c0 and x are scalars, c - vector, and p - scalar. If c is [ ], then p = c0. If c is a scalar, then p = c0 + c*x . Else, p =
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Subramanian Mahadevan
am 9 Nov. 2016
Beantwortet: Govind Mishra
am 14 Mär. 2018
function [p] = poly_val(c0,c,x)
N = length(c);
n=1:1:N;
if (N<=1)
if(isempty(c))
p=c0;
else
p= c0+(c*x);
end
end
if(N>1)
p = c0+(sum(c(n).*(power(x,n))));
end
end
2 Kommentare
Akzeptierte Antwort
Thorsten
am 14 Nov. 2016
function y = mypolyval(c0,c,x)
if isempty(c)
y = c0;
else
y = c0 + power(x, 1:numel(c))*c(:);
end
The case y = c0 + c*x is already covered by the else. And you can use matrix multiplication of a row and a column vector r*c instead of sum(r.*c').
2 Kommentare
Vijayramanathan B.tech-EIE-118006077
am 11 Feb. 2018
This is the easiest method! Well done Mr.Thorsten
*Usage of c(:) is appreciated! :)*
Raunil Raj
am 6 Mär. 2018
really! I went through the same problem. I however don't understand as to how c(:) can convert any row vector or column vector into a column vector. Can someone please explain?
Weitere Antworten (6)
Jorge Briceño
am 29 Jan. 2018
Here is my solution:
function p = poly_val(c0,c,x)
format long
n=(1:1:length(c));
c=c(:)' & This part converts any array/matrix into a colunm vector and transpose...
% it afterwards, since you are working with row vector properties.
if isempty(c)
p=c0;
elseif isscalar(c)
p=c0+sum(c.*x);
else
p=c0+sum((c.*(x.^n)));
end
end
0 Kommentare
Gabir Yusuf
am 8 Aug. 2017
if true
function p = poly_val(c0,c,x)
n=length(c);
if sum(size(c))==0
p = c0;
elseif isscalar(c)
p = c0 + c*x;
else
y=1:n;
z=x.^y;
if size(c)==[1 n]
p=sum(c.*z)+c0;
else
c=c';
p=sum(c.*z)+c0;
end
end
end
1 Kommentar
Anshuman Panda
am 19 Aug. 2017
function p=poly_val(c0,c,x) a=length(c); if a==0 p=c0; else if a==1 p=c0+c*x; else p=c0 + power(x , 1:a)*c(:); end end end
0 Kommentare
Darío Pascual
am 12 Mär. 2018
function p=poly_val(c0,c,x)
N = length(c);
n=1:1:N;
d=size(c);
if(isempty(c))
p=c0;
end
if N==1
p= c0+(c*x);
end
if N>1
if d(1)==1
p = c0+(sum(c(n).*(power(x,n))));
else
c=c'
p = c0+(sum(c(n).*(power(x,n))));
end
end
0 Kommentare
Govind Mishra
am 14 Mär. 2018
function [p] = poly_val(c0,c,x)
if(iscolumn(c)) c=transpose(c); end
N = length(c); n=1:1:N; if (N<=1) if(isempty(c)) p=c0; else p= c0+(c*x); end end if(N>1) p = c0+(sum(c(n).*(power(x,n)))); end end
0 Kommentare
Siehe auch
Kategorien
Mehr zu Polynomials finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!