR
, where Q is an orthogonal matrix
and R is an upper triangular matrix with non-negative entries on the diagonal.

答:
Any real square matrix $A$ may be decomposed as $A = QR$

where $Q$ is an orthogonal matrix and $R$ is an upper triangular matrix (also called right triangular matrix). This generalizes to a complex square matrix $A$ and a unitary matrix $Q$.

More generally, we can factor a complex $m\times n$ matrix $A$, with $m ≥ n$, as the product of an $m\times m$ unitary matrix $Q$ and an $m\times n$ upper triangular matrix $R$.

for more detail follow this link

Title: QR computation only in square matrix A?


问:
Proof of generalized open & closed Newton-Cotes formulas
I am looking for proof of generalized open & closed Newton-Cotes formulas. I couldn't find any reference which properly proves both theorems. Most books just state the theorem and do not provide any proof.

Suppose that $\sum_{i=0}^{n} a_{i} f\left(x_{i}\right)$ denotes the $(n+1)$ -point open Newton-Cotes formula with $x_{-1}=a, x_{n+1}=b,$ and $h=(b-a) /(n+2) .$ There exists $\xi \in(a, b)$ for which
$$
\int_{a}^{b} f(x) d x=\sum_{i=0}^{n} a_{i} f\left(x_{i}\right)+\frac{h^{n+3} f^{(n+2)}(\xi)}{(n+2) !} \int_{-1}^{n+1} t^{2}(t-1) \cdots(t-n) d t
$$
if $n$ is even and $f \in C^{n+2}[a, b],$ and
$$
\int_{a}^{b} f(x) d x=\sum_{i=0}^{n} a_{i} f\left(x_{i}\right)+\frac{h^{n+2} f^{(n+1)}(\xi)}{(n+1) !} \int_{-1}^{n+1} t(t-1) \cdots(t-n) d t
$$
$
\text { if } n \text { is odd and } f \in C^{n+1}[a, b] \text { . }
$

答:
You can find the proof in "Isaacson, E. and Keller. Analysis of Numerical Methods" chapter 7.

NOTE: To preview the book, one must have to borrow the book and to do that one must have an login account on Archive. You can create a for account for free.


问:
Simple question about the domain of the marginal density function

Let $X$ and $Y$ be continuous random variables with joint density function $f(x, y)=15y$ for $x^2\le y\le x$ and $0$ otherwise. Find the marginal density function of $Y$.

The answer I found was $15y^{3/2}(1-y^{1/2})$ for $x^2\le y\le x$. I used the same $y-$bounds as the joint density function because I thought that that's someone one needs to do. However, the correct $y-$bounds are $0\le y\le 1$. Why did we change these bounds? Could someone please explain what I am missing here?

答:
The marginal density function oof $Y$ is given by
$$
\int_{-\infty}^\infty f(x,y)\,dx
$$
It can not depends on $x$; this is a function in $y$.

The key thing is to figure out the region where $f$ is not zero:
$$
S=\{(x,y)\mid x^2<y<x\}
$$
It is very easy to visualize the set by drawing a picture. See below.

Observe that for any $y>1$ or $y<0$, $f(x,y)=0$. (This is why the correct "bounds" for $y$ is $0\le y\le 1$.)

Now, given $0\le y\le1$, you want to integrate on $x$. The bounds for $x$ should be given by
$$
y<x<\sqrt{y}
$$
and thus the integral is
$$
f_Y(y)=\int_y^{\sqrt{y}}15 y\,dx,\quad 0\le y\le 1
$$


答: Note: You might be conflating matrix minimal polynomial (linear algebra, square matrix $K$-algebras) and minimal polynomial of an element in a field extension (field theory). The latter is always irreducible (the key is that the overfield is also an integral domain), but the former doesn't have to be (a $K$-algebra may not be an integral domain).

Title: Minimal polynomial of the matrix

Body:
Recently I've encountered the following problem

Find minimal polynomial of $A =\begin{pmatrix}
2 & 3 & 4\\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}$ in $\mathbb{R}[x]$

Firstly, we can compute the characteristic polynomial which is obviously $c = (x - 2)^3$, then we know that the minimal polynomial is the one which divides $c$ and has the minimal degree. Straightforward computation yields that $m_A = (x - 2)^2$. And it makes sense, except the fact that minimal polynomial should be irreducible, and here there are two factors. Am I missing anything?


问:
Doubt on proof about integer basis of cubic field
I'm reading the following proof from the book Problems in algebraic number theory by M. Ray Murty and Jody Esmonde

I understand everything except from what I've marked in yellow.

I know that $-3^3(ab)^2|d_K$ if $3|r$ and $-3(ab)^2|d_K$ otherwise but how can I prove the equality?

EDITED: I add some notation:

$d_{K/\mathbb{Q}}(\alpha)$ is the discriminant of $\alpha$ that is
$$d_{K/\mathbb{Q}}(\alpha)=\operatorname{det}(\sigma_i(\alpha^j))^2$$
where $\sigma_i$ are the 3 homomorphisms from $K$ to $\mathbb{C}$ and $j$ goes from $0$ to $2$.

d_K is the discriminant of $K$ that is the discriminant of any integer base where the discriminant of $\alpha_1, \alpha_2, \alpha_3$ elements of $K$ is
$$d_{K/\mathbb{Q}}(\alpha_1, \alpha_2, \alpha_3)=\operatorname{det}(\sigma_i(\alpha_j))^2$$

An integer polynomial $f=a_n x^n + \cdots + a_1 x + a_0$ is called Eisenstenian with respect to a prime $p$ if $p \not| a_n$, $p|a_i$ for $i=0, \cdots, n-1$ and $p^2 \not | a_0$.

答:
I think I've solved it, and what I'm going to prove in the following is that
$$\begin{align}
 d_K&=-3(ab)^2 & \textrm{ if } \hspace{2mm} 3|m\\
d_K&=-3^3(ab)^2 & \textrm{ if } \hspace{2mm} 3 \not|m
\end{align}$$

To prove that $b^2|d_K$ what you do is to consider $\beta=\alpha^2/b$ and define
$$h=[\mathcal{O}_K:\mathbb{Z}+\mathbb{Z}\beta+\mathbb{Z}\beta^2]$$

Then, we can calculate that

$$d_{K/\mathbb{Q}}(\beta)=-3^3a^3r$$

and so

$$-3^3a^3r=h^2d_K$$

Then you see that for $g(x)=x^3-a^2b$ the minimal polynomial of $\beta$ we have that is Eisenstenian for every prime $p$ dividing $b$, so $p \not| h$ what implies that $b$ and $h$ are coprime and from last equation that $b^2|d_K$.

Now, we can say more because if $b \not = 1$ then for every prime $p$ dividing $b$ we have that $p$ does not divide $d_K/b^2$. In fact if this is not as
$$d_K/b^2 h^2=-3^3a^4$$
we have that $p|-3^3a^4$ but $a$ and $b$ are coprime (if not $ab$ cannot be squarefree), so $p|3$ and we have a contradiction because by hipotesis $3\not| b$.

Now, we know that in any situation $3a^2|d_K$ and $b^2|d_K$ and as $3a^2$ and $b^2$ are coprime, $3(ab)^2|d_K$.

Moreover as
$$m^2d_K=-3^3r^2=-3^3(ab)^2b^2$$
we have that $d_K<0$ and then there exists $c \in \mathbb{N}$ such that $d_K=-3(ab)^2c$ from what we get that
$$cm^2=3^2b^2$$

Now if $3|m$ then $m=3n$ for some natural number $n$, so
$$cn^2=b^2$$
and we get that $c|b^2$ so $c=1$ because if $c>1$ some of its prime factors divides $b$ and $d_K/b^2$ and we have prove that this is not possible. Because $c=1$ we have proven that $d_K=-3(ab)^2$.

If $3 \not | m$ then $3^2|c$ so $c=3^2n$ for some natural number $n$, so
$$nm^2=b^2$$
and as before we get that $n=1$, from what we get that $d_K=-3^3(ab)^2$.


问:
Finding invariant factors of finitely generated Abelian group
There is this question that I wasn't sure how to do but somehow got the answers partially correct (maybe).

Suppose that the abelian group $M$ is generated by three elements $x,y,z$ subject to the relations $4x+y+2z=0, 5x+2y+z=0,6y-6z=0$. Determine the invariant factors of $M$.

My attempt:

I wasn't sure what theorem or reasoning to use, but I have a feeling to write the equations in matrix form and find the equivalent matrix. My question is normally if we write in matrix form, we should be priorly given a map or homomorphism, but this question did not, or whether somehow we can interpret the equations as map? if so, what is the map?

Let me try to make one. Let $M$ be the $\mathbb{Z}$-module $F/N$ where $F=\mathbb{Z}^3$ and $N=\langle(4,5,0),(1,2,6),(2,1,-6)\rangle\le\mathbb{Z}^3$. Consider the homomorphism $\varphi:\mathbb{Z}^3\to\mathbb{Z}^3$ given by $(x,y,z)\to x(4,5,0)+y(1,2,6)+z(2,1,-6)$. Am I correct? Please correct me if I am wrong.

Then I find the equivalent matrix:

$\begin{bmatrix}4 & 1 & 2\\5 & 2 & 1\\0 & 6 & -6\end{bmatrix}$ $4R_2-5R_1$ $\begin{bmatrix}4 & 1 & 2\\0 & 3 & -6\\0 & 6 & -6\end{bmatrix}$ $R_3-2R_2$ $\begin{bmatrix}4 & 1 & 2\\0 & 3 & -6\\0 & 0 & 6\end{bmatrix}$ $3R_1+R_2$ $\begin{bmatrix}12 & 6 & 0\\0 & 3 & -6\\0 & 0 & 6\end{bmatrix}$ $R_2+R_3$ $\begin{bmatrix}12 & 6 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}$ $R_1-2R_2$ $\begin{bmatrix}12 & 0 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}$. Then by performing column and row swaps, the matrix becomes $\begin{bmatrix}3 & 0 & 0\\0 & 6 & 0\\0 & 0 & 12\end{bmatrix}$. Since 3,6,12 are non-units in $\mathbb{Z}$, then the invariant factors of M are 3, 6, 12.

However, the solution says that the invariant factors are 1,3,6.

Was my method correct? Where did I make the mistake?

Thanks!

答:
There is a bit of theory to go through to explain how the invariant factors of the matrix you wrote down is actually related to $M,$ you will have to go through your notes to see the precise connection.

As for the calculation, perhaps recheck your operations, or post them here so we might we able to say where you went wrong.

It turns out that for this matrix, we can get away with doing no work by using the fact that the product of the invariant factors is equal to the determinant of the matrix, which we can compute is $18 = 2\cdot 3^2.$ Since each invariant factor must divide the next, the only possibility is $1,3,6.$


问:
"Large" metric spaces
I'm studying Functional Analysis from Joseph Muscat's book. The author says the non separable spaces are very large. This makes sense too as finite metric spaces and countable spaces are trivially separable and so the only metric spaces which can be non-separable must be uncountable. This is what he says to justify to say that non-separable spaces are too large.

Can someone explain to me what he means by computing distances "precisely", "approximately" and "to any accuracy"? Given two points, we know distance between those points, so, where does the impreciseness occur here?

答:
It's useful to examine some concrete examples. Consider $\mathbb{Q}$ as a metric space. I can calculate the distance between two points exactly because $|y-x|$ is a rational number which takes up a finite amount of space as two relatively prime numbers are sufficient to describe it exactly. If the numbers are arbitrarily large this could take a long time to do.

Now if we consider $\mathbb{R}$ as a metric space we run into a problem, which is that most numbers are not computable. This means there is no finite length algorithm to describe the number, so encoding it would require an infinite amount of storage. However $\mathbb{Q}$ is a dense subset in $\mathbb{R}$ and because the space is separable we can approximate some real number by a rational number close to it. Formally this means that given some error term $\epsilon > 0$, typically small, that for every $x \in \mathbb{R}$ there exists a rational number $q \in \mathbb{Q}$ such that $|x-q| < \epsilon$. We can't describe the real number exactly but we can find a rational that is close enough for practical purposes.

Now, non-seperable spaces run into an overcrowding problem. Because points can't be separated we can't always tell them apart. This means there isn't some unique representation for the number so calculations using distances alone aren't sufficient to specify which is which. Consider a sequence of open sets given by $(-1/n,1/n)$ with $n \in \mathbb{N}$. The intersection of all these sets would be $\{0\}$ which is unique, but what if every open set contained other open number that we couldn't separate from zero, say $0'$. This means all the open sets would have to contain both of them and the intersection would be $\{0,0'\}$. Since the solution is not unique we can't tell if the answer is $0$ or $0'$ using topological methods. For metric spaces this means distances are insufficient to find unique solutions.

Title: "Large" metric spaces


问:
Calculate the maximum of $\sum_{1\le i<j<k\le 100}a_ia_ja_k$ where $\sum a_i = 2007$.

Calculate the maximum of $$\sum_{1\le i<j<k\le 100}a_ia_ja_k$$ where $\sum a_i = 2007$.

This is a question that I've seen long before and not been able to solve it.

My initial thought is that the more even $a_i$ is, the larger the sum will be. Hence, when there are $93$ $20$s and $7$ $21$s, the sum will supposedly become the biggest.

But I got stuck on proving this point. I'd appreciate some help.

PS. Can we extend the problem, e.g. $100$ to $n\in\mathbb{N}_+$ and $2007$ to $k>n\land k\in\mathbb{N}_+$?

答:
Your intuition is completely correct.

Suppose the maximum is achieved for a set of $a_i$ where two of them, $a$ and $b$ say, are such that $b\ge a+2$ and consider the effect of changing $a$ to $a+1$ and $b$ to $b-1$.

Split up the terms in the original summation into those involving neither $a$ nor $b$, involving $a$ not $b$, involving $b$ not $a$, involving both. Let these add up to $N,Xa,Xb,Yab$, respectively.

Then the sum $N+Xa+Xb+Yab$ is changed to $N+X(a+1)+X(b-1)+Y(a+1)(b-1)$ i.e. a positive change of $Y(b-a-1)$.

We can conclude that the $a_i$ take at most two values. Furthermore, we can see that this proof generalises very easily.

Over to you now?


问:
Finding all the solutions of the system
I'm struggling with the following problem:

Consider the system of equations of three unknowns $Ax=b$, for which we know that:

a) $A≠0$.

b) $(1, 2, 2)$ and $(0, 1, 1)$ are solutions,

c) $(0, a, b)$ is a solution if and only if $a=b=1$.

I'm asked to find all the solutions of the system.

The issue is that I'm not sure how to use c). Particularly the fact that no other solution of the form $(0, a, b)$ exists.

Also, the teacher uses $b$ first as a vector, and then as a scalar. My guess is that when $b$ is used as a vector it is meant to be $(b, b, b)$.

I'd appreciate if anyone could point me in the right direction.

I'm including the original problem here by petition of a user, it is however in Catalan.

I admit the teacher isn't clear in what is wanted, I believe he means us to solve

$$A\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} b \\ b \\ b \end{pmatrix}$$

such that for some fixed value $b=b_1$ we have that $(1, 2, 2)$ is a solution (similarly, for some fixed value $b=b_2$, we have that $(0,1,1)$ is a solution, in which case we know by c) that $b_2=1$), and such that $ay+bz=b$ if and only if $a=b=1$.

答:
(I only presume that a system whose coefficient matrix is $A$ has solutions satisfying the conditions (a), (b), and (c) regarding that "$AX=b$"-thing for a typo otherwise the problem would be senseless.)

Clearly the solution set's dimension is neither zero nor three.

Assuming the solution set has dimension $1$ (and assuming that $A$ has three columns, of course) the solution is a straight line given by
$$x=w+s\cdot u$$
with a suitable replacement-vector $w$ and a direction-vector $u$, where $u$ is an element of the kernel of $A$.  Now a candidate for $u$ is easily found: as $(1,2,2)$ and $(0,1,1)$ are solutions their difference $u=(1,1,1)$ is an element of the kernel.

For the replacement vector simply choose $w=(0,1,1)$; the solutions are given by
$$
\begin{pmatrix}
0\\1\\1
\end{pmatrix}
+s\begin{pmatrix}
1\\1\\1
\end{pmatrix},
$$
which may also be written as
$$
\begin{pmatrix}
-1\\0\\0
\end{pmatrix}
+r\begin{pmatrix}
-1\\-1\\-1
\end{pmatrix}.
$$

As $u$ and $w$ are not parallel the only solution who's first coordinate is zero is just $w$ as
$$\begin{pmatrix}0\\a\\b
\end{pmatrix}=\begin{pmatrix}0\\1\\1
\end{pmatrix}+s\begin{pmatrix}1\\1\\1
\end{pmatrix}$$
has only solution $s=0$.

In case the solution set's dimensions equals two, rref has the form
$$\begin{array}{ccc|c}
1 & \alpha & \beta & \gamma
\end{array}$$
hence the solutions are given by
$$
\begin{pmatrix}
\gamma\\0\\0
\end{pmatrix}
+r\begin{pmatrix}
\alpha\\-1\\0
\end{pmatrix}
+s\begin{pmatrix}
\beta\\0\\-1
\end{pmatrix}.
$$
As $(0,1,1)$ must be a solution we conclude that in this case $r=s=-1$ and hence $\gamma=\alpha+\beta$.

As $(1,2,2)$ must be a solution we conclude that in this case $r=s=-2$ and from here $\alpha+\beta=-1$, that is $\beta=-1-\alpha$.
Finally the solutions are given by
$$\begin{pmatrix}-1\\0\\0
\end{pmatrix}+s\begin{pmatrix}\alpha\\-1\\0
\end{pmatrix}+r\begin{pmatrix}-1-\alpha\\0\\-1\end{pmatrix}.$$

As obviously
$$\begin{pmatrix}-1\\0\\0
\end{pmatrix},\begin{pmatrix}\alpha\\-1\\0
\end{pmatrix}\text{ and} \begin{pmatrix}-1-\alpha\\0\\-1\end{pmatrix}$$
are linearly independent, $(0,a,b)$ is a solution iff $a=b=1$.

Title: Finding all the solutions of the system


问:
If vector $a = b$ and $a = c$, then $b = c$?
I have three vectors: $\vec{u}$, $\vec{v}$ and  $\vec{w}$ and this equality:
$\vec{u} ·  \vec{v} = \vec{u} ·  \vec{w}$. Then the question is if $\vec{v} =  \vec{w}$.
At first i tried to write the equality representing components:
$$\vec{u} · \vec{v} = u_1·v_1+\ldots+u_n·v_n$$
$$\vec{u} · \vec{w} = u_1·w_1+\ldots+u_n·w_n$$
Then:
$$u_1·v_1+\ldots+u_n·v_n = u_1·w_1+\ldots+u_n·w_n.$$
But I got stuck right there; I thought if $c_i, i \in [1,n]$ could be canceled from both terms, but don't know if that is possible somehow.

答:
$$\vec u\cdot v=\vec u\cdot w\implies \vec u\cdot(\vec v-\vec w)=0.$$

This expresses that $\vec v-\vec w$ is orthogonal to $\vec u$, and not necessarily $\vec0$.


问:
Limit $\lim_{n\rightarrow \infty} \delta_n$
Here is a solution to a limit of a sequence, but i feel like the squeeze theorem is really needed to get the right answer, but i cant see how excactly it should be used.Thank you for any comment $!$.
\begin{align}
\delta_{n} & =
\frac{1}{2n}\,\sqrt[n]{\,{1^{n} + 2^{n} + \cdots +
\left(2n\right)^{n}}\,}\,
\\[2mm] & =
\,\sqrt[n]{\,{\left(\frac{1}{2n}\right)^{n} +
\left(\frac{2}{2n}\right)^{n} + \cdots +
\left(\frac{2n}{2n}\right)^{n}}\,}\,\,\,\to 1
\end{align}

答:
By using $k\leq 2n$ we have

$$\frac{1}{2n}\sqrt[n]{(2n)^n}\leq\frac{1}{2n}\sqrt[n]{1^n+\dots+(2n)^n}\leq\frac{1}{2n}\sqrt[n]{(2n)^n+\dots +(2n)^n}$$

from where we get
$$ 1\leq S_n\leq\sqrt[n]{2n}.$$
Since $\lim_{n\to\infty} \sqrt[n]{2n}=1$, we have $$\lim_{n\to\infty} S_n=1.$$


答:
Let $\vec u=\vec v = \vec0$ and $\vec w\neq \vec0$ then $\vec u \cdot\vec v=\vec u\cdot\vec w =\vec0$.

More generally $\vec u \cdot\vec v=\vec u\cdot\vec w \implies \vec u\cdot(\vec v-\vec w)=0\implies\vec u$ perpendicular to $(\vec v-\vec w)$.

And the above example is true as $\vec0$ is perpendicular to every vector.

Title: If vector $a = b$ and $a = c$, then $b = c$?

Body:
I have three vectors: $\vec{u}$, $\vec{v}$ and  $\vec{w}$ and this equality:
$\vec{u} ·  \vec{v} = \vec{u} ·  \vec{w}$. Then the question is if $\vec{v} =  \vec{w}$.
At first i tried to write the equality representing components:
$$\vec{u} · \vec{v} = u_1·v_1+\ldots+u_n·v_n$$
$$\vec{u} · \vec{w} = u_1·w_1+\ldots+u_n·w_n$$
Then:
$$u_1·v_1+\ldots+u_n·v_n = u_1·w_1+\ldots+u_n·w_n.$$
But I got stuck right there; I thought if $c_i, i \in [1,n]$ could be canceled from both terms, but don't know if that is possible somehow.


答:
If $f'(x) = -f(x)$,
then
$0 = f'(x)+f(x)
= e^x(f'(x)+f(x))
= (e^x f(x))'
$
so $e^x f(x)$ is constant.

Set $e^x f(x) = c$
and set $x = 1$.
Then, since $f(1) = 1$,
$c = e f(1) = e$
so $f(x) = e e^{-x} = e^{1-x}$.

An alternative solution is to write
$f'(x)/f(x) = -1$
so $(\ln(f(x))'=-1$
so $\ln f(x) = c-x$
or $f(x) = e^{c-x}$
for some constant $c$.

This has the problem that $f(x) = 0$ causes problems,
and the first solution does not have this problem.

Title: If $f'(x) = -f(x)$ and $f(1)=1$, then what is $f(x)$?

Body:
If $f'(x) = -f(x)$ and $f(1)=1$, then $f(x)=$

(a) $1/2e^{-2x+2}$

(b) $e^{-x-1}$

(c) $e^{1-x}$

(d) $e^{-x}$

(e) $-e^{x}$

If you plug in $1$ for $x$ in option c it gives you $e^{1-1}$ or $e^0$ which is $1$ hence what is needed.


问:
Scientific Calculator without Memory or Exponents
My chemistry teacher requires us to use a scientific calculator that does not have memory, storage, or exponents, but I'm having trouble finding one. Does anyone know of a calculator that meets the requirements?

Also, sorry if this is the wrong exchange. This looks like the best place to ask a question about calculators, but I might be wrong. If there's a better place, let me know.

Edit:
My teacher approved an office calculator with memory. Turns out what my teacher put in the syllabus didn't match what she wanted. Hopefully that's the only inconsistency.

答:
My teacher approved an office calculator with memory. Turns out what my teacher put in the syllabus didn't match what she wanted. Hopefully that's the only inconsistency.

Also, can someone mark this answer as community wiki? I don't have enough reputation to do it myself.


答:
The answer is no if $\vec{u} \cdot (\vec{v}-\vec{w})=0$ which is possible even if $\vec{u}\neq 0$ that means let the $\vec{d}=\vec{v}-\vec{w}$ and assume angle between $\vec{u}$ and $\vec{d}$ 90°.So, it not necessary that $\vec{d}=\vec{0}$
Example let, $\vec{u}=\vec{i}+\vec{j}+\vec{k}$ and let $\vec{d}=-2\vec{i}+\vec{j}+\vec{k}$ Now there are many pairs with $a-b=-2$ ,$c-d=1$, $e-f=1$ where we set $\vec{v}=a\vec{i}+c\vec{j}+e\vec{k}$ and $\vec{u}=b\vec{i}+d\vec{j}+f\vec{k}$
Done.


问:
Confusion over the definition of disjoint union
From Wikipedia, the disjoint union of some sets, $\uplus A_i$, is a set $A$ with an injective function for each $A_i$ to $A$, such that the union of the images of these functions forms a partitions of $A$.

But the union of some sets is just a single set, whereas a partition is a set of sets, so what does it mean for this single set to "form a partition" of $A$?

From the Wikipedia example, for the sets $A_0=\{5,6,7\}$ and $A_1=\{5,6\}$ it gives the disjoint union as $A=\{(5,0),(6,0),(7,0),(5,1),(5,2)\}$ with the injective function for each $A_i$ as $f(x)=(x,i)$. Unless I'm mistaken, the union of the images of these functions is just $A$ itself, right? But $A$ is not a partition of $A$ - even for the $n=1$ case the partition would be $\{A\}$ rather than just $A$. So how is "the union of the images of the functions forms a partition of $A$" true?

答:
Yes, the union of these images is $A$, so that is maybe poorly phrased. What they mean is that the images of these injections partition $A$. That is
$$
P = \{ f_i(A_i) : 0 \leq i < n \}
$$
partitions $A = \biguplus_{0 \leq i < n} A_i$, where $f_i: A_i \to A$ is the injection corresponding to $0 \leq i < n$.

So in your example we have $n = 2$ and so
$$
P = \{f_0(A_0), f_1(A_1)\} = \{\{(5,0), (6,0), (7,0)\}, \{(5,1), (6,1)\}\},
$$
which is indeed a partition of
$$
A=\{(5,0),(6,0),(7,0),(5,1),(6,1)\}.
$$


答:
What that article says is just that the images $f(A_0)$ and $f(A_1)$ form a partition of $A$, which means
$$
A=f(A_0)\cup f(A_1)
$$
and $f(A_0)\cup f(A_1)=\emptyset$.

The phrase "the union of the images of these functions" should be understood as the collection of the images of these functions.

Title: Confusion over the definition of disjoint union

Body:
From Wikipedia, the disjoint union of some sets, $\uplus A_i$, is a set $A$ with an injective function for each $A_i$ to $A$, such that the union of the images of these functions forms a partitions of $A$.

But the union of some sets is just a single set, whereas a partition is a set of sets, so what does it mean for this single set to "form a partition" of $A$?

From the Wikipedia example, for the sets $A_0=\{5,6,7\}$ and $A_1=\{5,6\}$ it gives the disjoint union as $A=\{(5,0),(6,0),(7,0),(5,1),(5,2)\}$ with the injective function for each $A_i$ as $f(x)=(x,i)$. Unless I'm mistaken, the union of the images of these functions is just $A$ itself, right? But $A$ is not a partition of $A$ - even for the $n=1$ case the partition would be $\{A\}$ rather than just $A$. So how is "the union of the images of the functions forms a partition of $A$" true?


问:
Finding values that satisfy a differential equation

$y(x) = A\cos(3x) + B\sin(3x)$ that satisfies a differential equation:

$y'+y = 3\sin(3x)$,  how do I find values $A$ and $B$.

I have tried to derivate $y(x)$ and put the derivative and the original function to form:

$y = A\cos(3x) + B\sin(3x)$

$y' = 3B\cos(3x)-3A\sin(3x)$

$3B\cos(3x)-3A\sin(3x)+A\cos(3x)+B\sin(3x) = 3\sin(3x)$

I can not deduce the values or solve them.

答:
$$3B\cos(3x)-3A\sin(3x)+A\cos(3x)+B\sin(3x) = 3\sin(3x)$$
$$(3B+A)\cos(3x)+(B-3A)\sin(3x)= 3\sin(3x)+ \color {red}{0 \times \cos (3x)}$$
Now solve for $A,B$ the system of equations:
$$
\begin {cases}
3B+A=0 \\
B-3A=3
\end{cases}
$$