I need help with this proof.
Show that for any number $c$, a polynomial
$P(x)=b_0+b_1x+b_2x^2+…+b_nx^n$ can also be written
$P(x)=a_0+a_1(x-c)+a_2(x-c)^2+…+a_n(x-c)^n$ where $a_0=P(c)$. Show
that $a_n\ne 0$ if $b_n \ne 0$.
The solution manual indicates:
Let $y=x+c$. Then
$P(y)=b_0+b_1(x+c)+b_2(x+c)^2+…+b_n(x+c)^n$
Expanding all the binomials and collecting like terms results in a polynomial
$a_0+a_1x+a_2x^2+…+a_nx^n$ (This is where I get lost. Why the coefficients $b_i$ disappear and $a_i$ appear? How to see the expansion of binomials and the grouping of similar terms in an easy way?)
This is $P(x+c)$, so
$P(x)=a_0+a_1(x-c)+a_2(x-c)^2+…+a_n(x-c)^n$.
Note that $a_n=b_n$. (Why?)
Thanks.
Best Answer
The $a_i$'s are what you get after the expansion.
If, for example, $P(x)=b_0+b_1x+b_2x^2$, then
\begin{align}P(y)&=b_0+b_1(x+c)+b_2(x+c)^2\\&=b_0+b_1x+b_1c+b_2x^2+2b_2cx+b_2c^2\\&=\overbrace{b_0+b_1c+b_2c^2}^{\phantom{a_0}=a_0}+\overbrace{(b_1+2b_2c)}^{\phantom{a_1}=a_1}x+\overbrace{b_2}^{\phantom{a_2}=a_2}x^2.\end{align}And, as in this case, you always have $a_n=b_n$