numpy.polyder()

numpy.polyder(p, m=1) [source]

Return the derivative of the specified order of a polynomial.

Parameters:

Parameters:	p : poly1d or sequence Polynomial to differentiate. A sequence is interpreted as polynomial coefficients, see `poly1d`. m : int, optional Order of differentiation (default: 1)
Returns:	der : poly1d A new polynomial representing the derivative.

p : poly1d or sequence

Polynomial to differentiate. A sequence is interpreted as polynomial coefficients, see poly1d.

m : int, optional

Order of differentiation (default: 1)

Returns:

der : poly1d

A new polynomial representing the derivative.

The derivative of the polynomial $x^3 + x^2 + x^1 + 1$ is:

>>> p = np.poly1d([1,1,1,1])
>>> p2 = np.polyder(p)
>>> p2
poly1d([3, 2, 1])

which evaluates to:

>>> p2(2.)
17.0

We can verify this, approximating the derivative with (f(x + h) - f(x))/h:

>>> (p(2. + 0.001) - p(2.)) / 0.001
17.007000999997857

The fourth-order derivative of a 3rd-order polynomial is zero:

>>> np.polyder(p, 2)
poly1d([6, 2])
>>> np.polyder(p, 3)
poly1d([6])
>>> np.polyder(p, 4)
poly1d([ 0.])

Links:

doc_NumPy

2017-01-10 18:16:30

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