It took me a while to see what was required using the matrix, M and the
derived coefficients, a.
But this works ok. I haven't turned the explicit code into a function,
but that should be quite
straightforward.
NB. fixed point font best past here....
a =. y %. M =. 1,. }:"1 */\"1 (-/~ * +/\@:=@i.@#) x
M
1 0 0 0 0
1 3r4 0 0 0
1 3r2 9r8 0 0
1 9r4 27r8 81r32 0
1 3 27r4 81r8 243r32
a
_14.10142 17.559765 _10.878424 4.8348551 0
6{."1 [n =: 1,}:*/\ x-~/steps _1.5 1.5 30 NB. n is Newton basis
form of new x
1 1 1 1 1 1
0 0.1 0.2 0.3 0.4 0.5
0 _0.065 _0.11 _0.135 _0.14 _0.125
0 0.091 0.143 0.162 0.154 0.125
0 _0.19565 _0.29315 _0.3159 _0.2849 _0.21875
plot a +/ . * n
Earlier, I'd overlooked the (rather obvious!) fact that you need to
apply a to the Newton
basis form, n here, of your required prediction points, not the
untransformed points.
Mike
On 15/10/2020 14:50, Stefan Baumann wrote:
Dear all.
Recently I stumbled upon the Newton polynomial and took it as a practice to
implement it without using loops, but failed. I first didn't get a grab on
how to create the matrix used in
https://en.wikipedia.org/wiki/Newton_polynomial#Main_idea, and eventually
came up with code following
https://en.wikipedia.org/wiki/Newton_polynomial#Application:
NB. Newton polynomial
np=: 4 : 0
a=. {. y
for_i. }.>:i.#x do.
y=. (2 (-~)/\ y) % i ({:-{.)\ x NB. Divided differences
a=. a, {. y NB. Coefficients are the topmost entries
end.
NB. Convert the summands aᵢ(x-x₀)…(x-xᵢ₋₁) of the polynomial
NB. from multiplier-and-roots to coefficients form and add them up
+/@,:/ p."1 (;/a) ,. (<''), }:<\ x
)
x=: _3r2 _3r4 0 3r4 3r2
y=: 3&o. x
load'plot'
load'stats'
plot (];(x np y)&p.) steps _1.5 1.5 30
I also tried replacing the loop with fold F:. but again was not able to do
so. Anyone out there who can enlighten me?
Many thanks. Stefan.
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