Ok so I have been trying to do this proof of what seems to be a version of Factor theorem(I am now well versed in math so please forgive me if I am wrong).
Problem goes like this:
Prove that for any polynomial function f and any number a,there is polynomial g(x) such that f(x) = (x-a)g(x) + b .Proof is possible by induction on degree of polynomial
Proof for degree one is trivial.Now if we assume it is true for degree k we can proceed to prove for k+1.My proof for k+1 goes like this:
$f(x) = c_0 + c_1x + … c_{k+1}x^{k+1} $ which is of degree k+1
Then we can assume that :
$f(x) – c_{k+1}x^{k+1} = (x-a)g(x) + b $ which is of degree k
Now my issue is that from this form I can not get the form required.Also proof from the answer book is not of much help.It goes like this:
Assume the preposition is true for degree k and we have:
$$f(x) = a_{k+1}x^{k+1} + … + a_1x + a_0 $$Now polynomial $$h(x) = f(x) – a_{k+1}(x-a) $$ is of degree $\leq k$ so we can write:
$$ f(x) – a_{k+1}(x-1) = (x-a)g(x) + b $$ or
$$f(x)=(x-a)[g(x) + a_{k+1}] + b $$
My questions are :
1)How is h(x) of degree $\leq k $ when f(x) is k+1,should we not substract $a_{k+1}x^{k+1}$ to get lesser degree
2)How can we from first resulting form obtain the second one,since as far as I see in first form g(x) should be of degree k-1 ,and in second it should be k unless b or $a_{k+1}$ is of degree k+1 or k respectively
Please help .I have spent hours trying to figure this proof out,still does not make sense.
Problem can be found in Spivak Calculus vol.1 chapter 3 problem 7(a)
Best Answer
First, the answer as written is wrong; as you observe, the degree will only be reduced if we subtract the thing you recommended.
{In what follows, I'm using $c_i$ as the $i$th coefficient rather than $a_i$.}
But if you instead wrote $$ f(x)−c_{k+1}(x^{k+1} - a x^{k}) $$ you'd still have a degree-$k$ polynomial, and could by induction write $$ f(x)−c_{k+1}(x^{k+1} - a x^{k}) =(x−a)g(x)+b $$ where $g$ is a degree-$(k-1)$ polynomial. Now moving things to the other side, you get $$ f(x) = c_{k+1}(x^{k+1} - a x^{k}) + (x−a)g(x)+b \\ = c_{k+1}x^k (x- a) + (x−a)g(x)+b \\ $$ You can now combine the first two terms, and you're on your way!