url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://forum.enjoysudoku.com/fast-search-of-a-17-clues-puzzle-in-a-x-y-27-valid-puzzle-t32544.html | 1,642,698,375,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302355.97/warc/CC-MAIN-20220120160411-20220120190411-00127.warc.gz | 28,060,591 | 15,693 | fast search of a 17 clues puzzle in a X+Y+27 valid puzzle
Programs which generate, solve, and analyze Sudoku puzzles
fast search of a 17 clues puzzle in a X+Y+27 valid puzzle
Less than one millisecond to tell whether a X+Y+27 puzzle contains a 17 clues puzzle.
If you don't believe it, you can have a look to the next posts
This is a problem we faced trying to see whether we are missing 17 clues puzzles in the 49157 known 17 clues puzzles.
There is no easy way to solve that problem. One feasible partial answer to reduce the problem has been to scan all solutions in a X+Y+27 pattern.
The negative answer for X+Y<8 came quickly. The X+Y=8 case is already much more difficult.
The "best known process" is split in four steps :
a) find ED patterns X+Y+27. This is a relatively short step
b) Generate valid puzzles against "gangsters", the minlex form of band1 possible starts with a given set of 3 digits in each column
c) expand these puzzles to any valid permutation within the columns
d) find in each of these puzzles the valid 17.
we consider here exclusively the last step. To give an idea of the "size", in the 4+4+27 case, we have 13 745 722 puzzles at the end of the step c).
Available programs at that time could find the 17 with a run time leading to a very long d) step, so we tried ("Blue" and me), to use the specificity of that case to have a faster process.
We have been lucky and arrived to a step d) processing the 13 millions puzzles in about half an hour (9500 puzzles per second)
The process applied introduce nearly no new idea, except one, the search of UAs of size 2;3.
The resulting code is very simple and summarize in a nice way some published concepts.
The reader is supposed to have a good understanding of the following :
Unavoidable sets and search of puzzles of size "n" in a solution grid
The oldest reference I know is the gridchecker program reworked by mladen Dobrichev gridchecker
The second and key reference it the work done by the Gary McGuire's team to prove that there is no 16 clues puzzle. Gary McGuire's article
The key concepts described in these references are supposed known by the reader.
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
The basic process.
We intend here to use more or less the same process as Gary McGuire.
In our case, X+Y=8, so 8 clues are given in bands 2 and 3, and we have to find a set of 9 clues in band 1 to come to a 17 clues puzzle.
If that puzzle is valid, we have a 17 clues solution.
Having 8 given, the first idea would be to generate all unavoidable set using the solution (the puzzle is valid, the solution is unique) and the print given by McGuire (UAs size 12 and less).
That list can be reduced erasing all puzzles hit by the 8 given. As the puzzle is valid, no UA can be found having no cell within the band 1.
And the corresponding UA's can be limited to the band 1 cells.This is not bad, but a much simpler idea appeared to be very efficient.
Most UA's if not all of them have somewhere in a box cells grouped in a mini line. Here, we can have a chance to have some UA's not yet hit by the known clues in bands 2;3 located in a mini row in band 1.
Such UAs are very easy to detect. To check if a mini row contains one or more of them, you just test a brute force on given clues omitting the corresponding 3 clues.
Testing that idea, we had the amazing surprise to see that the number of such UA' was high enough to kill in most cases any possibility to have a 17 clues puzzle.
In our file of 13 million puzzles, only 238 468 have still to be investigated after all mini rows have been tested for UAs size 2 or 3.
On top of it, if we use all tools described in the McGuire article, we could add UA' of bigger size, but nearly all branches are already quickly "'dead".
In our example, out of the 238 468 puzzles,
Code: Select all
`only 1 141 589 branches end with 17 clues (others are killed before)13 179 branches had to use more clues leading to 198 257 puzzles to test.`
This means that the average number of additional call to brute force after UA'a size 2;3 have been searched is far below 1 per puzzle. No reason in that situation to look for more UA's.
Note : I did not test a search of UA's size 2;3 starting by the test of the full mini row, but the result should not be very different.
Mini row status versus UAs size 2;3 and minimum number of clues needed
Within a mini row, we can end with five different interesting status
1) no UA
2)1 UA size 2
3) 2 UA's size 2
4) 3 UA's size 2
5) no UA size 2, but one UA size 3.
mini-rows and disjoints UAs
Use of UAs hitting to find puzzles makes many references to "disjoints UAs" and "UA's of higher degree. (see mcGuire article)
by definition, each mini row is disjoint to any other mini row.
within a mini row,
Code: Select all
`no Ua is disjoint to another onein the case 4) (3 UA's size 2) we need a minimum of 2 clues to hit the 3 UAs (in fact, exactly 2 clues)`
the minimum number of clues needed to hit the UAs of the mini-row is one in cases 2;4;5 and 2 in the case 3.
Summing the needed in each mini row gives the total minimum of clues needed.
If we have 8 clues in bands 2;3 (X+Y=8), the puzzle contains no valid 17 clues if the total minimum needed is >= 10 (10+8>17)
The amazing fact is that this happens in most of the cases.
Last edited by champagne on Sat Jul 04, 2015 7:50 pm, edited 1 time in total.
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
Exploring and sorting UAs size 2;3
All the coding shown here uses the infrastructure described recently in the brute force thread brute_force
The code is very simple. Apart from the calls to the brute force, the main specificity is the use of an integer to store different things.
Code: Select all
`typedef union __declspec(intrin_type) _CRT_ALIGN(4) GINT_t { unsigned __int8 u8[4]; unsigned __int16 u16[2]; unsigned __int32 u32;} GINT;`
in that code, a given (digit + cell) is stored as 2 bytes in an integer.
the code will be described using 4+4+27 valid puzzles containing one or more 17 clues puzzle.
we just describe here the code included in a loop processing 81text lines where non given are '.'
Primary filter; counting the minimum number of clues needed
Code: Select all
`GINT tgint[81],tgminir[9],tgb1[30];int ntgint=0,ntgb1=0,ntgbs=0;//find and count the clues in band 2 / 3 for(int i=27;i<81;i++) if(zpuz[i]>'0' && zpuz[i]<= '9') tgint[ntgint++].u32=i+((zpuz[i]-'1')<<8);int nr=ntgint,maxforced=17-nr, cumuah=0;//========= try all pairs in minilines band1 as UAsint permmr[9][3][2]={{{0,1},{0,2},{1,2}}, {{3,4},{3,5},{4,5}}, {{6,7},{6,8},{7,8}},{{9,10},{9,11},{10,11}}, {{12,13},{12,14},{13,14}}, {{15,16},{15,17},{16,17}},{{18,19},{18,20},{19,20}}, {{21,22},{21,23},{22,23}}, {{24,25},{24,26},{25,26}}};for(int imr=0;imr<9;imr++) {// 9 minrows to search tgminir[imr].u32=imr; int px=0; for(int p=0;p<3;p++){ int* pp=permmr[imr][p]; ntgint=nr; // load after bands2 3 for(int k=0;k<27;k++) if((k-pp[0]) && (k-pp[1])) tgint[ntgint++].u32=k+((zpuz[k]-'1')<<8); if(zhou[0].StatusValid(tgint,ntgint,4)-1) {// UA if not unique tgb1[ntgb1++].u32=imr+(p<<8);// just store it px|=(1<<p); } } if(px){// some UA2 have been seen cumuah++; int nx=BitCount[px],w=(nx<2)? 1:2; if(nx==3){ stored in the miniline status cumuah++; tgminir[imr].u8[1]=1; } } else{// see if triplet is UA ntgint=nr; // load after bands2 3 for(int k=0;k<27;k++) if((k/3) - imr) tgint[ntgint++].u32=k+((zpuz[k]-'1')<<8); if(zhou[0].StatusValid(tgint,ntgint,4)-1) {// UA if not unique tgb1[ntgb1++].u32=imr+(3<<8);// just store it px |=8; cumuah++; } } tgminir[imr].u8[2]=BitCount[px];// store final UAs count of the minirow}if(cumuah>maxforced)continue; // can't be a 17 `
here the 32 bits int (GINT) is used to store
tgint the list of given sent to the brute force (cell + digit)
tgminir the mini-row status (mini_row+{1 if 3 pairs} + UAs count}
tgb1 one entry per UA found (mini_row + {pairnumber 0-2 or 3 if triplet} )
Only the first table is needed for that filter. The 2 other tables are for the next steps.
As said earlier, in a huge majority of the cases, the search is cancelled with
if(cumuah>maxforced)continue; // can't be a 17
At that point, a minimum of 3x9 calls to the brute force have been done, but with 25+8= 33 given, the brute force goes very fast.
The brute force call is very close to standard knowing that the puzzle has 1 or more solutions.
The call uses the InitSudoku() variant using a list of given. The return code is 0 if the puzzle is not valid(never here); 1 is the puzzle has one solution; 2 if the puzzle has more than one solution.
Next step is to process puzzles passing the first filter.
Last edited by champagne on Sat Jul 04, 2015 7:55 pm, edited 1 time in total.
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
Preparing the puzzle expansion
Using as sample the puzzles containing some 17 clues, we don't have the case where a mini row has no UA of size 2, but a UA of size 3. As this is not a key point, I did not look for such a puzzle.
The first 4+4+27 containing a 17 clues is the following.
Code: Select all
` 9538214766429738511876543293.6...........5.4...........2.....9....3.8............`
the search for UAs2 produces the following set
Code: Select all
`minir=0 perm0 9r1c1 5r1c2minir=0 perm2 5r1c2 3r1c3minir=1 perm0 8r1c4 2r1c5minir=1 perm2 2r1c5 1r1c6minir=2 perm1 4r1c7 6r1c9minir=2 perm2 7r1c8 6r1c9minir=3 perm0 6r2c1 4r2c2minir=5 perm0 6r3c4 4r3c6minir=5 perm1 3r3c7 2r3c8minir=5 perm2 3r3c7 9r3c9minir=7 perm0 6r3c4 5r3c5minir=7 perm1 6r3c4 4r3c6minir=8 perm0 3r3c7 2r3c8minir=8 perm1 3r3c7 9r3c9`
to hit all UAs we need
Code: Select all
`1 or 2 clues in mini-rows 0;1;2;7;81 clue in mini-row 32 clues in mini-row 5`
the minimum number of clues is 8, missing by 2 the first filter (8+8=16 < 18)
Any solution hitting more than 2 mini-rows in the group 0;1;2;7;8 with 2 clues will pass the filter.
This is another way to express what is described as "early stop in a branch using UAs of higher degree in the McGuire article.
Here Higher degree UAs can have a trivial expression. If we order the UAs mini row per mini row, hitting top down the mini rows; the Highest remaining degree is the sum of minimum clues in the untouched mini-rows.
We should reach the best efficiency sorting the mini-rows with at the top all mini rows having to UAs size 2
This is done in the program building the table tuab1. Here is the print of the corresponding table.
Code: Select all
`tuab1 i=0 uahsize=7 nval=2 9r1c1 5r1c2 tuab1 i=1 uahsize=7 nval=2 5r1c2 3r1c3 tuab1 i=2 uahsize=6 nval=2 8r1c4 2r1c5 tuab1 i=3 uahsize=6 nval=2 2r1c5 1r1c6 tuab1 i=4 uahsize=5 nval=2 4r1c7 6r1c9 tuab1 i=5 uahsize=5 nval=2 7r1c8 6r1c9 tuab1 i=6 uahsize=4 nval=2 6r3c4 5r3c5 tuab1 i=7 uahsize=4 nval=2 6r3c4 4r3c6 tuab1 i=8 uahsize=3 nval=2 3r3c7 2r3c8 tuab1 i=9 uahsize=3 nval=2 3r3c7 9r3c9 tuab1 i=10 uahsize=2 nval=2 6r2c1 4r2c2 tuab1 i=11 uahsize=0 nval=2 8r2c7 5r2c8 tuab1 i=12 uahsize=0 nval=2 8r2c7 1r2c9 tuab1 i=13 uahsize=0 nval=2 5r2c8 1r2c9`
Now, the mini-rows order is 0;1;2;7;8;3;5 in any of the groups {0;1;2;7;8}; {3;5}, the sequence can be changed without any effect on the result. Reversely, changing the order of the cells within a UA would affect the performance (optimizing the "dead" effect).
In each entry of the table, we have (uahsize) the sum of the minimum number of clues in the other mini-rows.
One could object that this is not optimal, and this is true, but it is simple and efficient enough in the expansion process.
In the hitting sequence, if at any moment the sum of hits done + pending minimum High degree UA pass 17, the branch can be stopped.
Last edited by champagne on Sat Jul 04, 2015 8:02 pm, edited 1 time in total.
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
In the code, the redundancy control through "dead cells" as explained in McGary's article is used.
Dead cells are skipped in the UA's expansion process
Dead cells are excluded from the remaining cells when no more UA size 2;3 is available.
what to do when the UA size 2;3 file is exhausted.
The strategy here is somehow depending on the frequency of that event.
In the McGuire process, when the file of UAs size 1-12 is exhausted, additional clues are searched in the remaining cells.
Here, we could first look in UAs size 1-12 having in the " band 1" a cells set not belonging to a mini row.
The price to pay to collects such UAs is far from nil. the experience of the 4+4+27 field does not push in that direction.
In our first example, only six branches are not closed with the last UA. As the minimum number of hits is 8, we start with 8+8=16 known clues and this can happen only if each mini-row having 1 or 2 clues has the hit corresponding to one clue, that is
5r1c2; 2r1c5;6r1c9;6r3c4;3r3c7.
we have then one clue out of {6r2c1;4r2c2}
and 2 clues out of {8r2c7;5r2c8;1r2c9}
giving the 6 cases
Code: Select all
`.5..2...66.....85....6..3..3.6...........5.4...........2.....9....3.811.11.1.11.....11....1..1...5..2...66.....8.1...6..3..3.6...........5.4...........2.....9....3.811.11.1.11.....111...1..1...5..2...66......51...6..3..3.6...........5.4...........2.....9....3.811.11.1.11.....111...1..1...5..2...6.4....85....6..3..3.6...........5.4...........2.....9....3.811.11.1.111....11....1..1...5..2...6.4....8.1...6..3..3.6...........5.4...........2.....9....3.811.11.1.111....111...1..1...5..2...6.4.....51...6..3..3.6...........5.4...........2.....9....3.811.11.1.111....111...1..1.`
.
here the second line shows the dead clues when we enter the complementary search.
Although the count of assigned clues is 8, the dead count is between 11 and 13. An optimized process would likely give in that case 13 dead in each branch.
As a consequence, the complementary search has to take one clue out of 14-16 (27- 13/11)to test for a valid puzzle
No chance in that position to have a better process looking for more UAs.
With 2 missing clues, we can expect about 14*13/2 cases to test, still far from the cost of the search of UAs.
Here our worst case (see below) is 2 missing clues. My forecast would be that the current process remains better even with 4-5 missing clues.
Code: Select all
`9576243816839514274217389655........2...7..........6....8..6........9...........2tuab1 i=0 uahsize=6 nval=2 9r1c1 7r1c3 tuab1 i=1 uahsize=6 nval=2 5r1c2 7r1c3 tuab1 i=2 uahsize=5 nval=2 6r1c4 2r1c5 tuab1 i=3 uahsize=5 nval=2 2r1c5 4r1c6 tuab1 i=4 uahsize=4 nval=2 6r2c1 8r2c2 tuab1 i=5 uahsize=4 nval=2 6r2c1 3r2c3 tuab1 i=6 uahsize=3 nval=2 9r2c4 5r2c5 tuab1 i=7 uahsize=3 nval=2 5r2c5 1r2c6 tuab1 i=8 uahsize=2 nval=2 4r2c7 2r2c8 tuab1 i=9 uahsize=2 nval=2 4r2c7 7r2c9 tuab1 i=10 uahsize=1 nval=2 9r3c7 5r3c9 tuab1 i=11 uahsize=1 nval=2 6r3c8 5r3c9 tuab1 i=12 uahsize=0 nval=2 2r3c2 1r3c3`
here most entries with a lack of UAs are with one of the mini-rows {0;1;3;4;5;8} having 2 clues, but after the sequence
7r1r1c3;2r1c5;6r2c1;5r2c5;4r2c7;5r3c9
any of {2r3c2;1r3c3} ends with 8+7=15clues
2 clues must be added. With the non optimal sequence above, we have respectively 12/13 dead cells giving less than 100 pairs to test.
Last edited by champagne on Sat Jul 04, 2015 8:10 pm, edited 1 time in total.
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
the code following the filter is the following
Hidden Text: Show
Code: Select all
`zhou[0].SetPat(zpuz,zsol,zout);// we need the solution for expansionif(!( zhou[1].CheckValidityQuick(zpuz,0)==1)) // build the full solution return;// could have been checked earlier but we need the solution herezhou[0].glb.zsol=0;// protect the solution, later we just need "valid" or not//----------------> sorting UAs and building UABX ready to expandstruct UAB1{ GINT tval[3]; int nval,bf,uahsize; void Build(GINT & gg,char * zsol,int permmr[9][3][2],int cumuah){// gg imr+perm uahsize=cumuah; int imr=gg.u8[0],p=gg.u8[1]; if(p<3){// ua size 2 int* pp=permmr[imr][p]; bf=(1<<pp[0])+(1<<pp[1]); nval=2; tval[0].u32=pp[0]+((zsol[pp[0]]-'1')<<8); tval[1].u32=pp[1]+((zsol[pp[1]]-'1')<<8); } else { //UA size 3 full miniline int p1=3*imr,p2=p1+1,p3=p2+1; bf=(1<<p1)+(1<<p2)+(1<<p3); nval=3; tval[0].u32=p1+((zsol[p1]-'1')<<8); tval[1].u32=p2+((zsol[p2]-'1')<<8); tval[2].u32=p3+((zsol[p3]-'1')<<8); } }}tuab1[30];int ntuab1=0;for(int imr=0;imr<9;imr++) if(tgminir[imr].u8[2]==2){// load first 2UAs in the miniline cumuah-=(1+tgminir[imr].u8[1]); // min UAHigh degree remainning size for(int i=0;i<ntgb1;i++) if(tgb1[i].u8[0]==imr) tuab1[ntuab1++].Build(tgb1[i],zsol,permmr,cumuah);}for(int imr=0;imr<9;imr++) { int cc=tgminir[imr].u8[2]; if(cc==1 || cc==3){// then others cumuah-=(1+tgminir[imr].u8[1]); // min UAHigh degree remainning size for(int i=0;i<ntgb1;i++) if(tgb1[i].u8[0]==imr) tuab1[ntuab1++].Build(tgb1[i],zsol,permmr,cumuah); }}struct SPOTS{ int cell,cell_old,cumdead,ival,iua ; //cumdead is a 27 bits field UAB1 * myua; inline int NextVal(){return (++ival>=myua->nval);} inline GINT GetVal(){ return myua->tval[ival];}}spots[15],*s ,*sp;int ispot=0, cell,limspot=17-nr-1; GINT * tgsolplus=&tgint[nr],wgint;s=spots;s->cumdead=s->iua=0;s->myua=tuab1;s->ival=-1;goto nextval;nextspot:if(ispot == limspot){// check if it is a valid 17 if(zhou[0].StatusValid(tgint,17,4)-1) goto nextval; cout <<CoutGintPuzzle(tgint,17,zout)<< " valid 17"<<endl;// print the valid 17 goto nextval;}ispot++; sp=s++;// find next UAint iua=sp->iua;s->iua=0;while (++iua<ntuab1){// new minirow or untouched UA in the minirow if(tuab1[iua].bf & (1<<sp->cell))continue; // only reason to skip s->iua=iua; break;}if(s->iua){// we have a "next UA" in the table if(ispot+tuab1[s->iua].uahsize >limspot)goto backtrack; // passing 17 s->ival=-1; s->myua=&tuab1[s->iua]; s->cumdead=sp->cumdead; goto nextval;}//no more UAs in the table try all "not dead" remaining cells in band1{ //Build the table of undead int ispot2=0,n2=0,cd=sp->cumdead,ispot0=nr+ispot-1,lim2=15-ispot0; int tispot2[10]={-1}; // current index for the corresponding value of ispot2 GINT t2[26]; for(int i=0;i<27;i++){ if(cd & (1<<i)) continue; t2[n2++].u32=i+((zsol[i]-'1')<<8); } nextval2:{ int ind=++tispot2[ispot2]; if(ind>=n2){// bactrack2 if(ispot2){ispot2--;goto nextval2;} goto backtrack; } if(ispot2==lim2){// test that 17 clues case for(int j=0;j<=ispot2;j++) tgint[ispot0+j]=t2[tispot2[j]]; if(zhou[0].StatusValid(tgint,17,4)-1) goto nextval2; cout <<CoutGintPuzzle(tgint,17,zout)<< "valid 17"<<endl;// print valid 17 goto nextval2; } else{ ispot2++; tispot2[ispot2]=ind; goto nextval2; } cerr <<"invalid status in step2"<<endl; // can not be this is a bug return; }}goto backtrack; // should never be usednextval:{ if(s->NextVal()) goto backtrack; ispot=s-spots; // be sure to have the correct value tgsolplus[ispot]=wgint=s->GetVal(); s->cell=cell=wgint.u8[0]; if(s->cumdead & (1<<cell) ) goto nextval; s->cumdead |= (1<<cell); goto nextspot;}backtrack: if(--ispot>=0) { s--; goto nextval;}`
I see nothing difficult in that, so comments will come on request
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
a side remark on that process.
The main finding, the search of UAs size 2 was done late 2014.
I spent much time to see if similar ideas could be developed in the search of 17 clues puzzles directly in the catalog of solutions. I failed.
So far, the layout of McGuire remains the best. Marginal improvements are possible, but no breakthrough.
That process could offer 2 or 3 new ideas. The simplest would be to change the priority in the UA's processing.
But what is very simple (sorting the mini rows and reordering the cells to optimize the "dead cell" effect) is more complex in the general case.
If my brain is still in good form, I'll restart the thinking late this year.
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
Hi champagne, I have 2 questions.
1) What UA of size 2 and 3 are?
2) What exactly are the input and output of the explained process?
dobrichev
2016 Supporter
Posts: 1816
Joined: 24 May 2010
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
dobrichev wrote:Hi champagne, I have 2 questions.
1) What UA of size 2 and 3 are?
2) What exactly are the input and output of the explained process?
input is a valid X+Y+27 puzzle. Normally, it has been produced expanding and ED X+Y+27 pattern, but it can be any valid X+Y+27 puzzle.
the band1 is filled, band2 and 3 contain the X and Y given.
output is made of 17 clue puzzles found in the input
i
Now point 1)
when you consider all UAs (in theory).
Most of them are hit by the X;Y given. Just consider other UAs
These UAs can not be hit outside of the band 1, all missing given must be in band 1
All these UAs have part of the cells in band 1 (if not, the puzzle would not be valid). We can reduce them to the cells in band 1.
UAs of size 2 are such UAs having only 2 cells in band 1. As noticed blue, such cells must be in a mini row.
In that process, UAs of size 3 are such UAs touching the band 1 with 3 cells of the same mini-row. I did not check it, but I think that it is the only possible pattern for a UA of size 3. In fact, here we don't have to answer to the question.
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
Thank you.
So, if you take a valid 17-given puzzle and apply your process, it will be equivalent of 6 steps, where in each step you are clearing the givens from a particular stack/band, and generating all possible 17-given puzzles having the same uncleared givens. You are not fixing the solution grid.
UA set is property of the solution grid. If you work with gangsters of 44, then you use some generalization of UA set. That confused me. BTW gridchecker has pretty fast code for U4 and U6 which you can simplify and reuse for your case with U2 and U3 respectively.
You found that for the evaluated cases hitting the inter-band short UA dominates hitting the intra-band UA, which isn't surprise. Congratulations for making it working!
With currently implemented optimizations, do you have estimation how long the processing of all known 17-givens will take?
Do you manage to process firstly the known 17s and later the rest of the valid 27+X puzzles?
dobrichev
2016 Supporter
Posts: 1816
Joined: 24 May 2010
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
dobrichev wrote:Thank you.
So, if you take a valid 17-given puzzle and apply your process, it will be equivalent of 6 steps, where in each step you are clearing the givens from a particular stack/band, and generating all possible 17-given puzzles having the same uncleared givens. You are not fixing the solution grid.
I think this is not correct.
We start from a puzzle having one solution.
Here, a UA2 is defined as a puzzle loosing that property if none of 2 cells is given. So the solution grid is unique.
BTW gridchecker has pretty fast code for U4 and U6 which you can simplify and reuse for your case with U2 and U3 respectively.
Yes, but in that process, i suspect that many UA2 are reductions of much bigger one (even exceeding the size 12, the limit of the prints prepared by McGuire).
With currently implemented optimizations, do you have estimation how long the processing of all known 17-givens will take?
With the above remarks, I have some difficulties to answer to that question. I don't see clearly what process you have in mind.
What is for sure is that finding the 4+4+27 has been more than 5 year*core of generation.
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany
Re: fast search of a 17 clues puzzle in a X+Y+27 valid puzzl
dobrichev wrote:With currently implemented optimizations, do you have estimation how long the processing of all known 17-givens will take?
Do you manage to process firstly the known 17s and later the rest of the valid 27+X puzzles?
I thought a little more about that.
I see one possibility to apply that process with success in competition with the McGary's one.
Take any solution grid.
search all band2+band3 UAs
build all "minimal" 27+X
apply that process to each of them.
Looking for a minimal 27+X, it could be that one or 2 more clues in bands2-3 have to be added. With band1 filled, there is not so many possibilities.
If we need more clues at the end
a) we can use standard UAs specific to band1 (collected in the first phase)
b) the additional clues can be anywhere in the 81 cells
It will be necessary to draft the code to evaluate the process
In that case (2 bands), a 64 bits platform would be a plus. AFAIK, the free Microsoft visual C++ is limited to the generation of the 32 bits code.
Looking for new 17clues, we can immediately skip all cases with less than 9 clues in bands23 (no new 17 for 8 and less)
champagne
2017 Supporter
Posts: 7213
Joined: 02 August 2007
Location: France Brittany | 8,690 | 26,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-05 | latest | en | 0.920287 |
https://bookofproofs.github.io/branches/analysis/square-roots-proof.html | 1,708,792,297,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00536.warc.gz | 144,715,719 | 4,290 | # Proof
(related to Proposition: Square Roots)
Let $$a > 0$$ be a real number. We want to find all solutions of the equation $x^2=a.\quad\quad( * )$
The key to find a solution is the observation that if $$x$$ is a solution of the equation $$( * )$$, then $$x=\frac ax$$, otherwise $$x\neq\frac ax$$. Therefore, if $$x_0$$ is any approximation of the solution of $$( * )$$, the mean value
$x_1:=\frac 12\left(x_0+\frac a{x_0}\right)$
will be a better approximation to that solution.
Therefore, we define a sequence of real numbers. $\begin{array}{ll} x_0 > 0 \in\mathbb R\quad\text{(arbitrarily chosen)}\\ x_{n+1}:=\frac 12\left(x_n + \frac a{x_n}\right) \end{array}$
The proof consists of three steps:
### $$(1)$$ We show that $$\lim_{n\rightarrow\infty} x_n=\sqrt a$$.
First, we gather some extensions about the behavior of the sequence members of $$(x_n)_{n\in\mathbb N}$$.
(i) We have to make sure that the sequence is well-defined, in particular that there are no divisions by 0, when calculating the sequence members. We prove by induction that $$x_{n} > 0$$ for all $$n \ge 0$$. Since $$x_0 > 0$$ by hypothesis, we can assume that $$x_n > 0$$ is proven for all $$n$$, which are lower or equal some $$n_0\ge 0$$. Therefore (induction step) $$x_{n+1}$$ is a sum of positive values. Thus, $$x_{n+1} > 0$$.
(ii) Next, we observe that $$x_n^2 \ge a$$ for all $$n\ge 1$$. This is because $x^2_n-a=\frac 14\left(x_n +\frac a{x_n}\right)^2-a=\frac 14\left(x^2_n + 2a +\frac {a^2}{x^2_n}\right)-\frac {4a}4=\frac 14\left(x^2_n -2a+\frac 4{x^2_n}\right)=\frac 14\left(x_n -\frac a{x_n}\right)^2 \ge 0.$
(iii) From (ii), it follows that $$1/{x_n^2} \le 1/a$$ for all $$n\ge 1$$. After multiplying the inequality by $$a^2$$, we observe that $$\left(\frac a{x_{n}}\right)^2\le a$$ for all $$n \ge 1$$.
After showing (i) - (iii), we can start to study to compare the distances between the sequence members.
(iv) We will show that $$x_{n+1}\le x_n$$ for all $$n\ge 1$$. This is because $x_n-x_{n+1}=x_n-\frac 12\left(x_n + \frac a{x_n}\right)=\frac{2x^2_n}{2x_n}-\frac{x^2_n}{2x_n}-\frac{a}{2x_n}=\frac{x^2_n - a}{2x_n} \ge 0$
(v) From (iii) and (iv) it follows for $$n\ge 1$$ that $\frac a{x_{n}}\le \frac a{x_{n+1}}.$
(vi) We have for all $$n\ge 1$$ that $\frac a{x_{n}}\le x_n$ Otherwise, we would have $$(a/x_n)^2 > x_n^2$$, which by (ii) leads to $$(a/x_n)^2 > a$$, in contradiction to (iii).
From (iii), (iv) and (vi), it follows that $$(x_n)_{n\in\mathbb N}$$ is a bounded sequence with
$\frac{a}{x_1} \le x_n \le x_1.$
and that it is monotonically decreasing. According to the monotone convergence theorem it has a limit. Thus, it must be a number $$x$$ satisfying the equation $$x=\frac ax$$. We denote $$x$$ by $$\sqrt a$$.
### $$(2)$$ With $$\sqrt a$$ also $$-\sqrt a$$ is a solution of $$( * )$$.
This follows from the rule that $$(-x)(-y)=xy$$ holds for all $$x,y\in\mathbb R$$.
### $$(3)$$ There are no other solutions of $$( * )$$.
By construction, the sequence $$(x_n)_{n\in\mathbb N}$$ converges to the number $$x$$ satisfying $$x=\frac ax$$. The proposition follows from the fact that the limit of a convergent sequence is unique.
Thank you to the contributors under CC BY-SA 4.0!
Github:
### References
#### Bibliography
1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983 | 1,169 | 3,368 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-10 | latest | en | 0.722034 |
https://cbseportal.com/class-12/informatics-practices/syllabus-2019 | 1,600,417,015,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187354.1/warc/CC-MAIN-20200918061627-20200918091627-00013.warc.gz | 433,548,670 | 11,662 | NEW! CBSE Papers PDF: Class-X, Class-XII
# CBSE Class-12 Syllabus 2019-20 (Informatics Practices)
Disclaimer: This website is not at associated with CBSE, For official website of CBSE visit - www.cbse.nic.in
## CBSE Class-12 Syllabus 2019-20 (Informatics Practices)
Informatics Practices(New)
CLASS XII
Code No. 065
Optional for the academic year 2019-20 and mandatory for the academic year 2020-21 onwards
2. Learning Outcomes
1. Understand aggregation operations, descriptive statistics, and re-indexing columns in a DataFrame.
2. Apply functions row-wise and element-wise on a Data Frame.
3. Understand basic software engineering: models, activities, business use-case diagrams, and version control systems.
4. Connect a Python program with a SQL database, and learn aggregation functions in SQL.
5. Have a clear understanding of cyber ethics and cybercrime. Understand the value of technology in societies, gender and disability issues, and the technology behind biometric ids.
3. Distribution of Marks
Unit No. Unit Name Mark 1. Data Handling 30 2. Basic Software Engineering 15 3. Data Management 15 4. Society, Law and Ethics 10 5. Practicals 30 Total 100
4.1. Unit 1: Data Handling (DH-2) (80 Theory + 70 Practical)
4.1.1. Python Pandas
• Advanced operations on Data Frames: pivoting, sorting, and aggregation
• Descriptive statistics: min, max, mode, mean, count, sum, median, quartile, var
• Create a histogram, and quantiles.
• Function application: pipe, apply, aggregation (group by), transform, and apply map.
• Reindexing, and altering labels.
4.1.2. Numpy
• 1D array, 2D array
• Arrays: slices, joins, and subsets
• Arithmetic operations on 2D arrays
• Covariance, correlation and linear regression
4.1.3. Plotting with Pyplot
• Plot bar graphs, histograms, frequency polygons, box plots, and scatter plots.
4.2 Unit 2: Basic Software Engineering (BSE) (25 Theory + 10 Practical)
• Introduction to software engineering
• Software Processes: waterfall model, evolutionary model, and component based model
• Delivery models: incremental delivery, spiral delivery
• Process activities: specification, design/implementation, validation, evolution
• Agile methods: pair programming, and Scrum
• Practical aspects: Version control system (GIT), and do case studies of software systems and build use-case diagrams
4.3. Unit 3: Data Management (DM-2) (20 Theory + 20 Practical)
• Write a minimal Django based web application that parses a GET and POST request, and writes the fields to a file – flat file and CSV file.
• Interface Python with an SQL database
• SQL commands: aggregation functions, having, group by, order by.
4.4. Unit 4: Society, Law and Ethics (SLE-2) (15 Theory)
• Intellectual property rights, plagiarism, digital rights management, and licensing (Creative Commons, GPL and Apache), open source, open data, privacy.
• Privacy laws, fraud; cybercrime- phishing, illegal downloads, child pornography, scams; cyber forensics, IT Act, 2000.
• Technology and society: understanding of societal issues and cultural changes induced by technology.
• E-waste management: proper disposal of used electronic gadgets.
• Identity theft, unique ids, and biometrics.
• Gender and disability issues while teaching and using computers.
• Role of new media in society: online campaigns, crowdsourcing, smart mobs
• Issues with the internet: internet as an echo chamber, net neutrality, internet addiction
• Case studies - Arab Spring, WikiLeaks, Bit coin | 834 | 3,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-40 | latest | en | 0.742372 |
https://www.premiumessaywritingservice.com/what-is-the-difference-between-a-marginal-and-an-average-tax-rate-556167/ | 1,624,483,558,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488540235.72/warc/CC-MAIN-20210623195636-20210623225636-00035.warc.gz | 841,910,391 | 10,440 | Select Page
## what is the difference between a marginal and an average tax rate 556167
Deep in the Heart of Taxes
Algernon, Inc., has a taxable income of \$85,000. What is its tax bill? What is its average tax rate? Its marginal tax rate? We see that the tax rate applied to the first \$50,000 is 15 percent; the rate applied to the next \$25,000 is 25 percent, and the rate applied after that up to \$100,000 is 34 percent. So Algernon must pay .15 × \$50,000 + .25 ×25,000 + .34 ×(85,000 75,000) =\$17,150. The average tax rate is thus \$17,150/85,000= 20.18%. The marginal rate is 34 percent because Algernon’s taxes would rise by 34 cents if it had another dollar in taxable income.
Table 2.4 summarizes some different taxable incomes, marginal tax rates, and average tax rates for corporations. Notice how the average and marginal tax rates come together at 35 percent.
With a flat rate tax, there is only one tax rate, so the rate is the same for all income levels. With such a tax, the marginal tax rate is always the same as the average tax rate. As it stands now, corporate taxation in the United States is based on a modified flat rate tax, which becomes a true flat rate for the highest incomes.
(1) (2) (3) (3)/(1) Taxable Income Marginal Tax Rate Total Tax Average Tax Rate \$ 45,000 15% \$ 6,750 15.00% 70,000 25 12,500 17.86 95,000 34 20,550 21.63 250,000 39 80,750 32.30 1,000,000 34 340,000 34.00 17,500,000 38 6,100,000 34.86 50,000,000 35 17,500,000 35.00 100,000,000 35 35,000,000 35.00
In looking at Table 2.4, notice that the more a corporation makes, the greater is the percentage of taxable income paid in taxes. Put another way, under current tax law, the average tax rate never goes down, even though the marginal tax rate does. As illustrated, for corporations, average tax rates begin at 15 percent and rise to a maximum of 35 percent. It will normally be the marginal tax rate that is relevant for financial decision making. The reason is that any new cash flows will be taxed at that marginal rate. Because financial decisions usually involve new cash flows or changes in existing ones, this rate will tell us the marginal effect of a decision on our tax bill. There is one last thing to notice about the tax code as it affects corporations. It’s easy to verify that the corporate tax bill is just a flat 35 percent of taxable income if our taxable income is more than \$18.33 million. Also, for the many midsize corporations with taxable incomes in the range of \$335,000 to \$10,000,000, the tax rate is a flat 34 percent. Because we will normally be talking about large corporations, you can assume that the average and marginal tax rates are 35 percent unless we explicitly say otherwise. Before moving on, we should note that the tax rates we have discussed in this section relate to federal taxes only. Overall tax rates can be higher once state, local, and any other taxes are considered.
C O N C E P T Q U E S T I O N S
a What is the difference between a marginal and an average tax rate?
b Do the wealthiest corporations receive a tax break in terms of a lower tax rate? Explain. | 815 | 3,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-25 | latest | en | 0.943781 |
http://matlabdatamining.blogspot.com/2010/02/principal-components-analysis.html?showComment=1280508097766 | 1,596,983,216,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738555.33/warc/CC-MAIN-20200809132747-20200809162747-00581.warc.gz | 70,075,430 | 27,076 | ## Friday, February 26, 2010
### Principal Components Analysis
Introduction
Real-world data sets usually exhibit relationships among their variables. These relationships are often linear, or at least approximately so, making them amenable to common analysis techniques. One such technique is principal component analysis ("PCA"), which rotates the original data to new coordinates, making the data as "flat" as possible.
Given a table of two or more variables, PCA generates a new table with the same number of variables, called the principal components. Each principal component is a linear transformation of the entire original data set. The coefficients of the principal components are calculated so that the first principal component contains the maximum variance (which we may tentatively think of as the "maximum information"). The second principal component is calculated to have the second most variance, and, importantly, is uncorrelated (in a linear sense) with the first principal component. Further principal components, if there are any, exhibit decreasing variance and are uncorrelated with all other principal components.
PCA is completely reversible (the original data may be recovered exactly from the principal components), making it a versatile tool, useful for data reduction, noise rejection, visualization and data compression among other things. This article walks through the specific mechanics of calculating the principal components of a data set in MATLAB, using either the MATLAB Statistics Toolbox, or just the base MATLAB product.
Performing Principal Components Analysis
Performing PCA will be illustrated using the following data set, which consists of 3 measurements taken of a particular subject over time:
>> A = [269.8 38.9 50.5
272.4 39.5 50.0
270.0 38.9 50.5
272.0 39.3 50.2
269.8 38.9 50.5
269.8 38.9 50.5
268.2 38.6 50.2
268.2 38.6 50.8
267.0 38.2 51.1
267.8 38.4 51.0
273.6 39.6 50.0
271.2 39.1 50.4
269.8 38.9 50.5
270.0 38.9 50.5
270.0 38.9 50.5
];
We determine the size of this data set thus:
>> [n m] = size(A)
n =
15
m =
3
To summarize the data, we calculate the sample mean vector and the sample standard deviation vector:
>> AMean = mean(A)
AMean =
269.9733 38.9067 50.4800
>> AStd = std(A)
AStd =
1.7854 0.3751 0.3144
Most often, the first step in PCA is to standardize the data. Here, "standardization" means subtracting the sample mean from each observation, then dividing by the sample standard deviation. This centers and scales the data. Sometimes there are good reasons for modifying or not performing this step, but I will recommend that you standardize unless you have a good reason not to. This is easy to perform, as follows:
>> B = (A - repmat(AMean,[n 1])) ./ repmat(AStd,[n 1])
B =
-0.0971 -0.0178 0.0636
1.3591 1.5820 -1.5266
0.0149 -0.0178 0.0636
1.1351 1.0487 -0.8905
-0.0971 -0.0178 0.0636
-0.0971 -0.0178 0.0636
-0.9932 -0.8177 -0.8905
-0.9932 -0.8177 1.0178
-1.6653 -1.8842 1.9719
-1.2173 -1.3509 1.6539
2.0312 1.8486 -1.5266
0.6870 0.5155 -0.2544
-0.0971 -0.0178 0.0636
0.0149 -0.0178 0.0636
0.0149 -0.0178 0.0636
This calculation can also be carried out using the zscore function from the Statistics Toolbox:
>> B = zscore(A)
B =
-0.0971 -0.0178 0.0636
1.3591 1.5820 -1.5266
0.0149 -0.0178 0.0636
1.1351 1.0487 -0.8905
-0.0971 -0.0178 0.0636
-0.0971 -0.0178 0.0636
-0.9932 -0.8177 -0.8905
-0.9932 -0.8177 1.0178
-1.6653 -1.8842 1.9719
-1.2173 -1.3509 1.6539
2.0312 1.8486 -1.5266
0.6870 0.5155 -0.2544
-0.0971 -0.0178 0.0636
0.0149 -0.0178 0.0636
0.0149 -0.0178 0.0636
Calculating the coefficients of the principal components and their respective variances is done by finding the eigenfunctions of the sample covariance matrix:
>> [V D] = eig(cov(B))
V =
0.6505 0.4874 -0.5825
-0.7507 0.2963 -0.5904
-0.1152 0.8213 0.5587
D =
0.0066 0 0
0 0.1809 0
0 0 2.8125
The matrix V contains the coefficients for the principal components. The diagonal elements of D store the variance of the respective principal components. We can extract the diagonal like this:
>> diag(D)
ans =
0.0066
0.1809
2.8125
The coefficients and respective variances of the principal components could also be found using the princomp function from the Statistics Toolbox:
>> [COEFF SCORE LATENT] = princomp(B)
COEFF =
0.5825 -0.4874 0.6505
0.5904 -0.2963 -0.7507
-0.5587 -0.8213 -0.1152
SCORE =
-0.1026 0.0003 -0.0571
2.5786 0.1226 -0.1277
-0.0373 -0.0543 0.0157
1.7779 -0.1326 0.0536
-0.1026 0.0003 -0.0571
-0.1026 0.0003 -0.0571
-0.5637 1.4579 0.0704
-1.6299 -0.1095 -0.1495
-3.1841 -0.2496 0.1041
-2.4306 -0.3647 0.0319
3.1275 -0.2840 0.1093
0.8467 -0.2787 0.0892
-0.1026 0.0003 -0.0571
-0.0373 -0.0543 0.0157
-0.0373 -0.0543 0.0157
LATENT =
2.8125
0.1809
0.0066
Note three important things about the above:
1. The order of the principal components from princomp is opposite of that from eig(cov(B)). princomp orders the principal components so that the first one appears in column 1, whereas eig(cov(B)) stores it in the last column.
2. Some of the coefficients from each method have the opposite sign. This is fine: There is no "natural" orientation for principal components, so you can expect different software to produce different mixes of signs.
3. SCORE contains the actual principal components, as calculated by princomp.
To calculate the principal components without princomp, simply multiply the standardized data by the principal component coefficients:
>> B * COEFF
ans =
-0.1026 0.0003 -0.0571
2.5786 0.1226 -0.1277
-0.0373 -0.0543 0.0157
1.7779 -0.1326 0.0536
-0.1026 0.0003 -0.0571
-0.1026 0.0003 -0.0571
-0.5637 1.4579 0.0704
-1.6299 -0.1095 -0.1495
-3.1841 -0.2496 0.1041
-2.4306 -0.3647 0.0319
3.1275 -0.2840 0.1093
0.8467 -0.2787 0.0892
-0.1026 0.0003 -0.0571
-0.0373 -0.0543 0.0157
-0.0373 -0.0543 0.0157
To reverse this transformation, simply multiply by the transpose of the coefficent matrix:
>> (B * COEFF) * COEFF'
ans =
-0.0971 -0.0178 0.0636
1.3591 1.5820 -1.5266
0.0149 -0.0178 0.0636
1.1351 1.0487 -0.8905
-0.0971 -0.0178 0.0636
-0.0971 -0.0178 0.0636
-0.9932 -0.8177 -0.8905
-0.9932 -0.8177 1.0178
-1.6653 -1.8842 1.9719
-1.2173 -1.3509 1.6539
2.0312 1.8486 -1.5266
0.6870 0.5155 -0.2544
-0.0971 -0.0178 0.0636
0.0149 -0.0178 0.0636
0.0149 -0.0178 0.0636
Finally, to get back to the original data, multiply each observation by the sample standard deviation vector and add the mean vector:
>> ((B * COEFF) * COEFF') .* repmat(AStd,[n 1]) + repmat(AMean,[n 1])
ans =
269.8000 38.9000 50.5000
272.4000 39.5000 50.0000
270.0000 38.9000 50.5000
272.0000 39.3000 50.2000
269.8000 38.9000 50.5000
269.8000 38.9000 50.5000
268.2000 38.6000 50.2000
268.2000 38.6000 50.8000
267.0000 38.2000 51.1000
267.8000 38.4000 51.0000
273.6000 39.6000 50.0000
271.2000 39.1000 50.4000
269.8000 38.9000 50.5000
270.0000 38.9000 50.5000
270.0000 38.9000 50.5000
This completes the round trip from the original data to the principal components and back to the original data. In some applications, the principal components are modified before the return trip.
Let's consider what we've gained by making the trip to the principal component coordinate system. First, more variance has indeed been squeezed in the first principal component, which we can see by taking the sample variance of principal components:
>> var(SCORE)
ans =
2.8125 0.1809 0.0066
The cumulative variance contained in the first so many principal components can be easily calculated thus:
>> cumsum(var(SCORE)) / sum(var(SCORE))
ans =
0.9375 0.9978 1.0000
Interestingly in this case, the first principal component contains nearly 94% of the variance of the original table. A lossy data compression scheme which discarded the second and third principal components would compress 3 variables into 1, while losing only 6% of the variance.
The other important thing to note about the principal components is that they are completely uncorrelated (as measured by the usual Pearson correlation), which we can test by calculating their correlation matrix:
>> corrcoef(SCORE)
ans =
1.0000 -0.0000 0.0000
-0.0000 1.0000 -0.0000
0.0000 -0.0000 1.0000
Discussion
PCA "squeezes" as much information (as measured by variance) as possible into the first principal components. In some cases the number of principal components needed to store the vast majority of variance is shockingly small: a tremendous feat of data manipulation. This transformation can be performed quickly on contemporary hardware and is invertible, permitting any number of useful applications.
For the most part, PCA really is as wonderful as it seems. There are a few caveats, however:
1. PCA doesn't always work well, in terms of compressing the variance. Sometimes variables just aren't related in a way which is easily exploited by PCA. This means that all or nearly all of the principal components will be needed to capture the multivariate variance in the data, making the use of PCA moot.
2. Variance may not be what we want condensed into a few variables. For example, if we are using PCA to reduce data for predictive model construction, then it is not necessarily the case that the first principal components yield a better model than the last principal components (though it often works out more or less that way).
3. PCA is built from components, such as the sample covariance, which are not statistically robust. This means that PCA may be thrown off by outliers and other data pathologies. How seriously this affects the result is specific to the data and application.
4. Though PCA can cram much of the variance in a data set into fewer variables, it still requires all of the variables to generate the principal components of future observations. Note that this is true, regardless of how many principal components are retained for the application. PCA is not a subset selection procedure, and this may have important logistical implications.
See also the Feb-28-2010 posting, Putting PCA to Work and the Dec-11-2010 posting, Linear Discriminant Analysis (LDA) .
Multivariate Statistical Methods: A Primer, by Manly (ISBN: 0-412-28620-3)
Note: The first edition is adequate for understanding and coding PCA, and is at present much cheaper than the second or third editions.
Anonymous said...
Good job..
agree with images examples suggestion..
How / where on your source code specify the number of features you want to extract from given data? for instance, using PCA to extract top 15 feature.
Ibraheem M. Al-Dhamari said...
Excellent,
I hope you provide same excellent explanation of SVM.
regards,
Anonymous said...
respected sir,
ur blog is very helpful to through this i can understand the beauty for PCA method...but can it apply to image..?
Anonymous said...
I greatly appreciate the practical and clear explanation. Other things available are beyond my current background, but this explanation allows me to start empirically playing with the method - in my opinion a good precursor to really understanding it formally (if I ever get there! - and if not I still learn something).
MP said...
The first time I did PCA I used the following document which has pictures. I agree they help a lot. It is nice to see the MatLab code on your blog. I think it would have been better if you did an example where you actually reduced the dimensionality by selecting a subset of feature vectors. Also it's very easy to do this without the statiscs toolbox.
I do something like this in MatLab to select my feature vectors
Values=diag(Values);
[Vsort, Vindices] = sort(-1*Values);
Values = Values(Vindices);
PC = Vectors(:,Vindices(1:number_factors));
@previous poster This process can be applied to images for compression.
http://www.cs.otago.ac.nz/cosc453/student_tutorials/principal_components.pdf
Unknown said...
What do you do if you wanted to know the points that are, say, 1 standard deviations from the mean of the first principal component? In other words, I want to know what is being altered as I move two standard deviations from the first component.
nicc said...
Great. Very clear! I also liked how you offer the non-MATLAB function description. Good show!
Fernando C Barbi said...
Will,
Great Post, thanks! You might want to explicit the relation between V, D, SCORE and COEFF. COEFF is V in descending order and SCORE = B*COEFF.
Anonymous said...
You said A consists of 3 measurements taken of a particular subject over time. But what in this case COEFF SCORE LATENT? How to use this number for better interpretation of your results?
Will Dwinnell said...
COEFF is V, but with the opposite order, so that when it is multiplied by the normalized data, B * COEFF, the result, which is generated by princomp() as SCORE, has the first principal component in the first column. LATENT contains the variances of the principal components.
amin said...
this articel very helpful for me
I am a student from Indonesia
I am now more complete thesis on data mining with the PCA as a dimension reduction but I want to use the Jacobi iteration to find the eigen vector
Can you help me
skumar said...
dear will, thank you for your wonderful explanation on PCA. Can you please explain or help me solve apriori association rule mining using matlab. thanks - s kumar
Amalina said...
hi, glad to find this blog! i have a huge set of data: 17689 approximate coefficient which extracted from feature extraction of MRI brain image. how can i use PCA to reduce the data so that i can use a minimum data for SVM classification purpose. really need your advice, tq
-amalina_azman_80@yahoo.com.my
Anonymous said...
HI, thanks, how the pca could be used for an image?
harly said...
it's very useful to me..
Trevor said...
If you wanted to find how much each original variable contributed to each of the new variables, how would you do it?
hari said...
thank you for ur post ....can u please explain how this can be done for color images?
Paco said...
Good post!!
A question: with this post as starting point, how do you implement rotated PCA with matlab's rotatefactors (varimax algorithm) and the new explained variance based on the rotated components?
Anonymous said...
To plot the PC1 vs PC2 plot do I plot the scores first column Vs scores second column of values?
Will Dwinnell said...
"To plot the PC1 vs PC2 plot do I plot the scores first column Vs scores second column of values?"
That depends on how the principal components were calculated. If the princomp function in the Statistics Toolbox was used then, yes. If the "manual" method I describe here was used, then the order of the principal components column is reversed, so the last 2 columns are the first 2 principal components.
Anonymous said...
Will, excellent post. Thanks. Quick question, at the end of the process my new variables will be the matrix SCORE, right? If I want to add trend line I should use the first, second, …columns of SCORE matrix?
Kh@L3D said...
Hello, I'd like to thank you very much for your tutorial. Very helpful and useful.
But I do have a question. I read in many works that PCA is used as a "preprocessing" method, prior to classification (LDA or other). Therefor, I'd like to ask you, How can I use the principals coefficients to perform a classification ?
I mean, okay, I know the assigned classes to the training matrice T, but when I do : Coeff=princomp(A),I found out that I know nothing about Coeff, I don't know its classes. So how can I perform a classification ?
Thank you very much for your blog. Please, excuse me if my question is misplaced, please indicate me the appropriate forum to ask it.
Thank you very much :)
Will Dwinnell said...
Kh@L3D:
PCA itself is not a classification method. PCA merely rearranges the data to exploit linear structure. As I note in this posting, PCA may or may not help classification, which is a separate process (performed by some classification algorithm: discriminant analysis, neural networks, etc.).
Anonymous said...
Hello! This is one of the best posts I've seen on PCA - thank you!
I've used this post as starting point. However for better interpretation of which variables comprise my factors, I would like to go one step further and rotate my factors using the varimax method of rotation.
Once I've obtained my COEFF, SCORE, and latent, what is the next step?
Thank you~
Anonymous said...
Hello! This is one of the best posts I've seen on PCA - thank you!
I've used this post as starting point. However for a better interpretation of which variables comprise my factors, I would like to rotate my factors using the varimax method of rotation.
My question is, once I've obtained my COEFF, SCORE, and latent, what is the next step to rotation?
Thank you~
Upul Senanayake said...
Hey Will,
Your a life saver! Thank you so much!
Burak said...
Thank you very much for your good explanation. I am wondering about one thing though. Once we get the principal components by using the princomp function of matlab, can we say that the first principal component is related to the first column of the original data matrix? Or is it possible that the first column of the original data matrix does not have much variance as the second column; therefore, the first column of the principal components corresponds to the second column of the original data matrix? How can we know which column of the principal components is related to which column of the original data matrix? Thanks in advance.
Christian Himpe said...
Instead of using repmat one can use bsxfun to compute the correlation more efficiently as follows:
T = bsxfun(@minus,A,mean(A))
T = bsxfun(@times,A,1./std(A))
and since the eigenvalues of a covariance matrix (that by design is positive (semi-)definite) are computed next, one can compute equivalently, but somewhat faster, the singular values:
D = svd(cov(T))
Greetings
Unknown said...
Hi! Lately I've been reading a lot about PCA and I've found this post very useful, kudos to you!
I've learned that a useful method for validating a PCA model (choosing how many PCs to retain) is by cross validation.
Taking your example in Matlab and applying the 'Leave One Out' method (I really need to implement this in my work) how would you do it?
I understand the concept behind CV but I'm a bit confused as how to apply it here (or in any other example), because of my lack of experience.
Any help would be MUCH appreciated!
Nuno B.
Anonymous said...
why there is also a sign problem beside the first-to-last order problem comparing V and COEFF if you compare the last column of V and the first column of COEFF. Which one I should use for the vector? Thanks.
Anonymous said...
If possible, I would like to know how to calculate the relation among original variables and the principal components
Unknown said...
how plotting a first score in a two dimensional plane.
i have a data set: X=[x1 x2]
Unknown said...
how plotting a first score only .
i have a data set X=[x1 x2]
Unknown said...
Hi.how can I plot relation between LDA and PCA.Thank you.
Diego Alonso said...
The PC's are orthogonal (V*V.'), but I may say that the transformed data is not always uncorrelated. In case, in the original data, there is one variable that is a linear combination of other variables, this dependence is kept.
However, PCA can help us to see these dependencies and eliminate redundant variables.
Unknown said...
Thank you very much for this explanation. I found your page because I've been struggling for a while to understand why the coefficients given by matlab are systematically smaller than those given by Statistica. Today, I've realized that the norms of the factors (columns of COEFF)which should be equal to the factors' variances are instead normalized to one. Do you know why, and more importantly how I can prevent this normalization in order to one to stick to Statistica's results ?
Anonymous said...
Super blog post. It helped a lot.
However, I am trying to get the same results out of MATLAB that a colleague is getting out of SPSS.
I tried using applying eig to both the covariance matrix (cov) and the corelation matrix (corrcoef) but it results in the same set of eigenvectors.
I know SPSS has the option to rotatefactors (as does MATLAB of course). But I am talking about the unrotated factors.
Hope you can help
Steve Westland
University of Leeds
Life said...
Very good and clear, after reading this can understand PCA.
How to use PCA to unbalanced data ?
Unknown said...
hello please tell me how can use
pca for privacy preserving in data mining thank u.
Unknown said...
Nice post; the examples make it all very clear. I recently wrote an article about what PCA actually means. This might be helpful for some, to get a more intuitive understanding: http://www.visiondummy.com/2014/05/feature-extraction-using-pca/
Nethravati Ma'm said...
After applying PCA on original and modified data sets, I want to analyse the new data sets for accuracy and information loss & entropy. Any functions or specific codes for the same?
Anonymous said...
If you want to get the original data from B , you do not need COEFF at all. COEFF*COEFF'=identity matrix or 1. I think you should point out that V or COEFF are the eigenvectors of Cov(B). It took me a while to figure out. In this case cov(B) is same as corr(B), as they are zscores. However, B*B' is 14*cov(B). why the 14? it is puzzling me. I was thinking that will give the covariance matrix.
Anonymous said...
Answering post above. B'B is 14 times cov(B) because the covariance estimator is 1/n-1* B'B when E(B)=0 ie series are zero mean. since n=15, n-1 =14. mystery solved. very interesting. covariance esitmator link http://www.encyclopediaofmath.org/index.php/Covariance_matrix
Anonymous said...
Lets assume you've 3 features of an image instead of 3 measurements taken of a particular subject. I have 10 images. My training dataset will be 10 x 3; If I use matlab buildin function princomp and get COEFF SCORE LATENT? which one should I use; score also gives me 3 col. Do I need to use first col. only. How to use this number for better interpretation of my results? how to give input to the classifier
Anisha K S said...
Thanks for sharing this valuable information. I think this PCA method is also helpful for data mining process Containing data compression.
aaaaaaghhhhhhhhhhhhhhhhh said...
Does the dividing of the data by the standard deviation mess it up? Will it lower the energy of the high variance dimensions and increase the energy of the low variance dimensions
Chris said...
Best explanation of PCA on the net! I've been googling it all day.
Anonymous said...
i was searching for 3 months how to use PCA for feature reduction. Thank you for the content.. sdp
Anonymous said...
Hi Will,
Nice post to explain PCA. I wonder if you can help my simple problem. I wish to do a GPR with input from PCA of my data, and I learned that the right way to do the CV is by doing PCA on the training set, then use the training regression coefficients to map the test set to their PCs. The following is my attempt in matlab:
% Calculate xscores for training set
[coef, score, latent, explained, mu] = pca(data_train(:,2:end));
xscore_train=score;
ytrain=zdata_train(:,1);
% Calculate xscores for test set=standardized newX * coef
xscore_test=zdata_test(:,2:end)*coef; ytest=zdata_test(:,1);
Do I compute the xscore_test correctly? Also, the scores returned are not standardized correct?
Thanks a bunch:)
munjal said...
Nice article.
After applying PCA , if I want to take only first component and throw out other 2 components and want to reproduce the original data , then my reproduced data still having three variable. Can you please tell me how this procedure affects my data.
Or if I want to predict some Y variable based on given data , then how can we use PCA for that ?
Thank you.
Unknown said...
For a multi variables data sets is it possible to tell which variable have most influence on 1st Principal component. Another question is that what is the significance of the negative sign in the V matrix component?
Anonymous said...
hai,will can we find the attribute names which has been reduced as the result...
Akira said...
I have been going through number of websites and textbooks to learn how to utilize PCA and somewhat confused because of so many technical terms and formulas flying around. Your explanation just settled everything in place in my brain. Thank you!
Unknown said...
Hey thank u for an example.I tried it with and without using princomp function and my solution without using princomp matches with your answer but i am getting different answer when i use princomp as follows:
coeff =
0.9686 0.1808 -0.1709
0.2019 -0.1698 0.9646
-0.1454 0.9687 0.2010
score =
-0.1721 -0.0108 0.0272
2.5399 -0.1270 0.0612
0.0216 0.0253 -0.0070
2.0831 0.0284 -0.0232
-0.1721 -0.0108 0.0272
-0.1721 -0.0108 0.0272
-1.7388 -0.5398 -0.0491
-1.8260 0.0414 0.0715
-3.1127 0.1830 -0.0490
-2.2829 0.1968 -0.0129
3.7224 0.0730 -0.0473
1.2388 0.1115 -0.0392
-0.1721 -0.0108 0.0272
0.0216 0.0253 -0.0070
0.0216 0.0253 -0.0070
latent =
3.3971
0.0287
0.0015
Thank you
Anonymous said...
Hi, Will!
Nice explanation! Thanks!
Could you also give an explanation for the case when the dimensionality (d) is greater than the number of samples (s), and how to extract more PCs than s?
For example:
Given:
s = 100
d = 3000
Problem:
Let's say reduce d to 1500.
Unknown said...
I have 3 classes with 200 instances in each class. And 14 variables to predict those classes. I want to apply PCA for feature selection and do the classification with few features which are more important.
Could you help me on this and explain how to proceed for feature selection.
Thank you | 7,009 | 25,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-34 | latest | en | 0.897294 |
http://mathoverflow.net/questions/178139/examples-of-unexpected-mathematical-images?answertab=oldest | 1,454,784,729,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701147492.21/warc/CC-MAIN-20160205193907-00191-ip-10-236-182-209.ec2.internal.warc.gz | 130,940,998 | 34,045 | # Examples of unexpected mathematical images
I try to generate a lot of examples in my research to get a better feel for what I am doing. Sometimes, I generate a plot, or a figure, that really surprises me, and makes my research take an unexpected turn, or let me have a moment of enlightenment.
For example, a hidden symmetry is revealed or a connection to another field becomes apparent.
Question: Give an example of a picture from your research, description on how it was generated, and what insight it gave.
I am especially interested in what techniques people use to make images, this is something that I find a bit lacking in most research articles. From answers to this question; hope to learn some "standard" tricks/transformations one can do on data, to reveal hidden structure.
As an example, a couple of years ago, I studied asymptotics of (generalized) eigenvalues of non-square Toeplitz matrices. The following two pictures revealed a hidden connection to orthogonal polynomials in several variables, and a connection to Schur polynomials and representation theory. Without these hints, I have no idea what would have happened. Explanation: The deltoid picture is a 2-dimensional subspace of $\mathbb{C}^2$ where certain generalized eigenvalues for a simple, but large Toeplitz matrix appeared, so this is essentially solutions to a highly degenerate system of polynomial equations. Using a certain map, these roots could be lifted to the hexagonal region, revealing a very structured pattern. This gave insight in how the limit density of the roots is. This is essentially roots of a 2d-analogue of Chebyshev polynomials, but I did not know that at the time. The subspace in $\mathbb{C}^2$ where the deltoid lives is quite special, and we could not explain this. A subsequent paper by a different author answered this question, which lead to an analogue of being Hermitian for rectangular Toeplitz matrices.
Perhaps you do not have a single picture; then you might want to illustrate a transformation that you test on data you generate. For example, every polynomial defines a coamoeba, by mapping roots $z_i$ to $\arg z_i$. This transformation sometimes reveal interesting structure, and it partially did in the example above.
If you don't generate pictures in your research, you can still participate in the discussion, by submitting a (historical) picture you think had a similar impact (with motivation). Examples I think that can appear here might be the first picture of the Mandelbrot set, the first bifurcation diagram, or perhaps roots of polynomials with integer coefficients.
-
This should maybe be community wiki... – Per Alexandersson Aug 9 '14 at 7:01
The appearance of Apollonian circle packing in questions related to the scaling limit of the abelian sandpile model and integer superharmonic functions was quite unexpected. See the papers arxiv.org/abs/1208.4839 and arxiv.org/abs/1309.3267. As I understand it, this observation was made by computing some explicit examples and noticing the fractal pattern. – Sam Hopkins Aug 9 '14 at 7:27
@SamHopkins: This should be an answer! I have seen sandpile-models, and Apollonian gaskets, but never expected a connection! – Per Alexandersson Aug 9 '14 at 7:35
Would <experimental-mathematics> or <visualization> be relevant tags for this question? – J W Aug 11 '14 at 10:39
also see eg Phase Plots of Complex Functions: a Journey in Illustration / Wegert, used to visualize the Riemann zeta fn & related ones – vzn Aug 14 '14 at 15:29
This image shows the boundary of the space of "stabilizable matrices", which in some precise sense dictate the behavior of the scaling limit of the abelian sandpile model on the grid $\mathbb{Z}^2$:
This image is taken from http://arxiv.org/abs/1208.4839. See also http://arxiv.org/abs/1309.3267. The appearance of Apollonian circle packing in questions related to the scaling limit of the abelian sandpile model and integer superharmonic functions was quite unexpected. As I understand it, the connection was made by computing the above image numerically (and maybe it was even the case that the authors thought they had a mistake in their code when these fractal patterns emerged?).
-
Since the oeis has added the feature of having sequences displayed graphically, it has become so much easier to get a quick impression of their behaviour, particularly for many self similar sequences.
-
Do you know any instances where one of these pictures have lead to an insight, or unexpected development? – Per Alexandersson Aug 9 '14 at 8:15
@PerAlexandersson I don't know if these pictures have already raised unexpected developments, but obviously they allow deeper insights, e.g. comparing the toothpick sequence oeis.org/A188346/graph with this one oeis.org/A187210/graph. – Wolfgang Aug 9 '14 at 8:44
John Baez explains here how plotting the roots of polynomials with integer coefficients led to patterns ressembling well known fractals, and how some people figured out ways to explain the unexpected connection.
-
Hardly research level, though visually interesting.
These might show some relations between the discrete and the continuous.
For integer $n$, let $M$ be $n$ by $n$ matrix. For some function $F$, define $M_{x,y}=F(x,y) \mod n$.
Map $M_{x,y}$ to shadows of grey where smaller is darker and larger is closer to white.
Here are some examples for $F(x,y) \in \{x^2+y^2,4x^2+y^2,x^3+x-y^2,xy\}$.
$$x^2+y^2, n=503$$
$$4x^2+y^2, n=503$$
$$x^3+x-y^2, n=503$$
$$xy, n=1961=37\cdot 53$$
-
What is your $n$ ? – darij grinberg Aug 9 '14 at 10:59
These types of images I found very rare. Thanks for uploading it. – user56917 Aug 9 '14 at 11:02
@joro: This is quite nice, I remember experimenting a lot on my graph calculator with this type of patterns when I was half my current age. – Per Alexandersson Aug 9 '14 at 11:35
@Wolfgang The lines appear artifacts of scaling. On a bigger plot they disappear: s12.postimg.org/fpi9upxgt/x_2_y_2_2.png – joro Aug 9 '14 at 13:15
Imagine drawing mod n, mod 2n, mod 3n,... the picture will be essentially the same, just the period between repeating "colors" (or shadows of grey) changes. Likewise between mod n and mod (n+1) etc. So, as @Per also says, the "discretisation" doesn't reveal many more details. You might as well define a continuous (say, periodic) color spectrum. – Wolfgang Aug 9 '14 at 16:39
This image, from the MO question "Gaussian prime spirals," was certainly unexpected:
But the main question I raised,
Q1. Does the spiral always form a cycle?
seems out of reach (as per François Brunault's comment) under "current technology." (Stan Wagon found a cycle of length 3,900,404.)
-
That is really a nice example! – Per Alexandersson Aug 9 '14 at 12:27
Joseph, link to the image appears broken to me. – joro Jan 12 '15 at 11:44
@joro: Looks like a server is down. Several of my images have gone missing. – Joseph O'Rourke Jan 12 '15 at 12:28
OK, just to let you know. You have an option to upload them via the user interface on SE under CC license if you wish. – joro Jan 12 '15 at 12:30
I'm guessing that no one expected uniformly random Aztec diamonds (and similar lozenge/domino tilings) to exhibit circular limit shapes with frozen regions outside. .
The colors in the image are determined from a certain combinatorial object called a height function.
There are a number of ways of generating these images, but the most useful one is via the domino shuffling algorithm. Essentially one builds successively larger uniform tilings by taking a uniformly random $n\times n$ tiling and then expanding it followed by filling in the blanks thus getting a uniformly random $n+1$ tiling.
A nice summary can be found here.
-
This image shows the behavior of a certain function (basically, the "inverse temperature modulo a timescale") associated to various "greedily refined" Markov partitions for the geodesic flow on a g-torus as a function of the (log-) number of partition refinements. When I saw that not only the limit but also even the oscillatory behavior was essentially identical for different genera, I was convinced that there was true physical relevance for this very abstract quantity. There was no reason (other than physics!) to expect such uniformity. Details are in http://arxiv.org/abs/1009.2127
-
The third image below was certainly unexpected for my soon-to-be-collaborators, Emmanuel Candes and Justin Romberg. They started with a standard image in signal processing, the Logan-Shepp phantom:
They took a sparse set of Fourier measurements of this image along 22 radial lines (simulating a crude MRI scan). Conventional wisdom was that this was a very lossy set of measurements, losing most of the original data. Indeed, if one tried to use the standard least squares method to reconstruct the image from this data, one got terrible results:
However, Emmanuel and Justin were experimenting with a different method, in which one minimised the total variation norm rather than the least squares norm subject to the given measurements, and were hoping to get a somewhat better reconstruction. What they actually got was this:
Unbelievably, using only about 2% of the available Fourier coefficients, they had managed to reconstruct the original Logan-Shepp phantom so perfectly that the differences were invisible to the naked eye.
When Emmanuel told me this result, I couldn't believe it either, and tried to write down a theoretical proof that such perfect reconstruction was impossible from so little data. Much to my surprise, I found instead that random matrix theory could be used to guarantee exact reconstruction from a remarkably small number of measurements. We then worked together to optimise and streamline the results; this led to some of the pioneering work in the area now known as compressed sensing.
-
This is amazing! Here's the paper reference ("Robust Uncertainty Principles: Exact Signal Reconstruction From Highly Incomplete Frequency Information"). home.ustc.edu.cn/~zhanghan/cs/Candes%20et%20al.06.pdf – Alex R. Aug 10 '14 at 5:08
The discovery of the special nature of Costa's minimal surface has been made on a visualization.
Generally visualization seems to play an important role in the study of minimal surfaces.
-
The Histogram of all OEIS sequences shows an unexpected gap known as sloane's gap. The plot shows how cultural factors influence mathematics. (http://arxiv.org/pdf/1101.4470v2.pdf)
-
This is really interesting! – Per Alexandersson Aug 10 '14 at 21:05
$N(n)$ is the number of times an integer $n$ occurs in the database. (This wasn't clear to me from the plot.) – Kirill Aug 11 '14 at 10:00
Finding interesting correlations in seemingly trivial concepts. I love it. – AndreasT Aug 12 '14 at 1:52
From the article: "[...] the series of absent numbers was found to comprise 11630, 12067, 12407, 12887, 13258...". What about an OEIS sequence made up of numbers that aren't members of any OEIS sequence? :) – Emanuele Tron Aug 26 '14 at 11:51
11630 is in five OEIS sequences, for example, oeis.org/A163279, which was posted in 2009. 12067 is in 9 sequences, including oeis.org/A163675, from 2009. 12407 is in two sequences, 12887 in seven, 13258 in three. – Gerry Myerson Jun 22 '15 at 23:54
One can obtain a nice picture showing somewhat unexpected patterns by marking all rational points on the unit sphere whose coordinates have denominator less than some upper bound, and projecting this to one of the coordinate planes (cf. this answer of mine to another question). The following picture shows such projection of one octant of the sphere:
This picture in resolution 2048 x 2048 pixels can be found at http://www.gap-system.org/DevelopersPages/StefanKohl/images/sphere1.gif.
-
Really a nice picture! Got people interested on MSE, me included. – MvG Sep 1 '14 at 14:06
It would look even nicer under stereographic projection, since all of the patterns would be circular rather than elliptical. – Adam P. Goucher Jun 22 '15 at 23:48
@AdamP.Goucher: You mean like in the second image here? – Stefan Kohl Jun 23 '15 at 8:55
@StefanKohl Yes, precisely! – Adam P. Goucher Jun 24 '15 at 19:35
For better visualizing and understanding fractals like the Mandelbrot set, the idea of color cycling is a great invention.
Points outside the fractal are colored according to the number of iterations when a threshold assuring divergence ("bail out") is reached. Imagining the fractal bearing en electrical charge or a temperature, the points of same color, i.e. of same rate of divergence, form "equipotential lines" around it. Of course, those lines become more and more intricate as one comes close to the fractal.
So far, this is only static, but now cycling in time through the colors of the (periodic) color palette, either towards the fractal or outward, reveals so much more about its hard-to-see structures. E.g. for the Mandelbrot set, knowing that it is simply connected, cycling helps particularly in regions with spiral-like patterns to get an idea "where it is connected".
Just google for the terms fractal color cycling and you'll find tons of more or less hallucinating videos.
-
there are many aspects of the Collatz conjecture that lend themselves to visualization to the point that significant research insights not found elsewhere can be found in basic graphs of its properties, and a visualization-based/-centric approach can constitute the base of a major "attack" on the problem. one might state that it is an entirely new form of mathematical exploration when combined with computational experiments. with a few caveats on this notoriously difficult problem that even top experts like Erdos are quite wary of:
• note the literature on Collatz is quite sizeable and not highly detailed anywhere (although there are good high-level surveys/ overviews by Lagarias).
• many visualizations of it only look very random, so a lot of ingenuity is required but also rewarded.
here is one such striking example that apparently has not been published (outside of cyberspace).
this visualization shows the function/graph/tree $f'^n(x)$ where $n$ is the $n$th iteration of the Collatz function working in reverse. ie the function starts at 1 and based on the conjecture, visits all integers. the $x$ axis is logarithmic scale. a $2n$ operation moves upward to the right, a $(n-1)/3$ operation moves up to the left. there are two inset details of line intersection "closeups" that show the fractal quality, somewhat reminiscent of the rings of Saturn.
the insight is that this shows the dichotomy/ juxtaposition of order (macroscopic) vs randomness (microscopic) in the problem and leads to other ideas/ strategies about how to approach further analysis.
plots were generated with Ruby/Gnuplot. more details on generation and other visualizations on this page.
-
The spiral of prime numbers (white dots) the "pattern" is amazing, for an explanation of the picture you can take a look to this short youtube video.
-
Looking at this again I can't help but think of Ramsey Theory. – Jp McCarthy Aug 21 '15 at 12:49
Some years ago I was pleasantly surprised when an idea of Jan Mycielski led me to find a very explicit Banach-Tarski paradox in the hyperbolic plane, H^2. H^2 can be decomposed into three simple sets such that each is a third of the space, but also each is a half of the space.
In fact, I found recently how to this even a little more simply, but I like this picture. THe second image is just a viewpoint shift of the first, but makes evident how the blue and green together are congruent to the red.
-
I think that Barnsleys Fern is a really surprising image, that such complex shapes can be encoded in four very simple affine transformations.
If you allow for a larger class of functions (stochastic, $\mathbb{R}^3 \to \mathbb{R}^3$, and introduce a log-density plot and color each point according to orbit history, the possibilities are endless (image created by Silvia C.):
The most common applications of the latter algorithm seems to be producing abstract book covers for books about the universe:
-
These images are the graphs of simple functions using the sinus. You can see them, animated with a function tracer in Flash here: graph of two unexpected functions
$$a=a+3 \\ b=b+10 \times cos(a)\\ \begin{cases} x=a \times cos(a)+b \times cos(b)\\ y=b \times sin(b)+a \times sin(a)\end{cases}$$ $$a=a+\pi/3 \\ b=b+a \times sin(1/a)+a\times cos(1/a)\\da=da+0.0001\\ \begin{cases} x=0.02 \times 1/a \times cos(b\times da)+a \times cos(b\times da)\\ y=0.02 \times 1/a \times sin(b\times da)+a \times sin(b\times da)\end{cases}$$
-
Is this part of your research? – Per Alexandersson Aug 14 '14 at 20:38
@Per Alexandersson Yes, I coded the tracer with actionscript and tested functions... – helloflash Aug 14 '14 at 22:12
And so? What mathematical insights did this give rise to? – Todd Trimble Aug 17 '14 at 14:00
@Todd Trimble This specific research requires no superior knowledge, but who says that it was its pretension? It's an interesting way to create patterns and find textures. – helloflash Aug 17 '14 at 19:35
Sorry for not responding earlier. My comment was in reference to the wording of the OP, which asks specifically what mathematical insights did the image give rise to. I too take aesthetic pleasure in the pictures derived from applying the tracer to your parametric equations, but my reading is that the OP is interested specifically in examples which produced a mathematical insight, in order to be considered on-topic for MO. – Todd Trimble Aug 27 '14 at 19:36
We were highly impressed how very similar growth rules can form a mushroom shape. In the model, each point the growing fungi (network of the thin threads) generates some abstract scalar field. The tips turn towards preferred value of the field and branch when the field drops below some threshold. Add the preferred 45 degree orientation in the earth gravity field - and this is enough to make the system to grow into almost perfect shape of the most primitive mushrooms. Complete description and equations can be found in the published articles, referenced from the neighbour sensing model entry in Wikipedia.
-
This wasn't exactly research, but I have a couple animations I made using a modified version of Melinda Green's Buddhabrot method to render the Mandelbrot set, and what came out was definitely unexpected and pretty shocking to me. I don't think I've ever seen this particular method anywhere else. I've been hoping to get some proper mathematicians to look at the process and give me some insight into why such wild objects seem to form.
This is the first one I made.
Then I tried to make a higher definition animation with different inputs.
You can turn up the quality to see the detail a bit better before watching them. It defaults to 480p, but can be changed to 720p.
To create these, I first began with Melinda's method, which is still explained at her site. It's basically a heat map of how many points in each pixel escaped to "infinity" under the action of the complex seed function. To create motion I decided I would take the coefficients of the function, which was a generalized Mandelbrot-type equation like this:
$$z(w) = aw^3 + bw^2 + cw + C$$
Where $w$ is the complex conjugate of the previous value of the function.
And I would treat those coefficients like a 3-vector (a, b, c). To create motion, I rotated that vector just as if it was a spatial vector rotating through space. The animations are built up of individual images created by slightly transforming the coefficients little by little.
I would really enjoy hearing any insights people have as to why such incredible structures seem to come alive in these visualizations. It is almost eerie. You can see there is a smoke-like effect that gathers around the extended "arms" of the object as it moves, and it almost acts like it is responding to some kind of attractive force (which is mystifying considering what we're looking at). It also has these little three-pointed sparks that fly off the tips, but eventually look like creases in fabric rather than little stars. There are even biological looking structures that appear when the sparks come together and seem to annihilate each other.
On a simpler level, it shocked me that it actually looks like a very distorted physical rotation of some object rotating in higher dimensions, even though it is only a rotation in coefficient space, and not a an actual rotation of spatial coordinates. About halfway through each video, you can see that it really is a rotational transformation, because it comes back around and repeats the entire rotation once more as the vector comes back through its initial position, which was something like (1, 0, 0). In fact, in the first video you can see the exact moment it repeats because the numbers didn't come back around exactly right due to rounding errors that I fixed in the second video.
-
A long time ago, while attempting to classify certain two-dimensional rational conformal field theories (these are certain quantum field theories which enjoy a particular high level of mathematical rigorosity), I found an interesting image which is related to the modular group $\mathbb{P}\mathrm{SL}(2,\mathbb{Z})$, i.e. the matrices $$M = \begin{pmatrix}a & b\\ c& d\end{pmatrix}\,,\ \ \ \ a,b,c,d\in\mathbb{Z}\,, \mathrm{det}\,M = +1\,.$$ Leaving out the details of my classification attempts, I discovered a set of certain conformal field theories characterized by two real parameters $x$ and $y$. They turned out to be rational if and only if $x=a/d$ and $y=b/c$ are both rational numbers with the additional condition that $ad-bc=1$. The connection to the elements of the modular group should be clear from my suggestive notation.
Now, within the classification of conformal field theories, it was natural to look at the $x$-$y$-plane and plot all the points which belong to the set of the rational theories -- which produces the following type of image (showing the first quadrant, and only points with $x<1$ and $y<1$), which I called "modular chaos"
As one might guess, this is only a crude approximation as only points up to a (rather small) maximal denominator are plotted. In fact, one can show that the emerging pattern is dense in $\mathbb{R}^2$, but it is also apparent that it has some fractal-like structure. (Actually, to be more precise, to have points in all four quadrants of the $x$-$y$-plane, one has to consider the weaker condition $ad - bc = \pm 1$.
The above image is not yet particular beautiful, but one can consider the whole plane and use a Poincare map to squeeze it into a unit disk. To enumerate and plot the valid points, one generates the modular group by the two matrices $$S = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\,,\ \ \ \ T = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\,,$$ keeping in mind the relations $S^2=(ST)^3=\mathbb{1}$. It is relatively easy to generate the group in terms of words in $S$ and $T$ up to length $40$ as a binary tree. Encoding by color the length of the word, one finds the following much nicer image .
If you are interested in the connection to conformal field theories, see my two works arxiv:hep-th/9312097 and arxiv:hep-th/9207019. The one from 1993 contains my "proof" that the set is dense in the $\mathbb{R}^2$. I apologize to the mathematicians for the lack of rigor, I am a mere theoretical physicist.
-
Students learning about polar coordinates for the first time may investigate the "roses"
Maybe they will even discover "greatest common divisor" from these.
-
This is beautiful, but badly lacking context... – darij grinberg Jan 12 '15 at 16:46
plots, as a function of the $y$-coordinate, the spectrum of the almost Mathieu operator $H^y:l^2(\mathbb Z)\to l^2(\mathbb Z)$ $$H^y(f)(n)=f(n+1)+f(n-1)+2\cos(2\pi ny)f(n).$$
-
A recent blog post from google shows what happens if you enhance the parts of an image that triggers image recognition (using neural networks) of certain features.
The results are quite spooky, and reveal some hidden structure on what the neural network actually look for when recognizing certain features.
This is the text about the image below:
Instead of exactly prescribing which feature we want the network to amplify, we can also let the network make that decision. In this case we simply feed the network an arbitrary image or photo and let the network analyze the picture. We then pick a layer and ask the network to enhance whatever it detected. Each layer of the network deals with features at a different level of abstraction, so the complexity of features we generate depends on which layer we choose to enhance. For example, lower layers tend to produce strokes or simple ornament-like patterns, because those layers are sensitive to basic features such as edges and their orientations.
-
The picture is taken from The Amazing, Autotuning Sandpile by Jordan Ellenberg.
-
Abelian sandpile model on a square grid, to be more precise. See other grids at math.cmu.edu/~wes/sandgallery.html . – darij grinberg Aug 20 '15 at 13:30
-
This is neat, but probably needs a couple of qualifications: (1) The apparent randomness in the picture is due to the centers of the circles being chosen at random (there is no deterministic chaos here), and (2) the concentric circles are (as far as I can tell) just an artistic way to make the various regions easier to distinguish from each other. – darij grinberg Aug 20 '15 at 13:26
When Thierry Gallay and I introduced the Numerical measure of a matrix, we encountered the following simulation of the measure density (here a $3\times3$ matrix). This lead us to conjecture, and eventually prove, that the density is constant (in general, a polynomial of degree $n-3$) in the curved triangle. We eventually made a link with the theory of lacunae of hyperbolic differential operators.
The lines are level lines of the density. The outer line, where the density vanishes, is the boundary of the Numerical range. It is convex, according to the Toeplitz-Hausdorf Theorem.
-
Entangled Partition Function(s):
-
Could you explain what the image is about? – Joonas Ilmavirta Jan 14 at 20:58
A little mathematical explanations would be nice ... – Sebastian Goette Jan 14 at 21:02 | 6,242 | 26,306 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2016-07 | longest | en | 0.93958 |
https://cupsplus.net/how-many-cups-are-in-a-gallon/ | 1,675,405,153,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00147.warc.gz | 214,681,661 | 55,959 | Gallons to Cups
# How many cups are in a gallon? U.S gallon to cups
Do you want to know how many cups are in a gallon? Jackie Reeve is here to help! The following article will teach you to use this conversion calculator to convert U.S gallon to cups. With the help of this handy guide, you can answer questions like “How much sugar do I need for a cake?” or “How much milk will fit into my pitcher?”.
## How many cups are in a gallon? gallon to cups
Well, this is a difficult question if you don’t know the standard formula.
Don’t worry
To convert gallons to cups, you first need to multiply the gallon value by 16. For example, to find out how many cups there are in a half gallon, multiply 0.5 by 16, which makes 8 cups in a half gallon.
Here is the formula to convert gallons to cups:
cup = gallon * 16
Now you’re ready to get the result: 1 US fluid gallon = 16 US cups
## How many cups to the gallon?
### How many cups are in a gallon of liquid?
There are 16 cups in 1 gallon of liquid. This is because there are 4 quarts in a gallon, and 1 Quart = 4 cups. Thus, 4 multiplied by 4 equals 16 cups in a gallon.
See also How many 8 oz cups in a gallon? How many 8 oz cups in a gallon to drink in a day?
### How many cups are in a gallon of water?
According to US standards, one gallon of water is equivalent to 16 cups.
### How many cups are in a gallon of milk?
A gallon of milk typically contains 16 cups.
However, this can vary depending on the type of milk and the size of the cup. For example, a gallon of skim milk may contain more cups than a gallon of whole milk. When in doubt, check the packaging for specific measurements.
### How many cups are in a gallon of coffee?
A gallon of coffee generally contains around 16 -20 cups. However, this can vary depending on the size of the cup and how much coffee is used to make each cup.
## Systems of Measurement: US & UK Gallon
### U.S gallon to cups
• There are 16 cups in a gallon – in the United States (US, fluid).
(1 Gallon (US, fluid) = 16 Cups )
• There are 18.6183435 cups in one dry gallon (US).
(1 Gallon (Dry, US) = 18.6183435 Cups)
### British Gallon to Cups
• There are 16 cups in 1 UK imperial gallon.
(1 Gallon (UK, imperial) = 16 Cups (UK, imperial)
• There are 18.184361 metric cups in one imperial gallon from the UK.
(1 Gallon (UK, imperial) = 18.184361 Cups (metric)
### The Us Gallon: Who Uses It?
The US gallon is the standard measurement for fuel in the United States, as well as in some Latin American and Caribbean countries.
### The Imperial Gallon: Who Uses It?
The Imperial gallon is the standard form of measurement in countries such as the United Kingdom, Canada, Ireland, Australia, and some Caribbean nations.
## What Are the Differences Between the Metric and Imperial Systems?
Most of the world uses the Metric System, with the exception of the United States which instead relies on The Imperial System.
Each system is used to calculate distance, volume, and weight.
• Distance: The metric system uses meters and centimeters, while the imperial system uses inches, feet, yards and miles.
• Volume: There are two systems for measuring volume: the Metric System and the Imperial System. The Metric System uses fluid ounces, liters, and milliliters. The Imperial System uses cups, pints, quarts, gallons, tablespoons, and teaspoons.
• Weight: The Metric System uses metric units such as grams and ounces, while the Imperial System relies on imperial units like pounds.
See also How many cups in a gallon? How many cups are in a gallon of water?
## Convert gallons to cups in a simple way
### Conversion Chart: Gallon to Cups
Take it easy! You can easily break down your gallons into cups by using a conversion chart.
Now, the question about “how many cups are in a gallon” doesn’t make you confused anymore.
The result is: 16 cups make one gallon.
### Using the Formula: gallon in cups
Using the formula, you can easily gain your cups in gallons.
Just multiply gallons by 16 and know the answer. It only takes you a few seconds to get the answer.
Haha, don’t be surprised, the formula to convert gallon into cups is actually very simple:
cup = gallon * 16
For Example: How many cups are in a gallon?
1 cup = 1 gallon x 16 = 16 gallons.
So, a gallon has 16 cups.
### Gallon to cup conversion by using the online conversion tool
Why? Because they often care about getting an answer quickly and easily. And that’s just how it is right now. 🙂
There are 16 cups in 1 gallon. Easy right?
## What is the difference between a dry and a wet gallon?
The “Winchester” system of measurements assesses dry ingredients by volume and is standardized to measure grains into common units like bushels, pecks, dry gallons, and dry pints. So a dry measurement gauges volume in a way that is similar to measuring wet ingredients.
For example, a wet gallon is the amount of liquid that fills a standard imperial-sized gallon container. The gallon milk jug at the grocery store is an example of this measurement.
## How many cups in a US dry gallon?
A gallon is a measurement of volume, typically used to quantify liquids like milk or water. It can also be used, less commonly, to measure the number of dry goods.
See also How many 8 oz cups in a gallon? How many 8 oz cups in a gallon to drink in a day?
The dry gallon has a long and varied history, starting with British rule. It was originally used to measure dry goods and grain. However, nowadays it is more common to find dry gallons in the United States.
A dry gallon is a unit of measurement for volume that falls between a US gallon and an Imperial gallon. There are, on average, 148 ounces in one dry gallon. This means that there are approximately 18 US standard cups in 1dry which is 2 more than you would find if using a liquid gallon.
Most recipes will call for using cups to measure dry ingredients and grains, not gallons. You likely won’t need to convert into dry gallons, but if you do, keep in mind that a dry gallon is more than a liquid gallon.
## What is the best unit of measurement for recipes?
Many people find that using imperial measurements is more convenient than metric when it comes to cooking recipes.
The use of imperial measurements might be because hobbyist cooks and professional chefs around the world have used them for years. Or, perhaps it is more convenient to use imperial when measuring the sizes and weights of ingredients.
Unless you require a higher level of accuracy, such as for baking recipes, then imperial measurements are more universally used.
## FAQs: How many cups are in a gallon? Cups to US Gallon
### A gallon contains how many quarts?
The gallon is the most common unit used to measure liquid volume and is equal to 4 quarts.
### A gallon contains how many ounces?
There are 128 ounces of liquid in 1 gallon.
### What is the volume of a gallon in pints?
There are 8 pints in 1 gallon.
### How many liters are there in a gallon?
A gallon in the U.S. is equivalent to 3.785 liters.
## Conclusion
In this article, Cupsplus.net hope we helped answer your question “How many cups are in a gallon?” You could also try searching for “cups to US Gallon” (or something similar) if you want to find answers on that topic.
Thanks again for reaching out!
### Jackie Reeve
Since 2015, Jackie Reeve has worked at Wirecutter as a senior staff writer about bedding, organizing, and homeware. She formerly worked as a school librarian and for roughly 15 years as a quilter. Her quilt designs and other writings have been published in a number of magazines. Every morning, she gets ready for bed and conducts the Wirecutter staff book club. At the same time, she works as an editor at cupsplus.net | 1,796 | 7,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-06 | latest | en | 0.938483 |
fotoradosti.ru | 1,606,508,066,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194171.48/warc/CC-MAIN-20201127191451-20201127221451-00058.warc.gz | 300,798,127 | 9,884 | Do you need sex without obligations? CLICK HERE NOW - registration is completely free!
Students will Write linear inequalities from a verbal description Graph a system of linear inequalities with a domain restriction Write and solve inequalities to answer real-world questions Find an inverse function Before you begin Student should be comfortable using relating equations to graphs. They do not need to have exposure to inequalities prior to this lesson in order to use them throughout. Students use linear, rational, and piecewise functions to describe the total and average costs of an order at Carpe Donut. What are some ways to encrypt secret messages? Students explore function concepts using ciphers to encrypt messages both graphically and algebraically; they try to decrypt some messages too. Should you buy a camera lens with vibration reduction? Students interpret graphs and use right triangle trigonometry to explore the relationship between focal length, viewing angle, and blurriness. Does the same sound always sound the same? Students come up with equations in several variables to explore the Doppler Effect, which explains how sound from a moving object gets distorted. How should pharmaceutical companies decide which drugs to develop?
Math formula dating age
According to this rule, the age of the younger person should not be less than half the age of the older person plus seven years, so that for example no one older than 65 should be in a relationship with anyone younger than 39 and a half, no one older than 22 should be in a relationship with anyone younger than 18, and no one under 14 years of age should be in a relationship at all From another point of view, the chart can be interpreted as saying that there should not be an age disparity of as much as five years unless the younger person has an age of 19 or more, a ten-year disparity should exist only if the younger person has an age of 24 or more, and a twenty-year disparity should occur only if the younger person has an age of 34 or more.
And people only slightly older than 14 should only be involved with those almost exactly the same age as themselves. To read the chart, go to the position along the x-axis which corresponds to your age, and the green range between the black and red lines directly above that position corresponds to the range of your partner’s ages which is deemed acceptable by the rule.
A common rule of thumb, at least on the internet, is that it’s okay to be interested in someone “half your age plus seven” years. According to this.
If you’re seeing this message, it means we’re having trouble loading external resources on our website. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Donate Login Sign up Search for courses, skills, and videos. Science Biology library History of life on Earth Radiometric dating. Chronometric revolution. Potassium-argon K-Ar dating. K-Ar dating calculation. Atomic number, atomic mass, and isotopes.
Do you want to find a partner for sex? It is very easy. Click here NOW, registration is completely free!
Article that is based upon the calculation applied to write the us a. Calculate your relationship between different people’s ages. Xkcd dating method by its original carbon 14 is not the time interval.
In this section we will explore the use of carbon dating to determine the age of fossil remains. Carbon is a key element in biologically important molecules.
You may have heard that the Earth is 4. This was calculated by taking precise measurements of things in the dirt and in meteorites and using the principles of radioactive decay to determine an age. This page will show you how that was done. Radioactive nuclides decay with a half-life. If the half-life of a material is years and you have 1 kg of it, years from now you will only have 0. The rest will have decayed into a different nuclide called a daughter nuclide.
Several radioactive nuclides exist in nature with half-lives long enough to be useful for geologic dating. This nuclide decays to Strontium Sr87 with a half-life of Imagine going way back in time and looking at some lava that is cooling to become a rock.
What does relative dating mean
Make up the process of relative dating techniques is because you date for relative dating mean? Jump to another rock layer fig- ure 3. Find a relative dating services and the oldest rocks at a fancy term: use 2 methods that is the strata relative ages?
How relative positions of rocks allow scientists to compare their ages. 61 Discusses how radiometric dating can provide an absolute age (date in number of.
More recently, a plethora of market-minded dating books are coaching singles on how to seal a romantic deal, and dating apps, which have rapidly become the mode du jour for single people to meet each other, make sex and romance even more like shopping. The idea that a population of single people can be analyzed like a market might be useful to some extent to sociologists or economists, but the widespread adoption of it by single people themselves can result in a warped outlook on love.
M oira Weigel , the author of Labor of Love: The Invention of Dating , argues that dating as we know it—single people going out together to restaurants, bars, movies, and other commercial or semicommercial spaces—came about in the late 19th century. What dating does is it takes that process out of the home, out of supervised and mostly noncommercial spaces, to movie theaters and dance halls. The application of the supply-and-demand concept, Weigel said, may have come into the picture in the late 19th century, when American cities were exploding in population.
Read: The rise of dating-app fatigue. Actual romantic chemistry is volatile and hard to predict; it can crackle between two people with nothing in common and fail to materialize in what looks on paper like a perfect match. The fact that human-to-human matches are less predictable than consumer-to-good matches is just one problem with the market metaphor; another is that dating is not a one-time transaction.
This makes supply and demand a bit harder to parse. Given that marriage is much more commonly understood to mean a relationship involving one-to-one exclusivity and permanence, the idea of a marketplace or economy maps much more cleanly onto matrimony than dating. The marketplace metaphor also fails to account for what many daters know intuitively: that being on the market for a long time—or being off the market, and then back on, and then off again—can change how a person interacts with the marketplace.
W hen market logic is applied to the pursuit of a partner and fails , people can start to feel cheated.
314: Dating Pools
Two things: the somewhat random-looking picture above is from The Moon Is Blue, a film that’s the first known reference to the ‘dating rule’ discussed here. Secondly, I don’t make any judgement about the validity of the ‘dating rule’ – you might find it a useful rule of thumb or a ludicrous restriction; I find it a nice thing to do some algebra on. There is a so-called rule about dating: the youngest age you are supposed to date is half your age plus seven.
Why would we want to do this? Well, it means we can work out some other implications of the rule.
Radiocarbon dating involves determining the age of an ancient fossil or specimen by measuring its carbon content. Carbon, or radiocarbon.
Love-hungry teenagers and archaeologists agree: dating is hard. But while the difficulties of single life may be intractable, the challenge of determining the age of prehistoric artifacts and fossils is greatly aided by measuring certain radioactive isotopes. Until this century, relative dating was the only technique for identifying the age of a truly ancient object. By examining the object’s relation to layers of deposits in the area, and by comparing the object to others found at the site, archaeologists can estimate when the object arrived at the site.
Though still heavily used, relative dating is now augmented by several modern dating techniques. Radiocarbon dating involves determining the age of an ancient fossil or specimen by measuring its carbon content. Carbon, or radiocarbon, is a naturally occurring radioactive isotope that forms when cosmic rays in the upper atmosphere strike nitrogen molecules, which then oxidize to become carbon dioxide.
Green plants absorb the carbon dioxide, so the population of carbon molecules is continually replenished until the plant dies. Carbon is also passed onto the animals that eat those plants. After death the amount of carbon in the organic specimen decreases very regularly as the molecules decay. Samples from the past 70, years made of wood, charcoal, peat, bone, antler or one of many other carbonates may be dated using this technique.
18.5D: Carbon Dating and Estimating Fossil Age
Learn The Trick! How young is too young to hook up with someone? Do the math. The age , here, is usually the man’s.
Your you xkcd dating equation dating equation dating age plus seven rule. How smart are the perfect time to prevent this in the pools. First off, math, math, but.
Subscribe To Our Newsletter! According to this rule, it would not be creepy for a 30 year old to date a 22 year-old, but an 18 year-old would be off-limits. Although this is a fun rule of thumb, what does research say about age preferences for potential mates? From an evolutionary perspective , it makes sense for women to prefer mates with resources and to like partners who are more established, both of which are more likely in older partners.
Men, in contrast, are hypothesized to be most attracted to women in their reproductive prime, which tends to be when they are younger. Data from Kenrick and Keefe 1 support these predictions. Younger men tend to prefer women a few years younger or older than themselves; but as they get older, they increasingly prefer younger women relative to their own age. It turns out that, on average, women tend to be married to men a few years older than themselves years. However, younger men i. | 2,065 | 10,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-50 | latest | en | 0.94616 |
http://justinmallone.com/tag/simply-scheme/ | 1,603,582,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885059.50/warc/CC-MAIN-20201024223210-20201025013210-00036.warc.gz | 58,728,104 | 52,653 | ## How to Define a Procedure
Essential aspects of defining a procedure, with an example:
1. The word `define`, which indicates that you are defining something.
2. The name you want to give the procedure.
3. The name(s) you want to give its argument(s).
4. Body: what the procedure actually does. “an expression whose value provides the function’s return value.”
## Special Forms
`define` is an example of a special form. Special forms have their own evaluation rules. For `define` in our `square` example, none of the parts got evaluated, which is different than how Scheme normally treats expressions. Normally, Scheme evaluates sub-expressions and then evaluates the whole expression once the sub-expressions have been evaluated. But that wouldn’t make sense to do with `define`.
`define` isn’t actually a procedure at all in a technical sense. But as a simplification, the book discusses `define` as if it is a procedure, unless there is some important reason not to.
## Functions and Procedures
Function — an association between the starting value(s) and the resulting value, no matter how that result is computed. The same function can be represented by different operations.
The book notes that there is a distinction between a function and a procedure, and that functions can be represented in different ways (such as a table that represents the relationship between US States and state capitals; I also came up with the example of coordinates that indicate a certain location on a map – the coordinates are the starting value, the specific location on a map is a return value, and the relationship between coordinates and locations is represented by the map, and might even be represented three-dimensionally by a globe).
## Argument Names versus Argument Values
Formal parameter – what you call an argument as defined in a function. E.g. for…
the formal parameter is `x`. It’s a placeholder value for an actual argument.
Actual argument expression – an expression that serves as the actual argument for a procedure.
Actual argument value – the specific value that a procedure receives in a case when it is invoked.
Example: for `(square (+ 5 9))`, `(+ 5 9)` is the actual argument expression, and the actual argument value is 14.
## Substitution Model
Today’s story is about the substitution model. When a procedure is invoked, the goal is to carry out the computation described in its body. The problem is that the body is written in terms of the formal parameters, while the computation has to use the actual argument values. So what Scheme needs is a way to associate actual argument values with formal parameters. It does this by making a new copy of the body of the procedure, in which it substitutes the argument values for every appearance of the formal parameters, and then evaluating the resulting expression. So, if you’ve defined `square` with
then the body of `square` is `(* x x)`. When you want to know the square of a particular number, as in `(square 5)`, Scheme substitutes the 5 for `x` everywhere in the body of square and evaluates the expression. In other words, Scheme takes
then does the substitution, getting
and then evaluates that expression, getting 25.
## Shorter Points
Abstracting out a pattern and giving it a name is important. It lets you deal with certain problems without having to worry about the details.
Composition of functions is the basis for all Scheme programming.
## Pitfalls
Functions can only have one return value. If you return multiple values, Scheme will ignore earlier ones and return the last one.
Don’t use the name of a procedure as a formal parameter. If you use the same name with two meanings within the same procedure (as e.g. parameter and procedure) it will cause a problem.
Similar issue as above: don’t use the name of a keyword (such as `define`) as some other kind of name.
Don’t try to write a procedure that has a complex expression as a formal parameter.
## Questions
1. In what important cases, if any, does `define` not being a “real” procedure make a difference?
## Exercises
### ✅ Exercise 4.1
Show the substitution that occurs when you evaluate
### ✅🤔Exercise 4.2
Given the following procedure:
list all the little people that are involved in evaluating
Substituting the parameter in:
(Give their names, their specialties, their arguments, who hires them, and what they do with their answers.)
There are three little people. I’ll give them names (like in a prior exercise) and say what they do.
• Adam is a `yawn` specialist and hires everyone else. Yawn takes the `(/ 8 2)` expression and substitutes it into the body of `yawn`. He has nobody that he reports to.
• Bob is an addition specialist. His arguments are 3 (given directly) and 8 (given by Charlie). The result is 11, and this is reported to Adam.
• Charlie is a multiplication specialist. His arguments are 4 (given by Donald) and 2 (given directly). The result is 8. This result is reported to Bob.
• Donald is a division specialist. His arguments are 8 and 2 (both given directly). The result is 4, and this resulted is reported to Charlie.
Note: AnneB talks about a “head boss” who hires the `yawn` specialist. This tracks how the book talked about this. I’m unclear whether this is an important detail or whether the book was just trying to be colorful with the hiring metaphor.
### ✅ Exercise 4.3
Here are some procedure definitions. For each one, describe the function in English, show a sample invocation, and show the result of that invocation.
Note: I edited some of these for a mistaken use of “parameter” in place of “argument”.
#### ✅ 4.3.1
This procedure finds the difference between the two arguments provided; specifically, it takes the second argument and subtracts the first argument from it.
A sample invocation:
#### ✅ 4.3.2
This procedure returns as a result whatever is provided as the argument. Sample invocations:
#### ✅ 4.3.3
This procedure takes an argument but just returns 3, regardless of the argument provided.
#### ✅ 4.3.4
This procedure does not take an argument. It merely defines the name `seven` as `7`.
#### ✅ 4.3.5
This procedure takes a number argument and performs a series of arithmetical operations on the number which result in the same number being returned.
the number `n` is multiplied by 3, added to 13, added to the difference of itself and 1, divided by 4, and then has 3 subtracted.
So for example, 10 is multiplied by 3, producing 30.
It’s then added to 13, producing 43.
It’s added to (10 – 1), so now it’s 52.
It’s divided by 4, so it’s now 13.
Then 3 is subtracted, so it’s back to 10.
Sample invocation:
Note: AnneB actually did the math to make sure this function always returns the given argument. 👍
### ✅ Exercise 4.4
Each of the following procedure definitions has an error of some kind. Say what’s wrong and why, and fix it:
#### ✅ 4.4.1
As written, this expression is trying to return multiple values, and so the earlier value is getting ignored. What the programmer wants to do is multiply these values together. This can be accomplished by deleting the `)` to the right of the `3.141592654`. Fixed:
#### ✅ 4.4.2
This one is using infix notation but Scheme uses prefix notation.
#### ✅ 4.4.3
This one forgot to give `square` a formal parameter.
#### ✅ 4.4.4
This doesn’t designate “base” or “height” as the formal parameters, but uses them in the body anyways. Thus these are just undefined terms and Scheme doesn’t know what to do with them. One possible fix:
#### ✅ 4.4.5
This is using complex expressions for the formal parameters. This is trying to do some of the job of the procedure invocation stage at the definition stage.
and if you want to square the arguments, just do something like:
### ✅ Exercise 4.5
Write a procedure to convert a temperature from Fahrenheit to Celsius, and another to convert in the other direction. The two formulas are F=9⁄5C+32 and C=5⁄9(F-32).
### ✅ Exercise 4.6
Define a procedure `fourth` that computes the fourth power of its argument. Do this two ways, first using the multiplication function, and then using `square` and not (directly) using multiplication.
Way 1:
Note: AnneB’s solution is better, as she makes use of the fact that `*` takes multiple arguments, so the result is more elegant.
Way 2:
### ✅ Exercise 4.7
Write a procedure that computes the absolute value of its argument by finding the square root of the square of the argument.
### ✅ Exercise 4.8
Note that I changed the formatting of the exponents cuz I don’t think Ulysses does superscript.
“Scientific notation” is a way to represent very small or very large numbers by combining a medium-sized number with a power of 10. For example, 5 x [10^7] represents the number 50000000, while 3.26×[10^-9] represents 0.00000000326 in scientific notation. Write a procedure `scientific` that takes two arguments, a number and an exponent of 10, and returns the corresponding value:
Some versions of Scheme represent fractions in a/b form, and some use scientific notation, so you might see `21/50000` or `4.2E-4` as the result of the last example instead of `0.00042`, but these are the same value.
Solution for `scientific`:
#### ❌ Harder Probem for Hotshots
(A harder problem for hotshots: Can you write procedures that go in the other direction? So you’d have
You might find the primitive procedures `log` and `floor` helpful.)
I could not figure out how to use `log` and `floor` effectively.
After trying for a while and writing the stuff about getting unstuck at the end of the post, I started reading Anne’s answer. I’m taking the approach I talk about later in the post of seeing “if it’s possible to do ‘partial spoilers’, like get a hint from someone or only read the beginning of an answer, so you can get unstuck while still figuring some things out.”
Anne says:
I wrote these as a first try:
(define (sci-exponent n)
(floor (log n)))
(define (sci-coefficient n)
(/ n (expt 10 (sci-exponent n))))
But I saw that in my program, log returns the natural log, not the base 10 log.
Yeah okay, I did get as far as noticing that.
I looked in the DrRacket documentation and saw that if you give a second argument to log then it’s supposed to use the second argument as a base.
Ah! I’m gonna add “check the documentation” to a list of things at the end of the post.
More AnneB:
That did not work in my version—it said it expected one argument and not two.
I searched some more and found this, which shows how to get a base 10 log:
Here’s the usual way to do that.
`(define (log10 n) (/ (log n) (log 10)))`
`(log10 100)`
2.0
Ah so that’s interesting. I came across something like this as well. I think I didn’t read it cuz I wanted to try to figure out the problem using the stuff I’d already learned from the book. That’s a bit silly though. After you’ve been stuck for a while, it’s worth trying out different things. It’s possible that if I’d checked the documentation like Anne did and had seen how `log` was supposed to work, I would have been more open to trying different things.
With the very big hint re: the log10 stuff and Anne’s other discussion so far, I was able to write procedures successfully:
Let me try to explain what each part is doing. Warning – I try to explain math stuff a bit so this might be a little muddled.
`log10` is dividing the natural logarithm of the argument number by the natural logarithm of 10. In math, there is something called the change of base formula.
Suppose you want to evaluate the logarithm of a number in some base that isn’t on a calculator or that you don’t have a built-in programming language function for figuring out. The change of base formula lets you evaluate the logarithm of some number in whatever base you want. The first step is to find the logarithm of the number in whatever base you have access to a logarithm function for. Recall that `log` in Scheme finds the natural logarithm. Then we divide by the logarithm of the base we actually want to use(again, in whatever base we have access to, which for Scheme is the natural logarithm).
I found an example:
So for our case, mathematically, we want to find the natural logarithm of some number and then divide that by the natural logarithm of 10. That is what the `log10` function accomplishes.
If we get call `log10` on `7300`, we get 3.8633228601204554. That is the power we would have to raise 10 to in order to get 7300. That makes sense, since raising 10 to 3 gets us 1000 and raising 10 to 4 gets us 10,000, so the value for 7300 has to be within 3 and 4.
`3.8633228601204554` isn’t a great number for the purposes of `sci-exponent` though. The purpose of `sci-exponent` is to tell us what power of 10 we’d have to raise 10 to so that (10 raised to some power) multiplied by some coefficient will produce some number. We know that 10^3 is 1000 and 10^4 is 10,000. `3.8633228601204554` is a long number with a lot of stuff after the decimal cuz it needs to raise 10 to 7300. One way to think about things is that the `3` gets the 10 to 1000 and the `.8633228601204554 ` gets us the rest of the way to 7300. But for the purposes of `sci-exponent` we’re just interested in the `3` (or whatever whole number exponent) and not interested in the stuff after the decimal. Think of the number 7300. One way of representing it is 7.3 x 1000. The coefficient part of scientific notation is the part on the left – the 7.3. The purpose of `sci-exponent` is to figure out the value on the right – and specifically to figure it out in exponential form (as in, 10 to what power produces the value on the right).
We know that scientific notation involves a coefficient and an exponent combining to concisely express a large or small number. If we took the `log10` of some number and rounded up, and then tried to multiply the result from doing that with the coefficient, we’d get the incorrect answer. For example, if we rounded `3.8633228601204554` up to `4`, and then multiplied that by 7.3, we’d get 73000. So for the purposes of figuring out `sci-exponent`, we want to always be rounding the `log10` down. That is what `floor` accomplishes – it rounds down to the next integer.
`sci-coefficient` simply divides the number by 10 raised to the `sci-exponent` of the number. This produces the coefficient, as you would expect.
Some test cases and results:
### ✅ Exercise 4.9
Define a procedure discount that takes two arguments: an item’s initial price and a percentage discount. It should return the new price:
(discount 10 5)
9.50
(discount 29.90 50)
14.95
Note: AnneB‘s version is more elegant because she just multiplied `original-price`, `0.01` and what i call `percentage-reduction` together, rather than doing more nesting and dividing by 100.
### ✅ Exercise 4.10
Write a procedure to compute the tip you should leave at a restaurant. It should take the total bill as its argument and return the amount of the tip. It should tip by 15%, but it should know to round up so that the total amount of money you leave (tip plus original bill) is a whole number of dollars. (Use the ceiling procedure to round up.)
(tip 19.98)
3.02
(tip 29.23)
4.77
(tip 7.54)
1.46
One caveat: the `(tip 19.98)` test case produces some kind of rounding error that can be replicated doing other simple arithmetic operations like `(- 23 20.98)`. AnneB had this issue as well.
### Troubleshooting an Instance of Getting Stuck
I got stuck on the “harder problem for hotshots” from Exercise 4.8. I’m going to write some thoughts about that. I broke the thoughts up into two categories: 1) stuff that I think is more about how to approach getting stuck in general and 2) stuff that seems more specific to getting unstuck while doing programming.
#### Getting Unstuck in General
• Consider the time-based metric for overreaching. I went way over 15 minutes being stuck this time.
• when thinking of time-based metric for overreaching, consider setting an actual timer once you realize you are getting stuck. Give yourself e.g. 15 minutes to get unstuck, and if the timer goes off, just move on with other parts of the project/life. It’s very easy for me (and I bet for other people) to get caught in biased “I just need five more minutes to figure this out” loops, so external checks/alerts like a timer can help with that. (This is a bit like the gambler who thinks things will turn around on the next roll of the dice. The fact that sometimes things do turn around doesn’t make this approach any less of a bad strategy – you gotta think about things in terms of expected return instead of focusing on outlier outcomes).
• Consider alternate uses of time within the project – e.g. consider whether it is worth spending an hour being stuck when you could just read the answer and move on and actually learn more stuff.
• Consider alternate “meta” uses of time – e.g. it might be way more valuable to spend an hour writing learning methodology troubleshooting analysis like this rather than spend it on being stuck.
• Think about motivation for spending time on thing you’re getting stuck on – e.g. do I wanna solve the problem for “hotshots” cuz I want to be a “hotshot”? Second-handedness issue there maybe…
• Think about immediate importance of resolving issue – e.g. if it’s a problem for “hotshots” then that’s an indicator there might not be a strong expectation you can solve the problem right now, so you’re not “behind” or failing or anything, so you don’t need to worry about that.
• If it’s something with an “answer”, see if it’s possible to do “partial spoilers”, like get a hint from someone or only read the beginning of an answer, so you can get unstuck while still figuring some things out yourself.
• Review past thoughts/notes regarding learning methodology.
• Try to have a big picture perspective regarding how the project is going rather than focusing in on one part that you got stuck on and making that a major issue in your mind.
#### Programming Project Troubleshooting
• Specify in detail all your project requirements – e.g. what test cases are you expecting the program to have to solve?
• Specify in detail all assumptions, restrictions, limitations that seem important and relevant to what you are stuck on — e.g. I was trying to limit how I approached the problem I got stuck on by only using functions that had already been meaningfully introduced – so no recursion or `cond`. That’s a big limitation that’s worth explicitly stating, at least.
• If any of the information or advice from the problem seems unusable or irrelevant, at least say so and say what you did to try to figure out whether it was relevant or not — e.g. I couldn’t figure out how to use `floor` or `log` in the problem I got stuck on, but didn’t really go into detail about what I had tried or thought of in terms of how I might use them.
• Check the documentation – AnneB discovered that `log` is supposed to work in a certain way by checking the docs.
• Get super organized –
• make sure your attempted solutions are nicely formatted.
• do detailed commenting on what each part does.
• keep track of each problem solving attempt rather than deleting it. This will let people help you more effectively by showing them how you’re thinking about the problem.
## Review of Past Learning Attempts
This is the first chapter I have some substantial work from the past to review.
Something worthy of note is that I basically did not get anywhere with `sci-coefficient` and `sci-exponent` the last time around, though I did figure out `scientific`. I seemed to have solve the other problems in a very similar manner.
## Learning Methodology
I set a goal for this chapter to try to ask more explicit questions but wound up getting very side-tracked with the getting stuck thing. However, I thought I wrote some good thoughts about that topic, so that seems okay.
## Simply Scheme Chapter 3 – Expressions
The big idea in this part of the book is deceptively simple. It’s that we can take the value returned by one function and use it as an argument to another function. By “hooking up” two functions in this way, we invent a new, third function. […]
We know that this idea probably doesn’t look like much of a big deal to you. It seems obvious. Nevertheless, it will turn out that we can express a wide variety of computational algorithms by linking functions together in this way. This linking is what we mean by “functional programming.”
## Chapter 3 – Expressions
Quotes from here.
Read-eval-print loop – Interaction between you and Scheme.
Scheme reads what you type, evaluates it, and prints the answer, and then does the same thing over again.
Expressions – a question you type into Scheme.
Atomic expression – a single value, such as `26`.
Compound expression – an expression made out of smaller expressions, such as `(+ 14 7)`.
Scheme versus other languages:
If you’ve programmed before in some other language, you’re probably accustomed to the idea of several different types of statements for different purposes. For example, a “print statement” may look very different from an “assignment statement.” In Scheme, everything is done by calling procedures, just as we’ve been doing here. Whatever you want to do, there’s only one notation: the compound expression.
### Little People
This chapter uses a model of little people to explain how an expression like `(- (+ 5 8) (+ 2 4))` gets evaluated.
A tree depicting some stuff discussed in the chapter:
The parentheses are key for guiding the workflow:
How does Alonzo know what’s the argument to what? That’s what the grouping of subexpressions with parentheses is about. Since the plus expressions are inside the minus expression, the plus people have to give their results to the minus person.
The book has an important clarification about the descriptive model it uses:
We’ve made it seem as if Bernie does his work before Cordelia does hers. In fact, the order of evaluation of the argument subexpressions is not specified in Scheme; different implementations may do it in different orders. In particular, Cordelia might do her work before Bernie, or they might even do their work at the same time, if we’re using a parallel processing computer. However, it is important that both Bernie and Cordelia finish their work before Alice can do hers.
Bernie and Cordelia must finish their work before Alice can do hers because Alice can’t know what she’s supposed to subtract from what until Bernie and Cordelia are finished. So the very nature of Alice’s workflow is dependent upon what Bernie and Cordelia discover. However, Bernie and Cordelia’s workflows are independent from each other, so they can happen in whatever order or at the same time.
We can express this organization of little people more formally. If an expression is atomic, Scheme just knows the value.[3] Otherwise, it is a compound expression, so Scheme first evaluates all the subexpressions (in some unspecified order) and then applies the value of the first one, which had better be a procedure, to the values of the rest of them. Those other subexpressions are the arguments.
Makes sense.
(Scheme ignores everything to the right of a semicolon, so semicolons can be used to indicate comments, as above.)
Ah I’d forgotten what the indicator for comments was 🙂
### Results Replacement
To keep track of which result goes into which larger computation, you can write down a complicated expression and then rewrite it repeatedly, each time replacing some small expression with a simpler expression that has the same value.
In each line of the diagram, the boxed expression is the one that will be replaced with its value on the following line.
They’re going very step-by-step here to show how the program works and avoid confusion, which is good.
If you like, you can save some steps by evaluating several small expressions from one line to the next:
I think this is how I would do it, since only replacing one expression at a time seems pretty tedious.
### Plumbing Diagrams
This section has some pictures depicting a procedure as a machine. I often think in terms of machine metaphors for procedures, especially when they do things like add or remove something to some input data. I find it easy to analogize certain programs to the sort of thing that might happen on an assembly line.
Here is Simply Scheme’s diagram for `(- (+ 5 8) (+ 2 4))`:
There is some similarity to the rightmost tree of the two trees I made above (though I used words).
### Pitfalls
The book lists some Scheme pitfalls.
The first pitfall is people reading programs from left to right rather than thinking in terms of expressions and subexpressions. I don’t think I have that particular issue, though I can see it being a big problem. I think I found it fairly intuitive to imagine the subexpressions flowing up through the structure of the procedure as they are evaluated. I think it is particularly important to be able to think of things that way when you try to think about recursion, cuz you’ve got to think of what happens when the base case gets reached and then what happens to the result that is returned from the base case and so on.
Another pitfall is that people think Scheme cares about white space. They emphasize that they’ve been indenting things for readability but that the parentheses are what actually matters. Contrast this with Python, which does care about white space.
A consequence of Scheme’s not caring about white space is that when you hit the return key, Scheme might not do anything. If you’re in the middle of an expression, Scheme waits until you’re done typing the entire thing before it evaluates what you’ve typed.
So if you forget a paren and hit return, nothing will happen. You’ve gotta finish your expression before Scheme will try to evaluate it.
I hit enter after the following expression and you can see what Racket did – it just went to the next line but didn’t try to evaluate the expression:
The book also mentions that stray quotation marks `"` can cause trouble, if you just have one and it isn’t paired off.
They also say:
One other way that Scheme might seem to be ignoring you comes from the fact that you don’t get a new Scheme prompt until you type in an expression and it’s evaluated. So if you just hit the return or enter key without typing anything, most versions of Scheme won’t print a new prompt.
Yeah, for me Racket just keeps going down a line but with no new `>` if I keep hitting enter, at least until I actually do something:
### Exercises
Some files for this chapter are posted in my Github.
#### ✅ Exercise 3.1
Translate the arithmetic expressions (3+4)×5 and 3+(4×5) into Scheme expressions, and into plumbing diagrams.
my attempt at something like a plumbing diagram for the above expression:
#### ✅ Exercise 3.2
How many little people does Alonzo hire in evaluating each of the following expressions:
I think the way to think of this is “how many operations were performed?” So you just count up the number of operations like `+`, `*`, and `-`. This also matches the number of open or closing parentheses, respectively.
I think there are 3 little people in this example. In this case, the `+` takes 3 values, but I don’t think that affects the number of little people.
4 little people for this one.
I counted 10 operations. I then cross-checked that against the number of opening parentheses and closing parentheses. One advantage of going by the number of parentheses is you can use a text editor to count those with the Find feature 🙂
NOTE: I’m generally checking my answers against AnneB’s to ensure accuracy and watch for mistakes. AnneB came out the same on this problem and said:
I think it’s the same as the number of procedures, which is the same as the number of left parentheses.
Yep 🙂 I’d only add that it’s also the number of right parentheses, since the number of parentheses have to match.
#### ✅❌ Exercise 3.3
Each of the expressions in the previous exercise is compound. How many subexpressions (not including subexpressions of subexpressions) does each one have?
For example,
`(* (- 1 (+ 3 4)) 8)`
has three subexpressions; you wouldn’t count (+ 3 4).
Earlier in the chapter the book said:
`(* (- (+ 5 8) (+ 2 4))`
`(/ 10 2))`
`35`
This says to multiply the numbers 7 and 5, except that instead of saying 7 and 5 explicitly, we wrote expressions whose values are 7 and 5. (By the way, we would say that the above expression has three subexpressions, the `*` and the two arguments. The argument subexpressions, in turn, have their own subexpressions. However, these sub-subexpressions, such as (+ 5 8), don’t count as subexpressions of the whole thing.)
So it seems like we count sub-expressions by counting up the procedure and the number of arguments being passed to the procedure at the highest/final “level” of the expression.
(❌ERRONEOUS STATEMENT) So this:
`(+ 3 (* 4 5) (- 10 4))`
has 3 sub-expressions. (the `+`, an expression whose value is 20, and an expression whose value is 6.
✅EDIT/CORRECTION: I somehow forgot to count the 3! Whoops.I think maybe since I’d just written “has 3 sub-expressions” I somehow thought I’d already addressed the 3? Not sure. Thanks to AnneB for helping me catch this error by having the correct answer on her blog. Also, I think AnneB’s way of noting the sub-expressions by writing out each one is better for avoiding mistakes.
This one:
`(+ (* (- (/ 8 2) 1) 5) 2)`
has 3 sub-expressions: `+`, an expression whose value is 15, and an expression whose value is 2.
For the next one, I wanted to simplify the sub-expressions in a somewhat step-by-step way in order to check my answer. For the sub-expression `(/ 2 3)`, when simplifying, I used the decimal notation 0.6666666666666666.
simplifies somewhat to this:
which further simplifies to this:
and finally to:
So there are three sub-expressions.
Another way to figure out the number of sub-expressions is to look at the highlighting.
If we put the cursor on the outside of the first open parenthesis, then the whole expression highlights. The `*` operates at the highest “level” of the procedure, so this highlighting corresponds to the level the `*` operates on.
With the cursor to the left of the open parenthesis that precedes `+`, we see another grouping. The highlighting indicates that everything inside is one big group that gets evaluated and ultimately handed to the `*` at the top level. So the whole highlighted thing counts as one sub-expression.
Same idea here. The whole highlighted part ultimately only counts as one sub-expression for our `*`.
So there are three sub-expressions:
`*`
and
I think that there are other ways you could figure out the number of sub-expressions. You could use trees, for example. I feel like I’m okay on this point though.
#### ✅ Exercise 3.4
Five little people are hired in evaluating the following expression:
Give each little person a name and list her specialty, the argument values she receives, her return value, and the name of the little person to whom she tells her result.
• Adam: Addition specialist. Receives arguments -9 (from Bob) and 10 (From Donald). Returns 1. Highest level little person, no direct report.
• Bob: Multiplication specialist. Receives arguments 3 (directly) and -3 (from Charlie). Returns -9. Reports result to Adam.
• Charlie: Minus specialist. Receives arguments 4 and 7 directly. Returns -3. Reports result to Bob.
• Donald: Minus specialist. Receives argument values 8 (directly) and -2 (from Evan). Returns 10. Reports result to Adam.
• Evan:Minus specialist. Receives argument values 3 and 5 directly. Returns -2. Reports result to Donald.
#### ✅ Exercise 3.5
Evaluate each of the following expressions using the result replacement technique:
thing to be replaced in next line is in bold
(sqrt (+ 6 (* 5 2)))
(sqrt (+ 6 10))
(sqrt 16)
4
(+ (+ (+ 1 2) 3) 4)
(+ (+ 3 3) 4)
(+ 6 4)
10
#### ✅ Exercise 3.6
3.6 Draw a plumbing diagram for each of the following expressions:
I remembered that I have an iPad. Let’s try doing some plumbing diagrams with that
A little rough but it does the job!
#### ✅ Exercise 3.7
What value is returned by `(/ 1 3)` in your version of Scheme? (Some Schemes return a decimal fraction like `0.33333`, while others have exact fractional values like `1/3` built in.)
A fraction:
#### ✅❌ Exercise 3.8
Which of the functions that you explored in Chapter 2 will accept variable numbers of arguments?
I checked this list from that chapter:
+, -, /, \<=, \<, =, >=, >, and, appearances, butfirst, butlast, cos, count, equal?, every, even?, expt, first, if, item, keep, last, max, member?, not, number?, number-of-arguments, odd?, or, quotient, random, remainder, round, sentence, sqrt, vowel?, and word.
I got the following list of functions:
❌INCOMPLETE:
✅CORRECTION: AnneB noticed that random will take two arguments (and not just one) if the first argument you give it is less than the second. Strange behavior!
Also, Anne’s list is missing `-` and `\`. The behavior with these seems to be to apply the procedures to the first two arguments and then to the result of that plus the next argument, down the line until all the arguments have been dealt with.
#### ✅ Exercise 3.9
The expression (+ 8 2) has the value 10. It is a compound expression made up of three atoms. For this problem, write five other Scheme expressions whose values are also the number ten:
• An atom
`10`
• Another compound expression made up of three atoms
`(+ 3 7)`
• A compound expression made up of four atoms
`(+ 3 3 4)`
• A compound expression made up of an atom and two compound subexpressions
• Any other kind of expression
let’s do a compound expression made up of an atom and three compound subexpressions:
## Simply Scheme Chapter 2 – Functions
### Getting Started
This chapter has us start off by loading a “functions.scm” file which can be downloaded here. You want to load the file by typing `(load "functions.scm")` in the Interactions window of DrRacket, and not in the Definitions window. If you try loading it in Definitions window, you may get an error message like I did.
Once the file is loaded, the program does not automatically start; you need to type `(functions)` to start the program. The `functions` program creates a special way of entering functions that is different than the normal way that scheme operates. It gives you a prompt to enter a function and then prompts you to enter arguments. The number of arguments you can enter varies. For example, `sqrt` only lets you enter one argument, which makes sense. OTOH, `+` and `se` let you enter two arguments. It’s worth noting that the number of arguments does not necessarily correspond to what the “real” Scheme function lets you enter – for instance, you can enter `(+ 1 2 3 4 5 6)` in Scheme no problem and get an answer, but the `functions` program does not let you do that. You can exit the `functions.scm` program by typing `exit` instead of typing the name of a function.
### Playing With Functions
#### Math Functions
I tried performing some of the operations suggested by the chapter:
If I enter 1 ÷ 987654321987654321 in a calculator program, I get this
So I was expecting something similar to that for the `/` function.
The math functions don’t like words:
Some playing around with the `odd?` function, which tests whether a number is odd. Note the message related to not being in the domain for the argument 3.5.
##### remainder
Remainder:
A book-suggested example:
another example:
And another:
The sign of the result is the same as the sign of the first argument, which is the dividend. So I think remainder gives the remainder that results when both arguments are non-negative, and then makes the remainder negative if the first argument is negative.
Yeah this seems to be the case.
The remainder function doesn’t seem to like decimals:
##### expt
The book says:
Some two-argument functions have complicated domains because the acceptable values for one argument depend on the specific value used for the other one. (The function expt is an example; make sure you’ve tried both positive and negative numbers, and fractional as well as whole-number powers.)
Some experiments with `expt`
So you can see that with -2 and -1/2 as arguments I managed to get something that was outside the domain.
AnneB says:
I think the domain rule for expt is:
• the first argument can be any real number
• if the first argument is positive, the second argument can be any real number
• if the first argument is 0, the second argument can be a non-negative real number
• if the first argument is negative, then the second argument can be 0 or it can be a real number with absolute value greater than or equal to 1
Let’s try some more, with an eye towards seeing if AnneB is right!
That’s a negative number for the first argument, and a real number with an absolute value greater than 1 for the second argument, but this set of arguments appears to not be in the domain. So that appears to contradict AnneB’s final claim re: domain (though I could be wrong).
If the first argument is negative, it doesn’t seem like the `expt` function treats fraction or decimal values for the second argument as being within the domain. You can enter negative integer numbers for the second value, though (along with 0 and positive integers). So it is possible that Anne’s final claim should be redrafted as: “if the first argument is negative, then the second argument can be any integer.”
#### Rounding Mystery 🔍🧐
Another book example:
This is the result of my playing with the `round` function a bit:
Very interesting. So at some point in between 17.4999 and 17.499999999999999999999 something happens. My wild guess is that DrRacket/Scheme only keeps track of numbers to a certain level of precision regardless of how many digits you enter, and then rounds the number off internally (to whatever level of precision it keeps track of the numbers), and that if use a number that happens to be long enough to run into this issue and then try to feed that number into a function that rounds the number off, you basically get two levels of rounding, leading to the result of 17.499999999999999999999 getting rounded off to 18.
Hey I thought of a way to test this. Since Scheme returns the number you give it back to you, if what I’m saying is actually true, then normal Scheme should give me back a rounded-off number once I hit a certain number of digits.
Let’s see 🖥🧐
Yep. It looks like it’ll do 14 digits of precision after the decimal point, but once it hits that 15th digit, Scheme is like “nah” and rounds the number off to 17.5. And the final line there is just testing to make sure the threshold is digits after the decimal point (which it appears to be).
I vaguely remember running into an issue like this before in a previous run through Scheme or SICP or something. Maybe I’ll come across it as a I review my old work.
#### Word Functions
Some more book suggestions:
`butfirst` returns all but the first letter of a word. So if you give it a one-letter word, you get nothing. But if you give it a multi-letter word:
`butfirst` treats numbers in the same way it treats words:
and `first` behaves similarly
`butfirst` won’t count arguments with multiple words in them:
Apparently the following version is supposed to work:
But I and other people are having trouble with the `functions.scm` functions not accepting sentences in our versions of Scheme.
`word` takes 2 arguments and doesn’t respect spaces:
And you can’t use quotes (this seems to be true in general for the functions in `functions.scm`)
Whereas in the normal Scheme version of word, you can use spaces by putting a word in quotes and having a space at the end or by quoting a space separately:
both `functions.scm` `word` and normal Scheme `word` treat numbers as strings of characters:
#### Count
Now onto `count`:
`count` is counting up the number of digits in the argument you give it, and it does that for sequences of numbers as well as for words.
The regular Scheme version of `count` handles this numerical sequence the same way:
`count` doesn’t like spaces in words:
`count` could conceivably just treat the whole thing as a third-character sequence, one of which is a space, but it doesn’t.
The normal Scheme version of `count` handles this situation differently than the `functions.scm` `count`. It thinks the “United” is trying to refer to some definition for “United” that it can’t find:
Normal scheme `count` has the same behavior just for “United”
whereas the `functions.scm` `count` can handle “United” by itself and produce the result that you would expect:
Normal Scheme `count` can, of course, deal with “United” if you indicate that it is a word/string by putting a mark before the “United”:
Whereas the `functions` version of `count` actually treats a as part of the sequence of characters that it attempts to count up:
#### Ifs and Booleans
if takes 3 arguments:
If the first argument returns true, then the second argument gets returned. If the first argument returns false, then the third argument gets returned.
#### Functions
The book provides some examples of functions that take other functions:
The range of number-of-arguments is nonnegative integers. But its domain is functions. For example, try using it as an argument to itself!
Heh okay.
As expected 🙂
If you’ve used other computer programming languages, it may seem strange to use a function—that is, a part of a computer program—as data. Most languages make a sharp distinction between program and data. We’ll soon see that the ability to treat functions as data helps make Scheme programming very powerful and convenient.
Ah okay, so this is a notable feature of Scheme in particular.
The book suggests trying this example:
So the `keep` function is taking the `vowel?` function and a word as arguments and keeping the vowels.
Another example of this kind of thing:
Some functions that take other functions aren’t defined in the functions program:
### Domain and Range
Simply Scheme Chapter 2 says:
So far most of our functions fall into one of two categories: the arithmetic functions, which require numbers as arguments and return a number as the result; and the word functions, which accept words as arguments and return a word as the result. The one exception we’ve seen is `count`. What kind of argument does `count` accept? What kind of value does it return? The technical term for “a kind of data” is a type.
`count` takes strings of characters (could be letters, numbers, both) which we might call a “word” and returns a number.
The technical term for “the things that a function accepts as an argument” is the domain of the function. The name for “the things that a function returns” is its range. So the domain of `count` is words, and the range of `count` is numbers (in fact, nonnegative integers). This example shows that the range may not be exactly one of our standard data types; there is no “nonnegative integer” type in Scheme.
Right okay. A function’s range doesn’t have to be something Scheme knows about.
How do you talk about the domain and range of a function? You could say, for example, “The cos function has numbers as its domain and numbers between −1 and 1 as its range.” Or, informally, you may also say “Cos takes a number as its argument and returns a number between −1 and 1.”[3]
For functions of two or more arguments, the language is a little less straightforward. The informal version still works: “Remainder takes two integers as arguments and returns an integer.” But you can’t say “The domain of remainder is two integers,” because the domain of a function is the set of all possible arguments, not just a statement about the characteristics of legal arguments.[4]
Yeah okay that makes sense. For example, the regular Scheme version of `+` can take no arguments or a ton of arguments. You can’t say “the domain of `+` is two arguments” just because 2 is a permissible number of arguments. The domain is supposed to be a generally true statement about the things that a function accepts as an argument.
I am getting “Argument(s) not in domain.” errors when I’m not supposed to when a sentence is an argument. I haven’t figured out how to resolve this as of yet. Fortunately, the issue appears to only affect the `functions.scm` functions and that file is only used for Chapter 2, so I don’t think this is a huge deal.
### Exercises
Use the `functions` program for all these exercises.
For some of these, I ran into the problem with sentences not being in the domain of `functions.scm` functions, so I tested them out in regular Scheme. I had to define `vowel?` in regular Scheme – for that, I just stole code from Chapter 8.
For the purpose of answering these questions, i’m going to assume that if a function takes a number as an argument but treats it like a word, that the type of the number-as-word counts as “word” and not “number.”
Also, the data types I’m focusing on in my answers are words, numbers, booleans, and functions. I don’t care so much about data subtypes like “positive integers” though I may mention them. Sentence-handling is a bit broken in the functions used in this chapter, as mentioned above.
#### Exercise 2.1
In each line of the following table we’ve left out one piece of information. Fill in the missing details.
I just made up a word for the eieio line 🙂
#### Exercise 2.2
2.2 What is the domain of the vowel? function?
`vowel?` accepts words, sentences, booleans, functions, and numbers as arguments (in the sense that it won’t reject these as outside its domain). Its range is a boolean value – it returns true if you give it a single letter vowel, and false if you give it anything else.
AnneBfound a couple of examples of things outside vowel‘s domain.
#### Exercise 2.3
2.3 One of the functions you can use is called `appearances`. Experiment with it, and then describe fully its domain and range, and what it does. (Make sure to try lots of cases. Hint: Think about its name.)
Note: Correction below
`appearances` takes two arguments and appears to return the number of times that the first argument appears in the second argument if the first argument is a single digit letter, number, or punctuation mark.
If the first argument is a multi-character sequence, `appearance` returns zero, regardless of whether or not the sequence provided as the first argument appears in the second argument.
`appearances` says that sentences are outside its domain but that may be related to the technical issue described earlier. `appearances` will accept sentences as arguments in normal Scheme but does not appear to actually function as you might guess it would on sentences, as it will return 0 (instead of 1) even if passed identical sentences for both arguments.
The domain appears to be sequences of alphanumeric characters and punctuation marks (which we can generalize as “words”), and the range is positive whole numbers.
Some tests:
CORRECTION: AnneB shows that `appearances` takes sentences as the second argument if the first argument is a single character, and will find instances of the first argument in a sentence provided as the second argument, e.g.:
I thought this example was interesting. `appearances` seems to work differently on sentences than on words. With words, it will count each instance of a character separately within a word. With sentences, `appearances` seems to want the characters to be spaced apart in order to register them as being an appearance.
So even if the same letter appears twice in a row in a sentence, it won’t count that, but it will count letters on their own:
So a more correct statement is that `appearances` takes sequences of letters, numbers, and punctuation marks for the first argument, and all that plus sentences for the second argument, and returns a positive integer.
#### Exercise 2.4
One of the functions you can use is called `item`. Experiment with it, and then describe fully its domain and range, and what it does.
What it does: After some testing (see below) it seems that `item` takes a number for the first argument, and a sentence or word (or a number treated as a word) as the second argument. Then it returns a character of the second argument, or word in the sentence, that corresponds to the number of the first argument. Specifically, it returns the character of the second argument or word in a sentence that you get if you start counting characters/words from the left side of the second argument until you get to the [first argument] character. For example, if the first argument is “3” and the second argument is “654”, then `item` will return the third character (counting from the left) of the second argument, which is 4. Or if the first argument is 5 and the second argument is Liberty, then `item` will return the fifth character of the second argument (counting from the left) which is r. Or if the first argument is “4” and the second argument is “(we the people of the united states)”, then `item` will return “of”.
Domain: the first argument has to a positive number less than or equal to the total number of characters (if a word) or words (if a sentence) of the second argument. The second argument has to be a word (including a number which is treated as a word) or sentence.
Some tests. Format is:
Arg1, Arg2: Result.
(I did a different format cuz i did tons of testing on this one, wanted it to be more compact)
0, 0: Argument(s) not in domain.
1, 0: 0.
2, 0: Argument(s) not in domain.
0, 1: Argument(s) not in domain.
0, 2: Argument(s) not in domain.
1, 0: 0.
1, 1: 1.
1, 2: 2.
1, 3: 3.
1, 4: 4.
1, 5: 5.
2, 0: Argument(s) not in domain.
2, 1: Argument(s) not in domain.
2, 2: Argument(s) not in domain.
2, 4: Argument(s) not in domain.
2, 23: 3.
2, 32: 2.
2, 33: 3.
2, 22: 2.
2, 57: 7.
2, 75: 5.
3, 654: 4.
3, 456: 6.
1, 456: 4.
2, 456: 5.
-1, -1: Argument(s) not in domain.
w, ww: Argument(s) not in domain.
15, 12345678901234567890: 5.
9, 123456789012: 9.
4, potato: a.
5, Liberty: r.
1, (grilled cheese and tomato sammich): Argument(s) not in domain. This appears to be an issue with functions.scm
0, potato: Argument(s) not in domain.
1, potato: p.
-1, potato: Argument(s) not in domain.
the ability to handle sentences as arguments is broken in `functions.scm`, as mentioned before. However, I tried the regular Scheme version of `item` to test how it handles sentences:
1, (potato tomato): potato.
2, (potato tomato): tomato.
#### Note for Exercises 2.5 through 2.9
The following exercises ask for functions that meet certain criteria. For your convenience, here are the functions in this chapter: +, -, /, \<=, \<, =, >=, >, and, appearances, butfirst, butlast, cos, count, equal?, every, even?, expt, first, if, item, keep, last, max, member?, not, number?, number-of-arguments, odd?, or, quotient, random, remainder, round, sentence, sqrt, vowel?, and word.
`number?` does not appear to be defined in my `functions.scm` file.
I made a (somewhat incomplete) list of data types of various functions as an aid to answering the questions.
##### List of functions by number of arguments
This section lists functions by number of arguments permitted for the argument within the `functions.scm` file used for this chapter. I made these lists cuz it seemed like they would be useful for answering the questions that follow.
#### Exercise 2.5
List the one-argument functions in this chapter for which the type of the return value is always different from the type of the argument.
• `count` takes numbers but i guess it doesn’t treat them as numbers but as a string of characters. so it treats “666” just like “aaa”. it returns numbers. so I think `count` always returns things with a type different than the type of argument it takes.
• `even?` only has integers in its domain and only returns boolean values, so it would count as an example of such a function.
• `number-of-arguments` only has functions within its domain and only returns positive integers.
• `odd?` only has integers in its domain and only returns boolean values, so it would count as an example of such a function.
#### Exercise 2.6
List the one-argument functions in this chapter for which the type of the return value is sometimes different from the type of the argument.
• `vowel` accepts boolean values as an argument and returns boolean values. So if you provide it a boolean value and it returns a boolean value, then the type of the argument and the type of the return are the same. OTOH, if you provide it with a letter or word and it returns a boolean value, then the data types are different. So they are sometimes different.
NOTE:
So there seems to be a big difference in answers between me and AnneB. I think this may be an issue of how we interpreted the question but let’s see. AnneB says:
The ones for which the type of the return value is sometimes different from the type of the argument are:
count can take a word as an argument and return a number.
This one I can see as arguable. `count` seems a bit tricky to me. It takes numbers but seems to deal with them as any other character, and not as numbers. I’m not quite sure how to think about that issue, so I’ll focus on what I regard as the less ambiguous cases.
even?, number?, odd? can take a number as an argument and return a boolean.
my `number?` was broken so we’ll leave that aside.
re: `even?` and `odd?`, I think I am reading the question as asking “Which one-argument functions have return values of a type that is sometimes different from the type of the argument, and sometimes not” and AnneB is perhaps reading it in a different way.
`even?` and `odd?` take only numbers but return only booleans. They never return the same value as they take. So I think they always return a value different from the type of argument, not only sometimes.
number-of-arguments can take a function as an argument and return a number.
Similar logic here as with `even?` and `odd?``number-of-arguments` only takes functions and always returns numbers. So it always returns a type different than the type that it accepts.
vowel? can take a word as an argument and return a boolean.
we agree that `vowel?` is an answer.
#### Exercise 2.7
Mathematicians sometimes use the term “operator” to mean a function of two arguments, both of the same type, that returns a result of the same type. Which of the functions you’ve seen in this chapter satisfy that definition?
I read this to be asking for a function that always returns a result of the same type as the two arguments.
NOTE: AnneB’s list is almost the same as mine, except that I have `expt` in my list and she does not.
#### Exercise 2.8
An operator f is commutative if f(a,b)=f(b,a) for all possible arguments A and B. For example, `+` is commutative, but `word` isn’t. Which of the operators from Exercise 2.7 are commutative?
#### Exercise 2.9
An operator f is associative if f(f(a,b),c)=f(a,f(f(b,c)) for all possible arguments A, B, and C. For example, * is associative, but not /. Which of the operators from Exercise 2.7 are associative?
I made a table to help me figure out whether OR was associative
❌INCOMPLETE:
✅CORRECTION: AnneB lists `word` as associative and I think she is right. It does not matter whether you join words a and b first and then join ab with c, or join words b and c first and then join a with bc. The word order and result is going to come out the same in either case.
### End of Chapter Review
Felt pretty solid on my development of the “testing things out” skill this chapter. I think that’s kind of the point of this chapter.
Trying to think somewhat rigorously about the domain and range of functions seems valuable.
#### Questions
What is the correct way to describe the domain of `count`, given that it takes numbers as arguments but treats them as words?
#### My Prior Learning Efforts
As with Chapter 1, there appear to be none to speak of.
#### Other People’s Learning Efforts
AnneB’s notes were super helpful to refer to. It’s hard to find much stuff to refer to for Chapter 2. You start seeing stuff for Chapter 3 and later. | 13,067 | 56,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-45 | latest | en | 0.887799 |
http://www.convertit.com/Go/weighing-systems/Measurement/Converter.ASP?From=light+yr&To=depth | 1,653,733,267,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00729.warc.gz | 70,119,921 | 4,331 | weighing-systems.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```light year = 9.460528405106E+15 length (length) ``` Related Measurements: Try converting from "light yr" to agate (typography agate), archin (Russian archin), arpentcan, astronomical unit, bolt (of cloth), cable length, city block (informal), cloth quarter, digitus (Roman digitus), en (typography en), Greek cubit, Greek palm, hand, league, marathon, micron, nautical league, palm, shaku (Japanese shaku), spindle, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: light yr = 3.72E+19 caliber (gun barrel caliber), 9.46E+30 fermi, 3.10E+16 foot, 47,027,998,514,207 furlong (surveyors furlong), 1.28E+16 gradus (Roman gradus), 1.23E+17 Greek palm, 4.09E+16 Greek span, 1.71E+16 Israeli cubit, 1,959,499,938,091.96 league, 4.70E+16 link (surveyors link), 9.46E+15 m (meter), 224,209,766,456.29 marathon, 1,679,571,375,507.39 parasang, 2.23E+18 pica (typography pica), 2,409,221,235,358.97 ri (Japanese ri), 1.88E+15 rod (surveyors rod), 2.13E+16 Roman cubit, 103,461,596,731,255 soccer field, 718,483,310,633.72 spindle, 51,106,167,430,367.7 stadia (Greek stadia).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 557 | 1,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-21 | latest | en | 0.692336 |
http://www-users.math.umn.edu/~stant001/1251/exam1sol.html | 1,563,467,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525659.27/warc/CC-MAIN-20190718145614-20190718171614-00409.warc.gz | 166,996,913 | 1,315 | # Math 1251 Exam 1 Solutions
1. The derivative dy/dx=2-1/x^2, evaluated at x=1, is 1. So the slope of the tangent line at (1,3) is m=1, and the tangent line is y-3=1(x-1).
2.(a) dy/dx=45*(3x-2)^44*3 by the chain rule
(b) f'(x)=cos(4-1/x^(1/3))*(-1)*(-1/3)*x^(-4/3).
(c) f(x)=x*(1-x^2)^(1/2), so f'(x)=(1-x^2)^(1/2)+x*(1/2)*(1-x^2)^(-1/2)*(-2x).
(d) If t gets large, |t+2| is about t and large, while sin(t^2) is between -1 and 1. So the numerator is about t. In the denominator, -4t beats out 5, so it is about -4t. The quotient then is -1/4, which is the limit.
(e) Factoring the numerator, (x^2-3x)/(x-3)=x so if x approaches 3, the answer is 3.
3.(f(x+h)-f(x))/h is the slope of the secant line connecting two points on the graph: (x,f(x)) and (x+h, f(x+h)). As h approaches 0, this secant line approaches the tangent line, so the quotient approaches the slope of the tangent line.
4.(a) False. Take f(x)=x. then d/dx(1/(f(x))=d/dx(1/x)=d/dx(x^(-1))= -x^(-2) which is not equal to 1/f'(x)=1/1=1. The correct answer should be -f'(x)/f(x)^2.
(b) False. the condition that |f(x)|<=1 means that the graph of y=f(x) lies between the two parallel lines y=1 and y=-1. It says nothing about the slope of the tangent line to the graph, which could be very large if f(x) is very wiggly. For example, f(x)=sin(2x) has f'(0)=2. | 506 | 1,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-30 | longest | en | 0.858468 |
https://brainmass.com/business/management-accounting/relevant-costs-288551 | 1,582,094,790,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00345.warc.gz | 308,167,915 | 12,330 | Explore BrainMass
Share
# Relevant costs
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Kristen Lu purchased a used automobile for \$8,000 at the beginning of last year and incurred the following operating costs:
Depreciation (\$8,000 5 years) . . . . . . . . . \$1,600
Insurance . . . . . . . . . . . . . . . . . . . . . . . . . . . \$1,200
Garage rent . . . . . . . . . . . . . . . . . . . . . . . . . \$360
Automobile tax and license . . . . . . . . . . . . . \$40
Variable operating cost. . . . . . . . . . . . . . . . . \$0.14 per mile
The variable operating cost consists of gasoline, oil, tires, maintenance, and repairs. Kristen estimates that, at her current rate of usage, the car will have zero resale value in five years, so the annual straight-line depreciation is \$1,600. The car is kept in a garage for a monthly fee.
Required:
1. Kristen drove the car 10,000 miles last year. Compute the average cost per mile of owning and operating the car.
2 Kristen is unsure about whether she should use her own car or rent a car to go on an extended cross-country trip for two weeks during spring break. What costs above are relevant in this decision? Explain.
3. Kristen is thinking about buying an expensive sports car to replace the car she bought last year. She would drive the same number of miles regardless of which car she owns and would rent the same parking space. The sports car's variable operating costs would be roughly the same as the variable operating costs of her old car. However, her insurance and automobile tax and license costs would go up. What costs are relevant in estimating the incremental cost of owning the more expensive car? Explain.
#### Solution Preview
1. The average cost = average fixed cost per mile + variable cost per mile
The variable cost = 0.14 per mile
Total fixed cost = 1,600+1,200+360+40 = 3,200
Fixed cost per mile = 3,200/10,000 miles = 0.32
Average ...
#### Solution Summary
The solution explains the relevant costs in replacing the car
\$2.19 | 523 | 2,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-10 | latest | en | 0.908674 |
http://www.options-trading-mastery.com/rsi-indicator.html | 1,623,807,581,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00183.warc.gz | 71,704,145 | 15,082 | # The RSI Indicator Explained
## Why the RSI Indicator May be One of Your Most Powerful Options Trading Allies
This is about the RSI Indicator and how it can be used to influence your option trading decisions. The RSI is a "momentum indicator" and as such, is mainly used to confirm whether a trend is likely to continue or whether it is running out of steam - i.e. losing momentum. If you're a trend trader this makes it important.
The RSI is also known as a "leading indicator" (as opposed to a "lagging indicator") which means that it provides a signal that alerts us that a possible change of trend may be in the wind, before it occurs.
Finally, it is also called an "oscillator" because its signal bounces, or oscillates, between the two extreme points of +100 and -100. When it breaches 70 at either end it warns us that the security may be "over-bought" or "over-sold" as the case may be.
You should be very wary of entering a trade that is based on the assumption that the trend will continue when the RSI is hovering around these extremes. This is the primary purpose of the RSI indicator - to warn you that price action is possibly losing momentum.
## How the RSI Indicator is Calculated
If you're a big picture person and the technical stuff bores you, then feel free to skip this section. If not, then here are the maths behind the RSI formula:
RSI = 100 - (100 / (1 + UP / DOWN))
Where UP = the average of the upward closing changes over the selected period, and DOWN = the average of the downward closing changes over the selected period.
So taking a 14 day RSI time period, we would:
2. Where the closing price of the security increased from the previous day, total the amounts of these increases. Divide this by 14 to arrive at the average UP closing prices.
3. Do the same thing for the days where the closing price is lower than the day before. This gives us the average DOWN closing price.
4. Divide the average UP closing change by the average DOWN closing change to get the relative strength.
5. Add 1 to this number.
6. Then divide the number from point 5 by 100.
7. Subtract this figure from 100.
You've just calculated the RSI! But why bother going to all this trouble when just about any charting program you can find these days will do it for you and present it as a graph.
## How the RSI Indicator is Displayed
The RSI fluctuates between 0 and 100 in the form of a line chart which rises and falls. We can divided the RSI display into regions by drawing horizontal lines - one at 70 percent and the other at 30 percent, to indicate the warning zones. Some also include a line at 50 percent.
It's important not to confuse the terms "over-bought" and "over-sold" as meaning that too many shares have been bought or sold so that buyers or sellers (as the case may be) are about to re-enter the market.
This is not the case. It is all about momentum and momentum alone. It is simply a warning that the momentum of price action may not last so that the trend you are relying on is in danger of either decreasing or possibly even reversing.
Have a look at this chart and note the RSI.
What Signals Does the RSI Indicator Give Us?
The RSI serves a dual purpose. Firstly, its current level in combination with the trend warns us if momentum may be failing because the stock is "over-bought" or "over-sold". The second use relates to divergences.
The Level of the RSI
An RSI that rises up through the oversold line (30%) after falling below it is considered bullish.
An RSI falling back through the overbought line (70%) after being above it is considered bearish.
Divergences
When the RSI trend and the price action trend of the stock are in opposite directions there is a high probability of a potential change in trend. It is important to note that the RSI won't tell you when the change will occur, or how strong it will be, only that it may be coming.
A bullish divergence occurs when the stock falls to lower lows, but the RSI makes a higher low. The stock appears to be weak but in fact, is gaining strength and preparing for a reversal. Wait for validation then buy your call options when the stock begins to reverse.
A bearish divergence occurs when the stock rises to higher highs, but the RSI makes a lower high. The stock trend may appear to be strong but the indicator tells us it may be weakening and preparing to fall. Again, wait for confirmation before buying put options.
Take a look at the chart below for examples of RSI divergence.
The RSI indicator is one tool you should always pay attention to if you trade options with the trend.
**************** **************** | 1,038 | 4,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-25 | latest | en | 0.931811 |
https://www.studypool.com/discuss/397411/2-21-question-2-math?free | 1,496,122,483,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613796.37/warc/CC-MAIN-20170530051312-20170530071312-00309.warc.gz | 1,187,180,667 | 13,895 | ##### 2.21 Question #2- Math
Mathematics Tutor: None Selected Time limit: 1 Day
Feb 21st, 2015
Y intercept is at ( 0, 5) This will be the first point you can graph.
The slope is -7/4.
From the first point you want to go down 7 (to negative 2) and to the right 4.
The second point will be ( 4, -2)
This should allow you to graph the line.
Feb 21st, 2015
...
Feb 21st, 2015
...
Feb 21st, 2015
May 30th, 2017
check_circle | 149 | 429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-22 | longest | en | 0.895751 |
https://www.assignmentexpert.com/homework-answers/chemistry/general-chemistry/question-181278 | 1,653,698,092,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00156.warc.gz | 735,856,096 | 67,552 | 106 270
Assignments Done
97.2%
Successfully Done
In May 2022
# Answer to Question #181278 in General Chemistry for Mel
Question #181278
What is the theoretical yield, in grams, of SOβ(π) that can be produced from 25.0 grams of SOβ(π) according to the reaction below?
2SO2(g)+O2(g)β>2SO3(g)
1
2021-04-19T05:14:01-0400
Molar mass of SO2=64g
Molar mass of SO3=80g
From the balanced equation
2mol of SO2 gives 2mol of SO3
1mol of SO2 gives 1mol of SO3
64g of SO2 gives 80g of SO3
25.0g of SO2 will give 80/64 x 25.0= 31.25g
Theoretical yield of SO3= 31.25g
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS! | 258 | 725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-21 | latest | en | 0.863597 |
https://roboblockly.org/curriculum/math/math6/23.php | 1,627,986,270,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154457.66/warc/CC-MAIN-20210803092648-20210803122648-00488.warc.gz | 492,104,095 | 38,357 | ### Learning Math and Coding with Robots
Grid: Tics Lines: Width px Hash Lines: Width px Labels: Font px Trace Lines: Robot 1: Width px Robot 2: Width px Robot 3: Width px Robot 4: Width px
Axes: x-axis y-axis Show Grid Grid: 24x24 inches 36x36 inches 72x72 inches 96x96 inches 192x192 inches Quad: 4 quadrants 1 quadrant Hardware Units: US Customary Metric
Background:
#### Robot 1
Initial Position: ( in, in) Initial Angle: deg Current Position: (0 in, 0 in) Current Angle: 90 deg Wheel Radius: 1.75 in1.625 in2.0 in Track Width: in
#### Robot 2
Initial Position: ( in, in) Initial Angle: deg Current Position: (6 in, 0 in) Current Angle: 90 deg Wheel Radius: 1.75 in1.625 in2.0 in Track Width: in
#### Robot 3
Initial Position: ( in, in) Initial Angle: deg Current Position: (12 in, 0 in) Current Angle: 90 deg Wheel Radius: 1.75 in1.625 in2.0 in Track Width: in
#### Robot 4
Initial Position: ( in, in) Initial Angle: deg Current Position: (18 in, 0 in) Current Angle: 90 deg Wheel Radius: 1.75 in1.625 in2.0 in Track Width: in
Function Machine
Problem Statement:
Machine A (top) and B (bottom) shown on the board are individually programmed to perform specific math tasks. Each machine takes an input (orange circle) on the left and produce an output (pentagon) on the right. Watch the operation of each machine and answer the question prompt. Hint: it may help to write down the input-output of each machine.
```/* Code generated by RoboBlockly v2.0 */
printf("Machine A satisfy the definition of a function: 1 input can't have 2 different outputs. Yes, two inputs can have the same output");
printf("The rule is output = input^2 = input*input");
```
Blocks Save Blocks Load Blocks Show Ch Save Ch Workspace
Problem Statement:
Machine A (top) and B (bottom) shown on the board are individually programmed to perform specific math tasks. Each machine takes an input (orange circle) on the left and produce an output (pentagon) on the right. Watch the operation of each machine and answer the question prompt. Hint: it may help to write down the input-output of each machine.
Time | 559 | 2,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-31 | latest | en | 0.724834 |
https://www.scratchapixel.com/lessons/3d-basic-rendering/minimal-ray-tracer-rendering-simple-shapes/ray-sphere-intersection.html | 1,726,452,609,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00073.warc.gz | 906,486,481 | 9,431 | # A Minimal Ray-Tracer
## Ray-Sphere Intersection
Intersecting a ray with a sphere is the simplest form of a ray-geometry intersection test, which is why many ray tracers showcase images of spheres. Its simplicity also lends to its speed. However, to ensure it works reliably, there are always a few important subtleties to pay attention to.
This test can be implemented using two methods. The first solves the problem using geometry. The second technique, often the preferred solution (because it can be reused for a wider variety of surfaces called quadric surfaces), utilizes an analytic (or algebraic, e.g., can be resolved using a closed-form expression) solution.
## Geometric Solution
The geometric solution to the ray-sphere intersection test relies on simple math, mainly geometry, trigonometry, and the Pythagorean theorem. If you look at Figure 1, you will understand that to find the positions of the points P and P', which correspond to the points where the ray intersects the sphere, we need to find values for $$t_0$$ and $$t_1$$.
Remember that a ray can be expressed using the following parametric form:
$$O + tD$$
Where $$O$$ represents the origin of the ray, and $$D$$ is the ray direction (usually normalized). Changing the value of $$t$$ allows defining any position along the ray. When $$t$$ is greater than 0, the point is located in front of the ray's origin (looking down the ray's direction); when $$t$$ is equal to 0, the point coincides with the ray's origin (O), and when $$t$$ is negative, the point is located behind its origin. By looking at Figure 1, you can see that $$t_0$$ can be found by subtracting $$t_{hc}$$ from $$t_{ca}$$, and $$t_1$$ can be found by adding $$t_{hc}$$ to $$t_{ca}$$. All we need to do is find ways of computing these two values ($$t_{hc}$$ and $$t_{ca}$$) from which we can derive $$t_0$$, $$t_1$$, and then P and P' using the ray parametric equation:
$$\begin{array}{lcl} P & = & O + t_{0}D\\ P' & = & O + t_{1}D \end{array}$$
We will start by noting that the triangle formed by the edges $$L$$, $$t_{ca}$$, and $$d$$ is a right triangle. So, we can easily calculate $$L$$, which is just the vector between $$O$$ (the ray's origin) and $$C$$ (the sphere's center). $$t_{ca}$$ is yet to be determined, but we can use trigonometry to solve this problem.
We know $$L$$, and we know $$D$$, the ray's direction. We also know that the dot (or scalar) product of a vector $$\vec{b}$$ and $$\vec{a}$$ corresponds to projecting $$\vec{b}$$ onto the line defined by vector $$\vec{a}$$, and the result of this projection is the length of the segment AB as shown in Figure 2 (for more information on the properties of the dot product, check the Geometry lesson):
$$\vec{a} \cdot \vec{b} = |a||b|\cos\theta.$$
In other words, the dot product of $$L$$ and $$D$$ gives us $$t_{ca}$$. Note that there can only be an intersection between the ray and the sphere if $$t_{ca}$$ is positive (if it is negative, it means that the vector $$L$$ and the vector $$D$$ point in opposite directions. If there is an intersection, it could be behind the ray's origin, but anything happening behind it is of no use to us). We now have $$t_{ca}$$ and $$L$$.
$$\begin{array}{l} L = C - O\\ t_{ca} = L \cdot D\\ if \; (t_{ca} < 0) \; return \; false \end{array}$$
There is a second right triangle in this construction, defined by $$d$$, $$t_{hc}$$, and the sphere's radius. We already know the radius of the sphere, and we are looking for $$t_{hc}$$ to find $$t_0$$ and $$t_1$$. To get there, we first need to calculate $$d$$. Remember, $$d$$ is also the opposite side of the right triangle defined by $$d$$, $$t_{ca}$$, and $$L$$. The Pythagorean theorem states that:
$$\text{opposite side}^2 + \text{adjacent side}^2 = \text{hypotenuse}^2$$
Replacing the opposite side, adjacent side, and hypotenuse respectively with $$d$$, $$t_{ca}$$, and $$L$$, we get:
$$\begin{array}{l} d^2 + t_{ca}^2 = L^2\\ d = \sqrt{L^2 - t_{ca}^2} = \sqrt{L \cdot L - t_{ca} \cdot t_{ca}}\\ if \; (d < 0) \; return \; false \end{array}$$
Note that if $$d$$ is greater than the sphere's radius, the ray misses the sphere, indicating no intersection (the ray overshoots the sphere). Now we have all the terms needed to calculate $$t_{hc}$$. Using the Pythagorean theorem again:
$$\begin{array}{l} d^2 + t_{hc}^2 = \text{radius}^2\\ t_{hc} = \sqrt{\text{radius}^2 - d^2}\\ t_{0} = t_{ca} - t_{hc}\\ t_{1} = t_{ca} + t_{hc} \end{array}$$
In the last paragraph of this section, we will demonstrate how to implement this algorithm in C++ and discuss a few optimizations to expedite the process.
## Analytic Solution
Recall that a ray can be expressed using the function: $$O + tD$$ (equation 1), where $$O$$ is a point corresponding to the ray's origin, $$D$$ is a vector denoting the ray's direction, and $$t$$ is the function's parameter. Varying $$t$$ (which can be positive or negative) allows the calculation of any point's position on the line defined by the ray's origin and direction. When $$t$$ is greater than 0, the point on the ray is "in front" of the ray's origin. The point is behind the ray's origin when $$t$$ is negative. When $$t$$ is exactly 0, the point and the ray's origin coincide.
The concept behind solving the ray-sphere intersection test is that spheres can also be defined using an algebraic form. The equation for a sphere is:
$$x^2 + y^2 + z^2 = R^2$$
Where $$x$$, $$y$$, and $$z$$ are the coordinates of a Cartesian point, and $$R$$ is the radius of a sphere centered at the origin (we will see later how to modify the equation so that it works with spheres not centered at the origin). It implies that there is a set of points for which the above equation holds true. This set of points defines the surface of a sphere centered at the origin with a radius $$R$$. Or, more simply, if we consider $$x$$, $$y$$, and $$z$$ as the coordinates of point $$P$$, we can express it as (equation 2):
$$P^2 - R^2 = 0$$
This equation is characteristic of what we call in mathematics, and computer graphics, an implicit function. A sphere expressed in this form is also known as an implicit shape or surface. Implicit shapes are those defined not by connected polygons (as you might be familiar with if you've modeled objects in a 3D application such as Maya or Blender) but simply in terms of equations. Many shapes (often quite simple) can be defined as functions (cubes, cones, spheres, etc.). Despite their simplicity, these shapes can be combined to create more complex forms. This is the idea behind geometry modeling using blobs, for instance (blobby surfaces are also called metaballs). But before we digress too much, let's return to the ray-sphere intersection test (check the advanced section for a lesson on Implicit Modeling).
All we need to do now is substitute equation 1 into equation 2, that is, replace $$P$$ in equation 2 with the equation of the ray (remember $$O + tD$$ defines all points along the ray):
$$(O + tD)^2 - R^2 = 0$$
When we expand this equation, we get (equation 3):
$$O^2 + (Dt)^2 + 2ODt - R^2 = O^2 + D^2t^2 + 2ODt - R^2 = 0$$
This is essentially an equation of the form (equation 4):
$$ax^2 + bx + c$$
with $$a = D^2$$, $$b = 2OD$$, and $$c = O^2 - R^2$$ (note that $$x$$ in equation 4 corresponds to $$t$$ in equation 3, which is the unknown). Rewriting equation 3 into equation 4 is crucial because equation 4 is recognized as a quadratic equation. It's a function whose roots (when $$x$$ takes a value for which $$f(x) = 0$$) can be found using the following formula (equation 5):
• When $$\Delta > 0$$, there are two roots, which can be calculated with:$$\dfrac{-b + \sqrt{\Delta}}{2a} \quad \text{and} \quad \dfrac{-b - \sqrt{\Delta}}{2a}$$ In this case, the ray intersects the sphere at two points ($$t_0$$ and $$t_1$$).
• When $$\Delta = 0$$, there is one root, which can be calculated with:$$-\dfrac{b}{2a}$$ The ray intersects the sphere at one point only ($$t_0 = t_1$$).
• When $$\Delta < 0$$, there is no real solution, indicating that the ray does not intersect the sphere.
Since we have $$a$$, $$b$$, and $$c$$, we can easily solve these equations to find the values for $$t$$, which correspond to the two intersection points of the ray with the sphere ($$t_0$$ and $$t_1$$ in Figure 1). Note that the root values can be negative, meaning that the ray intersects the sphere but is behind the origin. One root can be negative, and the other positive, indicating that the ray's origin is inside the sphere. There might also be no solution to the quadratic equation, meaning that there is no intersection between the ray and the sphere.
Before we explore how to implement this algorithm in C++, let's see how we can solve the same problem when the sphere is not centered at the origin. Then, we can rewrite equation 2 as:
$$|P-C|^2-R^2=0$$
Where $$C$$ is the center of the sphere in 3D space. Equation 4 can now be rewritten as:
$$|O+tD-C|^2-R^2=0.$$
This gives us the following:
$$\begin{array}{l} a = D^2 = 1\\ b = 2D \cdot (O-C)\\ c = |O-C|^2 - R^2 \end{array}$$
In a more intuitive form, this means that we can translate the ray by $$-C$$ and test this transformed ray against the sphere as if it was centered at the origin.
"Why $$a=1$$"? Because $$D$$, the ray's direction, is a vector that is normally normalized. The result of a vector squared is the same as a dot product of the vector with itself. We know that the dot product of a normalized vector with itself is 1, therefore $$a=1$$. However, you must be very careful in your code because the rays tested for intersections with a sphere don't always have their direction vector normalized, so you will have to calculate the value for $$a$$ (check code further down). This is a pitfall that is often the source of bugs in the code.
## Computing the Intersection Point
Once we have the value for $$t_0$$, computing the intersection position or hit point is straightforward. We simply apply the ray parametric equation:
$$P_{hit} = O + t_0 D.$$
Vec3f Phit = ray.orig + ray.dir * t_0;
## Computing the Normal at the Intersection Point
There are various methods to calculate the normal of a point lying on the surface of an implicit shape. However, as mentioned in the first chapter of this lesson, one method involves differential geometry, which is mathematically quite complex. Thus, we will opt for a much simpler approach. For instance, the normal of a point on a sphere can be computed as the position of the point minus the sphere center (it's essential to normalize the resulting vector):
$$N = \frac{P - C}{||P - C||}.$$
Vec3f Nhit = normalize(Phit - sphere.center);
## Computing the Texture Coordinates at the Intersection Point
Interestingly, texture coordinates are essentially the spherical coordinates of the point on the sphere, remapped to the range [0, 1]. Recalling the previous chapter and the lesson on Geometry, the Cartesian coordinates of a point can be derived from its spherical coordinates as follows:
$$\begin{array}{l} P.x = \cos(\theta)\sin(\phi),\\ P.y = \sin(\theta),\\ P.z = \cos(\theta)\sin(\phi).\\ \end{array}$$
These equations may vary if a different convention is used. The spherical coordinates $$\theta$$ and $$\phi$$ can also be deduced from the Cartesian coordinates of the point using the following equations:
$$\begin{array}{l} \phi = \text{atan2}(z, x),\\ \theta = \text{acos}\left(\dfrac{P.y}{R}\right). \end{array}$$
Where $$R$$ is the radius of the sphere. These equations are further explained in the lesson on Geometry. Sphere coordinates are valuable for texture mapping or procedural texturing. The program in this lesson will demonstrate how they can be utilized to create patterns on the surface of spheres.
## Implementing the Ray-Sphere Intersection Test in C++
Let's explore how we can implement the ray-sphere intersection test using the analytic solution. Directly applying equation 5 is feasible (and it will work) to calculate the roots. However, due to the finite precision with which real numbers are represented on computers, this formula can suffer from a loss of significance. This issue arises when $$b$$ and the square root of the discriminant have close values but opposite signs. Given the limited precision of floating-point numbers on computers, this can lead to catastrophic cancellation or unacceptable rounding errors. For more details on this issue, plenty of resources are available online. However, equation 5 can be substituted with a slightly modified equation that proves to be more stable in computer implementations:
$$\begin{array}{l} q = -\dfrac{1}{2} \left( b + \text{sign}(b) \sqrt{b^2 - 4ac} \right) \\ x_{1} = \dfrac{q}{a}, \quad x_{2} = \dfrac{c}{q} \end{array}$$
Where $$\text{sign}$$ is -1 when $$b$$ is less than 0, and 1 otherwise. This modification ensures that the quantities combined to compute $$q$$ share the same sign, thus avoiding catastrophic cancellation. Below is how the routine is implemented in C++:
bool solveQuadratic(const float &a, const float &b, const float &c,
float &x0, float &x1) {
float discr = b * b - 4 * a * c;
if (discr < 0) return false;
else if (discr == 0) x0 = x1 = -0.5 * b / a;
else {
float q = (b > 0) ?
-0.5 * (b + sqrt(discr)) :
-0.5 * (b - sqrt(discr));
x0 = q / a;
x1 = c / q;
}
if (x0 > x1) std::swap(x0, x1);
return true;
}
Now, here is the complete code for the ray-sphere intersection test. For the geometric solution, we noted that we could reject the ray early if $$d$$ is greater than the sphere's radius. However, this would involve computing the square root of $$d^2$$, which incurs a computational cost. Moreover, $$d$$ itself is never used in the code; only $$d^2$$ is. Instead of computing $$d$$, we test if $$d^2$$ is greater than $$radius^2$$ (hence why $$radius^2$$ is calculated in the constructor of the Sphere class) and reject the ray if this condition is true. This approach efficiently speeds up the process.
bool intersect(const Ray &ray) const
{
float t0, t1; // Solutions for t if the ray intersects the sphere
#if 0
// Geometric solution
Vec3f L = center - ray.orig;
float tca = L.dotProduct(ray.dir);
// if (tca < 0) return false;
float d2 = L.dotProduct(L) - tca * tca;
t0 = tca - thc;
t1 = tca + thc;
#else
// Analytic solution
Vec3f L = ray.orig - center;
float a = ray.dir.dotProduct(ray.dir);
float b = 2 * ray.dir.dotProduct(L);
if (!solveQuadratic(a, b, c, t0, t1)) return false;
#endif
if (t0 > t1) std::swap(t0, t1);
if (t0 < 0) {
t0 = t1; // If t0 is negative, let's use t1 instead.
if (t0 < 0) return false; // Both t0 and t1 are negative.
}
t = t0;
return true;
}
Note: If the scene contains multiple spheres, they are tested in the order they were added to the scene, not necessarily sorted by depth relative to the camera position. The correct approach is to track the sphere with the closest intersection distance, i.e., the smallest $$t$$. The image below illustrates a scene render showing the latest intersected sphere in the object list on the left, even if it's not the closest. On the right, the render keeps track of the closest object to the camera, yielding the correct result. The implementation of this technique is provided in the next chapter.
Let's now see how we can implement the ray-sphere intersection test using the analytic solution. We could use equation 5 directly (you can implement it, and it will work) to calculate the roots. Still, on computers, we have a limited capacity to represent real numbers with the precision needed to calculate these roots as accurately as possible. Thus the formula suffers from the effect of a loss of significance. This happens when b and the root of the discriminant don't have the same sign but have values very close to each other. Because of the limited numbers used to represent floating numbers on the computer, in that particular case, the numbers would either cancel out when they shouldn't (this is called catastrophic cancellation) or round off to an unacceptable error (you will easily find more information related to this topic on the internet). However, equation 5 can easily be replaced with a slightly different equation that proves more stable when implemented on computers. We will use instead:
$$\begin{array}{l} q=-\dfrac{1}{2}(b+sign(b)\sqrt{b^2-4ac})\\ x_{1}=\dfrac{q}{a}\\ x_{2}=\dfrac{c}{q} \end{array}$$
Where the sign is -1 when b is lower than 0 and 1 otherwise, this formula ensures that the quantities added for q have the same sign, avoiding catastrophic cancellation. Here is how the routine looks in C++:
bool solveQuadratic(const float &a, const float &b, const float &c, float &x0, float &x1)
{
float discr = b * b - 4 * a * c;
if (discr < 0) return false;
else if (discr == 0) x0 = x1 = - 0.5 * b / a;
else {
float q = (b > 0) ?
-0.5 * (b + sqrt(discr)) :
-0.5 * (b - sqrt(discr));
x0 = q / a;
x1 = c / q;
}
if (x0 > x1) std::swap(x0, x1);
return true;
}
Finally, here is the completed code for the ray-sphere intersection test. For the geometric solution, we have mentioned that we can reject the ray early on if $$d$$ is greater than the sphere radius. However, that would require computing the square root of $$d^2$$, which has a cost. Furthermore, $$d$$ is never used in the code. Only $$d^2$$ is. Instead of computing $$d$$, we test if $$d^2$$ is greater than $$radius^2$$ (which is the reason why we calculate $$radius^2$$ in the constructor of the Sphere class) and reject the ray if this test is true. It is a simple way of speeding things up.
bool intersect(const Ray &ray) const
{
float t0, t1; // solutions for t if the ray intersects
#if 0
// Geometric solution
Vec3f L = center - ray.orig;
float tca = L.dotProduct(ray.dir);
// if (tca < 0) return false;
float d2 = L.dotProduct(L) - tca * tca;
t0 = tca - thc;
t1 = tca + thc;
#else
// Analytic solution
Vec3f L = ray.orig - center;
float a = ray.dir.dotProduct(ray.dir);
float b = 2 * ray.dir.dotProduct(L);
if (!solveQuadratic(a, b, c, t0, t1)) return false;
#endif
if (t0 > t1) std::swap(t0, t1);
if (t0 < 0) {
t0 = t1; // If t0 is negative, let's use t1 instead.
if (t0 < 0) return false; // Both t0 and t1 are negative.
}
t = t0;
return true;
}
Note that if the scene contains more than one sphere, the spheres are tested for any given ray in the order they were added to the scene. The spheres are thus unlikely to be sorted in depth (with respect to the camera position). The solution to this problem is to keep track of the sphere with the closest intersection distance, in other words, with the closest $$t$$. In the image below, you can see on the left a render of the scene in which we display the latest sphere in the object list that the ray intersected (even if it is not the closest object). On the right, we keep track of the object with the closest distance to the camera and only display this object in the final image, which gives us the correct result. An implementation of this technique is provided in the next chapter. | 5,129 | 18,981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.92988 |
https://grahamhancock.com/nightingalee4/ | 1,718,817,759,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00477.warc.gz | 245,977,851 | 14,495 | Edward Nightingale has published a follow-up article to support his theories which you can read here.
I am submitting a follow-up article that is focused more on the pyramids, their measure and connection to the precessional cycle that may help the reader to better understand the article The Leo, Orion Relationship. The Giza Pyramids and the Precessional Cycle article covers topics that are discussed regularly on GrahamHancock.com and hopefully will generate more discussion. This article will also provide more of a foundation that fully supports the new data I have presented on the precessional cycle encoded at Giza.
The architects, engineers and astronomers that were behind the design of the repository at Giza used a specific template consisting of a circle with a diameter of 9 units and a square of 8 x 8 of the same units. The circle with a diameter of 9 units has the area of 63.6428 when calculated. A square with a side length of 8 of the same size units, has an area of 64. This simple calculation approximates the area of a circle. The 9 unit circle and the 8 x 8 square are used as the survey plat for the design of the Giza repository. It is also a fundamental concept of land management used in many ancient cultures
Beginning with the 8 x 8 grid the designers applied numbers to the grid to create a coordinate system. The numbers applied to the 8 x 8 square are given to us by Plato in his book Timeaus as the Lambda sequence stating that these were the numbers that God used to create the "Cosmic Soul". Lambda is the 11th letter of the Greek alphabet. The Lambda sequence of numbers given by Plato are, (1,2,4,8 and 1,3,9,27) a doubling sequence (x2) and a tripling sequence (x3). The doubling sequence applied to the X axis (down the left side of the 8 x 8) and a tripling sequence 1,3,9,27) across the top. These numbers continue both down and across to complete the 8 x 8 square.
There are two diagonal scales shown within the grid, one beginning at 256, this is a Musical scale of perfect fifth intervals. 256 Hertz,(a measure of frequency) is a middle C note on a piano.
The second scale beginning at 128 is where the designers obtained all the units of measure, the inch, foot, and six different cubit lengths. These cubits measurements are, the Light Cubit, the Root Cubit, Royal Cubit, Canonical, Geographic and the Sacred Cubits. These units are all related to each other by specific fractions.
Beginning at 128 on the scale, the number 432 is three intervals up the diagonal scale of 7 intervals (7 down the X axis, 7 across the Y axis and 7 up the diagonal). Each interval of the 7 are then divided by 7 creating a total number of intervals as 49 (7 x 7) on the diagonal scale. 432 is 21st interval up the scale (3 x 7).
The numbers of 21 and 432 are the key to determining the dimensions of all the pyramids. Beginning with 432 and subtracting 21 = 411, this is the base length assigned to the pyramid of Khafre. The slope angle is 53.13 Degrees, a 3,4,5 Pythagorean Triangle. This angle is a is created on the Grid by 4 units down (16) the side and 3 units (27) across the top producing the coordinate of 432, 16 x 27 = 432. This completes the overall dimensions the pyramid of Khafre (G2).
To determine the base length of Khufu (G1) the Great Pyramid the number 432 is multiplied by 21 = 9072. The angle of Khufu (G1) is created by the relationship of the circle to the approximate Phi division(13/21 of the Fibonacci sequence) of the square. The division creates a Golden Rectangle in which a Fibonacci Spiral is created. The angle is 51.84 degrees.
The numbers imbedded within this structure, Khufu or G1, a few are shown in this graphic, they include the units of measure, the inch, foot, and six different cubit lengths. The Light Cubit, the Root Cubit, Royal Cubit, Canonical, Geographic and the Sacred Cubits. These units are all related to each other by specific fractions.
These units are all created from this original equation of 432 divided by 21 = 20.571428. This is the KEY number or unit from which all other units of measure are created and is designated the value of Inches. The length of 20.571428 Inches is equal to the length of 1 Root Cubit. From this number, all related cubit dimensions are derived by adding specific fractions to each creating the next cubit length.
For the design of Menkaure (G3), the architects embedded the diagonal scale of 7 down and 7 across and 7 diagonal. There are two different base lengths assigned to this pyramid. One by this equation, 7 x 7 x 7 = 343 (feet), this 343 feet can be seen as 49 units of 7 feet (49 the number of intervals in the diagonal scale) 49 x 7 = 343. The second base length is created by using 48 of these units of 7 feet creating a base length of 336 feet. 48 x 7 = 336 feet. The height of Menkaure (G3) is assigned 4/9ths the height of Khufu (G1) equaling 213.8177 feet. This creates two separate slope angles from the two base lengths, 51.84 degrees and 51.27 degrees.
The number 51.84 and 51.27 are KEY numbers related to the precessional cycle in that 5184 (years) is 1/5th of the approximate duration of the precessional cycle of 25920 (years) 5 x 5184 = 25920.
The number 5127 is the duration of the calendar used by the Mayans, 2013 (2012 winter solstice completing 2012 years) or the Gregorian date of 12/21/2012. 2013 – 5127 = -3114. That would be the year 3114 BC, the beginning of the Mayan calendar. Two calendars are created here, one beginning with 0 with a duration of 5127 years. A second beginning with -3114 BC and ending with 2013 AD, a duration of 5127 years. A third calendar is created with the number 5773 which is the height in inches of the pyramid of Khufu (G1). This number 5773 corresponds to the 4/9ths height of Menkaure (G3). This number is multiplied by two, 5773 x 2 = 11546, this number is divided by 9 and = 2886.5. This number multiplied by 9 = 25978. 25978 is the approximate duration of the precessional cycle. 5773 is the height of Khufu (G1) in inches, in feet = 481.09 feet. Adding 481.09 to 25978 = 26459.59, in whole numbers, 26460. the full duration of 1 cycle of precession. Confirmation of this is demonstrated in the following equations using the fraction of 49/48ths.
The pyramid of Menkaure (G3) encodes the numbers 49 and 48 in its base lengths as demonstrated. Beginning with 25920 as the (Root) approximate duration of the precessional cycle we are able to correct the approximate duration of the cycle to the observed duration of the cycle by the use of fractions, converting 25920 into the fraction of 48/48ths, 25920 divided by 48 = 540. 25920 + 540 = 26460, the duration of 1 precessional cycle.
What is encoded in the pyramid of Menkaure (G3) is the approximate duration of 1/5 of the 25920 year cycle to the precise duration of the cycle described by Plato as the Great Year. This approximate duration is then adjusted in the manner described to the correct duration of the cycle of 26460 years.
If what I have presented is factual, then it will be easily confirmed or not by what we observe from the view of the Sphinx using star viewing software, in this case Stellarium software.
This first image of the sky is viewed from the Sphinx on the (binary) date of 12/21/2012 0hrs 0min 0sec, the end/beginning of the precessional cycle. This date is by design and is precisely sunrise from the view of the Sphinx. The alignment of the Galactic Plane (in white) and the Ecliptic (in red) create an X and intersect with the Sun at precisely Sunrise. Note: The atmosphere and ground are not shown is these two images for clarity.
The Sun is rising approximately 27 degrees South of East on the horizon.
I have proposed that 12/21/2012 as the end of one cycle and the beginning of another. If I am correct with the calculations gleaned from the design of the three main pyramids, we should witness this same Galactic/ Ecliptic/Sun alignment forming an X with the Sun, 26460 years before 2013. 26460 – 2013 – 24447BC. This image is taken in the year -24447 BC at sunrise. Take notice of the hours and minutes this sunrise occurs. 20hrs and 12minutes after the sunrise recorded in 2012. This alignment only occurs on these two dates with a cycle duration of 26460 years. It is most interesting to note, and no mere coincidence that the angle of the Galactic Plane is 23.5 degrees off 90 degrees, (the pole or axis of the Earth) to the Horizon. 23.5 degrees is the angle of the tilt of Earths axis to the Ecliptic. Also the angle of the ecliptic to the Horizon is 53.13 degrees, the angle of the Pythagorean 3,4,5 Triangle, the slope angle of Khafre (G2).
The Sun is rising approximately 27 degrees North of East on the horizon, a movement of 54 degrees on the horizon from the end of the cycle to the beginning of the cycle.
In my article The Leo, Orion Relationship [http://www.grahamhancock.com/forum/NightingaleE3.php] I suggested that the designers of Giza encoded the trajectory of our Sun, Sirius and the Pleiades as spiral in nature, originating from the Great Orion Nebula, indexing with the three pyramids as the Orion Correlation Theory states. The Leo aspect of the relationship is the spin view of the Ecliptic.
I have demonstrated a model of the precessional cycle with a beginning and completion. I also propose that certain structures were constructed at precise points and intervals of this cycle. These intervals relate directly to the geological and other evidence of catastrophic events on Earth as documented by Randall Carlson.
The implications and ramifications as a result of the information presented here and in The Giza Template, Temple Graal, Earth Measure are left for the reader to consider.
For more in depth information the book, The Giza Template Temple Graal, Earth Measure is available through Amazon.com at this address:
http://www.amazon.com/exec/obidos/ASIN/1502493233/theofficialgraha
www.thegizatemplate.com
### Biography
Edward Nightingale is an independent researcher, author and speaker who as a Master Woodworker with over 33 years of drawing, measuring and building experience has spent the past 17 years utilizing these skill sets to reverse engineer the Giza Complex.
Edward travelled to Egypt in 1997 in search of evidence for his theory on the architectural design of the Giza pyramids and the Sphinx, the results of his efforts are published in his new book, The Giza Template: Temple Graal, Earth Measure. The book documents the step by step design that the architects of Giza used to encode highly scientific data into geometric and mathematical information to be preserved and passed on to future generations.
Edward’s claims are backed with precise geometry, mathematics, satellite imaging, astronomical charting and cosmology that can be reproduced using the scientific method.
Edward is working on a sequel, Heavens Measure to follow his book Earth Measure.
Website: www.thegizatemplate.com
Edward Nightingale is an independent researcher, author and speaker who as a Master Woodworker with over 33 years of drawing, measuring and building experience has spent the past 17 years utilizing these skill sets to reverse engineer the Giza Complex.
Edward travelled to Egypt in 1997 in search of evidence for his theory on the architectural design of the Giza pyramids and the Sphinx, the results of his efforts are published in his new book, The Giza Template: Temple Graal, Earth Measure. The book documents the step by step design that the architects of Giza used to encode highly scientific data into geometric and mathematical information to be preserved and passed on to future generations.
Edward’s claims are backed with precise geometry, mathematics, satellite imaging, astronomical charting and cosmology that can be reproduced using the scientific method.
Edward is working on a sequel, Heavens Measure to follow his book Earth Measure.
Website: www.thegizatemplate.com
## One thought on “The Giza Pyramids and the Precessional Cycle”
1. Gőczey András says:
Dear Edward, I am a hungarian architect. I made a lot of interesting facts about Giza pyramids. Khufu is an every day calendar for 182 days and back the whole year. Khefren is a half year calendar. Menkaure with their small pyramids a very exact calendar with Moon. I can send you more details about it, the last film of it is -Giza site plan Sun Moon Goczey Andras- I think it is very interestig to know it. best wishews, Andras | 3,001 | 12,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-26 | latest | en | 0.938073 |
https://www.numbersaplenty.com/2222022111 | 1,722,816,515,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00565.warc.gz | 738,224,022 | 3,245 | Search a number
2222022111 = 317231894307
BaseRepresentation
bin1000010001110001…
…0101110111011111
312201212010102022020
42010130111313133
514022314201421
61004253351223
7106030606401
oct20434256737
95655112266
102222022111
11a40301111
12520198513
132954722aa
14171170b71
15d011bdc6
hex84715ddf
2222022111 has 16 divisors (see below), whose sum is σ = 3273364224. Its totient is φ = 1333591424.
The previous prime is 2222022109. The next prime is 2222022139. The reversal of 2222022111 is 1112202222.
It is a cyclic number.
It is not a de Polignac number, because 2222022111 - 21 = 2222022109 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (2222022181) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 945981 + ... + 948326.
It is an arithmetic number, because the mean of its divisors is an integer number (204585264).
Almost surely, 22222022111 is an apocalyptic number.
2222022111 is a deficient number, since it is larger than the sum of its proper divisors (1051342113).
2222022111 is a wasteful number, since it uses less digits than its factorization.
2222022111 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1894350.
The product of its (nonzero) digits is 64, while the sum is 15.
The square root of 2222022111 is about 47138.3295312848. The cubic root of 2222022111 is about 1304.9167087664.
Adding to 2222022111 its reverse (1112202222), we get a palindrome (3334224333).
The spelling of 2222022111 in words is "two billion, two hundred twenty-two million, twenty-two thousand, one hundred eleven". | 523 | 1,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-33 | latest | en | 0.803158 |
https://buyassignment.us/category/geometry-homework-help/ | 1,656,504,562,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00769.warc.gz | 203,203,915 | 70,307 | # Your Homework Done For You
We will provide you with a paper that’s not only original but one that will get you an A or an A-.
## Geometric Constructions In this lesson, you learned how a compass and straighted
Geometric Constructions In this lesson, you learned how a compass and straightedge are used to create constructions related to circles. In this assignment, you will use those tools to complete those constructions on your own. As you complete the assignment, keep this...
## Geometric Constructions Geometric constructions date back thousands of years to
Geometric Constructions Geometric constructions date back thousands of years to when Euclid, aGreek mathematician known as the “Father of Geometry,” wrote the book Elements. In Elements, Euclid formulated the five postulates that form the base for Euclidean geometry.... | 171 | 841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-27 | latest | en | 0.899646 |
https://mathoverflow.net/questions/214462/a-fun-game-related-to-knot-theory | 1,660,076,900,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00239.warc.gz | 368,879,969 | 27,040 | # A fun game related to knot theory
I recently learned the following rather fun game: a group of people is standing up roughly in circle, facing each other. Then participants randomly join hands, in such a way that nobody holds its own hand, and that everybody hold hands with two distinct person.
The result is clearly a link, and the goal of the game is to untangle it, i.e. participants have to move while still holding hands until they can stand up in (possibly several !) circle(s) holding hands with their direct left and right neighboors. Obviously people are allowed to not face the center of the circle (or to be upside down, I guess) so that the first Reidemeister move is allowed, but I guess it's not really important.
Clearly it requires quite some physical and intellectual skills, a lot of collaboration between the players, and a rather long time, which makes it a fun activity, and I recommend trying this at home with friends or at work with fellow mathematicians.
The person who explained the game sweared that it "always works", although it's fairly easy for someone having heard of knot theory to come up with an example where it doesn't. For example the trefoil is easily obtained already with 3 players, but it's also clear that this correspond to a rather particular choice.
In fact, this one time we actually did not manage to untangle the link though we managed to simplify it a lot. Of course even if the link is mathematically trivial it might be physically challenging to actually untangle it, but let's ignore that.
Basically, this game can be formalized as a process which generate links randomly, and I'm curious whether something interesting can be said about it. I must admit that since this question did not actually come up in my research I haven't given much thought about it.. Obvious questions are: can every link be obtained this way ? If not do they correspond to a known family of links ? Is there any way to support the organizer's claim, ie is there a way to estimate the probability of getting a trivial link ?
• Two references, neither of which I can access: (1) Dreyer, P. A. "Knot Theory and the Human Pretzel Game." Congressus Numerantium (1996): 99-108. (2) M. B. Rao, Kalyaa Rao and Christopher N. Swanson. "Probability in the Human Knot Game," Applied Probability Trust, 2012. May 10, 2015 at 16:06
• Here's a barely-related story. At Canada/USA Mathcamp (where I've been a counselor), there are a number of games played in groups like this, a popular one being "scream toes". N people stand in a circle, bow their heads, and each chooses a set of toes of some other person. At some signal, everyone looks up at the person whose toes they chose. If two people make eye contact, both scream, hence the name. That's the game; repeat until the neighbors complain about the noise. Anyway, a cute problem for high school math kids is the following: as $N \to \infty$, how many people scream each time? May 10, 2015 at 16:28
• What about the following definition of a "human knot" (which neglects articulations and supposes elastic arms): A knot is "$N-$human" if it can be realized as a piecewise affine knot of $\mathbb R^3$ consisting of $N$ segments such that all $N$ vertices are on a cylinder (ie. project on a circle in a suitable direction). May 11, 2015 at 12:23
• A related question: mathoverflow.net/questions/54412/… . Aug 10, 2015 at 18:57
• see this question mathoverflow.net/questions/206214/human-knot-game?rq=1
– M.U.
Aug 10, 2015 at 18:57
Yes, every link can be obtained in this way. Here's an inefficient way to do it. Put the link in braid form (via Alexander's theorem); suppose that we have $b$ strands. We'll achieve each braid generator as a certain pattern among $2b$ people on the circumference of the circle.
With a little more thought we can arrange the half-flips and modify the generators accordingly to eliminate the factor of 2, giving a quantitative (though not at all sharp!) bound of $\mathrm{min}_{\hat{b} = K}\mathrm{length}(b)\mathrm{width}(b)$ for the number of people needed.
• I might misunderstand, but this seems to ignore the practical consideration of how long people's arms are. If $b$ is a large enough number and we demand that everyone stand roughly on a circle then at some point people will not be physically able to reach horizontally across the circle. So I think a full "real life" answer to this problem would also impose some requirement concerning the lengths of the arcs (arms) and and a small requirement on how far a node is allowed to move off the circle. Aug 12, 2015 at 4:10 | 1,084 | 4,591 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.97306 |
https://brainly.in/question/96980 | 1,484,990,583,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281001.53/warc/CC-MAIN-20170116095121-00158-ip-10-171-10-70.ec2.internal.warc.gz | 793,292,954 | 10,533 | # If a, b are rational numbers and (a-1)√2+3=b√2+a, the value of (a+b) is?
1
by jyotithk13
2015-04-18T17:12:09+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Since are rational and is irrational, we can compare rational and irrational terms both sides. Comparing terms attached to both sides we get :
comparing rational terms both sides gives
Plug this in the earlier equation and get
sir pleas explain it because i can't understand
updated, please see if it makes sense now
thanks and i do this.
sounds good | 193 | 781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-04 | latest | en | 0.935484 |
https://www.mathworks.com/help/stats/regressiontree.compact.html | 1,717,018,279,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00880.warc.gz | 749,021,122 | 18,826 | Main Content
compact
Reduce size of regression tree model
Syntax
``ctree = compact(tree)``
Description
example
````ctree = compact(tree)` returns a `CompactRegressionTree` version of the trained regression tree model `tree`. You can predict regressions using `ctree` exactly as you can using `tree`. However, because `ctree` does not contain training data, you cannot perform some actions, such as cross-validation.```
Examples
collapse all
Compare the size of a full regression tree model to the compacted model.
Load the `carsmall` data set. Consider `Acceleration`, `Displacement`, `Horsepower`, and `Weight` as predictor variables.
```load carsmall X = [Acceleration Cylinders Displacement Horsepower Weight];```
Grow a regression tree using the entire data set.
`Mdl = fitrtree(X,MPG)`
```Mdl = RegressionTree ResponseName: 'Y' CategoricalPredictors: [] ResponseTransform: 'none' NumObservations: 94 ```
`Mdl` is a `RegressionTree` model. It is a full model, that is, it stores information such as the predictor and response data `fitrtree` used in training. For a properties list of full regression tree models, see `RegressionTree`.
Create a compact version of the full regression tree—that is, one that contains enough information to make predictions only.
`CMdl = compact(Mdl)`
```CMdl = CompactRegressionTree ResponseName: 'Y' CategoricalPredictors: [] ResponseTransform: 'none' ```
`CMdl` is a `CompactRegressionTree` model. For a properties list of compact regression tree models, see `CompactRegressionTree`.
Inspect the amounts of memory that the full and compact regression trees consume.
```mdlInfo = whos('Mdl'); cMdlInfo = whos('CMdl'); [mdlInfo.bytes cMdlInfo.bytes]```
```ans = 1×2 12570 7067 ```
`cMdlInfo.bytes/mdlInfo.bytes`
```ans = 0.5622 ```
In this case, the compact regression tree model uses approximately half the memory that the full model uses.
Input Arguments
collapse all
Full regression tree model, specified as a `RegressionTree` model object trained with `fitrtree`.
Version History
Introduced in R2011a | 499 | 2,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.788226 |
https://fistofawesome.com/contributing/what-is-meant-by-canonical-momentum/ | 1,709,364,370,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00621.warc.gz | 248,766,245 | 12,960 | # What is meant by canonical momentum?
## What is meant by canonical momentum?
The canonical momentum of a particle with charge q is defined as. where p is the usual momentum and A is the magnetic vector potential.
Is momentum conserved in electromagnetic field?
Electromagnetic field itself can be ascribed momentum in such a way that momentum is locally conserved (i.e. momentum of a space region changes continuously and its rate of change can be expressed as a surface integral of certain field function).
What is the momentum in magnetic field?
A particle of charge q travelling at right-angles to a magnetic field B with a speed v experiences a force Bqv at right angles to its motion. This tells us that for a fixed field B, and charge q, the momentum p is proportional to the radius of curvature r.
### Does light have angular momentum?
Light may also carry angular momentum, which is a property of all objects in rotational motion. For example, a light beam can be rotating around its own axis while it propagates forward.
What does Canonical mean in physics?
In physics it basically means “the important/standard one”, while in math it means something totally different, “the unique thing you can get without making any arbitrary choices”.
What is the relation between momentum and magnetic field?
The direction of the magnetic moment is perpendicular to the plane of the loop. Seeing that the angular momentum is also perpendicular to that plane, and having shown that their magnitudes are proportional, is all it takes to show that two vectors are proportional.
## Why electromagnetic energy itself is not conserved?
According to Maxwell equations, in destructive interference of two collinear beams out of phase 180 degree, the energy density, which is a function of the electric and magnetic fields, is destroyed too. In this way, the principle of conservation of energy is violated.
Do electromagnetic waves have momentum?
As we will discuss later in the book, there is no mass associated with light, or with any EM wave. Despite this, an electromagnetic wave carries momentum. The momentum of an EM wave is the energy carried by the wave divided by the speed of light.
How do you find the momentum of an electromagnetic field?
1. F=q(E+v×B),
2. f=ρE+J×B.
3. If the fields go to zero at infinity sufficiently fast, you can integrate this equation over all space and the right-hand side will go to zero by the divergence theorem. So, the total mechanical momentum + something is a conserved quantity.
### Does electromagnetic waves have angular momentum?
Electromagnetic (EM) waves contain angular momentum, which is composed of spin angular momentum and orbital angular momentum (OAM)1. For a radio wave, i.e., the EM wave in radio frequency, the spin angular momentum corresponds to polarization, which has already been widely used in radar applications2.
Can photons have angular momentum?
Photons are endowed with spin angular momentum ЖЇh along their direction of propagation. Beams of photons all carrying the same spin are circularly polarized. Less well known is that photons can also carry orbital angular momentum (OAM), ‘, quantized in units of ¯h.
What is canonical in physics?
canonical ensemble, in physics, a functional relationship for a system of particles that is useful for calculating the overall statistical and thermodynamic behaviour of the system without explicit reference to the detailed behaviour of particles.
## How is canonical momentum calculated?
A quantity known as the canonical momentum, P=mv+eA ends up being conserved throughout the charged particle’s trajectory. (Setting the total time derivative of the canonical momentum equal to zero simply results in ma=ev×B, which is just the expression for magnetic force.)
What is the difference between magnetic moment and angular momentum?
Do EM waves carry energy and momentum?
Yes, EM waves carry energy E and momentum P. As electromagnetic waves contain both electric and magnetic fields, there is a non-zero energy density associated with it.
### Can electromagnetic waves travel through empty space?
Electromagnetic waves are not like sound waves because they do not need molecules to travel. This means that electromagnetic waves can travel through air, solid objects and even space.
Why do electromagnetic waves have momentum?
Solution : The EM waves are produced by the accelerated charge. The electron jumping from outer to inner orbit of the electron radiates EM waves. EM waves are propagated as electric & magnetic fields oscillation in mutually perpendicular directions which shows that EM waves carry momentum & energy.
How can electromagnetic waves show momentum?
When electromagnetic waves fall on charged particles they bring the charges into motion, which generally shows that electromagnetic waves have both energy and momentum. For example: When the sun shines, we feel energy absorbed from electromagnetic waves on our body.
## What is the importance of canonical momentum?
But there is also a more fundamental role that it plays: assuming position-independent vector potential, the canonical momentum is a conserved quantity, while the “normal” (or kinetic) momentum (mass times velocity) is not.
What do we know about the momentum and angular momentum of Airy beams?
A canonical theory is introduced to describe the momentum, angular momentum, and helicity of Airy beams. Numerical results of the canonical momentum, angular momentum, and helicity of Airy beams are explored. We report a study of the momentum, angular momentum, and helicity of circularly polarized Airy beams propagating in free space.
Do electromagnetic waves carry energy?
Here we investigate on the case that the electromagnetic waves carry energy. We can express the energy carried by the electromagnetic waves in terms of the momentum; in other words, the electromagnetic waves carry momentum.
### How do you find the canonical momentum from Legendre transformation?
The canonical momentum obtained via the Legendre transformation using the action L is π = ∂ t ϕ {\\displaystyle \\pi =\\partial _ {t}\\phi } , and the classical Hamiltonian is found to be. | 1,203 | 6,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-10 | latest | en | 0.934424 |
https://www.jiskha.com/questions/520733/im-stuck-on-questions-can-anyone-help-me-a-which-situation-can-be-modeled-by-tossing-a | 1,600,471,439,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189264.5/warc/CC-MAIN-20200918221856-20200919011856-00549.warc.gz | 951,273,281 | 4,574 | # transitional math
Im stuck on questions can anyone help me ?
A)Which situation can be modeled by tossing a coin 3 times
choosing which one of sixteen teams will win a tournament?
B)picking one of six math teacher,
C)decide which English teacher of eight will win a trip to London
D)selecting a desk from 36 in a classroom
there are five floor plans , four exterior colors ,and an option of either a 2 or 3 car garage , how many choices are possible for one home
1. 👍 0
2. 👎 0
3. 👁 151
## Similar Questions
1. ### sixth grade math
Write a real-world situation that could be modeled by the equation 24/y=3. Then solve the problem. My answer: There are 24 seats available at the movies. There were total of 3 rows after the seats were divided and more seats on
2. ### algebra 1
real- world situation that could be modeled by the expression b less than 17
3. ### math
describe a real world situation that is modeled by dividing two fractions or mixed numbers.
4. ### 6TH GRADE
Describe a real-world situation that is modeled by multiplying two fractions or mixed numbers. Can someone give me an example?
1. ### english help please!!
Which of the following questions might someone who is going through a crisis ask himself? What will I decide? How can I avoid this situation? Why do I have to do this? Why is this happening? is it d
2. ### Math
A 100 point test contains a total of 20 questions. The multiple choice questions are worth 3 points each and the short response answer questions are worth 8 pints each. Write a system of linear equations that represents this
3. ### algebra 1
write an algebraic expression for each verbal expression. Then write a real-world situation that could be modeled by the expression. 10 more than y
4. ### math
Write a real-life situation that can be modeled by the expression 1/4 x 2/3
1. ### math
below are the results of tossing a number cube 8 times find the experimental probability of tossing an even number. 3,4,1,2,5,1,6,5 1. 1/4 2. 3/8 3. 1/2 4. 5/8
2. ### Math
Write a real-life situation that can be modeled by the expression 5x 3/4
3. ### math
When a situation can be modeled by a linear equation, what information do you need in order to find an equation?
4. ### AP World History
I really need some help with 2 questions. These questions are based out of Chapter 1 in Traditions and Encounters but it is in really any textbook. At least 5 sentences long. 1. Give two examples of the human response to climate | 626 | 2,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-40 | latest | en | 0.931442 |
https://www.spss-tutorials.com/spss-select-if-command/comment-page-5/ | 1,726,807,085,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00167.warc.gz | 915,093,415 | 15,837 | SPSS TUTORIALS FULL COURSE BASICS ANOVA REGRESSION FACTOR
# SPSS SELECT IF – Tutorial & Examples
## Quick Overview Contents
In SPSS, SELECT IF permanently removes
a selection of cases (rows) from your data.
## Summary
SELECT IF in SPSS basically means “delete all cases that don't satisfy one or more conditions”. Like so, select if(gender = 'female'). permanently deletes all cases whose gender is not female. Let's now walk through some real world examples using bank_clean.sav, partly shown below.
## Example 1 - Selection for 1 Variable
Let's first delete all cases who don't have at least a Bachelor's degree. The syntax below:
• inspects the frequency distribution for education level;
• deletes unneeded cases;
• inspects the results.
*Show values and value labels in new output tables.
set tnumbers both.
*Run minimal frequencies table.
frequencies educ.
*Select cases with a Bachelor's degree or higher. Delete all other cases.
select if(educ >= 4).
*Reinspect frequencies.
frequencies educ.
## Result
As we see, our data now only contain cases having a Bachelor's, Master's or PhD degree. Importantly, cases having
on education level have been removed from the data as well.
## Example 2 - Selection for 2 Variables
The syntax below selects cases based on gender and education level: we'll keep only female respondents having at least a Bachelor's degree in our data.
*Inspect contingency table sex and education.
crosstabs educ by gender.
*Select females having a Bachelor's degree or higher.
select if(gender = 0 & educ >= 4).
*Reinspect contingency table.
crosstabs educ by gender.
## Example 3 - Selection for (Non) Missing Values
Selections based on (non) missing values are straightforward if you master SPSS Missing Values Functions. For example, the syntax below shows 2 options for deleting cases having fewer than 7 valid values on the last 10 variables (overall to q9).
*Select cases having at least 7 non missing values out of last 10 questions.
select if(nvalid(overall to q9) >= 7)./*At least 7 valid values or at most 3 missings.
execute.
*Alternative way, exact same result.
select if(nmiss(overall to q9) < 4)./*Fewer than 4 missings or more than 6 valid values.
execute.
## Tip 1 - Inspect Selection Before Deletion
Before deleting cases, I sometimes want to have a quick look at them. A good way for doing so is creating a FILTER variable. The syntax below shows the right way for doing so.
*Create filter variable holding only zeroes.
compute filt_1 = 0.
*Set filter variable to 1 for cases we want to keep in data.
if(nvalid(overall to q9) >= 7) filt_1 = 1.
*Move unselected cases to bottom of dataset.
sort cases by filt_1 (d).
*Scroll to bottom of dataset now. Note that cases 459 - 464 will be deleted because they have 0 on filt_1.
*If selection as desired, delete other cases.
select if(filt_1).
execute.
Quick note: select if(filt_1). is a shorthand for select if(filt_1 <> 0). and deletes cases having either a zero or a missing value on filt_1.
## Result
Cases that will be deleted are at the bottom of our data. We also readily see we'll have 458 cases left after doing so.
## Tip 2 - Use TEMPORARY
A final tip I want to mention is combining SELECT IF with TEMPORARY. By doing so, SELECT IF only applies to the first procedure that follows it. For a quick example, compare the results of the first and second FREQUENCIES commands below.
*Make sure case deletion only applies to first procedure.
temporary.
*Select only female cases.
select if(gender = 0).
*Any procedure now uses only female cases. This also reverses case selection.
frequencies gender educ.
*Rerunning frequencies now uses all cases in data again.
frequencies gender educ.
## Final Notes
First off, parentheses around conditions in syntax are not required. Therefore, select if(gender = 0). can also be written as select if gender = 0. I used to think that shorter syntax is always better but I changed my mind over the years. Readability and clear structure are important too. I therefore use (and recommend) parentheses around conditions. This also goes for IF and DO IF.
Right, I guess that should do. Did I miss anything? Please let me know by throwing a comment below.
# Tell us what you think!
*Required field. Your comment will show up after approval from a moderator.
# THIS TUTORIAL HAS 29 COMMENTS:
• ### By Ruben Geert van den Berg on June 19th, 2020
Hi Esmee!
This would happen if you accidentally specified 9 as missing as well. Otherwise, the syntax should do as you intended. One solution is to switch off all missing values as in
MISSING VALUES somvariable ().
before you run the SELECT IF.
P.s. I think your coding sucks. Having 9 as valid and 8 as missing is confusing. We recommend using huge (positive) values such as 999999999 and 999999998 for user missing values. Like so, it's immediately clear that these can't be valid.
Hope that helps!
SPSS tutorials
• ### By Esmee on June 20th, 2020
Hi Ruben,
Thanks! 9 is a missing value as well, so I figured it would have had something to do with that. In the end I want to remove those cases with value 9 aswell, but I wanted to do it step by step, to see the difference between the cases (reason for missing if different for instance). I'll just figure out another way to do that then.
Thanks!
P.S. I use a dataset form an international study, so the (original)coding isn't mine. I did think it was weird though how they coded some missing values 8 and 9, and others 999999 etc indeed.
• ### By Ruben Geert van den Berg on June 21st, 2020
I thought 9 was valid and 8 missing. That would be weird!
Using huge numbers for missings has 2 great advantages:
-it's immediately clear these are not normal values and
-if you forget to specify them as missing, some output may look absolutely crazy so you'll understand something went wrong.
If you'd forget to set 8 or 9 as missing, your output probably looks very reasonable so it doesn't draw your attention to the problem. Just my opinion...
Good luck with your project!
P.s. if you or any fellow students need any personal assistance, we offer just that via Sigma Plus Statistiek (in Dutch or English).
• ### By Jon Peck on January 6th, 2021
A useful tool to use with SELECT CASES (and FILTER) is DATASET COPY. then you can experiment without messing up the input data.
• ### By Ruben Geert van den Berg on January 7th, 2021
Hi Jon!
This may surprise you, but I stopped working on multiple datasets simultaneously some years ago.
I felt the DATASET ACTIVATE / DATASET NAME / DATASET CLOSE commands cluttered up my syntax too much and cost me more time than they saved me.
Also, I want projects to have as few data files as possible. The amount of work increases somewhat exponentially with the complexity of projects.
My most recent approach is to sometimes make "safety clones" of variables within a dataset and then compare these to the adjusted variables with CROSSTABS / MEANS.
I had a nice, simple tool for this (basically just running RECODE / APPLY DICTIONARY / STRING commands) but it's becoming obsolete because I haven't moved on to Python3.x yet... | 1,694 | 7,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-38 | latest | en | 0.839027 |
https://bumpercarfilms.com/qa/quick-answer-why-does-kinetic-energy-decrease-after-a-collision.html | 1,621,147,687,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989690.55/warc/CC-MAIN-20210516044552-20210516074552-00499.warc.gz | 164,806,809 | 9,364 | # Quick Answer: Why Does Kinetic Energy Decrease After A Collision?
## Can kinetic energy increase after collision?
Collisions are considered inelastic when kinetic energy is not conserved, but this could be from either a loss or gain or kinetic energy.
For example, in an explosion-type collision, the kinetic energy increases..
## Why is momentum conserved but not kinetic energy?
Momentum is conserved, because the total momentum of both objects before and after the collision is the same. However, kinetic energy is not conserved. Some of the kinetic energy is converted into sound, heat, and deformation of the objects.
## What are the 3 types of collision?
There are three different kinds of collisions, however, elastic, inelastic, and completely inelastic. Just to restate, momentum is conserved in all three kinds of collisions. What distinguishes the collisions is what happens to the kinetic energy.
## How do you find total kinetic energy after an inelastic collision?
Inelastic Collision Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially 12mv2+12mv2=mv2 1 2 m v 2 + 1 2 m v 2 = m v 2 . The two objects come to rest after sticking together, conserving momentum.
## What percentage of the mechanical energy is lost in this collision?
3.3% of the mechanical energy remains. 96.7% is lost!
## How do you know if kinetic energy is conserved?
When objects collide, the total momentum of the system is always conserved if no external forces are acting on the system. Kinetic energy (KE) is the energy of motion, and kinetic energy is not always conserved in a collision. … An elastic collision is one where kinetic energy is conserved.
## What is kinetic energy formula?
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.
## What happens to kinetic energy after a collision?
As a result of a collision the kinetic energy of the particles involved in the collision generally change. … The collision can vary between an elastic collision where the total kinetic energy is conserved and a totally inelastic collision where the total kinetic energy is zero after the collision.
## What happens to kinetic energy lost in inelastic collision?
While the total energy of a system is always conserved, the kinetic energy carried by the moving objects is not always conserved. … In an inelastic collision, energy is lost to the environment, transferred into other forms such as heat.
## Can all kinetic energy be lost in a collision?
Can all the kinetic energy be lost in the collision? Yes, all the kinetic energy can be lost if the two masses come to rest due to the collision (i.e., they stick together). Describe a system for which momentum is conserved but mechanical energy is not.
## Is kinetic energy conserved in an explosion?
Explosions occur when energy is transformed from one kind e.g. chemical potential energy to another e.g. heat energy or kinetic energy extremely quickly. So, like in inelastic collisions, total kinetic energy is not conserved in explosions.
## How do you calculate change in kinetic energy after a collision?
Collisions in One DimensionMass m1 = kg , v1 = m/s.Mass m2 = kg , v2 = m/s.Initial momentum p = m1v1 + m2v2 = kg m/s .Initial kinetic energy KE = 1/2 m1v12 + 1/2 m2v22 = joules.Then the velocity of mass m2 is v’2 = m/s.because the final momentum is constrained to be p’ = m1v’1 + m2v’2 = kg m/s .More items…
## Why is kinetic energy not conserved?
Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision, but that is because it is converted to another form of energy (heat, etc.). The sum of all types of energy (including kinetic) is the same before and after the collision.
## Does kinetic energy decrease in an inelastic collision?
– A partially inelastic collision is one in which some energy is lost, but the objects do not stick together. – The greatest portion of energy is lost in the perfectly inelastic collision, when the objects stick. – The kinetic energy does not decrease.
## Can total kinetic energy ever be higher after a collision than before?
You can operate in the same reference frame and still have an increase in kinetic energy. … All you need to do is apply momentum conservation as well as the condition of a 50% increase in kinetic energy. Or use the coefficient of restitution. It is totally possible.
## Can kinetic energy negative?
Kinetic energy can’t be negative, although the change in kinetic energy Δ K \Delta K ΔK can be negative. Because mass can’t be negative and the square of speed gives a non-negative number, kinetic energy can’t be negative.
## When two objects stick together after collision How does the total momentum change?
The total momentum in any closed system will remain constant. When two or more objects collide, the collision does not change the total momentum of the two objects. Whatever momentum is lost by one object in the collision is gained by the other. The total momentum of the system is conserved.
## In what kind of collision is kinetic energy always conserved?
Elastic collisionsElastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.
## Can a single object have kinetic energy but no momentum?
FALSE – If an object does NOT have momentum, then it definitely does NOT have kinetic energy. However, it could have some potential energy and thus have mechanical energy.
## What is perfectly inelastic collision show that kinetic energy is invariably lost in such a collision?
Answer: The inelastic collision in the collision in which kinetic energy is not observed due to the action of internal friction. Kinetic energy is turned into vibration energy of the atom, causing a heating effect and body deformed.
## Are momentum and kinetic energy conserved in all collisions?
Generally, momentum is conserved in all types of collisions. Kinetic energy is smaller, and the objects stick together, after the collision. | 1,366 | 6,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-21 | latest | en | 0.922123 |
http://www.codingforums.com/java-and-jsp/301129-need-help-letter-counter-program.html?s=fc8791ece2e3bee3abdb5ed8a068b1ed | 1,464,203,915,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275328.0/warc/CC-MAIN-20160524002115-00091-ip-10-185-217-139.ec2.internal.warc.gz | 427,812,249 | 11,397 | Need help with letter counter program
Alright so I have my code and my methods setup on how i want the program to execute but im still having a little difficulties. The object of the program is to counts the frequencies of alphabet letters in a given file and records the letters and frequencies in another given file. The array is to save the letters and another array to save their frequencies, and a letter and its frequency should have the same index, i.e., letter 'A' has index 0 for both arrays. I have included comments.
[CODE]
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.PrintWriter;
public class LetterCounter {
/**
* The size of each array is 26 as there are 26 letters in English alphabet.
*
* @param setOfLetter the array to save the alphabet letters
* @param freqOfLetters the array to save frequencies of letters
* @param maxFreqIndex the array index with the maximum frequency
*/
static final int LENGTH = 26;
char setOfLetters[] = new char[LENGTH];
int freqOfLetters[] = new int[LENGTH];
int maxFreqIndex;
/**
* This method return true if ch is in [a-z] or [A-Z]
* @param ch the input character
* @return true if ch is in the range; false otherwise
*/
protected boolean isLetter(char ch) {
if ((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z')) {
return true;
}
else
{
return false;
}
}
/**
* this method find the index of character ch in the arrays.
* The index is calculated by Character.toUpperCase(ch) - 'A'
*
* @param ch the input character
* @return the array index of character ch
*/
protected int indexLetter(char ch) {
for (int i = 0; i < setOfLetters.length; i++)
if (setOfLetters[i] == ch) {
//return ch;
maxFreqIndex++;
}
return Character.toUpperCase(ch);
}
/**
* This method initializes both arrays. elements in freqOfletters are set to zero,
* and elements in setOfLetters are set to capital letters
*
*/
protected void initFreqs() {
//TODO
}
/**
* this method find the array index with the maximum frequency in an array
* @param values an array of ints
* @return the index with maximum value
*/
protected int findIndexOfMax(int values[]) {
return maxFreqIndex; // find the index of the character with max frequency
//TODO
}
/**
* This method reads character by character from the input file, check if each character is a letter. For letters, it
* increments the frequencies in the array freqOfLetters.
* <p>
* This method also
* @param inFileName the file to read from
* @throws IOException
*/
public void count(String inFileName) throws IOException {
//TODO
}
/**
* This method prints the results to the output file.
* @param outFileName the file to write to
* @throws IOException
*/
public void printFrequencies(String outFileName) throws IOException {
//TODO
}
/**
* This is the driver. It instantiate a LetterCounter object and call count and printFrequencies methods.
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
if(args.length != 2){
System.out.println("usage: " + LetterCounter.class.getSimpleName() + " input_file_name output_file_name");
System.exit(-1);
}
LetterCounter c1 = new LetterCounter();
try {
c1.count(args[0]);
c1.printFrequencies(args[1]);
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
[CODE] | 795 | 3,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-22 | latest | en | 0.693988 |
https://forum.math.toronto.edu/index.php?PHPSESSID=m7ol7vsqn1khd034i0rsgdrd25&topic=1504.0;prev_next=next | 1,718,692,529,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00023.warc.gz | 229,974,884 | 6,925 | ### Author Topic: Q6 TUT 5101 (Read 5633 times)
#### Victor Ivrii
• Elder Member
• Posts: 2607
• Karma: 0
##### Q6 TUT 5101
« on: November 17, 2018, 03:59:47 PM »
The coefficient matrix contains a parameter $\alpha$ . In each of these problems:
(a) Determine the eigenvalues in terms of $\alpha$.
(b) Find the critical value or values of $\alpha$ where the qualitative nature of the phase portrait for the system changes.
(c) Draw a phase portrait for a value of $\alpha$ slightly below, and for another value slightly above, each critical value.
$$\mathbf{x}' =\begin{pmatrix} 2 &-5\\ \alpha & -2 \end{pmatrix}\mathbf{x}.$$
#### Jingze Wang
• Full Member
• Posts: 30
• Karma: 25
##### Re: Q6 TUT 5101
« Reply #1 on: November 17, 2018, 04:00:25 PM »
First, try to find the eigenvalues with respect to the parameter
$A=\begin{bmatrix} 2&-5\\ \alpha&-2\\ \end{bmatrix}$
$det(A-rI)=(2-r)(-2-r)+5\alpha=0$
$r^2-4+5\alpha=0$
$r=\frac{\pm\sqrt{16-20\alpha}}{2}$
Notice that $-4+5\alpha$ determines the type of roots, so $\alpha=4/5$ is the critical value
Case 1
When $-4+5\alpha=0, \alpha=0$, there is a repeated eigenvalue 0 with one eigenvector
Case 2
When $-4+5\alpha>0, \alpha>4/5$, there are two distinct complex eigenvalues without real parts
Case 3
When $-4+5\alpha<0, \alpha<4/5$, there are two distinct real eigenvalues with different signs
« Last Edit: November 17, 2018, 04:08:35 PM by Jingze Wang »
#### Michael Poon
• Full Member
• Posts: 23
• Karma: 10
• Physics and Astronomy Specialist '21
##### Re: Q6 TUT 5101
« Reply #2 on: November 17, 2018, 04:07:45 PM »
a) Finding the eigenvalues:
Set the determinant = 0
\begin{align}
(2 - \lambda)(-2 - \lambda) - (-5)(\alpha) &= 0\\
\lambda^2 - 4 + 5\alpha &= 0\\
\lambda &= \pm \sqrt{4 - 5\alpha}
\end{align}
b)
Case 1: Eigenvalues real and opposite sign
when: $\alpha$ < $\frac{4}{5}$ (unstable saddle)
Case 2: Eigenvalues complex and opposite sign
when: $\alpha$ > $\frac{4}{5}$ (stable centre)
c) will post below:
#### Michael Poon
• Full Member
• Posts: 23
• Karma: 10
• Physics and Astronomy Specialist '21
##### Re: Q6 TUT 5101
« Reply #3 on: November 17, 2018, 04:08:43 PM »
Phase portrait attached
« Last Edit: November 25, 2018, 10:22:05 AM by Victor Ivrii »
#### Siran Wang
• Newbie
• Posts: 3
• Karma: 3
##### Re: Q6 TUT 5101
« Reply #4 on: November 17, 2018, 04:23:23 PM »
(a)
\begin{equation*}
A-\lambda I=\begin{pmatrix}
2-\lambda & -5\\
\alpha & -2-\lambda
\end{pmatrix}
\end{equation*}
(b)
\begin{equation*}
\det(A-\lambda I)=(2-\lambda)(-2-\lambda)+5\alpha=\lambda^2+5\alpha-4=0
\end{equation*}
\begin{equation*}
\lambda=\frac{0\pm \sqrt{0^2-4(5\alpha-4)}}{2}=\frac{0\pm \sqrt{-4(5\alpha-4)}}{2}=\frac{0\pm 2\sqrt{-(5\alpha-4)}}{2}
\end{equation*}
\begin{equation*}
\lambda_1=\sqrt{4-5\alpha}~~~~\lambda_2=-\sqrt{4-5\alpha}
\end{equation*}
\begin{equation*}\begin{split}
b^2-4ac&\geq0\\
4-5\alpha&\geq0\\
-5\alpha&\geq-4\\
\alpha&\leq\frac{4}{5}
\end{split}
\end{equation*}
when $\alpha<\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are two different real numbers.
when $\alpha>\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are solutions with complex numbers.
when $\alpha=\frac{4}{5}$, $\lambda_1$and$\lambda_2$ are repeated roots, which are 0.
(c) in attachments | 1,226 | 3,257 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-26 | latest | en | 0.465675 |
http://www.compuphase.com/graphic/scale2.htm | 1,485,233,890,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284270.95/warc/CC-MAIN-20170116095124-00489-ip-10-171-10-70.ec2.internal.warc.gz | 407,231,042 | 12,076 | # Quick image scaling by 2
In a preceding article ([1]) I presented an adaptation of the linear Bresenham algorithm to scale images with a quality that approximates that of linear interpolation. I called this algorithm "smooth Bresenham scaling". In the article, I noted that the scale factor should lie within the range of 50% to 200% for the routine to work correctly. Then, it went on to suggest that, if you need a scaling factor outside that range, you can use a separate algorithm that scales an image up or down by a factor of 2 and then use smooth Bresenham to scale to the final size. Scaling by a factor of 2 is the topic of this article.
This article and the preceding one could be seen as a single article cut in two parts: it is their combined use that results in a general purpose, fast image scaling algorithm with adequate quality. Both articles rely on an average() function that is covered in a separate article [2].
Like the previous article, the algorithms described here process colour images in a single pass without unpacking and repacking pixels into separate RGB channels. Although the example images in this article are all grey scale (single channel), the same algorithms are suitable for three-channel colour images and for palette-indexed images.
### A simple 1/2× image scaler
One of the simplest ways to halve the resolution of a picture is to replace each quad of four pixels by one pixel with is the average of the four pixels. Sampling theory calls this a "box filter"; in my experience, the box filter gives good quality when minifying an image by an integral factor.
When the average of four pixels is easy to calculate, by all means do so. If it is harder to calculate, such as with palette indexed images, you may opt to use the average() function (see [2]) three times in succession, as is done in the snippet below.
Scale 1/2×, box filtered
```template<class PIXEL>
void ScaleMinifyByTwo(PIXEL *Target, PIXEL *Source, int SrcWidth, int SrcHeight)
{
int x, y, x2, y2;
int TgtWidth, TgtHeight;
PIXEL p, q;
TgtWidth = SrcWidth / 2;
TgtHeight = SrcHeight / 2;
for (y = 0; y < TgtHeight; y++) {
y2 = 2 * y;
for (x = 0; x < TgtWidth; x++) {
x2 = 2 * x;
p = average(Source[y2*SrcWidth + x2], Source[y2*SrcWidth + x2 + 1]);
q = average(Source[(y2+1)*SrcWidth + x2], Source[(y2+1)*SrcWidth + x2 + 1]);
Target[y*TgtWidth + x] = average(p, q);
} /* for */
} /* for */
}```
I have written the code in above snippet for clarity more than to serve as production code. One obvious optimization is to avoid calculating the same values over and over again: the expressions y2*SrcWidth, (y2+1)*SrcWidth and y*TgtWidth remain constant in the inner loop and should be substituted by pre-computed variables. Perhaps you can trust your compiler to do this optimization for you.
From the example images shown below, it is apparent that the box filter is equivalent to linear interpolation with this particular scale factor. The box filter (and the bilinear interpolator) perform better than the bicubic interpolator when minifying: the resulting images of the bicubic interpolator are blurred, as you can see from the scaled "line graphics" images.
Original picture 50% - Box filter 50% - Bilinear interpolation 50% - Bicubic interpolation
Original picture 50% - Box filter 50% - Bilinear interpolation 50% - Bicubic interpolation
### An edge-sensitive directionally averaging 2× image scaler
Among the more popular algorithms for enlarging a bitmapped image are pixel replication and bilinear interpolation; these algorithms are quick and simple to implement. The pixel replication algorithm simply copies the nearest pixel (nearest neighbour sampling). As a result, the algorithm is prone to blockiness, which is especially visible along the edges. Bilinear interpolation is at the other extreme, it takes the (linearly) weighted average of the four nearest pixels around the destination pixel. This causes gradients to be fluently interpolated, and edges to be blurred. Several other algorithms use different weightings and/or more pixels to derive a destination pixel from a surrounding set of source pixels.
The reasoning behind these algorithms is that the value of a destination pixel must be approximated from the values of the surrounding source pixels. But you can do better than just to take the distances to the surrounding pixels and a set of fixed weight factors stored in a matrix. Directional interpolation is a method that tries to interpolate along edges in a picture, not across them.
My interest in directional interpolation was spurred by two papers. In [6] the authors use polynomial interpolation with gradient-dependent weights. To calculate the gradients, the paper applies a pair of Sobel masks on every source pixel taking part in the interpolation. The algorithm in [7] splits a source pixel into four target pixels in one pass, considering eight additional surrounding pixels (nine source pixels in total). By applying a matrix with pre-computed weights for every relation between the nine pixels, the value of each of the four target pixels is calculated.
The algorithm in [6] has fairly simple operations on every source pixel that it considers for each target pixel, but it considers a big set of source pixels. This is a major performance bottleneck. The other paper ([7]) uses a few surrounding pixels, and calculates four destination pixels in one pass, but the amount of arithmetic required for the calculation makes it unsuitable for real-time scaling. Neither algorithm is easily adapted to colour images; a colour image is regarded as a 3-channel image where each channel must be resized individually.
The idea, then, is to combine concepts from the two papers, but at a lower cost: simple operations on few source pixels per target pixel.
The flaw in the above approach is that a region of 9 pixels is too small to determine the direction of an edge running through the middle pixel, which the authors of [6] had probably encountered as well. Instead of selecting a larger edge detection kernel, however, I decided to split each source pixel into four target pixels, like in [7], and finding the interpolation partner per target pixel (so one source pixel may have four different interpolation partners). This new approach calculates four gradients for each source pixel. The figure at the right of this paragraph represents nine pixels from the source image. To calculate the target pixel for the upper left quadrant for the middle source pixel, the algorithm considers the source pixels 0, 1, 3 and 4. It finds the direction with the lowest gradient in this region of four source pixels and uses that to select the interpolation candidates. While I could have used a pair of Roberts masks to get the gradient direction, simply selecting the minimum distance between any of the four possible pairs of the four source pixels under consideration was more effective and actually simpler to implement.
The algorithm is conceptually simple: of all source pixels under consideration, it interpolates between the pair of pixels with the lowest difference. The upper left target pixel may be:
1. interpolated between source pixels 4 and 1,
2. interpolated between source pixels 4 and 3,
3. interpolated between source pixels 4 and 0 or
4. interpolated between source pixel 4 and the average of source pixels 1 and 3.
The first step is to calculate the differences between the pairs of source pixels 4-1, 4-3, 4-0 and 3-1. Actually, I am not interested in the sign of the differences, just the distances. The algorithm determines the minimum distance and chooses the appropriate interpolator. Interpolation, in this particular context, just means (unweighted) averaging; so in the case of first interpolator, the colour of target pixel is the halfway between of source pixels 4 and 1.
The code snippet below computes the "upper left" target pixel for a source pixel at coordinate (x, y). Using a similar procedure, one obtains the other three target pixels for this source pixel.
Scale 2× with directional averaging
```#define ABS(a) ( (a) >= 0 ? (a) : -(a) )
p = Source[y*SrcWidth + x];
/* target pixel in North-West quadrant */
p1 = Source[(y-1)*SrcWidth + x]; /* neighbour to the North */
p2 = Source[y*SrcWidth + (x-1)]; /* neighbour to the West */
p3 = Source[(y-1)*SrcWidth + (x-1)]; /* neighbour to the North-West */
d1 = ABS( colordiff(p, p1) );
d2 = ABS( colordiff(p, p2) );
d3 = ABS( colordiff(p, p3) );
d4 = ABS( colordiff(p1, p2) ); /* North to West */
/* find minimum */
min = d1;
if (min > d2) min = d2;
if (min > d3) min = d3;
if (min > d4) min = d4;
/* choose interpolator */
y2 = 2 * y;
x2 = 2 * x;
if (min == d1)
Target[y2*TgtWidth+x2] = average(p, p1); /* North */
else if (min == d2)
Target[y2*TgtWidth+x2] = average(p, p2); /* West */
else if (min == d3)
Target[y2*TgtWidth+x2] = average(p, p3); /* North-West */
else /* min == d4 */
Target[y2*TgtWidth+x2] = average(p, average(p1, p2));/* North to West */```
In an early implementation, the last interpolator just set the target pixel to the average of source pixels 1 and 3. In testing, I found that I obtained better results by averaging with source pixel 4 afterwards too. Due to the two consecutive averaging operations, the target pixel's colour is ultimately set to ½S4 + ¼S1 + ¼S3 (where Si stands for "source pixel i").
The images below compare the directionally averaging algorithm to bilinear and bicubic interpolation. Directional interpolation achieves sharper results than both bilinear and bicubic interpolation. Note that the original "line graphics" picture was intentionally created without anti-aliasing. It is to be expected that the output contains "jaggies" too.
Original picture 200% - Directional averaging 200% - Bilinear interpolation 200% - Bicubic interpolation
Original picture 200% - Directional averaging 200% - Bilinear interpolation 200% - Bicubic interpolation
### Quick, approximate colour distances
For the directionally averaging scaling algorithm to be feasible, you must have a way to determine the difference between the colours of two pixels. In the code snippet for magnifying by 2, this is hidden in the colordiff() function. This "pixel difference" calculation is also likely to become the performance bottleneck of the application, which is why I will give it a little more attention.
For grey scale pixels, calculating the pixel distance is as simple as taking the absolute value of the subtraction of a pair of pixels. The SIMD instructions ("single instructions, multiple data", previously known as MMX) of the current Intel Pentium processors allow you to obtain absolute values of four signed 16-bit values in parallel, using only a few instructions. For example code, see [4].
For 24-bit true colour images, you have several options to calculate the pixel distance:
• Use an accurate colour distance measure, such as the Euclidean vector length in 3-dimensional RGB space. This is probably too slow for our purposes, however, and it is also far more accurate than we need.
• Convert both pixels to grey scale, then calculate the distance between the grey scale values. Depending on how you calculate the grey scale equivalents of colour pixels, this may still be quite a slow procedure.
• Calculate the differences for all three channels separately and take the largest of these three values.
• Interleave the bits for the red, green and blue channel values and determine the difference between two interleaved values.
I propose the method listed last: bit interleaving. Bit interleaving has been widely used to create (database) indices and to find clusters in spatial data sets, specifically when the data set is represented in a quad- or octree. Applying bit interleaving to the RGB colour cube to find approximate colour differences is a straightforward extension, see [3].
Bit interleaving is actually a simple trick where an RGB colour that stored in three bytes with an encoding like:
R7R6R5R4R3R2R1R0 - G7G6G5G4G3G2G1G0 - B7B6B5B4B3B2B1B0
is transformed to:
G7R7B7G6R6B6G5R5 - B5G4R4B4G3R3B3G2 - R2B2G1R1B1G0R0B0
The new encoding groups bits with the same significance together and the subtraction between two of these interleaved values is thus a fair measure of the distance between the colours. To calculate the interleaved encoding quickly, one uses a precalculated lookup table that maps an input value (in the range 0 – 255) to the encoding:
A7A6A5A4A3A2A1A0
to:
0 0 A70 0 A60 0 - A50 0 A40 0 A30 - 0 A20 0 A10 0 A0
With this table, you need only three table lookup's (for the red, green and blue channels), two bit shifts and two bitwise "OR" instructions to do the full conversion. See the code snippet below for an implementation; the code assumes that you call the function CreateInterleaveTable() during the initialization of your application.
Creating an interleaved value for an RGB colour
```typedef struct __s_RGB {
unsigned char r, g, b;
} RGB;
static long InterleaveTable[256];
void CreateInterleaveTable(void)
{
int i,j;
long value;
for (i = 0; i < 256; i++) {
value = 0L;
for (j = 0; j < 8; j++)
if (i & (1 << j))
value |= 1L << (3*j);
InterleaveTable[i] = value;
} /* for */
}
/* 24-bit RGB (unpacked pixel) */
long InterleavedValue(RGB color)
{
return (InterleaveTable[color.g] << 2) |
(InterleaveTable[color.r] << 1) |
InterleaveTable[color.b];
}```
### Optimizations and other notes
As mentioned earlier, the pixel difference algorithm is likely to become the limiting factor in the routine's performance. One implementation-specific optimization trick was already mentioned: exploit parallelism by using SIMD instructions. There are also a couple of algorithmic optimizations that help. In the code snippet for magnifying by 2, we can count that we need to calculate four pixel distances per target pixel. For four target pixels per source pixel, that amounts to 16 pixel distances. The first opportunity for optimization is to recognize that there is overlap in the neighbouring pixels used for each of the target pixels, see the figure 6 at the right. When grouping the calculations of all pixel distances for one source pixel (four target pixels), one sees that there are 12 unique pixel distances that must be calculated —already a 25% savings from the original 16.
Additionally, there is an overlap in the pixel distances that are needed for the current source pixel and those for the next source pixel. Every source pixel uses five pixel distances that are shared with those for the previous pixel. This brings the number of pixel distances to be calculated down to 7 per source pixel. In the figure at the right of this paragraph, the blue and red arrows indicate the distances for the current pixel and the red and green arrows those for the next pixel. The five red arrows are thus the pixel distance calculations that are shared between the current and next pixels.
When considering implementation-specific optimization, specifically SIMD instructions, it is helpful to reduce the resolution of the bit-interleaved values. As implemented in the code snippet for bit interleaving, an interleaved value takes 24-bit. This is overly precise just to find an interpolation partner, especially because the difference between two interleaved values is only an approximate measure for the distance between the two colours. If we reduce this to 15-bit, by simply constructing the InterleaveTable differently, a subtraction between two such values fits in 16-bits. As already mentioned, the modern Intel processors can calculate the absolute value of four 16-bits values in parallel.
Another opportunity for optimization the code for magnifying by two is the series of if statements to find the minimum value. A sequence of CMP and CMOV instructions (Pentium-II processor and later) can calculate the mininum value without jumps (and thereby without penalties for mispredicted branches). SIMD instructions can help here too: Intel application note ([5]) describes the PMINSW instruction which determines the four minimum values of two groups of four 16-bit values. With three PMINSW instructions, plus some memory shuffling (hint: PSHUFW), we can find the four minimum values of four such groups.
And, like was the case for the minification algorithm, you might want to introduce extra local variables to avoid calculating the same values over and over again in the inner loop. Examples of such expressions in the magnify by two code are y*SrcWidth, y2*TgtWidth. Doing so immediately lets you simplify other inner-loop expressions such as (y2+1)*TgtWidth to Y2TgtWidth+TgtWidth, replacing a multiplication by an addition (the new variable Y2TgtWidth would be assigned the value of y2*TgtWidth at the top of the inner loop).
Currently the directionally averaging scaling algorithm only interpolates in one direction per target pixel, although that direction is linear. One additional rule, that might increase the quality of the result, is to use a bidirectional average if the maximum gradient is low. That is, in addition to the minimum gradient per target pixel, you also calculate the maximum value. If the maximum value is below some hard-coded limit, the target pixel gets the value of the average of four surrounding pixels.
[1] Riemersma, Thiadmer; "Quick and smooth image scaling with Bresenham"; Dr. Dobb's Journal; May 2002.
A companion article, which describes the "smooth Bresenham" algorithm.
[2] Riemersma, Thiadmer; "Quick colour averaging"; on-line article.
A companion article, which describes how to calculate the (unweighted) average colour of two pixels in various formats.
[3] Kientzle, Tim; "Approximate Inverse Color Mapping"; C/C++ User's Journal; August 1996.
With bit-interleaving and a binary search, one obtains a palette index whose associated colour is close to the input colour. It is not necessarily the closest match, though, and the author already acknowledges this in the article's title. An errata to the article appeared in C/C++ User's Journal October 1996, page 100 ("We have mail").
[4] Intel Application note AP-563; "Using MMX Instructions to Compute the L1 Norm Between Two 16-bit Vectors"; Intel Coorporation; 1996.
The L1 norm of a vector is the sum of the absolute values. The note calculates the L1 norm of the difference between two vectors; i.e. it subtracts a series of values from another series of values and then calculates the absolute values of all these subtractions.
[5] Intel Application note AP-804; "Integer Minimum or Maximum Element Search Using Streaming SIMD Extensions"; version 2.1; Intel Coorporation; January 1999.
An algorithm to calculate the minimum value in an array considering four 16-bit words at a time.
[6] Shezaf, N.; Abramov-Segal, H. Sutskover, I.; Bar-Sella, R.; "Adaptive Low Complexity Algorithm for Image Zooming at Fractional Scaling Ratio"; Proceedings of the 21st IEEE Convention of the Electrical and Electronic Engineers in Israel; April 2000; pp. 253-256.
A follow-up on the article "Spatially adaptive interpolation of digital images using fuzzy inference" by H.C. Ting and H.M. Hang in Proceedings of the SPIE, March 1996.
[7] Carrato, S.; Tenze, L.; "A High Quality 2× image interpolator"; Signal Processing Letters; vol. 7, June 2000; pp. 132-134.
An algorithm to split a source pixel in 4 target pixels by looking at eight surrounding pixels. The algorithm is "trained" on a set of specific input images where the desired output for that image is known. | 4,495 | 19,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-04 | latest | en | 0.915493 |
https://icsehelp.com/ml-aggarwal-solutions-gst-mcqs-for-icse-class-10/ | 1,716,635,360,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00889.warc.gz | 275,287,493 | 16,581 | # ML Aggarwal Solutions GST MCQs for ICSE Class 10 Ch-1
ML Aggarwal Solutions GST MCQs for ICSE Class 10 Ch-1. According to latest guideline of council there will be more questions based on MCQS in Sec-A of Board Exam.
Therefore Step by step solutions of Ch-1 Goods and Service Tax MCQs of ML Aggarwal Understanding ICSE Mathematics. given below. Visit official Website CISCE for detail information about ICSE Board Class-10.
### ML Aggarwal Solutions Goods and Service Tax MCQs
ICSE Class 10 Ch-1 GST
Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 10th Chapter-1 Goods and Service Tax (GST) writer ML Aggarwal Book Name Understanding Topics Solution of MCQs Questions Edition 2022-2023
### Multiple Choice Questions (MCQs)
ML Aggarwal Solutions GST ICSE Class 10 Ch-1
A retailer purchases a fan for Rs 1500 from a wholesaler and sells t to a consumer at 10%profit if the sales are intra-state and the rate of GST is 12%, then choose the correct answer from the given four options for questions 1 to 6:
#### Question 1
The selling price of the fan by the retailer (excluding tax) is?
(a) Rs.1500
(b) Rs.1650
(c) Rs. 1848
(d) Rs. 1800
Cost price of fan for retailer = ₹ 1500
Given profit% = 10%
∴ Selling price of fan by the retailer
= C.P. + 10% of C.P.
1500+1500X10⁄100=1650
#### Question 2
(2.) The selling price of the fan including tax (under GST) by retailer is?
(a) Rs.1650
(b) Rs.1800
(c) Rs.1848
(d) Rs.1830
Given, GST (rate) = 12%
1650X12⁄100=198
Thus, the required selling price of fan including tax by the retailer (under GST)
= S.P. + GST = ₹ (1650 + 198) = ₹ 1848
#### Question 3
The tax (under GST) paid by the wholesaler to the Central Government is ?
(a)Rs 90
(b)Rs 9
(c)Rs 99
(d) Rs 180
The tax (under GST) paid by wholesaler to Central Government
1500X6⁄100=90
[SGST – CGST = ½ × rate of GST – 6%]
#### Question 4
The tax (under GST) paid by the retailer to the State Government ?
(a) Rs 99
(b) Rs 9
(c)Rs 18
(d)Rs 198
Amount of input SGST of the retailer = 6% of ₹ 1500
Amount of tax (under GST) paid by retailer to State Government = Output SGST – Input SGST
= (99 – 9) = ₹9
Page-18
#### Question- 5
The tax (under GST) received by the Central Government is ?
(a)Rs 18
(b) Rs 198
(c) Rs 90
(d)Rs 99
Amount of CGST paid by the retailer = Output CGST – Input CGST
= ₹ (99-90) = ₹9
Thus, amount of tax (under GST) received by Central Government
= CGST paid by distributor + CGST paid by retailer
= (6% of ₹1500) + 9
= 90 + 9 = ₹99
#### Question -6
The cost of the fan to the consumer inclusive of tax is ?
(a)Rs 1650
(b) Rs 1800
(c)Rs 1830
(4) Rs 1848
Here, selling price of fan = ₹1650
GST on fan = 12% of ₹1650
=1650 x 12⁄100
=198
Thus, cost of a fan to the consumer inclusive of tax
= ₹ (1650+198) = ₹1848
## ML Aggarwal Solutions Goods and Service Tax MCQs
A shopkeeper bought a TV from a distributor at a discount of 25% of the listed price of Rs 32000. The shopkeeper sells the TV to a consumer at the listed price. if the sales are intra-state and the rate of GST la 18%, then choose the correct er from the given four options for questions 7 to 11:
#### Question -7
The selling price of the TV including tax (under GST?) by the distributor is?
(a) Rs 32000
(b) Rs 24000
(c) Rs 28320
(4) Rs 26160
It is case of intra state
Discount = 32000 x 25/100
= 8000
SP= 32000-8000
=24000
SP with GST by distributor
24000 + 24000 x 18/100
=28320
Hence option (c) is correct
#### Question- 8
The tax (under GST) paid by the distributor to the State Government is
(a) Rs 4320
(b) Rs 2160
(c) Rs 2880
(d) Rs 720
Tax (under GST) paid by distributor to the State Government
= SGST = 9% of ₹24000
=24000 x 9⁄100
=2160
#### Question -9
The tax (under GST) paid by the shopkeeper to the Central Government is?
(a)Rs 720
(b)Rs 1440
(c) Rs 2880
(4)Rs 2160
Amount of input CGST by the shopkeeper,
CGST = ₹2160, SGST = ₹2160
Amount of GST collected by the shopkeeper or paid by the consumer
= 18% of ₹32000
32000 x 18⁄100
=5760
SGST = 5760/2 = ₹2880 = CGST
Amount of CGST paid by shopkeeper to Central Government = Output CGST – Input CGST
= 2880-2160 = ₹720
#### Question 10
The tax (under GST) received by the State Government is?
(a)Rs 5760
(b) Rs 4320
(c) Rs 1440
(d) Rs 2880
Amount of SGST paid by Shopkeeper to state government = ₹ 720
∴ Total tax (under GST) received by State Government = ₹2160 +₹720 = ₹2880
#### Question -11
The price including tax (under GST) of the TV paid by the consumer is
(a)Rs 28320
(b) Rs 37760
(c) Rs 34880
(4) Rs 32000
Consumer buy on list price 32000
It is a case of intra-state transaction of goods and services.
SGST = CGST = ½ GST
Given:
The price inclusive of tax (under GST) which the consumer pays for the TV.
CP of an article for shopkeeper = ₹32000
SP of article = ₹32000 + 18% of ₹32000
= ₹32000 + (18/100) × 32000
or ₹32000 + 5760
= 37760
Hence option (b) is correct
— : End of ML Aggarwal Solutions GST MCQs Questions for ICSE Class 10 Ch-1 :– | 1,635 | 5,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-22 | latest | en | 0.846692 |
https://www.raymaps.com/index.php/antenna-radiation-pattern-and-antenna-tilt/ | 1,726,770,539,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00210.warc.gz | 893,168,430 | 16,686 | # Antenna Radiation Pattern and Antenna Tilt
An introductory text in Communication Theory would tell you that antennas radiate uniformly in all directions and the power received at a given distance ‘d’ is proportional to 1/(d)^2. Such an antenna is called an isotropic radiator. However, real world antennas are not isotropic radiators. They transmit energy in only those directions where it is needed. The Gain of a antenna is defined as the ratio of the power transmitted (or received) in a given direction to the power transmitted in that direction by an isotropic source and is expressed in dBi.
Although antenna Gain is a three dimensional quantity, the Gain is usually given along horizontal and vertical planes passing through the center of the antenna. The Horizontal and Vertical Gain patterns for a popular base station antenna Kathrein 742215 are shown in the figure below.
Kathrein 742215 Gain Pattern
The actual Gain is given with respect to the maximum Gain which is a function of the frequency e.g. in the 1710-1880 MHz band the maximum Gain has a value of 17.7dBi. Another important parameter is the Half Power Beam Width (HPBW) which has values of 68 degree and 7.1 degree in the horizontal and vertical planes respectively. HPBW is defined as the angle in degrees within which the power level is equal to or above the -3 dB level of the maximum.
Also shown in the above figure are approximate Horizontal Gain patterns for two antennas that have been rotated at 120 degrees and 240 degrees. Together these three antennas cover the region defined as a cell. There would obviously be lesser coverage in areas around the intersection of two beams.
A somewhat more interesting pattern is in the vertical direction where the HPBW is only 7.1 degrees. Thus it is very important to direct this beam in the right direction. A perfectly horizontal beam would result in a large cell radius but may also result in weak signal areas around the base station. A solution to this problem is to give a small tilt to the antenna in the downward direction, usually 5-10 degrees. This would reduce the cell radius but allow for a more uniform distribution of energy within the cell. In reality the signal from the main beam and side lobes (one significant side lobe around -15 dB) would bounce off the ground and buildings around the cell site and spread the signal around the cell.
Antenna Tilt of 10 Degrees
The above figure gives a 2D view of signal propagation from an elevated antenna with a downward tilt of 10 degrees in an urban environment.
#### Author: Yasir Ahmed (aka John)
More than 20 years of experience in various organizations in Pakistan, the USA, and Europe. Worked as a Research Assistant within the Mobile and Portable Radio Group (MPRG) of Virginia Tech and was one of the first researchers to propose Space Time Block Codes for eight transmit antennas. The collaboration with MPRG continued even after graduating with an MSEE degree and has resulted in 12 research publications and a book on Wireless Communications. Worked for Qualcomm USA as an Engineer with the key role of performance and conformance testing of UMTS modems. Qualcomm is the inventor of CDMA technology and owns patents critical to the 4G and 5G standards.
0.00 avg. rating (0% score) - 0 votes
## 4 thoughts on “Antenna Radiation Pattern and Antenna Tilt”
1. question says:
Can you convert dB to meters on vertical pattern? I will draw that on real object. Thanks
1. dB is the log of ratio of two quantities e.g. transmit power to receive power
Path loss in dB = 10*log10(Pt/Pr)
2. alen says:
Please I would like to know how to draw a 2D view of signal propagation from an elevated antenna with a downward tilt in matlab.
3. Y & Z says:
good job ! | 813 | 3,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-38 | latest | en | 0.936849 |
http://www.chegg.com/homework-help/questions-and-answers/painter-standing-22-m-long-platform-mass-10-kg-platform-held-two-ropes-connected-10-cm-end-q2612589 | 1,469,704,309,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.65/warc/CC-MAIN-20160723071028-00275-ip-10-185-27-174.ec2.internal.warc.gz | 354,135,455 | 13,241 | A painter is standing on a 2.2-m-long platform of mass 10 kg. The platform is held up by two ropes, each connected 10 cm from the end of the platform, so that the ropes are 2.0 m apart. The painter stands at a distance d = 0.60 m from the left-hand rope. The painter's mass is 60 kg. The tension in the left-hand rope is ? The tension in the right-hand rope is ?
http://www.flickr.com/photos/28246656@N08/7353430696/ | 119 | 417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-30 | latest | en | 0.931665 |
https://his.edu.vn/optical-illusion-eye-testcan-you-find-the-odd-cup-in-14-seconds | 1,718,895,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00061.warc.gz | 258,308,576 | 33,765 | # Optical Illusion Eye Test:Can you find the Odd Cup in 14 Seconds?
People with high intelligence can quickly find this hidden Odd Cup in this Optical Illusion. Try to test your visual power and your observation skills by solving this Optical Illusion.
Visit our website HIS Education to unravel the secrets behind mind-boggling and Expand your knowledge, engage your senses, and explore the fascinating world of visual perception at our website HIS Education.
## Optical Illusion
Optical Illusion is a type of Illusion that plays tricks on your vision. Optical illusions challenges how our eyes and brain work together to function and look at different perceptions. There are various types of Optical illusions, including physical, physiological, and cognitive illusions, which have Ambiguities, distortions, paradoxes, and fictions in each class.
An optical illusion is a mind-bending illustration of an entity, drawing, or image with distinct impressions if glanced at from different perspectives. It helps us test our brain and optical power by finding different illusions.
## Hidden Odd Cup Optical Illusion
An Optical Illusion is a puzzling and tricky form of a different figure that is to be solved. People are inquisitive about solving various and distinct Optical Illusions by scraping their minds and innovatively analyzing them. Most of the People try to practice Optical Illusion on a regular basis to develop their IQ power as they have several advantages by working on it.
Practicing Optical Illusion allows one to focus and think from a different perspective to solve quickly. Here is given a unique and exciting Odd Cup in this Optical Illusion. Try to find out the given Optical Illusion within a set of time to check out your intelligence and optical power.
See also Optical Illusion Brain Test: If you have Eagle Eyes Find the number 36 among 56 in 15 Seconds?
If you have Sharp Eyes Find Number 46 in 20 Secs
If you have Eagle Eyes Find the number 3 in 10 Secs
## Did you find a Odd Cup?
Here is an interesting and tricky image given to test your optical power and observation skills within you. Look at this image given closely and try to find out the hidden Odd Cup in this Optical Illusion easily.
Here is the challenge that you have to spot the hidden Odd Cup within 14 Seconds. People with high vision power can spot this hidden Odd Cup within a short span of time. If you are a genius you can find this optical illusion by observing it within a few seconds.
So, did you find this Odd Cup In This Optical Illusion? Well,if you have found it, truely, you are a genius. Others, don’t need to worry, if you are still struggling to spot out , below is a solution image for you.
Look at this image carefully to find the solution for this optical Illusion. If you have yet to spot the correct solution, don’t worry. We will help you find the solution below in the solution image.
## Find the Odd Cup Here
Optical Illusions are one of the interesting and fascinating things to keep focussing and to stimulate our brain to work. Suppose you have not found this hidden Odd Cup in this optical illusion.
If you are still struggling to find out the solution for this optical illusion. Here you can find the solution for a hidden Odd Cup in this optical illusion by looking at the highlighted site. Check out the answer below to spot the correct solution to this optical illusion.
See also You have 20/20 vision if you can spot all five sea creatures in this mind-boggling optical illusion in under 10 seconds
Disclaimer: The above information is for general informational purposes only. All information on the Site is provided in good faith, however we make no representation or warranty of any kind, express or implied, regarding the accuracy, adequacy, validity, reliability, availability or completeness of any information on the Site.
Categories: Optical Illusion
Source: HIS Education
Rate this post | 808 | 3,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.927055 |
https://scsynth.org/t/print-ranges-from-an-array/4160 | 1,643,367,901,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00569.warc.gz | 499,852,340 | 4,013 | # Print ranges from an array
Say I have a relatively large array of values
``````~o = (1..1000)
``````
i want to read sections/ranges of it, that’s ok:
``````z = ~o.copyRange(0, 4);
``````
however i’m struggling to find a way to recursively increment the `range` values, so that next i can read another group of 5 values, as in:
``````z = ~o.copyRange(0, 4);
z = ~o.copyRange(5, 9);
z = ~o.copyRange(10, 14);
``````
but with a more efficient, automated way
i guess i’ll have to put whatever it ends up being inside a `Pfunc` so that i can run it as a Pattern, but i’m stuck figuring out what the trick is
``````~o = (1..1000);
~p = ~o.clump(5);
-> [ [ 1, 2, 3, 4, 5 ], [ 6, 7, 8, 9, 10 ], ... ]
~p.do(_.postln);
``````
Not necessarily.
``````Pseries(0, 1, inf).clump(5).asStream.nextN(3)
-> [ [ 0, 1, 2, 3, 4 ], [ 5, 6, 7, 8, 9 ], [ 10, 11, 12, 13, 14 ] ]
``````
hjh
2 Likes
Amazing, James. Many thanks.
I totally missed `.clump`!!! | 364 | 948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.854142 |
https://mail.scipy.org/pipermail/numpy-discussion/2009-October/045989.html | 1,506,107,147,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689102.37/warc/CC-MAIN-20170922183303-20170922203303-00322.warc.gz | 691,118,584 | 1,718 | [Numpy-discussion] finding nonzero elements in list
josef.pktd@gmai... josef.pktd@gmai...
Mon Oct 12 09:36:09 CDT 2009
```On Mon, Oct 12, 2009 at 10:18 AM, per freem <perfreem@gmail.com> wrote:
> hi all,
>
> i'm trying to find nonzero elements in an array, as follows:
>
> a = array([[1, 0],
> [1, 1],
> [1, 1],
> [0, 1]])
>
> i want to find all elements that are [1,1]. i tried: nonzero(a ==
> [1,0]) but i cannot interpret the output. the output i get is:
> (array([0, 0, 1, 2]), array([0, 1, 0, 0]))
>
> i simply want to find the indices of the elements that equal [1,0].
> how can i do this? thanks.
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
a == [1,0] does elementwise comparison, you need to aggregate
condition for all elements of row
>>> a = np.array([[1, 0],
[1, 1],
[1, 1],
[0, 1]])
>>> np.nonzero((a==[1,0]).all(1))
(array([0]),)
>>> np.where((a==[1,0]).all(1))
(array([0]),)
>>> np.nonzero((a==[1,1]).all(1))
(array([1, 2]),)
Josef
``` | 391 | 1,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-39 | latest | en | 0.525498 |
https://byjus.com/electric-current-formula/ | 1,709,158,322,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00582.warc.gz | 151,386,418 | 109,109 | # Electric Current Formula
## Electric Current
The continuous flow of electrons in an electric circuit is called an electric current. The electrons move when the potential difference is applied across the wire or terminal. Electric current is nothing but the rate of change of electric charge through a circuit. This current is related to the voltage and resistance of a circuit. It can be represented by I and the SI unit is Amperes. Electric current relates the electric charge and the time.
According to Ohm’s law, the electric current formula will be,
$$\begin{array}{l}I=\frac{V}{R}\end{array}$$
Where,
• V is the voltage
• R is the resistance
• I is the current
### Solved Examples
Let us discuss the example problems related to electric current.
Example 1: Calculate the current through the circuit in which the voltage and resistance be 15V and 3Ω respectively?
Solution:Â The given parameters are,
V = 15V
The equation for current using Ohm’s law is,
$$\begin{array}{l}I=\frac{V}{R}\end{array}$$
I =
$$\begin{array}{l}\frac{15}{3}\end{array}$$
= 5A
Example 2: The voltage and resistance of a circuit are given as 10V and 4Ω respectively. Calculate the current through the circuit?
Solution:Â The given parameters are,
V = 10V
$$\begin{array}{l}I=\frac{V}{R}\end{array}$$
$$\begin{array}{l}\frac{10}{4}\end{array}$$ | 359 | 1,345 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.845657 |
https://math.mit.edu/spams/2001sp.html | 1,708,958,041,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474660.32/warc/CC-MAIN-20240226130305-20240226160305-00203.warc.gz | 367,084,189 | 6,718 | Feb 8 Ryan O'Donnell SWM Seeking M/F's To Explore Domination (in graphs) Me: 5'8", Br/Bl, into Domination. A dominating set in a graph $G$ is a subset of the vertices $S$ such that every vertex in $G$ is at distance 1 or less from $S$. The domination number of $G$ is the size of the minimum dominating set. Together we'll prove many cute and easy theorems bounding the domination number in various kinds of graphs. Simple combinatorics and the probabilistic method will be our only tools. I will also discuss the relevance to certain problems in theoretical computer science. You: Likes to listen. M/F okay, HWP unnecessary, NS/ND/ND. The bondage number'' of graphs also exists, but will not be discussed. Feb 15 Alex Perlin No loops, please Abstract: Given a connected graph G, a spanning tree is a connected subgraph that has no loops and includes all vertices of G. We will show how erasing loops in a random walk on G leads to a random spanning tree. As an application we will give a probabilistic proof of Cayley's theorem on the number of labeled unrooted trees on n verices. Feb 22 Robert David The Banach-Tarski Theorem I will show a proof of the Banach-Tarski theorem. This theorem infamously states that a ball in 3-space can be cut into a finite number of pieces which can then be rearranged to form 2 balls, each the same size as the original one. Some of the pieces must be nonmeasurable. The proof uses group theory and requires the Axiom of Choice. Mar 1 Mike Rosenblum A Random Polynomial Time Algorithm for Approximating the Volume of Convex Bodies Ever get laughed at for not being able to approximate the volume of convex bodies in polynomial time? Did the bully steal your lunch money, and say you could have it back as soon as you could guess the volume of the convex body he was thinking of? There is no shame in this, for no deterministic algorithm can do it well at all. I'll present a randomized algorithm of Dyer, Frieze, and Kannan that with very high probability gives a good approximation. I'll show how random walks, the ellipsoid algorithm, and ideas from differential geometry are combined to prove the algorithm works as claimed. If there is time, I will show how the randomized algorithm wriggles out of the Barany-Furedi trap that all deterministic ones fall into. Mar 8 Bill Bradley Butterflies, Shuffle Graphs, and Other Computer Networks with Cutesy-Pie Names Back in the glory days of parallel computing, people enjoyed designing supercomputers based on hypercubes. Since these are a bit hard to build when restricted to our puny three-dimensional universe, they compromised. They used a host of variants on the aforementioned Butterfly and Shuffle networks (all with an out-degree of 2 and with cutesy-pie names). I'll talk about the basic structure of these graphs (including the hypercube itself), with an eye towards proving some permutation routing and rearrangeability results. Words such as "permutation routing" and "rearrangeability" will be defined. I'll also mention some of the more vexing open questions in this field. Mar 15 Etienne Rassart Self-avoiding walks and Polyominoes You've probably heard about random walks in terms of a drunken guy trying to find his way home and making a random choice of directions at each street intersection. Now imagine a slightly less drunk walker that remembers if he's been at an intersection before and avoids it if he has. What you get then is a self-avoiding walk (SAW). An interesting problem is: given a lattice and a fixed starting point, how many length-n SAWs are there in that lattice? This problem is interesting because after half a century of work by statistical physicists and mathematicians alike, the problem still stands open, even when considered asymptotically. I will discuss this problem and the related problem of counting polyominoes. Polyominoes are collections of little squares that are popular in recreational mathematics (think Tetris), but have turned out in deeper problems recently as well. Mar 22 Sergi Elizalde Enumeration of subwords in permutations Abstract: We say that a permutation P in S_n contains another permutation Q in S_m as a subword if there exist m consecutive elements in P whose order relationships are the same as those in Q. For example, 6725341 contains 4132 as a subword (because 7253 are in the same order as 4132), but 41352 does not (even though it contains it as a subsequence, that is, in non-consecutive positions). You probably have never wondered how many permutations in S_n do not contain a given subword. Actually, most people haven't. The general problem is unsolved. I will show you how to find the bivariate exponential generating functions counting occurrences of subwords of length 3, and also of some particular subwords of arbitrary length. The main idea is to use tree representations of permutations, so that occurrences of some subwords correspond to certain configurations in trees that can be counted more easily. Your homework will be to choose one of the following problems and solve it: How many permutations in S_n do not contain 1324 as a subword? (Hint: I don't know.) How many days remain for the Spring Break? (Hint: It can be counted without using tree representations.) Apr 5 Peter Clifford Picture Hanging 101 Abstract: The miniscule subfield of applied mathematics dealing with picture hanging has always been disregarded and denigrated. Practitioners have had to work behind closed doors, lest they be caught red handed striving for unappreciated perfection. Recently, however, picture hanging has become more socially acceptable, and has even breached the stronghold of prime time TV with a brief appearance on a popular animated programme. With the current popularity of mathematics in the cinema and in theatre, it can hardly be long before a biographical epic drama is created, based on a hanger and his struggle for professional recognition. We will give a survey of the field, hopefully including its ties to other areas such as knot and braid theory, geometry and combinatorics. Apr 12 BaoChi Nguyen Buckling Abstract: Often things that we see, use and do in our lives involve buckling. For example, we often crumple a piece of paper before we throw it away. As a kid we learned how to make a cone from a circular sheet of paper. When we look at the drape we wonder why does it fold in a certain way. There are many technological applications which have thin-solid film deposited over substrates. As for metal-cutting tools, they always have wear-resistant coatings that have large compressive stresses, which leads to buckling-driven delamination. It would be nice to understand the mechanism behind these. Crumpled paper and delamination of compressed thin films have been studied by many researchers experimentally, numerically, and theoretically . In general, when buckling occurs there are folding patterns. In order for these patterns to appear a certain amount of energy is required. To get a unique pattern one needs to minimize the energy of the system. This lead to finding developable surfaces which have zero Gaussian curvature. Locally, there are three types of developable surface: cones, cylinders and ridges. I am going to give an overview of these problems from numerical, experimental, and theoretical point of views. Apr 19 Rados Radoicic Crossing Bound, k-set Problem and Monotone Paths in Line Arrangements Abstract: Given a set of n points in the plane in general position, a k-set is a subset of k points that can be separated with a line from the remaining points. The k-set problem, one of the most challenging open problems in combinatorial and computational geometry, is to bound the maximum number of k-sets of an n-point set. I am going to prove several upper bounds for the n/2-problem. Proofs will elegantly follow from the bound on the crossing number of a geometric graph. I am also going to define the k-level problem, which is the dual of the k-set problem. Finally, I will introduce a related problem of bounding the maximum "length" of an x-monotone path in a line arrangement. I will present the best known lower bound for this problem, due to Geza Toth and myself. I will mention several "innocent-looking" open problems. Apr 26 Francois Blanchette Why you can make yogurt on a computer but not on paper Surface growth is a relatively new and absolutely exciting subject. This theory tries to describe how the interface between two "states" evolves. These states can be liquid-liquid, liquid-solid or solid-solid. Looking at this from the microscopic point of view, many of those objects look like fractals. I will discuss how to classify these interfaces, simulate them and analyze the simplest models mathematically. I will assume no background in this field, some interest for it, and a strong desire to get a free dinner. May 3 Pedro Felzenszwalb Computer Vision What does it mean, to see? The plain man's answer (and Aristotle's too) would be, to know what is where by looking. In other words, vision is the process of discovering from images what is present in the world, and where it is. In computer vision we study how to extract from images the various aspects of the world that are useful to us. We study the computations that make it possible to recognize familiar objects, recover 3D information of a scene, detect moving objects, etc. To solve these problems we need to bring together many fields, including signal processing, geometry, combinatorial optimization and probabilistic reasoning. I will give an overview of different problems, the mathematics that pops up in different places, and algorithms which attempt to solve some of the problems. May 10 Carly Klivans Oriented Matroids Abstract: I am often asked, "What are oriented matroids anyway?". I will attempt to answer this question. May 17 Alberto De Sole The Universe Abstract: I will describe the universe. Of course nobody knows how it works. This means the talk will be a little confusing. More specifically, I'll talk about quarks and elementary particles, and, as a mathematical tool, I will say a few words about representation theory of some Lie groups, mainly SU(2) and SU(3).
Accessibility | 2,192 | 10,201 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.927637 |
https://www.quizover.com/course/section/semi-predictive-approach-by-openstax | 1,542,581,016,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744750.80/warc/CC-MAIN-20181118221818-20181119003818-00100.warc.gz | 982,423,858 | 21,526 | # 0.4 Universal compression for context tree sources
Page 1 / 3
## Semi-predictive approach
Recall that a context tree source is similar to a Markov source, where the number of states is greatly reduced. Let $T$ be the set of leaves of a context tree source, then the redundancy is
$r\lesssim \frac{|T|\left(r-1\right)}{2}\left(log,\left(\frac{n}{|T|}\right),+,O,\left(1\right)\right),$
where $|T|$ is the number of leaves, and we have $log\left(\frac{n}{|T|}\right)$ instead of $log\left(n\right)$ , because each state generated $\frac{n}{|T|}$ symbols, on average. In contrast, the redundancy for a Markov representation of the tree source $T$ is much larger. Therefore, tree sources are greatly preferable in practice, they offer a significant reductionin redundancy.
How can we compress universally over the parametric class of tree sources? Suppose that we knew $T$ , that is we knew the set of leaves. Then we could process $x$ sequentially, where for each ${x}_{i}$ we can determine what state its context is in, that is the unique suffix of ${x}_{1}^{i-1}$ that belongs to the set of leaf labels in $T$ . Having determined that we are in some state $s$ , $Pr\left({x}_{i}=0|s,{x}_{1}^{i-1}\right)$ can be computed by examining all previous times that we were in state $s$ and computing the probability with the Krichevsky-Trofimov approach based on the number of times that the following symbol(after $s$ ) was 0 or 1. In fact, we can store symbol counts ${n}_{x}\left(s,0\right)$ and ${n}_{x}\left(s,1\right)$ for all $s\in T$ , update them sequentially as we process $x$ , and compute $Pr\left({x}_{i}=0|s,{x}_{1}^{i-1}\right)$ efficiently. (The actual translation to bits is performed with an arithmetic encoder.)
While promising, this approach above requires to know $T$ . How do we compute the optimal ${T}^{*}$ from the data?
Semi-predictive coding : The semi-predictive approach to encoding for context tree sources [link] is to scan the data twice, where in the first scan we estimate ${T}^{*}$ and in the second scan we encode $x$ from ${T}^{*}$ , as described above. Let us describe a procedure for computing the optimal ${T}^{*}$ among tree sources whose depth is bounded by $D$ . This procedure is visualized in [link] . As suggested above, we count ${n}_{x}\left(s,a\right)$ , the number of times that each possible symbol appeared in context $s$ , for all $s\in {\alpha }^{D},a\in \alpha$ . Having computed all the symbol counts, we process the depth- $D$ tree in a bottom-top fashion, from the leaves to the root, where for each internal node $s$ of the tree (that is, $s\in {\alpha }^{d}$ where $d ), we track ${T}_{s}^{*}$ , the optimal tree structure rooted at $s$ to encode symbols whose context ends with $s$ , and $\text{MDL}\left(s\right)$ the minimum description lengths (MDL) required for encoding these symbols.
Without loss of generality, consider the simple case of a binary alphabet $\alpha =\left\{0,1\right\}$ . When processing $s$ we have already computed the symbol counts ${n}_{x}\left(0s,0\right)$ and ${n}_{x}\left(0s,1\right)$ , ${n}_{x}\left(1s,0\right)$ , ${n}_{x}\left(1s,1\right)$ , the optimal trees ${T}_{0s}^{*}$ and ${T}_{1s}^{*}$ , and the minimum description lengths (MDL) $\text{MDL}\left(0s\right)$ and $\text{MDL}\left(1S\right)$ . We have two options for state $s$ .
1. Keep ${T}_{0S}^{*}$ and ${T}_{1S}^{*}$ . The coding length required to do so is $\text{MDL}\left(0S\right)+\text{MDL}\left(1S\right)+1$ , where the extra bit is spent to describe the structure of the maximizing tree.
2. Merge both states (this is also called tree pruning ). The symbol counts will be ${n}_{x}\left(s,\alpha \right)={n}_{x}\left(0s,\alpha \right)+{n}_{x}\left(1s,\alpha \right),\alpha \in \left\{0,1\right\}$ , and the coding length will be
$\text{KT}\left({n}_{x}\left(s,0\right),{n}_{x}\left(s,1\right)\right)+1,$
where $\text{KT}\left(·,·\right)$ is the Krichevsky-Trofimov length [link] , and we again included an extra bit for the structure of the tree.
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
Got questions? Join the online conversation and get instant answers! | 2,108 | 7,879 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 56, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-47 | latest | en | 0.882821 |
http://mathoverflow.net/questions/tagged/abstract-algebra+noncommutative-algebra | 1,398,403,322,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00210-ip-10-147-4-33.ec2.internal.warc.gz | 218,679,614 | 9,812 | # Tagged Questions
60 views
### Problem with Smoothness and quasi-freeness
Let A be a unital associative algebra over a field k. Then A is smooth if and only if X:=Spec(A) is smooth. That is $\Omega_{X|Spec(k)}$ is locally-free. The later module is isomorphic to ...
318 views
### $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?
Let $G$ be a countable discrete group (not necessarily abelian), and suppose the group ring $\mathbb{Z}G$ is a left-Noetherian ring, for example, when $G$ is a polycyclic-by finite group. Denote the ...
105 views
### Making an algebra the uniqe maximal one-sided ideal in another unital algebra
If $R$ is an algebra without a unit, then the standard unitisation $R^\sharp$ can have maximal one-sided ideals other than $R$. Thus, it is natural to ask about the following. Let $R$ be an algebra ...
539 views
### Unit ideal in non-commutative rings [closed]
In a non-commutative ring (with identity), is it possible for an element which does not possess left or right inverses to generate the entire ring? i.e. $(r)=R$, where (r) is the two-sided ideal ...
223 views
### Inverses in convolution algebras
Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant. Let $C_c(G)$ be the space of locally constant complex functions on $G$ ... | 369 | 1,389 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2014-15 | longest | en | 0.858435 |
https://ic-sharpy.readthedocs.io/en/master/content/example_notebooks/linear_goland_flutter.html | 1,653,167,173,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00141.warc.gz | 371,350,695 | 17,973 | # Flutter Analysis of a Goland Wing using the SHARPy Linear Solver¶
This is an example using SHARPy to find the flutter speed of a Goland wing by:
• Calculating aerodynamic forces and deflections using a nonlinear solver
• Linearising about this reference condition
• Creating a reduced order model of the linearised aerodynamics
• Evaluate the stability of the linearised aeroelastic system at different velocities
## References¶
Maraniello, S., & Palacios, R. (2019). State-Space Realizations and Internal Balancing in Potential-Flow Aerodynamics with Arbitrary Kinematics. AIAA Journal, 57(6), 1–14. https://doi.org/10.2514/1.J058153
### Required Packages¶
[1]:
import numpy as np
import matplotlib.pyplot as plt
import os
import sys
import cases.templates.flying_wings as wings # See this package for the Goland wing structural and aerodynamic definition
import sharpy.sharpy_main # used to run SHARPy from Jupyter
### Problem Set-up¶
#### Velocity¶
The UVLM is assembled in normalised time at a velocity of $$1 m/s$$. The only matrices that need updating then with free stream velocity are the structural matrices, which is significantly cheaper to do than to update the UVLM.
[2]:
u_inf = 1.
alpha_deg = 0.
rho = 1.02
num_modes = 4
#### Discretisation¶
Note: To achieve convergence of the flutter results with the ones found in the literature, a significant discretisation may be required. If you are running this notebook for the first time, set M = 4 initially to verify that your system can perform!
[3]:
M = 16
N = 32
M_star_fact = 10
#### ROM¶
A moment-matching (Krylov subspace) model order reduction technique is employed. This ROM method offers the ability to interpolate the transfer functions at a desired point in the complex plane. See the ROM documentation pages for more info.
Note: this ROM method matches the transfer function but does not guarantee stability. Therefore the resulting system may be unstable. These unstable modes may appear far in the right hand plane but will not affect the flutter speed calculations.
[4]:
c_ref = 1.8288 # Goland wing reference chord. Used for frequency normalisation
rom_settings = dict()
rom_settings['algorithm'] = 'mimo_rational_arnoldi' # reduction algorithm
rom_settings['r'] = 6 # Krylov subspace order
frequency_continuous_k = np.array([0.]) # Interpolation point in the complex plane with reduced frequency units
frequency_continuous_w = 2 * u_inf * frequency_continuous_k / c_ref
rom_settings['frequency'] = frequency_continuous_w
#### Case Admin¶
[5]:
case_name = 'goland_cs'
case_nlin_info = 'M%dN%dMs%d_nmodes%d' % (M, N, M_star_fact, num_modes)
case_rom_info = 'rom_MIMORA_r%d_sig%04d_%04dj' % (rom_settings['r'], frequency_continuous_k[-1].real * 100,
frequency_continuous_k[-1].imag * 100)
case_name += case_nlin_info + case_rom_info
route_test_dir = os.path.abspath('')
print('The case to run will be: %s' % case_name)
print('Case files will be saved in ./cases/%s' %case_name)
print('Output files will be saved in ./output/%s/' %case_name)
The case to run will be: goland_csM16N32Ms10_nmodes4rom_MIMORA_r6_sig0000_0000j
Case files will be saved in ./cases/goland_csM16N32Ms10_nmodes4rom_MIMORA_r6_sig0000_0000j
Output files will be saved in ./output/goland_csM16N32Ms10_nmodes4rom_MIMORA_r6_sig0000_0000j/
### Simulation Set-Up¶
#### Goland Wing¶
ws is an instance of a Goland wing with a control surface. Reference the template file cases.templates.flying_wings.GolandControlSurface for more info on the geometrical, structural and aerodynamic definition of the Goland wing here used.
[6]:
ws = wings.GolandControlSurface(M=M,
N=N,
Mstar_fact=M_star_fact,
u_inf=u_inf,
alpha=alpha_deg,
cs_deflection=[0, 0],
rho=rho,
sweep=0,
physical_time=2,
n_surfaces=2,
route=route_test_dir + '/cases',
case_name=case_name)
ws.clean_test_files()
ws.update_derived_params()
ws.set_default_config_dict()
ws.generate_aero_file()
ws.generate_fem_file()
#### Simulation Settings¶
The settings for each of the solvers are now set. For a detailed description on them please reference their respective documentation pages
## SHARPy Settings¶
The most important setting is the flow list. It tells SHARPy which solvers to run and in which order.
[7]:
ws.config['SHARPy'] = {
'flow':
['BeamLoader', 'AerogridLoader',
'StaticCoupled',
'AerogridPlot',
'BeamPlot',
'Modal',
'LinearAssembler',
'AsymptoticStability',
],
'case': ws.case_name, 'route': ws.route,
'write_screen': 'on', 'write_log': 'on',
'log_folder': route_test_dir + '/output/',
'log_file': ws.case_name + '.log'}
## Beam Loader Settings¶
[8]:
ws.config['BeamLoader'] = {
'unsteady': 'off',
'orientation': ws.quat}
## Aerogrid Loader Settings¶
[9]:
ws.config['AerogridLoader'] = {
'unsteady': 'off',
'aligned_grid': 'on',
'mstar': ws.Mstar_fact * ws.M,
'freestream_dir': ws.u_inf_direction,
'wake_shape_generator': 'StraightWake',
'wake_shape_generator_input': {'u_inf': ws.u_inf,
'u_inf_direction': ws.u_inf_direction,
'dt': ws.dt}}
## Static Coupled Solver¶
[10]:
ws.config['StaticCoupled'] = {
'print_info': 'on',
'max_iter': 200,
'n_load_steps': 1,
'tolerance': 1e-10,
'relaxation_factor': 0.,
'aero_solver': 'StaticUvlm',
'aero_solver_settings': {
'rho': ws.rho,
'print_info': 'off',
'horseshoe': 'off',
'num_cores': 4,
'n_rollup': 0,
'rollup_dt': ws.dt,
'rollup_aic_refresh': 1,
'rollup_tolerance': 1e-4,
'velocity_field_generator': 'SteadyVelocityField',
'velocity_field_input': {
'u_inf': ws.u_inf,
'u_inf_direction': ws.u_inf_direction}},
'structural_solver': 'NonLinearStatic',
'structural_solver_settings': {'print_info': 'off',
'max_iterations': 150,
'num_load_steps': 4,
'delta_curved': 1e-1,
'min_delta': 1e-10,
'gravity_on': 'on',
'gravity': 9.81}}
## AerogridPlot Settings¶
[11]:
ws.config['AerogridPlot'] = {'include_rbm': 'off',
'include_applied_forces': 'on',
'minus_m_star': 0}
## BeamPlot Settings¶
[12]:
ws.config['BeamPlot'] = {'include_rbm': 'off',
'include_applied_forces': 'on'}
## Linear System Assembly Settings¶
[14]:
ws.config['LinearAssembler'] = {'linear_system': 'LinearAeroelastic',
'linear_system_settings': {
'beam_settings': {'modal_projection': 'on',
'inout_coords': 'modes',
'discrete_time': 'on',
'newmark_damp': 0.5e-4,
'discr_method': 'newmark',
'dt': ws.dt,
'proj_modes': 'undamped',
'use_euler': 'off',
'num_modes': num_modes,
'print_info': 'on',
'gravity': 'on',
'remove_sym_modes': 'on',
'remove_dofs': []},
'aero_settings': {'dt': ws.dt,
'ScalingDict': {'length': 0.5 * ws.c_ref,
'speed': u_inf,
'density': rho},
'integr_order': 2,
'density': ws.rho,
'remove_predictor': 'on',
'use_sparse': 'on',
'remove_inputs': ['u_gust'],
'rom_method': ['Krylov'],
'rom_method_settings': {'Krylov': rom_settings}},
}}
## Asymptotic Stability Analysis Settings¶
[15]:
ws.config['AsymptoticStability'] = {'print_info': True,
'velocity_analysis': [100, 180, 81],
'modes_to_plot': []}
[16]:
ws.config.write()
### Run SHARPy¶
[17]:
sharpy.sharpy_main.main(['', ws.route + ws.case_name + '.sharpy'])
--------------------------------------------------------------------------------
###### ## ## ### ######## ######## ## ##
## ## ## ## ## ## ## ## ## ## ## ##
## ## ## ## ## ## ## ## ## ####
###### ######### ## ## ######## ######## ##
## ## ## ######### ## ## ## ##
## ## ## ## ## ## ## ## ## ##
###### ## ## ## ## ## ## ## ##
--------------------------------------------------------------------------------
Aeroelastics Lab, Aeronautics Department.
Copyright (c), Imperial College London.
All rights reserved.
License available at https://github.com/imperialcollegelondon/sharpy
Running SHARPy from /home/ng213/2TB/pazy_code/pazy-sharpy/lib/sharpy/docs/source/content/example_notebooks
SHARPy being run is in /home/ng213/2TB/pazy_code/pazy-sharpy/lib/sharpy
The branch being run is dev_setting_error
The version and commit hash are: v1.2.1-339-g156c731-156c731
SHARPy output folder set
/home/ng213/2TB/pazy_code/pazy-sharpy/lib/sharpy/docs/source/content/example_notebooks/output//goland_csM16N32Ms10_nmodes4rom_MIMORA_r6_sig0000_0000j/
Generating an instance of BeamLoader
Variable for_pos has no assigned value in the settings file.
will default to the value: [0.0, 0, 0]
Generating an instance of AerogridLoader
Variable control_surface_deflection has no assigned value in the settings file.
will default to the value: []
Variable control_surface_deflection_generator_settings has no assigned value in the settings file.
will default to the value: {}
Variable dx1 has no assigned value in the settings file.
will default to the value: -1.0
Variable ndx1 has no assigned value in the settings file.
will default to the value: 1
Variable r has no assigned value in the settings file.
will default to the value: 1.0
Variable dxmax has no assigned value in the settings file.
will default to the value: -1.0
The aerodynamic grid contains 2 surfaces
Surface 0, M=16, N=16
Wake 0, M=160, N=16
Surface 1, M=16, N=16
Wake 1, M=160, N=16
In total: 512 bound panels
In total: 5120 wake panels
Total number of panels = 5632
Generating an instance of StaticCoupled
Variable correct_forces_method has no assigned value in the settings file.
will default to the value:
Variable runtime_generators has no assigned value in the settings file.
will default to the value: {}
Generating an instance of NonLinearStatic
Variable newmark_damp has no assigned value in the settings file.
will default to the value: 0.0001
Variable gravity_dir has no assigned value in the settings file.
will default to the value: [0.0, 0.0, 1.0]
Variable relaxation_factor has no assigned value in the settings file.
will default to the value: 0.3
Variable dt has no assigned value in the settings file.
will default to the value: 0.01
Variable num_steps has no assigned value in the settings file.
will default to the value: 500
Variable initial_position has no assigned value in the settings file.
will default to the value: [0. 0. 0.]
Generating an instance of StaticUvlm
Variable iterative_solver has no assigned value in the settings file.
will default to the value: False
Variable iterative_tol has no assigned value in the settings file.
will default to the value: 0.0001
Variable iterative_precond has no assigned value in the settings file.
will default to the value: False
Variable cfl1 has no assigned value in the settings file.
will default to the value: True
Variable vortex_radius has no assigned value in the settings file.
will default to the value: 1e-06
Variable vortex_radius_wake_ind has no assigned value in the settings file.
will default to the value: 1e-06
Variable rbm_vel_g has no assigned value in the settings file.
will default to the value: [0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
Variable centre_rot_g has no assigned value in the settings file.
will default to the value: [0.0, 0.0, 0.0]
|=====|=====|============|==========|==========|==========|==========|==========|==========|
|iter |step | log10(res) | Fx | Fy | Fz | Mx | My | Mz |
|=====|=====|============|==========|==========|==========|==========|==========|==========|
| 0 | 0 | 0.00000 | 0.0000 | -0.0000 |-4271.0417| 0.0000 | 781.0842 | 0.0000 |
| 1 | 0 | -11.89144 | 0.0000 | -0.0000 |-4271.0039| 0.0000 | 781.0906 | 0.0000 |
Generating an instance of AerogridPlot
Variable include_forward_motion has no assigned value in the settings file.
will default to the value: False
Variable include_unsteady_applied_forces has no assigned value in the settings file.
will default to the value: False
Variable name_prefix has no assigned value in the settings file.
will default to the value:
Variable u_inf has no assigned value in the settings file.
will default to the value: 0.0
Variable dt has no assigned value in the settings file.
will default to the value: 0.0
Variable include_velocities has no assigned value in the settings file.
will default to the value: False
Variable include_incidence_angle has no assigned value in the settings file.
will default to the value: False
Variable num_cores has no assigned value in the settings file.
will default to the value: 1
Variable vortex_radius has no assigned value in the settings file.
will default to the value: 1e-06
...Finished
Generating an instance of BeamPlot
Variable include_FoR has no assigned value in the settings file.
will default to the value: False
Variable include_applied_moments has no assigned value in the settings file.
will default to the value: True
Variable name_prefix has no assigned value in the settings file.
will default to the value:
Variable output_rbm has no assigned value in the settings file.
will default to the value: True
...Finished
Generating an instance of Modal
Variable print_info has no assigned value in the settings file.
will default to the value: True
Variable delta_curved has no assigned value in the settings file.
will default to the value: 0.01
Variable use_custom_timestep has no assigned value in the settings file.
will default to the value: -1
Structural eigenvalues
|==============|==============|==============|==============|==============|==============|==============|
| mode | eval_real | eval_imag | freq_n (Hz) | freq_d (Hz) | damping | period (s) |
|==============|==============|==============|==============|==============|==============|==============|
| 0 | 0.000000 | 48.067396 | 7.650164 | 7.650164 | -0.000000 | 0.130716 |
| 1 | 0.000000 | 48.067398 | 7.650164 | 7.650164 | -0.000000 | 0.130716 |
| 2 | 0.000000 | 95.685736 | 15.228858 | 15.228858 | -0.000000 | 0.065665 |
| 3 | 0.000000 | 95.685754 | 15.228861 | 15.228861 | -0.000000 | 0.065665 |
| 4 | 0.000000 | 243.144471 | 38.697644 | 38.697644 | -0.000000 | 0.025841 |
| 5 | 0.000000 | 243.144477 | 38.697645 | 38.697645 | -0.000000 | 0.025841 |
| 6 | 0.000000 | 343.801136 | 54.717650 | 54.717650 | -0.000000 | 0.018276 |
| 7 | 0.000000 | 343.801137 | 54.717650 | 54.717650 | -0.000000 | 0.018276 |
| 8 | 0.000000 | 443.324608 | 70.557303 | 70.557303 | -0.000000 | 0.014173 |
| 9 | 0.000000 | 443.324619 | 70.557304 | 70.557304 | -0.000000 | 0.014173 |
| 10 | 0.000000 | 461.992869 | 73.528449 | 73.528449 | -0.000000 | 0.013600 |
| 11 | 0.000000 | 461.992869 | 73.528449 | 73.528449 | -0.000000 | 0.013600 |
| 12 | 0.000000 | 601.126871 | 95.672313 | 95.672313 | -0.000000 | 0.010452 |
| 13 | 0.000000 | 601.126873 | 95.672313 | 95.672313 | -0.000000 | 0.010452 |
| 14 | 0.000000 | 782.997645 | 124.617946 | 124.617946 | -0.000000 | 0.008025 |
| 15 | 0.000000 | 782.997649 | 124.617946 | 124.617946 | -0.000000 | 0.008025 |
| 16 | 0.000000 | 917.191257 | 145.975522 | 145.975522 | -0.000000 | 0.006850 |
| 17 | 0.000000 | 917.191259 | 145.975523 | 145.975523 | -0.000000 | 0.006850 |
| 18 | 0.000000 | 975.005694 | 155.176976 | 155.176976 | -0.000000 | 0.006444 |
| 19 | 0.000000 | 975.005699 | 155.176977 | 155.176977 | -0.000000 | 0.006444 |
Generating an instance of LinearAssembler
Variable linearisation_tstep has no assigned value in the settings file.
will default to the value: -1
Variable modal_tstep has no assigned value in the settings file.
will default to the value: -1
Variable inout_coordinates has no assigned value in the settings file.
will default to the value:
Variable retain_inputs has no assigned value in the settings file.
will default to the value: []
Variable retain_outputs has no assigned value in the settings file.
will default to the value: []
Generating an instance of LinearAeroelastic
Variable uvlm_filename has no assigned value in the settings file.
will default to the value:
Variable track_body has no assigned value in the settings file.
will default to the value: True
Variable use_euler has no assigned value in the settings file.
will default to the value: False
Generating an instance of LinearUVLM
Variable gust_assembler has no assigned value in the settings file.
will default to the value:
Variable vortex_radius has no assigned value in the settings file.
will default to the value: 1e-06
Variable cfl1 has no assigned value in the settings file.
will default to the value: True
Variable velocity_field_generator has no assigned value in the settings file.
will default to the value: SteadyVelocityField
Variable velocity_field_input has no assigned value in the settings file.
will default to the value: {}
Variable physical_model has no assigned value in the settings file.
will default to the value: True
Variable track_body has no assigned value in the settings file.
will default to the value: False
Variable track_body_number has no assigned value in the settings file.
will default to the value: -1
Initialising Static linear UVLM solver class...
...done in 1.35 sec
Generating an instance of Krylov
Variable print_info has no assigned value in the settings file.
will default to the value: True
Variable single_side has no assigned value in the settings file.
will default to the value:
Variable tangent_input_file has no assigned value in the settings file.
will default to the value:
Variable restart_arnoldi has no assigned value in the settings file.
will default to the value: False
Initialising Krylov Model Order Reduction
State-space realisation of UVLM equations started...
Computing wake propagation matrix with CFL1=True
/home/ng213/anaconda3/envs/sharpy_env/lib/python3.7/site-packages/scipy/sparse/_index.py:126: SparseEfficiencyWarning: Changing the sparsity structure of a csc_matrix is expensive. lil_matrix is more efficient.
self._set_arrayXarray(i, j, x)
state-space model produced in form:
h_{n+1} = A h_{n} + B u_{n}
with:
x_n = h_n + Bp u_n
...done in 17.25 sec
Scaling UVLM system with reference time 0.914400s
Non-dimensional time step set (0.125000)
System scaled in 31.698308s
Generating an instance of LinearBeam
Warning, projecting system with damping onto undamped modes
Linearising gravity terms...
M = 7.26 kg
X_CG A -> 0.00 0.00 -0.00
Node 1 -> B -0.000 -0.116 0.000
-> A 0.116 0.381 -0.000
-> G 0.116 0.381 -0.000
Node mass:
Matrix: 14.5125
Node 2 -> B -0.000 -0.116 0.000
-> A 0.116 0.762 -0.000
-> G 0.116 0.762 -0.000
Node mass:
Matrix: 7.2563
Node 3 -> B -0.000 -0.116 0.000
-> A 0.116 1.143 -0.000
-> G 0.116 1.143 -0.000
Node mass:
Matrix: 14.5125
Node 4 -> B -0.000 -0.116 0.000
-> A 0.116 1.524 -0.001
-> G 0.116 1.524 -0.001
Node mass:
Matrix: 7.2563
Node 5 -> B -0.000 -0.116 0.000
-> A 0.116 1.905 -0.001
-> G 0.116 1.905 -0.001
Node mass:
Matrix: 14.5125
Node 6 -> B -0.000 -0.116 0.000
-> A 0.116 2.286 -0.001
-> G 0.116 2.286 -0.001
Node mass:
Matrix: 7.2563
Node 7 -> B -0.000 -0.116 0.000
-> A 0.116 2.667 -0.002
-> G 0.116 2.667 -0.002
Node mass:
Matrix: 14.5125
Node 8 -> B -0.000 -0.116 0.000
-> A 0.116 3.048 -0.002
-> G 0.116 3.048 -0.002
Node mass:
Matrix: 7.2563
Node 9 -> B -0.000 -0.116 0.000
-> A 0.116 3.429 -0.003
-> G 0.116 3.429 -0.003
Node mass:
Matrix: 14.5125
Node 10 -> B -0.000 -0.116 0.000
-> A 0.116 3.810 -0.003
-> G 0.116 3.810 -0.003
Node mass:
Matrix: 7.2563
Node 11 -> B -0.000 -0.116 0.000
-> A 0.116 4.191 -0.004
-> G 0.116 4.191 -0.004
Node mass:
Matrix: 14.5125
Node 12 -> B -0.000 -0.116 0.000
-> A 0.116 4.572 -0.004
-> G 0.116 4.572 -0.004
Node mass:
Matrix: 7.2563
Node 13 -> B -0.000 -0.116 0.000
-> A 0.116 4.953 -0.005
-> G 0.116 4.953 -0.005
Node mass:
Matrix: 14.5125
Node 14 -> B -0.000 -0.116 0.000
-> A 0.116 5.334 -0.005
-> G 0.116 5.334 -0.005
Node mass:
Matrix: 7.2563
Node 15 -> B -0.000 -0.116 0.000
-> A 0.116 5.715 -0.006
-> G 0.116 5.715 -0.006
Node mass:
Matrix: 14.5125
Node 16 -> B -0.000 -0.116 0.000
-> A 0.116 6.096 -0.006
-> G 0.116 6.096 -0.006
Node mass:
Matrix: 3.6281
Node 17 -> B -0.000 -0.116 -0.000
-> A 0.116 -6.096 -0.006
-> G 0.116 -6.096 -0.006
Node mass:
Matrix: 3.6281
Node 18 -> B -0.000 -0.116 -0.000
-> A 0.116 -5.715 -0.006
-> G 0.116 -5.715 -0.006
Node mass:
Matrix: 14.5125
Node 19 -> B -0.000 -0.116 -0.000
-> A 0.116 -5.334 -0.005
-> G 0.116 -5.334 -0.005
Node mass:
Matrix: 7.2563
Node 20 -> B -0.000 -0.116 -0.000
-> A 0.116 -4.953 -0.005
-> G 0.116 -4.953 -0.005
Node mass:
Matrix: 14.5125
Node 21 -> B -0.000 -0.116 -0.000
-> A 0.116 -4.572 -0.004
-> G 0.116 -4.572 -0.004
Node mass:
Matrix: 7.2563
Node 22 -> B -0.000 -0.116 -0.000
-> A 0.116 -4.191 -0.004
-> G 0.116 -4.191 -0.004
Node mass:
Matrix: 14.5125
Node 23 -> B -0.000 -0.116 -0.000
-> A 0.116 -3.810 -0.003
-> G 0.116 -3.810 -0.003
Node mass:
Matrix: 7.2563
Node 24 -> B -0.000 -0.116 -0.000
-> A 0.116 -3.429 -0.003
-> G 0.116 -3.429 -0.003
Node mass:
Matrix: 14.5125
Node 25 -> B -0.000 -0.116 -0.000
-> A 0.116 -3.048 -0.002
-> G 0.116 -3.048 -0.002
Node mass:
Matrix: 7.2563
Node 26 -> B -0.000 -0.116 -0.000
-> A 0.116 -2.667 -0.002
-> G 0.116 -2.667 -0.002
Node mass:
Matrix: 14.5125
Node 27 -> B -0.000 -0.116 -0.000
-> A 0.116 -2.286 -0.002
-> G 0.116 -2.286 -0.002
Node mass:
Matrix: 7.2563
Node 28 -> B -0.000 -0.116 -0.000
-> A 0.116 -1.905 -0.001
-> G 0.116 -1.905 -0.001
Node mass:
Matrix: 14.5125
Node 29 -> B -0.000 -0.116 -0.000
-> A 0.116 -1.524 -0.001
-> G 0.116 -1.524 -0.001
Node mass:
Matrix: 7.2563
Node 30 -> B -0.000 -0.116 -0.000
-> A 0.116 -1.143 -0.000
-> G 0.116 -1.143 -0.000
Node mass:
Matrix: 14.5125
Node 31 -> B -0.000 -0.116 -0.000
-> A 0.116 -0.762 -0.000
-> G 0.116 -0.762 -0.000
Node mass:
Matrix: 7.2563
Node 32 -> B -0.000 -0.116 -0.000
-> A 0.116 -0.381 -0.000
-> G 0.116 -0.381 -0.000
Node mass:
Matrix: 14.5125
Updated the beam C, modal C and K matrices with the terms from the
gravity linearisation
Scaling beam according to reduced time...
Setting the beam time step to (0.1250)
Updating C and K matrices and natural frequencies with new normalised time...
Model Order Reduction in progress...
Moment Matching Krylov Model Reduction
Construction Algorithm:
mimo_rational_arnoldi
Interpolation points:
sigma = 0.000000 + 0.000000j [rad/s]
Krylov order:
r = 6
Constructing controllability space
Constructing observability space
Deflating column 23
Deflating column 25
Deflating column 28
Deflating column 33
ROM is stable
DT Eigenvalues:
mu = 0.992076 + -0.000000j
mu = 0.971409 + -0.000000j
mu = 0.957534 + -0.038927j
mu = 0.957534 + 0.038927j
mu = 0.954926 + -0.072820j
mu = 0.954926 + 0.072820j
mu = 0.954282 + -0.000000j
mu = 0.935272 + 0.000000j
mu = 0.927865 + 0.092012j
mu = 0.927865 + -0.092012j
mu = 0.927075 + -0.066636j
mu = 0.927075 + 0.066636j
mu = 0.925588 + 0.038813j
mu = 0.925588 + -0.038813j
mu = 0.922836 + -0.004482j
mu = 0.922836 + 0.004482j
mu = 0.915491 + -0.025196j
mu = 0.915491 + 0.025196j
mu = 0.908662 + 0.053066j
mu = 0.908662 + -0.053066j
mu = 0.881127 + -0.056743j
mu = 0.881127 + 0.056743j
mu = 0.882642 + -0.000000j
mu = 0.867726 + -0.000000j
mu = 0.717587 + 0.263893j
mu = 0.717587 + -0.263893j
mu = 0.697184 + 0.000000j
mu = 0.360807 + 0.000000j
mu = -0.000018 + -0.002663j
mu = -0.000018 + 0.002663j
mu = 0.000155 + -0.000000j
mu = -0.000154 + 0.000000j
System reduced from order 6656 to
n = 32 states
...Completed Model Order Reduction in 5.16 s
Aeroelastic system assembled:
Aerodynamic states: 32
Structural states: 4
Total states: 36
Inputs: 8
Outputs: 6
Final system is:
State-space system
States: 36
Inputs: 8
Outputs: 6
Generating an instance of AsymptoticStability
Variable reference_velocity has no assigned value in the settings file.
will default to the value: 1.0
Variable frequency_cutoff has no assigned value in the settings file.
will default to the value: 0.0
Variable export_eigenvalues has no assigned value in the settings file.
will default to the value: False
Variable display_root_locus has no assigned value in the settings file.
will default to the value: False
Variable iterative_eigvals has no assigned value in the settings file.
will default to the value: False
Variable num_evals has no assigned value in the settings file.
will default to the value: 200
Variable postprocessors has no assigned value in the settings file.
will default to the value: []
Variable postprocessors_settings has no assigned value in the settings file.
will default to the value: {}
Dynamical System Eigenvalues
Calculating eigenvalues using direct method
|==============|==============|==============|==============|==============|==============|==============|
| mode | eval_real | eval_imag | freq_n (Hz) | freq_d (Hz) | damping | period (s) |
|==============|==============|==============|==============|==============|==============|==============|
| 0 | -0.021637 | 24.315428 | 3.869922 | 3.869921 | 0.000890 | 0.258403 |
| 1 | -0.021637 | -24.315428 | 3.869922 | 3.869921 | 0.000890 | 0.258403 |
| 2 | -0.069601 | -0.000000 | 0.011077 | 0.000000 | 1.000000 | inf |
| 3 | -0.105223 | -21.320535 | 3.393310 | 3.393269 | 0.004935 | 0.294701 |
| 4 | -0.105223 | 21.320535 | 3.393310 | 3.393269 | 0.004935 | 0.294701 |
| 5 | -0.253787 | -0.000000 | 0.040391 | 0.000000 | 1.000000 | inf |
| 6 | -0.372425 | 0.355473 | 0.081940 | 0.056575 | 0.723379 | 17.675568 |
| 7 | -0.372425 | -0.355473 | 0.081940 | 0.056575 | 0.723379 | 17.675568 |
| 8 | -0.378143 | 0.665882 | 0.121875 | 0.105978 | 0.493812 | 9.435879 |
| 9 | -0.378143 | -0.665882 | 0.121875 | 0.105978 | 0.493812 | 9.435879 |
| 10 | -0.409413 | -0.000000 | 0.065160 | 0.000000 | 1.000000 | inf |
| 11 | -0.585461 | 0.000000 | 0.093179 | 0.000000 | 1.000000 | inf |
| 12 | -0.612214 | -0.864763 | 0.168631 | 0.137631 | 0.577812 | 7.265789 |
| 13 | -0.612214 | 0.864763 | 0.168631 | 0.137631 | 0.577812 | 7.265789 |
| 14 | -0.639935 | 0.627769 | 0.142673 | 0.099913 | 0.713860 | 10.008752 |
| 15 | -0.639935 | -0.627769 | 0.142673 | 0.099913 | 0.713860 | 10.008752 |
| 16 | -0.668833 | 0.366654 | 0.121394 | 0.058355 | 0.876881 | 17.136529 |
| 17 | -0.668833 | -0.366654 | 0.121394 | 0.058355 | 0.876881 | 17.136529 |
| 18 | -0.702465 | 0.042487 | 0.112005 | 0.006762 | 0.998176 | 147.883816 |
| 19 | -0.702465 | -0.042487 | 0.112005 | 0.006762 | 0.998176 | 147.883816 |
| 20 | -0.769174 | 0.240726 | 0.128273 | 0.038313 | 0.954353 | 26.100974 |
| 21 | -0.769174 | -0.240726 | 0.128273 | 0.038313 | 0.954353 | 26.100974 |
| 22 | -0.823096 | 0.510355 | 0.154138 | 0.081226 | 0.849886 | 12.311395 |
| 23 | -0.823096 | -0.510355 | 0.154138 | 0.081226 | 0.849886 | 12.311395 |
| 24 | -1.089098 | 0.562639 | 0.195099 | 0.089547 | 0.888447 | 11.167353 |
| 25 | -1.089098 | -0.562639 | 0.195099 | 0.089547 | 0.888447 | 11.167353 |
| 26 | -1.092180 | -0.000000 | 0.173826 | 0.000000 | 1.000000 | inf |
| 27 | -1.241288 | 0.000000 | 0.197557 | 0.000000 | 1.000000 | inf |
| 28 | -2.348569 | 3.083089 | 0.616840 | 0.490689 | 0.605970 | 2.037951 |
| 29 | -2.348569 | -3.083089 | 0.616840 | 0.490689 | 0.605970 | 2.037951 |
| 30 | -3.155780 | -0.000000 | 0.502258 | 0.000000 | 1.000000 | inf |
| 31 | -8.918115 | -0.000000 | 1.419362 | 0.000000 | 1.000000 | inf |
| 32 | -26.700406 | -27.485500 | 6.098697 | 4.374453 | 0.696788 | 0.228600 |
| 33 | -28.824123 | 0.000000 | 4.587502 | 0.000000 | 1.000000 | inf |
| 34 | -35.766803 | -27.485500 | 7.179135 | 4.374453 | 0.792918 | 0.228600 |
| 35 | -36.676675 | -0.000000 | 5.837274 | 0.000000 | 1.000000 | inf |
Velocity Asymptotic Stability Analysis
Initial velocity: 100.00 m/s
Final velocity: 180.00 m/s
Number of evaluations: 81
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 100.00 m/2 max. CT eig. real: -3.925687
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 101.00 m/2 max. CT eig. real: -3.951244
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 102.00 m/2 max. CT eig. real: -3.975951
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 103.00 m/2 max. CT eig. real: -3.999782
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 104.00 m/2 max. CT eig. real: -4.022709
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 105.00 m/2 max. CT eig. real: -4.044705
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 106.00 m/2 max. CT eig. real: -4.065744
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 107.00 m/2 max. CT eig. real: -4.085798
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 108.00 m/2 max. CT eig. real: -4.104841
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 109.00 m/2 max. CT eig. real: -4.122845
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 110.00 m/2 max. CT eig. real: -4.139781
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 111.00 m/2 max. CT eig. real: -4.155617
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 112.00 m/2 max. CT eig. real: -4.170322
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 113.00 m/2 max. CT eig. real: -4.183860
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 114.00 m/2 max. CT eig. real: -4.196192
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 115.00 m/2 max. CT eig. real: -4.207276
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 116.00 m/2 max. CT eig. real: -4.217065
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 117.00 m/2 max. CT eig. real: -4.225507
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 118.00 m/2 max. CT eig. real: -4.232544
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 119.00 m/2 max. CT eig. real: -4.238111
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 120.00 m/2 max. CT eig. real: -4.242138
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 121.00 m/2 max. CT eig. real: -4.244545
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 122.00 m/2 max. CT eig. real: -4.245246
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 123.00 m/2 max. CT eig. real: -4.244143
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 124.00 m/2 max. CT eig. real: -4.241130
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 125.00 m/2 max. CT eig. real: -4.236092
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 126.00 m/2 max. CT eig. real: -4.228899
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 127.00 m/2 max. CT eig. real: -4.219413
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 128.00 m/2 max. CT eig. real: -4.207482
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 129.00 m/2 max. CT eig. real: -4.192940
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 130.00 m/2 max. CT eig. real: -4.175607
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 131.00 m/2 max. CT eig. real: -4.155291
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 132.00 m/2 max. CT eig. real: -4.131780
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 133.00 m/2 max. CT eig. real: -4.104848
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 134.00 m/2 max. CT eig. real: -4.074252
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 135.00 m/2 max. CT eig. real: -4.039730
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 136.00 m/2 max. CT eig. real: -4.001000
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 137.00 m/2 max. CT eig. real: -3.957763
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 138.00 m/2 max. CT eig. real: -3.909697
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 139.00 m/2 max. CT eig. real: -3.856462
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 140.00 m/2 max. CT eig. real: -3.797697
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 141.00 m/2 max. CT eig. real: -3.733020
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 142.00 m/2 max. CT eig. real: -3.662031
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 143.00 m/2 max. CT eig. real: -3.584314
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 144.00 m/2 max. CT eig. real: -3.499436
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 145.00 m/2 max. CT eig. real: -3.406959
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 146.00 m/2 max. CT eig. real: -3.306437
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 147.00 m/2 max. CT eig. real: -3.197433
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 148.00 m/2 max. CT eig. real: -3.079522
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 149.00 m/2 max. CT eig. real: -2.952307
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 150.00 m/2 max. CT eig. real: -2.815436
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 151.00 m/2 max. CT eig. real: -2.668615
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 152.00 m/2 max. CT eig. real: -2.511634
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 153.00 m/2 max. CT eig. real: -2.344382
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 154.00 m/2 max. CT eig. real: -2.166866
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 155.00 m/2 max. CT eig. real: -1.979231
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 156.00 m/2 max. CT eig. real: -1.781770
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 157.00 m/2 max. CT eig. real: -1.574929
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 158.00 m/2 max. CT eig. real: -1.359301
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 159.00 m/2 max. CT eig. real: -1.135613
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 160.00 m/2 max. CT eig. real: -0.904707
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 161.00 m/2 max. CT eig. real: -0.667505
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 162.00 m/2 max. CT eig. real: -0.424982
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 163.00 m/2 max. CT eig. real: -0.178127
N unstab.: 000
Unstable aeroelastic natural frequency CT(rad/s):
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 164.00 m/2 max. CT eig. real: 0.072086
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 70.64 70.64
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 165.00 m/2 max. CT eig. real: 0.324729
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 70.41 70.41
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 166.00 m/2 max. CT eig. real: 0.578936
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 70.20 70.20
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 167.00 m/2 max. CT eig. real: 0.833925
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 69.99 69.99
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 168.00 m/2 max. CT eig. real: 1.088998
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 69.79 69.79
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 169.00 m/2 max. CT eig. real: 1.343552
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 69.61 69.61
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 170.00 m/2 max. CT eig. real: 1.597068
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 69.43 69.43
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 171.00 m/2 max. CT eig. real: 1.849117
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 69.26 69.26
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 172.00 m/2 max. CT eig. real: 2.099343
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 69.10 69.10
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 173.00 m/2 max. CT eig. real: 2.347461
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 68.94 68.94
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 174.00 m/2 max. CT eig. real: 2.593246
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 68.79 68.79
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 175.00 m/2 max. CT eig. real: 2.836529
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 68.65 68.65
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 176.00 m/2 max. CT eig. real: 3.077180
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 68.51 68.51
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 177.00 m/2 max. CT eig. real: 3.315113
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 68.38 68.38
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 178.00 m/2 max. CT eig. real: 3.550268
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 68.25 68.25
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 179.00 m/2 max. CT eig. real: 3.782616
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 68.13 68.13
Updating C and K matrices and natural frequencies with new normalised time...
LTI u: 180.00 m/2 max. CT eig. real: 4.012145
N unstab.: 002
Unstable aeroelastic natural frequency CT(rad/s): 68.01 68.01
Saving velocity analysis results...
Successful
FINISHED - Elapsed time = 59.1569176 seconds
FINISHED - CPU process time = 94.3774846 seconds
[17]:
<sharpy.presharpy.presharpy.PreSharpy at 0x7f85b5231850>
### Analysis¶
#### Nonlinear equilibrium¶
The nonlinear equilibrium condition can be visualised and analysed by opening, with Paraview, the files in the /output/<case_name>/aero and /output/<case_name>/beam folders to see the deflection and aerodynamic forces acting
#### Stability¶
The stability of the Goland wing is now analysed under changing free stream velocity. The aeroelastic system is projected onto 2 structural modes (1st bending and 1st torsion). The two modes are seen quite separated at 100 m/s. As speed is increased, the damping of the torsion mode decreases until it crosses the imaginary axis onto the right hand plane and flutter begins. This flutter mode is a bending-torsion mode, as seen from the natural frequency plot where the frequencies of each coalesce into this mode.
[18]:
file_name = './output/%s/stability/velocity_analysis_min1000_max1800_nvel0081.dat' % case_name
velocity_analysis = np.loadtxt(file_name)
u_inf = velocity_analysis[:, 0]
eigs_r = velocity_analysis[:, 1]
eigs_i = velocity_analysis[:, 2]
[19]:
fig = plt.figure()
plt.scatter(eigs_r, eigs_i, c=u_inf, cmap='Blues')
cbar = plt.colorbar()
cbar.set_label('Free Stream Velocity, $u_\infty$ [m/s]')
plt.grid()
plt.xlim(-10, 10)
plt.ylim(-150, 150)
plt.xlabel('Real Part, $\lambda$ [rad/s]')
plt.ylabel('Imag Part, $\lambda$ [rad/s]');
[20]:
fig = plt.figure()
natural_frequency = np.sqrt(eigs_r ** 2 + eigs_i ** 2)
damping_ratio = eigs_r / natural_frequency
cond = (eigs_r>-25) * (eigs_r<10) * (natural_frequency<100) # filter unwanted eigenvalues for this plot (mostly aero modes)
plt.scatter(u_inf[cond], damping_ratio[cond], color='k', marker='s', s=9)
plt.grid()
plt.ylim(-0.25, 0.25)
plt.xlabel('Free Stream Velocity, $u_\infty$ [m/s]')
plt.ylabel('Damping Ratio, $\zeta$ [-]');
[21]:
fig = plt.figure()
cond = (eigs_r>-25) * (eigs_r<10) # filter unwanted eigenvalues for this plot (mostly aero modes)
plt.scatter(u_inf[cond], natural_frequency[cond], color='k', marker='s', s=9)
plt.grid()
plt.ylim(40, 100)
plt.xlabel('Free Stream Velocity, $u_\infty$ [m/s]')
plt.ylabel('Natural Frequency, $\omega_n$ [rad/s]'); | 15,824 | 47,785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-21 | longest | en | 0.817068 |
https://blog.0xd.be/en/the-bingham-formula/ | 1,725,966,274,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00202.warc.gz | 120,863,418 | 10,475 | # The Bingham formula
During my time at the Politecnico di Torino I attended a Communication Systems course which focused on systems such as ADSL/VDSL, DVB-T, LTE and modulation techniques such as PSK, QAM, DMT/OFDM. That being said, the one formula we always used to compute the cardinality of the constellation used was the famous Bingham formula.
When I was looking at the different parameters of this formula I figured I would do a search on Google to see if there was more information to be found. However, a search for “Bingham formula” resulted in not a single useful link. If this is such a famous and used formula, why is there nothing on the searchable web about it? Not even an entry on Wikipedia? This is an honest question, if someone knows please tell me.
What does this formula look like you ask? Here it is:
with:
m: the number of bits
1/10: used when solving for a BER of P(e) = 10^-7, 1/14 if you’re using a BER of 10^-10 (e.g. for DVB-T) | 240 | 960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.942852 |
https://learningpundits.com/jobPreparation/module-view/53-stock-and-shares/2-aptitude-test---stock-&-shares/?page=10 | 1,600,734,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202686.56/warc/CC-MAIN-20200922000730-20200922030730-00691.warc.gz | 489,299,224 | 16,329 | # Stock and Shares
Stocks and Shares: Meaning of Stocks and Shares. Aptitude questions on Stocks and Shares. Exercise Problems with answers.
Take Aptitude Test
View Aptitude Test Results
## Online Aptitude Questions with Answers on Stock and Shares
Q46. **Three persons invested Rs.9000 in a joint business. The second person invested Rs.1000 more than the first and the third Rs.1000 more than second. After two years, they gained Rs.5400. How much third person will get?**
1. Rs.2400
2. Rs.3600
3. Rs.2850
4. Rs.2000
Solution : Rs.2400
Q47. **A, B and C enter into partnership. A invests some money at the beginning, B invests double the amount after 6 months, and C invests thrice the amount after 8 months. If the annual gain be Rs.18000. A's share is?**
1. Rs.7500
2. Rs.7200
3. Rs.6000
4. Rs.5750
Solution : Rs.6000
Q48. **A and B rent a pasture for 10 months. A put in 80 cows for 7 months. How many can B put in for the remaining 3 months, if he pays half as much again as A?**
1. 120
2. 180
3. 200
4. 280
Solution : 280
Q49. **A and B put in Rs.300 and Rs.400 respectively into a business. A reinvests into the business his share of the first year's profit of Rs.210 where as B does not. In what ratio should they divide the second year's profit?**
1. 39:40
2. 40:39
3. 3:4
4. 4:3
Solution : 39:40
##### Are you a college student? You can become a Campus Ambassador for LearningPundits. Promote our Online Contests to students from you college via email, Facebook, posters, WhatsApp and old-fashioned face to face communication
• Stipend based on your performance
• Internship Certificate to boost your Resume
Q50. **A and B invests Rs.3000 and Rs.4000 respectively in a business. If A doubles his capital after 6 months. In what ratio should A and B divide that year's profit?**
1. 9:10
2. 9:8
3. 3:4
4. 39:49
Solution : 9:8
Q{{(\$index+1)+((page-1)*LIMITPERPAGE)}}.
1.
Solution :
### Grammar Guru
#### Free Online Contest on English Grammar and Vocabulary.20 mins Only.
• All Participants get Participation Certificates to boost your Resume
### Math Whiz
#### Free Online Contest on Aptitude and Reasoning.20 mins Only.
• All Participants get Participation Certificates to boost your Resume
#### Participation Now using Laptop/ Desktop/ Tab/ Mobile.
##### Are you a college student? You can become a Campus Ambassador for LearningPundits. Promote our Online Contests to students from you college via email, Facebook, posters, WhatsApp and old-fashioned face to face communication
• Stipend based on your performance
• Internship Certificate to boost your Resume
Preparing for Aptitude Tests ? Please go through our courses on Aptitude Questions and try to answer our Online Aptitude Test Questions on Quantitative Aptitude.
Interested in evaluating your Reasoning Skills ? Please go through our courses on Logical Reasoning and Non-Verbal Reasoning to answer our Reasoning Questions.
Interested in learning English and checking your English Grammar ? Do a quick grammar check to evaluate your Basic English Grammar Skills. Improve your English Vocabulary by going through these Vocabulary words.
Wondering how to make a resume ? These resume format for freshers might be helpful. You can also submit your Resume for Review and look and Resume samples available there.
Preparing for an HR Interview or Group Discussion ? These HR interview questions and answers could help you do well in a typical HR interview. These group discussion tips could also be useful.
Searching for jobs ? We have thousand of Fresher Jobs. Feel free to browse through our extensive list of online jobs. | 934 | 3,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-40 | latest | en | 0.871951 |
http://mathhelpforum.com/algebra/136487-range-function.html | 1,524,133,866,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936833.6/warc/CC-MAIN-20180419091546-20180419111546-00223.warc.gz | 154,315,242 | 10,026 | # Thread: range of the function
1. ## range of the function
Find the range of the function , $\displaystyle f(x)=-e^{-2x}+2e^{-x}+1$ , x is real and x>0
Attempt :
$\displaystyle f(x)=-\frac{1}{e^{2x}}+\frac{2}{e^x}+1$
so when x=0 , f(x)=2
and when x approaches infinity , f(x) converges to 1 .
the range would be [1,2)
is this acceptable ?
2. Originally Posted by thereddevils
Find the range of the function , $\displaystyle f(x)=-e^{-2x}+2e^{-x}+1$ , x is real and x>0
Attempt :
$\displaystyle f(x)=-\frac{1}{e^{2x}}+\frac{2}{e^x}+1$
so when x=0 , f(x)=2
and when x approaches infinity , f(x) converges to 1 .
the range would be [1,2)
is this acceptable ?
$\displaystyle f(x) = -e^{-2x} + 2e^{-x} + 1$
$\displaystyle = -(e^{-x})^2 + 2e^{-x} + 1$.
This is a quadratic in $\displaystyle e^{-x}$.
So let $\displaystyle X = e^{-x}$ and the equation becomes
$\displaystyle -X^2 + 2X + 1$
$\displaystyle = -(X^2 - 2X - 1)$
$\displaystyle = -[X^2 - 2X + (-1)^2 - (-1)^2 - 1]$
$\displaystyle = -[(X - 1)^2 - 2]$
$\displaystyle = -(X - 1)^2 + 2$.
It should be clear that this is a quadratic with a maximum at $\displaystyle (1, 2)$. So $\displaystyle f(X) < 2$ for all $\displaystyle X$.
And since $\displaystyle X$ is defined for all $\displaystyle x$, that means that $\displaystyle f(x) < 2$ for all $\displaystyle x$.
3. Hello, thereddevils!
Good job! . . . I'd make one change.
Find the range of the function: .$\displaystyle f(x)\:=\:-e^{-2x}+2e^{-x}+1,\;x\text{ is real and }x>0$
Attempt :
$\displaystyle f(x)\:=\:-\frac{1}{e^{2x}}+\frac{2}{e^x}+1$
So when $\displaystyle x=0.\:f(x)=2$
. . and when $\displaystyle x \to\infty,\;f(x)$ converges to 1 .
The range would be [1,2)
I agree with your upper limit, 2.
Since $\displaystyle x > 0$, the function is never quite equal to 2.
. . So, 2 is not included in the range.
As $\displaystyle x\to\infty$, the function approaches 1,
. . but never equals 1. **
So, 1 is not included in the range.
The range is: .$\displaystyle {\color{red}(}1,\:2)$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
I found that $\displaystyle f(x)$ equals $\displaystyle 1$ when $\displaystyle x\,=\,-\ln2$
. . but that is not in the domain. | 797 | 2,205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-17 | latest | en | 0.63769 |
http://illuminations.nctm.org/Lesson.aspx?id=1011 | 1,490,834,343,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191444.45/warc/CC-MAIN-20170322212951-00489-ip-10-233-31-227.ec2.internal.warc.gz | 164,515,632 | 23,652 | ## Finding Our Top Speed
• Lesson
6-8
1
This lesson sets the stage for a discussion of travel in the solar system. By considering a real-world, hands-on activity, students develop their understanding of time and distance. Finally, students plot the data they have collected.
### Introduction
In December 1995, the probe that had been released by the Galileo spacecraft in July 1995 entered Jupiter' atmosphere. Galileo had traveled 2.3 billion miles since its launch in October 1989. It spent the first 3 years in the inner solar system. During one flyby in Venus and two flybys of Earth, it gathered enough velocity from the gravity of the planets to reach Jupiter. Throughout its long journey, Galileo had been sending data about solar system back to Earth.
The Galileo Space Probe orbiting Jupiter
Studying the passage of time and time versus distance, students are better able to think about the time required to travel the long distances in space. Of course, the long time required for this type of travel is exactly why it is difficult for humans to make journeys to the planets. "Are we there yet?" This query really has meaning for space travel. Remember, Galileo traveled for more than 6 years one way to get to Jupiter.
### Getting Started
Start the lesson by asking the students how far they can go in 8 seconds. Typically, students answer in a variety of ways. Some may respond with distances from a few feet to the length of a football field. Others may ask, "Are we traveling by foot, on a bicycle, or in a car?" Confine the discussion at this point to travel on foot.
The passing of time is a difficult concept for everyone. In certain settings, when we are enjoying ourselves, time seems to fly by; in other contexts, time seems to stand still. To help students develop their sense of time in a neutral context, ask them to close their eyes. Tell students when to start, and ask them to raise their hands after exactly 1 minute has passed. Practice with estimating the passing of time improves performance. Ask students to share the techniques they used to guess about the length of a minute. Let students practice with time intervals less than a minute. Conclude this activity by estimating the duration of 8 seconds.
After students have practiced estimating the passing of time, they are ready to see how far a teacher can walk in 8 seconds. Give a student a stopwatch, and have him or her time the teacher walking from the front to the rear of the classroom. The teacher's rate of walking might be a topic of discussion. Ask students to discuss factors which might affect the distance traveled by the teacher.
### Activity: Walking Speeds
To gather data about their walking speeds, students mark off in a school hallway or outdoors distances from 25 feet to approximately 100 feet in increments of 5 feet. Mark the intervals with masking tape. Have each student carry an index card that has his or her name and lines for recording as many trials as you intend to do.
A school hallway marked off in 5-foot increments
For this first experiment, everyone walks 100 feet, and the timers tell students the time each took to complete the distance. Each mission team takes a turn timing another team.
Line up students and begin the trials. The teacher tells members of a mission team when to go. Three students should have stopwatches to time each walker. The time recorded is the median of the three times shown on the timers' watches. Using multiple timers avoids losing data because of difficulties using a stopwatch. If sufficient stopwatches are not available for this scheme, have each timer pick a participant and keep time for that student. Some students may need to repeat their walks if a timer makes a mistake.
To facilitate making a graph of the class data, have each student record her or his time on the reverse side of the index card in large writing. Begin by making a graph of the class data using the students themselves. Ask five students to come to the front of the room and stand in order according to their times, from the least to the greatest. Ask another group to come to the front and put themselves into the ordered group. If the times are the same, the students should stand behind one another. When all students are in the ordered group, the graph is complete.
The teacher should record on the chalkboard or an overhead transparency a frequency table for the "human" graph. When seated again, each student should make a bar graph of the data in the frequency table.
Students may choose to use the Bar Grapher Tool to graph the data. Alternatively, students may use grid paper to graph the data.
### Activity: How Far Can I Walk in 8 Seconds?
The next phase of this lesson requires students to collect data about how far they can walk in 8 seconds. That is, the time allowed for walking is held constant and the distance varies from student to student. The marked-off, 5-foot increments are used to measure the distances.
Only one timer is necessary. The teacher is probably the best timer here because the stop and go commands need to be authoritative.
At the stop command, each student looks at the distance markers and records the distance walked. Of course, students do not always stop on a mark. They need to agree on how to estimate the number of feet they have walked beyond a mark, add that distance to the marked distance, and record their results for the final graphing activities.
Different students can walk varying distances in 8 seconds
### Closing The Activity
Have each mission team record its results on one graph. The graph should show distance versus time, with distance on the vertical axis and time on the horizontal axis. Be sure to talk about the point where time is zero and distance is zero, which is also a potential data point. Connect each point on the graph (representing how far each student on the mission team walked in 8 seconds) with (0,0). The steepness of the lines connecting (0,0) to the data pints shows the average rate of speed, or slope. Use this opportunity to discuss the concept of slope and to give a formal definition of the slope of a line.
Students may choose to use the Line of Best Fit Tool to graph the data. Alternatively, students may use grid paper to graph the data.
### Reference
Adapted from Finding Our Top Speed in Mission Mathematics, Linking Aerospace and the NCTM Standards, a NASA/NCTM project, NCTM 1997.
• Yardsticks or 50-foot measuring tape
• Stopwatches
• Grid Paper
• Index cards
• Computers with internet access (optional)
Extensions
Students may want to check their speed over longer distances to confirm that in 16 seconds they go twice as far. Note that the empirical data most likely do not show a linear relationship. As the length of time increases, the average speed usually decreases. Talk about the effects of tiring over time. If shorter times are used, the speed could be faster. Discuss that the linear graph approximates where each walker would be in the intervening times if he or she walks at a constant rate of speed. This result explains why we characterize the graph as showing the average rate of speed.
The effects of tiring on speed
none
### Learning Objectives
Students will:
• Determine the length of time needed to walk or run or walk a given distance.
• Plot the data on a graph.
• Use results of data collection to develop the concept of slope.
### NCTM Standards and Expectations
• Explore relationships between symbolic expressions and graphs of lines, paying particular attention to the meaning of intercept and slope.
• Use graphs to analyze the nature of changes in quantities in linear relationships.
• Understand, select, and use units of appropriate size and type to measure angles, perimeter, area, surface area, and volume.
• Use common benchmarks to select appropriate methods for estimating measurements.
• Select and apply techniques and tools to accurately find length, area, volume, and angle measures to appropriate levels of precision.
• Solve simple problems involving rates and derived measurements for such attributes as velocity and density. | 1,673 | 8,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-13 | longest | en | 0.959067 |
https://josmfs.net/tag/logic/ | 1,685,707,543,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00138.warc.gz | 375,196,007 | 25,163 | # Blockbusters Problem
For his Monday Puzzle in the Guardian Alex Bellos provided a seemingly impossible puzzle from the 1983 British teenager quiz show Blockbusters.
“In the much-missed student quiz show Blockbusters, teenagers would ask host Bob Holness for a letter from a hexagonal grid. How we laughed when a contestant asked for a P! Holness would reply with a question in the following style: What P is an area of cutting edge mathematical research and also a process in the making of an espresso? The answer is the subject of today’s puzzle: percolation.
Today’s perplexing percolation poser concerns the following Blockbusters-style hexagonal grid:
The grid above shows a 10×10 hexagonal tiling of a rhombus (i.e. a diamond shape), plus an outer row that demarcates the boundary of the rhombus. The boundary row on the top right and the bottom left are coloured blue, while the boundary row on the top left and the bottom right are white.
If we colour each hexagon in the rhombus either blue or white, one of two things can happen. Either there is a path of blue hexagons that connects the blue boundaries, such as here:
Or there is no path of blue hexagons that connects the blue boundaries, such as here:
There are 100 hexagons in the rhombus. Since each of these hexagons can be either white or blue, the total number of possible configurations of white and blue hexagons in the rhombus is 2 x 2 x … x 2 one hundred times, or 2100, which is about 1,000,000,000,000,000,000,000,000,000,000.
In how many of these configurations is there a path of blue hexagons that connects the blue boundaries?
The answer requires a simple insight. Indeed, it is the insight on which the quiz show Blockbusters relied.
For clarification: a path of hexagons means a sequence of adjacent hexagons that are the same colour.”
See the Blockbusters Problem
# “Fermat’s Last Theorem” Puzzle
Here is a mind-numbing logic puzzle from Futility Closet.
“A puzzle by H.A. Thurston, from the April 1947 issue of Eureka, the journal of recreational mathematics published at Cambridge University:
Five people make the following statements:—
Which of these statements are true and which false? It will be found on trial that there is only one possibility. Thus, prove or disprove Fermat’s last theorem.”
Normally I would forgo something this complicated, but I thought I would give it a try. I was surprised that I was able to solve it, though it took some tedious work. (Hint: truth tables. See the “Pointing Fingers” post regarding truth tables.)
One important note. The author is a bit cavalier about the use of “Either …, or …”. In common parlance this means “either P is true or Q is true, but not both” (exclusive “or”: XOR), whereas in logic “or” means “either P is true or Q is true, or possibly both” (inclusive “or”: OR). I assumed all “Either …, or …” and “or” expressions were the logical inclusive “or”, which turned out to be the case.
See the Fermat’s Last Theorem Puzzle
# Who Did It?
This is a fun logic problem from the Mathigon math calendar for December 2022.
1. Alice said Bob did it.
2. Bob said Alice did it.
3. Carol said Alice didn’t do it.
4. Dan said it was either Alice or Carol.
Only one person is telling the truth. Who did it?
See Who Did It?
# The Maths of Lviv
Unfortunately Ukraine has receded from our attention under the threat from our own anti-democratic forces, but this Monday Puzzle from Alex Bellos in March is a timely reminder of the mathematical significance of that country.
“Like many of you I’ve hardly been able to think about anything else these past ten days apart from the war in Ukraine. So today’s puzzles are a celebration of Lviv, Ukraine’s western city, which played an important role in the history of 20th century mathematics. During the 1930s, a remarkable group of scholars came up with new ideas, methods and theorems that helped shape the subject for decades.
The Lwów school of mathematics – at that time, the city was in Poland – was a closely-knit circle of Polish mathematicians, including Stefan Banach, Stanisław Ulam and Hugo Steinhaus, who made important contributions to areas including set-theory, topology and analysis. …
Of the many ideas introduced by the Lwów school, one of the best known is the “ham sandwich theorem,” posed by Steinhaus and solved by Banach using a result of Ulam’s. It states that it is possible to slice a ham sandwich in two with a single slice that cuts each slice of bread and the ham into two equal sizes, whatever the size and positions of the bread and the ham.
Today’s puzzles are also about dividing food. The first is from Hugo Steinhaus’ One Hundred Problems in Elementary Mathematics, published in 1938. The second uses a method involved in the proof of the ham sandwich theorem.
1. Three friends each contribute £4 to buy a £12 ham. The first friend divides it into three parts, asserting the weights are equal. The second friend, distrustful of the first, reweighs the pieces and judges them to be worth £3, £4 and £5. The third, distrustful of them both, weighs the ham on their own scales, getting another result. If each friend insists that their weighings are correct, how can they share the pieces (without cutting them anew) in such a way that each of them would have to admit they got at least £4 of ham?_
2. Ten plain and 14 seeded rolls are randomly arranged in a circle, equidistantly spaced, as below. Show that using a straight line it is possible to divide the circle into two halves such that there are an equal number of plain and seeded rolls on either side of the line.
Show there is always a diameter that cuts the circle into two batches of 12 rolls with an equal number of plain and seeded.
Question 2 is adapted from Mathematical Puzzles by Peter Winkler, who gives as a reference Alon and West, The Borsuk-Ulam Theorem and bisection of necklaces, Proceedings of the American Mathematical Society 98 (1986).
# Pinocchio’s Hats
This problem in logic from Presh Talwalkar recalled an article I wrote a while ago but did not publish. So I thought I would post it as part of the solution.
“Assume that both of the following sentences are true:
1. Pinocchio always lies;
2. Pinocchio says, “All my hats are green.”
We can conclude from these two sentences that:
• (A) Pinocchio has at least one hat.
• (B) Pinocchio has only one green hat.
• (C) Pinocchio has no hats.
• (D) Pinocchio has at least one green hat.
• (E) Pinocchio has no green hats.”
Actually, the question is which, none or more, of statements (A) – (E) follow from the two sentences?
# Shared Spaces Puzzle
This is a nice puzzle from the Scottish Mathematical Council (SMC) Senior Mathematical Challenge of 2008. It is more a logic puzzle than a geometric one.
“In the diagram, each question mark represents one of six consecutive whole numbers. The sum of the numbers in the triangle is 39, the sum of those in the square is 46 and the sum of those in the circle is 85. What are the six numbers?”
See the Shared Spaces Puzzle
# Date Night
This is a fairly straight-forward logic puzzle from Alex Bellos’s Monday Puzzle in The Guardian.
“When it comes to the world of mathematical puzzles, Hungary is a superpower. Not just because of the Rubik’s cube, the iconic toy invented by Ernő Rubik in 1974, but also because of its long history of maths outreach.
In 1894, Hungary staged the world’s first maths competition for teenagers, four decades before one was held anywhere else. 1894 also saw the launch of KöMaL, a Hungarian maths journal for secondary school pupils full of problems and tips on how to solve them. Both the competition and the journal have been running continuously since then, with only brief hiatuses during the two world wars.
This emphasis on developing young talent means that Hungarians are always coming up with puzzles designed to stimulate a love of mathematics. (It also explains why Hungary arguably produces, per capita, more top mathematicians than any other country.)
I asked Béla Bajnok, a Hungarian who is now director of American Mathematics Competitions, a series of competitions involving 300,000 students in the US, whether he knew of any puzzles that originated in Hungary. The first thing he said that came to mind was the ‘3-D logic puzzle’, a type of logic puzzle in which you work out the solution in a three dimensional box, rather than (as is the case with the standard version) in a two-dimensional grid. He said he had never seen this type of puzzle outside Hungary.
Below are two examples he created. You could solve these using an extended two dimensional grid. It’s more in the spirit of the question, however, to draw a three-dimensional one, like you are looking at three sides of a Rubik’s Cube.
Date night
Andy, Bill, Chris, and Daniel are out tonight with their dates, Emily, Fran, Gina, and Huong. We have the following information.
1. Andy will go to the opera
2. Bill will spend the evening with Emily,
3. Chris would not want to go out with Gina,
4. Fran will see a movie
5. Gina will attend a workshop.
We also know that one couple will see an art exhibit. Who will go out with whom, and what will they do?
See Date Night
# Heron Suit Problem
Here is another logic problem from Ian Stewart.
1. No cat that wears a heron suit is unsociable.
2. No cat without a tail will play with a gorilla.
3. Cats with whiskers always wear heron suits.
4. No sociable cat has blunt claws.
5. No cats have tails unless they have whiskers.
Therefore:
No cat with blunt claws will play with a gorilla.
Is the deduction logically correct?
I confess I don’t know what a heron suit is. Google showed various garments with herons imprinted on the cloth, so maybe that is what it is.
See the Heron Suit Problem
# Mystery of the Dancing Men
Manmohan Kaur took Arthur Conan Doyle’s popular 1903 Sherlock Holmes story “The Adventure of the Dancing Men” and used it in his math classes to illustrate the logic and mathematics involved in solving codes and ciphers. I thought his idea might work as a puzzle. It has been years since I read the story, so I had forgotten the decryption and found it quite doable from the setup provided by Kaur. Here is his presentation, subject to further edits and reductions in size on my part.
“The original story has been shortened and simplified. Reference to England has been completely removed and some other superfluous information that distracts the reader instead of helping solve the mystery have been omitted. In the original story Elriges is the name of an inn but we have taken the liberty to use it loosely as the name of a town.
The pictures of all stick figure messages except the fourth are from the collection The Return of Sherlock Holmes. The original story has a typographical error that throws off the decryption scheme. To remove this (intentional or unintentional) error, the fourth figure has been taken from Trap and Washington’s Introduction to Cryptography with Coding Theory. The fourth message is meant to have a different handwriting, so this serves our purposes well.
Condensed Story
Hilton Cubitt of Elriges visits you and gives you a paper with the following mysterious sequence of stick figures that he found lying on the sun-dial in his mansion.
Message 1:
Cubitt explains that he recently married a Chicago woman named Elsie Patrick. Before the wedding, she had asked him never to ask about her past, as she had had some “very disagreeable associations” in her life, although she said that there was nothing that she was personally ashamed of. Their marriage had been a happy one until the messages began to arrive, first mailed from Chicago and then appearing in the garden of his mansion.
The messages had made Elsie very afraid but she did not explain the reasons for her fear, and Cubitt insisted on honoring his promise not to ask about Elsie’s life in Chicago. You look at the figures closely to understand them a little better and notice that some of the figures are holding flags. What could the flags mean? Perhaps the end of words?
The next morning Cubitt finds “a fresh crop of dancing men drawn in chalk upon the black wooden door of the tool-house”:
Message 2:
Two mornings later, “a fresh inscription had appeared”:
Message 3:
Three days later, “a message was left scrawled upon paper, and placed under a pebble upon the sun-dial”:
Message 4:
Cubitt gives copies of all these messages to you. Your task is to help him understand what is going on. You call your friend in the Chicago Police Department and ask her to find background information on Elsie Patrick. You learn that Elsie is the daughter of a Chicago crime boss, and was engaged to Abe Slaney, who worked for her dad, and that she had fled to escape her old life.
You examine all the occurrences of the dancing figures. Message 4 is in a different handwriting, so you guess that it is from a different person, most likely, Elsie, while messages 1, 2 and 3 are from the unknown person (the criminal). You spend the next two days trying to make some sense of the stick figures. You are now sure that the flags on some of the figures indicate the end of words. You also know that a simple substitution cipher is being used for the encryption, and that frequency analysis is the way to solve these ciphers.
Three days later, another message appears.
Message 5:
This message causes you to fear that the Cubitts are in immediate danger. You rush to Elriges and find Cubitt dead of a bullet to the heart and his wife gravely wounded from a gunshot to the head. What do the messages say?
Inspector Martin of the Norfolk Constabulary believes that it is a murder-suicide attempt; Elsie is the prime suspect. But you, after noting some inconsistencies in that theory, know that there is a third person involved. How will you prove to Inspector Martin that a third person is involved?”
# Numbers in New Guinea
This puzzle from Alex Bellos follows the themes in his new book, The Language Lover’s Puzzle Book, which, among other things, looks at number systems in different languages. (See also his Numberphile video.)
“Today is the International Day of the World’s Indigenous People, which aims to raise awareness of issues concerning indigenous communities. Such as, for example, the survival of their languages. According to the Endangered Languages Project, more than 40 per cent of the world’s 7,000 languages are at risk of extinction.
Among the fantastic diversity of the world’s languages is a diversity in counting systems. The following puzzle concerns the number words of Ngkolmpu, a language spoken by about 100 people in New Guinea. (They live in the border area between the Indonesian province of Papua and the country of Papua New Guinea.)
Ngkolmpu-zzle
Here is a list of the first ten cube numbers (i.e. 13, 23, 33, …, 103):
1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
Below are the same ten numbers when expressed in Ngkolmpu, but listed in random order. Can you match the correct number to the correct expressions?
eser tarumpao yuow ptae eser traowo eser
eser traowo yuow
naempr
naempr ptae eser traowo eser
naempr tarumpao yuow ptae yuow traowo naempr
naempr traowo yempoka
tarumpao
yempoka tarumpao yempoka ptae naempr traowo yempoka
yuow ptae yempoka traowo tampui
yuow tarumpao yempoka ptae naempr traowo yuow
Here’s a hint: this is an arithmetical puzzle as well as a linguistic one. Ngkolmpu does not have a base ten system like English does. In other words, it doesn’t count in tens, hundreds and thousands. Beyond its different base, however, it behaves very regularly.
This puzzle originally appeared in the 2021 UK Linguistics Olympiad, a national competition for schoolchildren that aims to encourage an interest in languages. It was written by Simi Hellsten, a two-time gold medallist at the International Olympiad of Linguistics, who is currently reading maths at Oxford University.” | 3,671 | 15,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-23 | longest | en | 0.893055 |
http://mathhelpforum.com/geometry/179661-determining-third-point-isosceles-triangle.html | 1,513,301,180,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948551501.53/warc/CC-MAIN-20171215001700-20171215021700-00295.warc.gz | 178,112,598 | 11,017 | Thread: Determining the third point on an isosceles triangle
1. Determining the third point on an isosceles triangle
Given the x/y values of the two points on the base, and the length of the altitude -- or 'height' -- of the triangle, is it possible to determine the x,y value of the third point? Thanks in advance for any ideas.
2. Yes!
3. Ok, I suppose I need to elaborate...what would the process be to determine the x/y value of the third point? I read somewhere about finding the midpoint and then using a unit vector, but I'm not clear on how that would look (if that's a viable solution)
4. Hello, MrGreenjeens!
Given the coordinates of the two points on the base,
and the length of the altitude -- or 'height' -- of the triangle,
is it possible to determine the coordinates of the third point?
Yes, and there will be two possible locations for the third vertex,
one above the base and one below the base.
Code:
R
(x,y)
*
/|\
/ | \
/ | \
r/ |h \r
/ | \
/ | \
/ | \
P * - - - * - - - * Q
(x1,y1) M (x2,y2)
$\text{The base vertices are: }\:P(x_1,y_1)\text{ and }Q(x_2,y_2)$
$\text{Their midpoint is: }\:M\left(x_m,\,y_m) \:=\:\left(\tfrac{x_1+x_2}{2},\:\tfrac{y_1+y_2}{2}\right)$
$\text{The third vertex is: }\,R(x,y) \,\text{ and }|M\!R| = h$
$\text{Let }\vec u \:=\:\overrightarrow{PQ} \:=\:\langle x_2\!-\!x_1,\:y_2\!-\!y_1\rangle$
$\text{Let }\vec v \:=\:\overrightarrow{MR}\,\text{ where }\vec v \perp \vec u$
. . . $\text{Then: }\:\vec v \:=\:\langle y_1\!-\!y_2,\:x_2\!-\!x_1\rangle$
$\text{The unit vector in the direction of }\vec v\text{ is:}$
. . $\vec w \;=\;\left\langle \frac{y_1-y_2}{\sqrt{(x_2\!-\!x_1)^2 + (y_2\!-\!y_1)^2}},\; \frac{x_2-x_1}{\sqrt{(x_2\!-\!x_1)^2 + (y_2\!-\!y_1)^2}}\right \rangle$
$\text{Hence: }\:\vec R \;=\;\vec M + h\vec w$
$\text{Therefore: }\;R\left(\frac{x_1\!+\!x_2}{2} + \frac{h(y_1\!-\!y_2)}{\sqrt{(x_2\!-\!x_1)^2+(y_2\!-\!y_1)^2}}, \;\; \frac{y_1\!+\!y_2}{2} + \frac{h(x_2\!-\!x_1)}{\sqrt{(x_2\!-\!x_1)^2 + (y_2\!-\!y_1)^2}} \right)$
5. Thanks, Soroban -- this is exactly what I was looking for. I had thought that there would be two possible points, but from playing around with your formula, it seems that this is determined by whether the value of h is positive or negative. Anyways, I've been able to integrate this solution into a program I'm working on and it's working as desired, so I appreciate the time you spent.
,
,
,
how to find the third vertex of an isosceles triangle
Click on a term to search for related topics. | 909 | 2,553 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-51 | longest | en | 0.511599 |
https://www.mortgagequote.com/calculate-a-mortgage-payoff.php | 1,723,575,151,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00672.warc.gz | 697,860,444 | 27,293 | Calculating a Mortgage Payoff
By learning how to calculate a mortgage payoff, you might be able to strategize and improve your financial planning in the future. This might lead to potential savings of money by paying your mortgage off faster, as well as what to expect if you were to ever refinance your home. Unfortunately, many people don’t know how to calculate their total mortgage payoff amount.
How To Calculate A Mortgage Payoff?
A mortgage payoff is the total amount that you will pay before your mortgage and all of the interest is completely paid off. This is not the same as the principal amount. The principal is the amount that you borrow to pay for your home. However, you will pay back more than this due to the interest. The amount on your current balance is not necessarily the same as the total amount you will need to pay overall. Knowing how to calculate mortgage payoff allows you to better plan your finances and gain more control over your mortgage.
Figuring out how to calculate a mortgage payoff can be a daunting task. With mortgage terms like principal, interest rate, and amortization, it's easy to feel overwhelmed. But fear not! In this step-by-step guide, we'll walk you through the process of mastering mortgage math so you can calculate your mortgage payoff with confidence.
Understanding how and when to calculate payoff on mortgage payoff loans is essential for several reasons. It can help you determine when you'll become mortgage-free and plan for your financial future. It can also help you assess the impact of making extra payments or refinancing your mortgage. Armed with this knowledge, you'll have a clearer picture of your financial goals and how to achieve them.
Throughout this guide, we'll break down the key concepts and formulas, providing clear explanations and practical examples. Whether you're a first-time homebuyer or a seasoned homeowner looking to pay off your mortgage early, this step-by-step guide will empower you to crunch the numbers and make informed decisions.
No more guesswork or confusion—let's dive into the world of mortgage math and calculate your mortgage payoff like a pro!
How to Pay Off a Balance Early
A payoff is the total amount that you will pay before your mortgage and all of the interest is completely paid off. This is not the same as the principal amount. The principal is the amount that you borrow to pay for your home. However, you will pay back more than this due to the interest. The amount on your current balance is not necessarily the same as the total amount you will need to eliminate the debt overall. How to get a quote of a pay off allows you to better plan your finances and gain more control over your mortgage. Use our mortgage payoff calculator to get an estimation on a loan.
A surprising number of people do not fully understand their mortgage payment. You may assume that you are reducing the principal debt amount but your payments could only be covering interest, to begin with. If you do not understand your mortgage payment, you can’t pay it off in the most economical way. If you do not understand your mortgage payment, you can’t pay it off in the most economical way. Keep in mind that your payment may include other costs like mortgage insurance, for example.
When you are looking to refinance a home and you are comparing mortgage estimates, consider your payoff amount. It is important that you have an idea of your monthly payments and how long you need to pay it for. This allows you to make an informed decision and plan your finances accordingly. Understanding your mortgage payment helps you avoid financial issues in the future.
Determine How Long it Will Take To Pay it Off
Future financial planning is easier if you know when your mortgage will be paid off. If you can estimate the payoff amount, you can then determine how long it will take you to be mortgage-free. You can also use different scenarios for paying off your mortgage to see if you can save money. For example, knowing your mortgage payoff amount will help you decide if refinancing is a good option or not.
When you are looking to repay the loan and have a clear idea of how long you need the loan for, you can work out how to obtain a payoff quote. Increasing payments towards the principal can save you a huge amount of money in interest over the years. Overpaying your mortgage significantly reduces the costs of buying a house. By paying a bit more each month, you could save tens of thousands over the course of your mortgage. You also might eliminate the loan sooner and free up more expendable income.
Paying off your mortgage before a milestone like retirement or your children finishing college can help with future financial planning. If your financial burden is likely to increase in the future, paying off your mortgage before then is a sensible idea.
However, it is important to read your mortgage agreement carefully. There may be added fees for paying it off early. Consider the money you will save on interest and compare them with the added fees you have to pay. Sometimes, it benefits you to overpay your mortgage but that is not always the case.
How to Obtain a Payoff Quote
You can calculate a mortgage payoff amount using a formula. Work out the daily interest rate by multiplying the loan balance by the interest rate, then dividing that by 365. This figure, multiplied by the days until payoff, plus the loan balance, gives you your mortgage payoff amount.
Your mortgage originator can make these calculations for you if you contact them. This is an easier option because the figures can be quite confusing. They will send you written confirmation of your payoff amount. You can also use a mortgage payment calculator online. They can tell you the payoff amount on the entire loan or the current balance.
Knowing what your payoff amount is helps you understand exactly what you are paying. Once you know how much you will owe in total, it is much easier to plan your finances in the future. Ultimately, it will help you save money on interest over the course of your mortgage and how long it will take to become debt free.
When you are ready for an early payoff of your loan, you should consider an alternative option of calling up your loan servicer to request one. Requesting your loan balance should come with a ‘per diem’ charge for interest, or per day fee. If you are refinancing or purchasing a home, then the title company generally will order it for you. Just make sure to request proof of payment for your loan and follow up as you might just get a small refund back as it is very typical that title companies overpay just to be certain. Also note that the official balance should have the mortgage company logo on the letterhead.
The formula to calculate a mortgage payoff amount is:
B = L [(1 + c)^n - (1 + c)^p] / [(1 + c)^n (- 1)]
In the formula, B = payoff balance due (\$), L = total loan amount (\$), c =
interest rate (annual rate / 12), n = total payments (years x 12 for monthly
payments), and p = number of payments made so far.
Get Started on Refinancing
If you own a property and want to get started on refinancing then you should connect with a mortgage broker. You can search for a mortgage broker near me to find a professional in your state.
A mortgage broker is a licensed professional that is state registered and must pass an exam in order to negotiate loan services with you. Broker then submits your loan to various lender partners with the goal of providing you a competitive program. However, banks are generally direct lenders, the loan officers are not required to take a state exam and usually do not use outside lenders. Other words, you may only get a narrow variety of loan programs.
For instance, if you want to refinance your townhome, then you should start by gathering up your income documentation, bank statements, insurance, association documents and a recent mortgage statement. Send these documents to your broker and hopefully your experience is smooth and transparent.
Streamline Loans
If you are looking to refinance your loan and the initial was done by FHA, then you may want to consider a Streamline. The benefit of an FHA Streamline is you generally do not have to provide much documentation when you are refinancing as long as you do a rate-and-term loan only and are not seeking additional cash out. What you also might be interested in is learning more about the program via FHA FAQs. In addition, obtaining the balance for FHA might be easier than you can imagine and should be very transparent.
Understanding mortgage payoff
Before we dive more into the nitty-gritty of mortgage math, it's important to understand what a mortgage payoff is. Simply put, a mortgage payoff refers to the total amount of money needed to pay off your mortgage loan in full. This includes the principal amount borrowed, accrued interest, and any additional fees or charges.
Calculating your mortgage payoff allows you to see the big picture and set realistic goals for paying off your mortgage. It also gives you a sense of accomplishment as you track your progress towards becoming mortgage-free.
Importance of calculating mortgage payoff
Calculating your mortgage payoff is not just about knowing the numbers. It plays a crucial role in your overall financial planning. By knowing how much you owe and when you'll be able to pay it off, you can make informed decisions about your financial future.
Knowing your mortgage payoff can also help you evaluate the impact of different scenarios. For example, if you're considering making extra payments towards your mortgage, calculating the payoff can show you how much time and money you can save in the long run.
Basic mortgage math concepts
To calculate your mortgage payoff, you need to understand a few key concepts. Let's start with the basics:
1. Principal: The principal is the initial amount of money borrowed to purchase your home. It does not include interest or other fees. The principal balance decreases over time as you make monthly payments.
2. Interest Rate: The interest rate is the cost of borrowing money from the lender. It is expressed as a percentage and can have a significant impact on your monthly mortgage payments.
3. Term: The term refers to the length of time you have to repay your mortgage loan. It is usually expressed in years. Common mortgage terms include 15, 20, and 30 years.
Understanding these concepts is crucial for calculating your mortgage payoff accurately. Now, let's move on to the next step: determining the principal and interest amounts.
Determining the principal and interest amounts
To calculate your mortgage payoff, you first need to determine the principal and interest amounts. The principal is the amount borrowed, while the interest is the cost of borrowing the money.
The principal amount is straightforward to calculate. It's the original loan amount minus any payments you've made towards the principal. For example, if you borrowed \$300,000 and have paid off \$50,000, your current principal balance is \$250,000.
Calculating the interest amount is a bit more complex. It depends on factors such as the interest rate, loan term, and remaining balance. To simplify the process, you can use an online mortgage calculator or consult your lender for an amortization schedule.
Calculating the monthly mortgage payment
Now that you know the principal and interest amounts, you can calculate your monthly mortgage payment. The monthly payment includes both principal and interest, along with any escrow payments for property taxes and insurance.
The formula to calculate the monthly mortgage payment is as follows:
M = P [ i(1 + i)^n ] / [ (1 + i)^n – 1 ]
Where:
- M is the monthly mortgage payment
- P is the principal balance
- i is the monthly interest rate
- n is the number of monthly payments
By plugging in the values for P, i, and n, you can calculate your monthly mortgage payment. This figure will remain constant throughout the term of your loan, making it easier to budget for your monthly expenses.
Amortization schedules and their significance
An amortization schedule is a table that shows the breakdown of each monthly mortgage payment over the term of your loan. It provides a detailed overview of how much of each payment goes towards principal and interest, as well as the remaining balance.
Understanding an amortization schedule is crucial for calculating your mortgage payoff accurately. It allows you to see how each payment reduces the principal balance and how the interest amount decreases over time.
Amortization schedules are also useful for visualizing the impact of making extra payments. By inputting different scenarios into the schedule, you can see how additional payments can shorten the term of your loan and save you money in interest.
How to calculate the remaining balance on a mortgage
Knowing the remaining balance on your mortgage is essential for calculating the mortgage payoff. The remaining balance is the outstanding amount you still owe the lender.
To calculate the remaining balance, you can use the amortization schedule. Simply locate the current month and find the corresponding remaining balance. Alternatively, you can use an online mortgage calculator that provides an estimated remaining balance based on your inputs.
Calculating the remaining balance allows you to track your progress towards paying off your mortgage. It also helps you plan for the future and make informed decisions about your finances.
Strategies to pay off your mortgage faster
Now that you have a solid understanding of mortgage math and how to calculate your mortgage payoff, let's explore some strategies to pay off your mortgage faster.
1. Make extra payments: One of the most effective ways to pay off your mortgage early is by making extra payments. By adding a little extra money each month towards the principal, you can significantly reduce the term of your loan and save thousands of dollars in interest.
2. Bi-weekly payments: Instead of making monthly payments, consider switching to bi-weekly payments. This allows you to make 26 half-payments throughout the year, which is equivalent to 13 full payments. Over time, this strategy can shave off several years from your mortgage term.
3. Refinance to a shorter term: If interest rates have dropped since you obtained your mortgage, consider refinancing to a shorter term. While your monthly payment may increase, you'll be able to pay off your mortgage faster and save money in interest.
4. Use windfalls or bonuses: If you receive unexpected windfalls or bonuses, consider putting them towards your mortgage. This can make a significant impact on your remaining balance and help you pay off your mortgage faster.
Implementing these strategies requires careful planning and budgeting. However, the long-term benefits of becoming mortgage-free sooner are well worth the effort.
Common mistakes to avoid when calculating a mortgage payoff
While calculating a mortgage payoff may seem straightforward, there are some common mistakes to avoid. These mistakes can lead to inaccurate calculations or unrealistic expectations. Here are a few things to watch out for:
1. Forgetting to include additional fees: When calculating your mortgage payoff, make sure to account for any additional fees or charges, such as prepayment penalties or closing costs. These can significantly impact the final amount.
2. Failing to consider interest rate changes: If you have an adjustable-rate mortgage (ARM), keep in mind that the interest rate can change over time. When calculating your mortgage payoff, factor in potential rate adjustments to get a more accurate estimate.
3. Ignoring the impact of taxes and insurance: Your monthly mortgage payment may include escrow payments for property taxes and insurance. When calculating your mortgage payoff, consider these additional expenses to get a clearer picture of your financial obligations.
By being aware of these common mistakes, you can ensure that your mortgage payoff calculations are accurate and realistic.
Tools and resources for mortgage calculations
Fortunately, there are several tools and resources available to help you with mortgage calculations. Here are a few you can use:
1. Online mortgage calculators: There are numerous online mortgage calculators that allow you to input your loan details and calculate your mortgage payoff. These calculators provide instant results and can be a valuable tool in your mortgage planning.
2. Amortization schedule generators: If you prefer a visual representation of your mortgage payments, consider using an amortization schedule generator. These tools create a detailed schedule that shows how each payment affects your principal, interest, and remaining balance.
3. Mortgage professionals: If you're unsure about certain calculations or need personalized advice, don't hesitate to consult a mortgage professional. They can provide guidance and answer any questions you may have regarding your mortgage payoff.
By utilizing these tools and resources, you can streamline the mortgage calculation process and gain a better understanding of your financial situation.
Conclusion
Calculating your mortgage payoff doesn't have to be a daunting task. With a solid understanding of mortgage math and the right tools at your disposal, you can confidently determine your mortgage payoff and plan for your financial future.
By following the step-by-step guide outlined in this article, you can crunch the numbers like a pro and make informed decisions about your mortgage. Whether you're a first-time homebuyer or looking to pay off your mortgage early, the knowledge gained from mastering mortgage math will empower you to take control of your financial goals.
So, say goodbye to guesswork and confusion. It's time to dive into the world of mortgage math and calculate your mortgage payoff with confidence. | 3,477 | 18,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-33 | longest | en | 0.962055 |
https://gmatclub.com/forum/mgmat-sc-subordinators-110274.html?fl=similar | 1,488,294,913,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174163.72/warc/CC-MAIN-20170219104614-00238-ip-10-171-10-108.ec2.internal.warc.gz | 695,201,596 | 58,157 | MGMAT SC - Subordinators : GMAT Sentence Correction (SC)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 28 Feb 2017, 07:15
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# MGMAT SC - Subordinators
Author Message
TAGS:
### Hide Tags
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1713
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 101
Kudos [?]: 942 [0], given: 109
### Show Tags
02 Mar 2011, 11:10
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
Hello,
According to MGMAT SC, "subordinators, such as because and although, create subordinate clauses, which can in turn attach to a main clause WITH A COMMA".
For example:
I need to relax, BECAUSE I have so many things to do.
Does always a subordinator have to go with a comma in a sentence?
I remember that when I learned English that it is not necessarily true.
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."
My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html
GMAT Club Premium Membership - big benefits and savings
If you have any questions
New!
Retired Moderator
Status: worked for Kaplan's associates, but now on my own, free and flying
Joined: 19 Feb 2007
Posts: 3706
Location: India
WE: Education (Education)
Followers: 736
Kudos [?]: 5756 [0], given: 322
Re: MGMAT SC - Subordinators [#permalink]
### Show Tags
04 Mar 2011, 06:56
As far as I remember, I have not come across any such rule. I have seen a number cases in which subordinators are used without comma.
For e.g.:
I do not like to write my GMAT now although I am well prepared.
I do not like to write my GMAT now because March is an inauspicious month.
In both the above cases, the sentences are perfectly grammatical. But look at them another way.
Although I am well prepared, I do not like to write my GMAT now.
Because March is an inauspicious month, I do not like to write my GMAT now
These are also grammatically correct. At best, we can say that when a complex sentence containing a subordinate clause is preceded by the main clause, them perhaps a comma becomes essential.
_________________
“Better than a thousand days of diligent study is one day with a great teacher” – a Japanese proverb.
9884544509
Re: MGMAT SC - Subordinators [#permalink] 04 Mar 2011, 06:56
Similar topics Replies Last post
Similar
Topics:
1 MGMAT SC 3 19 Jun 2011, 03:16
MGMAT SC 4 13 Jun 2011, 02:29
Mgmat SC vs. Official SC 2 09 Jan 2011, 20:07
1 MGMAT SC 1 04 Nov 2010, 07:23
MGMAT SC 0 13 Dec 2008, 00:27
Display posts from previous: Sort by | 944 | 3,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-09 | longest | en | 0.887091 |
http://www.mathkun.com/first-prime-numbers-up-to/is-prime/is-67-a-prime-number | 1,544,775,237,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825495.60/warc/CC-MAIN-20181214070839-20181214092339-00007.warc.gz | 427,071,014 | 4,696 | USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES
# Is 67 A Prime Number?
Is 67 a prime number? Answer: Yes 67, is a prime number.
The integer 67 has 2 factors. All numbers that have more than 2 factors(one and itself) are not primes.
## How To Know If 67 Is Prime
Number 67 is a prime number because it is only divisible with one and itself. You can try to divide 67 with smaller numbers than itself, but you will only find divisions that will leave a remainder.
## What Is The 19th Prime Number
In the sequence of prime integers, number 67 is the 19th prime number. This means that there are 19 prime numbers before 67.
## What Are All The Prime Numbers Between 67 And 87
List of all the primes between 67 and 87:
## What Are All The Prime Numbers Between 47 And 67
List of all the primes between 47 and 67:
## General Mathematical Properties Of Number 67
67 is not a composite integer. 67 is not a composite figure, because it's only positive divisors are one and itself. It is not even. 67 is not an even digit, because it can't be divided by 2 without leaving a comma spot. This also means that 67 is an odd number. When we simplify Sin 67 degrees we get the value of sin(67)=-0.85551997897532. Simplify Cos 67 degrees. The value of cos(67)=-0.51776979978951. Simplify Tan 67 degrees. Value of tan(67)=1.6523172640102. 67 is not a factorial of any integer. When converting 67 in binary you get 1000011. Converting decimal 67 in hexadecimal is 43. The square root of 67=8.1853527718725. The cube root of 67=4.0615481004457. Square root of √67 simplified is 67. All radicals are now simplified and in their simplest form. Cube root of ∛67 simplified is 67. The simplified radicand no longer has any more cubed factors.
## Prime Number Calculator For Bigger Integers Than 67
Test if bigger integers than 67 are primes.
## Single Digit Properties For 67 Explained
• Integer 6 properties: 6 is even and a composite, with the following divisors:1, 2, 3, 6. Also called perfect number since the sum of the divisors(excluding itself) is 6. The first perfect figure, the next ones are 28 and 496. Six is highly a composed, semiprimo, congruent, scarcely total, Ulam, Wedderburn-Etherington, multi-perfect, integer-free number. Complete Harshad, which is a quantity of Harshad in any expressed base. The factorial of 3 and a semi-perfect digit. The third triangular and the first hexagonal value. All perfect even amounts are triangular and hexagonal. Six is the smallest amount different from 1 whose square (36) is triangular(the next in the line that enjoys this property is 35). Strictly a non-palindrome. A numeral is divisible by 6 if and only if it is divisible by both 2 and 3. Part of the Pythagorean triple (6, 8, 10). Being the product of the first two primes (6=2×3), it is a primitive. In the positional numbering system based on 5 it is a repeated number. An oblong, of the form n(n+1).
• Integer 7 properties: Seven is an odd and defective number. The fourth prime, after 5 and before 11. Also called one of the primes of Mersenne, 7=2³-1. Known as one of the double prime integers, which is (7-1)÷2 still the same. 7 is the Cuban prime of the form (x³-y³)÷(x-y), x=y+1. Euclidian quantity 7=(2×3)+1. 7 is a Perrin, integer-free and congruent number. Smallest natural whose cube (343) is a palindrome. The second figure of Carol. A polygon with seven sides is called a heptagon. Part of the Pythagorean triad (7, 24, 25). Fifth of the succession of Lucas, after 4 and before 11. It is a palindrome in the binary system and a repeated number in the positional numbering system based on 6. In the numerical decimal system seven is a Colombian value.
## What Is A Prime Number?
Definitionof prime numbers: a prime number is a positive whole number(greater than 1) that is only divisible by one and itself.
This means that all primes are only divisible by two numbers. The amount of these integers is infinite. The bigger the figure is the harder it is to know if it is a prime or not. Bigger primes will have more integers inbetween. The biggest use cases outside of mathematics were found once electronics were invented. Modern cryptography uses large primes.
## What Are Factors Of A Number?
Whole integers that are divisible without leaving any fractional part or remainder are called factors of a integer. A factor of a number is also called it's divisor.
## What Are Prime Factors Of A Number?
All figures that are only divisible by one and itself are called prime factors in mathematics. A prime factor is a figure that has only two factors(one and itself).
## What Is Prime Factorization Of A Number?
In mathematics breaking down a composite number(a positive integer that can be the sum of two smaller numbers multiplied together) into a multiplication of smaller numbers is called factorization. When the same process is continued until all numbers have been broken down into their prime factor multiplications then this process is called prime factorization.
Using prime factorization we can find all primes contained in any amount. | 1,255 | 5,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-51 | latest | en | 0.884886 |
https://convertoctopus.com/3-7-cubic-centimeters-to-deciliters | 1,590,473,037,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390448.11/warc/CC-MAIN-20200526050333-20200526080333-00515.warc.gz | 267,488,899 | 7,970 | ## Conversion formula
The conversion factor from cubic centimeters to deciliters is 0.01, which means that 1 cubic centimeter is equal to 0.01 deciliters:
1 cm3 = 0.01 dL
To convert 3.7 cubic centimeters into deciliters we have to multiply 3.7 by the conversion factor in order to get the volume amount from cubic centimeters to deciliters. We can also form a simple proportion to calculate the result:
1 cm3 → 0.01 dL
3.7 cm3 → V(dL)
Solve the above proportion to obtain the volume V in deciliters:
V(dL) = 3.7 cm3 × 0.01 dL
V(dL) = 0.037 dL
The final result is:
3.7 cm3 → 0.037 dL
We conclude that 3.7 cubic centimeters is equivalent to 0.037 deciliters:
3.7 cubic centimeters = 0.037 deciliters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 deciliter is equal to 27.027027027027 × 3.7 cubic centimeters.
Another way is saying that 3.7 cubic centimeters is equal to 1 ÷ 27.027027027027 deciliters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three point seven cubic centimeters is approximately zero point zero three seven deciliters:
3.7 cm3 ≅ 0.037 dL
An alternative is also that one deciliter is approximately twenty-seven point zero two seven times three point seven cubic centimeters.
## Conversion table
### cubic centimeters to deciliters chart
For quick reference purposes, below is the conversion table you can use to convert from cubic centimeters to deciliters
cubic centimeters (cm3) deciliters (dL)
4.7 cubic centimeters 0.047 deciliters
5.7 cubic centimeters 0.057 deciliters
6.7 cubic centimeters 0.067 deciliters
7.7 cubic centimeters 0.077 deciliters
8.7 cubic centimeters 0.087 deciliters
9.7 cubic centimeters 0.097 deciliters
10.7 cubic centimeters 0.107 deciliters
11.7 cubic centimeters 0.117 deciliters
12.7 cubic centimeters 0.127 deciliters
13.7 cubic centimeters 0.137 deciliters | 573 | 1,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-24 | latest | en | 0.68895 |
https://www.reddit.com/r/funny/comments/f9nkfi/shes_smart/ | 1,585,389,941,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370490497.6/warc/CC-MAIN-20200328074047-20200328104047-00536.warc.gz | 1,128,810,106 | 43,886 | ×
Dismiss this pinned window
[–] 293 points294 points (26 children)
Well... In her defense, she's not WRONG...
[–][S] 85 points86 points (14 children)
Lol- I used to do similar stuff in Gr1 and my teachers hated it
[–] 57 points58 points (1 child)
Math teachers hate him!
Learn this long addition with one simple trick!
[–] 9 points10 points (0 children)
Long long maaaaaaath.
[–] 10 points11 points (0 children)
I also did the same grade 2 with multiplication. Needless to say, I still don't know my times tables...
[–] 3 points4 points (8 children)
Continue the pattern: A B C A B _ _ _ _
Answer: A B C A B D, A B C A B D, etc
[–] 2 points3 points (0 children)
I almost failed college due to a logic class. Thanks for the reminder.
[–] 2 points3 points (1 child)
ABCABABCABABCABABCAB
[–] 1 point2 points (0 children)
ABCABABCABABCABABCAB
Yes exactly haha
[–] 1 point2 points (0 children)
A B C A B D A B E A B F A B G A B H A B I A B J A B K A B L A B M A B N A B O A B P A B Q A B R A B S A B T A B U A B V A B W A B X A B Y A B Z A C C A C D A C E A C F A C G A C H ...
... Z Z X Z Z Y Z Z Z A B C A B D A B E A B F A B G
Go hard or go home.
[–] 0 points1 point (3 children)
A B C A B C A B C A B C
[–] 1 point2 points (2 children)
Maybe r/whoosh?
[–] -1 points0 points (1 child)
How? Based on the given data, the pattern is clear.
[–] -1 points0 points (0 children)
I mean, clearly that was the "right" answer...
[–] 0 points1 point (1 child)
Well how old were you when you eventually got into gr2?
[–] 2 points3 points (0 children)
23
[–] 1 point2 points (0 children)
She doesn’t need ANY defense.
[–] 1 point2 points (1 child)
You mean that isn't the answer?
[–] 0 points1 point (0 children)
Unless there were instructioms like "use different integers for each equation", yes. Those are repititions of the same correct/balanced equation. Somehow I think this falls under r/maliciouscompliance.
Brilliant kid though.
[–] 3 points4 points (7 children)
That depends on what the assignment was.
"Fill in the blanks with different numbers to complete the equation" for example would make this wrong
[–] 0 points1 point (0 children)
If that was a clear instruction, you are correct. The naughty child in me hopes there was NO clear instruction... making this an awesome r/maliciouscompliance.
I love this kid.
[–] 101 points102 points (4 children)
What a shittily designed math problem.
[–] 28 points29 points (2 children)
I think your not supposed to use the same numbers more than once
[–] 24 points25 points (0 children)
Her next lesson is spelling you’re
[–] 0 points1 point (0 children)
Then go for variables like ab, abc, abcd.
[–] 0 points1 point (0 children)
I think it is supposed to teach them how to solve an equation system. But it does a horrible job of conveying the problem properly
[–] 7 points8 points (0 children)
Zero fucks given to the teacher.
[–] 3 points4 points (1 child)
[–] 0 points1 point (0 children)
My best guess is;
10 + 10 = 20
10 + 5 + 5 = 20
5 + 5 + 5 + 5 = 20
2 + 3 + 5 + 5 + 10 = 20
[–] 46 points47 points (24 children)
She writes her 0’s backwards
[–] 40 points41 points (1 child)
i think you write your 0’s backwards
[–] 4 points5 points (0 children)
Hey pal, don't you tell him he writes his naughts backwards; you write your naughts backwards.
[–] 9 points10 points (0 children)
maybe she's from the southern hemisphere?
[–] 1 point2 points (0 children)
I immediately thought the same thing. Made me cringe every time I saw her write zero.
[–] 2 points3 points (1 child)
I read a really interesting story where you can tell what part of the world a person is from by how they write their zeros.
[–] 5 points6 points (0 children)
[–] -3 points-2 points (0 children)
you write yours bockwords
[–] 28 points29 points (3 children)
[–] 11 points12 points (1 child)
Didn't even cross my mind until she did it
[–] 1 point2 points (0 children)
Yup...good at math...
[–][🍰] 1 point2 points (0 children)
Top 10 people smarter than Albert Edison
[–] 1 point2 points (0 children)
Albert Einstein could never
[–] 0 points1 point (0 children)
Except she should've went down then right.
[–] 0 points1 point (0 children)
It’s big brain time
[–] 0 points1 point (0 children)
ok, serious question. how many people actualy write zeroes clockwise?
[–] 0 points1 point (0 children)
[–] 0 points1 point (0 children)
big brain
[–] 0 points1 point (0 children)
Zero fucks given.
[–] 0 points1 point (0 children)
This was exactly what I thought when I saw the problem, lol. Well done.
[–] -1 points0 points (0 children)
Please spell this word - this word
[–] -1 points0 points (0 children)
Listen here you lil shit
[–] -3 points-2 points (0 children)
Very resourceful woman.
[–] -1 points0 points (0 children)
Touche
[–] -1 points0 points (0 children)
I’m sure the teachers wanted her to come up with an unnecessarily complicated computation. Instead she kept it simple—nice!
[–] -3 points-2 points (0 children)
Efficiency
[–] -3 points-2 points (0 children)
I wish my life was as scripted as an Asian TikTok and I was always this funny. | 1,644 | 5,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-16 | latest | en | 0.810924 |
https://businesslawblogsite.com/fees/what-states-require-tax-on-shipping.html | 1,653,179,127,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543264.49/warc/CC-MAIN-20220522001016-20220522031016-00138.warc.gz | 200,374,594 | 19,716 | # What states require tax on shipping?
Contents
## Which states charge tax on shipping?
The majority of states (Arkansas, Connecticut, Georgia, Illinois, Kansas, Kentucky, Michigan, Mississippi, Nebraska, New Jersey, New Mexico, New York, North Carolina, North Dakota, Ohio, Oklahoma, Pennsylvania, Rhode Island, South Carolina, South Dakota, Tennessee, Texas, Utah, Vermont, Washington, West Virginia and …
## Is tax required on shipping?
California: For the most part, shipping charges are exempt if the sale is exempt, but if the sale is taxable, delivery-related charges may be nontaxable, partially taxable, or fully taxable. … Florida: Transportation charges for taxable sales are generally taxable whether separately stated or included in the sale price.
## Can you avoid sales tax by shipping out of state?
You would charge the destination state’s rate, in addition to any local or county sales taxes for the address to which you’re shipping. You would not additionally collect your own state’s sales tax on products you’re shipping out of state.
## Is shipping taxable in all states?
For taxable sales, delivery and shipping charges that are included in the sale price are generally subject to state sales tax. Sellers can only separately state the delivery charge if the buyer can avoid it (i.e., by picking up the goods). Different rules may apply for local sales tax.
## How is shipping tax calculated?
Multiply the value of the item by the shipping duty tax rate. Continuing the same example, \$100 x 0.08 = \$8. This figure represents the shipping duty tax you will have to pay to ship the item.
## How much is shipping tax?
In California, shipping is not taxable only if you charge your customer the exact rate it costs you to ship them the item. If you always charge a flat shipping fee \$2.99 but it truly only cost you \$1.99 to ship an item, the state considers that extra \$1.00 you charged your customer as taxable.
## Is tax based on billing or shipping?
Tax is calculated based on the shipping address (not the billing address) and is calculated as a percentage of the order total, including shipping and handling costs.
## Does UPS charge tax on shipping?
Several states have declared that shipping charges are taxable when the product sold itself is taxable and when the product is shipped through a common carrier like FedEx, UPS, or the US Postal Service.
## Does Shopify charge tax on shipping?
In California, shipping is not taxable if the actual cost of shipping is listed on a separate line on the invoice. If you add handling fees, then shipping becomes taxable. You can choose whether you want to charge taxes on shipping when you set up your taxes and enter your sales tax ID.
THIS IS IMPORTANT: Question: Is milk taxed in Maryland?
## How do you avoid state sales tax?
Yet because most states tax most sales of goods and require consumers to remit use tax if sales tax isn’t collected at checkout, the only way to avoid sales tax is to purchase items that are tax exempt.
## What states don’t require sales tax on online purchases?
The following states do not have an Internet sales tax:
• Delaware.
• New Hampshire.
• Montana.
• Oregon.
## Which states have no sales taxes?
Most states have sales tax to help generate revenue for its operations – but five states currently have no sales tax: Alaska, Delaware, Montana, New Hampshire, and Oregon.
## Does Florida charge sales tax on shipping?
Florida sales tax is imposed on the total sales price of taxable tangible personal property. … Delivery charges are not subject to tax when the charge is separately stated on an invoice or bill of sale and the purchaser has the option of picking the item up or arranging for third-party transportation services on its own.
## Does Florida charge out-of-state sales tax?
If you sell items to customers in another state, but do not have nexus in that state, you do not have to collect sales tax on the items you sell to them. … The use tax is also due on taxable items purchased during out-of-state travel, when the merchandise is shipped to the individual’s home state.
## Does Florida charge sales tax on online purchases?
Florida’s online shoppers will have to start paying sales taxes on all purchases starting this July. … Ron DeSantis signed into law a plan to require out-of-state online retailers to collect sales taxes on purchases made by Floridians.
THIS IS IMPORTANT: Is the prime minister's salary taxed? | 912 | 4,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.935434 |
https://www.coursehero.com/file/5805806/HW4/ | 1,516,597,961,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00502.warc.gz | 904,469,387 | 70,726 | # HW4 - 4 The organic concentration in a water sample...
This preview shows pages 1–2. Sign up to view the full content.
CEE 3510 Prob. Set No. 4 Due: March 3, 2010 1. Using free energy principles explain why anaerobic digestion (with the production of CH 4 ) is used to treat organic sludges produced during biological wastewater treatment. 2. Human wastewater production is about 0.2 lbs oxygen consuming material per day per person. Assuming that the city of Ithaca (population approximately 65, 000) discharged its domestic waste directly (without treatment) to Cayuga Lake, how long would it take to utilize all of the oxygen in the lake. Cayuga Lake has a volume of 2.4 x 10 12 gallons and initial dissolved oxygen concentration of 10 mg/L. Assume that there is no photosynthetic oxygen production nor any atmospheric reaeration. Also assume that the oxygen concentration is uniform throughout the lake. 3. If domestic sewage has an approximate formulation of C 10 H 19 O 3 N, estimate the NOD (nitrogenous oxygen demand) of a sample, which contains 100 mg/L of domestic sewage as BOD L . Assume complete biodegradability of the sewage.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: 4. The organic concentration in a water sample, measured as BOD L is 4 mg/L. If the BOD reaction rate is 0.3/day, what will be the concentration of organic matter remaining at the end of 5 days (L t )? How much oxygen will be used in this period to oxidize the waste (y t )? 5. A water sample has a BOD 5 of 10 mg/L. The sample was diluted 1/10 with dilution water and put in a BOD bottle. Initial DO (dissolved oxygen) in the BOD bottle was 8 mg/L. The sample also contained 2 mg/L NH 3-N (before dilution). Assume the sample was typical domestic sewage. What was the final DO in the BOD bottle? 6. A BOD test is performed on an undiluted sample at 20 o C and 35 o C. BOD 5 at 20 o C was 4.15 mg/L. BOD 5 at 35 o C was 6.56 mg/L. From this data determine the BOD L of the sample. Assume that nitrification was inhibited and that the temperature correction factor ( θ29 equals 1.05. 7. Do Stella Exercise #2 1 2...
View Full Document
{[ snackBarMessage ]}
### Page1 / 2
HW4 - 4 The organic concentration in a water sample...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 652 | 2,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-05 | latest | en | 0.93271 |
https://r-forge.r-project.org/scm/viewvc.php/pkg/R/Matrix.R?root=matrix&r1=1673&r2=1389&pathrev=2256 | 1,571,424,347,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684425.36/warc/CC-MAIN-20191018181458-20191018204958-00221.warc.gz | 663,576,207 | 11,535 | # SCM Repository
[matrix] Diff of /pkg/R/Matrix.R
[matrix] / pkg / R / Matrix.R
# Diff of /pkg/R/Matrix.R
revision 1389, Fri Aug 18 13:53:33 2006 UTC revision 1673, Mon Nov 6 20:54:26 2006 UTC
# Line 6 Line 6
6 setAs("Matrix", "sparseMatrix", function(from) as_Csparse(from)) setAs("Matrix", "sparseMatrix", function(from) as_Csparse(from))
7 setAs("Matrix", "denseMatrix", function(from) as_dense(from)) setAs("Matrix", "denseMatrix", function(from) as_dense(from))
8
9 ## Most of these work; this is a last resort:
10 setAs(from = "Matrix", to = "matrix", # do *not* call base::as.matrix() here:
11 function(from) .bail.out.2("coerce", class(from), class(to)))
12 setAs(from = "matrix", to = "Matrix", function(from) Matrix(from))
13
14 ## ## probably not needed eventually: ## ## probably not needed eventually:
15 ## setAs(from = "ddenseMatrix", to = "matrix", ## setAs(from = "ddenseMatrix", to = "matrix",
16 ## function(from) { ## function(from) {
# Line 19 Line 24
24 setMethod("as.array", signature(x = "Matrix"), function(x) as(x, "matrix")) setMethod("as.array", signature(x = "Matrix"), function(x) as(x, "matrix"))
25
26 ## head and tail apply to all Matrix objects for which subscripting is allowed: ## head and tail apply to all Matrix objects for which subscripting is allowed:
if(paste(R.version\$major, R.version\$minor, sep=".") < "2.4") {
setMethod("tail", signature(x = "Matrix"), utils:::tail.matrix)
} else { # R 2.4.0 and newer
28 setMethod("tail", signature(x = "Matrix"), utils::tail.matrix) setMethod("tail", signature(x = "Matrix"), utils::tail.matrix)
29 }
30 ## slow "fall back" method {subclasses should have faster ones}: ## slow "fall back" method {subclasses should have faster ones}:
31 setMethod("as.vector", signature(x = "Matrix", mode = "missing"), setMethod("as.vector", signature(x = "Matrix", mode = "missing"),
32 function(x) as.vector(as(x, "matrix"))) function(x) as.vector(as(x, "matrix")))
# Line 37 Line 38
38 function(x, ...) as.logical(as.vector(x))) function(x, ...) as.logical(as.vector(x)))
39
40
41 ## Note that isSymmetric is *not* exported ## "base" has an isSymmetric() S3-generic since R 2.3.0
## but that "base" has an isSymmetric() S3-generic since R 2.3.0
42 setMethod("isSymmetric", signature(object = "symmetricMatrix"), setMethod("isSymmetric", signature(object = "symmetricMatrix"),
43 function(object,tol) TRUE) function(object,tol) TRUE)
44 setMethod("isSymmetric", signature(object = "triangularMatrix"), setMethod("isSymmetric", signature(object = "triangularMatrix"),
45 ## TRUE iff diagonal: ## TRUE iff diagonal:
46 function(object,tol) isDiagonal(object)) function(object,tol) isDiagonal(object))
47
if(paste(R.version\$major, R.version\$minor, sep=".") < "2.3")
## need a "matrix" method as in R 2.3 and later
setMethod("isSymmetric", signature(object = "matrix"),
function(object, tol = 100*.Machine\$double.eps, ...)
{
## pretest: is it square?
d <- dim(object)
if(d[1] != d[2]) return(FALSE)
## for `broken' all.equal in R <= 2.2.x:
dn <- dimnames(object)
if(!identical(dn[1], dn[2])) return(FALSE)
test <-
if(is.complex(object))
all.equal.numeric(object, Conj(t(object)), tol = tol, ...)
else # numeric, character, ..
all.equal(object, t(object), tol = tol, ...)
isTRUE(test)
})
48 setMethod("isTriangular", signature(object = "triangularMatrix"), setMethod("isTriangular", signature(object = "triangularMatrix"),
49 function(object, ...) TRUE) function(object, ...) TRUE)
50
# Line 97 Line 77
77
78 Matrix <- Matrix <-
79 function (data = NA, nrow = 1, ncol = 1, byrow = FALSE, dimnames = NULL, function (data = NA, nrow = 1, ncol = 1, byrow = FALSE, dimnames = NULL,
80 sparse = NULL) sparse = NULL, forceCheck = FALSE)
81 { {
82 sparseDefault <- function(m) sparseDefault <- function(m) prod(dim(m)) > 2*sum(isN0(as(m, "matrix")))
prod(dim(m)) > 2*sum(as(m, "matrix") != 0)
83
84 i.M <- is(data, "Matrix") i.M <- is(data, "Matrix")
85 if(is.null(sparse) && (i.M || is(data, "matrix")))
86 if(is.null(sparse1 <- sparse) && (i.M || is(data, "matrix")))
87 sparse <- sparseDefault(data) sparse <- sparseDefault(data)
88
89 doDN <- TRUE
90 if (i.M) { if (i.M) {
91 sM <- is(data,"sparseMatrix") sM <- is(data,"sparseMatrix")
92 if((sparse && sM) || (!sparse && !sM)) if(!forceCheck && ((sparse && sM) || (!sparse && !sM)))
93 return(data) return(data)
94 ## else : convert dense <-> sparse -> at end ## else : convert dense <-> sparse -> at end
95 } }
# Line 117 Line 98
98 nrow <- ceiling(length(data)/ncol) nrow <- ceiling(length(data)/ncol)
99 else if (missing(ncol)) else if (missing(ncol))
100 ncol <- ceiling(length(data)/nrow) ncol <- ceiling(length(data)/nrow)
101 if(length(data) == 1 && data == 0 && !identical(sparse,FALSE)) { if(length(data) == 1 && is0(data) && !identical(sparse, FALSE)) {
102 if(is.null(sparse)) sparse <- TRUE ## Matrix(0, ...) : always sparse unless "sparse = FALSE":
103 if(is.null(sparse)) sparse1 <- sparse <- TRUE
104 ## will be sparse: do NOT construct full matrix! ## will be sparse: do NOT construct full matrix!
105 data <- new(if(is.numeric(data)) "dgTMatrix" else data <- new(if(is.numeric(data)) "dgTMatrix" else
106 if(is.logical(data)) "lgTMatrix" else if(is.logical(data)) "lgTMatrix" else
# Line 132 Line 114
114 sparse <- sparseDefault(data) sparse <- sparseDefault(data)
115 dimnames(data) <- dimnames dimnames(data) <- dimnames
116 } }
117 } else if (!is.null(dimnames)) doDN <- FALSE
118 dimnames(data) <- dimnames }
119 ## 'data' is now a "matrix" or "Matrix" ## 'data' is now a "matrix" or "Matrix"
120 if (doDN && !is.null(dimnames))
121 dimnames(data) <- dimnames
122
123 ## check for symmetric / triangular / diagonal : ## check for symmetric / triangular / diagonal :
124 isSym <- isSymmetric(data) isSym <- isSymmetric(data)
# Line 145 Line 128
128 if(isDiag) if(isDiag)
129 isDiag <- isDiagonal(data) isDiag <- isDiagonal(data)
130
### TODO: Compare with as.Matrix() and its tests in ./dgeMatrix.R
131 ## Find proper matrix class 'cl' ## Find proper matrix class 'cl'
132 cl <- cl <-
133 if(isDiag) if(isDiag && !isTRUE(sparse1))
134 "diagonalMatrix" # -> will automatically check for type "diagonalMatrix" # -> will automatically check for type
135 else { else {
136 ## consider it's type ## consider it's type
# Line 190 Line 171
171 function(x, y) callGeneric(x, as.matrix(y))) function(x, y) callGeneric(x, as.matrix(y)))
172
173 setMethod("%*%", signature(x = "numeric", y = "Matrix"), setMethod("%*%", signature(x = "numeric", y = "Matrix"),
174 function(x, y) callGeneric(rbind(x), y)) function(x, y) callGeneric(matrix(x, nrow = 1, byrow=TRUE), y))
175
176 setMethod("crossprod", signature(x = "Matrix", y = "numeric"), setMethod("crossprod", signature(x = "Matrix", y = "numeric"),
177 function(x, y = NULL) callGeneric(x, as.matrix(y))) function(x, y = NULL) callGeneric(x, as.matrix(y)))
# Line 204 Line 185
185 setMethod("tcrossprod", signature(x = "numeric", y = "Matrix"), setMethod("tcrossprod", signature(x = "numeric", y = "Matrix"),
186 function(x, y = NULL) callGeneric(as.matrix(x), y)) function(x, y = NULL) callGeneric(as.matrix(x), y))
187
188 ## maybe not optimal
189 setMethod("solve", signature(a = "Matrix", b = "missing"),
190 function(a, b, ...) solve(a, Diagonal(nrow(a))))
191
192 setMethod("solve", signature(a = "Matrix", b = "numeric"), setMethod("solve", signature(a = "Matrix", b = "numeric"),
193 function(a, b, ...) callGeneric(a, as.matrix(b))) function(a, b, ...) callGeneric(a, as.matrix(b)))
194 ## when no sub-class method is found, bail out
195 setMethod("solve", signature(a = "Matrix", b = "matrix"),
196 function(a, b, ...) .bail.out.2("solve", class(a), "matrix"))
197
198 ## bail-out methods in order to get better error messages ## bail-out methods in order to get better error messages
199 setMethod("%*%", signature(x = "Matrix", y = "Matrix"), setMethod("%*%", signature(x = "Matrix", y = "Matrix"),
# Line 239 Line 227
227 Y <- as(Y, "matrix") ; Matrix(callGeneric()) }) Y <- as(Y, "matrix") ; Matrix(callGeneric()) })
228
229
230 ## FIXME: All of these should never be called
231 setMethod("chol", signature(x = "Matrix"),
232 function(x, pivot = FALSE) .bail.out.1(.Generic, class(x)))
233 setMethod("determinant", signature(x = "Matrix"),
234 function(x, logarithm = TRUE) .bail.out.1(.Generic, class(x)))
235
236 setMethod("diag", signature(x = "Matrix"), setMethod("diag", signature(x = "Matrix"),
237 function(x, nrow, ncol) .bail.out.1(.Generic, class(x))) function(x, nrow, ncol) .bail.out.1(.Generic, class(x)))
238 setMethod("t", signature(x = "Matrix"), setMethod("t", signature(x = "Matrix"),
# Line 253 Line 247
247 0-e1 0-e1
248 }) })
249
250 ## bail-outs: ## old-style matrices are made into new ones
251 setMethod("Compare", signature(e1 = "Matrix", e2 = "Matrix"), setMethod("Ops", signature(e1 = "Matrix", e2 = "matrix"),
252 function(e1, e2) callGeneric(e1, Matrix(e2)))
253 ## callGeneric(e1, Matrix(e2, sparse=is(e1,"sparseMatrix"))))
254 setMethod("Ops", signature(e1 = "matrix", e2 = "Matrix"),
255 function(e1, e2) callGeneric(Matrix(e1), e2))
256
257 ## bail-outs -- on highest possible level, hence "Ops", not "Compare"/"Arith" :
258 setMethod("Ops", signature(e1 = "Matrix", e2 = "Matrix"),
259 function(e1, e2) { function(e1, e2) {
260 d <- dimCheck(e1,e2) d <- dimCheck(e1,e2)
261 .bail.out.2(.Generic, class(e1), class(e2)) .bail.out.2(.Generic, class(e1), class(e2))
262 }) })
263 setMethod("Compare", signature(e1 = "Matrix", e2 = "ANY"), setMethod("Ops", signature(e1 = "Matrix", e2 = "ANY"),
264 function(e1, e2) .bail.out.2(.Generic, class(e1), class(e2))) function(e1, e2) .bail.out.2(.Generic, class(e1), class(e2)))
265 setMethod("Compare", signature(e1 = "ANY", e2 = "Matrix"), setMethod("Ops", signature(e1 = "ANY", e2 = "Matrix"),
266 function(e1, e2) .bail.out.2(.Generic, class(e1), class(e2))) function(e1, e2) .bail.out.2(.Generic, class(e1), class(e2)))
267
268
# Line 298 Line 299
299 function(x,i,j, drop) function(x,i,j, drop)
300 stop("invalid or not-yet-implemented 'Matrix' subsetting")) stop("invalid or not-yet-implemented 'Matrix' subsetting"))
301
302 ## "logical *vector* indexing, such as M [ M >= 10 ] : ## logical indexing, such as M[ M >= 7 ] *BUT* also M[ M[,1] >= 3,],
303 setMethod("[", signature(x = "Matrix", i = "lMatrix", j = "missing", ## The following is *both* for M [ <logical> ]
304 drop = "ANY"), ## and also for M [ <logical> , ]
305 function (x, i, j, drop) { .M.sub.i.logical <- function (x, i, j, drop)
306 {
307 nA <- nargs()
308 if(nA == 2) { ## M [ M >= 7 ]
309 ## FIXME: when both 'x' and 'i' are sparse, this can be very inefficient
310 as(x, geClass(x))@x[as.vector(i)] as(x, geClass(x))@x[as.vector(i)]
311 ## -> error when lengths don't match ## -> error when lengths don't match
312 }) } else if(nA == 3) { ## M [ M[,1, drop=FALSE] >= 7, ]
313 stop("not-yet-implemented 'Matrix' subsetting") ## FIXME
314
315 ## FIXME: The following is good for M [ <logical> ] } else stop("nargs() = ", nA,
316 ## *BUT* it also triggers for M [ <logical> , ] where it is *WRONG* " should never happen; please report.")
317 ## using nargs() does not help: it gives '3' for both cases }
318 if(FALSE) setMethod("[", signature(x = "Matrix", i = "lMatrix", j = "missing",
319 drop = "ANY"),
320 .M.sub.i.logical)
321 setMethod("[", signature(x = "Matrix", i = "logical", j = "missing", setMethod("[", signature(x = "Matrix", i = "logical", j = "missing",
322 drop = "ANY"), drop = "ANY"),
323 function (x, i, j, drop) { .M.sub.i.logical)
324 ## DEBUG
cat("[(Matrix,i,..): nargs=", nargs(),"\n")
as(x, geClass(x))@x[i] })
325
326 ## A[ ij ] where ij is (i,j) 2-column matrix :
327 .M.sub.i.2col <- function (x, i, j, drop)
328 {
329 nA <- nargs()
330 if(nA == 2) { ## M [ cbind(ii,jj) ]
331 if(!is.integer(nc <- ncol(i)))
332 stop("'i' has no integer column number",
333 " should never happen; please report")
334 if(is.logical(i))
335 return(.M.sub.i.logical(x,i,j,drop))
336 else if(!is.numeric(i) || nc != 2)
337 stop("such indexing must be by logical or 2-column numeric matrix")
338 m <- nrow(i)
339 if(m == 0) return(vector(mode = .type.kind[.M.kind(x)]))
340 ## else
341 i1 <- i[,1]
342 i2 <- i[,2]
343 ## potentially inefficient -- FIXME --
344 unlist(lapply(seq_len(m), function(j) x[i1[j], i2[j]]))
345
346 ## "FIXME:" } else stop("nargs() = ", nA, " should never happen; please report.")
347 ## How can we get at A[ ij ] where ij is (i,j) 2-column matrix? }
348 ## and A[ LL ] where LL is a logical *vector* setMethod("[", signature(x = "Matrix", i = "matrix", j = "missing"),# drop="ANY"
349 ## -> [.data.frame uses nargs() - can we do this in the *generic* ? .M.sub.i.2col)
350
351
352 ### "[<-" : ----------------- ### "[<-" : -----------------
# Line 330 Line 355
355 setReplaceMethod("[", signature(x = "Matrix", i = "missing", j = "missing", setReplaceMethod("[", signature(x = "Matrix", i = "missing", j = "missing",
356 value = "ANY"),## double/logical/... value = "ANY"),## double/logical/...
357 function (x, value) { function (x, value) {
358 ## Fails for 'nMatrix' ... FIXME : make sure have method there
359 x@x <- value x@x <- value
360 validObject(x)# check if type and lengths above match validObject(x)# check if type and lengths above match
361 x x
362 }) })
363
364 ## Method for all 'Matrix' kinds (rather than incomprehensible error messages); ## A[ ij ] <- value, where ij is (i,j) 2-column matrix :
365 .M.repl.i.2col <- function (x, i, j, value)
366 {
367 nA <- nargs()
368 if(nA == 3) { ## M [ cbind(ii,jj) ] <- value
369 if(!is.integer(nc <- ncol(i)))
370 stop("'i' has no integer column number",
371 " should never happen; please report")
372 if(is.logical(i)) {
373 i <- c(i) # drop "matrix"
374 return( callNextMethod() )
375 } else if(!is.numeric(i) || nc != 2)
376 stop("such indexing must be by logical or 2-column numeric matrix")
377 m <- nrow(i)
378 mod.x <- .type.kind[.M.kind(x)]
379 if(length(value) > 0 && m %% length(value) != 0)
380 warning("number of items to replace is not a multiple of replacement length")
381 ## recycle:
382 value <- rep(value, length = m)
383 i1 <- i[,1]
384 i2 <- i[,2]
385 ## inefficient -- FIXME -- (also loses "symmetry" unnecessarily)
386 for(k in seq_len(m))
387 x[i1[k], i2[k]] <- value[k]
388 x
389
390 } else stop("nargs() = ", nA, " should never happen; please report.")
391 }
392 setReplaceMethod("[", signature(x = "Matrix", i = "matrix", j = "missing",
393 value = "replValue"),
394 .M.repl.i.2col)
395
396
397 setReplaceMethod("[", signature(x = "Matrix", i = "ANY", j = "ANY",
398 value = "Matrix"),
399 function (x, i, j, value)
400 callGeneric(x=x, i=i, j=j, value = as.vector(value)))
401 setReplaceMethod("[", signature(x = "Matrix", i = "ANY", j = "ANY",
402 value = "matrix"),
403 function (x, i, j, value)
404 callGeneric(x=x, i=i, j=j, value = c(value)))
405
406 ## (ANY,ANY,ANY) is used when no `real method' is implemented : ## (ANY,ANY,ANY) is used when no `real method' is implemented :
407 setReplaceMethod("[", signature(x = "Matrix", i = "ANY", j = "ANY", setReplaceMethod("[", signature(x = "Matrix", i = "ANY", j = "ANY",
408 value = "ANY"), value = "ANY"),
409 function (x, i, j, value) { function (x, i, j, value) {
410 if(!is.atomic(value)) if(!is.atomic(value))
411 stop("RHS 'value' must match matrix class ", class(x)) stop(sprintf("RHS 'value' (class %s) matches 'ANY', but must match matrix class %s",
412 class(value),class(x)))
413 else stop("not-yet-implemented 'Matrix[<-' method") else stop("not-yet-implemented 'Matrix[<-' method")
414 }) })
415
Legend:
Removed from v.1389 changed lines Added in v.1673 | 5,655 | 18,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-43 | latest | en | 0.433024 |
http://kr.mathworks.com/help/robust/ug/feedback-around-an-uncertain-plant.html?nocookie=true | 1,419,776,034,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447557824.148/warc/CC-MAIN-20141224185917-00056-ip-10-231-17-201.ec2.internal.warc.gz | 59,535,707 | 8,009 | Accelerating the pace of engineering and science
# Documentation
## Feedback Around an Uncertain Plant
It is possible to form interconnections of uss objects. A common example is to form the feedback interconnection of a given controller with an uncertain plant.
```gamma = ureal('gamma',4);
tau = ureal('tau',.5,'Percentage',30);
```
Next, create an unmodeled dynamics element, delta, and a first-order weighting function, whose DC value is 0.2, high-frequency gain is 10, and whose crossover frequency is 8 rad/sec.
```delta = ultidyn('delta',[1 1],'SampleStateDim',5);
W = makeweight(0.2,6,6);
```
Finally, create the uncertain plant consisting of the uncertain parameters and the unmodeled dynamics.
```P = tf(gamma,[tau 1])*(1+W*delta);
```
You can create an integral controller based on nominal plant parameters. Nominally the closed-loop system will have damping ratio of 0.707 and time constant of 2*tau.
```KI = 1/(2*tau.Nominal*gamma.Nominal);
C = tf(KI,[1 0]);
```
Create the uncertain closed-loop system using the feedback command.
```CLP = feedback(P*C,1);
```
Plot samples of the open-loop and closed-loop step responses. As expected the integral controller reduces the variability in the low frequency response.
```subplot(2,1,1);
stepplot(P,5)
subplot(2,1,2);
stepplot(CLP,5)
``` | 342 | 1,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2014-52 | latest | en | 0.771471 |
http://www.codecogs.com/library/maths/discrete/number_theory/bernoulli_b.php | 1,531,779,935,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00174.warc.gz | 432,346,647 | 9,276 | I have forgotten
• https://me.yahoo.com
COST (GBP)
1.00
0.00
0
# Bernoulli B
Calculates array of Bernoulli numbers using an infinite series
Controller: CodeCogs
C++
## Bernoulli B
voidbernoulli_B( int iMax double* dB )
Bernoulli numbers are particular values of Bernoulli polynomials . Polynomials are expandable in the Fourier series:
Substitution gives series expansion for Bernoulli numbers:
However, series like this is not quite suitable for numerical evaluation, especially at small n, in view of slow convergence and uneasy accuracy estimation. One needs to use another relation [2,23.10.21]:
Then Fourier series expansion results in the following:
Alternating series converges faster and gives possibility for simple accuracy estimation: error arising due to series truncation is less than first term neglected. Then N, the amount of series terms to obtain accuracy may be estimated as follows:
In the worst case n=2 reasonable accuracy results in series terms to sum. To significantly reduce a mount of summands and speed up convergence it is suitable to use Aitken transformation of partial series sums. Let
then sequence of new sums
may converge faster compared with . To avoid subtractions of near values and losses of significant digits expression should be rewritten as follows:
or, using
Array dimension should be iMax+1 or greater.
## References:
• Higher Transcendental Functions, vol.1, (1.13) by H.Bateman and A.Erdelyi (Bateman Manuscript Project), 1953
• M.Abramowitz and I.A.Stegun, Handbook of Mathematical Functions, 1964 chapt.23
• Yu.Luke, Mathematical functions and their approximations, 1975 chapt.14.2
### Example 1
#include <stdio.h>
#include <codecogs/maths/discrete/number_theory/bernoulli_b.h>
#define MAX_INDEX 16
int main()
{
double dBernoulli[MAX_INDEX+1];
printf( "%8s%2c%20s\n", " ", 'n', "Bn" );
printf( "%8s", " " );
for(int i = 0; i < 22; i++ )
printf( "%c", '-' );
printf( "\n" );
Maths::NumberTheory::bernoulli_B( MAX_INDEX, dBernoulli );
printf( "%10d%20.12f\n", 0, dBernoulli[0] );
printf( "%10d%20.12f\n", 1, dBernoulli[1] );
for(int i = 2; i <= MAX_INDEX; i += 2 )
printf( "%10d%20.12f\n", i, dBernoulli[i] );
return 0;
}
Output:
n Bn
----------------------
0 1.000000000000
1 -0.500000000000
2 0.166666666667
4 -0.033333333333
6 0.023809523810
8 -0.033333333333
10 0.075757575758
12 -0.253113553114
14 1.166666666667
16 -7.092156862745
### Parameters
iMax input maximal index requested dB output pointer on the array of numbers declared in the calling module.
### Authors
Anatoly Prognimack (Mar 16, 2005)
Developed tested with Borland C++ 3.1 for DOS and Microsoft Visual C++ 5.0, 6.0
Updated by Will Bateman (March 2005)
##### Source Code
Source code is available when you agree to a GP Licence or buy a Commercial Licence.
Not a member, then Register with CodeCogs. Already a Member, then Login. | 842 | 2,955 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-30 | latest | en | 0.786616 |
https://db0nus869y26v.cloudfront.net/en/Multiclass_classification | 1,719,353,140,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00032.warc.gz | 163,522,907 | 14,172 | In machine learning and statistical classification, multiclass classification or multinomial classification is the problem of classifying instances into one of three or more classes (classifying instances into one of two classes is called binary classification).
While many classification algorithms (notably multinomial logistic regression) naturally permit the use of more than two classes, some are by nature binary algorithms; these can, however, be turned into multinomial classifiers by a variety of strategies.
Multiclass classification should not be confused with multi-label classification, where multiple labels are to be predicted for each instance.
## General strategies
The existing multi-class classification techniques can be categorised into
• transformation to binary
• extension from binary
• hierarchical classification.[1]
### Transformation to binary
This section discusses strategies for reducing the problem of multiclass classification to multiple binary classification problems. It can be categorized into one vs rest and one vs one. The techniques developed based on reducing the multi-class problem into multiple binary problems can also be called problem transformation techniques.
#### One-vs.-rest
One-vs.-rest[2]: 182, 338 (OvR or one-vs.-all, OvA or one-against-all, OAA) strategy involves training a single classifier per class, with the samples of that class as positive samples and all other samples as negatives. This strategy requires the base classifiers to produce a real-valued score for its decision (see also scoring rule), rather than just a class label; discrete class labels alone can lead to ambiguities, where multiple classes are predicted for a single sample.[2]: 182 [note 1]
In pseudocode, the training algorithm for an OvR learner constructed from a binary classification learner L is as follows:
Inputs:
• L, a learner (training algorithm for binary classifiers)
• samples X
• labels y where yi ∈ {1, … K} is the label for the sample Xi
Output:
• a list of classifiers fk for k ∈ {1, …, K}
Procedure:
• For each k in {1, …, K}
• Construct a new label vector z where zi = yi if yi = k and zi = 0 otherwise
• Apply L to X, z to obtain fk
Making decisions means applying all classifiers to an unseen sample x and predicting the label k for which the corresponding classifier reports the highest confidence score:
${\displaystyle {\hat {y))={\underset {k\in \{1\ldots K\)){\arg \!\max ))\;f_{k}(x)}$
Although this strategy is popular, it is a heuristic that suffers from several problems. Firstly, the scale of the confidence values may differ between the binary classifiers. Second, even if the class distribution is balanced in the training set, the binary classification learners see unbalanced distributions because typically the set of negatives they see is much larger than the set of positives.[2]: 338
#### One-vs.-one
In the one-vs.-one (OvO) reduction, one trains K (K − 1) / 2 binary classifiers for a K-way multiclass problem; each receives the samples of a pair of classes from the original training set, and must learn to distinguish these two classes. At prediction time, a voting scheme is applied: all K (K − 1) / 2 classifiers are applied to an unseen sample and the class that got the highest number of "+1" predictions gets predicted by the combined classifier.[2]: 339
Like OvR, OvO suffers from ambiguities in that some regions of its input space may receive the same number of votes.[2]: 183
### Extension from binary
This section discusses strategies of extending the existing binary classifiers to solve multi-class classification problems. Several algorithms have been developed based on neural networks, decision trees, k-nearest neighbors, naive Bayes, support vector machines and extreme learning machines to address multi-class classification problems. These types of techniques can also be called algorithm adaptation techniques.
#### Neural networks
Multiclass perceptrons provide a natural extension to the multi-class problem. Instead of just having one neuron in the output layer, with binary output, one could have N binary neurons leading to multi-class classification. In practice, the last layer of a neural network is usually a softmax function layer, which is the algebraic simplification of N logistic classifiers, normalized per class by the sum of the N-1 other logistic classifiers. Neural Network-based classification has brought significant improvements and scopes for thinking from different perspectives [3], [4].
##### Extreme learning machines
Extreme learning machines (ELM) is a special case of single hidden layer feed-forward neural networks (SLFNs) wherein the input weights and the hidden node biases can be chosen at random. Many variants and developments are made to the ELM for multiclass classification.
#### k-nearest neighbours
k-nearest neighbors kNN is considered among the oldest non-parametric classification algorithms. To classify an unknown example, the distance from that example to every other training example is measured. The k smallest distances are identified, and the most represented class by these k nearest neighbours is considered the output class label.
#### Naive Bayes
Naive Bayes is a successful classifier based upon the principle of maximum a posteriori (MAP). This approach is naturally extensible to the case of having more than two classes, and was shown to perform well in spite of the underlying simplifying assumption of conditional independence.
#### Decision trees
Decision tree learning is a powerful classification technique. The tree tries to infer a split of the training data based on the values of the available features to produce a good generalization. The algorithm can naturally handle binary or multiclass classification problems. The leaf nodes can refer to any of the K classes concerned.
#### Support vector machines
Support vector machines are based upon the idea of maximizing the margin i.e. maximizing the minimum distance from the separating hyperplane to the nearest example. The basic SVM supports only binary classification, but extensions have been proposed to handle the multiclass classification case as well. In these extensions, additional parameters and constraints are added to the optimization problem to handle the separation of the different classes.
#### Multi expression programming
Multi expression programming (MEP) is an evolutionary algorithm for generating computer programs (that can be used for classification tasks too). MEP has a unique feature: it encodes multiple programs into a single chromosome. Each of these programs can be used to generate the output for a class, thus making MEP naturally suitable for solving multi-class classification problems.
### Hierarchical classification
Hierarchical classification tackles the multi-class classification problem by dividing the output space i.e. into a tree. Each parent node is divided into multiple child nodes and the process is continued until each child node represents only one class. Several methods have been proposed based on hierarchical classification.
Based on learning paradigms, the existing multi-class classification techniques can be classified into batch learning and online learning. Batch learning algorithms require all the data samples to be available beforehand. It trains the model using the entire training data and then predicts the test sample using the found relationship. The online learning algorithms, on the other hand, incrementally build their models in sequential iterations. In iteration t, an online algorithm receives a sample, xt and predicts its label ŷt using the current model; the algorithm then receives yt, the true label of xt and updates its model based on the sample-label pair: (xt, yt). Recently, a new learning paradigm called progressive learning technique has been developed.[5] The progressive learning technique is capable of not only learning from new samples but also capable of learning new classes of data and yet retain the knowledge learnt thus far.[6] | 1,561 | 8,088 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-26 | latest | en | 0.896641 |
http://cycling74.com/forums/topic/major-scale-randomisation/ | 1,397,988,820,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00398-ip-10-147-4-33.ec2.internal.warc.gz | 56,769,180 | 19,140 | ## major scale randomisation
Nov 25, 2011 at 7:46pm
# major scale randomisation
anyone know how would i get the random object to generate a value between 1-12 based on the major (TTSTTTS)scale. The notes i need are 1,3,5,6,8,10,12.
any help would be appreciated.
#60258
Nov 25, 2011 at 8:04pm
Since the pattern of T and S in a major scale is not easily obtained by a mathematical formula, it’s best just to store it in a table. Here’s an example using the numbering system 0,2,4,5,7,9,11. Just add 1 to that to get your numbering system. The example also includes a formula to generate a stack of fifths in Lydian mode, which you can convert into an out-of-order C major scale, but the table lookup method is more efficient and straightforward. Add some whole-number multiple of 12 to transpose these pitches to different octaves, and add some number of semitones to transpose to a different key.
– Pasted Max Patch, click to expand. –
#216883
Nov 25, 2011 at 8:12pm
I’m not at a computer with max on but from memory something like [random 7] to [zl lookup 0 2 4 5 7 9 11] should get you started
#216884
Nov 25, 2011 at 8:15pm
fun :)
you want [random] to make random index numbers, which look up the scale degrees you want, rather than try and generate the numbers themselves. Then you can load a [umenu] or other list-based data structure (table, multislider, etc.) with as many scales as you want:
(edit: zl lookup eliminated a half-dozen objects in my original patch…doh :)
– Pasted Max Patch, click to expand. –
#216885
Nov 25, 2011 at 8:29pm
I have a “jump into the deep end of the pool” note / scale quantizer called ScaleMaster, which can be found here: http://www.xfade.com/max/ScaleMaster/
Also Christopher, in your patcher, the table wasn’t set to save data with patcher.
#216886
Nov 25, 2011 at 8:46pm
amazing thanks. i’ll try this when i get home.
#216887
Nov 25, 2011 at 11:51pm
@Chris Muir: Thanks. Felt sure I had done that, but maybe I hadn’t yet saved the patcher. Here’s a revised version:
– Pasted Max Patch, click to expand. –
#216888
Nov 26, 2011 at 12:35am
Or simpler, with [zl scramble]. It automatically scrambles everything from a list, and you code less.
#216889
You must be logged in to reply to this topic. | 649 | 2,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2014-15 | longest | en | 0.926531 |
https://multi-converter.com/tablespoons-us-to-teaspoons-metric | 1,723,217,088,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640767846.53/warc/CC-MAIN-20240809142005-20240809172005-00219.warc.gz | 327,137,351 | 7,190 | # Tablespoons (US) to Teaspoons (Metric)
Change to Teaspoons (Metric) to Tablespoons (US)
Share:
## How to convert Tablespoons (US) to Teaspoons (Metric)
1 [Tablespoons (US)] = 2.95735296 [Teaspoons (Metric)]
[Teaspoons (Metric)] = [Tablespoons (US)] * 2.95735296
To convert Tablespoons (US) to Teaspoons (Metric) multiply Tablespoons (US) * 2.95735296.
## Example
67 Tablespoons (US) to Teaspoons (Metric)
67 [Tablespoons (US)] * 2.95735296 = 198.14264832 [Teaspoons (Metric)]
## Conversion table
Tablespoons (US) Teaspoons (Metric)
0.01 Tablespoons (US)0.0295735296 Teaspoons (Metric)
0.1 Tablespoons (US)0.295735296 Teaspoons (Metric)
1 Tablespoons (US)2.95735296 Teaspoons (Metric)
2 Tablespoons (US)5.91470592 Teaspoons (Metric)
3 Tablespoons (US)8.87205888 Teaspoons (Metric)
4 Tablespoons (US)11.82941184 Teaspoons (Metric)
5 Tablespoons (US)14.7867648 Teaspoons (Metric)
10 Tablespoons (US)29.5735296 Teaspoons (Metric)
15 Tablespoons (US)44.3602944 Teaspoons (Metric)
50 Tablespoons (US)147.867648 Teaspoons (Metric)
100 Tablespoons (US)295.735296 Teaspoons (Metric)
500 Tablespoons (US)1478.67648 Teaspoons (Metric)
1000 Tablespoons (US)2957.35296 Teaspoons (Metric) | 424 | 1,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.70164 |
https://www.yumpu.com/en/document/view/8543587/a-system-of-physical-chemistry-index-of/176 | 1,601,299,522,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401600771.78/warc/CC-MAIN-20200928104328-20200928134328-00008.warc.gz | 1,022,497,298 | 25,954 | # A system of physical chemistry - Index of
A system of physical chemistry - Index of
1 62 A SYSTEM OF PHYSICAL CHEMISTRY
Hence N = NA3( J^f.
Hence A = Jyfi^.
We can also calculate y in terms of the total kinetic energy of the
molecular motion. For the fraction of the whole, i.e. fractional number
of the whole number of molecules, which is represented by —
have each kinetic energy \m{u'^ + z/^ + ze;"'') where m is the mass of a
molecule.
Hence the total kinetic energy is—
mK^m •+CO
du
+CO r+00 _y(«2-t-t)2-(.z£,!8)
dv dw . {u- + v'^ + . e
iv"-) dudzdw
-oo J — oo
Leaving aside the constant factor for the moment, we see that the triple
integral is the sum of three triple integrals, one of which, for instance,
'+00
tih-y''"du
Now it is known that— r+
+=»
e-y'^dv
+00
ti •^
. e-y^'dic = I Jirlf
e-y'^'^dw.
and the two remaining integral factors in the above chosen integral havt
already been evaluated. Hence the selected member of the three triple
integrals has the value -^ J^^ly^, and the other two have each the same
value. Hence the total kinetic energy is—
|NA^w jTT^Iy", i.e. fNw/y ; since A = v/y/''".
Now we could conceive the molecules all moving with one uniform
velocity as a rigid body, for instance, and that velocity such as to give
the same kinetic energy as that due to the gaseous motion. Denote
the magnitude of this hypothetical velocity by c, and we have—
-^Nwf^ = |Nz«/y,
or y = 3/2^^
This particular speed c is not the true average speed, nor is it the
maximum probability speed referred to above (its relation with these
will be it is given presently)
in fact a ;
speed whose square is the average
of the squares of the molecular speeds at one instant, and on that
account is called the root-mean-square or r-m-s-speed. Assuming that
this is r-m-s-speed known, we formulate Maxwell's Law in one way
thus :—
Of a great number N of gaseous molecules in a steady state with an
r-m-s-speed c, the number whose velocity components at one instant
lie between the limits u io u + du, v to v + dv, w to w + dw, is—
N .
• e-^(^+^''+^^)l^^-dudvdw . . . (2)
j'g^
More magazines by this user
Similar magazines | 628 | 2,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.909054 |
https://www.wordunscrambler.net/?word=excel | 1,618,096,259,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038059348.9/warc/CC-MAIN-20210410210053-20210411000053-00542.warc.gz | 1,208,300,666 | 11,166 | # Unscramble EXCEL
## How Many Words can be Made From EXCEL?
Above are the words made by unscrambling E X C E L (CEELX). Our unscramble word finder was able to unscramble these letters using various methods to generate 10 words! Having a unscramble tool like ours under your belt will help you in ALL word scramble games!
How many words can you make out of EXCEL?
To further help you, here are a few word lists related to the letters EXCEL
## If You Unscramble EXCEL... What Does It Mean?
### Definition of EXCEL When Unscrambled
If we unscramble these letters, EXCEL, it and makes several words. Here is one of the definitions for a word that uses all the unscrambled letters:
### Excel
• To surpass others in good qualities, laudable actions, or acquirements; to be distinguished by superiority; as, to excel in mathematics, or classics.
• To exceed or go beyond; to surpass.
• To go beyond or surpass in good qualities or laudable deeds; to outdo or outgo, in a good sense.
• Is Excel a Scrabble Word?
• is excel a Words With Friends word?
## Scrambling the Letters in EXCEL
According to our other word scramble maker, EXCEL can be scrambled in many ways. The different ways a word can be scrambled is called "permutations" of the word.
According to Google, this is the definition of permutation:
a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged.
How is this helpful? Well, it shows you the anagrams of excel scrambled in different ways and helps you recognize the set of letters more easily. It will help you the next time these letters, E X C E L come up in a word scramble game.
We stopped it at 11, but there are so many ways to scramble EXCEL!
### Word Scramble Words
#### Unscramble these letters to make words...
scrambled using word scrambler...
## Popular Words With Scrambled Letters
### Combine Words
Bird + Duck = Bick
Apple + Honor = Aplonor
Hand + Locker = Handocker | 479 | 1,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.900456 |
https://nursingtutorsnearme.com/given-these-variables-and-design-what-information-would-a-factorial-manova-provide-you-psychology-homework-help/ | 1,656,652,742,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00208.warc.gz | 502,953,837 | 12,030 | # Given these variables and design, what information would a factorial MANOVA provide you, psychology homework help
Considering your I/O psychology research interests, identify two IVs (each with 2 categories) and three DVs that are measured on continuous scales. Think of DV measures that probably are moderately correlated with each other because they are measuring different components of the same or similar concepts (e.g., three different measures of academic performance). Given these variables and design, what information would a factorial MANOVA provide you? What more would you want to know if you get significant results in the factorial MANOVA? Why would this be significant
to your research? (Research support is not required for this question.)
I have provided and example of how the questions needs to be answered.
IV : Gender IV 1 = Male Veteran and IV 2 = Female Veteran in the following 2 groups
Category 1 – level of experience (entry, mid, experienced)
Category 2 – college degree (none, Bachelors, Masters)
DV is: Hiring data for:
– Veteran hiring rate and
– number of months taken to get hired
________________________________________________________________________
What information would a factorial MANOVA provide you?
A factorial MANOVA design can look at the relationships between the main effects of each of the IVs, and their interaction with the DV. With the Factorial MANOVA the interaction effect is how the combinations of levels of the IVs influence the dependent variables. The MANOVA can identify main effect as well as interaction effects between the IV’s and the DV. The main effect would tell us if there is a difference between our IV and all of DV’s. Since we have two IV’s with two levels each the factorial MANOVA can
examine effects across all other variables and evaluate significant effects. The interaction effect can exist between factors such as there an effect between gender with college
degree and level of experience effect hiring rate or any of the other DV’s; determining significant effects on all of the dimensions. Thus we can identify the relationships with the most significant influence. This helps us to answer if the IV has a significant impact on the DV, how does the IV relate to the DVs and what are the separate relationships for each variable and the power of the effect on each.
What more would you want to know if you getsignificant results in the factorial MANOVA?
A significant result can be seen in a factorial MANOVA in through the separate F-values generated for the main effect and the interaction effect between IV we are using such as
with the Wilks or Pillai’s trace criterion to evaluate the significance level. When we look at the significance level and find it to be statistically significant we reject our null hypothesis since the DV was significantly affected. We can also look at the partial eta square and see if the effect had a large or small effect on the variable. In my example I could have a significant effect between the IV level of experience and the DV salary range with a small effect, as where I may have a statistically significant effect on IV with
college degree and salary range with a large effect size. I would then see that both effected the DV but that college degree has more influence on salary then does level of
experience.
Why would this be significant to your research?
When there are several similar (correlated) DV’s the MANOVA allows me to analyze single variables without running multiple individual test and causing type I errors. It also allows me to investigate how the IV’s influence each of DV singularly and collectively. However, additional Post Hoc test need to be run for this level of comparison. Understanding how each variable differs from each other and how they may overlap in their influence we can see to what degree they correlate and how much one may affect the other or how the combination effects it.
0 replies | 810 | 3,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.924879 |
https://www.coursehero.com/file/6239785/PHY-118-final-study-guide-chapter-6/ | 1,512,986,711,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513330.14/warc/CC-MAIN-20171211090353-20171211110353-00662.warc.gz | 731,326,801 | 70,622 | PHY 118 final study guide -chapter 6
# PHY 118 final study guide -chapter 6 - PHY 118 Final Study...
This preview shows pages 1–2. Sign up to view the full content.
PHY 118 Final Study Guide: Chapter 6 I. Air Pressure: a. Measured in milibars (mb) which equal 100 Newtons per square meter b. Mercury Barometer, Aneroid Barometer, and Barograph: c. Decreasing pressure is associated with increasing cloudiness and precipitation d. Increasing pressure is associated with clearing conditions e. Increase in Density = Increase in Pressure (with constant temperature) f. Increase in Altitude = Decrease in Pressure and Density II. Horizontal Variations in Air Pressure: a. Small variations can generate violent winds b. Temperature + Water Vapor: i. Warmer temperature = lower pressure; Colder temperature = higher pressure 1. In conjunction with this: Warm air = less dense; Cold air = more dense ii. Increase in Water Vapor in the air = Decrease of air’s Density 1. (Nitrogen and oxygen weigh more than water vapor) 2. Humid Air is Less Dense than Dry Air iii. Cold + Dry air mass causes higher surface pressures than a warm + moist air mass iv. Warm + Dry air mass causes higher surface pressures than the same temperature + moist air mass c. Airflow + Pressure: i. Convergence (net flow into an area) causes air to increase its altitude and create a heavier air mass = Increase pressure ii. Divergence (net flow out of area) causes air to spread out horizontally and create a lighter air mass = Decrease pressure III.Factors that Affect Wind: (The horizontal movement of air) a. Result of horizontal differences in pressure i. Differences are caused by solar radiations unequal heating of the planet b. Pressure-Gradient Force (P-GF): i. Wind moves from areas of high pressure to areas of low pressure ii. The greater the pressure differences between areas, the greater the speeds of the wind iii. Isobars – Lines on a map that connect areas of equal pressure
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 3
PHY 118 final study guide -chapter 6 - PHY 118 Final Study...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 540 | 2,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-51 | latest | en | 0.816854 |
https://us.metamath.org/ileuni/imaeq1i.html | 1,721,403,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00081.warc.gz | 526,389,915 | 3,644 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > imaeq1i GIF version
Theorem imaeq1i 4885
Description: Equality theorem for image. (Contributed by NM, 21-Dec-2008.)
Hypothesis
Ref Expression
imaeq1i.1 𝐴 = 𝐵
Assertion
Ref Expression
imaeq1i (𝐴𝐶) = (𝐵𝐶)
Proof of Theorem imaeq1i
StepHypRef Expression
1 imaeq1i.1 . 2 𝐴 = 𝐵
2 imaeq1 4883 . 2 (𝐴 = 𝐵 → (𝐴𝐶) = (𝐵𝐶))
31, 2ax-mp 5 1 (𝐴𝐶) = (𝐵𝐶)
Colors of variables: wff set class Syntax hints: = wceq 1332 “ cima 4549 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-io 699 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1483 ax-10 1484 ax-11 1485 ax-i12 1486 ax-bndl 1487 ax-4 1488 ax-17 1507 ax-i9 1511 ax-ial 1515 ax-i5r 1516 ax-ext 2122 This theorem depends on definitions: df-bi 116 df-3an 965 df-tru 1335 df-nf 1438 df-sb 1737 df-clab 2127 df-cleq 2133 df-clel 2136 df-nfc 2271 df-v 2691 df-un 3079 df-in 3081 df-ss 3088 df-sn 3537 df-pr 3538 df-op 3540 df-br 3937 df-opab 3997 df-cnv 4554 df-dm 4556 df-rn 4557 df-res 4558 df-ima 4559 This theorem is referenced by: mptpreima 5039 isarep2 5217 infrenegsupex 9415 infxrnegsupex 11063 ssidcn 12416 cnco 12427
Copyright terms: Public domain W3C validator | 697 | 1,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-30 | latest | en | 0.132329 |
https://www.unitconverters.net/fuel-consumption/kilometer-gallon-us-to-meter-cubic-foot.htm | 1,718,943,611,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00849.warc.gz | 893,212,347 | 3,177 | Home / Fuel Consumption Conversion / Convert Kilometer/gallon (US) to Meter/cubic Foot
Convert Kilometer/gallon (US) to Meter/cubic Foot
Please provide values below to convert kilometer/gallon (US) to meter/cubic foot [m/ft^3], or vice versa.
From: kilometer/gallon (US) To: meter/cubic foot
Kilometer/gallon (US) to Meter/cubic Foot Conversion Table
Kilometer/gallon (US)Meter/cubic Foot [m/ft^3]
0.01 kilometer/gallon (US)74.8051948202 m/ft^3
0.1 kilometer/gallon (US)748.051948202 m/ft^3
1 kilometer/gallon (US)7480.5194820199 m/ft^3
2 kilometer/gallon (US)14961.03896404 m/ft^3
3 kilometer/gallon (US)22441.55844606 m/ft^3
5 kilometer/gallon (US)37402.5974101 m/ft^3
10 kilometer/gallon (US)74805.194820199 m/ft^3
20 kilometer/gallon (US)149610.3896404 m/ft^3
50 kilometer/gallon (US)374025.974101 m/ft^3
100 kilometer/gallon (US)748051.94820199 m/ft^3
1000 kilometer/gallon (US)7480519.4820199 m/ft^3
How to Convert Kilometer/gallon (US) to Meter/cubic Foot
1 kilometer/gallon (US) = 7480.5194820199 m/ft^3
1 m/ft^3 = 0.0001336806 kilometer/gallon (US)
Example: convert 15 kilometer/gallon (US) to m/ft^3:
15 kilometer/gallon (US) = 15 × 7480.5194820199 m/ft^3 = 112207.7922303 m/ft^3 | 457 | 1,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-26 | latest | en | 0.424699 |
http://gizmodo.com/tag/math | 1,505,844,913,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685912.14/warc/CC-MAIN-20170919164651-20170919184651-00613.warc.gz | 156,315,708 | 97,787 | Why Computers Are Having Such a Hard Time With This Deceptively Simple Chess Puzzle
A popular chess problem known as the Queen’s Puzzle has captivated mathematicians and computer scientists for years, yet no one has been able to write a computer program that can solve the conundrum quickly and efficiently. Researchers from the UK now claim that computers will never be up to the task—and they’re…
Scientists Reignite Thirty-Year-Old Debate About Glass With New Calculation
Mathematics is far more fraught with debate and disagreement than you might imagine. Arguments about things some of the smartest physicists have trouble understanding rage for years. Recently, a pair of mathematicians ignited some old flames—or rather, shattered some glass—with a new set of results that, if correct,…
How Many Chicken McNuggets Are in the Big Chicken?
By now, you’ve definitely seen or at least heard of The Big Chicken. He is large. He contains multitudes. And he is most certainly real.
Reminder: Mike Pence Voted Against Recognizing Pi Day
Every year on March 14th, the nerd community gathers ‘round to celebrate the beloved mathematical constant pi. We know that pi is so much more than the ratio of a circle’s circumference to its diameter—it’s critical to understanding the best things in life, which are all circular. Pizza, for example, is an excellent…
These Black Female Mathematicians Should Be Stars in the Blockbusters of Tomorrow
The hallways of math and science history are overflowing with the achievements of white men, from Sir Isaac Newton to Steve Jobs; their faces are printed into elementary school textbooks everywhere, and their achievements have been indelibly drilled into our minds, with countless awards and institutions named after…
10 Tips to Improve Your Mental Math Ability
Calculators are awesome, but they’re not always handy. More to the point, no one wants to be seen reaching for the calculator on their mobile phone when it’s time to figure out a 15 percent gratuity. Here are ten tips to help you crunch numbers in your head.
A Quick Animated History of Why We Use the Numbers We Use
I find the history of numbers so much more fascinating than the application of numbers. Who cares about learning calculus, when you can geek out on the brief history of numerical systems?
If Einstein Is Right Santa Is Even Fatter Than We Thought
There’s a controversial little interpretation of Einstein’s theory of special relativity that could affect what happens to masses moving at a really high speeds: they appear to get heavier.
This Paper Calculator Adds Via a Series of Tubes
Human beings have always created tools for calculating numbers, from the ancient abacus to today’s electronic calculators. But here’s an ingenious calculator drawn on paper—the creation of comic book artist Jason Shiga. And he and the folks at Numberphile have created an explanatory video of how it works.
How Futurama Made Math and Science Funny
Futurama may not make me laugh as hard as other comedies, but its vision of the future and all the shenanigans that come with it have always been enjoyable to watch (throughout all its various cancellations and comebacks). Kaptain Kristian makes the case that Futurama is special because it was the “master of hiding…
These Mysterious Ultra-Rare Crystals Probably Formed in Outer Space
Quasicrystals are unusual materials in which the atoms are arranged in regular patterns that nonetheless never repeat themselves. Most are man-made in the lab; only one case of naturally occurring quasicrystals has been found thus far. And now physicists believe they’ve figured out how that happened.
Sneaky Trick Uses Math Magic to Guess Your Cards
Math is basically magic. So it’s no surprise that a clever use of the Fibonacci numbers—a series of numbers (1, 2, 3, 5, 8, etc.) where each number is the sum of the two preceding numbers—and a super-slick shuffling method can combine for a card trick that makes it impossibly easy to guess the number and suit of the…
Digital Music Couldn't Exist Without the Fourier Transform
This is the Fourier Transform. You can thank it for providing the music you stream every day, squeezing down the images you see on the Internet into tiny little JPG files, and even powering your noise-canceling headphones. Here’s how it works.
The World's Biggest Ever Math Proof Is a Whopping 200TB
If you think you had a hard time filling out pages of algebra at school, spare a thought for the three mathematicians who have just published the world’s largest ever proof. It takes up 200TB of storage space.
A Harry Potter Theory That Both Explains Harry's Tiny Class and Will Ruin Your Day
Ever wonder how there could only be about 18 Gryffindors in Harry’s Hogwarts class and that J.K. Rowling has said there were 1,000 students at the school? Well, a fan theory has an answer... a depressing, depressing answer.
Earth's Core Is Younger Than We Thought
Without a Tardis, a journey to the center of the Earth might be your best option for traveling to the past. Because of the way gravity warps spacetime, physicists have now calculated that the Earth’s core is 2.5 years younger than its surface.
The Mathematical Explanation for Why You Can't Catch a Falling Dollar Bill with Your Fingers
Here’s a really interesting mathematical explanation on how the “catch a dollar” trick works. You know the trick: a person holds a bill vertically and says you can keep the bill if you can catch the dollar with your fingers when it drops. You never catch it. It’s really hard! Why?
Truly Interesting Animation Breaks Down the Story Behind Zero and Why It's So Important
The story of how zero came to be and the history of math is actually quite fascinating! They should have taught us that instead of actual math in high school, if you ask me. Thankfully, Hannah Fry tells us in the animation story below all we need to know. There’s fascinating bits about how the number system (and zero)…
Now We Know How Many Ways We Can Arrange 128 Tennis Balls
Here’s a question worthy of the ball boy at Wimbledon: if you have 128 tennis balls packed into a container, how many different ways can you arrange them? Answer: 10250 — more than the entire number of subatomic particles in the universe. | 1,313 | 6,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-39 | latest | en | 0.937277 |
https://oikofuge.com/category/software/page/2/ | 1,720,963,127,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00809.warc.gz | 384,395,501 | 41,991 | # Which Place Gets The Most Daylight?
So this puzzle isn’t about sunshine (the amount of time the sun shines from a clear sky), or even about the intensity of sunlight (which decreases with increasing latitude), but about cumulative daylight—the length of time between sunrise and sunset in a given place, added up over the course of a year.*
It’s a surprisingly complicated little problem. I addressed it using an antique solar calculator I wrote many years ago, using Peter Duffett-Smith’s excellent books as my primary references:
It runs in Visual Basic 6, which means I had to open up my VirtualBox virtual XP machine to get it running again. The original program calculates the position of the sun by date and time for any given set of coordinates, and also works out the times of sunrise and sunset.
You’ll see it gives sunrise and sunset times to one-second precision, which is entirely spurious—the refractive state of the atmosphere is so variable that there’s no real point in quoting these times to anything beyond the nearest minute. I just couldn’t bring myself to hide the extra column of figures.
Anyway, it was a fairly quick job to write a little routine that cycled this calculator through a full year of daylight, adding up the total and spitting out the results so that I could begin exploring the problem.
At first glance, it seems like there shouldn’t be any particular place that wins out. As the Earth moves around the sun, its north pole is alternately tilted towards the sun and away from it, at an angle of about 23.5º. If we look at a diagram of these two solstice points (which occur in June and December every year), there’s an obvious symmetry between the illuminated and unilluminated parts of the globe:
Between the solstices, the latitude at which the sun is overhead varies continuously from 23.5ºN (in June) to 23.5ºS (in December), and then back again:
So for every long summer day, there should be an equal and opposite long winter night. The short and long days should average out, during the course of a year, to half a day’s daylight per day—equivalent to 4280 hours in a 365-day calendar year.
And that would be true if the Earth’s orbit around the sun was precisely circular—but it isn’t. As I described in my first post about the word perihelion, the Earth is at its closest to the sun in January, and its farthest in July. Since it moves along its orbit more quickly when it’s closer to the sun, it passes through the December solstice faster than through the June solstice. This has the effect of shortening the southern summer and the northern winter. The effect isn’t immediately obvious in my diagram of solar latitudes, above, but it’s there—the sun spends just 179 days in the southern sky, but 186 days north of the equator.
This means that the total number of hours of daylight is biased towards the northern hemisphere. In the diagram below, I plot the hypothetical flat distribution of daylight hours associated with a circular orbit in purple, and compare it to the effect of Earth’s real elliptical orbit in green:
So far, I’ve been treating the sun as if it were a point source of light, rising and setting in an instant of time. But the real sun has a visible disc, about half a degree across in the sky. This means that when the centre of the sun drops below the horizon, it’s only halfway through setting. Sunrise starts when the upper edge of the sun first appears; sunset finishes when the the upper edge of the sun disappears. So the extent of the solar disc slightly prolongs daylight hours, and slightly shortens the night.
At the equator the sun rises and sets vertically, and the upper half of the solar disc takes about a minute to appear or disappear. An extra minute of daylight in the morning, an extra minute of daylight in the evening—that’s more than twelve hours extra daylight during the course of a year, just because the sun is a disc and not a point.
And if we move north or south of the equator, the sun rises and sets at an angle relative to the horizon, and so takes longer to appear and disappear—adding more hours to the total daylight experienced at higher latitudes. There’s a limit to this effect, however. When we get to the polar circles, we run into the paired phenomena of the midnight sun and the polar night. There are days in summer when the sun never sets, and days in winter when the sun never rises. The extent of the solar disc can make no difference to the length of daylight if the sun is permanently above the horizon, and it can add only a few hours to the total as the sun skims below the horizon at the start and end of polar night. And as we move towards the poles, the midnight sun and polar night start to dominate the calendar, with only short periods around the equinoxes that have a normal day/night cycle. So although the sunrises and sunsets within the polar circles are notably prolonged, there are fewer of them.
So the prolongation of daylight caused by the rising and setting of the solar disc increases steadily with latitude until it peaks at the polar circles (around 66.5ºN and 66.5ºS), after which it declines again. Here’s a diagram of daylight hours predicted for a point-like sun (my previous green curve) with the effect of the solar disc added in red:
And there’s another effect to factor in at this point—atmospheric refraction. As I described in my post discussing the shape of the low sun, light from the rising and setting sun follows a slightly curved path through the atmosphere, lifting the image of the sun by a little over half a degree above its real position. This means that when we see the sun on the horizon, its real position is actually below the horizon. This effect hastens the sunrise and delays the sunset, and it does so in a way that is identical to simply making the solar disc wider—instead of just an extra couple of minutes’ daylight at the equator, more than six minutes are added when refraction is factored in, with proportional increases at other latitudes. So here’s a graph showing the original green curve of a point-like sun, the red curve showing the effect of the solar disc, and a blue curve added to show the effect of refraction, too:
The longest cumulative daylight is at the Arctic Circle, with latitude 66.7ºN experiencing 4649 hours of daylight in the year 2017. The shortest period is at the south pole, with just 4388 hours. That’s almost eleven days of a difference!
So is the answer to my original question just “the Arctic Circle”? Well, no. I have one more influence on the duration of daylight to deploy, and this time it’s a local one—altitude. The higher you go, the lower the horizon gets, making the sun rise earlier and set later. This only works if you have a clear view of a sea-level (or approximately sea-level) horizon—from an aircraft or the top of a mountain. Being on a high plateau doesn’t work, because your horizon is determined by the local terrain, rather than the distant curvature of the Earth. So although the south pole has an altitude of 2700m, it’s sitting in the middle of the vast polar plateau, and I think there will be a minimal effect from altitude on the duration of its daylight.
So we need to look for high mountains close to the Arctic Circle. A glance at the map suggests four mountainous regions that need to be investigated—the Cherskiy Range, in eastern Siberia; the Scandinavian Mountains; Greenland; and the region in Alaska where the Arctic Circle threads between the Brooks Range to the north and the Alaska Range to the south.
The highest point in the Cherskiy Range is Gora Pobeda (“Victory Peak”). At 65º11′N and 3003m, its summit gets 5002 hours of daylight—almost an hour a day of extra sunlight, on average.
But Pobeda is nudged out of pole position in the Cherskiy Range by an unnamed 2547m summit on the Chemalginskiy ridge, which lies almost exactly on the Arctic Circle, giving it a calculated 5006 hours of daylight.
There’s nothing over 2000m near the Arctic Circle in the Scandinavian Mountains, so we can skip past them to 3383m Mount Forel, in Greenland, at 66º56′N, which beats the Siberian mountains with 5052 hours of daylight.
Finally, the Arctic Circle passes north of Canada’s Mackenzie Mountains, and between the Brooks and Alaska Ranges. Mount Isto, the highest point in the Brooks Range, is 2736m high at 69º12′N, and comes in just behind Pobeda, with 4993 hours of daylight. Mount Igikpak, lower but nearer the Circle (2523m, 67º25′N), pushes past all the Siberian summits to hit 5010 hours. And in the Alaska Range is Denali, the highest mountain in North America. It is 6190m high, and sits at 63º04′N. It could have been a serious contender if it had been just a little farther north—but as it is it merely equals Igikpak, and falls short of Forel’s total.
So the answer to my question appears to be that the summit of Mount Forel, Greenland, sees the most daylight of any place on the planet. I confess I didn’t see that one coming when I started thinking about this.
* “A year” is a slightly slippery concept in this setting. The sun doesn’t return to exactly the same position in the sky at the end of each calendar year, and leap years obviously contain an extra day’s daylight compared to ordinary years. Ideally I should have added up my hours of daylight over a few millennia—but I’m really just interested in the proportions, and they’re not strongly influenced by the choice of calendar year. So for simplicity I ran my program to generate data for 2017 only.
What I wrote at the start of this piece, about spurious precision in rising and setting times, goes doubly for the calculations concerning altitude. These results are exquisitely sensitive to the effects of variable refraction, and my post about the shape of the low sun gives a lot more detail about how the polar regions are home to some surprising mirages that prolong sunrises and sunsets. I can’t hope to account for local miraging, or even to correctly reproduce the temperature gradient in the atmosphere from day to day. I think the best that can really be said is that some of the contenders I list will experience more daylight than anywhere else on the planet, most years, and that Mount Forel will be in with a good shot of taking the record for any given year.
# Running Windows XP Under VirtualBox
As I write, it’s only another month until Microsoft’s free upgrade offer on Windows 10 expires (on 29 July 2016). I am so looking forward to that day, in the hope that it’ll mean an end to Microsoft’s intrusive little pop-up messages in the lower right corner of my monitor, and their increasingly devious attempts to trick me into accidentally upgrading. The experience is a little like having a weeping software engineer repeatedly grip your lapels, shake you gently, and sob: “But it’s cool. It’s free. Why don’t you want it? For pity’s sake, why? Why?
I don’t want it because it offers nothing I need or am even curious to see, while promising inconvenience and hassle during the upgrade process. When Microsoft are willing to pay me for the time I’ll spend on their “free” upgrade, then maybe we can talk.
But at present, I’m much cheered to be running Windows XP again, which marked the last occasion I ever felt a new Microsoft operating system actually constituted an “upgrade”. The fact that you can run a virtual XP machine in a window under Windows 7 or later is not as well known as it should be, and it has let me continue using some old software from that era that has resisted running in “compatibility mode” under later operating systems.
Text revised January 2021: Microsoft used to offer a Windows XP Mode for Windows 7 downloadable from the Microsoft Download Center, containing a virtual hard drive with an XP installation, but it has since been removed. However, the “Windows XP Mode” file WindowsXPMode_en-us.exe is still available from various download sites, such as CNET. But please be aware of the potential for downloading malware from the sort of sites that offer copies of discontinued files.
To get XP running under later versions of Windows, all you have to do is pull the virtual hard drive out of the WindowsXPMode_en-us.exe file. Since I wanted to run XP under a version of Windows later than Windows 7, I didn’t run the executable—I tucked it away in my Downloads folder. Embedded inside two layers of compression is the virtual hard drive I needed.
I used 7-Zip to find and extract the necessary file. 7-Zip is a handy, open-source archive manipulation program, which adds a couple of options to the Windows menu you see when you right-click on a file. So I navigated to where I’d stored WindowsXPMode_en-us.exe, right clicked on the file, selected “7-Zip” and then “Open archive”. 7-Zip then gave me a view of the contents of the archive file, which includes a folder named sources. In that folder, there’s a file called xpm. This is also compressed, so I right-clicked it and opened it in 7-Zip. In the archive listing for xpm, there’s a file called VirtualXPVHD. That’s the virtual hard drive containing the XP installation. I right-clicked it, and then selected “Copy to …”, telling 7-Zip to extract and decompress VirtualXPHD. I put it in a new folder called XP.
So that Windows could recognize VirtualXPVHD as a virtual hard drive, I now edited the file-name by adding a .vhd extension to it.
For an environment in which to run my new VHD, I downloaded VirtualBox, and installed it with the default options. Then I ran the program, told it I wanted to set up a new virtual XP machine using an existing virtual hard disk, and pointed it to the location of my VirtualXPVHD.vhd file.
And that was that. Now you I could launch an XP machine in a window on my Windows 8.1 desktop.
VirtualBox does a lot of handy things. There are a couple worth knowing about if you’re setting up your own virtual machine:
1) You can scale the window in which the XP machine runs, using the menu option View/Scale Factor. This is useful if you find yourself peering at a tiny window in the middle of your high resolution monitor.
2) At first, you’re going to need to let the XP window “capture” your mouse pointer. When you click inside the window, a dialogue box appears offering to capture the pointer. When you accept, you find your mouse is trapped inside the XP window. You can release it again by tapping a “hot key”, which defaults to the right control key on your keyboard. It’s worth checking that you actually have a right control key before you fire up XP for the first time, otherwise your mouse will be permanently trapped. If you don’t have one, you can change to a new hot key in the VirtualBox menu, File/Preferences/Input/Virtual Machine/Host Key Combination. Select the “Shortcut” box and press whichever key you want to designate as the new hot key.
I found the virtual machine was a bit crashy when XP was going through its initial set-up on my laptop (but not on my desktop). A couple of times I had to press the hot key to free up my mouse, and send a “power off” signal to the virtual machine using the VirtualBox interface. On each occasion it rebooted happily and let me proceed further with the set-up.
The first priority once you have the XP desktop on display is to add some additional function. Free up your mouse pointer so that you can use the VirtualBox menu at the top of the window, click on “Devices”, and then “Insert Guest Additions CD Image …” This makes the XP virtual machine think you’ve inserted a software CD, and it opens a set-up dialogue to let you install the new software. Accept this (and choose “Install Anyway” each time XP objects to the certification of the software that’s being installed).
Now you don’t need to go through the business of capturing and releasing the mouse pointer! The XP window is integrated into your desktop and behaves like any other windowed software. (On occasion, you may find that the mouse pointer behaves a little oddly in some programs—flickering, or scanning too quickly. On those occasions, capturing the mouse is useful. You can turn mouse capture on and off using Input/Mouse Integration in the VirtualBox menu at the top of the window.) I also found that installing the Guest Additions CD eliminated the crashes I’d been having during setup.
One hitch in all this was that XP soon announced that it needed activation, and demanded a Product Key. Now, back in the day, computers used to come with a copy of their operating system on a disc, and I still had the old XP installation disc for a long-defunct computer tucked away in a cupboard. I offered the Product Key from that disc, and XP was happy. In fact, I now have three virtual XP machines all happily registered with the same Product Key. Which is not unreasonable, given that Microsoft haven’t actually been supporting this operating system for a couple of years now.
A couple of the programs I installed under XP are so old they’ll only run if the parent disc is in a drive for them to access. That’s not exactly convenient, so I copied the necessary discs to .iso image files using DVD to ISO , and put the .iso files into a directory on my virtual XP machine.
Then I installed the Virtual CD-ROM Control Panel from Microsoft, which lets XP mount .iso files as if they were physical discs. (Note added, 2020: Sadly, this is no longer available to download from Microsoft. You can still find it on various external sites, but I can’t vouch for the cleanliness of the downloaded files, so I’m not posting any links here.) It’s slightly finicky to install, but the readme file talks you through the process. You need to move the .sys driver file to your Windows XP drivers directory, and then run VCdControlTool.exe to install the driver. Once the driver is installed, running VCdControlTool again lets you create an unused drive letter and then mount an .iso file to that drive. I set my .iso files up as “persistent mounts”, so they’re always available.
Finally, although I’m no great fan of cloud storage (sure, I’ll let a multinational corporation store my personal documents and photographs at some random location on the internet; what could possibly go wrong with that idea?) I do like to share a few program and configuration files between my various devices. The small storage capacity offered by the free version of DropBox has been more than enough for this—and anyone who cares to hack into my DropBox account isn’t going to find much that’s comprehensible to them, let alone useful. But DropBox discontinued support for XP recently, so I needed a way to transfer files from my XP virtual machine to a directory on its parent machine, where DropBox could then take over.
VirtualBox lets you set up folders that are shared between the XP virtual machine and the parent machine (go through Devices/Shared Folders) but some of my XP software is so ancient it refused to recognize the network drive on which these shared folders resided. So I tried using the free version of Tonido. Tonido synchronizes files through their server without ever storing them—I installed the server software on the parent machine, and the client software on the virtual XP machine. Presto! My shared XP files were transferred to the parent machine, where they could be DropBoxed or backed up as required. This had the advantage of being easy to set up, but the rather bizarre and deeply unsatisfying consequence of a computer transferring files to itself over the internet. It also has to be said that the Tonido file synchronization was often slow, and on occasion delayed for hours.
Anyway, once I’d got that straightened out, I used the Devices menu in VirtualBox to make the DropBox folder on the parent machine a shared folder, ticking the boxes to make it “auto-mount” and “permanent”. It then appeared as a network drive the next time I booted the virtual XP machine. Using Link Shell Extension in XP, I could pick folders inside the DropBox network drive, and drop copies of them as symbolic links on the XP c:\ drive, where my ancient programs could see them. When the programs modify those files, the modification is then reflected in the DropBox folder on the parent machine, and that cascades off to my other devices, and their virtual XP machines. Joy!
So now, every time Microsoft offers me a free upgrade, I can sit back and enjoy my free downgrade instead.
# Pennycook et al.: On the reception and detection of pseudo-profound bullshit
This from the November 2015 issue of Judgment And Decision Making. Here are links to the original paper (pdf) and its supplementary tables (pdf).
The authors seek to find a preliminary answer to the questions, “Are people able to detect blatant bullshit? Who is most likely to fall prey to bullshit and why?” Their study is therefore of the characteristics of the bullshittee, rather than the bullshitter, or of bullshit itself.
They suggest that bullshit occupies a sort of halfway house between lie and truth. Bullshit is “something that is designed to impress but […] constructed absent direct concern for the truth.” (That is, the author of bullshit doesn’t care whether it’s true or not, in contrast to the liar, who is deliberately subverting the truth.) And “bullshit, in contrast to mere nonsense, is something that implies but does not contain adequate meaning or truth.”
I’m indebted to them for providing links to two sources of pseudo-profound bullshit, used in their study.
One, Wisdom of Chopra, uses random words taken from the Twitter feed of Deepak Chopra to construct novel sentences. Here’s an example of its output:
The unexplainable arises and subsides in the doorway to energy
The other, Seb Pearce‘s New-Age Bullshit Generator, generates an entire, beautiful page of random bullshit. Here’s one headline:
You and I are entities of the quantum matrix. By evolving, we believe
So that’s all pseudo-profound bullshit.
According to Pennycook et al., reasons you might mistake that for actual profundity include:
• A deficiency of analytic thinking
• Ontological confusion (confusing different categories of existence, such as the mental and the physical)
• Epistemically suspect beliefs (such as paranormal or supernatural ideas)
Four studies are reported in the paper. They all look for correlations between the particular cognitive biases listed above with a “Bullshit Receptivity” scale—a measure of an individual’s tendency to rate randomly generated bullshit as “profound” on a five-point scale ranging from “not at all profound” to “very profound”.
I haven’t even counted the number of separate correlation measures to which the authors assign significance values; I’ll leave that as an exercise for the Interested Reader.
But what we seem to see is that:
• Participants tended to score random nonsense as moderately profound.
• Participants scored selected real Deepak Chopra Tweets as a little more profound than random nonsense, but less profound than some motivational quotations.
• Some participants scored even mundane statements like “Most people enjoy some sort of music” as having some level of profundity. These participants tended to give high profundity scores across the board.
• To quote the authors: “Those more receptive to bullshit are less reflective, lower in cognitive ability (ie. verbal and fluid intelligence, numeracy), are more prone to ontological confusions and conspiratorial ideation, are more likely to hold religious and paranormal beliefs, and are more likely to endorse complementary and alternative medicine.”
• Waterloo University undergraduates (or at least, those who sign up for this sort of study) are catastrophically gullible, assigning various levels of profundity to some quite astonishing twaddle (Table 1). Snake-oil salesmen are presumably converging on the campus even as I type.
So it’s good to have all that sorted out.
# Book Collector
When you have more than 4000 books scattered around the house, it gets difficult to find the one you’re looking for. Especially if you’re hunting for a short story and you can’t quite remember which book you read it in. This used to happen a lot, chez Oikofuge. But not any more.
Book Collector is a book cataloguing program from Collectorz.com, and it’s the best I’ve run into. I can’t now imagine life without it.
Data entry is easy, and highly automated. If your book has an ISBN, you can type it in or scan it with a barcode reader, and the basic book data are pulled down off the Collectorz.com central database. These days, the ISBN is readily visible on the back cover, but with older books (1967 to the mid-70s) you may have to look for numbers written on the spine, or listed in the front matter. Some UK publications of that vintage have nine-digit SBNs instead of ten-digit ISBNs, but the conversion is easy—just add a zero at the left end.
Before 1967, there were no ISBNs, but Book Collector also lets you add books automatically by entering the author and title. This option will bring up all matching entries in the Collectorz.com database, so you might need to do a little poking around to find the entry that matches your specific edition.
That will get the basic data into your database, including a version of the cover art if it’s available. But the software offers a huge number of additional relevant data-fields, which you can fill in or ignore according to your wishes. (You’ll probably want to make use of the “book location” field, unless you have a memory much better than mine.) It even lets you create custom fields.
The on-line cover art comes from a variety of sources—it varies in size and quality, and can occasionally be for the wrong edition of your book. But Book Collector has an automated search facility that lets you look for more cover art on-line. It also lets you add your own art by scanning the cover. If you’re of an obsessive nature (who, me?) you may find yourself scanning a lot of book covers to get precisely the right edition.
You can view your database in various ways, usually splitting the view between some sort of overview of the books, and a detailed view of a specific volume. The overviews available are a “bookshelf” depiction of cover art (which I find useful when browsing for a specific book) and a spreadsheet-type display of multiple customizable columns. There’s also a “cover flow” option available, but the less said about that the better—it’s the sort of triumph of style over utility that could only appeal to an Apple user.
You can choose from a growing number of different formats for viewing your book details; or, with a little knowledge of HTML/ XML, and some digging around in the file structure, you can customize up your own view.
Searching is easy. There’s a quick-and-dirty search option that just looks for your chosen text anywhere in the book’s description. It’ll bring up false hits, but often it lets you narrow down the display enough to zero in visually on the book you want. But you can also create moderately sophisticated “filter” views, using simple Boolean logic functions, to pull out the books you want.
As someone who has a lot of short stories in my book collection, I particularly value the fact that I can enter a contents list for my books. Book Collector offers you a cut-down database and user-defined fields for each short story in a book. Unfortunately, the contents list won’t come down to you automatically from Collectorz.com—manual entry is required, which can be tedious if you have a lot of “complete works” volumes on your shelves. And I would appreciate it if Book Collector some day offered a detailed view by short story as well as by book. But that’s a minor niggle when I can choose to display anything I want in the columns of the spreadsheet view.
What else? Collectorz.com offers a cloud storage facility, so you can be sure you have the same data on all your devices. There’s a responsive Support team (I’ve only ever had to use them once) and an active users’ forum where people are happy to help out with minor queries. And there’s a free try-before-you-buy download.
If you’re in the market for book cataloguing software, do give it go.
# Software: Introduction
I started computer programming on punched tape and IBM 80-column punched cards, using Fortran, back in 1974.
One of my first teachers was a young woman who could read the program directly off the punched tape, and debug it using a hole punch and some sticky dots. I fixated on her utterly, like a baby duckling.
Later, I moved on to Teletype terminals, on which I programmed primitive astronomy calculations. Then my very own computer (who saw that coming?) the Sinclair ZX Spectrum, plugged into the television, on which I built a primitive model of the solar system using early editions of Peter Duffett-Smith’s fine astronomy programming books.
Apple IIs and PCs after that, mainly working in various dialects of Basic, and then making the switch to event-driven Visual Basic when it came along in the early 90s.
I produced and published various utility programs for the medical specialty I worked in, and even sold a couple of units of a reference-manager program I wrote as an add-on for the early scientific word processor, ChiWriter. But it was all essentially recreational until I made the mistake of writing a rostering program for the department I worked for in 1997 (out of frustration with the very poor quality of commercial rostering software at that time). That put me on a maintenance treadmill for the next fourteen years.
So I rather lost interest in developing any more complete software packages at that point! My other programming activity therefore became confined to writing little routines to get some particular piece of data-handling or calculation done, as part of some other project. I did a lot of that when I was a developer on the open-source 3D planetarium software, Celestia, and also when accessing NASA’s Shuttle Radar Topography Mission data to research mountains for Ginge Fullen’s Africa’s Highest Challenge, in which he climbed the highest point of every country in Africa.
I have a number of little software projects in mind for the future, some of which I may share here. But I think initially this category of blog post will be devoted to a couple of reports on software I find useful and/or fun, and want to recommend. | 6,502 | 30,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.947394 |
https://blog.thelifeofkenneth.com/2008/12/physics-9b-practice-final-problem-1.html | 1,726,780,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00039.warc.gz | 114,838,595 | 20,736 | ### Physics 9B Practice Final Problem #1
Water (density 103 kg/m3) is being siphoned out of a very large tank. The pipe extends a distance 3 m into the water and rises to a height 4 m above the water before dropping a distance 8 m. On the surface of the water, the pressure is 105 N/m2.
a) What is the flow speed in the pipe?
b) Is there a limit on how high the siphon top can rise above the surface of the water and still work? Explain. If so, what is the limit?
Question so kindly provided by Joe Kiskis
Part A
To solve this, simply apply Bernoulli's principle:
• P + ρgy + 1/2 ρv2 = C
• P - pressure
• ρ - Density
• g - acceleration due to gravity
• y - height from arbitrary reference point
• v - velocity of the fluid
• C - arbitrary constant
We're trying to find the velocity of the fluid in the tube, so we can choose any point in the tube. We also need another point to calculate C; the surface of the water should do, since we know everything about it.
• 105N/m2 + 103 kg/m3*9.8m/s2*0m + 1/2*103 kg/m3*(0m/s)2 = 105N/m2
• We can arbitrarily set the origin for y as the surface of the water, but really any other point would do, just not as easily.
• We can assume the surface of the water isn't moving, since this is a lousy lower division physics class.
• C = 105N/m2
Now that we know C, we can plug it into the equation anywhere else. The easiest place would be at the mouth of the tube, since at that point it equalizes pressure with air, who's pressure was given.
• 105N/m2 + 103 kg/m3*9.8m/s2*(-4m) + 1/2*103 kg/m3*v2 = 105N/m2
• y = -4m since the tube goes up 4m, then down 8m.
• Now just solve for v
• v = 8.85 m/s
Part B
There is an upper limit to how high the siphon can operate, since it's the surface air pressure that is pushing the water up the tube. Once the weight of the water exceeds the air pressure, the water can't raise any farther. Above this upper limit, the inside of the tube is a vacuum, so to find this limit, set P=0. At the same time, energy has to be conserved, and we know the water is moving through the tube at 8.85 m/s, so we need to factor the kinetic energy into the use of Bernoulli's principle When the siphon fails, the water stops moving, which means that it's only the height of the tube that matters:
• 0 + ρgy + 0 = C
• Solve for y
• y = C / ρg
• y = 9.75 m 10.2 m
Note: The use of KE in this equation originally seemed iffy to me. Consider this: the water in the tank is compressed to 105 N/m2 by the air. The water then needs to be accelerated into the tube, which will translate some of this pressure into KE, and will lower the maximum height from the stationary column of water height of 10.2m.
Note: I was all kinds of confused on this problem. Thankfully, the professor was kind enough to drop into the classes chat room and point out all my failures in logic.
Now the real question is what happens between 9.75m and 10.2m? On the one hand, the siphon still works as long as the water is moving, but we've already calculated how fast the water has to exit the tube to conserve the lost potential energy of dropping 4m. How can the water slow down at the top of the tube? | 873 | 3,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.911137 |
https://www.gurufocus.com/term/ChangeInInventory/WRLD/Change+In+Inventory/World+Acceptance+Corporation | 1,511,253,491,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806327.92/warc/CC-MAIN-20171121074123-20171121094123-00280.warc.gz | 828,189,373 | 40,274 | Switch to:
World Acceptance Corp (NAS:WRLD) Change In Inventory: \$0.0 Mil (As of Sep. 2017)
World Acceptance Corp's change in inventory for the quarter that ended in Sep. 2017 was \$0.0 Mil. It means World Acceptance Corp's inventory stayed the same from Jun. 2017 to Sep. 2017 .
World Acceptance Corp's change in inventory for the fiscal year that ended in Mar. 2017 was \$0.0 Mil. It means World Acceptance Corp's inventory stayed the same from Mar. 2016 to Mar. 2017 .
World Acceptance Corp's Total Inventories for the quarter that ended in Sep. 2017 was \$0.0 Mil.
Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
Total Inventories can be measured by Days Sales of Inventory (DSI). World Acceptance Corp's days sales of inventory (DSI) for the quarter that ended in Sep. 2017 was 0.00.
Inventory Turnover measures how fast the company turns over its inventory within a year.
Inventory-to-Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. World Acceptance Corp's Inventory-to-Revenue for the quarter that ended in Sep. 2017 was 0.00.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
World Acceptance Corp Annual Data
Mar08 Mar09 Mar10 Mar11 Mar12 Mar13 Mar14 Mar15 Mar16 Mar17 Change In Inventory 0.00 0.00 0.00 0.00 0.00
World Acceptance Corp Quarterly Data
Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Sep17 Change In Inventory 0.00 0.00 0.00 0.00 0.00
Calculation
Change In Inventory is the difference between last period's ending inventory and the current period's ending inventory.
Explanation
1. Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
World Acceptance Corp's Days Inventory for the quarter that ended in is calculated as:
Days Inventory = Total Inventories / Cost of Goods Sold * Days in Period = 0 / 0 * 365 / 4 = N/A
2. Total Inventories can be measured by Days Sales of Inventory (DSI).
World Acceptance Corp's Days Sales of Inventory for the quarter that ended in Sep. 2017 is calculated as
Days Sales of Inventory (DSI) = Total Inventories / Revenue * Days in Period = 0 / 126.215 * 365 / 4 = 0.00
3. Inventory Turnover measures how fast the company turns over its inventory within a year.
World Acceptance Corp's Inventory Turnover for the quarter that ended in Sep. 2017 is calculated as
Inventory Turnover = Cost of Goods Sold / Total Inventories = 0 / 0 = N/A
4. Inventory-to-Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
World Acceptance Corp's Inventory to Revenue for the quarter that ended in Sep. 2017 is calculated as
Inventory-to-Revenue = Total Inventories / Revenue = 0 / 126.215 = 0.00
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms | 803 | 3,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-47 | latest | en | 0.88845 |
https://support.nag.com/numeric/nl/nagdoc_27/clhtml/g07/g07bfc.html | 1,708,579,937,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00291.warc.gz | 569,894,062 | 6,303 | # NAG CL Interfaceg07bfc (estim_genpareto)
## 1Purpose
g07bfc estimates parameter values for the generalized Pareto distribution by using either moments or maximum likelihood.
## 2Specification
#include
void g07bfc (Integer n, const double y[], Nag_OptimOpt optopt, double *xi, double *beta, double asvc[], double obsvc[], double *ll, NagError *fail)
The function may be called by the names: g07bfc, nag_univar_estim_genpareto or nag_estim_gen_pareto.
## 3Description
Let the distribution function of a set of $n$ observations
$yi , i=1,2,…,n$
be given by the generalized Pareto distribution:
$Fy = 1- 1+ ξy β -1/ξ , ξ≠0 1-e-yβ , ξ=0;$
where
• $\beta >0$ and
• $y\ge 0$, when $\xi \ge 0$;
• $0\le y\le -\frac{\beta }{\xi }$, when $\xi <0$.
Estimates $\stackrel{^}{\xi }$ and $\stackrel{^}{\beta }$ of the parameters $\xi$ and $\beta$ are calculated by using one of:
• method of moments (MOM);
• probability-weighted moments (PWM);
• maximum likelihood estimates (MLE) that seek to maximize the log-likelihood:
$L = -n ln β^ - 1+ 1ξ^ ∑ i=1 n ln 1+ ξ^yi β^ .$
The variances and covariance of the asymptotic Normal distribution of parameter estimates $\stackrel{^}{\xi }$ and $\stackrel{^}{\beta }$ are returned if $\stackrel{^}{\xi }$ satisfies:
• $\stackrel{^}{\xi }<\frac{1}{4}$ for the MOM;
• $\stackrel{^}{\xi }<\frac{1}{2}$ for the PWM method;
• $\stackrel{^}{\xi }<-\frac{1}{2}$ for the MLE method.
If the MLE option is exercised, the observed variances and covariance of $\stackrel{^}{\xi }$ and $\stackrel{^}{\beta }$ is returned, given by the negative inverse Hessian of $L$.
## 4References
Hosking J R M and Wallis J R (1987) Parameter and quantile estimation for the generalized Pareto distribution Technometrics 29(3)
McNeil A J, Frey R and Embrechts P (2005) Quantitative Risk Management Princeton University Press
## 5Arguments
1: $\mathbf{n}$Integer Input
On entry: the number of observations.
Constraint: ${\mathbf{n}}>1$.
2: $\mathbf{y}\left[{\mathbf{n}}\right]$const double Input
On entry: the $n$ observations ${y}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$, assumed to follow a generalized Pareto distribution.
Constraints:
• ${\mathbf{y}}\left[i-1\right]\ge 0.0$;
• $\sum _{\mathit{i}=1}^{n}{\mathbf{y}}\left[i-1\right]>0.0$.
3: $\mathbf{optopt}$Nag_OptimOpt Input
On entry: determines the method of estimation, set:
${\mathbf{optopt}}=\mathrm{Nag_PWM}$
For the method of probability-weighted moments.
${\mathbf{optopt}}=\mathrm{Nag_MOM}$
For the method of moments.
${\mathbf{optopt}}=\mathrm{Nag_MOMMLE}$
For maximum likelihood with starting values given by the method of moments estimates.
${\mathbf{optopt}}=\mathrm{Nag_PWMMLE}$
For maximum likelihood with starting values given by the method of probability-weighted moments.
Constraint: ${\mathbf{optopt}}=\mathrm{Nag_PWM}$, $\mathrm{Nag_MOM}$, $\mathrm{Nag_MOMMLE}$ or $\mathrm{Nag_PWMMLE}$.
4: $\mathbf{xi}$double * Output
On exit: the parameter estimate $\stackrel{^}{\xi }$.
5: $\mathbf{beta}$double * Output
On exit: the parameter estimate $\stackrel{^}{\beta }$.
6: $\mathbf{asvc}\left[4\right]$double Output
On exit: the variance-covariance of the asymptotic Normal distribution of $\stackrel{^}{\xi }$ and $\stackrel{^}{\beta }$. ${\mathbf{asvc}}\left[0\right]$ contains the variance of $\stackrel{^}{\xi }$; ${\mathbf{asvc}}\left[3\right]$ contains the variance of $\stackrel{^}{\beta }$; ${\mathbf{asvc}}\left[1\right]$ and ${\mathbf{asvc}}\left[2\right]$ contain the covariance of $\stackrel{^}{\xi }$ and $\stackrel{^}{\beta }$.
7: $\mathbf{obsvc}\left[4\right]$double Output
On exit: if maximum likelihood estimates are requested, the observed variance-covariance of $\stackrel{^}{\xi }$ and $\stackrel{^}{\beta }$. ${\mathbf{obsvc}}\left[0\right]$ contains the variance of $\stackrel{^}{\xi }$; ${\mathbf{obsvc}}\left[3\right]$ contains the variance of $\stackrel{^}{\beta }$; ${\mathbf{obsvc}}\left[1\right]$ and ${\mathbf{obsvc}}\left[2\right]$ contain the covariance of $\stackrel{^}{\xi }$ and $\stackrel{^}{\beta }$.
8: $\mathbf{ll}$double * Output
On exit: if maximum likelihood estimates are requested, ll contains the log-likelihood value $L$ at the end of the optimization; otherwise ll is set to $-1.0$.
9: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
## 6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_INT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}>1$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NE_OPTIMIZE
The optimization of log-likelihood failed to converge; no maximum likelihood estimates are returned. Try using the other maximum likelihood option by resetting optopt. If this also fails, moments-based estimates can be returned by an appropriate setting of optopt.
Variance of data in y is too low for method of moments optimization.
NE_REAL_ARRAY
On entry, ${\mathbf{y}}\left[〈\mathit{\text{value}}〉\right]=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{y}}\left[i-1\right]\ge 0.0$ for all $i$.
NE_ZERO_SUM
The sum of y is zero within machine precision.
NW_PARAM_DIST
The asymptotic distribution of parameter estimates is invalid and the distribution of maximum likelihood estimates cannot be calculated for the returned parameter estimates because the Hessian matrix could not be inverted.
NW_PARAM_DIST_ASYM
The asymptotic distribution is not available for the returned parameter estimates.
NW_PARAM_DIST_OBS
The distribution of maximum likelihood estimates cannot be calculated for the returned parameter estimates because the Hessian matrix could not be inverted.
Not applicable.
## 8Parallelism and Performance
g07bfc is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
g07bfc makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The search for maximum likelihood parameter estimates is further restricted by requiring
$1+ ξ^yi β^ > 0 ,$
as this avoids the possibility of making the log-likelihood $L$ arbitrarily high.
## 10Example
This example calculates parameter estimates for $23$ observations assumed to be drawn from a generalized Pareto distribution.
### 10.1Program Text
Program Text (g07bfce.c)
### 10.2Program Data
Program Data (g07bfce.d)
### 10.3Program Results
Program Results (g07bfce.r) | 2,123 | 7,315 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 78, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-10 | latest | en | 0.489998 |
https://www.mathworks.com/matlabcentral/cody/problems/57-summing-digits-within-text/solutions/674375 | 1,575,713,219,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540497022.38/warc/CC-MAIN-20191207082632-20191207110632-00207.warc.gz | 776,731,259 | 15,610 | Cody
# Problem 57. Summing Digits within Text
Solution 674375
Submitted on 23 May 2015 by Andrea Ramirez
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% str = '4 and 20 blackbirds baked in a pie'; total = 24; assert(isequal(number_sum(str),total))
2 Pass
%% str = '2 4 6 8 who do we appreciate?'; total = 20; assert(isequal(number_sum(str),total))
3 Pass
%% str = 'He worked at the 7-11 for \$10 an hour'; total = 28; assert(isequal(number_sum(str),total))
4 Pass
%% str = 'that is 6 of one and a half dozen of the other'; total = 6; assert(isequal(number_sum(str),total)) | 211 | 707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-51 | latest | en | 0.737618 |
https://answers.yahoo.com/question/index?qid=20201029143441AAAXr2D | 1,606,885,447,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00216.warc.gz | 185,700,229 | 24,697 | # Tow lengths are in the ratio 7:8.If the first length is 273 m,what amount is the second length ?
Relevance
• 1 month ago
Let 7x be the first length
Let 8x be the second length
The first length is 273:
7x = 273
x = 273/7
x = 39
Now calculate the second length:
8x = 8(39) = 312
312 m
P.S. If you want a shortcut, multiply 273 by 8/7.
• David
Lv 7
1 month ago
The 2nd length is 312 m because 273/312 is a ratio of 7/8 and (273+312)/15 = 39
So: 7*39 = 273 and 8*39 = 312 | 181 | 483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-50 | latest | en | 0.830714 |
https://www.studypool.com/discuss/1268975/Probability-and-mathematics?free | 1,508,504,274,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824104.30/warc/CC-MAIN-20171020120608-20171020140608-00719.warc.gz | 1,022,437,033 | 14,396 | ##### Probability and mathematics
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Telephone sales have skyrocketed this past month. The probability that 1 of the 4 telephone lines is busy is 75%. Therefore, currently the probability of all 4 lines being busy at the same time is about 32%. I want to reduce the probability of all telephone lines being busy at the same time. If I add 2 more telephone lines, how much do I reduce the probability of all telephone lines being busy at the same time? (Assume the probability that any telephone line is busy remains at 75%.)
Comment
Nov 13th, 2015
4phones = 32% one phone being busy is 8%
6 phone not busy= 32- 8 =24%
Nov 13th, 2015
...
Nov 13th, 2015
...
Nov 13th, 2015
Oct 20th, 2017
check_circle | 211 | 789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-43 | latest | en | 0.947222 |
https://www.developmaths.com/numbers/hcf_lcm/answers/HCF.php | 1,701,176,827,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00170.warc.gz | 815,074,690 | 2,737 | ### Answers − Highest Common Factor or Greatest Common Factor
1. What is the highest common factor of 8 and 12? The factors of 8 are 1, 2, 4 and 8 The factors of 12 are 1, 2, 3, 4, 6 and 12 So the highest common factor is 4
2. What is the highest common factor of 16 and 24? The factors of 16 are 1, 2, 4, 8 and 16 The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 So the highest common factor is 8
3. What is the highest common factor of 28 and 42? The factors of 28 are 1, 2, 4, 6, 14 and 28 The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42 So the highest common factor is 14
4. What is the highest common factor of 8 and 24? The factors of 8 are 1, 2, 4 and 8 The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 So the highest common factor is 8
5. What is the highest common factor of 15 and 18? The factors of 15 are 1, 3, 5 and 15 The factors of 18 are 1, 2, 3, 6, 9 and 18 So the highest common factor is 3
6. What is the highest common factor of 20 and 32? The factors of 20 are 1, 2, 4, 5, 10 and 20 The factors of 32 are 1, 2, 4, 8, 16 and 32 So the highest common factor is 4
7. What is the highest common factor of 18 and 30? The factors of 18 are 1, 2, 3, 6, 9 and 18 The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30 So the highest common factor is 6
8. What is the highest common factor of 15, 24 and 30? The factors of 15 are 1, 3, 5 and 15 The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30 So the highest common factor is 3
back to: | 618 | 1,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-50 | latest | en | 0.955459 |
http://oeis.org/A329907 | 1,618,846,418,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038887646.69/warc/CC-MAIN-20210419142428-20210419172428-00095.warc.gz | 69,046,871 | 4,195 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A329907 Number of iterations of A329904 needed to reach 1. 4
0, 1, 2, 2, 3, 3, 4, 4, 3, 5, 3, 5, 4, 6, 4, 6, 5, 7, 5, 4, 7, 4, 4, 6, 8, 6, 5, 8, 5, 5, 7, 9, 7, 6, 9, 6, 6, 4, 8, 10, 5, 8, 5, 5, 7, 10, 7, 7, 5, 9, 11, 6, 9, 5, 6, 6, 8, 11, 8, 8, 6, 10, 12, 7, 10, 6, 7, 7, 5, 9, 12, 5, 6, 9, 9, 7, 6, 11, 6, 13, 8, 11, 7, 8, 8, 6, 10, 13, 6, 7, 10, 10, 6, 8, 7, 12, 7, 14, 9, 12, 8, 9, 9, 7, 11 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Equally, starting from A025487(n), number of iterations of A329899 needed to reach 1. Any k > 0 occurs 2^(k-1) times in total in this sequence. LINKS Antti Karttunen, Table of n, a(n) for n = 1..10000 FORMULA a(1) = 0; for n > 1, a(n) = 1 + a(A329904(n)). a(1) = 0; for n > 1, a(n) = A070939(A329905(n)). a(n) = A252464(A181815(n)). For all n >= 1, a(n) >= A061394(n). PROG (PARI) A329907(n) = if(1==n, 0, 1+A329907(A329904(n))); (PARI) A329907(n) = #binary(A329905(n)); CROSSREFS Cf. A025487, A061394, A070939, A181815, A252464, A329899, A329904, A329905. Sequence in context: A339731 A234475 A339082 * A329958 A309969 A036041 Adjacent sequences: A329904 A329905 A329906 * A329908 A329909 A329910 KEYWORD nonn AUTHOR Antti Karttunen, Dec 24 2019 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified April 19 11:31 EDT 2021. Contains 343114 sequences. (Running on oeis4.) | 786 | 1,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-17 | latest | en | 0.689191 |
https://oeis.org/A118541 | 1,620,580,133,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989006.71/warc/CC-MAIN-20210509153220-20210509183220-00402.warc.gz | 451,425,617 | 4,348 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A118541 Product of digits of prime factors of n, with multiplicity. 1
1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 12, 3, 14, 15, 16, 7, 18, 9, 20, 21, 2, 6, 24, 25, 6, 27, 28, 18, 30, 3, 32, 3, 14, 35, 36, 21, 18, 9, 40, 4, 42, 12, 4, 45, 12, 28, 48, 49, 50, 21, 12, 15, 54, 5, 56, 27, 36, 45, 60, 6, 6, 63, 64, 15, 6, 42, 28, 18, 70, 7, 72, 21, 42, 75, 36, 7, 18 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS See also: A007954 Product of digits of n. See also: A118503 Sum of digits of prime factors of n, with multiplicity. LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 FORMULA Completely multiplicative with a(p) = A007954(p) for prime p. EXAMPLE a(22) = 2 because 22 = 2 * 11 and the digital product of 2 * the digital product of 11 = 2 * ! * 1 = 2. a(121) = 1 because 121 = 11^2 = 11 * 11, multiplying the digits of the prime factors with multiplicity gives A007954(11) +A007954(11) = 1 * 1 = 1. MATHEMATICA Table[Times @@ Flatten@ Map[IntegerDigits, Table[#1, {#2}] & @@@ FactorInteger@ n], {n, 0, 78}] (* Michael De Vlieger, Jun 16 2016 *) PROG (PARI) \\ here b(n) is A007954. b(n)={my(v=digits(n)); prod(i=1, #v, v[i])} a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); b(p)^e)} \\ Andrew Howroyd, Jul 23 2018 CROSSREFS Cf. A001221, A001222, A007953, A007954, A095402, A118503. Sequence in context: A203814 A280505 A262401 * A084905 A180409 A320485 Adjacent sequences: A118538 A118539 A118540 * A118542 A118543 A118544 KEYWORD base,easy,nonn,mult AUTHOR Jonathan Vos Post, May 06 2006 EXTENSIONS a(36) corrected by Giovanni Resta, Jun 16 2016 Keyword:mult added by Andrew Howroyd, Jul 23 2018 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified May 9 13:03 EDT 2021. Contains 343742 sequences. (Running on oeis4.) | 856 | 2,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-21 | latest | en | 0.682497 |
https://www.isnt.org.in/lognormal-distribution-matlab-with-code-examples.html | 1,723,135,728,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00093.warc.gz | 674,624,607 | 50,923 | # Lognormal Distribution - Matlab With Code Examples
In this article, we will look at how to get the solution for the problem, Lognormal Distribution - Matlab With Code Examples
## What is lognormal distribution formula?
Lognormal distribution of a random variable If X is a random variable and Y=ln(X) is normally distributed, then X is said to be distributed lognormally. Similarly, if Y has a normal distribution, then the exponential function of Y will be having a lognormal distribution, i.e. X=exp(Y).
````pd = makedist('Lognormal','mu',log(20000),'sigma',1)`
```
We were able to fix the Lognormal Distribution - Matlab problem by looking at a number of different examples.
## How do you generate a lognormal distribution in Matlab?
Compute the mean of the logarithmic values. The mean of the log of x is close to the mu parameter of x , because x has a lognormal distribution. Construct a histogram of logx with a normal distribution fit. The plot shows that the log values of x are normally distributed.
## How do you simulate lognormal distribution?
The method is simple: you use the RAND function to generate X ~ N(μ, σ), then compute Y = exp(X). The random variable Y is lognormally distributed with parameters μ and σ. This is the standard definition, but notice that the parameters are specified as the mean and standard deviation of X = log(Y).
## What is lognormal distribution formula?
Lognormal distribution of a random variable If X is a random variable and Y=ln(X) is normally distributed, then X is said to be distributed lognormally. Similarly, if Y has a normal distribution, then the exponential function of Y will be having a lognormal distribution, i.e. X=exp(Y).
## What is the CDF of a lognormal distribution?
The CDF function for the lognormal distribution returns the probability that an observation from a lognormal distribution, with the log scale parameter θ and the shape parameter λ, is less than or equal to x.
## How do you convert normal distribution to lognormal distribution?
f(z;μ,σ)dz=ϕ(log(z)−μσ)d(log(z)−μσ)=1zσϕ(log(z)−μσ)dz. For z>0, this is the PDF of a Normal(μ,σ) distribution applied to log(z), but divided by z. That division resulted from the (nonlinear) effect of the logarithm on dz: namely, dlogz=1zdz.
## Why do we use lognormal distribution?
Lognormal distribution plays an important role in probabilistic design because negative values of engineering phenomena are sometimes physically impossible. Typical uses of lognormal distribution are found in descriptions of fatigue failure, failure rates, and other phenomena involving a large range of data.
## What is the difference between lognormal and normal distribution?
The lognormal distribution differs from the normal distribution in several ways. A major difference is in its shape: the normal distribution is symmetrical, whereas the lognormal distribution is not. Because the values in a lognormal distribution are positive, they create a right-skewed curve.
## How do you calculate lognormal distribution parameters?
Lognormal distribution formulas
• Mean of the lognormal distribution: exp(μ + σ² / 2)
• Median of the lognormal distribution: exp(μ)
• Mode of the lognormal distribution: exp(μ - σ²)
• Variance of the lognormal distribution: [exp(σ²) - 1] ⋅ exp(2μ + σ²)
• Skewness of the lognormal distribution: [exp(σ²) + 2] ⋅ √[exp(σ²) - 1]
## What is a lognormal distribution for dummies?
A log-normal distribution is a continuous distribution of random variable whose natural logarithm is normally distributed. For example, if random variable y = exp { y } has log-normal distribution then x = log ( y ) has normal distribution.
## How do you convert lognormal to normal?
If your data follows a lognormal distribution and you transform it by taking the natural log of all values, the new values will fit a normal distribution. In other words, when your variable X follows a lognormal distribution, Ln(X) fits a normal distribution.
## How do you simulate lognormal distribution?
The method is simple: you use the RAND function to generate X ~ N(μ, σ), then compute Y = exp(X). The random variable Y is lognormally distributed with parameters μ and σ. This is the standard definition, but notice that the parameters are specified as the mean and standard deviation of X = log(Y).
## Why do we use lognormal distribution?
Lognormal distribution plays an important role in probabilistic design because negative values of engineering phenomena are sometimes physically impossible. Typical uses of lognormal distribution are found in descriptions of fatigue failure, failure rates, and other phenomena involving a large range of data.
## How do you convert normal distribution to lognormal distribution?
f(z;μ,σ)dz=ϕ(log(z)−μσ)d(log(z)−μσ)=1zσϕ(log(z)−μσ)dz. For z>0, this is the PDF of a Normal(μ,σ) distribution applied to log(z), but divided by z. That division resulted from the (nonlinear) effect of the logarithm on dz: namely, dlogz=1zdz.
## What is a lognormal distribution for dummies?
A log-normal distribution is a continuous distribution of random variable whose natural logarithm is normally distributed. For example, if random variable y = exp { y } has log-normal distribution then x = log ( y ) has normal distribution.
## How do you generate a lognormal distribution in Matlab?
Compute the mean of the logarithmic values. The mean of the log of x is close to the mu parameter of x , because x has a lognormal distribution. Construct a histogram of logx with a normal distribution fit. The plot shows that the log values of x are normally distributed.
## How do you convert lognormal to normal?
If your data follows a lognormal distribution and you transform it by taking the natural log of all values, the new values will fit a normal distribution. In other words, when your variable X follows a lognormal distribution, Ln(X) fits a normal distribution.
## How do you calculate lognormal distribution parameters?
Lognormal distribution formulas
• Mean of the lognormal distribution: exp(μ + σ² / 2)
• Median of the lognormal distribution: exp(μ)
• Mode of the lognormal distribution: exp(μ - σ²)
• Variance of the lognormal distribution: [exp(σ²) - 1] ⋅ exp(2μ + σ²)
• Skewness of the lognormal distribution: [exp(σ²) + 2] ⋅ √[exp(σ²) - 1]
## What is the difference between lognormal and normal distribution?
The lognormal distribution differs from the normal distribution in several ways. A major difference is in its shape: the normal distribution is symmetrical, whereas the lognormal distribution is not. Because the values in a lognormal distribution are positive, they create a right-skewed curve.
## What is the CDF of a lognormal distribution?
The CDF function for the lognormal distribution returns the probability that an observation from a lognormal distribution, with the log scale parameter θ and the shape parameter λ, is less than or equal to x.
## Object.Entries In Javascript With Code Examples
In this article, we will look at how to get the solution for the problem, Object.Entries In Javascript With Code Examples What is object assign in JavaScript? Object.assign() The Object.assign() method copies all enumerable own properties from one or more source objects to a target object. It returns the modified target object. Object.entries(obj).forEach(([key, value]) => { console.log(key, value); }); myObject = { "key": "value", "key2":"value2" } Object.keys(myObject); //console.log(Obje
## Appsettings In Console Application C# With Code Examples
In this article, we will look at how to get the solution for the problem, Appsettings In Console Application C# With Code Examples Where is Appsettings json? appsettings. json is one of the several ways, in which we can provide the configuration values to ASP.NET core application. You will find this file in the root folder of our project. We can also create environment-specific files like appsettings. <ItemGroup> <PackageReference Include="Microsoft.Extensions.Configuration" Version="3.1.3" /
## Placeholder Font-Family With Code Examples
In this article, we will look at how to get the solution for the problem, Placeholder Font-Family With Code Examples How do you use placeholder? something that marks or temporarily fills a place (often used attributively): I couldn't find my bookmark, so I put a coaster in my book as a placeholder. We're using placeholder art in this mock-up of the ad layout. .mainLoginInput::placeholder { /* Chrome, Firefox, Opera, Safari 10.1+ */ font-family: 'myFont', Arial, Helvetica, s
## Jquery Set Checkbox With Code Examples
In this article, we will look at how to get the solution for the problem, Jquery Set Checkbox With Code Examples How do I check if a Radiobutton is selected in jQuery? We can check the status of a radio button by using the :checked jQuery selector together with the jQuery function is . For example: \$('#el').is(':checked') . It is exactly the same method we use to check when a checkbox is checked using jQuery. //jQuery 1.6+ use \$('.checkbox').prop('checked'
## Boto3 Delete Bucket Object With Code Examples
In this article, we will look at how to get the solution for the problem, Boto3 Delete Bucket Object With Code Examples How do I connect my S3 bucket to boto3? Basic Operations: Now that we are ready, let's start exploring some basic operations. Creating a Bucket. Listing all Buckets in S3. Uploading an Object. Downloading an Object from S3. List all Objects of a specific Bucket. Copy an Object. Deleting an Object. Deleting a Bucket. s3 = boto3.resource('s3') s3.Object('yo | 2,227 | 9,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.80823 |
https://git.kernel.dk/cgit/fio/tree/lib/ieee754.c?id=3d2d14bcb844e72809192311369a642c5d415472 | 1,606,689,519,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00254.warc.gz | 317,313,415 | 3,085 | summaryrefslogtreecommitdiff log msg author committer range
path: root/lib/ieee754.c
blob: 2154065cc136e2c15c514865baa2d3262e78853a (plain)
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 ``` ``````/* * Shamelessly lifted from Beej's Guide to Network Programming, found here: * * http://beej.us/guide/bgnet/output/html/singlepage/bgnet.html#serialization * * Below code was granted to the public domain. */ #include "ieee754.h" uint64_t pack754(long double f, unsigned bits, unsigned expbits) { long double fnorm; int shift; long long sign, exp, significand; unsigned significandbits = bits - expbits - 1; // -1 for sign bit // get this special case out of the way if (f == 0.0) return 0; // check sign and begin normalization if (f < 0) { sign = 1; fnorm = -f; } else { sign = 0; fnorm = f; } // get the normalized form of f and track the exponent shift = 0; while (fnorm >= 2.0) { fnorm /= 2.0; shift++; } while (fnorm < 1.0) { fnorm *= 2.0; shift--; } fnorm = fnorm - 1.0; // calculate the binary form (non-float) of the significand data significand = fnorm * ((1LL << significandbits) + 0.5f); // get the biased exponent exp = shift + ((1 << (expbits - 1)) - 1); // shift + bias // return the final answer return (sign << (bits - 1)) | (exp << (bits-expbits - 1)) | significand; } long double unpack754(uint64_t i, unsigned bits, unsigned expbits) { long double result; long long shift; unsigned bias; unsigned significandbits = bits - expbits - 1; // -1 for sign bit if (i == 0) return 0.0; // pull the significand result = (i & ((1LL << significandbits) - 1)); // mask result /= (1LL << significandbits); // convert back to float result += 1.0f; // add the one back on // deal with the exponent bias = (1 << (expbits - 1)) - 1; shift = ((i >> significandbits) & ((1LL << expbits) - 1)) - bias; while (shift > 0) { result *= 2.0; shift--; } while (shift < 0) { result /= 2.0; shift++; } // sign it result *= (i >> (bits - 1)) & 1 ? -1.0 : 1.0; return result; } `````` | 787 | 2,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-50 | latest | en | 0.45963 |
http://somewhatabnormal.blogspot.com/2010/07/why-isnt-tomorrow-ever-yesterday-part.html | 1,531,999,949,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590866.65/warc/CC-MAIN-20180719105750-20180719125750-00269.warc.gz | 343,512,499 | 15,520 | ## Thursday, July 1, 2010
### Why Isn't Tomorrow Ever Yesterday? Part II
Last time, I introduced the question of the arrow of time and Sean Carroll's proposed solution of it. Now I want to ask what it is that this solution solves.
First, a little about the nature of physical models. There are the equations of the theory - say, general relativity, or the Standard Model. The equations (we suppose) apply to a wide variety of different situations. In addition, there are boundary conditions that apply to some specific situation that we are trying to model. For instance, if I want to model the water in a beaker, I need the equations of fluid dynamics, but I also need to specify the conditions at the boundary of the water: where the water meets the beaker. In this case, I would need to specify the shape of the beaker and give some condition for what the water does when it hits the edge.
The equations of the theory will respect some symmetries, like the time reversal symmetry I mentioned last time. But the solutions of the equations will not necessarily respect the same symmetries. Take, for example, the simple case of a freely moving object in an otherwise empty spacetime. The equation of motion is:
Acceleration = O
This equation has rotational symmetry: it doesn't single out any particular direction in space. Now consider a solution to the equation for a specific object, say a rock. The solution is that the rock's velocity remains constant. But if the rock is moving at all, it is moving in some particular direction. The solution singles out a special direction in space: the direction of motion. The solution for a particular situation will not, in general, display the same symmetries as the equations of motion.
From this point of view, there's no need to worry about the fact that the universe starts out with low entropy: the low entropy is simply a boundary condition we need to impose in order to produce a model that reflects the universe we live in. The lack of time reversal symmetry is not mysterious: the equations of physics are time reversal invariant, but the boundary conditions are not.
What, then, does Carroll's "explanation" of time reversal asymmetry actually explain? Carroll cannot avoid the issue of boundary conditions. His slab of de Sitter space acts as the boundary condition for the droplets that are the baby universes. He has just replaced one boundary condition with another. True, his choice of boundary condition restores some kind of overall time symmetry to the solution. But that symmetry is completely unobservable, because the other baby universes are causally disconnected from our universe. If Carroll is right, there is no way we will ever know it!
Now, maybe I'm not being fair to Carroll's model. Maybe there is something in the process of spawning baby universes that gives specific, testable predictions. And maybe some day those predictions will be confirmed. But even if they are, the overall picture can never be tested experimentally. Specifically, the time symmetry aspect - which, after all, is what Carroll claims to be explaining - can never be checked.
To me, this removes the model from the realm of science. The existence of other baby universes is a purely metaphysical question: do you prefer a picture which has an overall time symmetry, or can you live without it? Does it bother you to postulate an infinite number of unobservable universes, or are you OK with that? Certainly esthetic principles have played a role in theoretical physics in the past and have led to deeper understanding. But when your esthetic principle leads to a picture that is completely untestable, I wonder whether it has any scientific value. | 758 | 3,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-30 | latest | en | 0.928355 |
https://ciestateagentsltd.com/qa/what-is-refraction-and-its-causes.html | 1,627,576,345,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153860.57/warc/CC-MAIN-20210729140649-20210729170649-00069.warc.gz | 183,768,581 | 8,364 | # What Is Refraction And Its Causes?
## What are 3 materials that refract light?
Three examples of materials that refract light rays are water, glass and diamond.
When light rays (travelling in air) enter these materials, their speed decreases..
## What is a real life example of refraction?
What are some examples of refraction of light? “The process of bending of light as it passes from air into glass and vice versa is called refraction of light.”Mirage,bent pencil in glass of water,rainbow,sunset are some examples of refraction of light.
## What is refraction explain with an example?
1. Refraction is the bending of a light or sound wave, or the way the light bends when entering the eye to form an image on the retina. An example of refraction is a bending of the sun’s rays as they enter raindrops, forming a rainbow. An example of refraction is a prism. noun.
## What is refraction class 10th?
Refraction of light is the phenomenon of change in the path of light in going from one medium to another. • In going from a rarer to a denser medium, the ray of light bends towards normal and in going from a denser to a rarer medium, the ray of light bends away from normal.
## What is refraction with diagram?
Draw the diagram of refraction of light in glass slab. Write the laws of refraction. Refraction is the bending of a wave when it enters a medium where its speed is different. We can define it as the change in direction of a wave passing from one medium to another or from a gradual change in the medium.
## What is the main cause for refraction?
The cause of refraction of light is that light travels with different speeds in different media. This Change in the speed of light when it moves from one medium to another causes it to bend.
## Why does refraction occur Class 10?
When the light rays are on the boundary of two media (let’s say water and air) there is not only a change in the speed of light but also in the change in the wavelength. This results in the change in the direction of light. This change in speed and wavelength of light causes refraction of light.
## What are the uses of refraction?
Refraction has many applications in optics and technology. A lens uses refraction to form an image of an object for many different purposes, such as magnification. A prism uses refraction to form a spectrum of colors from an incident beam of light.
## What is difference between reflection and refraction?
This phenomenon usually occurs in mirrors. This phenomenon usually occurs in Lenses. Reflection can simply be defined as the reflection of light when it strikes the medium on a plane. Refraction can be defined as the process of the shift of light when it passes through a medium leading to the bending of light.
## What is the root word of refraction?
refraction (n.) 1570s, from Late Latin refractionem (nominative refractio) “a breaking up,” noun of action from past participle stem of Latin refringere “to break up,” from re- “back” (see re-) + combining form of frangere “to break” (from PIE root *bhreg- “to break”).
## What is refraction state laws of refraction?
the principle that for a ray, radar pulse, or the like, that is incident on the interface of two media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocity of the ray in the first medium to the velocity in the second medium and the incident ray, …
## What are the rules of refraction?
The two laws followed by a beam of light traversing through two media are: The incident ray refracted ray, and the normal to the interface of two media at the point of incidence all lie on the same plane. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.
## What are effects of refraction?
Effects of refraction of light An object appears to be raised when paced under water. Pool of water appears less deep than it actually is. If a lemon is kept in a glass of water it appears to be bigger when viewed from the sides of glass. It is due to refraction of light that stars appear to twinkle at night.
## What are the types of refraction?
Types of Refractive ErrorMyopia. Myopia, also called nearsightedness, is the inability to see distant objects clearly. … Hyperopia. Hyperopia, also called farsightedness, occurs when distant objects are easier to see clearly than nearby objects. … Astigmatism. Astigmatism is blurred vision caused by an unusually shaped cornea. … Presbyopia.
## What is the first law of refraction?
First law of refraction states that the incident ray, the refracted ray and the normal to the interface all lie in the same plane.
## What is the cause of reflection?
Light is reflected when there is a mismatch between materials through which the light is travelling. … Normally the index of refraction is used to determine the angle that light propagates in a material when it “bends” or refracts, but we can also use it to calculate how much light reflects.
## How do glasses refract light?
Bending Light with Refraction Lenses are pieces of glass that bend light. … Those glasses have specially ground lenses that bend the rays of light just enough to focus the image for the person to see properly. All lenses bend and refract rays of light.
## Why does refraction occur for kids?
The bending is called refraction. It happens because light travels at different speeds in different materials. If light rays travel through air and enter a more dense material, such as water, they slow down and bend into the more dense material.
## What is refraction explain?
In physics, refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium. Refraction of light is the most commonly observed phenomenon, but other waves such as sound waves and water waves also experience refraction.
## Which is the best definition of refraction?
refraction. A change of direction that light undergoes when it enters a medium with a different density from the one through which it has been traveling — for example, when, after moving through air, it passes through a prism. (Compare reflection.) | 1,353 | 6,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-31 | latest | en | 0.928679 |
https://code.ascend4.org/ascend/trunk/models/johnpye/cavity.a4c?sortby=log&r1=354&r2=353&pathrev=354 | 1,718,744,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00809.warc.gz | 150,688,552 | 4,257 | # Diff of /trunk/models/johnpye/cavity.a4c
revision 353 by johnpye, Wed Mar 1 13:53:45 2006 UTC revision 354 by johnpye, Fri Mar 3 04:47:59 2006 UTC
# Line 16 ATOM energy_flux REFINES solver_var Line 16 ATOM energy_flux REFINES solver_var
16 nominal := 1000{W/m^2}; nominal := 1000{W/m^2};
17 END energy_flux; END energy_flux;
18
19 ATOM heat_transfer_coefficient REFINES solver_var
20 DIMENSION M/T^3/TMP
21 DEFAULT 5{W/m^2/K};
22
23 lower_bound := 0{W/m^2/K};
24 upper_bound := 1e50{W/m^2/K};
25 nominal := 5{W/m^2/K};
26
27 END heat_transfer_coefficient;
28
29 MODEL cavity; MODEL cavity;
30 W,B,S,N,C,E,D IS_A distance; W,B,S,N,C,E,D IS_A distance;
31 theta, phi, psi IS_A angle; theta, phi, psi IS_A angle;
# Line 33 MODEL cavity; Line 43 MODEL cavity;
43 C^2 = E^2 + S^2 - 2* E * S * cos(psi); C^2 = E^2 + S^2 - 2* E * S * cos(psi);
44
45 F_WN = (W+N-C)/2/W; F_WN = (W+N-C)/2/W;
46 F_WW = (2*E-2*D)/2/W; F_WW = (2*E-2*D)/2/W; (* from top to directly opp part of bottom *)
47 F_WB = 1 - 2 * F_WN; F_WB = 1 - 2 * F_WN;
48 F_WS = (1 - 2 * F_WN - F_WB)/2; F_WS = (1 - 2 * F_WN - F_WB)/2;
49 F_NB = (N+B-C)/2/N; F_NB = (N+B-C)/2/N;
# Line 42 MODEL cavity; Line 52 MODEL cavity;
52 F_BN = F_NB*N/B; F_BN = F_NB*N/B;
53 F_NN = 1 - F_NW - F_NB; F_NN = 1 - F_NW - F_NB;
54 F_BW = F_WB*W/B; F_BW = F_WB*W/B;
55
56 n IS_A set OF symbol_constant; n IS_A set OF symbol_constant;
57 n :== ['W','B','L','R']; n :== ['W','B','L','R'];
58
# Line 91 MODEL cavity; Line 101 MODEL cavity;
101 FOR i IN n CREATE FOR i IN n CREATE
102 q[i] * (1-eps[i]) = (E_b[i] - J[i]) * (eps[i]*A[i]); q[i] * (1-eps[i]) = (E_b[i] - J[i]) * (eps[i]*A[i]);
103 END FOR; END FOR;
104
105 METHODS METHODS
106 METHOD default_self; METHOD default_self;
107 RUN reset; RUN values; RUN bound_self; RUN reset; RUN values; RUN bound_self;
# Line 126 METHOD values; Line 136 METHOD values;
136
137 END values; END values;
138
END cavity;
139 END cavity;
140
141 (*========================================
142 This model adds external convection
143 coefficients to the model, and an
144 ambient temperature.
145
146 We also calculate the F_rad correlation
147 parameter.
148 *)
149 MODEL cavity_losses REFINES cavity;
150 h_B, h_N IS_A heat_transfer_coefficient;
151 T_amb IS_A temperature;
152
153 - q['B'] = h_B * B * (T['B'] - T_amb);
154 - q['L'] = h_N * N * (T['L'] - T_amb);
155 - q['R'] = h_N * N * (T['L'] - T_amb);
156
159
160 - q['B'] = F_rad * eps['W'] * 1{SIGMA_C} * (T['W']^4 - T['B']^4);
161
162 - q['B'] = F_rad_1 * 1{SIGMA_C} * (T['W']^4 - T['B']^4) /
163 (1/eps['B'] + 1/eps['W'] - 1);
164
165 METHODS
166 METHOD specify;
167 FIX T['W'], T_amb;
168 FIX h_B, h_N;
169 FIX W,D,theta;
170 FIX eps[n];
171 END specify;
172 METHOD values;
173 T['W'] := 550 {K};
174 T_amb := 290 {K};
175 W := 500 {mm};
176 D := 300 {mm};
177 theta := 30 {deg};
178 eps['W'] := 0.49;
179 eps['B'] := 0.9;
180 eps['L','R'] := 0.1;
181 h_B := 10 {W/m^2/K};
182 h_N := 0.5 {W/m^2/K};
183 (* free values *)
184 T['L','R'] := 500 {K};
185 T['B'] := 400 {K};
186 END values;
187 END cavity_losses;
Legend:
Removed from v.353 changed lines Added in v.354 | 1,454 | 3,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.328513 |
http://www.convertit.com/Go/BusinessConductor/Measurement/Converter.ASP?From=gradus&To=depth | 1,660,337,029,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00051.warc.gz | 75,283,967 | 3,703 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: Roman gradus = 0.740664 length (length) Related Measurements: Try converting from "gradus" to barleycorn, Biblical cubit, cable length, chain (surveyors chain), earth to moon (mean distance earth to moon), ell, Greek fathom, hand, Israeli cubit, league, li (Chinese li), link (surveyors link), marathon, mil, nautical league, naval shot, rod (surveyors rod), sazhen (Russian sazhen), soccer field, yard, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: gradus = .02087629 actus (Roman actus), 408.24 agate (typography agate), 7,406,640,000 angstrom, 87.48 barleycorn, 2,916 caliber (gun barrel caliber), 6.48 cloth finger, 39.95 digitus (Roman digitus), .648 ell, 4,214.79 en (typography en), 9.6 Greek palm, .34964029 ken (Japanese ken), 7.83E-17 light yr (light year), 740,664 micron, .972 pace, 174.96 pica (typography pica), 1.67 Roman cubit, .00675 skein, 2.43 survey foot, .88390421 vara (Mexican vara), .81 yard.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 454 | 1,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-33 | latest | en | 0.666216 |
https://www.w3resource.com/python-exercises/puzzles/python-programming-puzzles-69.php | 1,653,822,198,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662644142.66/warc/CC-MAIN-20220529103854-20220529133854-00726.warc.gz | 1,222,681,832 | 29,803 | Python: Create a new string by taking a string, and word by word rearranging its characters in ASCII order - w3resource
# Python: Create a new string by taking a string, and word by word rearranging its characters in ASCII order
## Python Programming Puzzles: Exercise-69 with Solution
Write a Python program to create a new string by taking a string, and word by word rearranging its characters in ASCII order.
Input: Ascii character table
Output:
Aciis aaccehrrt abelt
Input: maltos won
Output:
almost now
Pictorial Presentation:
Sample Solution-1:
Python Code:
def test(strs):
return " ".join("".join(sorted(w)) for w in strs.split(' '))
strs = "Ascii character table"
print("Original string:",strs)
print("New string by said string, and word by word rearranging its characters in ASCII order:")
print(test(strs))
strs = "maltos won"
print("\nOriginal string:",strs)
print("New string by said string, and word by word rearranging its characters in ASCII order:")
print(test(strs))
Sample Output:
Original string: Ascii character table
New string by said string, and word by word rearranging its characters in ASCII order:
Aciis aaccehrrt abelt
Original string: maltos won
New string by said string, and word by word rearranging its characters in ASCII order:
almost now
Flowchart:
## Visualize Python code execution:
The following tool visualize what the computer is doing step-by-step as it executes the said program:
Sample Solution-2:
Python Code:
def test(strs):
words = strs.split(' ')
rwords = []
for word in words:
occurrences = {}
for c in word:
occurrences[c] = occurrences.get(c, 0) + 1
subsequence = []
for c, count in occurrences.items():
subsequence += [c]*count
subsequence.sort()
subsequence = subsequence*(len(word)//len(subsequence)) + subsequence[:len(word)%len(subsequence)]
rwords.append(''.join(subsequence))
return ' '.join(rwords)
strs = "Ascii character table"
print("Original string:",strs)
print("New string by said string, and word by word rearranging its characters in ASCII order:")
print(test(strs))
strs = "maltos won"
print("\nOriginal string:",strs)
print("New string by said string, and word by word rearranging its characters in ASCII order:")
print(test(strs))
Sample Output:
Original string: Ascii character table
New string by said string, and word by word rearranging its characters in ASCII order:
Aciis aaccehrrt abelt
Original string: maltos won
New string by said string, and word by word rearranging its characters in ASCII order:
almost now
Flowchart:
## Visualize Python code execution:
The following tool visualize what the computer is doing step-by-step as it executes the said program:
Python Code Editor :
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
## Python: Tips of the Day
Clamps num within the inclusive range specified by the boundary values x and y:
Example:
def tips_clamp_num(num,x,y):
return max(min(num, max(x, y)), min(x, y))
print(tips_clamp_num(2, 4, 6))
print(tips_clamp_num(1, -1, -6))
Output:
4
-1 | 771 | 3,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-21 | longest | en | 0.864059 |
https://www.teachoo.com/8180/2502/Ex-2.2--8/category/Ex-2.2/ | 1,723,278,314,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00135.warc.gz | 781,912,810 | 21,575 | Ex 2.1
Chapter 2 Class 7 Fractions and Decimals
Serial order wise
### Transcript
Ex 2.1, 8 Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 2/5 of the water. Pratap consumed the remaining water. (i) How much water did Vidya drink?Total water = 5 litres Given that, Vidhya drank 𝟐/𝟓 of the water ∴ Water Vidhya drank = 2/5 × Total Water = 2/5 × 5 = 2 litres Ex 2.1, 8 Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed 2/5 of the water. Pratap consumed the remaining water. (ii) What fraction of the total quantity of water did Pratap drink ?Now, Water Pratap drank = Total water − Water Vidhya drunk Water Pratap drank = 5 − 2 Water Pratap drank = 3 litres Fraction of total water Pratap drunk = (𝑊𝑎𝑡𝑒𝑟 𝑃𝑟𝑎𝑡𝑎𝑝 𝑑𝑟𝑎𝑛𝑘)/(𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑒𝑟) = 𝟑/𝟓 | 342 | 885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.92835 |
https://franklin.dyer.me/entries/2017-4-17.html | 1,545,036,605,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828448.76/warc/CC-MAIN-20181217065106-20181217091106-00186.warc.gz | 602,675,497 | 8,822 | ## n-Involutory Rational Functions
2017 April 17
Find a function with the property $$f^3(x)=x$$.
Find a function with the property $$f^4(x)=x$$.
Find a function with the property $$f^6(x)=x$$.
Lately I’ve been fascinated with a different class of functions - involutory functions, or functions that invert themselves. These functions have the property $$f^2(x)=x$$. A few of these functions include:
$f_1(x)=-x$
$f_2(x)=1-x$
$f_3(x)=\frac{1}{x}$
$f_4(x)=-\frac{1}{x}$
$f_5(x)=\frac{x+1}{x-1}$
In fact, there are infinitely many such functions, because if $$f$$ is in the form
$f(x)=(g\circ h\circ g^{-1})(x)$
then $$f$$ is an involution whenever $$h$$ is. A very special type of function to consider is a function of the type
$f(x)=\frac{ax+b}{cx+d}$
which does some interesting things when iterated. Notice what happens when we compose two rational functions of the form
$f(x)=\frac{ax+b}{cx+d}$
$g(x)=\frac{ex+f}{gx+h}$
$(f\circ g)(x)=\frac{a\frac{ex+f}{gx+h}+b}{c\frac{ex+f}{gx+h}+d}$
$(f\circ g)(x)=\frac{a(ex+f)+b(gx+h)}{c(ex+f)+d(gx+h)}$
$(f\circ g)(x)=\frac{(ae+bg)x+(af+bh)}{(ce+dg)x+(cf+dh)}$
This may not seem remarkable at first, but notice what happens when we multiply two matrices:
$\begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}e & f\\g & h\end{bmatrix}$
$\begin{bmatrix}ae+bg & af+bh\\ce+dg & cf+dh\end{bmatrix}$
Which is analogous to what we got for our iterated rational function. Therefore, if we map the rational function
$f(x)=\frac{ax+b}{cx+d}$
onto the matrix
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$
so that we map the coefficients of $$x$$ on the numerator and denominator to the leftmost entries and the constants to the rightmost entries, then $$f^n(x)$$ maps onto the matrix
$\begin{bmatrix}a & b\\c & d\end{bmatrix}^n$
Notice then that if such a function is involutory, then the matrix that it maps onto has the property that
$\begin{bmatrix}a & b\\c & d\end{bmatrix}^2=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
which is the $$2$$ x $$2$$ identity matrix. Interestingly, this holds even for fractional and negative iterates of $$f$$. Using this result, we can find a couple very interesting types of rational functions. For example, functions of the form
$f(x)=\frac{ax-a^2}{x}$
are 3-involutory, meaning that $$f^3(x)=x$$. The function
$f(x)=\frac{ax-a^2}{x+a}$
is 4-involutory. We can even find a function that is $$6-involutory$$ by finding the halfth iterate of our 3-involutory function. To find the halfth iterate or functional square root of a rational function
$f(x)=\frac{ax+b}{cx+d}$
we map it onto a matrix
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$
and use the matrix square root formula:
$\sqrt{\begin{bmatrix}a & b\\c & d\end{bmatrix}}=\frac{1}{t}\begin{bmatrix}a+s & b\\c & d+s\end{bmatrix}$
where $$s$$ is the positive or negative root of the determinant. We don’t have to worry about what $$t$$ is, because when we revert this back to a rational function, it will cancel out. So the square root of our rational function is
$f(x)=\frac{(a\pm\sqrt{ad-bc})x+b}{cx+(d\pm\sqrt{ad-bc})}$
and using this, we can find the functional square root of our 3-involutory function:
$f(x)=\frac{(a\pm\sqrt{a^2})x+b}{cx+(d\pm\sqrt{a^2})}$
$f(x)=\frac{(a\pm a)x-a^2}{x\pm a}$
and so now we know that functions of the form
$f(x)=\frac{2ax-a^2}{x+a}$
$f(x)=\frac{a^2}{a-x}$
are 6-involutory. We could go even further and find a 12-involutory function or a 24-involutory function, but… nah. There’s one more interesting property about these types of functions. If
$f(x)=\frac{ax+b}{cx+d}$
and
$f^n(x)=\frac{ex+f}{gx+h}$
then if
$g(x)=\frac{ax+c}{bx+d}$
it must be true that
$g^n(x)=\frac{ex+g}{fx+h}$
Why is this true? Well, try mapping the functions $$f$$ and $$g$$ onto matrices $$F$$ and $$G$$. Then it is clear that $$F$$ is the transpose of $$G$$, or that $$F^T=G$$. There is a property of matrices stating that for any matrices $$A$$ and $$B$$,
$(AB)^T=B^TA^T$
so it follows from this that for any matrix $$A$$,
$(A^{n})^T=(A^{T})^n$
Since we are likening exponentiation of matrices to the iteration of functions, this means that the “transpose” of $$f$$ iterated $$n$$ times is equal to the the nth iterate of its transpose, proving our original statement. Furthermore, if a function of this type is n-involutory, then so is its “transpose”, telling us that in addition to our above 6-involutory functions, we also have the functions
$f(x)=\frac{2ax+1}{a-a^2x}$
$f(x)=\frac{-1}{a^2x+a}$
And this theorem can be applied to all types of n-involutory functions of this type.
Edit: 2017 June 3
Okay, there is a way to find an n-involutory rational function for any positive integer $$n$$. Such a function is given by
$f(x)=\frac{x\cos\zeta-\sin\zeta}{x\sin\zeta-\cos\zeta}$
where
$\zeta=\frac{2\pi}{n}$
To prove this, we will have to employ the trigonometric sum angle formulas for the sine and cosine:
$\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi$
$\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi$
This can be derived by, once again, using matrices in place of actual rational functions. Suppose we are multiplying the matrices
$\begin{bmatrix}\cos a\zeta & -\sin a\zeta\\\sin a\zeta & \cos a\zeta\end{bmatrix}\begin{bmatrix}\cos b\zeta & -\sin b\zeta\\\sin b\zeta & \cos b\zeta\end{bmatrix}$
When we carry out the multiplication, we get
$\begin{bmatrix}\cos a\zeta\cos b\zeta-\sin a\zeta\sin b\zeta & -\sin a\zeta\cos b\zeta-\cos a\zeta\sin b\zeta\\\sin a\zeta\cos b\zeta+\cos a\zeta\sin b\zeta & \cos a\zeta\cos b\zeta-\sin a\zeta\sin b\zeta\end{bmatrix}$
Look! These are the sum angle formulas, and this can be simplified to
$\begin{bmatrix}\cos (a+b)\zeta & -\sin (a+b)\zeta\\\sin (a+b)\zeta & \cos (a+b)\zeta\end{bmatrix}$
Now that we know this, we can say that
$\begin{bmatrix}\cos \zeta & -\sin \zeta\\\sin \zeta & \cos \zeta\end{bmatrix}^n=\begin{bmatrix}\cos n\zeta & -\sin n\zeta\\\sin n\zeta & \cos n\zeta\end{bmatrix}$
And, if $$\zeta=\frac{2\pi}{n}$$,
$\begin{bmatrix}\cos \zeta & -\sin \zeta\\\sin \zeta & \cos \zeta\end{bmatrix}^n=\begin{bmatrix}\cos 2\pi & -\sin 2\pi\\\sin 2\pi & \cos 2\pi\end{bmatrix}$
$\begin{bmatrix}\cos \zeta & -\sin \zeta\\\sin \zeta & \cos \zeta\end{bmatrix}^n=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
Thus it is proven. Because of the relationship previously established between matrices and rational functions, if
$f(x)=\frac{x\cos\zeta-\sin\zeta}{x\sin\zeta-\cos\zeta}$
and
$\zeta=\frac{2\pi}{n}$
then
$f^n(x)=x$ | 2,232 | 6,454 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-51 | latest | en | 0.709302 |
https://www.studiestoday.com/tags/cbse-class-12-mathematics-chapter-vbqs-7620.html | 1,652,763,835,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00106.warc.gz | 1,211,952,187 | 15,886 | # CBSE Class 12 Mathematics Chapter VBQs
• ### CBSE Class 12 Mathematics Value Based Questions Set A
VALUE BASED QUESTIONS NOTE: Model answers are being given to the value based questions. These are only suggestive answers. The students can answer these questions according to their own thinking. CHAPTER 1 Q.1. Determine whether each of the following relations in the set A of students in a school...
• ### CBSE Class 12 Mathematics Value Based Questions Set B
Q.1 Prove that f: R → R is a bijection given by f(x) = x3+3. Find f -1(x). Does the truthfulness and honesty may have any relation? Q.2 Set A ={ a1.a2.a3,a4,a5 }and B= {b1,b2,b3,b4 } when ai’s and bi’s Are school going students. Define a relation from a set A to set B by x R y iff is a true...
• ### CBSE Class 12 Mathematics Value Based Questions Set C
MATHEMATICS – STD.XII VALUE BASED QUESTIONS RELATIONS AND FUNCTIONS 1. Let R be a relation defined as R = { (x,y) : x and y study in the same class}. Show that R is an equivalence relation.If x is a brilliant student and y is a slow learner and x helps y in his studies, what quality does x...
# CBSE Class 12 Mathematics Value Based Questions Set A
VALUE BASED QUESTIONS
Click for more Mathematics Study Material
CBSE Class 12 Mathematics Value Based Questions Set A CBSE Class 12 Mathematics Value Based Questions Set B CBSE Class 12 Mathematics Value Based Questions Set C
# CBSE Class 12 Mathematics Value Based Questions Set B
Q.1 Prove that f: R → R is a bijection given by f(x) = x3+3. Find f -1(x). Does the truthfulness and honesty may have any relation?
Click for more Mathematics Study Material
CBSE Class 12 Mathematics Value Based Questions Set A CBSE Class 12 Mathematics Value Based Questions Set B CBSE Class 12 Mathematics Value Based Questions Set C
# CBSE Class 12 Mathematics Value Based Questions Set C
MATHEMATICS – STD.XII
Click for more Mathematics Study Material
CBSE Class 12 Mathematics Value Based Questions Set A CBSE Class 12 Mathematics Value Based Questions Set B CBSE Class 12 Mathematics Value Based Questions Set C
## Latest NCERT & CBSE News
Read the latest news and announcements from NCERT and CBSE below. Important updates relating to your studies which will help you to keep yourself updated with latest happenings in school level education. Keep yourself updated with all latest news and also read articles from teachers which will help you to improve your studies, increase motivation level and promote faster learning
### All India Children Educational Audio Video Festival
The Central Institute of Educational Technology (CIET), a constituent unit of National Council of Educational Research and Training (NCERT), is inviting entries for the 26th All India Children’s Educational Audio Video Festival (AICEAVF). This festival showcases the...
### Board Exams Date Sheet Class 10 and Class 12
Datesheet for CBSE Board Exams Class 10 (Scroll down for Class 12 Datesheet) Datesheet for CBSE Board Exams Class 12
### Online courses for classes XI and XII offered by NCERT
Ministry of Education (MoE), Government of India has launched a platform for offering Massive Open Online Courses (MOOCs) that is popularly known as SWAYAM (Study Webs of Active learning for Young Aspiring Minds) on 9 th July, 2017. NCERT now offers online courses for... | 779 | 3,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-21 | latest | en | 0.874731 |
https://it.scribd.com/document/434612821/Experiment1-ZN | 1,603,731,271,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891428.74/warc/CC-MAIN-20201026145305-20201026175305-00267.warc.gz | 367,869,695 | 77,636 | Sei sulla pagina 1di 6
# Experiment 1
## ZIEGLER NICHOLS TUNING
1 OBJECTIVE
Design the P, PI, PID controller for a given transfer function by using Ziegler
Nichols Tuning method.
2 THEORY
Ziegler and Nichols proposed rules for determining values of the proportional
gain, integral time and derivative time based on the transient response characteristics of
a given plant. Such determination of the parameters of PID controllers or tuning of PID
controllers can be made by engineers on-site by experiments on the plant.
There are two methods called Ziegler-Nichols tuning rules: the first method and
second method.
## 2.1 First Method
In the first method, we obtain step response of a plant experimentally. By using
that output curve we have to calculate delay time L and time constant T. The delay time
and time constant are determined by drawing a tangent line at the inflection point of the
(S-shaped) curve.
## Fig.1. Response curve for ZN first method
TABLE.I Ziegler-Nichols tuning rule based on step response of plant
## 2.2 Second Method
In this method we have to calculate critical gain and corresponding time period
of a plant by using Routh Hurwitz Criteria.
3 ZIEGLER NICHOLS MATLAB CODE
## %Hemant Kumar 19530006
%ZIEGLER NICHOLS TUNING
clear all;
close all;
num=[6];
din=[1 6 11 6];
G=tf(num,din) %plant transfer function
figure(1)
step(G)
CL_G=feedback(G,1)
figure(2)
step(CL_G)
Kc=margin(G)% using routh hurwitz criteria we can calculate
Kc=10
T=1.8945
figure(3)
% P controller design
Kp1=0.5*Kc con1=Kp1
com1=series(G,con1);
P_sys=feedback(com1
,1)
hold on
step(P_sys)
% PI controller design
Kp2=0.45*Kc
Ki2=(1/(0.83*T))*Kp2
con2=tf([Kp2 Ki2],[1 0])
com2=series(G,con2);
PI_sys=feedback(com2
,1)
hold on
step(PI_sys)
## % PID controller design
Kp3=0.6*Kc
Ki3=(1/(0.5*T))*Kp3
Kd3=(0.125*T)*Kp3
con3=tf([Kd3 Kp3 Ki3],[1
0]) com3=series(G,con3);
PID_sys=feedback(com3,1
) step(PID_sys)
4 ZIEGLER NICHOLS MATLAB SIMULINK MODEL
Type of
Controller
P 5 ∞ 0
PI 4.5 2.862 0
## Fig.3. unit step response of the system
Fig.4.Unit step response with critical gain Kc=10
## Fig.5. P controller output response
Fig.6. PI controller output response | 679 | 2,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-45 | latest | en | 0.70642 |
http://www.asktheinnercoach.com/spiritual-development/physical-time-space | 1,513,332,812,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00274.warc.gz | 305,399,021 | 13,516 | # Physical Time & Space
We measure physical time as distance divided by speed.
I Day (of time) is equal to the distance of one revolution of the Earth divided by its speed of revolution. [ 1 day = 24,000 miles / 1000 mph = 24 hours].
We measure physical space or the space betwen physical objects as a volume when it is 3 dimensional; as an area when it is 2 dimensional; and as a distance when it is 1 dimensional. Therefore: distance = length; area = length x breadth; and volume = length x breadth x height (or depth).
We measure the space inside physical matter, which is the space between molecules and atoms, as its density. Density or specific gravity = weight (mass) / volume.
We measure physical time in hours, minutes and seconds.
We measure physical distance of rotation in space as degrees, minutes and seconds.
We measure physical distance in space in a straight line using time as a light-year (the distance light travels in a year).
Energy only becomes subject to time and space once it slows down sufficiently to materialise as matter. We can therefore only measure time and space relative to physical matter.
Spiritual Energy exists in Spiritual Time & Space and requires a different method of calculation. | 260 | 1,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-51 | latest | en | 0.945417 |
https://www.physicsforums.com/threads/atlas-of-complex-projective-3-space-mathbb-c-p-3.663674/ | 1,532,107,625,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591718.31/warc/CC-MAIN-20180720154756-20180720174756-00598.warc.gz | 974,116,686 | 12,627 | # Atlas of complex projective 3-space $\mathbb{C}P^3$
1. Jan 11, 2013
### kloptok
I'm constructing an atlas for complex projective 3-space $\mathbb{C}P^3$. I use the construction with inhomogeneous coordinates $z_{i}/z_{j}$ and a chart is given by the points $\mathcal{U}_j=\{z_j\ne 0\}$. At the intersections $\mathcal{U}_i\cap \mathcal{U}_j$ I should specify (holomorphic) transition functions between coordinates. So far I'm fine. Now my problem is, what do I do with the intersections between three charts, $\mathcal{U}_i\cap \mathcal{U}_j\cap \mathcal{U}_k$ ? I have a feeling that I can skip them since $\mathcal{U}_i\cap\mathcal{U}_j \cap \mathcal{U}_k=\mathcal{U}_i \cap(\mathcal{U}_j\cap \mathcal{U}_k)$ and I know what to do at each of the intersections $\mathcal{U}_i\cap\mathcal{U}_j$. After all, isn't the point that at this intersection I can use either of the three coordinate systems and I only have to require holomorphic transitions between them?
Is this a correct way of thinking? It's probably a silly question but I can't get my head around this. All I can find when I search on the web is what to do at intersections between two charts, and nothing about multiple intersections.
2. Jan 11, 2013
### quasar987
You only need to check holomorphicity of the transition functions between each pair of charts since if you do that, then on the intersection of triplets, you're just restricting the domain of a map you already know is holomorphic.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 431 | 1,551 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-30 | latest | en | 0.865561 |
https://www.jakesonline.org/bear-stearns-pzzvpo/mixed-fraction-calculator-28e249 | 1,627,197,816,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151638.93/warc/CC-MAIN-20210725045638-20210725075638-00643.warc.gz | 847,930,518 | 9,462 | A mixed fraction is a fraction of the form \$$c {n \over d}\$$, where \$$c\$$ is an integer and \$$n d\$$. Convert the fraction to a mixed number by using long division to find the quotient and remainder 5 ÷ 3 = 1 R2 The quotient will be the whole number in the fraction, and the remainder will be the numerator in the mixed fraction 5 3 = 1 2 3 You can use … Multiply Mixed Numbers Calculator. We also offer step by step solutions. Formulas. Equivalent decimals (D) and reduced Fractions (R) will appear underneath. Study of mathematics online. Here is an adding and subtracting mixed numbers calculator to find the addition and subtraction of mixed fractions. Whether it is addition, subtraction, multiplication, or division. The mixed fraction is the combination of whole number and the proper fraction. Online Calculators for Operations with Simple and Mixed Fractions, including comparison. It is therefore the sum of a whole number and a proper fraction. A mixed fraction is a whole number and a proper fraction combined, i.e. The two fractions … DOSSIER.NET sommario. See how each example is made up of a whole number anda proper fraction together? First select if you want to use the default or mixed fraction calculator. The procedure to use the adding mixed fractions calculator … Fraction Calculators: Mixed Fraction Calculator, Mixed Number Calculator, Adding Fractions Calculator, Multiplying Fractions Calculator, Dividing Fractions Calculator, and Subtracting Fractions Calculator. One and three-half should be typed as 1 3/2. Dividing fractions calculator. Make your calculation easy by this mixed fraction calculator. The calculator performs basic and advanced operations with mixed numbers, fractions, integers, decimals. The calculator provided returns fraction inputs in both improper fraction form, as well as mixed number form. For example, to convert 5 4/5, we will do the following. Supports evaluation of mixed fractions (e.g. Yes, with this fraction calculating tool you can operate with both mixed and … An improper fraction is a fraction where the numerator (top number) is larger than the denominator (bottom number). To divide two fractions, MiroCalc.net mixed calculator for fractions uses the multiplication algorithm… with a twist: If your provided a mixed fraction, the calculator for fractions will convert it to an improper one. It is composed by the Numerator placed above a dash (called Fraction … You just need to enter your two fractions, then select the operation you want to perform. Mixed Numbers Calculator (also referred to as Mixed Fractions): This online calculator handles simple operations on whole numbers, integers, mixed numbers, fractions and improper fractions by adding, subtracting, dividing or multiplying. The mixed fraction calculator simplifies the result and displays it. Fraction Calculator. Learn to add, subtract, multiply and divide fractions. Adding Mixed Fractions Calculator is a free online tool that displays the sum of two mixed fractions. Calculate: enter 2 or 3 Fractions, select arithmetic operators using drop-downs and click on [=] button to get result. To advance your math, read Everything You Need to Ace Math in One Big Fat Notebook. Visual Fractions. Step 2) We add, subtract, multiply, or divide those improper fractions together from Step 1. The improper fraction can be converted to the mixed fraction. Keep exactly one space between the whole number and fraction and use a forward slash to input fractions. The primary reason is that the code utilizes Euclid's Theorem for reducing fractions which can be found at How MiroCalc.net Fractions Calculator is Dividing Fractions. Mixed Fraction to Improper Fraction Calculator is a free online tool that displays the conversion of mixed fraction into an improper fraction. It accepts proper, improper, mixed fractions and whole number inputs. For example, \$${5 \over 4}\$$. "Simple and Mixed". Fractions Calculator A simple, but powerful, fraction calculator to solve multiple types of fraction math problems. It is therefore the sum of a whole number and a proper fraction. A mixed fraction is a whole number and a proper fraction combined, i.e. Converting between fractions and decimals: Converting from decimals to fractions … An __improper fraction__ is a fraction whose nominator is greater than its denominator. Fraction calculator This calculator supports the following operations: addition, subtraction, multiplication, division and comparison of two fractions or mixed numbers. one and three-quarters. Electrical Calculators Real Estate Calculators When it will convert fraction to percent, a step by step procedure, results in percentage and result in decimal will be displayed separately. The combination of a proper fraction and whole number is called as mixed fraction. This unique tool will simplify your mixed number to its lowest form. © 2006 -2020CalculatorSoup® There general steps to subtract fractions are described below. This online calculator handles simple operations on whole numbers, integers, mixed numbers, fractions and improper fractions by adding, subtracting, dividing or multiplying. You can enter up to 3 digits in length for each the numerators and denominators (e.g., 456/789). Unless you got the gift of scientific mind, you need calculator to make sure you will get the correct computation during the mathematical operation with mixed fractions. Simplify Mixed Numbers. Example: Convert the following mixed number to an improper fraction. All rights reserved. Multiplying 3 Fractions Calculator is a handy tool that performs the multiplication of given three fractions in a short span of time. Sign in Log in Log out Use the online mixed fraction calculator to practice calculations with mixed fractions. You have to provide the numerator, denominator numbers of those three fractions in the respective input sections and tap on the calculate button to find the exact result with in seconds. So, in this case, you will need to divide the fractions: 1 (3/5) and 1 (4/7). The fraction to percent calculator is used to convert proper or improper fractions and mixed number into a percent corresponding to the given fraction. Do math calculations with mixed numbers (mixed fractions) performing operations on fractions, whole numbers, integers, mixed numbers, mixed fractions and improper fractions. A mixed number is a combination of a whole number and a fraction. This step-by-step comparing fractions calculator will help you understand how to Compare fractions or mixed numbers. It accepts proper, improper, mixed fractions and whole number inputs. Step 1) Convert mixed number to an improper fraction. This calculator for simply mixed fractions and allows you to change a mixed number to an improper fraction/proper fraction or vice versa. Mixed number refers to any whole number next to a fraction, for example 1 2/4. Use this fraction calculator to do the four basic operations: add fractions, subtract fractions, multiply and divide fractions. The slash separates the numerator (number above a fraction line) and denominator (number below). You can enter up to 3 digits in length for each whole number, numerator or denominator (123 456/789). If you want to simplify an individual fraction into lowest terms use our It also shows detailed step-by-step information about the fraction calculation procedure. Please, input values in this format: a b/c or b/c. Use this fractions calculator to easily perform calculations with fractions. BYJU’S online mixed fraction to improper fraction calculator tool performs the calculation faster and it displays the conversion value in a fraction of seconds. It can also convert mixed numbers to fractions and vice versa. It's not just a tool but a complete package of utilities that will make your fractional calculation simpler. It can also convert mixed numbers to fractions and vice versa. Fraction Calculator is an easy way to solve complex to complex fraction problems. BYJU’S online mixed fraction to improper fraction calculator tool performs the calculation faster and it displays the conversion value in a fraction of seconds. To multiply two or more mixed numbers we need to first convert the mixed numbers into improper fractions, multiply the improper fractions and reduce the resultant value to the lowest terms to get the answer. How to Use the Adding Mixed Fractions Calculator? Read our. For simplifying, our mixed number simplifier calculator converts the mixed number into the improper fraction. Here we have presented you a mixed fraction calculator. Sign in Log in Log out About. With this free App you can solve fraction related problems and exercices such as: Addition, Subtraction, Multiplication, Division, Simplification, Comparasion between Proper, Improper and Mixed Fractions. Example (Click to view) 1 1/3 + 2 1/4 Try it now. Adding Mixed Fractions Calculator is a free online tool that displays the sum of two mixed fractions. Use this calculator to convert your improper fraction to a mixed fraction. Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. The fraction calculator will generate a step-by-step explanation on how to obtain the results in the REDUCED FORM! TOGGLE 2-OR 3. Convert the fraction part of the mixed number to a decimal 2. A mixed fraction (also called mixed number) is a whole number and a proper fraction combined. Cite this content, page or calculator as: Furey, Edward "Mixed Numbers Calculator"; CalculatorSoup, 1/2 + 1/3 = (1×3+1×2) / (2×3) = 5/6 No rights can be derived from the information and calculations on this website. Use the algebraic formula for addition of fractions: Reduce fractions and simplify if possible, Use the algebraic formula for subtraction of fractions: a/b - c/d = (ad - bc) / bd, Use the algebraic formula for multiplying of fractions: a/b * c/d = ac / bd, Use the algebraic formula for division of fractions: a/b ÷ c/d = ad / bc. Fraction format button is used to work with all fractions. Enter fractions and press the = button. Simplify Online Fractions. That is why it is called a "mixed" fraction (or mixed number). 3 FRACTIONS, DECIMALS AND MIXED NUMBERS. Online Comparing Fractions Calculator… Mixed fractions are also called mixed numbers. The answer is provided in a reduced fraction and a mixed number if it exists. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. Mixed numbers: Enter as 1 1/2 which is one and one half or 25 3/32 which is twenty five and three thirty seconds. Compare: subtract second Fraction … Practical Example #1: Let’s say that you are looking to subtract: 4 (1/8) – 1 (1/5) So, as we mentioned, the first thing that you need to do is to make sure that you have the mixed fractions selected and not the regular fractions. Look for a button that has a black box over a white box, x/y, or b/c. For example, \$${5 \over 4}\$$. Push this button to open the fraction feature on your calculator. $$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{(a \times d) + (b \times c)}{b \times d}$$, $$1 \dfrac{2}{6} + 2 \dfrac{1}{4} = \dfrac{8}{6} + \dfrac{9}{4}$$, $$= \dfrac{(8 \times 4) + (9 \times 6)}{6 \times 4}$$, $$= \dfrac{32 + 54}{24} = \dfrac{86}{24} = \dfrac{43}{12}$$, $$\dfrac{a}{b} - \dfrac{c}{d} = \dfrac{(a \times d) - (b \times c)}{b \times d}$$, $$\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$$, $$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a \times d}{b \times c}$$. Greatest Common Factor Calculator. With this online mixed fraction calculator (or mixed number calculator) with whole numbers and fractions you can easily add mixed fractions, subtract mixed fractions, multiply mixed fractions and divide mixed fractions. With this calculator you can practice calculations with mixed fractions (mixed numbers). Dividing fractions calculator online. Fraction Operations and Manipulations Click on the decimal format button, enter a fraction or mixed number, then click equals. A mixed number is a combination of a whole number and a fraction. This calculator divides two fractions. … BYJU’S online adding mixed fractions calculator tool makes the calculation faster and it displays the sum of two mixed fractions in a fraction of seconds. If you would like to convert an improper fraction to a mixed fraction, see our improper to mixed fraction calculator. Free Fractions calculator - Add, Subtract, Reduce, Divide and Multiply fractions step-by-step This website uses cookies to ensure you get the best experience. Adding and subtracting mixed number is confusing. This calculator allows you to add, subtract, multiply, or divide simple fractions and mixed fractions online. This video will show you how to convert from mixed to improper fractions as well as improper to mixed fractions using the calculator Casio fx-991ms Mixed number to fraction conversion calculator that shows work to represent the mixed number in impropoer fraction. Fractions Calculator. Mixed fractions are also called mixed numbers. Add, Subtract, Multiply, Divide, Compare Mix or Compound Fractions … Simplify Fractions Calculator. If you are simplifying large fractions by hand you can use the If they exist, the solutions and answers are provided in simplified, mixed and whole formats. Order. The fraction calculator is very helpful in real life. This online fraction calculator will help you understand how to add, subtract, multiply or divide fractions, mixed numbers (mixed fractions), whole numbers and decimals.The fraction calculator will compute the sums, differences, products or quotients of fractions. A mixed fraction is a fraction of the form c n d, where c is an integer and n < d. For example, 11 4 = 2 3 4. Mixed fraction … This mixed fraction calculator by calculator-online is the smart tool that helps you in adding, subtracting, multiplying, and dividing mixed numbers fraction. You have 6 sections: 1. The Fraction Calculator will reduce a fraction to its simplest form. If the fraction or mixed number is only part of the calculation then omit clicking equals and continue with the calculation per usual. You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. Study math with us and make sure that "Mathematics is easy!" Fill in two fractions and choose if you want to add, subtract, multiply or divide and click the "Calculate" button. There general steps to divide fractions are described below. 1/2 + 1/3 = (1×3+1×2) / (2×3) = 5/6 The answer will be in fraction form and whole/decimal form. Fraction calculator This calculator supports the following operations: addition, subtraction, multiplication, division and comparison of two fractions or mixed numbers. Advatages: It is easy to use. You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. Mixed Fraction 1 : Operator : Mixed Fraction 2 . Read on to learn about fraction percent and its formula and how to turn a fraction … Use this calculator to convert your improper fraction to a mixed fraction. Adding mixed numbers, converting fraction to whole number, multiplying fractions by whole numbers, subtracting mixed numbers, and multiplying mixed fractions are among the processes this calculator can do. Enter the value of two mixed numbers and click calculate, it displays the step by step solution with the answer. When it will convert fraction to percent, a step by step procedure, results in percentage and result in decimal will be displayed separately. The Fraction Calculator will reduce a fraction to its simplest form. Here you can transform any improper fraction into a mixed number by using our 'Online Improper Fraction to Mixed Number Calculator'. Open the fraction calculator to practice calculations with mixed numbers calculator to solve types... Lowest forms by dividing both numerator and denominator, i.e., for,. Free online tool that displays the conversion of mixed fraction is a number! Would like to convert an improper fraction to power ( fraction or mixed numbers improper! Return the result and displays it the numerator ( top number ) is larger than the denominator ( 123 )! Into a percent corresponding to the mixed fraction… the fraction feature is on, should... Faster than others you might find number simplifier calculator converts the mixed fraction is a,... Special thing about it is therefore the sum of two mixed numbers calculator to proper... Number ) is larger than the denominator ( bottom number ) is larger than the denominator ( above... With the calculation then omit clicking equals and continue with the answer provided. The relationship between two integers ; practically it 's a division are mixed. Reduced fraction and a mixed fraction ( also called mixed number if it exists and negative (... Are using mixed numbers in the boxes above, with this fraction calculator to multiple... Log out fractions calculator step 4 ) if the fraction to a mixed fraction will return the result step. R ) will appear underneath refers to any whole number and a fraction template on calculator... On how to obtain the results in the boxes above, and press calculate and Manipulations the fraction to fraction... Fraction to a decimal 2 improper, mixed fractions and a proper fraction combined view ) 1 1/3 + 1/4... Fraction where the numerator and denominator ( 123 456/789 ) ( fraction vice... Types of fraction calculation procedure operators using drop-downs and click on [ = ] button to the... Slash “ / ” between the whole number and a proper fraction combined math with us make! With integers, decimals, and press calculate greater than its denominator will! Any mixed number may be written by improper fraction to improper fraction to power ( fraction or number. ( D ) and 1 ( 4/7 ) for all operations and Manipulations fraction! A tool but a complete package of utilities that will make your easy... To complex fraction problems both improper fraction to power ( fraction or vice versa solutions and answers are provided a... And advanced operations with mixed numbers, fractions are described below the of. A mixed '' fraction ( also called mixed number problems like.! Fraction represents the relationship between two integers ; practically it 's not just a tool but a complete package utilities. Also called mixed number to an improper fraction/proper fraction or vice versa no way guarantees that information... Four tenths should be typed as 1 3/2 to do the following this fractions calculator to find the addition subtraction! Log out fractions calculator will help you understand how to factor numbers to find the addition subtraction. Can add, subtract fractions are presented in their lowest forms by dividing both numerator and denominator by their common! In no way guarantees that the information and calculations on this website are correct = ( 1×3+1×2 ) / 2×3! For all operations and Manipulations the fraction calculator ) see the greatest common factor look for button. Reduced to it ’ s simplest form why it is that it supports two kinds of fraction math.! Simplifies the result fraction converted in to decimals an percent is shown a proper.... Use this fraction calculating tool you can use the default or mixed numbers in the form. Operations on simple proper or improper fractions together from step 2 if necessary as 3/4 which is fourths. You might find fraction ( also called mixed number to an improper fraction.... Written by improper fraction answers into mixed numbers to find the greatest factor. On, you will need to Ace math in one Big Fat Notebook use a forward slash input. Form, as well as raise a fraction line ) and denominator, i.e. for. Their greatest common factor calculator typed as 1 3/2 arithmetic operators using drop-downs and click on the format! Convert your improper fraction to power ( fraction or vice versa click calculate, it displays conversion. Click to view ) 1 1/3 + 2 1/4 Try it now a tool but a complete package of that! With Remainders calculator to easily perform calculations with fractions to 3 digits in length for each the numerators and mixed fraction calculator. Three fourths or 3/100 which is three fourths or 3/100 which is twenty and... Is easy! drop-downs and click calculate, it displays the conversion of mixed fractions and you! Called mixed number is only part of the figure fraction feature is on, you will need enter... You would like to convert your improper fraction, but with the calculation per usual __improper is. Fraction or mixed numbers whole formats should see a fraction over a white box x/y. To divide the fractions: enter as 3/4 which is three one hundredths convert it to mixed. Calculation easy by this mixed fraction is a ( mixed ) fraction reduced to ’. It supports two kinds of fraction math problems sure that Mathematics is easy! view ) 1 +. The numerator ( number above a fraction to improper fraction power ( fraction or mixed numbers improper. Helpful in real life including comparison fraction 1: Operator: mixed fraction to its simplest form it shows. In real life easy!, 456/789 ) thirty seconds fraction converted in to decimals an percent shown! + 1/3 = ( 1×3+1×2 ) / ( 2×3 ) = 5/6 learn to add, subtract,,... Use the calculator provided returns fraction inputs in both cases, fractions are presented their. 'S a division as 1 3/2 easily subtract mixed fractions and mixed fractions, select operators! Answers are provided in simplified, mixed fractions, select arithmetic operators drop-downs! To a mixed number calculator ' study math with us and make sure that Mathematics easy. Division and comparison of two fractions … multiply mixed numbers with space fraction represents the relationship between two ;! Cases, fractions, including comparison in this case, you can enter to! \\ ( { 5 \over 4 } \\ ) multiply, and press calculate change a mixed fraction calculator solve... Integers, decimals, and press calculate factor ( GCF ) see the greatest common factor calculator 1 ) mixed. A complete package of utilities that will make your calculation easy by this mixed fraction ( called., \\ ( { 11 \over 4 mixed fraction calculator \\ ) operation you want to perform math operations on proper., input values in this case, you will need to divide fractions are described below hand can. To change a mixed number to a mixed fraction ( also called number. Negative fractions ( e.g., 456/789 ), Compare and order fractions multiply mixed numbers find! Want to use the slash “ / ” between the whole number is part. Fractions online math solver with step-by-step solutions of whole number, be sure to leave a space between whole. Is that it supports two kinds of fraction math problems it ’ s simplest form than you! Fraction math problems terms use our simplify fractions calculator, simplify, Compare Mix or Compound fractions a! { 11 \over 4 } \\ ) simplifies the result fraction converted in to decimals percent... Find the addition and subtraction of mixed fraction calculator to do the four operations... By hand you can easily subtract mixed fractions online five and three thirty seconds 2 or fractions. Represents the relationship between two integers ; practically it 's a division you. { 11 \over 4 } \\ ) should be typed as 1 3/2 answers... Can operate with both mixed and … online Calculators for operations with simple and mixed number is a template! ’ s simplest form number plus a fractional part 3 fractions, including comparison reduced and. Package of utilities that will make your calculation easy by this mixed fraction 2 drop-downs click. Fraction is the combination of whole number and a mixed number to fraction conversion calculator shows! To subtract fractions, subtract fractions, expressions with fractions the fractions 1... Math in one Big Fat Notebook between the the calculations on this are. On how to Compare fractions or mixed fraction number ) you can the! Keep exactly one space between the whole number and the proper fraction 1 1/2 which is one and should... The greatest common factor ( GCF ) see the greatest common factor numbers to and! Faction calculator handles mixed fractions, including comparison it also shows detailed step-by-step information about the fraction procedure. Number into the improper fraction answers into mixed numbers, with this calculator simplifies the in... Fraction where the numerator ( top number ) is larger than the denominator ( above., we will do the four basic operations: add fractions, subtract, multiply and divide fractions then!, input values in this format: a b/c or b/c fraction … adding subtracting!: enter as 1 1/2 which is three fourths or 3/100 which is one and three-half should be typed 1... Fractions online calculator ' example: 1/2 ÷ 1/3 enter mixed numbers.... Displays the step by step solution with the lesser numerator and denominator by their greatest common factor ( GCF see! … with this fraction calculator simplifies the result is a free online tool that displays the sum of a number! Thirty seconds a forward slash to input fractions not just a tool but a complete package of utilities will... Fractional calculation simpler denominator ( bottom number ) is larger than the denominator ( 123 456/789 ) between... | 5,314 | 25,334 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-31 | latest | en | 0.84123 |
https://legacy.voteview.com/POLI279_2006_3.htm | 1,701,319,996,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00266.warc.gz | 408,027,994 | 8,039 | This site is an archived version of Voteview.com archived from University of Georgia on May 23, 2017. This point-in-time capture includes all files publicly linked on Voteview.com at that time. We provide access to this content as a service to ensure that past users of Voteview.com have access to historical files. This content will remain online until at least January 1st, 2018. UCLA provides no warranty or guarantee of access to these files.
POLI 279 MEASUREMENT THEORY
Third Assignment
Due 5 May 2006
1. The aim of this problem is to show you how to use R to simulate a complex model of stimulus evaluation. The idea is that people use a simple mental model to compare two stimuli. The distinguishing features of the stimuli are assumed to be represented by dimensions in a simple geometric model. The stimuli are positioned on the dimensions according to the levels of the attributes represented by the dimensions. People are assumed to perceive the stimuli correctly with some random error. When asked to perform a stimulus comparison, people draw a momentary psychological value from a very tight error distribution around the locations of each of the stimuli. Their judgement of similarity is assumed to be an exponential function of the psychological distance between the two stimuli -- e-kd -- where d is the distance between the two momentary psychological values expressed as points in psychological space, and k is a scaling constant (see the figure below).
What we are going to do is use a program to simulate this process and numerically compute the expected value of e-kd for 40 values of the true distance (from 0 to 2) between the true locations of the stimuli.
Use the following R program below to do this experiment:
Shepard-Ennis-Nofosky_1.r -- Program to test the Shepard-Ennis_Nofosky Model of Stimulus Comparison
Here is what the program looks like:
```
#
#
# normal_2.r -- Test of Shepard_Ennis_Nofosky model of Stimulus
# Comparison
#
# Exemplars have bivariate normal distributions. One point is
# drawn from each distribution and the distance computed.
# This distance is then exponentiated -- exp(-d) -- and
# this process is repeated m times to get a E[exp(-d)].
#
# Note that when STDDEVX is set very low then this is equivalent
# to a legislator's ideal point so that the stimulus comparison
# is between the ideal point and the momentary draw from the
# Exemplar distribution.
#
# Set Means and Standard Deviations of Two Bivariate Normal Distributions
#
# Remove all objects just to be safe
#
rm(list=ls(all=TRUE))
#
STDDEVX <- .001 Initialize Means and Standard
STDDEVY <- .4 Deviations of the Two Normal Distributions
XMEAN <- 0
YMEAN <- 0
#
# Set Number of Draws to Get Expected Value
#
m <- 10000
#
# Set Number of Points
#
n <- 40
#
# Set Increment for Distance Between Two Bivariate Normal Distributions
#
XINC <- 0.05 Note that n*XINC = 2.0
#
x <- NULL Initialize vectors
y <- NULL
z <- NULL
xdotval <- NULL
ydotval <- NULL
zdotval <- NULL
iii <- 0
while (iii <= n) { ****Start of Outer Loop -- it Executes Until y has n=40 Entries
x <- 0
i <- 0
iii <- iii + 1
while (i <= m) { ****Start of Inner Loop -- it Executes m=10000 Times
#
# Draw points from the two bivariate normal distributions
# Draw two numbers randomly from Normal Distribution with Mean XMEAN and stnd. dev. of STDDEVX and put them in the vector xdotval
xdotval <- c(rnorm(2,XMEAN,STDDEVX))
ydotval <- c(rnorm(2,YMEAN,STDDEVY))
# Draw two numbers randomly from Normal Distribution with Mean YMEAN and stnd. dev. of STDDEVY and put them in the vector ydotval
# Compute distance Between two randomly drawn points
# Calculate Euclidean Distance Between xdotval and ydotval
zdotval <- sqrt((xdotval[1]-ydotval[1])**2 + (xdotval[2]-ydotval[2])**2)
#
# Exponentiate the distance
#
zdotval <- exp(-zdotval)
#
# Compute Expected Value
#
x <- x + zdotval/m Store exp(-d) in x -- note the division by m so that x will contain
i <- i + 1 the mean after m=10000 trials
} ****End of Inner Loop
#
# Store Expected Value and Distance Between Means of Two Distributions
#
y[iii] <- x
z[iii] <- sqrt(2*YMEAN^2)
YMEAN <- YMEAN + XINC Increment YMEAN by .05
} ****End of Outer Loop
#
# The type="n" tells R to not display anything
#
plot(z,y,xlim=c(0,3),ylim=c(0,.7),type="n",
main="",
xlab="",
ylab="",font=2)
#
# Another Way of Doing the Title
#
title(main="Test of Shepard-Ennis-Nofosky, .001, .4\nTest of Ideal Point vs. Outcome Model")
# Main title
mtext("Test of Shepard-Ennis-Nofosky, .001, .4\nTest of Ideal Point vs. Outcome Model",side=3,line=1.00,cex=1.2,font=2)
# x-axis title
mtext("True Distance Between Stimuli",side=1,line=2.75,font=2,cex=1.2)
# y-axis title
mtext("Expected Value of Observed Similarity",side=2,line=2.75,font=2,cex=1.2)
##
# An Alternative Way to Plot the Points and the Lines
#
points(z,y,pch=16,col="blue",font=2)
lines(z,y,lty=1,lwd=2,font=2,col="red")
#```
The program will produce a graph that looks similar to this:
1. Run the Program and turn in the resulting plot.
2. Use Epsilon to change STDDEVX and STDDEVY in Shepard-Ennis-Nofosky_1.R, re-load it into R, and produce a plot like the above. For example, change STDDEVX to .1 and STDDEVY to .4 and re-run the program. Change the "title" each time to reflect the different values! Pick 3 pairs of reasonable values of STDDEVX and STDDEVY and produce plots for each experiment.
2. The aim of this problem is to show you how to double-center a matrix in R and extract the eigenvalues and eigenvectors of the matrix. Use the R program below to do this exercise:
Double_Center_Drive_Data.r -- Program to Double-Center a Matrix of Squared Distances
```#
# double_center_drive_data.r -- Double-Center Program
#
# Data Must Be Transformed to Squared Distances Below
#
# ATLANTA 0000 2340 1084 715 481 826 1519 2252 662 641 2450
# BOISE 2340 0000 2797 1789 2018 1661 891 908 2974 2480 680
# BOSTON 1084 2797 0000 976 853 1868 2008 3130 1547 443 3160
# CHICAGO 715 1789 976 0000 301 936 1017 2189 1386 696 2200
# CINCINNATI 481 2018 853 301 0000 988 1245 2292 1143 498 2330
# DALLAS 826 1661 1868 936 988 0000 797 1431 1394 1414 1720
# DENVER 1519 891 2008 1017 1245 797 0000 1189 2126 1707 1290
# LOS ANGELES 2252 908 3130 2189 2292 1431 1189 0000 2885 2754 370
# MIAMI 662 2974 1547 1386 1143 1394 2126 2885 0000 1096 3110
# WASHINGTON 641 2480 443 696 498 1414 1707 2754 1096 0000 2870
# CASBS 2450 680 3160 2200 2330 1720 1290 370 3110 2870 0000
#
# Remove all objects just to be safe
#
rm(list=ls(all=TRUE))
#
library(MASS)
library(stats)
#
#
T <- matrix(scan("C:/ucsd_course/drive2.txt",0),ncol=11,byrow=TRUE)
#
nrow <- length(T[,1])
ncol <- length(T[1,])
# Another way of entering names -- Note the space after the name
names <- c("Atlanta ","Boise ","Boston ","Chicago ","Cincinnati ","Dallas ","Denver ",
"Los Angeles ","Miami ","Washington ","CASBS ")
#
# pos -- a position specifier for the text. Values of 1, 2, 3 and 4,
# respectively indicate positions below, to the left of, above and
# to the right of the specified coordinates
#
namepos <- rep(2,nrow)
#
TT <- rep(0,nrow*ncol)
dim(TT) <- c(nrow,ncol)
TTT <- rep(0,nrow*ncol)
dim(TTT) <- c(nrow,ncol)
#
xrow <- NULL
xcol <- NULL
xcount <- NULL
matrixmean <- 0
matrixmean2 <- 0
#
# Transform the Matrix
#
i <- 0
while (i < nrow) {
i <- i + 1
xcount[i] <- i Ignore -- Just a diagnostic
j <- 0
while (j < ncol) {
j <- j + 1
#
# Square the Driving Distances
#
TT[i,j] <- (T[i,j]/1000)**2 Note the division by 1000 for scale purposes
#
}
}
#
# Put it Back in T
#
T <- TT
TTT <- sqrt(TT) This creates distances -- see below
#
#
# Call Double Center Routine From R Program
# cmdscale(....) in stats library
# The Input data are DISTANCES!!! Not Squared Distances!!
# Note that the R routine does not divide
# by -2.0
#
ndim <- 2
#
#
# returns double-center matrix as dcenter\$x if x.ret=TRUE
#
# Do the Division by -2.0 Here
#
TTT <- (dcenter\$x)/(-2.0) Here is the Double-Centered Matrix
#
#
# Below is the old Long Method of Double-Centering
#
# Compute Row and Column Means
#
i <- 0
while (i < nrow) {
i <- i + 1
xrow[i] <- mean(T[i,])
}
i <- 0
while (i < ncol) {
i <- i + 1
xcol[i] <- mean(T[,i])
}
matrixmean <- mean(xcol)
matrixmean2 <- mean(xrow)
#
# Double-Center the Matrix Using old Long Method
# Compute comparison as safety check
#
i <- 0
while (i < nrow) {
i <- i + 1
j <- 0
while (j < ncol) {
j <- j + 1
TT[i,j] <- (T[i,j]-xrow[i]-xcol[j]+matrixmean)/(-2) This is the Double-Center Operation
}
}
#
# Run some checks to make sure everything is correct
#
xcheck <- sum(abs(TT-TTT)) This checks to see if both methods are the same numerically
#
#
# Perform Eigenvalue-Eigenvector Decomposition of Double-Centered Matrix
#
ev <- eigen(TT)
#
# Find Point furthest from Center of Space
#
aaa <- sqrt(max((abs(ev\$vec[,1]))**2 + (abs(ev\$vec[,2]))**2)) These two commands do the same thing
bbb <- sqrt(max(((ev\$vec[,1]))**2 + ((ev\$vec[,2]))**2))
#
# Weight the Eigenvectors to Scale Space to Unit Circle
#
torgerson1 <- ev\$vec[,1]*(1/aaa)*sqrt(ev\$val[1])
torgerson2 <- -ev\$vec[,2]*(1/aaa)*sqrt(ev\$val[2]) Note the Sign Flip
#
plot(torgerson1,torgerson2,type="n",asp=1,
main="",
xlab="",
ylab="",
xlim=c(-3.0,3.0),ylim=c(-3.0,3.0),font=2)
points(torgerson1,torgerson2,pch=16,col="red",font=2)
text(torgerson1,torgerson2,names,pos=namepos,offset=0.2,col="blue",font=2) Experiment with the offset value
#
#
# Main title
mtext("Double-Centered Driving Distance Matrix \n Torgerson Coordinates",side=3,line=1.25,cex=1.5,font=2)
# x-axis title
mtext("West --- East",side=1,line=2.75,cex=1.2,font=2)
# y-axis title
mtext("South --- North",side=2,line=2.5,cex=1.2,font=2)
#```
The program will produce a graph that looks similar to this:
1. Run the Program and turn in the resulting plot.
2. Change the values in "namepos" until you have a nice looking plot with no overlapping names and no names cut off by the margins and turn in the resulting plot.
3. Use Epsilon to add a 12th city (you will have to get a driving atlas to do this) to Drive2.txt and call the matrix Drive3.txt. Modify the R program to read and plot the 12 by 12 driving distance matrix. Turn in Drive3.txt, the resulting plot, and the modified R program.
3. In this problem we are going to analyze Eckman's famous color circle data. Use the R program below to do this exercise:
Double_Center_Color_Circle.r -- Program to Double-Center a Matrix of Squared Distances
```#
#
# double_center_color_circle.r -- Double-Center Program
#
# Data Must Be Transformed to Squared Distances Below
#
# **************** The Eckman Color Data *****************
#
# 434 INDIGO 100 86 42 42 18 6 7 4 2 7 9 12 13 16
# 445 BLUE 86 100 50 44 22 9 7 7 2 4 7 11 13 14
# 465 42 50 100 81 47 17 10 8 2 1 2 1 5 3
# 472 BLUE-GREEN 42 44 81 100 54 25 10 9 2 1 0 1 2 4
# 490 18 22 47 54 100 61 31 26 7 2 2 1 2 0
# 504 GREEN 6 9 17 25 61 100 62 45 14 8 2 2 2 1
# 537 7 7 10 10 31 62 100 73 22 14 5 2 2 0
# 555 YELLOW-GREEN 4 7 8 9 26 45 73 100 33 19 4 3 2 2
# 584 2 2 2 2 7 14 22 33 100 58 37 27 20 23
# 600 YELLOW 7 4 1 1 2 8 14 19 58 100 74 50 41 28
# 610 9 7 2 0 2 2 5 4 37 74 100 76 62 55
# 628 ORANGE-YELLOW 12 11 1 1 1 2 2 3 27 50 76 100 85 68
# 651 ORANGE 13 13 5 2 2 2 2 2 20 41 62 85 100 76
# 674 RED 16 14 3 4 0 1 0 2 23 28 55 68 76 100
#
#
# Remove all objects just to be safe
#
rm(list=ls(all=TRUE))
#
library(MASS)
library(stats)
#
#
rcx.file <- "c:/ucsd_course/color_circle.txt"
#
# Standard fields and their widths
#
rcx.fields <- c("colorname","x1","x2","x3","x4","x5",
"x6","x7","x8","x9","x10","x11","x12","x13","x14")
rcx.fieldWidths <- c(18,4,4,4,4,4,4,4,4,4,4,4,4,4,4)
#
# Read the vote data from fwf
#
dim(U)
ncolU <- length(U[1,]) Note the "trick" I used here
T <- U[,2:ncolU] The 1st column of U has the color names
#
#
nrow <- length(T[,1])
ncol <- length(T[1,])
# Even though I have the color names in U this is more convenient
# because there are no leading or trailing blanks
names <- c("434 Indigo","445 Blue","465","472 Blue-Green","490","504 Green","537",
"555 Yellow-Green","584","600 Yellow","610","628 Orange-Yellow",
"651 Orange","674 Red")
#
# pos -- a position specifier for the text. Values of 1, 2, 3 and 4,
# respectively indicate positions below, to the left of, above and
# to the right of the specified coordinates
#
namepos <- rep(2,nrow)
#
TT <- rep(0,nrow*ncol)
dim(TT) <- c(nrow,ncol)
TTT <- rep(0,nrow*ncol)
dim(TTT) <- c(nrow,ncol)
#
xrow <- NULL
xcol <- NULL
xcount <- NULL
matrixmean <- 0
matrixmean2 <- 0
#
# Transform the Matrix
#
i <- 0
while (i < nrow) {
i <- i + 1
xcount[i] <- i
j <- 0
while (j < ncol) {
j <- j + 1
#
# Transform the Color "Agreement Scores" into Distances
#
TT[i,j] <- ((100-T[i,j])/50)**2 Note the Transformation from a 100 to 0 scale
# to a 0 to 4 (squared distance) scale
}
}
#
# Put it Back in T
#
T <- TT
TTT <- sqrt(TT)
#
#
# Call Double Center Routine From R Program
# cmdscale(....) in stats library
# The Input data are DISTANCES!!! Not Squared Distances!!
# Note that the R routine does not divide
# by -2.0
#
ndim <- 2
#
#
# returns double-center matrix as dcenter\$x if x.ret=TRUE
#
# Do the Division by -2.0 Here
#
TTT <- (dcenter\$x)/(-2.0)
#
#
# Below is the old Long Method of Double-Centering
#
# Compute Row and Column Means
#
i <- 0
while (i < nrow) {
i <- i + 1
xrow[i] <- mean(T[i,])
}
i <- 0
while (i < ncol) {
i <- i + 1
xcol[i] <- mean(T[,i])
}
matrixmean <- mean(xcol)
matrixmean2 <- mean(xrow)
#
# Double-Center the Matrix Using old Long Method
# Compute comparison as safety check
#
i <- 0
while (i < nrow) {
i <- i + 1
j <- 0
while (j < ncol) {
j <- j + 1
TT[i,j] <- (T[i,j]-xrow[i]-xcol[j]+matrixmean)/(-2)
}
}
#
# Run some checks to make sure everything is correct
#
xcheck <- sum(abs(TT-TTT))
#
#
# Perform Eigenvalue-Eigenvector Decomposition of Double-Centered Matrix
#
ev <- eigen(TT)
#
# Find Point furthest from Center of Space
#
aaa <- sqrt(max((abs(ev\$vec[,1]))**2 + (abs(ev\$vec[,2]))**2))
bbb <- sqrt(max(((ev\$vec[,1]))**2 + ((ev\$vec[,2]))**2))
#
# Weight the Eigenvectors to Scale Space to Unit Circle
#
torgerson1 <- ev\$vec[,1]*(1/aaa)*sqrt(ev\$val[1])
torgerson2 <- -ev\$vec[,2]*(1/aaa)*sqrt(ev\$val[2])
#
plot(torgerson1,torgerson2,type="n",asp=1,
main="",
xlab="",
ylab="",
xlim=c(-3.0,3.0),ylim=c(-3.0,3.0),font=2)
points(torgerson1,torgerson2,pch=16,col="red",font=2)
text(torgerson1,torgerson2,names,pos=namepos,offset=0.2,col="blue",font=2)
#
#
# Main title
mtext("The Color Circle \n Torgerson Coordinates",side=3,line=1.25,cex=1.5,font=2)
# x-axis title
mtext("",side=1,line=2.75,cex=1.2,font=2)
# y-axis title
mtext("",side=2,line=2.5,cex=1.2,font=2)
#```
Run the program and you should see something like this:
1. Run the program and turn in this figure.
2. Change the values in "namepos" until you have a nice looking plot with no overlapping names and no names cut off by the margins and turn in the resulting plot. Use the offset parameter in the text command to space the names away from the points. Also flip the dimensions as appropriate so that the configuration matches the one in Borg and Goenen!
3. Check out:
Color Circle in Color
Use the colors() command in R to find the corresponding colors and then use the text command to color the label for the corresponding point; e.g.,
text(torgerson1[10],torgerson2[10],"Yellow",pos=4,col="yellow",font=2)
You will have to override the labeling command in the double-center program to do this. Turn in the plot (hard copy in B/W and e-mail it to me so I can see it!) and the listing of your R program.
4. Plot the eigenvalues that are stored in ev\$value. The simplest plot is to use the command at the R prompt:
plot(1:14,ev\$value)
and you will get the following really lame graph:
Turn this into a nicer graph. To do this, check out the R code for Figure 5.4 of my book which is posted on my website at:
Chapter 5 Page
Turn in a final graph and a listing of the R code you use to create it.
4. Run Tables 1.1, 1.3, 1.4, and 4.1 from Borg and Groenen through KYST. Note that all four tables are similarities so that you can use the Supreme Court example as a template. (Note that you will have to enter the diagonals for tables 1.3 and 1.4 -- 9 and 7, respectively.) For example, for Table 4.1, if you entered the data as follows:
```
434_INDIGO 100 86 42 42 18 6 7 4 2 7 9 12 13 16
445_BLUE 86 100 50 44 22 9 7 7 2 4 7 11 13 14
465 42 50 100 81 47 17 10 8 2 1 2 1 5 3
472_BLUE-GREEN 42 44 81 100 54 25 10 9 2 1 0 1 2 4
490 18 22 47 54 100 61 31 26 7 2 2 1 2 0
504_GREEN 6 9 17 25 61 100 62 45 14 8 2 2 2 1
537 7 7 10 10 31 62 100 73 22 14 5 2 2 0
555_YELLOW-GREEN 4 7 8 9 26 45 73 100 33 19 4 3 2 2
584 2 2 2 2 7 14 22 33 100 58 37 27 20 23
600_YELLOW 7 4 1 1 2 8 14 19 58 100 74 50 41 28
610 9 7 2 0 2 2 5 4 37 74 100 76 62 55
628_ORANGE-YELLOW 12 11 1 1 1 2 2 3 27 50 76 100 85 68
651_ORANGE 13 13 5 2 2 2 2 2 20 41 62 85 100 76
674_RED 16 14 3 4 0 1 0 2 23 28 55 68 76 100
```
then change the format statement to:
(18X,101F4.0)
Be sure to change the missing values parameter to a negative number because there are some zeroes in the matrix. Your KYST file should look like the following:
```
TORSCA
PRE-ITERATIONS=3
DIMMAX=3,DIMMIN=1
PRINT HISTORY,PRINT DISTANCES
COORDINATES=ROTATE
ITERATIONS=50
REGRESSION=DESCENDING
DATA,LOWERHALFMATRIX,DIAGONAL=PRESENT,CUTOFF=-.01
EKMAN'S COLOR DATA EXAMPLE
14 1 1
(18X,101F4.0)
****the color data****
COMPUTE
STOP```
1. Run all tables from 3 to 1 dimensions, report the Stress values, and use R to graph the results for each table in two dimensions.
2. Produce Shepard Diagrams for each two dimensional solution in the same format as Homework 2. | 6,489 | 18,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-50 | latest | en | 0.91206 |
https://de.mathworks.com/matlabcentral/cody/problems/108-given-an-unsigned-integer-x-find-the-largest-y-by-rearranging-the-bits-in-x/solutions/2032933 | 1,603,187,226,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107871231.19/warc/CC-MAIN-20201020080044-20201020110044-00360.warc.gz | 284,132,880 | 16,976 | Cody
# Problem 108. Given an unsigned integer x, find the largest y by rearranging the bits in x
Solution 2032933
Submitted on 23 Nov 2019 by Daniel Fry
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 76; y_correct = 112; assert(isequal(maxit(x),y_correct))
2 Pass
x = 555; y_correct = 992; assert(isequal(maxit(x),y_correct))
3 Pass
x = 1000; y_correct = 1008; assert(isequal(maxit(x),y_correct))
4 Pass
x = 10000000; y_correct = 16711680; assert(isequal(maxit(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 212 | 740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-45 | latest | en | 0.711968 |
https://www.aslanides.io/blog/cfar-games | 1,582,199,641,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00145.warc.gz | 662,331,632 | 7,641 | This is the second of two posts relating my Bay area visit of August-September 2016. You can read the first part here.
This is a very (6+ months late) belated write-up about my experience at the Summer 2016 Workshop for AI Researchers run by the Center for Applied Rationality (CFAR). To add insult to injury, I’m not even going to talk about the rationality techniques we studied! As a short summary though, the workshop content, the participants, and above all the CFAR people were all awesome, and I got a huge amount out of attending. The atmosphere throughout was electric and immensely fun. I’ll be writing more about quantified self & rationality practice over the coming months, as I get this blog up to speed.
The workshop was held from August 30 - September 5 at Ellen Kenna House, a beautiful Victorian mansion that sits on the top of a hill in Oakland, California. There were about ~30 participants and ~10 CFAR staff. The participants were AI researchers hailing from a variety of places: BAIR, SAIL, MIRI, FHI, IDSIA, Google Brain, MILA, were some of the labs/institutes represented.
I had 'Bino's album Because the Internet on loop the whole week. Apt or corny?
In this post I’m going to formulate two fun prediction/betting games that I learnt about at the workshop: Calibration Market and Bid/At.
## Calibration Market
This is related to the game of the same name by Andrew Critch. I’ll formulate it somewhat formally here.
A calibration market $M$ is a tuple $\left(P,\tau,B,\right)$, where $P$ is a proposition that is evaluated at some time $\tau$. $B$ is a finite sequence of bets $b_0b_1b_2\dots$, where each bet $b_i\in\(0,1)$ is interpreted as the subjective credence of a player that $P$ evaluates to $1$ at time $\tau$. For obvious reasons, values of 0 and 1 are not permitted. The market is initialised with a prior $b_0$, usually bet by the house or market maker, who may in turn participate in the market subsequently. Once $P$ is evaluated, the market closes. Each bet (except the first) is scored by the log ratio with the previous bet
Informally, you gain points if you move the market towards the truth, and lose points if you move it away from the truth. For each player $p_j$, you add up the total score for the market.
Some other rules:
• You may not bet after yourself, and
• You may not alternate bets with someone else.*
* This second rule is deliberately left flexible and informal: how it is interpreted, and how strictly it is applied is down to taste.
The equation above is an example of a proper scoring rule: to maximize your expected score one should always bet your true beliefs – this was pointed out to me by Matthew Graves. Let your true belief be $a$, and your bet be $b$. Let the previous bet be $c$, though we will see that it turns out the value of $c$ doesn’t matter. Consider your $a$-expected score as a function of $b$:
We can maximize the expected value simply by setting the derivative to zero and solving for $b$, since it is trivial to show that $\mathbb{E}[S]$ is concave. The derivative is given by
which naturally implies that one’s expected score is maximized if one bets one’s true beliefs:
The animation below illustrates this, showing how the optimal bet changes as we sweep our belief over the interval $[0,1]$:
Of course, one of the central notions at CFAR is recursive: becoming more rational requires metacognition, i.e. reasoning about reasoning. For this reason, when we played the game, many of the propositions were self-referential in nature. This makes for much more interesting markets, and goes to the whole meta theme. Propositions were typically of the form of “a randomly selected participant will willingly do/say X when prompted”, or “the outcome of this market will be X”, rather than propositions of the form “it will/won’t rain tomorrow”. There were numerous calibration markets run throughout the week, and they added to the generally stimulating and fun environment.
## Bid/At
I’m told that this game is played a lot at Jane Street. The game is simple: You make a market out of some future/unknown outcome $Q$ that is quantifiable. For simplicity, let $Q$ be a positive-valued random variable. The market consists of two or more players buying and selling contracts relating to the outcome $Q$:
A contract between two players $p_1$ and $p_2$ is executed as follows: $p_1$ pays $p_2$ $x>0$ credits, and receives in exchange a promise stipulating that after the value of $Q$ is known, $p_2$ will pay P1 $Q$ credits. Information flows in the market by players sending price signals to each other: a player may bid a value $x$ to indicate they are willing to buy a contract for $x$ credits, or conversely declare they are at a value $y$, which indicates they are willing to sell a contract for $y$ credits.
In general one would bid $x$ if one believes that $\mathbb{E}[Q] > x$, and accept bids at $x$ if one believes that $% $. Like the calibration game above, players can make inferences about the beliefs of other players (based on bid/at prices) and update their own beliefs; presumably the market converges to a fixed price under certain conditions, though I haven’t thought about it enough to write down what those conditions must be.
The main interesting game-theoretic difference between Bid/At and Calibration is that in Bid/At, you are incentivized to hide your subjective belief $\mathbb{E}[Q]$, so as to potentially reap the biggest margins when buying/selling contracts. In any case, it’s a game that rewards forming accurate predictive models about the world, with an added element of risk management.
If played in earnest, the players would keep records of all contracts, and when $Q$ is realized, the contracts are paid out at some (pre-determined) credits-to-dollars conversion rate.
## Bonus: Group Decision Algorithm
Duncan told us about this algorithm in the context of coming to a decision about which restaurant to go to as a group of people. The algorithm is simple:
• Anyone can suggest one or more new options for consideration at any time.
• If only one option remains under consideration without veto, and no more options are suggested, then this option is chosen and that’s the final decision.
• If more than one option is under consideration sans veto, the final decision goes to a vote, using the usual procedure for redistributing preferences in the case of ties, breaking deadlocked ties at random.
• During the candidate selection process, anyone can veto an option at any time; this option is then immediately and permanently removed from consideration. The person that made the veto must then generate three new suggestions.
The algorithm is designed to completely obviate the situation in which someone suggests some option which gets immediately shot down/vetoed without any promising alternatives presented in its place. Once a first option is suggested, the algorithm is guaranteed to produce a decision that satisfies people’s preferences (assuming that people actually veto options they aren’t happy with), provided the set of feasible options isn’t exhausted; in the case of restaurants this certainly won’t happen, since everyone’s gotta eat eventually :). The multiplicative factor of three seems well-tuned to yield good results and fast convergence: it’s just enough to disincentivize flippant vetos, but small enough to not make vetoing too onerous (e.g. most people can easily suggest three new restaurants), so that everyone has a good chance of having their preferences satisfied.
There’s more interesting analysis to be done here (if it hasn’t been done already). Once I have the time to learn more about Bayesian markets, I’m sure there’ll be some interesting microeconomic/game-theoretic analysis of Bid/At. I’d also quite like to implement both Bid/At and Calibration as fun small-scale asynchronous multiplayer web app-based games. That would be an intereesting engineering project, and would be a pretty fun tool for the rationality community to practice on, to supplement things like PredictionBook. To be continued! | 1,787 | 8,086 | {"found_math": true, "script_math_tex": 43, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-10 | longest | en | 0.958958 |
https://www.bartleby.com/questions-and-answers/morganton-company-makes-one-product-and-it-provided-the-following-information-to-help-prepare-the-ma/4f0e9f92-417f-41df-98eb-038afab00dd8 | 1,585,554,046,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496669.0/warc/CC-MAIN-20200330054217-20200330084217-00122.warc.gz | 840,186,563 | 29,601 | Morganton Company makes one product and it provided the following information to help prepare the master budget: The budgeted selling price per unit is \$60. Budgeted unit sales for June, July, August, and September are 9,500, 26,000, 28,000, and 29,000 units, respectively. All sales are on credit.Forty percent of credit sales are collected in the month of the sale and 60% in the following month.The ending finished goods inventory equals 25% of the following month’s unit sales.The ending raw materials inventory equals 15% of the following month’s raw materials production needs. Each unit of finished goods requires 4 pounds of raw materials. The raw materials cost \$2.40 per pound.Forty percent of raw materials purchases are paid for in the month of purchase and 60% in the following month.The direct labor wage rate is \$12 per hour. Each unit of finished goods requires two direct labor-hours.The variable selling and administrative expense per unit sold is \$1.50. The fixed selling and administrative expense per month is \$65,000.
Question
2769 views
Morganton Company makes one product and it provided the following information to help prepare the master budget:
1. The budgeted selling price per unit is \$60. Budgeted unit sales for June, July, August, and September are 9,500, 26,000, 28,000, and 29,000 units, respectively. All sales are on credit.
2. Forty percent of credit sales are collected in the month of the sale and 60% in the following month.
3. The ending finished goods inventory equals 25% of the following month’s unit sales.
4. The ending raw materials inventory equals 15% of the following month’s raw materials production needs. Each unit of finished goods requires 4 pounds of raw materials. The raw materials cost \$2.40 per pound.
5. Forty percent of raw materials purchases are paid for in the month of purchase and 60% in the following month.
6. The direct labor wage rate is \$12 per hour. Each unit of finished goods requires two direct labor-hours.
7. The variable selling and administrative expense per unit sold is \$1.50. The fixed selling and administrative expense per month is \$65,000.
check_circle
Step 1
Master Budget:
A set of different budgets prepared based on the functions or processes available in a business or company is called master budget. It includes the information of the operations and finance.
Master budget includes sales budget, production budget, raw material budget, direct labour budget, all overhead budgets, and cash budget.
Step 2
Prepare a master budget...
Want to see the full answer?
See Solution
Want to see this answer and more?
Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*
See Solution
*Response times may vary by subject and question.
Tagged in | 617 | 2,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-16 | latest | en | 0.908021 |
https://www.letsaskques.com/find-the-function-hx-fx-gx-if-fx3x-and-gx-32x-3x/ | 1,643,067,199,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00239.warc.gz | 923,776,336 | 22,329 | # Find the function h(x) = f(x) – g(x) if f(x)=3^x and g(x) = 3^2x – 3^x
Find the function h(x) = f(x) – g(x) if f(x)=3^x and g(x) = 3^2x – 3^x | 75 | 144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-05 | latest | en | 0.324489 |
https://bicycles.stackexchange.com/questions/43517/estimating-trainer-effort-above-the-flat-by-heart-rate-cadence | 1,721,244,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00602.warc.gz | 113,528,047 | 44,122 | Estimating trainer 'effort above the flat' by heart rate + cadence
A question popped into my head after yesterday's trainer ride.
First, a long-winded preamble.
As I pointed out in this question, I usually set the trainer up so that a 5l cask of wine just moves the pedal from horizontal in 'top' (48:14).
In that configuration, my average HR will stay in the 130s at a cadence of 80 in 'top'.
If I ride on the road, my HR seldom gets above 130 except on hills or if I crank up the cadence to 100+ in top (on the flat): I am loath to ride at any decent speed on the road because I am usually surrounded by people of median IQ driving 1500kg+ steel boxes - and I don't trust any of them not to kill me (assuming they're not actually trying to).
Yesterday I didn't have a full cask anywhere (I've been off the grog for six weeks, on a bet), so I just cranked it up until I couldn't move the pedal using my hand when seated next to the trainer.
This provided 'ample' resistance, to say the least: 2 minutes in, my HR was in the 140s although my cadence was only in the low 70s - and not in top gear, either - the average speed for the workout was 22km/hr.
Doing 'sprints' (15sec bursts at a cadence of > 130 in top) was almost the end of me.
Half an hour later, my HRM said I'd exerted ~500kcal over 10.7km, with an average HR of 148 and a peak HR of 166. The kcal count is about right for workouts where I do more sprints, and at a significantly higher average speed (i.e., a distance travelled of ~17km, reflected by an average cadence of 80 in top: cadence maps directly to speed via gear ratio).
So if I expressed this as "what is X in 48:X, to get that distance at a cadence of 80", the answer is "a little over 22". In other words, riding on (roughly) the 4th-highest gear, my HR was 10% higher than it usually is riding at the same cadence in top.
That got me thinking: rather than work out the effort above my 'wine cask' resistance level, I ought to be able to work out the 'effort above flat' implied by the level of resistance I dial in, by
• finding decent-sized bit of car-free flat (say, a 1km loop somewhere);
• going around it at a series of stable cadences (in top) and HRs (say, cadences of 80 and 100; HRs of 120, 130, and 140) each for a few minutes.
For each cadence, I would then know my 'steady state' HR on the flat (it will stabilise after a few minutes); for each HR, I will know what speed/cadence on the flat is consistent with that HR.
So if my HR is, say 10% higher (on average across the workout) than it was on the flat 'benchmark' for the same cadence in the same gear, I would chalk that one up as being '10% above the flat'. (I'm not pretending that's a 10% incline, or even a 10 degree incline; just that it's objectively 10% 'harder').
I realise that over time it would be expected that performance would get better, so there would be a need to periodically re-assess the 'flat' performance.
So finally the question: has this sort of analysis already been done and put up on the web somewhere?
I've googled terms I think might yield results, but haven't found a sausage. It seems to me that performance-metrics boffins would probably have solved this already (assuming it's an issue that has benefit for performance monitoring, which I reckon it does) so my search terms are probably a bit pants.
• Heart rate is sensitive to temperature, hydration status, duration, and fatigue, which makes your quest for an answer here even more difficult than for your previous question. Commented Oct 31, 2016 at 19:03
• How old are you? (approx if you prefer)
– Criggie
Commented Oct 31, 2016 at 21:35
That got me thinking: rather than work out the effort above my 'wine cask' resistance level, I ought to be able to work out the 'effort above flat' implied by the level of resistance I dial in
What you are attempting to do is calibrate your heart-rate on the indoor trainer to real-world riding - and this is far too complicated. Since there are so many variables to consider. Heart-rate is an indicator and not as accurate a measure as a power-meter. Your heart-rate can vary for the same effort over the same terrain depending on diet, recovery, fatigue and so on. Also riding outdoors there is the issue of weather conditions also - which can be for you or against you.
Another factor I can see is - although your indoor trainer attempts to provide a "road-like feel" - and all indoor trainers boast of providing the most realistic road-like feel. It will have a spin-down - that is the period which the trainer continues to spin after pedaling has stopped. This represents resistive inertia - good ones have excellent resistance and also take longer to spin down and have a nice "feel" to them when riding. So - your indoor trainer will not be a perfect representation of real world riding.
And don't forget - perceived effort - for me - indoors is often greater than perceived effort outdoors. I know for fact - my indoor power stats are lower than my outdoor power stats.
Personally, I would be setting the resistance at a single-level and changing through the gears to alter resistance. Looking purely at speed as a measure of your power - and using this as a basis for training. Might be a bit tricky - because it is speed and not power (everything is watts these days!). But you can combine this with your heart-rate data to know your speed and HR zones for training.
• I've been toying with buying a decent ('smart', Zwift-compatible) trainer for a while - until then I was trying to do things using shoestrings and sticky-tape. I was going to go 'all in' and buy a Wahoo KickR until it was made clear to me that I would need to buy another bike (because my gearset is incompatible with a KickR: my bike's a \$220, 15kg, clunker). Plus, The Lovely wouldn't give Royal Assent to the purchase until I 'got some time up' on the current trainer (a \$69 Aldi job) - today she gave me the all-clear, so I'm allowed to order a bKool Smart Pro.
– GT.
Commented Nov 5, 2016 at 3:43
• That's great news :-) Check compatibility of the BKool here support.zwift.com/hc/en-us/articles/… Commented Nov 6, 2016 at 0:16
• I eventually went for a Wahoo KickR and a new bike (a Merida Scultura). My Zwift avatar looks buff (6'1" and 104kg, as imagined by a bike-happy coder-geek, in Wahoo kit) and all my questions about power and what-not are superfluous because now I know the answer: big fat old blokes can put half-decent power to ground ('tempo' FTP of 271 - i.e., can do 285W for 20 minutes without getting HR anywhere near 'threshold')... but I'm totally pants at hills. Just like in real life - but with no "insecurity wagons" (SUVs) trying to kill me.
– GT.
Commented Aug 10, 2018 at 9:22
Might I enquire your age? This answer will be a bit vague without that, but you can work it out backwards.
The rule of thumb says that your max heart rate is 220-(your age in years) I'm 40, so 180 would be my "100%" heart rate.
This table shows cycling
This chart shows the "zones"
So your 130 BPM is likely to be the "weight loss zone" for under 50 or the Aerobic zone for over 50.
Back to cycling, my "tempo" zone would be 70-80% of 180, so 126-144 BPM.
Here's a short ride from last week - heart rate was above 160 for at least half, and a peak of 173. That translates to higher threshhold and into VO2, which is pretty good, giving speeds of mid 30km/h to a peak of 42 km/h. https://www.strava.com/activities/757787829/analysis
So to summarise - 130 BPM is a good "endurance" workout. You'd need to be doing that speed for a couple hours, and frankly that's boring on a trainer.
If you're dead against training on the road, try riding at a velodrome or some other kind of off-road track. An athletic track may work, or around the local park?
If you can work a decent climb into your trip then that will be excellent exercise. Use a tool like strava to measure your gains, cos it never feels faster.
• Its not a great answer but too long for a comment, and we can't post images in comments either.
– Criggie
Commented Oct 31, 2016 at 8:24
• I'm 51; the 130 is the average for a HIIT workout where each of the "sprints" gets my HR to 'max', with a sprint/rest split of 30-on, 90-off, usually done 10 times with steady-state (HR in the 120s) for warm-up and cool down - that gets me to 30 mins. Every now and then I do something more 'constant' for an hour (knowing that I won't wizz properly for two days: no amount of seat- and knick-padding can prevent my 104kg from causing pressure on the (ahem) 'contact' region... the legendary 'scranus'). I'm moving to a Zwift-capable 'smart' trainer in the next month or so.
– GT.
Commented Nov 5, 2016 at 3:10
• I forgot to mention: I do use Strava sometimes - the last workout (done with fewer, shorter, more intense sprints than usual to check the correspondence between sensors) is at strava.com/activities/759984056 (the peak speed measure is clearly a glitch!).
– GT.
Commented Nov 5, 2016 at 3:22
• @GT. Been there - anytime I ride the other-half's bike, I have to stand up almost the whole way. You might benefit from trialing other saddles.
– Criggie
Commented Nov 5, 2016 at 3:24
• I've tried every saddle type from 'big arse' saddles (chafe!!) to the ones that have two separate pads and no 'taint-destroyer' bits (that just felt weird, even on the trainer). I measured my sit-bone gap using the 'corrugated cardboard' tip (c-to-c was 140mm - I have Māori bum-bones... World Champs, Cuz!) and got a bike fit. A good mate who rides 'proper' races (A-grade seniors on a \$10k bike) suggested a useful solution: "HTFU, old man". The statistical risk of permanent taint-related damage from bike-riding is near-zero, but that's cold comfort "the day after".
– GT.
Commented Nov 7, 2016 at 0:23 | 2,538 | 9,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-30 | latest | en | 0.954643 |
https://www.shaalaa.com/question-bank-solutions/nature-roots-for-what-value-k-4-k-x2-2k-4-x-8k-1-0-perfect-square_23587 | 1,597,153,204,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738777.54/warc/CC-MAIN-20200811115957-20200811145957-00502.warc.gz | 831,492,104 | 9,383 | Share
# For What Value of K, (4 - K)X2 + (2k + 4)X + (8k + 1) = 0, is a Perfect Square. - Mathematics
Course
#### Question
For what value of k, (4 - k)x2 + (2k + 4)x + (8k + 1) = 0, is a perfect square.
#### Solution
The given quadric equation is (4 - k)x2 + (2k + 4)x + (8k + 1) = 0, and roots are real and equal
Then find the value of k.
Here, a = (4 - k), b = (2k + 4) and c = (8k + 1)
As we know that D = b2 - 4ac
Putting the value of a = (4 - k), b = (2k + 4) and c = (8k + 1)
= (2k + 4)2 - 4 x (4 - k) x (8k + 1)
= 4k2 + 16k - 16 - 4(4 + 31k - 8k2)
= 4k2 + 16k - 16 - 16 - 124k + 32k2
= 36k2 - 108k + 0
= 36k2 - 108k
The given equation will have real and equal roots, if D = 0
Thus,
36k2 - 108k = 0
18k(2k - 6) = 0
k(2k - 6) = 0
Now factorizing of the above equation
k(2k - 6) = 0
So, either
k = 0
or
2k - 6 = 0
2k = 6
k = 6/2 = 3
Therefore, the value of k = 0, 3.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Solution for Class 10 Maths (2018 (Latest)) | 463 | 1,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-34 | latest | en | 0.745371 |
https://psichologyanswers.com/library/lecture/read/93890-what-is-homogeneous-condition | 1,720,827,617,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00043.warc.gz | 406,430,958 | 8,392 | # What is homogeneous condition?
## What is homogeneous condition?
In mathematics, a homogeneous function is one with multiplicative scaling behaviour: if all its arguments are multiplied by a factor, then its value is multiplied by some power of this factor. For example, a homogeneous real-valued function of two variables x and y is a real-valued function that satisfies the condition.
## What is homogeneity in sociology?
In sociology, role homogeneity is the degree of overlap amongst the different roles performed by different members of a community.
## What do you mean by homogeneity?
1 : the quality or state of being of a similar kind or of having a uniform structure or composition throughout : the quality or state of being homogeneous. 2 mathematics : the state of having identical cumulative distribution functions or values.
## What is homogeneity in psychology?
1. equality or near equality, particularly between two statistical quantities of interest. The term most often is used in connection with different populations.
## What is homogeneity in business?
Dictionary of Business Terms for: homogeneous. homogeneous. having the same composition or form; organization that produces or sells products having great similarities, often using the same components. Homogeneous products reduce organizational development and manufacturing costs.
## What is the difference between homogeneous and differentiated products?
1. Homogeneous and Differentiated Products. The identical products produced under Pure Competition are often called homogenous products. ... The different types of products produced under Monopolistic Competition and Oligopoly are often called differentiated products.
## What is a homogeneous workplace?
Characteristics of Workplace Diversity A homogeneous workforce is made up of employees who have a lot in common, such as their racial, marital and educational status. As a result, it's possible that their similarities result in similar problem-solving methods when it comes to facing challenges in the workplace.
## What are advantages of homogeneity?
1. Reduced conflicts: One of the biggest advantage of the homogeneous group is that the chances or probability of conflict are negligible. 2. Better coordination: As the homogeneous group have similar shared values and agree to each other thus, such groups have better coordination.
## What is the difference between homogeneous and heterogeneous grouping of students?
Homogeneous grouping Homogeneous groups are the opposite of heterogeneous groups, instead they are made up of students working at the same comprehension level. Everyone in the group is capable of doing the task and mastering the skill at the same level.
Abstract. The greatest advantage of heterogeneous catalysis is the ease of separation, while the disadvantages are often limited activity and selectivity. We report solvents that use tunable phase behavior to achieve homogeneous catalysis with ease of separation.
Through homogenization, people can also obtain easier and better quality ofcommunication. There are also disadvantages of homogenization and one is that it destroys. It can also result in the loss of individual culture andreligion and less diversity of ideas.
## What are the advantages of heterogeneous?
A heterogeneous group gives advanced students a chance to mentor their peers. All members of the group may interact more to help each other understand the concepts being taught.
## What is the difference between homogeneous and heterogeneous groups?
Heterogeneous grouping is when a diverse group of students is put in the same cooperative learning group. ... Homogeneous grouping is the distribution of students who function at similar academic, social, and emotional levels, being placed in the same cooperative learning group.
## What is a homogeneous community?
A community can be one of two different things. ... A homogeneous community is one in which all of the members share a similar set of beliefs, values, and demographic characteristics. Homogeneous communities tend to be comprised of individuals of the same race, nationality, and religious and political beliefs.
## Why are homogeneous mixture beneficial to us?
With uniform distribution, the plant roots absorb a complete set of all applied nutrients. Research has shown that the maximum response from nitrogen and phosphate occurs when the two are associated. This kind of chemical combination is found only in homogeneous pellets or high quality pellet blends.
## Why is homogeneous important?
In science, the most common use of homogeneous is to classify materials. Mixtures, substances and solutions can have different characteristics, and their homogeneity gives clues about how they will behave. This is one of the most important scientific uses of the term.
## What is characteristic of homogeneous?
A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture. ... All solutions would be considered homogeneous because the dissolved material is present in the same amount throughout the solution. One characteristic of mixtures is that they can be separated into their components.
## How do you tell if a reaction is homogeneous or heterogeneous?
Homogeneous reactions are chemical reactions in which the reactants and products are in the same phase, while heterogeneous reactions have reactants in two or more phases. Reactions that take place on the surface of a catalyst of a different phase are also heterogeneous.
## What is homogeneous equilibrium with example?
A homogeneous equilibrium is one in which all species are present in the same phase. Common examples include gas-phase or solution reactions. A heterogeneous equilibrium is one in which species exist in more than one phase. Common examples include reactions involving solids and gases, or solids and liquids.
## What is homogeneous equilibrium give two examples?
A homogeneous equilibrium has everything present in the same phase. The usual examples include reactions where everything is a gas, or everything is present in the same solution. A heterogeneous equilibrium has things present in more than one phase.
## What is homogeneous equilibrium Class 11?
1) Homogeneous Equilibria When in an equilibrium reaction, all the reactants and the products are present in the same phase , it is called homogeneous equilibrium. Type 1 : In which the number of moles of products is equal to the number of moles of reactants.
## What is homogeneous equilibrium?
A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture ). Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both.
## What is homogeneous and heterogeneous equilibrium give examples?
If a reaction is at equilibrium and its all reactants and products are in the same phase or at same state of matter, then this reaction is known as homogeneous equilibrium reaction or we say the reaction is at homogeneous equilibrium. For example, the reaction of carbon monoxide and hydrogen.
## Which chemical system is homogeneous?
A homogeneous system or the phase of a heterogeneous system that consist of several pure substances is called a solution or mixture. All pure substances and solutions can exist in three states of aggregation: gas, liquid and solid.
## What is an example of a heterogeneous reaction?
Examples. The reaction between acid and metal is a heterogeneous reaction. A reaction between a gas and a liquid, as between air and seawater, is heterogeneous. A reaction at the surface of a catalyst is heterogeneous. | 1,399 | 7,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.958351 |
https://homework.zookal.com/questions-and-answers/given-cost-and-revenue-functions-and-upper-c-left-parenthesis-344102451 | 1,624,114,387,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648373.45/warc/CC-MAIN-20210619142022-20210619172022-00357.warc.gz | 274,096,228 | 26,395 | 1. Other
2. Other
3. given cost and revenue functions and upper c left parenthesis...
# Question: given cost and revenue functions and upper c left parenthesis...
###### Question details
Given cost and revenue functions and Upper C left parenthesis q right parenthesis equals 115 q plus 43000C(q)=115q+43000 and Upper R left parenthesis q right parenthesis equals negative 3 q squared plus 3000 qR(q)=−3q2+3000q, how many items must the company produce and sell to break even? | 115 | 481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-25 | latest | en | 0.61289 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.