Azzera filtri
Azzera filtri

calculate the function in vector.

2 visualizzazioni (ultimi 30 giorni)
JaeSung Choi
JaeSung Choi il 12 Giu 2017
Commentato: JaeSung Choi il 13 Giu 2017
I want to calculate my 'poly' function for domain of linspace(0,1,100) so I tried ---------------------------------
%make poly function
function [output] = poly(input)
output= ([input^0 input^1 input^2 input^3 input^4 input^5]*transpose([1.0000 1.0001 0.4991 0.1703 0.0349 0.0139]) )
end
----------------------------------
x = linsapce(0,1,100)
poly(x)
----------------------------------
but it doesn't work. I found that for sin(x) it does. I want to know what's different between to func. and how to solve the problem.
  1 Commento
KSSV
KSSV il 12 Giu 2017
What is the input you used? You have to take care of element by element operations.

Accedi per commentare.

Risposta accettata

Andrei Bobrov
Andrei Bobrov il 13 Giu 2017
Modificato: Andrei Bobrov il 13 Giu 2017
function [output] = AsPolyvalForJaeSung(input)
output = bsxfun(@power,input(:),0:5)*[1.0000;1.0001;0.4991;0.1703;0.0349;0.0139];
end
  1 Commento
JaeSung Choi
JaeSung Choi il 13 Giu 2017
Oh my god, you exactly catched what i wanted. Thank you very much!!

Accedi per commentare.

Più risposte (2)

KSSV
KSSV il 12 Giu 2017
For
input = linspace(0,1,100) ;
In the line
output= ([input.^0 input.^1 input.^2 input.^3 input.^4 input.^5]*transpose([1.0000 1.0001 0.4991 0.1703 0.0349 0.0139]) )
The size of term in square braces would be 1X600 where as the term transpose i.e second term got only 6X1 terms. How you expect them to multiply? You need to rethink on your code.
  1 Commento
JaeSung Choi
JaeSung Choi il 12 Giu 2017
That's what I'm in problem. I want to derive y = [poly(0) poly(0.01) ...... poly(1)] (i.e. calculate for each domain) As for 'sin' function If we take x = linspace(0,1,100) y = sin(x) then y = [sin(0) sin(0.01) sin(0.02)..... sin(1)] I want to do same for my own function

Accedi per commentare.


Torsten
Torsten il 12 Giu 2017
output= ([(input.').^0 (input.').^1 (input.').^2 (input.').^3 (input.').^4 (input.').^5]*([1.0000 1.0001 0.4991 0.1703 0.0349 0.0139]).'
Best wishes
Torsten.
  2 Commenti
JaeSung Choi
JaeSung Choi il 12 Giu 2017
I've already tested for the same code. Thanks for your answer but that's not what I needed.
Torsten
Torsten il 13 Giu 2017
??
According to your question, I think this is exactly what you needed.
Best wishes
Torsten.

Accedi per commentare.

Categorie

Scopri di più su 수학 in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!