Hello Raul.
My work is not differential equations. It is simpler.
It is about generalizing the concept of one mean value into two or more values,
characteristics of a dataset.
Here are some simple examples.
] a =. (i.11) , 5#20 NB. some test data. 16 numbers
0 1 2 3 4 5 6 7 8 9 10 20 20 20 20 20
0j1":b =. 3 summary@tomb a NB. 16 numbers summarized into 3 numbers.
1.9 7.4 19.7
0j1":3 summary@tomb b NB. 3 numbers summarized into the same 3 numbers.
1.9 7.4 19.7
0j1":2 summary@tomb a NB. 16 numbers summarized into 2 numbers
2.3 17.1
0j1":2 summary@tomb b NB. 3 numbers summarized into the same 2 numbers
2.3 17.1
0j1":1 summary@tomb a NB. 16 numbers summarized into 1 number
9.7
0j1":1 summary@tomb b NB. 3 numbers summarized into the same number
9.7
Thank you!
Bo
Den lørdag den 22. august 2020 18.57.38 CEST skrev Raul Miller
<[email protected]>:
Hmm...
Trying to bring myself up to speed on what you seem to be saying here
first lead me to https://mathworld.wolfram.com/PowerMean.html and
https://en.wikipedia.org/wiki/Newton%27s_identities and then to
https://www.jsoftware.com/papers/MSLDE1.htm
I don't have anything meaningful to add right now -- I am not even
sure if I can form a relevant question. Maybe if I spend a half hour a
day on this for long enough, I'll be able to come back to it
intelligently.
Until then, good luck.
--
Raul
On Sat, Aug 22, 2020 at 11:20 AM 'Bo Jacoby' via Programming
<[email protected]> wrote:
>
> Thanks to Raul and Brian!
> tomb=:(^/ i.@>:)~ , NB. It works miracuously! Thank you!
>
> Raul wrote: "It sort of smells like you were working with matrices designed
> for inversion and polynomial fitting"
> Well, I want to find n numbers such that their mean powers are equal to the
> mean powers of my input data set. I have n nonlinear equations in n
> unknowns. The trick is to use the symmetry of the power sums. (A power sum
> does not depend on the order of the variables). So the n unknowns are roots
> in a polynomial of degree n. The coefficients of this polynomial are
> solutions to n linear equations in n unknowns. See Newton's identities
>
> |
> |
> | |
> Newton's identities
>
> Let x1, ..., xn be variables, denote for k ≥ 1 by pk(x1, ..., xn) the k-th
> power sum:
> |
>
> |
>
> |
>
>
> So yes, I compute the n*n-matrix m from the n power sums s, and solve the
> linear equation m +/ .* e = s (by e=.(%.m)s), and the the nonlinear equation
> e p. x = 0 (by p. e). It is not polynomial fitting. Thank you for your help
> and for your interest!
> Bo
>
>
>
> Den lørdag den 22. august 2020 15.44.33 CEST skrev Raul Miller
><[email protected]>:
>
> I don't really follow what you are doing here -- for example, I look
> at stuff like 11 summary@hist i.11 and do not understand how that
> result corresponds to "size of the dataset" -- but even without
> knowing your intentions, I can point out that you can simplify tomb:
>
> tomb=:(^/ i.@>:)~ ,
>
> It sort of smells like you were working with matrices designed for
> inversion and polynomial fitting, but that doesn't tell me much (if
> anything) about your destination.
>
> Good luck,
>
> --
> Raul
>
> On Sat, Aug 22, 2020 at 5:35 AM 'Bo Jacoby' via Programming
> <[email protected]> wrote:
> >
> > Thanks to Michael and to ethiejiesa!
> > The program is now:
> >
> > tomb=.[ (^/ i.@>:)~ ,@] NB. Powers of a tombola
> >
> > hist=.]* ((^/&i. >:)~ #) NB. Powers of a histogram
> > s=.(*#)@(}.%{.)@(+/) NB. summation
> > f1=.$~2 # # NB. n*n matrix
> > f2=.>:@i.@#,}: NB. insert S0, remove Sn
> > f3=.*_1^i.@# NB. change signs
> > f4=.|:|.!.0"0 1~-@i.@# NB. shift in zeroes
> > m=.|:@f4@(f3"1@f2@|:@f1) NB. Matrix
> > e=.1,(%.m) NB. solve lin.eqns.
> > pol=._1 x: -@|.@>@{:@p.@|. NB. solve alg.eqn.
> > hp=.x: :: ] NB. high precision
> > summary=.pol@e@s@hp f. NB. complete program
> > (I need the comma in tomb, so now it is smelly again.)
> > Michael wrote: "Any clue as to the input to summary, or what you're
> > summarising? Sorry if it's obvious!"
> > No sir, it is not obvious if it is not obvious to you.
> > Left input to summary@tomb and summary@hist is the size of the output
> > dataset.
> > Right input to summary@tomb is the input dataset described as a tombola.
> > The same number may occur on many tickets. The order of the tickets is
> > immaterial.
> > Right input to summary@hist is the input dataset described as a histogram.
> > the number of zeroes, the number of ones, the number og twos &c.
> > Examples:
> >
> > 11 summary@tomb i.11
> >
> >
> > 0 1 2 3 4 5 6 7 8 9 10
> >
> > 11 summary@tomb 11 NB. this wouldn't work without the comma in tomb
> > 11 11 11 11 11 11 11 11 11 11 11
> >
> > 1 summary@tomb i.11 NB. mean value
> > 5
> >
> > 2 summary@tomb i.11 NB. mean plus/minus std.dev.
> > 1.83772 8.16228
> >
> >
> > 3 summary@tomb i.11 NB. generalized into 3 numbers
> >
> > 1.12702 5 8.87298
> >
> >
> >
> > Explanations:
> >
> >
> >
> > 5 tomb i.11
> >
> > 1 0 0 0 0 0
> >
> > 1 1 1 1 1 1
> >
> > 1 2 4 8 16 32
> >
> > 1 3 9 27 81 243
> >
> > 1 4 16 64 256 1024
> >
> > 1 5 25 125 625 3125
> >
> > 1 6 36 216 1296 7776
> >
> > 1 7 49 343 2401 16807
> >
> > 1 8 64 512 4096 32768
> >
> > 1 9 81 729 6561 59049
> >
> > 1 10 100 1000 10000 100000
> >
> > +/ hp 5 tomb i.11
> > 11 55 385 3025 25333 220825
> >
> > (}.%{.) +/ hp 5 tomb i.11
> > 5 35 275 2303 20075
> >
> > (*#) (}.%{.) +/ hp 5 tomb i.11
> > 25 175 1375 11515 100375
> >
> > s hp 5 tomb i.11
> > 25 175 1375 11515 100375
> >
> > f1 s hp 5 tomb i.11
> > 25 175 1375 11515 100375
> >
> > 25 175 1375 11515 100375
> >
> > 25 175 1375 11515 100375
> >
> > 25 175 1375 11515 100375
> >
> > 25 175 1375 11515 100375
> >
> > |: f1 s hp 5 tomb i.11
> > 25 25 25 25 25
> >
> > 175 175 175 175 175
> >
> > 1375 1375 1375 1375 1375
> >
> > 11515 11515 11515 11515 11515
> >
> > 100375 100375 100375 100375 100375
> >
> > f2 |: f1 s hp 5 tomb i.11
> > 1 2 3 4 5
> >
> > 25 25 25 25 25
> >
> > 175 175 175 175 175
> >
> > 1375 1375 1375 1375 1375
> >
> > 11515 11515 11515 11515 11515
> >
> > f3"1 f2 |: f1 s hp 5 tomb i.11
> > 1 _2 3 _4 5
> >
> > 25 _25 25 _25 25
> >
> > 175 _175 175 _175 175
> >
> > 1375 _1375 1375 _1375 1375
> >
> > 11515 _11515 11515 _11515 11515
> >
> > f4 f3"1 f2 |: f1 s hp 5 tomb i.11
> > 1 25 175 1375 11515
> >
> > 0 _2 _25 _175 _1375
> >
> > 0 0 3 25 175
> >
> > 0 0 0 _4 _25
> >
> > 0 0 0 0 5
> >
> > |: f4 f3"1 f2 |: f1 s hp 5 tomb i.11
> > 1 0 0 0 0
> >
> > 25 _2 0 0 0
> >
> > 175 _25 3 0 0
> >
> > 1375 _175 25 _4 0
> >
> > 11515 _1375 175 _25 5
> >
> > m s hp 5 tomb i.11
> > 1 0 0 0 0
> >
> > 25 _2 0 0 0
> >
> > 175 _25 3 0 0
> >
> > 1375 _175 25 _4 0
> >
> > 11515 _1375 175 _25 5
> >
> > (%.m) s hp 5 tomb i.11
> > 25 225 875 1340 450
> >
> > |.1,(%.m) s 5 tomb i.11
> > 450 1340 875 225 25 1
> >
> > p.|.1,(%.m) s hp 5 tomb i.11
> > ┌─┬─────────────────────────────────────┐
> >
> > │1│_9.54306 _7.0882 _5 _2.9118 _0.456938│
> >
> > └─┴─────────────────────────────────────┘
> >
> > _1 x: - |. > {: p.|.1,(%.m) s hp 5 tomb i.11
> > 0.456938 2.9118 5 7.0882 9.54306
> >
> > 5 summary@tomb i.11
> > 0.456938 2.9118 5 7.0882 9.54306
> >
> >
> >
> > Thanks!
> > Bo. Den fredag den 21. august 2020 17.01.49 CEST skrev 'Michael Day' via
> > Programming <[email protected]>:
> >
> > Any clue as to the input to summary, or what you're summarising?
> >
> > Sorry if it's obvious!
> >
> > Note these (display ?) errors:
> >
> > pol =. _1 [space] x: ...
> >
> > hp =. x [space] :: ] ...
> >
> > Cheers,
> >
> > Mike
> >
> >
> > On 21/08/2020 15:41, 'Bo Jacoby' via Programming wrote:
> > >
> > > tomb=.(^/i.@>:)~ NB. tombola lottery powers
> > > hist=.({."1~>:)~(*[:^/~i.@#@,)NB. histogram powers
> > > s=.(*#)@ (}.%{.) @ (+/) NB. summation
> > > f1=.$~2# # NB. n*n matrix
> > > f2=.>:@i.@#, }: NB. insert S0, remove Sn
> > > f3=.*_1^i.@# NB. change sign
> > > f4=.|:|.!.0"01~-@i.@# NB. shift zeroes in
> > > m=.|:@f4@(f3"1@f2@|:@f1) NB. matrix
> > > e=.1,(%.m) NB. solve linear equations
> > > pol=._1x: -@|.@>@{:@p.@|. NB.solve alg.eqn.
> > > hp=.x::: ] NB. high precision
> > > summary=.pol@e@s@hp f. NB. complete program
> > > sorry for the still missing carriage returns.
> > > Den fredag den 21. august 2020 16.37.19 CEST skrev Bo Jacoby
> > ><[email protected]>:
> > >
> > > tomb=.(^/i.@>:)~ NB. tombola lottery
> > >powershist=.({."1~>:)~(*[:^/~i.@#@,)NB. histogram powerss=.(*#)@ (}.%{.) @
> > >(+/) NB. summationf1=.$~2# # NB. n*n matrixf2=.>:@i.@#, }: NB. insert S0,
> > >remove Snf3=.*_1^i.@# NB. change signf4=.|:|.!.0"01~-@i.@# NB. shift
> > >zeroes inm=.|:@f4@(f3"1@f2@|:@f1) NB. matrixe=.1,(%.m) NB. solve linear
> > >equationspol=._1x: -@|.@>@{:@p.@|. NB.solve alg.eqn.hp=.x::: ] NB. high
> > >precisionsummary=.pol@e@s@hp f. NB. complete program
> > > This should be better. Sorry.
> > > Bo. Den fredag den 21. august 2020 16.29.35 CEST skrev 'Bo Jacoby' via
> > > Programming <[email protected]>:
> > >
> > > Thank you!
> > > I was trying to accomplish this
> > > tomb=.(^/i.@>:)~ NB. tombola lottery
> > > powershist=.({."1~>:)~(*[:^/~i.@#@,)NB. histogram powerss=.(*#)@ (}.%{.)
> > > @ (+/) NB. summationf1=.$~2# # NB. n*n matrixf2=.>:@i.@#, }: NB. indsæt
> > > S0, fjern Snf3=.*_1^i.@# NB. change signf4=.|:|.!.0"01~-@i.@# NB. shift
> > > zeroes inm=.|:@f4@(f3"1@f2@|:@f1) NB. matrixe=.1,(%.m) NB. solve linear
> > > equationspol=._1x: -@|.@>@{:@p.@|. NB.solve alg.eqn.hp=.x::: ] NB. high
> > > precisionsummary=.pol@e@s@hpf. NB. complete program
> > > 1&(summary@tomb) computes the mean value of a dataset.
> > >
> > > 1 summary@tomb i.11
> > > 5
> > >
> > >
> > > The mean value of 99 zeroes and 1 one is
> > >
> > > 1 summary@hist 99 1
> > >
> > > 0.01
> > >
> > >
> > >
> > > The mean value plus/minus the standard deviation is computed like this
> > >
> > > 2 summary@tomb i.11
> > >
> > > 1.83772 8.16228
> > >
> > > 2 summary@hist 99 1
> > > _0.0894987 0.109499
> > >
> > >
> > >
> > > Of course these two numbers have the correct mean value
> > >
> > > 1 summary@tomb 2 summary@tomb i.11
> > >
> > > 5
> > >
> > > 1 summary@tomb 2 summary@hist 99 1
> > > 0.01
> > >
> > >
> > >
> > > The computation is generalized to 3 numbers
> > >
> > > 3 summary@tomb i.11
> > >
> > > 1.12702 5 8.87298
> > >
> > > 3 summary@hist 99 1
> > > _0.108204j0.16452 _0.108204j_0.16452 0.246409
> > >
> > > having the correct mean values and standarddeviations.
> > >
> > > 2 summary@tomb 3 summary@tomb i.11
> > >
> > > 1.83772 8.16228
> > >
> > > 2 summary@tomb 3 summary@hist 99 1
> > > _0.0894987 0.109499
> > >
> > >
> > >
> > > Thus complicated probability distributions functions are summarized by a
> > > few numbers. That is what I accomplished!
> > > Thanks!
> > > Bo.
> > > Den fredag den 21. august 2020 15.33.50 CEST skrev ethiejiesa via
> > >Programming <[email protected]>:
> > >
> > > Bo Jacoby <[email protected]> wrote:
> > >> How to de-smell this:
> > >> 3([* i.@#@,@[ ^/ i.@>:@])~ 0 0 1 1 1
> > > 3 (]* ((^/&i. >:)~ #)) 0 0 1 1 1
> > >
> > > Maybe? That comes from just a little mechanical algebra:
> > >
> > >> 3([* i.@#@,@[ ^/ i.@>:@])~ 0 0 1 1 1
> > > First notice the repeated (i.) on either argument of (^/). There is a
> > > general
> > > pattern, (f@u v f@w <--> u v&f w), which in this case specializes to
> > >
> > > 3([* #@,@[ ^/&i. >:@])~ 0 0 1 1 1
> > >
> > > Then, at least in this example the (@,) is superfluous, so we elide it:
> > >
> > > 3([* #@[ ^/&i. >:@])~ 0 0 1 1 1
> > >
> > > Finally, here is an idiom: ( (u@[ v w@]) <--> ((v w)~ u)~ ), and since
> > > (x v~~ y <--> x v y), we have
> > >
> > > 3 (]* ((^/&i. >:)~ #)) 0 0 1 1 1
> > >
> > > The last idiom is really a matter of preference. I like the bare verbs,
> > > but
> > > code golf-wise it doesn't win.
> > >
> > > More importantly, however, I really have no idea what you are trying to
> > > accomplish. The above just performs some J syntax manipulations; however,
> > > this
> > > doesn't at all demonstrate the main point of avoiding code smell as I see
> > > it,
> > > which is more about finding places to improve the *algorithm* rather than
> > > only
> > > cleaning up the syntax.
> > > ----------------------------------------------------------------------
> > > For information about J forums see http://www.jsoftware.com/forums.htm
> > >
> > > ----------------------------------------------------------------------
> > > For information about J forums see http://www.jsoftware.com/forums.htm
> > >
> > > ----------------------------------------------------------------------
> > > For information about J forums see http://www.jsoftware.com/forums.htm
> >
> > --
> > This email has been checked for viruses by Avast antivirus software.
> > https://www.avast.com/antivirus
> >
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
> >
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
