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Evaluate a Polynomial without polyval

Asked by Louis Vertosnik on 5 May 2017
Latest activity Commented on by Dillen.A on 27 Feb 2018
I need to create a user defined function that evaluates a vector of the coeficients of a polynomial and a given x value without using the polyval function.
This is what I have so far:
function accum =mypolyval(p,x)
accum = 0;
orderofp = length(p)-1;
n=1:length(p);
for i=1:length(p)
for q=orderofp:-1:0
xtothepower= x^q;
end
y(i) = accum+p(i)*xtothepower;
accum = y(i);
end
end

  1 Comment

Guillaume
on 23 Feb 2018
Note that even without polyval you can evaluate a polynomial in just one line, without the need for a loop (the only functions needed are sum, numel, .^, .* and :)
However, I assume that you have to use a loop for your assignment.

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3 Answers

Answer by Walter Roberson
on 5 May 2017

In your line
for q=orderofp:-1:0
xtothepower= x^q;
end
you are overwriting all of xtothepower each iteration.

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Answer by Dillen.A on 23 Feb 2018
Edited by Dillen.A on 23 Feb 2018

For anyone still interested in this, my quick solution:
pv=@(p,x) sum(permute(p,[1,3,2]).*(x.^permute(0:length(p)-1,[1,3,2])),3)
y=pv(p,x)
Should work for any 2D matrix, I used permute rather than reshape assuming it is faster. for 3D data you can use [1,2,4,3] in your permute, and then sum over the 4th dim, and so on. polyval is still faster.

  3 Comments

Guillaume
on 23 Feb 2018
you made a small mistake in that you're evaluating the polynomial the wrong way round, p(end) is the polynomial constant, not p(1), therefore the exponents should be numel(p)-1:-1:0, not 0:numel(p)-1.
Note that permute is never faster reshape. reshape only affects the header of a matrix (the part that says this matrix has n dimensions of size x, y, z...) without ever touching the actual elements of the matrix whereas permute needs to move all these elements around since their order change. In the case of a vector, the two probably take the same time.
Guillaume
on 23 Feb 2018
A truly generic version (works with any n-dimension x array), using the same algorithm:
pv = @(p, x) sum(shiftdim(p(:), -ndims(x)) .* x .^ shiftdim((numel(p)-1:-1:0).', -ndims(x)), ndims(x)+1)
Thanks for the knowledge. I simply forgot that p was "the wrong way around" as I always remember the expansion as a0 + a1.*x + a2.*x.^2 + ...
Also good to know that reshape is quicker.

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Answer by Roger Stafford on 24 Feb 2018

Note that you can avoid the necessity of computing powers of x in the following. It works even if x is an array. It should save some computing time.
y = repmat(p(1),size(x));
for k = 2:length(p)
y = y.*x+p(k);
end

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