@asa12 

the web site marked as correct,

but i follow the radian in maple, 

it can not be solved with solve function

with(MTM):
evalf(solve(atan(-1) - 45*Pi/180 = atan(x), x));

@asa12 

but when over 180 degree , tangent seems wrong.

evalf(tan(45+270)) = 1.117282703 != -1

@vv 

after think again, i think my googled equation from rotation matrix is wrong

i use tangent already enough

L:=NULL;
kmax:=0; # increase it to see more of the spiral
for ii from 0 by 5 to 365 do
print('ii'=ii);
for k from 0 to kmax do
eq1 := sqrt(x^2 + y^2) - arctan(y,x) = 2*k*Pi;
eq2 := y = tan(ii)*x;
xy := fsolve({eq1, eq2});
#print(xy):
L:=L,eval([x,y],xy);
od:od:
plot([L],style=point);

@taro 

i think the googled equation from matlab is absolutely a wrong thing

y/x = slope = tan(theta)

when think from rotation matrix, just make complex

i use tangent must correct

but when over 180 degree , tangent seems wrong.

evalf(tan(45+270)) = 1.117282703 != -1/2

actually i search from

https://www.mathworks.com/matlabcentral/answers/139548-finding-the-angle-of-rotation-given-the-slope-of-the-line-connecting-two-rotated-points?requestedDomain=www.mathworks.com

for ii from 0 by 5 to 365 do
eq1 := x^2 + y^2 = 1;
eq2 := y = tan(ii)*x;
xy := fsolve({eq1, eq2});
print("M",rhs(xy[1]),rhs(xy[2]),","):
od:

@vv 

for ii from 0 by 5 to 365 do
eq1 := frem(sqrt(x^2 + y^2) - arctan(y,x), 2.*Pi) = 0;
eq2 := y = evalf((atan(-1) - ii*Pi/180)/Pi)*(-1.0)*x;
xy := fsolve({eq1, eq2});
print("M",rhs(xy[1]),rhs(xy[2]),","):
od:

Error, invalid input: frem received (x^2+y^2)^(1/2)-arctan(y, x), which is not valid for its 1st argument, x

i run your equation got error, which syntax is wrong?

@taro 

my version of maple is 2015 

it show 45 degree is 0.5, this is correct

but i am not sure whether other values correct or not

it seems your version has problem

@Carl Love 

https://drive.google.com/file/d/0B7fHc_dTzkY_NjNwdC1VakxIbUE/view?usp=sharing

here is the example

first column is algebra (feature)
second column is decimal number (feature)
third column is original time series data (data)
fourth column is two decimal data (feature)

top 1 row is the latest
top 2 row is the previous day

first column sometimes it has pretty(integer) coefficient algebra consecutively
so i use intersection of terms in previous question
for previous and current row and the nops

using eyes and trial by editing script is time consuming.
and there are some decimal is 2.99999 which can be said like 3.0000 in algebra equation
so i choices become two, one is round it, another is see it as another different equation
then i think to use nops(nops(op() intersect op()))/max(nops(op(..)), nops(op()))
to find percentage of the same, if over 80% of terms are equal then i see it as the same equation
this will become fifth column (feature)

sixth column will be use original time series data if current row x(t) higher than previous day's data x(t+3)
then see algebra equation is a feature of this

since t+3, 3 is just my guess, i feel it is also time consuming to try

and the performance of accuracy may change depend on competitors change

this is why i am curious about the effective of data mining algorithm

the goal is identify useful algebra equation represent the change of data

and how algebra equation in first columns mapped with four column which are two decimal numbers

for example, would like to find whether decimal numbers in fourth columns are grouped by algebra equation in first column

and if grouped, how do it affect the data

@Kitonum 

really awesome

@Carl Love 

thank you 

actually the goal is

m := -t*(-1+t)^3/(t^3+2);
op(m);

actually change this result
-1, t, (-1+t)^3, 1/(t^3+2)

to

-1, t, (-1+t), (-1+t), (-1+t), 1/(t^3+2)

i tried

m := -t*(-1+t)^3/(t^3+2);

g := map(e->`if`(e::`^`, `$`(op(e)), e), {op(m)});

however this try, need one more unknown function repeat the polynomial with number of power

@Carl Love 

position may not be the same

for example

x^2*y + x 

x^2*y at position 1

y + x^2*y

x^2*y at position 2

however, it should return x^2*y after intersect them

@Carl Love 

[op(x^2*y+x)] ∩ [op(x^2*y+y)]

it should return x^2*y

but

nops([op(x^2*y+x)] ∩ [op(x^2*y+y)])

Error, invalid input: `intersect` received [x^2*y, x], which is not valid for its 1st argument

@Joe Riel 

would like to find a function represent a set of function

originally 

i have some point mapping one to many

for example

x+1  map to [1,2,3,4,5]

x+2 map to [10,11,12,13,14]

however, would like to generalize x+1 and x+2 like interpolate of functions

in another words, you can think that i am finding and verify neuron function like neural network

such as 1/(1+exp(-x))

would like to verify that after generatlize x+1 and x+2, is it really equal to 1/(1+exp(-x))

@vv 

your matrix is wrong,

because i use PermutationMat in gap system to calculate aa and bb

and it is obvious to see with eyes

that

(1, 2)(3, 5)(4, 7)

=

[1,2], [2,1], [3,5],[5,3],[4,7],[7,4]

do you mean that my result should use aa*bb that is correct solution?

moreover , your matrix only use g[1], how about g[2]?

if your solution is correct, does it mean that use your result from g[1] 

and then matrix multiply the result from g[2] with your code?

@vv 

yes, there is matrix

((1, 2)(3, 5)(4, 7),7);

aa := Matrix([ [ 0, 1, 0, 0, 0, 0, 0 ], [ 1, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 1, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 1 ], [ 0, 0, 1, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 1, 0 ], [ 0, 0, 0, 1, 0, 0, 0 ] ])

((1, 3, 6)(2, 4, 8),8);

bb := Matrix([ [ 0, 0, 1, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 1, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 1, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 1 ],
[ 0, 0, 0, 0, 1, 0, 0, 0 ], [ 1, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 1, 0 ], [ 0, 1, 0, 0, 0, 0, 0, 0 ] ])

two generators , does it mean two different solutions?

if only one solution, 

should it add aa + bb to form one matrix as solution?

@vv 

you said only two groups have permutation group representation in maple,

how are the rest group used in real industry?