I'm currently working on writing $3(x)_4 – 12(x)_3 + 4(x)_1 – 17$ as a linear combination of $(x)_4,\ldots,(x)_0$ and am having difficulty understanding where the conversion comes from. I have the answer as $3x^4 – 30x^3 + 69x^2 -38x – 17$ and need help with how to work these problems. Or how to work it the other way such as writing $(x)_4 – 2(x)_2$ as a linear combination of the power $x$?
Those numbers following the $x$'s should be falling factorials
Best Answer
It will help a little to write out the falling factorials in a more familiar way:
\begin{align*} (x)_4 &= x^4 - 6x^3 + 11x^2 -6 x\\ (x)_3 &= x^3-3 x^2+2 x\\ (x)_2 &= x^2 - x\\ (x)_1 &= x\\ (x)_0 &= 1 \end{align*}
To write $x^4 - 2x$ as a linear combination of these, we must start with $(x)_4$, since this is the only way to get a term of $x^4$. This term brings with it, however, an unwanted $-6x^3$. This can only be cancelled by adding $6(x)_3$. Continuing in this way, we end up with $$ x^4 - 2x = (x)_4 + 6(x)_3 +7(x)_2 - (x)_1. $$
For the other direction, just refer to our chart to get $$ (x)_4 - 2(x)_2 = x^4 - 6x^3 + 9x^2 -8x. $$