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 =

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

[John D'Errico]

What exactly is your question?

[Subramanian Mahadevan]

My question is to create a function which performs in the above mentioned way.

Answer 1

[Thorsten]

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').

[Vijayramanathan B.tech-EIE-118006077]

This is the easiest method! Well done Mr.Thorsten

 *Usage of c(:) is appreciated! :)*

[Raunil Raj]

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?

Answer 2

[Jorge Briceño]

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

Answer 3

[KSSV]

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

[Subramanian Mahadevan]

hello KSSV; This doesn't work for a column matrix!

[KSSV]

Change 'n' to N in the vector case. I have edited, the above code.

[Subramanian Mahadevan]

on changing n to N,isn't the polynomial evaluated for only the last value of n i.e N?

[KSSV]

N is the number of elements in C.

Answer 4

[Gabir Yusuf]

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

[Vijayramanathan B.tech-EIE-118006077]

This is such a long code

Answer 5

[Anshuman Panda]

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

Answer 6

[Darío Pascual]

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

Answer 7

[Govind Mishra]

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