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 =

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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

Accepted Answer

Thorsten
Thorsten on 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 Comments
Raunil Raj
Raunil Raj on 6 Mar 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?

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More Answers (6)

Jorge Briceño
Jorge Briceño on 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

KSSV
KSSV on 10 Nov 2016
Edited: KSSV on 10 Nov 2016
function [p] = poly_val(c0,c,x)
N = length(c); % length of c
if N == 0 % if c is empty
p = c0 ;
elseif N == 1 % if c is a scalar
p = c0+c*x ;
else % if c is a vector
p = c0+(sum(c.*(power(x,N))));
end
  4 Comments

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Gabir Yusuf
Gabir Yusuf on 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

Anshuman Panda
Anshuman Panda on 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

Darío Pascual
Darío Pascual on 12 Mar 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

Govind Mishra
Govind Mishra on 14 Mar 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

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