SumOfSquaresRepresentations
in [Mathematica]
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From: janos on 23 Sep 2009 23:51 To be on the safe side: ClearAll[d, n, x, a, SumOfSquaresRepresentations]; Take this from the help: SumOfSquaresRepresentations[d_, n_] := Module[{x, a}, (a = Array[x, d]) /. {ToRules[Reduce[a.a == n, a, Integers]]}] Let's apply it: SumOfSquaresRepresentations[3, 15] {x$110[1], x$110[2], x$110[3]} And this is the only case that you get this funny result writing any integer between 1 and 200 in the place of 15. Any idea? Thank you. J=E1nos
From: Andrzej Kozlowski on 24 Sep 2009 07:50 First of all, you do not need to define yourself a function that does this, since one already exists: PowersRepresentations[15, 3, 2] {} Now, this also tells you the solution to your problem! The solution is that there are no solutions and you have written your program assuming that there always are going to be any. Really, to see that all you needed was to look at the output of: With[{a = Array[x, 3]}, Reduce[a.a == 15, a, Integers]] False Actually, your program will also not work correctly in other cases when there are no solutions. Here are the numbers of ways of representing the first 10 integers as sum of squares SquaresR[3, Range[10]] {6,12,8,6,24,24,0,12,30,24} We can see that for n = 7 there are no solutions and your code will produce: Module[{x, a}, (a = Array[x, 3]) /. {ToRules[Reduce[a.a == 7, a, Integers]]}] {x$2267(1),x$2267(2),x$2267(3)} So how come you claim that 15 is the only integer between 1 and 200 for which this happens?? Andrzej Kozlowski On 24 Sep 2009, at 12:51, janos wrote: > To be on the safe side: > ClearAll[d, n, x, a, SumOfSquaresRepresentations]; > > Take this from the help: > SumOfSquaresRepresentations[d_, n_] := > Module[{x, > a}, (a = Array[x, d]) /. {ToRules[Reduce[a.a == n, a, Integers]]}] > > Let's apply it: > SumOfSquaresRepresentations[3, 15] > {x$110[1], x$110[2], x$110[3]} > > And this is the only case that you get this funny result > writing any integer between 1 and 200 in the place of 15. > > Any idea? > > Thank you. > > J=E1nos >
From: Bob Hanlon on 24 Sep 2009 07:51 You just need to handle the case when there is no solution SumOfSquaresRepresentations[d_, n_] := Module[{x, a, rules}, a = Array[x, d]; rules = {ToRules[Reduce[a.a == n, a, Integers]]}; If[rules == {}, {}, a /. rules]] Bob Hanlon ---- janos <janostothmeister(a)gmail.com> wrote: ============= To be on the safe side: ClearAll[d, n, x, a, SumOfSquaresRepresentations]; Take this from the help: SumOfSquaresRepresentations[d_, n_] := Module[{x, a}, (a = Array[x, d]) /. {ToRules[Reduce[a.a == n, a, Integers]]}] Let's apply it: SumOfSquaresRepresentations[3, 15] {x$110[1], x$110[2], x$110[3]} And this is the only case that you get this funny result writing any integer between 1 and 200 in the place of 15. Any idea? Thank you. J=E1nos
From: Carl Woll on 24 Sep 2009 07:52 janos wrote: > To be on the safe side: > ClearAll[d, n, x, a, SumOfSquaresRepresentations]; > > Take this from the help: > SumOfSquaresRepresentations[d_, n_] := > Module[{x, > a}, (a = Array[x, d]) /. {ToRules[Reduce[a.a == n, a, Integers]]}] > > Let's apply it: > SumOfSquaresRepresentations[3, 15] > {x$110[1], x$110[2], x$110[3]} > > And this is the only case that you get this funny result > writing any integer between 1 and 200 in the place of 15. > > Any idea? > > Thank you. > > J=E1nos > > Things are going wrong because there are no such representations. If you have version 6/7, then you can use the built in function PowersRepresentations instead. For example: In[36]:= PowersRepresentations[15, 3, 2] Out[36]= {} showing you that there are no such representations. Carl
From: janos on 25 Sep 2009 05:57 Thank to All, To All, This is not MY program, I have taken it from the help, more precisely from here: Compatibility/tutorial/NumberTheory/NumberTheoryFunctions To Andrzej, You are right in the last point: there are 32 integers with no decompositions between 1 and 200, and not one, my claim was well founded by a mistake. Still, a function like the one I have taken from the help, IS needed, because PowersRepresentations only gives the numbers of different representations. To Bob, Thanks for fixing the problem, it should be included into the help and into the corresponding item of mathworld as well. Best wishes, J=E1nos On szept. 24, 13:50, Andrzej Kozlowski <a...(a)mimuw.edu.pl> wrote: > First of all, you do not need to define yourself a function that does > this, since one already exists: > > PowersRepresentations[15, 3, 2] > > {} > > Now, this also tells you the solution to your problem! The solution is = > that there are no solutions and you have written your program assuming = > that there always are going to be any. Really, to see that all you > needed was to look at the output of: > > With[{a = Array[x, 3]}, Reduce[a.a == 15, a, Integers]] > > False > > Actually, your program will also not work correctly in other cases > when there are no solutions. Here are the numbers of ways of > representing the first 10 integers as sum of squares > > SquaresR[3, Range[10]] > > {6,12,8,6,24,24,0,12,30,24} > > We can see that for n = 7 there are no solutions and your code will > produce: > > Module[{x, > a}, (a = Array[x, 3]) /. {ToRules[Reduce[a.a == 7, a, Intege= rs]]}] > > {x$2267(1),x$2267(2),x$2267(3)} > > So how come you claim that 15 is the only integer between 1 and 200 > for which this happens?? > > Andrzej Kozlowski > > On 24 Sep 2009, at 12:51, janos wrote: > > > To be on the safe side: > > ClearAll[d, n, x, a, SumOfSquaresRepresentations]; > > > Take this from the help: > > SumOfSquaresRepresentations[d_, n_] := > > Module[{x, > > a}, (a = Array[x, d]) /. {ToRules[Reduce[a.a == n, a, Integer= s]]}] > > > Let's apply it: > > SumOfSquaresRepresentations[3, 15] > > {x$110[1], x$110[2], x$110[3]} > > > And this is the only case that you get this funny result > > writing any integer between 1 and 200 in the place of 15. > > > Any idea? > > > Thank you. > > > J=E1nos
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