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# numpy.polyval

`numpy.``polyval`(p, x)

  Evaluate a polynomial at specific values.

  If p is of length N, this function returns the value:

  > `p[0]*x**(N-1) + p[1]*x**(N-2) + ... + p[N-2]*x + p[N-1]`

  If x is a sequence, then p(x) is returned for each element of x. If x is another polynomial then the composite polynomial p(x(t)) is returned.

  Parameters : | p : array_like or poly1d object 1D array of polynomial coefficients (including coefficients equal to zero) from highest degree to the constant term, or an instance of poly1d. x : array_like or poly1d object A number, a 1D array of numbers, or an instance of poly1d, “at” which to evaluate p.
  Returns : | values : ndarray or poly1d If x is a poly1d instance, the result is the composition of the two polynomials, i.e., x is “substituted” in p and the simplified result is returned. In addition, the type of x - array_like or poly1d - governs the type of the output: x array_like => values array_like, x a poly1d object => values is also.

  See also

  `poly1d`
    A polynomial class.

  Notes

  Horner’s scheme [R63] is used to evaluate the polynomial. Even so, for polynomials of high degree the values may be inaccurate due to rounding errors. Use carefully.

  References

  [R63] | (1, 2) I. N. Bronshtein, K. A. Semendyayev, and K. A. Hirsch (Eng. trans. Ed.), Handbook of Mathematics, New York, Van Nostrand Reinhold Co., 1985, pg. 720.

  Examples

      >>> np.polyval([3,0,1], 5)  # 3 * 5**2 + 0 * 5**1 + 1
      76
      >>> np.polyval([3,0,1], np.poly1d(5))
      poly1d([ 76.])
      >>> np.polyval(np.poly1d([3,0,1]), 5)
      76
      >>> np.polyval(np.poly1d([3,0,1]), np.poly1d(5))
      poly1d([ 76.])

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