UTM Ideals Varieties and Algorithm - Chapter 1 Section 5 Exercise 13

Problem:

Solution:

Throughout the problem, we let $f(x) = \sum\limits_{i = 0}^{n} c_i x^i$, $g(x) = \sum\limits_{i = 0}^{n} d_i x^i$,

Part (a)

$\begin{eqnarray*} & & (bf)' \\ &=& \sum\limits_{i = 1}^{n} i c_i b x^{i-1} \\ &=& b \sum\limits_{i = 1}^{n} i c_i x^{i-1} \\ &=& bf' \end{eqnarray*}$

Part (b) - suppose the two polynomials do not have same degree, just pad the shorter one with 0 coefficients.

$\begin{eqnarray*} & & (f + g)' \\ &=& \sum\limits_{i = 1}^{n} i (c_i + d_i) x^{i-1} \\ &=& \sum\limits_{i = 1}^{n} i c_i x^{i-1} + \sum\limits_{i = 1}^{n} i d_i x^{i-1}\\ &=& f' + g' \end{eqnarray*}$

Part (c)

$\begin{eqnarray*} & & (fg)' \\ &=& ((\sum\limits_{i = 0}^{n}c_i x^i)(\sum\limits_{j = 0}^{n}d_j x^j))' \\ &=& (\sum\limits_{i = 0}^{n}\sum\limits_{j = 0}^{n}c_i d_j x^{i+j})' \\ &=& \sum\limits_{i = 0}^{n}\sum\limits_{j = 0}^{n}(i + j)c_i d_j x^{i+j-1} \\ &=& \sum\limits_{i = 0}^{n}\sum\limits_{j = 0}^{n}ic_i d_j x^{i+j-1} + \sum\limits_{i = 0}^{n}\sum\limits_{j = 0}^{n}jc_i d_j x^{i+j-1} \\ &=& (\sum\limits_{i = 1}^{n}ic_i x^{i-1})(\sum\limits_{j = 0}^{n}d_j x^j) + (\sum\limits_{i = 0}^{n}c_ix^{i})(\sum\limits_{j = 1}^{n}j d_j x^{j-1}) \\ &=& f'g + fg' \end{eqnarray*}$

Note - in the third and fourth equal sign, the double summation should exclude $i + j = 0$ case, that would explain the switching of summation limits on the fifth equal sign.