Htykuut

This question is based on the discussion in B4-1.2 of Fundamentals of Mathematics, Volume 1 Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle.

See the bold text for the part of the proof I am seeking help with. My question is: how do I use the method suggested to obtain the needed result?

Technically this proof is applied to $mathcal{R}$ which is an arbitrary
commutative ring with unit element 1, and without null divisors. And the expression

$$
fleft[xright]=sum_{i=0}^{n}a_{i}x^{i}
$$

is equivalent to an entire rational function of one argument
in $mathcal{R}.$ Here we have defined $0^{0}equiv1$ to support
our notation. For the purpose of this question we can simply call
$f$ a polynomial of degree $n$ in $mathbb{R}.$

Our objective is to prove the following theorem: If $f$ is expressible
in the form

$$
fleft[xright]=sum_{i=0}^{n}a_{i}x^{i}
$$

with $a_{n}ne0,$ then $f$ has at most $n$ roots.

Proof: First we introduce the constant expressions

$$
bar{a}_{k}=sum_{i=k+1}^{n}a_{i}alpha^{i-k-1},
$$

and the function

$$
f_{1}left[xright]equivsum_{k=0}^{n-1}x^{k}bar{a}_{k}=sum_{k=0}^{n-2}x^{k}bar{a}_{k}+a_{n}x^{n-1}.
$$

The right-most expression follows from $bar{a}_{n-1}=a_{n},$ and
shows that $f_{1}$ has at most $n-1$ roots by the induction hypothesis.

Next we observe that for the real number constant $alpha$ we have

$$
(x-alpha)sum_{k=0}^{i-1}x^{k}alpha^{i-k-1}=sum_{k=0}^{i-1}x^{k+1}alpha^{i-k-1}-sum_{k=0}^{i-1}x^{k}alpha^{i-k}
$$

$$
=sum_{k=1}^{i}x^{k}alpha^{i-k}-sum_{k=0}^{i-1}x^{k}alpha^{i-k}=x^{i}-alpha^{i},
$$

and apply this result to obtain, for $n>0$

$$
fleft[xright]-fleft[alpharight]=left(x-alpharight)f_{1}left[xright].
$$

The case of $n=0$ is the constant function $fleft[xright]=a_{0}x^{0},$
which is of the required form. If $alpha$ is a root of $f,$ we have
$$
fleft[xright]-fleft[alpharight]=left(x-alpharight)f_{1}left[xright]=0.
$$

If $xnealpha,$ is a root of $f$ then $f_{1}left[xright]=0.$ It follows that $f$
has at most one more root than $f_{1}$ Since $f_{1}$ has at most $n-1$ roots this which completes the proof.

The step I am not following is the establishment of $fleft[xright]-fleft[alpharight]=left(x-alpharight)f_{1}left[xright]$
using

$$(x-alpha)sum_{k=0}^{i-1}x^{k}alpha^{i-k-1}=x^{i}-alpha^{i}.$$
I am able to establish the result using brute force as follows: Expand
$f_{1}$

$$
f_{1}left[xright]equivsum_{k=0}^{n-1}left(sum_{i=k+1}^{n}a_{i}alpha^{i-k-1}right)x^{k}
$$

$$
=left(a_{1}+a_{2}alpha+a_{3}alpha^{2}+a_{4}alpha^{3}+ldots+a_{n}alpha^{n-1}right)
$$

$$
+left(a_{2}+a_{3}alpha+a_{4}alpha^{2}+a_{5}alpha^{3}+ldots+a_{n}alpha^{n-2}right)x
$$

$$
+left(a_{3}+a_{4}alpha+a_{5}alpha^{2}+a_{6}alpha^{3}+ldots+a_{n}alpha^{n-3}right)x^{2}
$$

$$
+dots+a_{n}x^{n-1}
$$

$$
=a_{1}+a_{2}(alpha+x)+a_{3}left(alpha^{2}+alpha x+x^{2}right)
$$

$$
+a_{4}left(alpha^{3}+alpha^{2}x+alpha x^{2}+x^{3}right)
$$

$$
+ldots+a_{i}left(alpha^{i-1}+alpha^{i-2}x+ldots+alpha x^{i-2}+x^{i-1}right)
$$

$$
+ldots+a_{i}left(alpha^{n-1}+alpha^{n-2}x+ldots+alpha x^{n-2}+x^{n-1}right).
$$

Multiplying by $left(x-alpharight)$ and canceling like terms of opposite
sign produces the desired result

$$
left(x-alpharight)f_{1}left[xright]=a_{1}left(x-alpharight)+a_{2}(alpha^{2}-x^{2})
$$

$$
+a_{3}left(alpha^{2}x+alpha x^{2}+x^{3}-alpha^{3}-alpha^{2}x-alpha x^{2}right)
$$

$$
+ldots+a_{i}left(alpha^{i-1}x+alpha^{i-2}x^{2}+ldots+alpha^{2}x^{i-2}+alpha x^{i-1}+x^{i}right)
$$

$$
-a_{i}left(alpha^{i}+alpha^{i-1}x+alpha^{i-2}x^{2}+ldots+alpha^{2}x^{i-2}+alpha x^{i-1}right)
$$

$$
+ldots+a_{n}left(alpha^{n-1}x+alpha^{n-2}x^{2}+ldots+alpha^{2}x^{n-2}+alpha x^{n-1}+x^{n}right)
$$

$$
-a_{n}left(alpha^{n}+alpha^{n-1}x+alpha^{n-2}x^{2}+ldots+alpha^{2}x^{n-2}+alpha x^{n-1}right)
$$

$$
=fleft[xright].
$$

Well,
begin{align*}
f[x]-f[alpha]&=sum_{i=0}^na_ix^i-sum_{i=0}^na_ialpha^i \
&=sum_{i=0}^na_i(x^i-alpha^i) \
&=sum_{i=0}^na_i(x-alpha)sum_{k=0}^{i-1}x^{k}alpha^{i-k-1} \
&=(x-alpha)sum_{i=0}^nsum_{k=0}^{i-1}a_ialpha^{i-k-1} x^{k}.
end{align*}

Now we interchange the two sums to get
begin{align*}
f[x]-f[alpha]&=(x-alpha)sum_{k=0}^{n-1}sum_{i=k+1}^{n}a_ialpha^{i-k-1}x^{k} \
&=(x-alpha)sum_{k=0}^{n-1}bar{a}_kx^k \
&=(x-alpha)f_1[x].
end{align*}