问:
Subset of Permutations of distinct symbols, with positional count constraint/
I am working a personal project, and I need to derive subsets of permutations of unique elements.

For example:  I have a set of 5 unique elements {A, B, C, D, E}.   There are 60 permutations tuples of size 3 of this: {A,B,C}, {A,B,D}...{E,C,D}, {E,D,C}.  What I want is to select tuples from this list such that the count of each symbol in the same position across all tuples is constrained as 0 < A# < B# < C# < D# < E# the count of each symbol is the same in each position.

E.g.

```<code>{A, B, C}
{A, B, D}
⋮
{E, C, D}
{E, D, C}
</code>```

For the full set of tuples, there are 12 of each symbol in each position within the tuple, or specifically for what is explicitly written above there are quantity A2 and E2 in position 0, B2 C1 D1 in position 1, and C2 and D2 in position 2, so this does not meet the criteria.

With the example of 5 permute 3, as there are only 60 permutation tuples I was able to create such a desired subset by brute force, finding a collection of 20 permutation tuples resulting in counts of A2, B3, C4, D5, E6 in each position.

My question is:   Is there a known algorithmic way of selecting subsets with this type of criteria, and is there a predictive formula to calculate how many valid solutions can be derived?  I will need this to be for quantities of symbols up to 15 and up to tuple size 5 so brute force is sub-optimal.

For completeness, the solution subset I found for 5 symbols permut 3 is the following set:

```<code>AEC
AED
BCE
BDE
BED
CBE
CDB
CDE
CED
DBC
DBE
DCE
DEB
DEC
EAC
EAD
ECA
ECD
EDA
EDB
</code>```

Edit:  Using logic inspired by Ross Millikan, I verified that 4 permut 3 has no solution, as the permut domain is too small.  I did find via brute force a solution for 6 permut 3: A3,B4,C5,D6,E7,F8 for a total of 33 tuples:

```<code>ABC
ABE
ACD
BCF
BDF
BEF
BFE
CAD
CED
CEF
CFD
CFE
DCE
DCF
DEC
DEF
DFC
DFE
ECF
EDB
EDC
EDF
EFB
EFC
EFD
FAD
FAE
FBA
FBE
FDA
FDB
FEA
FEB
</code>```

答:
Not a complete answer, however some interesting research results

Using the setup of 5 distinct symbols {A,B,C,D,E} permut 3, 60 permutations.  I have run exhaustive checks on all permutations for the positional symbol count criteria form of A#1, B#2, C#3, D#4, E#5.

For ease of computation I arbitrarily grouped the set based upon the symbol in position 0, thus having 5 subgroups of 12 tuples each, e.g. ABC ABD ABE ACB ACD ACE ADB ADC ADE AEB AEC AED ...
With this, the maximum number of checks can be calculated as
$$
\binom{12}{1}\binom{12}{2}\binom{12}{3}\binom{12}{4}\binom{12}{5}=68,309,049,600
$$
For optimization, I recursively worked through the initial symbol sub groups starting with symbol A, and could abandon a recursion before reaching symbol E subgroup as soon as the remaining symbol count violated the constraint.   With this optimization, when I ran this the code only had to perform 12,788,832 checks to find there were in total only 130 solutions that met the criteria.  Since it is such a small set, below is the exhaustive list by row of the valid tuple subsets.

```<code>ACE BDE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
ACE BDE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
ACE BDE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
ACE BDE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
ACE BDE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
ACE BDE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
ACE BDE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
ACE BDE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
ACE BDE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
ACE BDE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
ACE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
ACE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
ACE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
ACE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
ADE BCE BDE CDE CEB CED DCE DEA DEB DEC EAD EBC EBD ECD EDC
ADE BCE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
ADE BCE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
ADE BCE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
ADE BCE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
ADE BCE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
ADE BCE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
ADE BCE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
ADE BCE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
ADE BCE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
ADE BCE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
ADE BDE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECA ECD EDC
ADE BDE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECB ECD EDC
ADE BDE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECB ECD EDC
ADE BDE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECB ECD EDC
ADE BDE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECB ECD EDC
ADE BDE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECB ECD EDC
ADE BDE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECA ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECB ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECA ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECA ECD EDC
ADE BEC BED CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
ADE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
ADE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
ADE BEC BED CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
ADE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
ADE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
ADE BEC BED CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
ADE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
ADE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
ADE BEC BED CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
ADE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
ADE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AEC BCE BDE CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
AEC BCE BED CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
AEC BCE BED CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
AEC BCE BED CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
AEC BCE BED CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
AEC BDE BED CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
AEC BDE BED CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
AEC BDE BED CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
AEC BDE BED CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
AEC BDE BED CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
AEC BDE BED CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
AEC BDE BED CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
AEC BDE BED CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
AEC BDE BED CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AED BCE BDE CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
AED BCE BDE CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
AED BCE BDE CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
AED BCE BDE CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
AED BCE BDE CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
AED BCE BDE CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
AED BCE BDE CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
AED BCE BDE CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
AED BCE BDE CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
AED BCE BEC CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
AED BCE BEC CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
AED BCE BEC CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
AED BCE BEC CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
AED BCE BED CAE CDE CED DBE DCE DEB DEC EBC ECD EDA EDB EDC
AED BCE BED CBE CDE CED DAE DCE DEB DEC EBC ECD EDA EDB EDC
AED BCE BED CBE CDE CED DBE DCE DEB DEC EAC ECD EDA EDB EDC
AED BCE BED CDE CEB CED DAE DBE DCE DEC EBC ECD EDA EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
AED BDE BEC CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
AED BDE BEC CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
AED BDE BEC CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
AED BDE BEC CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
AED BDE BEC CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
AED BDE BEC CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
AED BDE BEC CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
AED BDE BEC CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
AED BDE BEC CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AED BDE BED CAE CDE CED DBE DCE DEA DEC EBC ECB ECD EDB EDC
AED BDE BED CAE CDE CED DBE DCE DEB DEC EBC ECA ECD EDB EDC
AED BDE BED CAE CDE CED DBE DCE DEB DEC EBC ECB ECD EDA EDC
AED BDE BED CBE CDE CED DAE DCE DEA DEC EBC ECB ECD EDB EDC
AED BDE BED CBE CDE CED DAE DCE DEB DEC EBC ECA ECD EDB EDC
AED BDE BED CBE CDE CED DAE DCE DEB DEC EBC ECB ECD EDA EDC
AED BDE BED CBE CDE CED DBE DCE DEA DEC EAC ECB ECD EDB EDC
AED BDE BED CBE CDE CED DBE DCE DEB DEC EAC ECA ECD EDB EDC
AED BDE BED CBE CDE CED DBE DCE DEB DEC EAC ECB ECD EDA EDC
AED BDE BED CDE CEA CED DAE DBE DCE DEC EBC ECB ECD EDB EDC
AED BDE BED CDE CEB CED DAE DBE DCE DEC EBC ECA ECD EDB EDC
AED BDE BED CDE CEB CED DAE DBE DCE DEC EBC ECB ECD EDA EDC
AED BEC BED CBE CDE CED DAE DBE DCE DEC ECB ECD EDA EDB EDC
</code>```

Counting the total tuples in this collection of the 130 sets of 15 tuples that met the criteria revealed some interesting insights.  Below is this count, sorted by first position symbol and then by quantity found in the total collection of tuples:

```<code>ACE:          20
AEC:          20
ADE:          45
AED:          45
--------------------
BCE:          45
BEC:          45
BDE:          85
BED:          85
--------------------
CAE:          20
CEA:          20
CBE:          45
CEB:          45
CDE:         130
CED:         130
--------------------
DAE:          45
DEA:          45
DBE:          85
DEB:          85
DCE:         130
DEC:         130
--------------------
EAC:          20
ECA:          20
EAD:          45
EBC:          45
ECB:          45
EDA:          45
EBD:          85
EDB:          85
ECD:         130
EDC:         130
</code>```

The first insight that jumped out to me is clearly there is a reflexive pair pattern.  This makes sense, since effectively one could swap all the symbols in position 1 and 2 to yield the mirror result.

The next insight that surprised me is that subgroups C, D, and E shows tuples pairs that must appear in a valid solution, e.g. CDE/CED, DCE/DEC, ECD/EDC, and some tuple pairs in every sub group that never appear.

To see how the pattern extends, I repeated this process with the same initial collection of 5 permute 3, changing the criteria to A#2 B#3 C#4 D#5 E#6.  This had a possibility of requiring 5.2E12 comparisons, though the optimizations reduced this to only ~1E9 tests required to find 50,202 valid solutions.  The tuple count in the valid solutions, below, shows the same reflective pattern, as well as perhaps some other insights.

```<code>ABC:         561
ACB:         561
ABD:       1,144
ADB:       1,144
ABE:       3,131
AEB:       3,131
ACD:       5,140
ADC:       5,140
ACE:      13,792
AEC:      13,792
ADE:      26,434
AED:      26,434
--------------------
BAC:         561
BCA:         561
BAD:       1,144
BDA:       1,144
BAE:       3,131
BEA:       3,131
BCD:       9,889
BDC:       9,889
BCE:      24,351
BEC:      24,351
BDE:      36,227
BED:      36,227
--------------------
CAB:         561
CBA:         561
CAD:       5,140
CDA:       5,140
CBD:       9,889
CDB:       9,889
CAE:      13,792
CEA:      13,792
CBE:      24,351
CEB:      24,351
CDE:      46,671
CED:      46,671
--------------------
DAB:       1,144
DBA:       1,144
DAC:       5,140
DCA:       5,140
DBC:       9,889
DCB:       9,889
DAE:      26,434
DEA:      26,434
DBE:      36,227
DEB:      36,227
DCE:      46,671
DEC:      46,671
--------------------
EAB:       3,131
EBA:       3,131
EAC:      13,792
ECA:      13,792
EBC:      24,351
ECB:      24,351
EAD:      26,434
EDA:      26,434
EBD:      36,227
EDB:      36,227
ECD:      46,671
EDC:      46,671
</code>```

Research is ongoing to see if I can derive a predictive formula, as well as ongoing optimizations to the exhaustive checks (taking advantage of the mirror reflection at least).

Body:
Not a complete answer, however some interesting research results

Using the setup of 5 distinct symbols {A,B,C,D,E} permut 3, 60 permutations.  I have run exhaustive checks on all permutations for the positional symbol count criteria form of A#1, B#2, C#3, D#4, E#5.

For ease of computation I arbitrarily grouped the set based upon the symbol in position 0, thus having 5 subgroups of 12 tuples each, e.g. ABC ABD ABE ACB ACD ACE ADB ADC ADE AEB AEC AED ...
With this, the maximum number of checks can be calculated as
$$
\binom{12}{1}\binom{12}{2}\binom{12}{3}\binom{12}{4}\binom{12}{5}=68,309,049,600
$$
For optimization, I recursively worked through the initial symbol sub groups starting with symbol A, and could abandon a recursion before reaching symbol E subgroup as soon as the remaining symbol count violated the constraint.   With this optimization, when I ran this the code only had to perform 12,788,832 checks to find there were in total only 130 solutions that met the criteria.  Since it is such a small set, below is the exhaustive list by row of the valid tuple subsets.

ACE BDE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
ACE BDE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
ACE BDE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
ACE BDE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
ACE BDE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
ACE BDE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
ACE BDE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
ACE BDE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
ACE BDE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
ACE BDE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
ACE BDE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
ACE BDE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
ACE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
ACE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
ACE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
ACE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
ADE BCE BDE CDE CEB CED DCE DEA DEB DEC EAD EBC EBD ECD EDC
ADE BCE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
ADE BCE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
ADE BCE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
ADE BCE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
ADE BCE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
ADE BCE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
ADE BCE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
ADE BCE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
ADE BCE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
ADE BCE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
ADE BCE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
ADE BCE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
ADE BDE BEC CBE CDE CED DCE DEA DEB DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEA CED DBE DCE DEB DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEB CED DBE DCE DEA DEC EAD EBD ECB ECD EDC
ADE BDE BEC CDE CEB CED DBE DCE DEB DEC EAD EBD ECA ECD EDC
ADE BDE BED CAE CDE CED DCE DEA DEB DEC EBC EBD ECB ECD EDC
ADE BDE BED CBE CDE CED DCE DEA DEB DEC EAC EBD ECB ECD EDC
ADE BDE BED CBE CDE CED DCE DEA DEB DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEA CED DAE DCE DEB DEC EBC EBD ECB ECD EDC
ADE BDE BED CDE CEA CED DBE DCE DEB DEC EAC EBD ECB ECD EDC
ADE BDE BED CDE CEA CED DBE DCE DEB DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEB CED DAE DCE DEA DEC EBC EBD ECB ECD EDC
ADE BDE BED CDE CEB CED DAE DCE DEB DEC EBC EBD ECA ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEA DEC EAC EBD ECB ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEA DEC EAD EBC ECB ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEB DEC EAC EBD ECA ECD EDC
ADE BDE BED CDE CEB CED DBE DCE DEB DEC EAD EBC ECA ECD EDC
ADE BEC BED CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
ADE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
ADE BEC BED CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
ADE BEC BED CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
ADE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
ADE BEC BED CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
ADE BEC BED CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
ADE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
ADE BEC BED CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
ADE BEC BED CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
ADE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
ADE BEC BED CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AEC BCE BDE CBE CDE CED DCE DEA DEB DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEA CED DBE DCE DEB DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEB CED DBE DCE DEA DEC EAD EBD ECD EDB EDC
AEC BCE BDE CDE CEB CED DBE DCE DEB DEC EAD EBD ECD EDA EDC
AEC BCE BED CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
AEC BCE BED CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
AEC BCE BED CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
AEC BCE BED CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
AEC BDE BED CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
AEC BDE BED CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
AEC BDE BED CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
AEC BDE BED CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
AEC BDE BED CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
AEC BDE BED CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
AEC BDE BED CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
AEC BDE BED CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
AEC BDE BED CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
AEC BDE BED CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AED BCE BDE CAE CDE CED DCE DEA DEB DEC EBC EBD ECD EDB EDC
AED BCE BDE CBE CDE CED DCE DEA DEB DEC EAC EBD ECD EDB EDC
AED BCE BDE CBE CDE CED DCE DEA DEB DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEA CED DAE DCE DEB DEC EBC EBD ECD EDB EDC
AED BCE BDE CDE CEA CED DBE DCE DEB DEC EAC EBD ECD EDB EDC
AED BCE BDE CDE CEA CED DBE DCE DEB DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEB CED DAE DCE DEA DEC EBC EBD ECD EDB EDC
AED BCE BDE CDE CEB CED DAE DCE DEB DEC EBC EBD ECD EDA EDC
AED BCE BDE CDE CEB CED DBE DCE DEA DEC EAC EBD ECD EDB EDC
AED BCE BDE CDE CEB CED DBE DCE DEA DEC EAD EBC ECD EDB EDC
AED BCE BDE CDE CEB CED DBE DCE DEB DEC EAC EBD ECD EDA EDC
AED BCE BDE CDE CEB CED DBE DCE DEB DEC EAD EBC ECD EDA EDC
AED BCE BEC CAE CDE CED DBE DCE DEB DEC EBD ECD EDA EDB EDC
AED BCE BEC CBE CDE CED DAE DCE DEB DEC EBD ECD EDA EDB EDC
AED BCE BEC CBE CDE CED DBE DCE DEB DEC EAD ECD EDA EDB EDC
AED BCE BEC CDE CEB CED DAE DBE DCE DEC EBD ECD EDA EDB EDC
AED BCE BED CAE CDE CED DBE DCE DEB DEC EBC ECD EDA EDB EDC
AED BCE BED CBE CDE CED DAE DCE DEB DEC EBC ECD EDA EDB EDC
AED BCE BED CBE CDE CED DBE DCE DEB DEC EAC ECD EDA EDB EDC
AED BCE BED CDE CEB CED DAE DBE DCE DEC EBC ECD EDA EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEA DEC EBD ECB ECD EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEB DEC EBD ECA ECD EDB EDC
AED BDE BEC CAE CDE CED DBE DCE DEB DEC EBD ECB ECD EDA EDC
AED BDE BEC CBE CDE CED DAE DCE DEA DEC EBD ECB ECD EDB EDC
AED BDE BEC CBE CDE CED DAE DCE DEB DEC EBD ECA ECD EDB EDC
AED BDE BEC CBE CDE CED DAE DCE DEB DEC EBD ECB ECD EDA EDC
AED BDE BEC CBE CDE CED DBE DCE DEA DEC EAD ECB ECD EDB EDC
AED BDE BEC CBE CDE CED DBE DCE DEB DEC EAD ECA ECD EDB EDC
AED BDE BEC CBE CDE CED DBE DCE DEB DEC EAD ECB ECD EDA EDC
AED BDE BEC CDE CEA CED DAE DBE DCE DEC EBD ECB ECD EDB EDC
AED BDE BEC CDE CEB CED DAE DBE DCE DEC EBD ECA ECD EDB EDC
AED BDE BEC CDE CEB CED DAE DBE DCE DEC EBD ECB ECD EDA EDC
AED BDE BED CAE CDE CED DBE DCE DEA DEC EBC ECB ECD EDB EDC
AED BDE BED CAE CDE CED DBE DCE DEB DEC EBC ECA ECD EDB EDC
AED BDE BED CAE CDE CED DBE DCE DEB DEC EBC ECB ECD EDA EDC
AED BDE BED CBE CDE CED DAE DCE DEA DEC EBC ECB ECD EDB EDC
AED BDE BED CBE CDE CED DAE DCE DEB DEC EBC ECA ECD EDB EDC
AED BDE BED CBE CDE CED DAE DCE DEB DEC EBC ECB ECD EDA EDC
AED BDE BED CBE CDE CED DBE DCE DEA DEC EAC ECB ECD EDB EDC
AED BDE BED CBE CDE CED DBE DCE DEB DEC EAC ECA ECD EDB EDC
AED BDE BED CBE CDE CED DBE DCE DEB DEC EAC ECB ECD EDA EDC
AED BDE BED CDE CEA CED DAE DBE DCE DEC EBC ECB ECD EDB EDC
AED BDE BED CDE CEB CED DAE DBE DCE DEC EBC ECA ECD EDB EDC
AED BDE BED CDE CEB CED DAE DBE DCE DEC EBC ECB ECD EDA EDC
AED BEC BED CBE CDE CED DAE DBE DCE DEC ECB ECD EDA EDB EDC

Counting the total tuples in this collection of the 130 sets of 15 tuples that met the criteria revealed some interesting insights.  Below is this count, sorted by first position symbol and then by quantity found in the total collection of tuples:

ACE:          20
AEC:          20
ADE:          45
AED:          45
--------------------
BCE:          45
BEC:          45
BDE:          85
BED:          85
--------------------
CAE:          20
CEA:          20
CBE:          45
CEB:          45
CDE:         130
CED:         130
--------------------
DAE:          45
DEA:          45
DBE:          85
DEB:          85
DCE:         130
DEC:         130
--------------------
EAC:          20
ECA:          20
EAD:          45
EBC:          45
ECB:          45
EDA:          45
EBD:          85
EDB:          85
ECD:         130
EDC:         130

The first insight that jumped out to me is clearly there is a reflexive pair pattern.  This makes sense, since effectively one could swap all the symbols in position 1 and 2 to yield the mirror result.

The next insight that surprised me is that subgroups C, D, and E shows tuples pairs that must appear in a valid solution, e.g. CDE/CED, DCE/DEC, ECD/EDC, and some tuple pairs in every sub group that never appear.

To see how the pattern extends, I repeated this process with the same initial collection of 5 permute 3, changing the criteria to A#2 B#3 C#4 D#5 E#6.  This had a possibility of requiring 5.2E12 comparisons, though the optimizations reduced this to only ~1E9 tests required to find 50,202 valid solutions.  The tuple count in the valid solutions, below, shows the same reflective pattern, as well as perhaps some other insights.

ABC:         561
ACB:         561
ABD:       1,144
ADB:       1,144
ABE:       3,131
AEB:       3,131
ACD:       5,140
ADC:       5,140
ACE:      13,792
AEC:      13,792
ADE:      26,434
AED:      26,434
--------------------
BAC:         561
BCA:         561
BAD:       1,144
BDA:       1,144
BAE:       3,131
BEA:       3,131
BCD:       9,889
BDC:       9,889
BCE:      24,351
BEC:      24,351
BDE:      36,227
BED:      36,227
--------------------
CAB:         561
CBA:         561
CAD:       5,140
CDA:       5,140
CBD:       9,889
CDB:       9,889
CAE:      13,792
CEA:      13,792
CBE:      24,351
CEB:      24,351
CDE:      46,671
CED:      46,671
--------------------
DAB:       1,144
DBA:       1,144
DAC:       5,140
DCA:       5,140
DBC:       9,889
DCB:       9,889
DAE:      26,434
DEA:      26,434
DBE:      36,227
DEB:      36,227
DCE:      46,671
DEC:      46,671
--------------------
EAB:       3,131
EBA:       3,131
EAC:      13,792
ECA:      13,792
EBC:      24,351
ECB:      24,351
EAD:      26,434
EDA:      26,434
EBD:      36,227
EDB:      36,227
ECD:      46,671
EDC:      46,671

Research is ongoing to see if I can derive a predictive formula, as well as ongoing optimizations to the exhaustive checks (taking advantage of the mirror reflection at least).


问:
Given the sequence $a_1=1$,$a_{n+1}=1+\frac{n}{a_n}$, does the sequence $a_n+n-a_n^2$ converges?

Given the sequence $a_1=1$,$a_{n+1}=1+\frac{n}{a_n}$, does the sequence $a_n+n-a_n^2$ converges? If it converges, find the limit.

Now, I proved
$\frac{1+\sqrt{1+4(n-1)}}{2}\leq a_n\leq \frac{1+\sqrt{1+4n}}{2}$,
and
$0\leq p_n=a_n+n-a_n^2\leq 1,p_{n+1}+\frac{n}{a_n^2}p_n=1$,which implies $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n-1}$,
but I cannot prove $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n}$ which means the squence $a_n+n-a_n^2$ limit equals to ${1\over2}$

so the question is: how can I prove $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n}$ ?

答:
I don't like my following solution. By giving it, I just want to tell to others that this sequence is indeed convergent (contradicting to what I'm used to think).

I'm still looking forward to seeing other approaches to this question.

Lemma(Some sort of controlled time series)

Let $(a_n)$ and $(b_n)$ be two real sequence such that:

$\lim_{n \rightarrow+ \infty} a_n= \lim_{n \rightarrow +\infty} b_n= 0$

$a_n \ge 0$ and $ \sum_{n \ge 0} a_n =+\infty$

Then for any sequence $(q_n)$ such that:
$$ q_{n+1}=q_n+a_n( b_n-q_n)$$
we have: $$\lim q_n=0$$
$\square$

Back to our main question

Main solution( If your work is true)

We have :
$$ p_{n+1}=1-\frac{n}{a_n^2}p_n=1-\frac{n}{a_n^2}(1-\frac{n-1}{a_{n-1}^2}p_{n-1})=\frac{n(n-1)}{a_n^2a_{n-1}^2}p_{n-1}+(1-\frac{n}{a_n^2})$$
Thus by choosing:

$r_n:=p_{n}-\frac{1}{2}$

$c_n:=1-\frac{n(n-1)}{a_n^2a_{n-1}^2}$

$d_n:= \dfrac{1-\frac{n}{a_n^2} }{1-\frac{n(n-1)}{a_n^2a_{n-1}^2}}-\frac{1}{2}$

We have:

$r_{n+1}= r_{n-1}-c_n( d_n-r_{n-1})$

$\lim c_n= \lim d_n= 0$

$c_n \ge 0$ and $\sum_{k} c_{2k} = \sum_{k} c_{2k+1}=+\infty$

(Note that:  $1- \frac{n}{a_n^2} \sim \frac{1}{\sqrt{n}}$)
From which, we apply my lemma at the very beginning for two sequence $(r_{2n})$ and $(r_{2n-1})$.

Hence, we have the conclusion that :
$$ \lim p_n=\frac{1}{2}$$
$\square$

Comments

Though my solution is not complicated, this question should not be this complicated.

I think this question is a simple one until I have to manually craft a  lemma for it.

It would be nice if Mr.Grape can extend my knowledge by telling me where did you find this question.

I'm still looking forward to nicer solution.

Updated comment

Oh, so this question is not a simple question. And so my lemma provides a general method to deal with this kind of question and even proves (apparently) the convergence when $a_0$ arbitrary. (which is some bit more general than the newled provided link)

ContentLicense: CC BY-SA 4.0


问:
What real number is exactly one less than its cube?
And does it have any of the special properties that the golden ratio (one less than its square) has?</p>

答:
A Pisot–Vijayaraghavan number is an algebraic integer greater than 1 where all its conjucates have modulus strictly less than 1.

The number you are describing is the smallest Pisot–Vijayaraghavan number (proved by Carl Ludwig Siegel here) called plastic number.

The golden ratio $\phi$ is the smallest limit point of the set of Pisot–Vijayaraghavan numbers.

For more see this.

ContentLicense: CC BY-SA 3.0


问:
Closed-form expression for a beta-type sum
I was looking for a closed-form expression for the following sum:

$$
\sum_{k=0}^{n-1}(-1)^kC(n-1,k)\frac{x^k}{a+k}
$$
where $C(n-1,k)=\frac{(n-1)!}{k!(n-1-k)!}$ is the binomial coefficient, $0<x<1$, and $a>1$. In case there is no closed-form expression for this sum, is there any approximation or upper-bound?

Thank you very much in advance.

答:
Without a loss of generality we can write $$S_n(x)=\sum_{k=0}^{n} (-1)^k {n \choose k} \frac{x^k}{a+k}= \sum_{k=0}^{n} (-1)^k {n \choose k} x^k \int_{0}^{1} t^{a+k-1} dt= \int_{0}^{1} t^{a-1}\sum_{k=0}^{n}(-1)^{k} {n \choose k} (xt)^k$$
$$\implies S_n(x)=\int_{0}^{1}t^{a-1} (1-xt)^n dt = \frac{1}{a}~_2F_1(a,-n;1+a;x)$$
the Gauss Hypergeometric function is written as $$~_2F_1(A,B;C,Z)=1+\frac{AB}{C}\frac{x}{1!}+\frac{A(A+1)B(B+1)}{C(C+1)}\frac{x^2}{2!}+...$$
Serveral interesting properties/approximations  of $_2F_1$ can be used from the literature. For example: $$_2F_1(A,B;C;1)=\frac{\Gamma(C)\Gamma(C-A-B)}{\Gamma(C-A)
 \Gamma(C-B)}$$ will give $$S_n(1)= \frac{\Gamma(1+a) \Gamma(1+n)}{a\Gamma(1+a+n)}=B(a,n+1)=\frac{1}{a{a+n \choose n}}$$ B is the $\beta$ function.


问:
Evaluate the line integral of the square C
The four corners of the square C are $(0,0), (0,2), (2,2) , (0,2) $ and the given integral is $$\int_C y^4dx+5x^2dy$$
Parameterizing the square is where I am stuck.

I have $$\left<2t,0 \right> for \space 0 \leq t \leq 1$$

$$\left<2, 2(t-1) \right> for \space 1\leq t\leq 2$$
$$\left<6-2t, 2 \right> for \space 2\leq t\leq 3$$
$$\left<0,0 \right> for\space 3\leq t\leq 4$$

答:
There are two ways to go about this -

Apply Green's theorem if you know about it already

Take individuals paths and do line integral. Given the way vector field is set up and your path, you really have to find integral along two segments only.

The four corners of the square are $A(0,0),B(2,0),C(2,2),D(0,2)$ (assuming there was a typo on the second one. Then, we have a path with counterclockwise orientation.

If you are going by the second method,

Most of your parametrization looked ok but you do not necessarily have to choose increasing values for $t$. These are all individual line segments.

For $AB,$ we have $(0,0) + (2-0, 0-0)t = (2t,0) \,$ for $0 \leq t \leq 1$

For $BC,$ we have $(2,0) + (2-2, 2-0)t = (2,2t) \,$ for $0 \leq t \leq 1$

Similarly for $CD,$ we have $(2-2t,2) \,$ for $0 \leq t \leq 1$

For $DA,$ we have $(0,2-2t) \,$ for $0 \leq t \leq 1$

Now I will take one of these line integrals as an example and you can do the rest and add,

Over path $BC (2, 2t)$,

$\frac{dx}{dt} = 0, \frac{dy}{dt} = 2$

Line integral $I_{BC} = \displaystyle \int_0^1 ((2t)^4 \frac{dx}{dt} + 5(2)^2 \frac{dy}{dt}) dt = \int_0^1 40 \, dt = 40$

Now the line integral for $AB$ and $DA$ will be zero the way vector field is set up (you should check that). So the only other one you will have to find is along $CD$.

The other way is to apply Green's theorem which should make the job easier -

As it is a smooth simple closed curve, we can directly apply Green's theorem according to which, for a counterclockwise oriented curve,

$\int_C P \, dx + Q \, dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA$

Where $A$ is the area of the region. As our region is a square with $0 \leq x \leq 2, 0 \leq y \leq 2, \,$ and

$P = y^4, Q = 5x^2$

Our line integral is converted to

$\displaystyle \int_0^2 \int_0^2 (10x - 4y^3) \, dx \, dy$

which is easy to calculate.


问:
Integrating $\text{sech}(x)$ using a hyperbolic substitution method
I have been tasked to find $\int{\text{sech}(x)dx}$ using both hyperbolic and trig substitutions, for the trig substitution method I did the following.
$$I=\int{\frac{2e^x}{e^{2x}+1}dx} $$ $$\text{Let} \space u=e^x \implies dx=\frac{1}{e^x}du $$
Then applying the first substitution and using the trig substitution of $u=\tan(t)$:
$$\therefore I=\int\frac{2u(\frac{1}{u})}{u^2+1}du \iff \int\frac{2}{u^2+1}du$$
$$\text{Let}\space u=\tan(t) \implies du=\sec^2(t)dt$$
And simplifying:
$$\therefore I=2\int{\frac{\sec^2(t)}{tan^2(t)+1}dt \iff 2\int{1dt}}$$
$$I=2t$$
And finally resubstituting variables to bring it back to in terms of $x$:
$$\because t=\arctan(u) ,  \space u=e^x$$
$$\therefore I=2\arctan(e^x) + c$$

Which checks out on wolfram alpha, however for hyperbolic substitutions I have tried using $u=\text{sinh}(t)$ which just returns the original integral back:

$$\text{Let} \space u=\text{sinh}(t) \iff du=\text{cosh}(t)dt$$
$$\therefore I=2\int{\frac{\text{cosh}(t)}{\text{sinh}^2(t)+1}dt} \iff 2\int{\frac{1}{\text{cosh}(t)}dt}$$

I have also tried using the substitution of $u=\text{csch}(t)$ which also led back to the original integral, too my knowledge there happen to be no other useful hyperbolic substitutions to carry out on this integral.

Have I made a mistake in my integration or am I missing some other useful substitution which may be carried out here?

答:
Note that for any $ x\in\mathbb{R} $, $ \cosh{x}=\frac{\cosh{\left(2\times\frac{x}{2}\right)}}{1}=\frac{\cosh^{2}{\left(\frac{x}{2}\right)}+\sinh^{2}{\left(\frac{x}{2}\right)}}{\cosh^{2}{\left(\frac{x}{2}\right)}-\sinh^{2}{\left(\frac{x}{2}\right)}}=\frac{1+\tanh^{2}{\left(\frac{x}{2}\right)}}{1-\tan^{2}{\left(\frac{x}{2}\right)}} $.

Substituting $ \left\lbrace\begin{matrix}y=\tanh{\left(\frac{x}{2}\right)}\\ \mathrm{d}x=\frac{2\,\mathrm{d}y}{1-y^{2}}\ \ \ \ \ \ \ \ \ \ \end{matrix}\right. $, we get : \begin{aligned}\int{\frac{\mathrm{d}x}{\cosh{x}}}&=2\int{\frac{\mathrm{d}y}{1+y^{2}}}\\&=2\arctan{y}+C\\&=2\arctan{\left(\tanh{\left(\frac{x}{2}\right)}\right)}+C\end{aligned}


问:
The Monte Carlo Method For Simulation Example Problem
Q: A person has 5 neckties and randomly selects one tie each workday.
In a given workweek of 5 days, find the probability that the person
will wear the same tie two or more days a week.

A: Use the numbers 1, 2, 3, 4, and 5 on a die to represent the ties.
Ignore any sixes. Roll the die five times for each week and count
any time you get the same number twice as selecting the same tie
twice.

In the context of answer, is counting means writing down a number(tie) that has appeared multiple times in each week or counting how many times a certain number(tie) has appeared?

答:
In your stated problem, "counting" is counting weeks where any tie was chosen twice.

After you do your trials, rolling the dice for X number of week sequences, you will end up with a number of weeks that a tie was chosen twice / number of week sequences you ran for your Monte Carlo series, and this value will be your estimated probability that in a future week a tie will be chosen twice.

The idea of Monte Carlo simulations is that when you have some sort of random set of events that are too large (or perhaps infinite) to exhaustively calculate all finite probabilities, so by running a number of random tests you can start to approximate the probability.   The larger your number of tests, the better the approximation is.

Your particular assigned problem is pretty straightforward to calculate exhaustively, though to list out the 3125 different permutations would be a bit tedious.  What the assignment is trying to show you is how many "weeks" you have to simulate to start to approach the final probability.


问:
Relation between chords length and radius of circle
Two chords of a circle, of lengths $2a$ and $2b$ are mutually perpendicular. If the distance of the point at which the chords intersect,from the centre of the circle is $c$($c<$radius of the circle),then find out the radius of the circle in terms of $a,b$ and $c$.Show some short-cut to do it quickly</p>

答:
Let $P$ be the point where the two chords (and a diameter) meet. Let $h$ (and $k$) be the distance from $P$ to the midpoint of the $2a$ chord (respectively, the $2b$ chord); that is, say $P$ divides the chord into sub-segments of length $a+h$ and $a-h$ (respectively, $b+k$ and $b-k$). Note that $P$ divides a diameter into sub-segments of length $r+c$ and $r-c$ (where $r$ is the radius of the circle); note also that $c$ is the hypotenuse of a right triangle with legs $h$ and $k$: so, $c^2 = h^2 + k^2$.

The Power of a Point principle says that every chord through a particular point of a circle is divided into sub-segments such that the product of the lengths of those sub-segments is a constant (the so-called "power" of the point in question). Thus,

$$(a+h)(a-h) = (b+k)(b-k) = (r+c)(r-c)$$

More succinctly,

$$a^2 - h^2 \;\;=\;\; b^2 - k^2 \;\;=\;\; r^2 - c^2$$

With an eye towards combining an $h^2$ with a $k^2$, I'll add the left-hand and "middle-hand" sides together; their sum is necessarily twice the right-hand side:

$$\begin{align}
( a^2 - h^2 ) + ( b^2 - k^2 ) &= 2 (r^2 - c^2) \\
\implies a^2 + b^2 - ( h^2 + k^2 ) &= 2 r^2 - 2 c^2 \\
\implies a^2 + b^2 - c^2 &= 2 r^2 - 2 c^2 \\
\implies a^2 + b^2 + c^2 &= 2 r^2
\end{align}$$


问:
Prove that there exists a unique differentiable function
Prove that $\exists$ a unique differentiable function $f: R \to R$ s.t. $f'(x) = e^{x^2} \forall x \in R$ and $f(0) = 0$

I'm having trouble showing the uniqueness part. This is what I have so far.
Write
$F(x) = \int_{a}^{x} e^{t^2} dt$

Choose $a = 0$

$F(x) = \int_{0}^{x} e^{t^2} dt$

We are told $F(0) = 0$, so

$F(0) = 0 = \int_{0}^{0} e^{t^2} dt = 0$

So the integral exists. Now we must show it is unique. I am really not sure where to go from here. I think the $e^{x^2}$ is throwing me off because there is no anti-derivative. So I'm not sure how I would show it is unique here. Any help is appreciated!

答:
Assume $G$ is another such function. Then $H:=G-F$ is differentiable with $H'\equiv 0$ and $H(0)=0$.
Then $H(x)=0$ for all $x$. Indeed, by the Mean Value Theorem, there exists $\xi $ between $0$ and $x$ such that $\frac{H(x)-H(0}{x-0}=H'(\xi)=0$.


问:
Show that the following function defined by an integral is uniformly continuous
Consider $\phi: C([a,b]) \to R$ given by
$$f \to \int_{a}^ b f dx$$
Show that $\phi$ is uniformly continuous.

Not really sure where to start here. It seems like I'm given very little. I'm given information about f, not $\phi$, and $f$ is not equal to that integral, it is mapped to there because of the arrow. So I'm not really sure what to do with what is given. Any help is appreciated!

答:
The inequality

$$\vert \phi(f) - \phi (g) \vert =\left\vert \int_a^b \ f(x)\ dx - \int_a^b \ g(x)\ dx \right\vert \le (b-a) \Vert f - g \Vert$$

where $$ \Vert f-g\Vert= \sup_{x \in [a,b]} \vert f(x) -g(x) \vert$$

proves that $\phi$ is $(b-a)$-Lipschitz and is sufficient to get your desired conclusion. Providing that $\mathcal C([a,b], \mathbb R)$ is equipped with the $\sup$ norm.


问:
Why is it meaningless for a closed set to be polygonal path connected?
My textbook (Complex Analysis by Saff & Snider) defines connectedness for open sets; the given definition of a connected open set is: a set in which every pair of points can be joined by a polygonal path that lies entirely in the set.

Using the given definition of connected set, I don't understand why it isn't similarly defined for closed sets too?

答:
Because there is a difference between path connected and conected. And there is even a difference between polygonal connected and path connected. For example taking the set
\[ \{ z \in \mathbb{C}: |z|=1\}\]
has no polygonal paths connecting two points. But you intuitively says that this set should be connected. Connected in the pure topologic version sound this way:

A Topologic Space $(T,\tau)$ is called connected when there aren't any nonempty sets $O_1 \in \tau, O_2\in \tau$ with $O_1\cap O_2=\varnothing$ and $O_1 \cup O_2=T$. A subset
  of a Topological space is connected when it is connected as a topological space with the trace topology.

Every path connected space is connected but not every connected space is path connected.
In $\mathbb{R}^n$ path connection, polygonal path connection and connection fall together for open sets, they are all equivalent when you know your set is already open.

This defintion is for sure much more complicated than your special case, and as you
are interested in complex analysis about meromorphic functions most time you usually only use open sets anyway. so he wanted to avoid a complicated definition by using the special case of a more general one.


问:
Prove that determinant is zero iff $x= 2, 3$

Prove that
$$\begin{vmatrix}
1 & x & x^{2}\\
1 & 2 & 2^{2}\\
1 & 3 & 3^{2}
\end{vmatrix}=0\Leftrightarrow x= 2, 3$$

I see that if determinant is zero, two lines must equal, in this problem, there are between line 1 and line 2, 3, but how can I prove there is no other solutions ?? And what for $x^{3}, x^{4}, etc$ instead ?? Thks

答: The Leibniz formula for determinants implies that the determinant in question is a quadratic polynomial in $x$. Since you have found two different roots, they are all the roots.


问:
Random probability for internet connection
An internet service provider has two connection lines for its customers.

$80\%$ of customers are connected through Line $I$, the rest through Line $II$.

Line $I$ has a $\operatorname{Gamma}(3, 1/2)$ connection time (in minutes), while Line $II$ has a $U(20, 50)$ connection time (in seconds).

Find the probability that it takes a randomly selected customer more than $30$ seconds to connect to the internet.

Can somebody give me some ideas, please?

答:
What have  you tried? It is simply to apply Total Probability theorem.

$$0.80\times p_1+0.20\times p_2$$

where $p_1,p_2$ are calculated with the given probability distributions.

note that Gamma distribution can be expressed in terms of $\chi^2$ to get a numerical result.

Please show also what do you mean as Gamma density because there is more than one parametrization.


答:
$$p(x) = \begin{vmatrix}
1 & x & x^{2}\\
1 & 2 & 2^{2}\\
1 & 3 & 3^{2}
\end{vmatrix}= (x-2)(x-3)$$

Why? See that $p(x)$ has degree at most $2$, and it has two roots $2,3$ (by inspection). So it must be of the form $\alpha(x-2)(x-3)$ for some $\alpha \in \mathbb{R}$. $\alpha = 1$, but you don't really need this information to conclude that $2$ and $3$ are the only roots of $p(x)$, since we know that any polynomial $f(x)$ has roots $x_1,...,x_n$, iff $\lambda f(x)$ has roots $x_1,...,x_n$ for some scalar $\lambda\in\mathbb{R}$.

It should be clear now!

As mentioned in the comments, you can look at Vandermonde determinants for a generalization.

Title: Prove that determinant is zero iff $x= 2, 3$

Body:
Prove that
$$\begin{vmatrix}
1 & x & x^{2}\\
1 & 2 & 2^{2}\\
1 & 3 & 3^{2}
\end{vmatrix}=0\Leftrightarrow x= 2, 3$$

I see that if determinant is zero, two lines must equal, in this problem, there are between line 1 and line 2, 3, but how can I prove there is no other solutions ?? And what for $x^{3}, x^{4}, etc$ instead ?? Thks


问:
Series and Sequence converges or diverges
Define the sequence $a_n$ as $a_n=1/2$ as $n $ is a multiple of $3$ or $a_n=1/n^2$ otherwise.
Determine whether $\sum_{1}^{\infty}a_n$ and $S_n$ converges or diverges, where $S_n$ is the sequence of partial sums corresponding to $a_n$.

For this question may I just use the test of divergence to show it?
Like series diverges, since $\lim_{n \rightarrow \infty} a_n \ne0$ or DNE.

答:
Yes, if you pick $0<\epsilon<\frac{1}{2}$, the for all $n\in \mathbb{N}$, pick the index $3n>n$, the term $a_{3n}=\frac{1}{2}$, so $|a_{3n}|>\epsilon$, this says that $\lim~a_n\neq 0$.

That`s enough to say that $S_n$ diverge, by the contrapositive of the Theorem that says that if a series converge, then the sequence of terms goes to zero.


问:
Prove that $\det(\lambda I-A)\le\lambda^n-1$
Let $A$ be an $n\times n$ nonnegative matrix with spectral radius $1$. Suppose $\lambda>1$, then prove $\det(\lambda I-A)\le\lambda^n-1$.

答:
There should be a simpler proof, but at least the inequality can be proved by applying Jacobi's formula twice.

Suppose $A\ge0$ (i.e. it is entrywise nonnegative) and $\rho(A)=1$. In the following list, each statement is either equivalent to or weaker than the subsequent one. Therefore, it suffices to prove the last statement in the list.

$\det(xI-A) \le x^n-1$ for all $x>1$.

$\frac{d}{dx}\det(xI-A) \le \frac{d}{dx}(x^n-1)$ for all $x>1$. This implies statement 1 because $\det(xI-A)=0=x^n-1$ when $x=1$.

$\operatorname{tr}\big(\operatorname{adj}(xI-A)\big) \le nx^{n-1}$ for all $x>1$. This is equivalent to statement 2, by Jacobi's formula.

$\det(xI-B) \le x^{n-1}$ for all $x>1$ and all principal $(n-1)\times(n-1)$ submatrices $B$ of $A$.

$\det(I-tB)\le1$ for all $0<t<1$ and all nonnegative matrices $B$ with $\rho(B)\le1$.

$\det(I-tB)$ is decreasing on $(0,1)$ when $B\ge0$ and $\rho(B)\le1$. This implies the previous statement because $\det(I-tB)=1$ when $t=0$.

$\operatorname{tr}\big(\operatorname{adj}(I-tB)\,(-B)\big) \le 0$ for all $0<t<1$ when $B\ge0$ and $\rho(B)\le1$. This is equivalent to statement 6, by Jacobi's formula.

$\operatorname{tr}\big(\operatorname{adj}(I-tB)\,(I-tB)\big) \le \operatorname{tr}\big(\operatorname{adj}(I-tB)\big)$ for all $0<t<1$ when $B\ge0$ and $\rho(B)\le1$.

$\det(I-tB) \le \frac1m\operatorname{tr}\big(\operatorname{adj}(I-tB)\big)$ for all $0<t<1$ when $M_m(\mathbb R)\ni B\ge0$ and $\rho(B)\le1$.

$\det(I-tB) \le \sqrt[m]{\prod_{i=1}^m\big(\operatorname{adj}(I-tB)\big)_{ii}}$ for all $0<t<1$ when $M_m(\mathbb R)\ni B\ge0$ and $\rho(B)\le1$. Note that the conditions on $B$ imply that each principal minor of $I-tB$ is positive. Therefore the inequality here implies statement 9, by the AM-GM inequality.

$1\le\prod_i\big(\frac{\operatorname{adj}(I-tB)}{\det(I-tB)}\big)_{ii}$ for all $0<t<1$ when $B\ge0$ and $\rho(B)\le1$.

$1\le\prod_i \big((I-tB)^{-1}\big)_{ii} $ for all $0<t<1$ when $B\ge0$ and $\rho(B)\le1$.

Now statement 12 is true because all diagonal elements of $(I-tB)^{-1}=I+tB+(tB)^2+(tB)^3+\cdots$ are bounded below by $1$.


答:
Why not to expand the determinant and find all the solutions?
$$\begin{vmatrix}
1 & x & x^{2}\\
1 & 2 & 2^{2}\\
1 & 3 & 3^{2}
\end{vmatrix} =
\begin{vmatrix}
1 & x & x^{2}\\
0 & 2-x & 2^{2}-x^2\\
0 & 1 & 5
\end{vmatrix} =
\begin{vmatrix}
2-x & 2^{2}-x^2\\
1 & 5
\end{vmatrix}$$

$$=10-5x-(4-x^2) = x^2-5x+6=(x-2)(x-3)$$
Which equals to zero iff $x=2,3$


问:
How to optimize a rational function
Just a calculus problem:

As a function of $K \geq 1$, what is the minimum value of $f/a + f/b + f/c + f/d + f/e$ subject to the following constraints? $$\begin{cases} 1 \leq a \leq c \\ 1 \leq b \leq c \\ 1 \leq d \\ 1 \leq e \\ f = \frac{a^2 b^2 d e}{c} \\ f = K \end{cases}$$

I am fine with a reasonably detailed "here is how you do this calculus" answer, or "here is how to ask wolfram alpha/maple" answer (that works).  I need to be able to handle variations (the formula for f is always a "monomial" but the powers on the variables can change, and the inequalities amongst the $a,b,c,d,e$ variables might change slightly, though all of them are always at least 1).

A version I can do:  As a function of K ≥ 1, find the minimum value of $f/a + f/b + f/c + f/d + f/e$ subject to the following constraints: $$\begin{cases} 1 \leq a,b,c,d,e \\ f = abcde \\ f = K \end{cases}$$

This version is highly symmetric and I basically understand the region I am optimizing over. Setting any variable to 1 results in a highly non-optimal solution, so the minimum occurs in the middle of the surface where abcde = K, and so the gradient of the objective function is a scalar multiple of the normal to the surface.  Both are very symmetric and the algebra involved in solving them is almost silly.  The answer is the expected $a=b=c=d=e=K^{1/5}$ due to symmetry.

Motivation: In the background, $a,b,c,d,e,f$ are all positive odd integers describing the structure of an unknown group.  In the previous incarnations of this problem, I assumed they were real numbers bounded below by 1, and the calculus minimum was in fact the group theory minimum.

On the new problem, I asked maple to give the unconstrained (except for the "f=" constraints) problem a shot, and it claims there is only one local extrema, and it involves a lot of negative numbers.  I guess that means the minimum is at a "corner" (discontinuity of the function defining the boundary of the feasible region), but I have no idea what that means in more than 2 dimensions, and I am a little nervous that such an answer is wrong, at least from the group theory standpoint.

答:
Now solved (more or less).  In case anyone wants to see it:

Solution as stated: First off, as stated the problem is likely silly.  Set a=b=c=d=e=∞, f=1, (so make sure c goes to infinity at the right rate), and the minimum is 0.

Corrections: As I tell my students: it is possible for a real life problem to be unbounded, but usually it just means someone forgot a constraint.

The forgotten constraints are that a,b,c,d,e ≤ f.  What a difference this makes!

Also while I'm being honest, the objective function was wrong too!  Obj = 2f/a + 2f/b + f/c + f/d + f/e.  I forgot some 2s.  This will repair some symmetry in the answer.

Solution:

Now we have a compact polytope to start with: Fix f=K.  Now 1 ≤ a ≤ c ≤ f, 1 ≤ b ≤ c ≤ f, 1 ≤ d ≤ f, 1 ≤ e ≤ f.  Of course we slice a hypersurface out of this: f = aabbde/c.

The resulting set is compact and so the minimum will really be obtained.  It could happen in two kinds of places: in the "middle" of the hypersurface, so that the polytope doesn't matter, or on the boundary of the polytope, so that in some weak sense the hypersurface doesn't matter.

Which is it?  If it were in the middle of the hypersurface, then Lagrange multipliers would apply:

∇obj = ( -2f/(aa), -2f/(bb), -f/(cc), -f/(dd), -f/(ee) )

∇sur = ( 2abbde/c, 2aabde/c, -aabbde/(cc), aabbe/c, aabbd/c )

The direction of greatest decrease in the objective function should be normal to the hypersurface, and so there is some real number λ such that: λ ∇obj = ∇sur.

We check what this means in the first and third components.  On the one hand, -2λf/(aa) = 2abbde/c so that -λf/a = aabbde/c = f, and so a = -λ.  On the other hand, -λf/(cc) = -aabbde/(cc) so that -λf/c = -aabbde/c = -f, and c = λ.  In particular, one cannot have both a and c are positive!

This contradicts our polytope constraints, and so the minimum occurs on the boundary.  Where?  Well, this is a little shifty, but I think it is "clear" that the inequality that needs to become an equality is c ≤ f.  We want c big, so that a,b,d,e can be big.

Ok, so set c=f and redo the problem:

For fixed f, minimize obj = 2f/a + 2f/b + f/d + f/e + 1 subject to each variable a,b,d,e bounded between 1 and f, and subject to ff = aabbde.

We again have the intersection of a polytope (a hypercube!) with a hypersurface, so we check for middle extrema first:

∇obj = (-2f/(aa), -2f/(bb), -f/(dd), -f/(ee) )

∇sur = ( 2abbde, 2aabde, aabbe, aabbd )

There must be a number λ such that λ∇obj = ∇sur.  We examine the components: -2fλ/(aa) = 2abbde, -fλ/a = f, and a = -λ.  Similarly b = -λ.  Then -fλ/(dd) = aabbe, -fλ/d = aabbde = f, and again d = -λ.  Similarly e = -λ.

Hence our solution is:

$$ a=b=d=e = f^{(1/3)}, c=f, \min = 6f^{(2/3)} + 1$$

Caveat: One should be a lot more cautious checking the boundary, but this answer agrees with my preconceived notion and so I am happy for now.

Title: How to optimize a rational function


答:
The connectedness at here is path connectedness, which means any two points can be connected by a path traveling across the domain. For open sets it is not difficult to show polygon path connectedness and path connected is the same, because any polgyon path is a path, and any path can be approximated by a polygon path given the domain is open. Using some topology you can show path connected is the same as connected for open sets.

For closed sets the situation is different. What we normally regard as "connected" can differ from "path connected". For example the graph of $\sin[\frac{1}{x^{2}}]$ with $0$ when $x=0$ is connected, but not path connected. So complications arise this way. The book probably tried to sweep these issues under the rug when it introduce the concept to a first time learner. For more details on this you are welcome to read this:

http://en.wikipedia.org/wiki/Topologist%27s_sine_curve

Title: Why is it meaningless for a closed set to be polygonal path connected?

Body:
My textbook (Complex Analysis by Saff & Snider) defines connectedness for open sets; the given definition of a connected open set is: a set in which every pair of points can be joined by a polygonal path that lies entirely in the set.

Using the given definition of connected set, I don't understand why it isn't similarly defined for closed sets too?


问:
Probability Q on rolling two 8-sided dice
Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?

So this is my answer and please correct me if I am wrong:

Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%

Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%

But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?

Thanks!

EDIT

From the answers provided below, we have three different results:

84%
49%
60%

Which one would be the correct answer?

答:
Here is another approach, this time using an exponential generating function.  Readers interested in learning about generating functions can find many resources in the answers to this question: How can I learn about generating functions?

We want to find the probability that a sequence of rolls by two 8-sided dice contains at least one 7 and at least one 11 before it contains two 9s.  For now, let's assume that the final roll is a 7.  Since 11 has the same probability of 7, the probability of ending in 11 is the same as the probability of ending in 7.

Consider a successful sequence of rolls just before rolling the final 7.  Such a sequence must contain at least one 11, no 7, at most one 9, and any number of other rolls (not 7, 9, or 11).  The probability of rolling a 7 is $6/64$, the probability of rolling an 11 is $6/64$, the probability of rolling  9 is $8/64$, and the probability of rolling anything else is $44/64$. So the exponential generating function for the probability of such a sequence is
$$f(x) = (e^{(6/64) x} -1) \;(1+ (8/64)x)\; e^{(44/64)x}$$
Then the next roll must be a 7.  So the EGF for a successful sequence of $n+1$ rolls ending in 7 is
$$g(x) = (e^{(6/64) x} -1) \;(1+ (8/64)x)\; e^{(44/64)x}\;(6/64) \tag{1}$$
That is to say,
$$g(x) = \sum_{n=0}^{\infty} p_n \frac{x^n}{n!}$$
where $p_n$ is the probability of a successful series of $n+1$ rolls ending in 7. We want to know the sum of all the $p_n$'s. The trick here is that
$$\sum_{n=0}^{\infty} p_n = \int_0^{\infty} e^{-x}g(x) \;dx \tag{2}$$
which is based on the identity
$$\int_0^{\infty} e^{-x} x^n \;dx = n!$$
Substituting $(1)$ into equation $(2)$ and evaluating the integral, we find
$$\sum_{n=0}^{\infty} p_n = \frac{621}{2450}$$
This is the probability of a successful sequence of rolls ending in 7. The probability of a successful sequence ending in either 7 or 11 is twice that,
$$\frac{621}{1225}$$


答:
Assuming by the "test of divergence" you mean this test:

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

then yes, quoting this theorem and explaining why your series satisfies the conditions of this theorem is this is sufficient to show that the series diverges.

That is, assuming you're teacher is allowing you to quote such a test in your proofs...

Title: Series and Sequence converges or diverges

Body:
Define the sequence $a_n$ as $a_n=1/2$ as $n $ is a multiple of $3$ or $a_n=1/n^2$ otherwise.
Determine whether $\sum_{1}^{\infty}a_n$ and $S_n$ converges or diverges, where $S_n$ is the sequence of partial sums corresponding to $a_n$.

For this question may I just use the test of divergence to show it?
Like series diverges, since $\lim_{n \rightarrow \infty} a_n \ne0$ or DNE.


问:
If $Y_n\rightarrow c$ in probability does $E(\lvert Y_n - c\rvert)\rightarrow 0$?
Let $Y_n$ be real-valued, $c$ a real constant.

If $Y_n\rightarrow c$ in probability, does $E(\lvert Y_n - c\rvert)\rightarrow 0$?

I need this in a larger proof. It seems obvious, but that is often misleading...
I tried
$$E(\lvert Y_n-c\rvert) = E(\lvert Y_n-c\rvert1_{\lvert Y_n-c\rvert > \delta}) + E(\lvert Y_n-c\rvert1_{\lvert Y_n-c\rvert \le \delta}) \le E(\lvert Y_n-c\rvert1_{\lvert Y_n-c\rvert > \delta}) + \delta$$
but I'm having trouble controlling the first term.

How can it be argued?

答: In general no. As a simple example, consider $Y_n=c$ with probability $1-1/n$ and $c+n$ with probability $1/n$ then $E(|Y_n-c|) \equiv 1$.


问:
Isomorphism between $V^* \otimes V$ and End$(V)$
I am studying Humphreys "Introduction to Lie algebras and Representation theory" and I am stuck with a remark in fact concerning linear algebra more (see a piece of text below).

I need to show that for $V$ finite dimensional vector space it is true that $V^* \otimes V$ is isomorphic to End$(V)$.
In Humphreys they suggest to show that a map sending a generator $f \otimes v$ to an endomorphism $w \mapsto f(w)v$  is an epimorphism, i.e. that $ \forall a\in End(V) \ \exists \  b \in V^* \otimes V$ such that it maps to $a$, and then use the fact, that spaces have equal dimensions.

I am having troubles with tensor products, and maybe this is the problem of my attempt to show this, but here is what I tried: let $\operatorname{dim}V = n$, and $\{e_j\}_{j=1}^n$ be a basis of $V$. For every $a\in End(V)$ I need to find an element of $V \otimes V^*$, that maps to it. Every element of $V^* \otimes V$ can be written as $\sum_{i} \alpha_{i_1, i_2} f_{i_1} \times v_{i_2}$ (right?), so we need:
$$\sum_{i} \alpha_i f_i(w)v_i = a(w)$$
We write everything in coordinates:  $v_i=\sum v_i^{(k)} e_k$, $w = \sum w_j e_j$, $f_i=\sum f_i^{(j)} e^j$ (standard dual basis associated to $\{e_j\}_{j=1}^n$); then we obtain a large sum:
$$\sum_{i_1,i_2,j,k} \alpha_{i_1,i_2} f_{i_1}^{(j)} w_j v_{i_2}^{(k)} e_k = \sum_{j,k} a_{j,k} w_j e_k$$
Hence
$$\sum_{i_1,i_2,j} \alpha_{i_1,i_2} f_{i_1}^{(j)} w_j v_{i_2}^{(k)}= \sum_{j} a_{j,k} w_j, \ \forall k$$
which has to be true for every $w$. That's why we need
$$\sum_{i_1,i_2} \alpha_{i_1,i_2} f_{i_1}^{(j)} v_{i_2}^{(k)}= \sum_{j,k} a_{j,k}, \ \forall k,j$$
which seems to be true, because there so many things I can control (I mean coefficients $a_{j,k}$ are the only one given),but I don't know how to justify it properly.

Also, if there is a simpler way to do this, I would like to know.

答:
I guess you know that the $e^i \otimes e_j$ with $i, j = 1,\ldots n$, form a basis of $V^* \otimes V$. You send $e^i \otimes e_j$ to
$$e_{ij} : V \to V, e_{ij}(w) = e^i(w)e_j = w_ie_j ,$$
where $w = \sum_{k=1}^n w_k e_k$.

A linear map $f : V \to V$ is determined by its values on the basis $e_1,\ldots, e_n$. Writing $f(e_i) =  \sum_{j=1}^n f_{ij}e_j$ we get
$$f(w) = \sum_{i=1}^n w_if(e_i) =  \sum_{i=1}^n w_i(\sum_{j=1}^n f_{ij}e_j) = \sum_{i,j=1}^n f_{ij} w_ie_j = \sum_{i,j=1}^n f_{ij} e_{ij}(w) .$$
This shows that $f$ is a linear combination of the $e_{ij}$ and shows that your assigment is an epimorphism.

Alternatively you can also directly show that the $e_{ij}$ are linearly independent and therefore form a basis of $\text{End}(V)$. Let $\sum_{i,j=1}^n f_{ij} e_{ij} = 0$.  We have $e_{ij}(e_k) = \delta_{ik}e_j$ with the usual $\delta_{ik}$-notation. Then for each $k$
$$(\sum_{i,j=1}^n f_{ij} e_{ij})(e_k) = \sum_{i,j=1}^n f_{ij} \delta_{ik}e_j = \sum_{j=1}^n f_{kj} e_j = 0 ,$$
thus $f_{kj} = 0$ for $j= 1, \ldots, n$.


答:
Polygonal connectedness (which is a stronger condition than topological connectedness) can be defined for any subset of the complex plane. The definition will be, as you'd expect, that a set $A\subseteq \mathbb C$ is polygonally connected if any two points in it can be connected by a polygonal path inside $A$. </p>

<p>The authors simply did not bother with the most general possibility. They will mostly care about subsets of the plane that are open and polygonally connected. The reason is that in complex analysis such subsets are very important. Openness is important because you want to be able to talk about a function being holomorphic in the given region, and if the region is not open this leads to annoying technical difficulties. Polygonal connectedness is important because it means any function defined on the region can be integrated between any two points in the region. These are things that happen often in complex analysis, and so you will see lots and lots of theorems that say something like "if $f$ is holomorphic on an open polygonally connected set, then ....". </p>

Body:
Polygonal connectedness (which is a stronger condition than topological connectedness) can be defined for any subset of the complex plane. The definition will be, as you'd expect, that a set $A\subseteq \mathbb C$ is polygonally connected if any two points in it can be connected by a polygonal path inside $A$.

The authors simply did not bother with the most general possibility. They will mostly care about subsets of the plane that are open and polygonally connected. The reason is that in complex analysis such subsets are very important. Openness is important because you want to be able to talk about a function being holomorphic in the given region, and if the region is not open this leads to annoying technical difficulties. Polygonal connectedness is important because it means any function defined on the region can be integrated between any two points in the region. These are things that happen often in complex analysis, and so you will see lots and lots of theorems that say something like "if $f$ is holomorphic on an open polygonally connected set, then ....".


问:
$W=\left( W\cap V_1 \right) \oplus \left( W\cap V_2 \right) $ if and only if $W=\left( W+V_1 \right) \cap \left( W+V_2 \right) $
Let $V_1,V_2$ be two subspaces of the finite dimensional vector space $V$, such that $V=V_1\oplus V_2$. Let $W$ be any subspace of $V$.
Prove that, $W=\left( W\cap V_1 \right) \oplus \left( W\cap V_2 \right) $ if and only if $W=\left( W+V_1 \right) \cap \left( W+V_2 \right) $.

I have showed that $\left( W+V_1 \right) \cap \left( W+V_2 \right)$ is a subspace of $W$. But how to proceed further?

答:
“$\Rightarrow$”: Given $W=(W\cap V_1)\oplus (W\cap V_2)$ we want to show $W=(W+V_1)\cap (W+V_2)$. Let $W_i=W\cap V_i$ and choose complements $U_i$ so that $V_i=W_i\oplus U_i$. Thus we have decomposed $V$ as
$$
V = \underbrace{W_1 \oplus U_1}_{V_1} \oplus \underbrace{W_2\oplus U_2}_{V_2} = \underbrace{W_1\oplus W_2}_W \oplus \underbrace{U_1 \oplus U_2}_U.
$$
Hence, every vector $v\in V$ has a unique decomposition as
$$
v = w_1 + u_1 + w_2 + u_2
$$
with $w_i\in W_i$ and $u_i\in U_i$. The elements of $W+V_1$ are those with $u_2=0$ and the elements of $W+V_2$ are those with $u_1=0$. Hence the intersection consists of those elements with $u_1=u_2=0$, which are exactly the elements of $W$.

“$\Leftarrow$”: In the other direction, we are given $V=V_1\oplus V_2$ and $W=(W+V_1)\cap(W+V_2)$. We want to show that $W=W_1\oplus W_2$, where $W_i=W\cap V_i$ as before. Since $W_1,W_2\subseteq W$, we have $W_1+W_2\subseteq W$. Furthermore, the sum is direct since
$$
W_1\cap W_2 \subseteq V_1\cap V_2 = \{0\}.
$$
To show that $W_1\oplus W_2$ is not a proper subspace of $W$ but in fact equal to $W$, we need the finite dimensionality of the involved spaces: From $W_i=W\cap V_i$ we have the following identities of dimension:
\begin{align*}
\dim(W+V_i) = \dim(W) + \dim(V_i) - \dim(W_i).
\end{align*}
Since we are given $W=(W+V_1)\cap(W+V_2)$ this yields
\begin{align*}
\dim(W) &= \dim(W+V_1) + \dim(W+V_2) - \dim(\underbrace{W+V_1+W+V_2}_V)
\\&= \dim(W) + \dim(V_1) - \dim(W_1)
\\&\quad+ \dim(W) + \dim(V_2) - \dim(W_2)
-\dim(V)
\\&= \dim(W) - \dim(W_1)
+ \dim(W)  - \dim(W_2).
\end{align*}
This is equivalent to $\dim(W) = \dim(W_1)+\dim(W_2) = \dim(W_1\oplus W_2)$ and hence $W=W_1\oplus W_2$.


问:
Range of quadratic function using discriminant

Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.

Let $2x^2-2xy+y^2=a$.

Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.

We divide both side by $y^2$ and let $t=\frac{x}{y}$.

Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.

Since its discriminant is not negative, $\frac{\Delta}{4}\ge 0\implies a^2-7a-4\ge 0$. It gives us $a$ can have negative values like $-1$. But if clearly contracts $a-4=x^2+4y^2\ge 0\implies a\ge 4$. Where did I mistake?

答:
Method$\#1:$

For real $t,$  $$a^2-7a-4\ge0$$

If  $x=ty$

$$4=x^2-2xy-3y^2=y^2(t^2-2t-3)$$

$$t^2-2t-3=\dfrac4{y^2}>0$$

$$\implies(t-3)(t+1)>0$$

Either $t<-1$ or $t>3$

So, the values of $a$ must satisfy this condition as well

Method$\#2:$

$$4=(x-y)^2-(2y)^2$$

WLOG $y=\tan t, x-y=2\sec t\implies x=2\sec t+\tan t$

$$a=2(2\sec t+\tan t)^2-2(2\sec t+\tan t)\tan t+\tan^2t$$

Multiplying both sides  by $\cos^2t=1-\sin^2t,$

$$(1+a)\sin^2t+4\sin t+8-a=0$$

What if $a+1=0?$

Else

$$\sin t=\dfrac{-2\pm\sqrt{a^2-7a-4}}{a+1}$$

As $\sin t$ is real, the discriminant must be $\ge0$

But that is not sufficient, we need $$-1\le\dfrac{-2\pm\sqrt{a^2-7a-4}}{a+1}\le1$$ as well for real $t$

Also, $\dfrac xy=2\csc t+1$  $\implies\dfrac xy\ge2+1$ or $\dfrac xy\le-2+1$


问:
Prove that $\gcd(a, b) = \gcd(a, b, ax + by)$ for any given integers $a, b, x, y$
How can I prove that $\gcd(a,b) = \gcd(a,b,ax+by)$?

I know that $\gcd(a,b)$ can be written in linear equation like this:
$$ \gcd(a,b) = ax + by $$
Does that mean that I can substitute it to $\gcd(a,b,ax+by)$?
If yes, what's next?

答:
First you should look at the hints and try to do it yourself, believe me if will feel much better if you do it by yourself.

But if you can't figure it out here is the solution.

 First thing you should notice is that $gcd(a,b,ax+by) \leq gcd(a,b)$

 Let $gcd(a,b)=g$ then $g|a,g|b \Rightarrow g|(ax+by)$

 Hence $gcd(a,b)=gcd(a,b,ax+by)$


问:
Determine the common period
Let $V = \mathrm{Vect}( e^{ 4 \pi it } , e^{ 5 \pi i t} , e^{ 6 \pi i t} ) $ and $f \in V$.

I want to determine the common period between the functions : $t \mapsto e^{ 4 \pi it } $, $t \mapsto e^{ 5 \pi i t }$ and $t \mapsto e^{ 6 \pi i t}$ ; i.e the period of $f \in V$.

We have

$ t \mapsto e^{ 4 \pi i t} $ is periodic with period $\frac{1}{2}$

$ t \mapsto e^{ 5 \pi i t} $ is periodic with period $\frac{2}{5}$

$ t \mapsto e^{ 6 \pi i t} $ is periodic with period $\frac{1}{3}$

and I don't know how to pursue it. Is there a particular method to found a common period? I need a hint.

Thank you

答:
Hint: You can use that

$$\operatorname{lcm}\biggl(\frac pq,\frac rs\biggr)=\frac{\operatorname{lcm}(p,r)}{\gcd(q,s)}.$$

Title: Determine the common period


问:
Reconciling my understanding of real numbers as Dedekind cuts with Dedekind and Penrose
Penrose (The Road to Reality, Section 3.2) describes Dedekind as defining real numbers via a "knife-edge" cut in the size-ordered sequence of rationals, separating them into two sets; where a cut does not fall on a rational, it must therefore fall on a kind of number which we call real.

I find two flaws in this. In what follows I shall use the term "two sets of rationals" to indicate a division of all rationals into two sets, ordered by size.

The first flaw appears to be circular reasoning. On what basis does Dedekind allow that a gap between the two sets of rationals has any meaning? It seems to me that he implicitly assumes the existence of reals, for which he must already have a definition in order to assume anything about them. Assuming your result is not valid logic.

Next, consider a convergent series and its termination, such as 1/2 + 1/4 + 1/8 ... = 2. The shortfall in the series gets smaller and smaller (in this example, at any point in the series it is equal to the latest term in the series). Yet, in the calculus at least, we are happy to accept that in the limit, the shortfall vanishes and the equality expressed above is valid; this is how we justify a continuous derivative of a continuous curve, for example. Now apply this to the rational number line. As more rationals are included, the gaps between the two sets of rationals shrinks. In the limit, by the reasoning of the calculus the gap also vanishes, as do all gaps where other cuts might fall, and the rational number line becomes continuous. This appears to undermine the assumption of a gap between two sets of rationals.

But I am somewhat confused by this question on Proof on density of rationals and irrationals in R via dedekind cut, which appears to treat a Dedekind cut as a much blunter instrument than Penrose does, by talking of "infinitely many rationals that are greater than other rational in our dedekind cut".

Who offers the flawed argument; Dedekind, Penrose or me, and where is the flaw?

答:
The first flaw appears to be circular reasoning. On what basis does Dedekind allow that a gap between the two sets of rationals has any meaning?

It only relies on the existence of the total order on rational numbers. If you have two nonempty sets $A,B$ of rationals with the property that $a<b$ for all $a\in A$ and $b\in B$, and $A\cup B=\mathbb Q$, this partition of the rationals has a suggestive "void" between them, and those are the Dedekind cuts that contribute new nonrational points.

Now apply this to the rational number line. As more rationals are included, the gaps between the two sets of rationals shrinks. In the limit, by the reasoning of the calculus the gap also vanishes, as do all gaps where other cuts might fall, and the rational number line becomes continuous.

I would accept this as a heuristic explanation of why the addition of Dedekind cuts fill up the gaps in the rational number line, but you're speaking as if it were a justification that the rational number line is already continuous. It isn't, at least it isn't complete in the ways we want it to be.  There is an increasing sequence of rationals converging to $\sqrt{2}$, but the limit is not in $\mathbb Q$ and that is the point.  There's a hole where the limit should be.  Indeed, calculus with rationals isn't worth doing until you've filled up these holes with real numbers.

It sounds a little like you are confusing density for completeness.  The first one means points get in all the open sets, but the second means that the set is maximally packed, in a sense.

Certainly there aren't any intervals of positive length between rational numbers. (Perhaps that is what you assumed a gap would be.) But that reflects their density not their completeness.  The irrationals are also dense, but somehow the two sets, while disjoint, intermingle in the entirety of the real line.

Who offers the flawed argument; Dedekind, Penrose or me, and where is the flaw?

There are no flaws in Dedekind's argument and as far as I can tell Penrose's account is an accurate description of Dedekind cuts. By process of elimination, it seems the issue is currently with your understanding of the matter.

Please continue to clarify what exactly you see is an important difference. Perhaps the most important ones were already dealt with above, but if not, please provide some more concrete definitions and things you see as opposing each other.  I can tell something's bugging you, but until you formulate what it is exactly, it'll be hard to help any more.


问:
Equations with negative exponents
This may be standard matter, but how an equation of the form $a x + b + c x^{-1} = 0$ is solved?

答: Verify that $x=0$ cannot be a solution (note the negative power). Then multiply throughout by $x$ giving a quadratic.

Title: Equations with negative exponents


答:
$$ax+b+cx^{-1}=0\Leftrightarrow \begin{cases} x\ne 0\\ \frac{ax^2+bx+c}{x}=0\end{cases}\Leftrightarrow\begin{cases} x\ne 0\\ ax^2+bx+c=0\end{cases}$$

and so forth. First step is the same for inequations as well, but for those you would have to evaluate the sign of the whole fraction, not just the numerator.

Added: It might be a matter of debate what happens for $c=0$ and, specifically, if for $c=0$ the quantity $cx^{-1}$ is to be interpreted as the constant $0$ function or as $0$ multiplied by the inverse of $x$: i.e. as if I asked a machine that computes according to PEDMAS to evaluate in $x=0$, $c=0$ and $a,b=\text{something}$ the expression $ax+b+cx^{-1}$, under the additional hypothesis that the machine returns an error as soon as it is asked to evaluate a function outside of its domain. My discussion is in the second case.

Title: Equations with negative exponents

Body: This may be standard matter, but how an equation of the form $a x + b + c x^{-1} = 0$ is solved?


答:
Hint:

You can use that $\gcd(a,b)$ is the positive generator $d$ of the ideal $I=(a,b)$. If this positive generator satisfies  the Bézout's identity $d=ax+by$, what can you say of the ideal $J=(a,b, ax+by)$?

Title: Prove that $\gcd(a, b) = \gcd(a, b, ax + by)$ for any given integers $a, b, x, y$

Body:
How can I prove that $\gcd(a,b) = \gcd(a,b,ax+by)$?

I know that $\gcd(a,b)$ can be written in linear equation like this:
$$ \gcd(a,b) = ax + by $$
Does that mean that I can substitute it to $\gcd(a,b,ax+by)$?
If yes, what's next?


问:
Norm inequality of Fourier transform
Let $f\in L^1(R^d)$, where
$$\hat{f}(\xi) = \int f(x) e^{-2\pi\xi i x}dx$$
and
$$\hat{f} = f$$
How may we prove
$$\|\hat{f}\|_p \leq \|f\|_1$$
holds?

I tried to use Holder's inequality, but end up in opposite direction, and the Hausdorff-Young was unable to deal with the gap between 1 and 2.

And insights are greatly appreciated!

Title: Norm inequality of Fourier transform

Body:
We have : $\| \hat{f} \|_{\infty} \le  \| f\|_1$

Thus,
$$ \int_{\mathbb{R}} | \hat{f}(t)|^p dt \le \| f\|_1^{p-1} \int_{\mathbb{R}} | \hat{f}(t)|dt= \| f\|_1^p$$
Done.


答:
Assuming $c \neq 0$,

$$a x + b + c x^{-1} = 0 \\
\iff ax^2 + bx + c = 0$$

and solve as you would normally solve a quadratic.

Title: Equations with negative exponents


问:
$x^2+x+1$ is the cube of a prime.
Please help me find all natural numbers $x$ so that $x^2+x+1$ is the cube of a prime number.
(Used in here)

答:
This is never a complete solution, but larger than the size that a comment can afford

If $x^2+x+1=p^3$ where $p$ is prime.

Clearly, $p\ne2$

As $p$ divides $x^2+x+1,p$ will divide $(x-1)(x^2+x+1)=x^3-1$

$\implies x^3\equiv1\pmod p\implies ord_px=3\implies 3$ divides $p-1$

So, prime $p=3q+1$ for some natural number $q$

As, $p>2$ i.e., odd, $p$ can be written as $6r+1$ for some natural number $r$

Now, $x^2+x=p^3-1=(6r+1)^3-1=18r(12r^2+6r+1)$

Observe that $(18r,12r^2+6r+1)=1$

As $(x,x+1)=1,$

$x,x+1$ can be $18r,12r^2+6r+1$

$\implies 12r^2+6r+1-(18r)=x+1-x=1$

$\implies  12r^2-12r=0\implies r=1$ as $r>0$

Now, we can test for other obvious combinations unless we can factorize $12r^2+6r+1$

$x=2r,x+1=9(12r^2+6r+1)$

$x=9r,x+1=2(12r^2+6r+1)$

etc.

Title: $x^2+x+1$ is the cube of a prime.


问:
Identity component and rational point over algebraically closed field
For an algebraic group $G$ over $\mathbb{C}$ (in general, an algebraically closed field with character $0$), let the identity component of G be denoted by $G_{0}$. I wonder whether the rational point of $G_0$ equals the identity component of $G$'s rational point, namely $G_0(\mathbb{C})=G(\mathbb{C})_0$, holds.

Here, the topology is defined with regard to the fact that rational point of an algebraic group over $\mathbb{C}$ forms an Lie group over $\mathbb{C}$.

答:
This is true if, as reuns suggests, what you mean is the following:

Fact: Let $G$ be an algebraic group over $\mathbb{C}$. Then, $(G^\circ)^\mathrm{an}=(G^\mathrm{an})^\circ$.

Here I am using $\circ$ to denote the connected component (of an algebraic group or a topological group) I am using $X^\text{an}$ to denote the analytification of a variety $X$ over $\mathbb{C}$ (i.e. equipping $X(\mathbb{C})$ with the usual structure of a complex analytic space).

The reason for this is simple. To show that $(G^\circ)^\text{an}$ is the connected component of $G^\text{an}$ we need to show that"

$(G^\circ)^\text{an}$ is connected.

$G^\text{an}/(G^\circ)^\text{an}$ is finite.

The first of these is a general property of the analytification functor in general (e.g. see this post) and for the second we note that $G/G^\circ$ is finite and eviently we have the following equality of sets

$$G^\text{an}/(G^\circ)^\text{an}=G(\mathbb{C})/G^\circ(\mathbb{C})=(G/G^\circ)(\mathbb{C})$$

from where the conclusion follows.


问:
Proving a statement in Well-formed formula language
I'm trying to prove the following question:

Prove that $\Sigma\subseteq WFF$ is a set so for every $\alpha,\beta\in WFF$, $\Sigma\models\left(\alpha\to\beta\right)$ or $\Sigma\models\left(\beta\to\left(\lnot\alpha\right)\right)$ if and only if for every $\gamma\in WFF$, $\Sigma\models\gamma$ or $\Sigma\models\lnot\gamma$.

I tried to start from $\Rightarrow$ part. Let there be $\Sigma$ so for every $\alpha,\beta\in WFF$, $\Sigma\models\left(\alpha\to\beta\right)$ or $\Sigma\models\left(\beta\to\left(\lnot\alpha\right)\right)$. Let $\gamma\in WFF$. Now I want to split into cases but not sure if the cases should be $\Sigma\models\left(\alpha\to\beta\right)$ and $\Sigma\models\left(\beta\to\left(\lnot\alpha\right)\right)$. Assuming that I used those two cases, I says that if $\Sigma\models\left(\alpha\to\beta\right)$ then for every $z$ so $z\models\Sigma$ you get $z\models\left(\alpha\to\beta\right)$ (by definition). But how to continue from here?

答: Hint: take $\alpha$ to be a true sentence, e.g., $\forall x(x = x)$, and take $\beta$ to be a given WFF $\gamma$. Then see if you can simplify $\alpha \to \beta$ and $\beta \to \lnot \alpha$.


问:
How to determine if something is a discrete random variable.
This seems very trivial. I am trying to determine which of these are discrete random variables on the sample space $\Omega = \{1,2,3,4\}$ with an event space defined as the power set of the sample space.

$X: \Omega \rightarrow \mathbb{R}$ where $X(\omega)$ = $-0.5\omega$

$X: \Omega \rightarrow \mathbb{R}$ where $X(\omega)$ = $-\omega$

$X: \Omega \rightarrow \mathbb{R}$ where $X(\omega)$ = $1$

$X: \Omega \rightarrow \mathbb{R}$ where $X(\omega)$ = $\sqrt\omega$

I seem to  think all of them are discrete random variables on my event space and sample space as they all take on discrete values, but me and my friend disagree...

答: A discrete random variable has a countable number of possible values, which in your case is satisfied by all examples since you have a finite sample space, and hence a finite event space.</p>


答:
Every common divisor of the two numbers $a$ and $b$ is also a divisor of $ax + by$, so is a common divisor of the three numbers $a$, $b$ and $ax + by$. Of course also every common divisor of the three numbers $a$, $b$ and $ax + by$ is a common divisor of just the two numbers $a$ and $b$.

So the set of common divisors of $a$ and $b$ is the same set as the set of common divisors of $a$, $b$ and $ax + by$. Therefore, the greatest elements of these two sets are also equal.


问:
Matrix and inner product
Suppose that $K$ is symmetric and positive-definite, and $v_k$ is eigenvector for $\lambda_k$ eigenvalue of $K$.

I can not understand why the following equation is correct:

$$\langle\vec v_k,K\vec v_k\rangle=\lambda_k||\vec v||^2$$
(standard matrix inner product).

I know that $K\vec v_k=λ\vec v_k$, but how do I "open" the inner product to prove the correctness of the equation?

答: @Daniel and @MichaelHoppe are right to encourage you to use$$K|v_k\rangle=\lambda_k|v_k\rangle\implies\langle v_k,\,Kv_k\rangle=\langle v_k,\,\lambda_kv_k\rangle=\lambda_k\langle v_k,\,v_k\rangle,$$provided we work with a linear-in-right-argument convention (which physicists favour; mathematicians don't). If you must see it done in coordinates (which is a pedagogically valuable exercise), let's choose an orthonormal basis viz. $\langle i,\,j\rangle=\delta_{ij}$. Then the identity operator is $\Bbb I=\sum_i|i\rangle\langle i|$, so$$|v_k\rangle=\Bbb I|v_k\rangle=\sum_i\langle i,\,v_k\rangle|i\rangle,\,\langle v_k|=|v_k\rangle^\dagger=\sum_i\langle v_k,\,i\rangle\langle i|.$$Since$$\langle u,\,v\rangle=\sum_{ij}\langle u,\,i\rangle\langle j,\,v\rangle\langle i,\,j\rangle=\sum_i\langle u,\,i\rangle\langle i,\,v\rangle$$recovers the usual sum-of-products formulation,$$\langle v_k,\,Kv_k\rangle=\sum_i\langle v_k,\,i\rangle\langle i,\,Kv_k\rangle=\sum_i\langle v_k,\,i\rangle\lambda_k\langle i,\,v_k\rangle=\lambda_k\langle v_k,\,v_k\rangle.$$

Title: Matrix and inner product


问:
Why is it meaningless for a closed set to be polygonal path connected?
My textbook (Complex Analysis by Saff & Snider) defines connectedness for open sets; the given definition of a connected open set is: a set in which every pair of points can be joined by a polygonal path that lies entirely in the set.

Using the given definition of connected set, I don't understand why it isn't similarly defined for closed sets too?

答:
Besides to @Dominic's neat answer, I am noting a good Theorem below. However it is about $\mathbb R$, you can get the points in another dramatized answer better:

Theorem: Let $E\subset\mathbb R$ including at least two points. Then $E$ is connected if and only if it is an interval.

In fact, if $E$ is not an interval,so $$\exists~p\notin E, a,b\in E~~ a<p<b$$ Now if you set $$G=(-\infty,p),~~H=(p,+\infty)$$ then $a\in G, b\in H$ and then $E\cap G, H\cap E$ are disjoint different non empty sets that can make $E$. This shows that $E$ is not connected!


答:
Consider the curves $x^2-2xy-3y^2=4 \ (C)$ and $2x^2-2xy+y^2=k^2+4 \ (E)$ At the points where they intersect, we can substitute the value of $-2xy$ from the first into the second to get $$x^2+4y^2=k^2 \ (E)$$ This represents an ellipse centered at the origin. Now, either with a bit of graph sketching, or by considering $(C)$ as a quadratic in $x^2$, deduce that $y$ can range anywhere in the real numbers, and that $(C)$ is a hyperbola. Increasing $k$ in $(E)$ just enlarges the ellipse, and it’s not hard to see that there is no upper bound on $k$ for $(E)$ and $(C)$ to intersect. For obtaining a lower bound, we need to find $k$ such that the two curves touch. Note that if $(x,y)$ lies on the two curves, then so does $(-x,-y)$. So we’re looking for exactly two intersections. Then substituting $x=\pm\sqrt{k^2-4y^2}$ in $(C)$ gives a quadratic in $y^2$: $$65y^4-y^2(18k^2-56)+(k^2-4)^2 = 0 $$ Setting the discriminant equal to zero will give $k^2=\frac{\sqrt{65}-1}{2}$ and hence we have the desired range: $$\frac{\sqrt{65}+7}{2} \le 2x^2-2xy+y^2 =k^2+4\lt \infty $$

What was wrong in your approach? What you stated was a necessary condition on $a$, but that doesn’t mean that $a$ could really equal anything in that range.

Title: Range of quadratic function using discriminant

Body:
Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.

Let $2x^2-2xy+y^2=a$.

Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.

We divide both side by $y^2$ and let $t=\frac{x}{y}$.

Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.

Since its discriminant is not negative, $\frac{\Delta}{4}\ge 0\implies a^2-7a-4\ge 0$. It gives us $a$ can have negative values like $-1$. But if clearly contracts $a-4=x^2+4y^2\ge 0\implies a\ge 4$. Where did I mistake?


问:
What is special about the numbers 9801, 998001, 99980001 ..?
I just saw this post, and realized that

1/9801 =
0.(000102030405060708091011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495969799)(repeat)

That is, after the decimal point, the number $00$, $01$, $02$, up to $97$ appear in order, (followed by $99$) and then repeat.

Similar properties are also exhibited by numbers 998001, 99980001, .. and so on.

Edit: fixed periodic start and end

It is not very obvious to me why this happens. Is there some simple explanation to this property?

答:
It happens in every base that $1/(n-1)^2$ is equal to the series of numbers in increasing order, dropping just the number n-2

```<code>  1/5²  = 0.0 1 2 3 5 0 1 2 3 5 0 ...    base 6
 1/15²  = 0.0 1 2 3 4 5 6 7 8 9 A B C D F 0 1 2  ... base 16
</code>```

The examples quoted are for bases 10, 100, 1000 &c.

Similar series occur with eg these.  These are calculated in base 1000, with leading zeros in each place suppressed.  It's true for all bases and their powers.

```<code> 1/(n-1)³ =   0. 0 0  1 3 6 10,15 21 28,36 45 55,66 78 91,105 ... (trianguar numbers)
 1/(n²-n-1) = 0. 0 1 1 2 3 5 8 13 21 34 55 89 144 &c.
 1/(n²-2n-1) = 0.0 1 2,5 12 29,70 169 408,987   (approxmates to sqrt(2)).
</code>```

Body:
It happens in every base that $1/(n-1)^2$ is equal to the series of numbers in increasing order, dropping just the number n-2

  1/5²  = 0.0 1 2 3 5 0 1 2 3 5 0 ...    base 6
 1/15²  = 0.0 1 2 3 4 5 6 7 8 9 A B C D F 0 1 2  ... base 16

The examples quoted are for bases 10, 100, 1000 &c.

Similar series occur with eg these.  These are calculated in base 1000, with leading zeros in each place suppressed.  It's true for all bases and their powers.

 1/(n-1)³ =   0. 0 0  1 3 6 10,15 21 28,36 45 55,66 78 91,105 ... (trianguar numbers)
 1/(n²-n-1) = 0. 0 1 1 2 3 5 8 13 21 34 55 89 144 &c.
 1/(n²-2n-1) = 0.0 1 2,5 12 29,70 169 408,987   (approxmates to sqrt(2)).


问:
Show that a sequence of functions that converge uniformly are Riemann integrable. What if they only converge point wise?
Let $f_n$ be a sequence of Riemann integrable functions on $[a,b]$ that converge uniformly to a function f. Show that f is also Riemann integrable. What happens if $f_n$ only converges pointwise?

Considering this scenerio, show that

$$\lim_{n \to \infty} \int_{a}^{b} f_n(x) dx = \int_{a}^{b} f(x) dx$$

Not really sure how to start this problem. If they already converge to a function, uniformly, that function has to be continuous right? So then it is trivially Riemann integrable. The pointwise case I'm not sure about. And then the second part with the limits I'm not sure how to go about. Any help is appreciated!

答:
Let $\{q_n\}_{n \in \mathbb{N}}$ be the rational numbers in the interval $[0,1]$, and let us consider the functions
$$f_n(x)=
\begin{cases}
1 &\text{if}\quad x \in \{q_1, \ldots, q_n\},\\
0 &\text{otherwise.}
\end{cases}
$$

The $f_n(x)$ are Riemann-integrable but they converge to the Dirichlet function, which is not Riemann-integrable.


问:
Nonlinear Implicit Finite Difference
How do you produce an implicit finite difference system with a nonlinear term in the pde?

For example, if you have the reaction-diffusion equation:
$$\frac{\partial{u}}{\partial{t}} = \Delta u + f(u)$$

with $f$ being non-linear, e.g. $f(u) = u^2$, how do you form a finite difference system to solve?

I understand how to derive if $f(u)=0$. For example in a 1D setting, you have the form:
$$w_{i}^{n-1} = (1+2\lambda)w_{i}^n - \lambda(w_{i+1}^n + w_{i-1}^n)$$
producing a system in the form:
$$Aw^n=w^{n-1}$$

答:
In this case, a common approach is make the nonlinear part explicit. You start with the initial condition $u^{(0)}$ and , for each $k\ge 0$, solve a linear problem. The following example uses a simple forward difference for the time derivative, but you can replace it for something fancier.

$$
\frac{u^{(k+1)}- u^{(k)}}{\delta t} = \Delta u^{(k+1)}+f(u^{(k)})
$$

or, equivalently,

$$
-\Delta u^{(k+1)}+\frac{1}{\delta t} u^{(k+1)} = \frac{1}{\delta t} u^{(k)}+f(u^{(k)})
$$

Title: Nonlinear Implicit Finite Difference


答:
The periods are
$\frac{1}{2}, \frac{2}{5},\frac{1}{3}$.

Since $\operatorname{lcm}(2,5,3)=30$, we can write these periods as
$\frac{15}{30}, \frac{12}{30},\frac{10}{30}$.

Since $\operatorname{lcm}(15,12,10)=60$, the lcm of the periods is $\frac{60}{30}=2$.

Title: Determine the common period

Body:
Let $V = \mathrm{Vect}( e^{ 4 \pi it } , e^{ 5 \pi i t} , e^{ 6 \pi i t} ) $ and $f \in V$.

I want to determine the common period between the functions : $t \mapsto e^{ 4 \pi it } $, $t \mapsto e^{ 5 \pi i t }$ and $t \mapsto e^{ 6 \pi i t}$ ; i.e the period of $f \in V$.

We have

$ t \mapsto e^{ 4 \pi i t} $ is periodic with period $\frac{1}{2}$

$ t \mapsto e^{ 5 \pi i t} $ is periodic with period $\frac{2}{5}$

$ t \mapsto e^{ 6 \pi i t} $ is periodic with period $\frac{1}{3}$

and I don't know how to pursue it. Is there a particular method to found a common period? I need a hint.

Thank you


问:
Bellman Criteria (Graphtheory)
Lemma (Bellman Criteria)  [Brandes 2001]

Let $G$ be a graph and $s, t, v \in V(G)$.

A  vertex $v \in V(G)$ lies on a shortest path between vertices $s, t \in  V$, if and only if dist(s,t) = dist(s,u)+dist(u,t).

Can i show by the definition that dist(s,v)+dist(v,t) is the shortest path from $s$ to $t$ via $v$ and then show by contradiction that ">" "<" is valid and thus equality must apply? Or is it possible to show that only by equivalence direction?

答:
The distance between two vertices is by definition the length of a shortest path between them. So if $d(s,v)+d(v,t)=d(s,t)$ there is an $st$-path containing $v$, which is a shortest path from $s$ to $t$.

Conversely assume that $d(s,v)+d(v,t)\neq d(s,t)$ and note that by definition of the latter distance we indeed have $d(s,v)+d(v,t)>d(s,t)$. Moreover any $st$-path $P$ containing $v$ satisfies $\operatorname{len} P \geq d(s,v) + d(v,t)$, since it decomposes into an $sv$-path $P_s$ and $vt$-path $P_t$, which by definition satisfy $\operatorname{len} P_s \geq d(s,v)$ and $\operatorname{len} P_t \geq d(v,t)$. But this way $v$ cannot possibly lie on a shortest $st$-path.

Title: Bellman Criteria (Graphtheory)


问:
Find greatest lower bound and least upper bound
Let $\mathbb{R}$ be the set of real numbers and $≤$ be the partial order on $\mathbb{R}$.

Consider the subset
$$ A = \{ x ∈ \mathbb{R}: 1<x<2 \} $$
Find the greatest lower bound and least upper bound of $A$.

答:
$$ A = \{ x ∈ \mathbb{R}: 1<x<2 \} $$
Claim: The least upper bound (l.u.b) of $A$ is $2$, and the greatest lower bound (g.l.b) of $A$ is $1$.

Outline of proof:

To show that $2$ is the l.u.b, show that it is an upper bound for $A$, and that if $u\in\mathbb{R}$ is an upper bound for $A$, then $u\geq 2$.

Similarly to show that $1$ is the g.l.b, show that it is a lower bound for $A$, and that if $l \in\mathbb{R}$ is a lower bound for $A$ then $l\leq 1$.

The above outline really just follows from the definitions, and I hope you can fill in the details.


答:
Over and over you say that if we assume X then things are inconsistent or circular, when in fact nobody ever assumed X! I bet I know what happened; I bet you're confusing the "motivation" with the actual official definition-theorem-proof exposition.

Huh? For example, if I were writing about Dedekind cuts I'd probably start here:

(*) For each $x\in\Bbb R$ define $S_x=\{r\in\Bbb Q: r>x\}$. It's easy to see that the mapping $x\mapsto S_x$ is injective. In fact knowing $S_x$ tells us everything there is to know about $x$, so we may as well go ahead and define the reals in terms of the sort of sets that arise as S_x.

Then I'd go ahead and define Dedekind cuts and prove that they do what they need to do. What's important to realize is that (*) above is NOT part of the "official" math. If (*) were included then yes, things would be circular. But officially (*) has nothing to do with anything; it's just "motivation", trying to see how someone might have come up with that definition. (In particular you will never see the notation "$S_x$" in the official part...)

Title: Reconciling my understanding of real numbers as Dedekind cuts with Dedekind and Penrose

Body:
Penrose (The Road to Reality, Section 3.2) describes Dedekind as defining real numbers via a "knife-edge" cut in the size-ordered sequence of rationals, separating them into two sets; where a cut does not fall on a rational, it must therefore fall on a kind of number which we call real.

I find two flaws in this. In what follows I shall use the term "two sets of rationals" to indicate a division of all rationals into two sets, ordered by size.

The first flaw appears to be circular reasoning. On what basis does Dedekind allow that a gap between the two sets of rationals has any meaning? It seems to me that he implicitly assumes the existence of reals, for which he must already have a definition in order to assume anything about them. Assuming your result is not valid logic.

Next, consider a convergent series and its termination, such as 1/2 + 1/4 + 1/8 ... = 2. The shortfall in the series gets smaller and smaller (in this example, at any point in the series it is equal to the latest term in the series). Yet, in the calculus at least, we are happy to accept that in the limit, the shortfall vanishes and the equality expressed above is valid; this is how we justify a continuous derivative of a continuous curve, for example. Now apply this to the rational number line. As more rationals are included, the gaps between the two sets of rationals shrinks. In the limit, by the reasoning of the calculus the gap also vanishes, as do all gaps where other cuts might fall, and the rational number line becomes continuous. This appears to undermine the assumption of a gap between two sets of rationals.

But I am somewhat confused by this question on Proof on density of rationals and irrationals in R via dedekind cut, which appears to treat a Dedekind cut as a much blunter instrument than Penrose does, by talking of "infinitely many rationals that are greater than other rational in our dedekind cut".

Who offers the flawed argument; Dedekind, Penrose or me, and where is the flaw?


问:
$C[0,1]$ is complete w.r.t. which norm(s)
$C[0,1]$ is complete w.r.t. which norm(s)

$\displaystyle\|f\|_\infty=\sup_{t\in[0,1]}|f(t)|$

$\displaystyle\|f\|_1=\int_0^1|f(t)| \, dt$

$\displaystyle\|f\|_\infty^{0,1}=\|f\|_\infty+|f(0)|+|f(1)|$

$\displaystyle\|f\|=\sqrt{\int_0^1|f(t)| \, dt}$

I know 1 is true. But I don't know how to check completeness of norm base on know complete norm. How can I solve it?

答:
Norm $1$ and  $3$ are equivalent, so $C[0,1]$ is also complete in the norm $3$.

$||f||_{\infty}^{0,1}=||f||_{\infty}+|f(0)|+|f(1)|\le 3||f||_{\infty}$


问:
solution of a sequence in a given range
Given an equation $$x^{\frac{1}{n}}=\sqrt{1-x}$$ has a unique solution $x_n$ in $[0,+\infty)$ for each $n \in\mathbb{N}$. One can see that $$x_n=1-x^{\frac{2}{n}}$$ But how do we show that the solutions are unique?

答:
Consider $$f(x) = x^a$$ for some $a>0$. $f(x)$ is clearly increasing since $f'(x) \geq 0$ for $x\geq 0$.

Consider $$g(x) = \sqrt{1-x}$$ and see that $$g'(x) = \frac{-1}{2\sqrt{1-x}} < 0$$ for all $x\geq 0$. So $g(x)$ is decreasing.

Can you now conclude that $f(x)$ and $g(x)$ cannot have more than one intersection? In other words, $f(x) = g(x)$ cannot have more than one solution for a fixed $a>0$.

Title: solution of a sequence in a given range


问:
Products of simple $R$-modules are semisimple?
Let $R$ be a ring with identity element such that $R/J(R)$ is right Artinian. It is equivalent to say that $R$ is a semilical ring, i. e., $R$ has finitely many maximal right ideals. I want to know whether any direct product of simple $R$- modules is a semisimple module.
Thanks for any help or answer.

答:
Yes, but for trivial reasons.

The semisimple $R$ modules are in 1-1 correspondence with the semisimple $R/J(R)$ modules, which in this case are all $R/J(R)$ modules, since the ring is semisimple Artinian.

Therefore the product of simple modules - taken in $R$-mod - is exactly the same module as the one in $R/J(R)$-mod, and since it is semisimple there, it is semisimple in $R$-mod.