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.
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.
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