polynom coeff integ

Integration of polynomial: Coefficients operation

polynomial as list

A polynomial in the form of

$$\tag{1} y = \sum_{i=0}^n b_i t^i$$

can be presented in a column matrix

$$\tag{2} y \equiv [b_0 \ \ b_1 \ \ b_2 \ \ b_3 \ \ \dots \ \ b_{n-1} \ \ b_n]$$

or simply a list in Python.

differential

Let us define other polynomial

$$\tag{3} x = \sum_{j=0}^m a_j t^j$$

and use it to represent differential of (1)

$$\tag{4} \frac{dy}{dt} = \sum_{i=1}^n ib_i t^{i-1}.$$

Equate (3) and (4) will give

$$\tag{5} a_j = (j+1) b_{j+1}$$

since

$$\tag{6} \begin{array}{rcl} x & = & [b_1 \ \ 2 b_2 \ \ 3b_3 \ \ \dots \ \ (n-1) b_{n-1} \ \ \ \ n b_n \ ] \newline & = & [a_0 \ \ \ a_1 \ \ \ a_2 \ \ \ \dots \ \ \ \ \ \ \ \ a_{n-2} \ \ \ \ \ \ \ \ \ a_{n-1}]. \end{array}$$

Notice that in this case $m = n - 1$.

integral

Again another polynomial

$$\tag{7} z = \sum_{j=0}^m c_j t^j$$

to represent integral of (1)

$$\tag{8} \int y \ dt = c_0 + \sum_{i=0}^n \frac{1}{i+1} b_i t^{i+1}.$$

Equate (7) and (8) will give

$$\tag{9} c_j = \frac{1}{j-1} b_{j-1}$$

since

$$\tag{10} \begin{array}{rcl} z & = & [c_0 \ \ b_0 \ \ \frac12 b_1 \ \ \frac13 b_2 \ \ \dots \ \ (n-1) b_{n-1} \ \ \ \ n b_n \ ] \newline & = & [c_0 \ \ \ c_1 \ \ \ c_2 \ \ \ \ c_3 \ \ \ \dots \ \ \ \ \ \ \ \ c_n \ \ \ \ \ \ \ \ \ c_{n+1}]. \end{array}$$