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copied!<p>The problem statement is slightly unclear, so first I will clarify my own interpretation of it:</p> <p>You have a polynomial function</p> <p><strong>f(x) = C<sub>n</sub>x<sup>n</sup> + C<sub>n-1</sub>x<sup>n-1</sup> + ... + C<sub>0</sub></strong></p> <p>[I changed A, B, ... Z into C<sub>n</sub>, C<sub>n-1</sub>, ..., C<sub>0</sub> to more easily work with linear algebra below.]</p> <p>Then you also have a transformation such as:   <strong>z = ax + b</strong>   that you want to use to find coefficients for the <em>same</em> polynomial, but in terms of <em>z</em>:</p> <p><strong>f(z) = D<sub>n</sub>z<sup>n</sup> + D<sub>n-1</sub>z<sup>n-1</sup> + ... + D<sub>0</sub></strong></p> <p>This can be done pretty easily with some linear algebra. In particular, you can define an (n+1)×(n+1) matrix <em>T</em> which allows us to do the matrix multiplication</p> <p>  <strong><em>d</em> = <em>T</em> * <em>c</em></strong> ,</p> <p>where <em>d</em> is a column vector with top entry <em>D<sub>0</sub></em>, to last entry <em>D<sub>n</sub></em>, column vector <em>c</em> is similar for the <em>C<sub>i</sub></em> coefficients, and matrix <em>T</em> has (i,j)-th [i<sup>th</sup> row, j<sup>th</sup> column] entry <em>t<sub>ij</sub></em> given by</p> <p>  <strong><em>t<sub>ij</sub></em> = (<em>j</em> choose <em>i</em>) <em>a<sup>i</sup></em> <em>b<sup>j-i</sup></em></strong>.</p> <p>Where (<em>j</em> choose <em>i</em>) is the binomial coefficient, and = 0 when <em>i</em> > <em>j</em>. Also, unlike standard matrices, I'm thinking that i,j each range from 0 to n (usually you start at 1).</p> <p>This is basically a nice way to write out the expansion and re-compression of the polynomial when you plug in z=ax+b by hand and use the <a href="http://en.wikipedia.org/wiki/Binomial_theorem" rel="nofollow noreferrer">binomial theorem</a>.</p>

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