How Does Polynomial Long Division Validate the Existence of Remainders?

[osnarf]

Spivak's "Calculus," chapter 3 - problem 7 - a

Homework Statement

Prove that for any polynomial function f, and any number a, there is a polynomial function g, anad a number b, such that f(x) = (x - a)*g(x) + b for all x. (The idea is simply to divide (x - a) into f(x) by long division, until a constant remainder is left. For example, the calculation : see below - section 2 A formal proof is possible by induction on the degree of f.

Homework Equations

<here spivak divides x³ - 3x + 1 by x - 1 using long division>
<he arrives at x³ -3x +1 = (x - 1)(x² + x -2) - 1>

The Attempt at a Solution

Proving a polynomial of degree 1 is easy enough for the induction proof.

For proving it for a degree k+1, spivak writes :
suppose the result is true for polynomials of degree <= k. If f has degree k + 1, then f has the form:

f(x) = a_{k + 1}x^k+1 + ... + a₁ + a ₀.


agreed so far...
then he writes, immediately following the previous quote:
Now the polynomial function h(x) = f(x) - a_k+1(x-a) has degree <k, so we can...

The rest is simple, assuming the previous statement is correct. However, I don't understand how subtracting
a_k+1(x-a)
from a function of degree k+1 is going to make it <k, unless k was 0.

Could anyone explain this to me please?

 

[Mark44]

Spivak's "Calculus," chapter 3 - problem 7 - a

Homework Statement

Prove that for any polynomial function f, and any number a, there is a polynomial function g, anad a number b, such that f(x) = (x - a)*g(x) + b for all x. (The idea is simply to divide (x - a) into f(x) by long division, until a constant remainder is left. For example, the calculation : see below - section 2


A formal proof is possible by induction on the degree of f.

Homework Equations

<here spivak divides x³ - 3x + 1 by x - 1 using long division>
<he arrives at x³ -3x +1 = (x - 1)(x² + x -2) - 1>

The Attempt at a Solution

Proving a polynomial of degree 1 is easy enough for the induction proof.

For proving it for a degree k+1, spivak writes :
suppose the result is true for polynomials of degree <= k. If f has degree k + 1, then f has the form:

f(x) = a_{k + 1}x^k+1 + ... + a₁ + a ₀.


agreed so far...
then he writes, immediately following the previous quote:
Now the polynomial function h(x) = f(x) - a_k+1(x-a) has degree <k, so we can...

I think you are missing an exponent. The only way this makes sense is by subtracting a_k+1x^k+1. Then, h(x) is a polynomial of degree <= k.
Edit: changed the above from a_k+1(x - a)^k+1.

The rest is simple, assuming the previous statement is correct. However, I don't understand how subtracting
a_k+1(x-a)
from a function of degree k+1 is going to make it <k, unless k was 0.

Could anyone explain this to me please?

 

[osnarf]

okay i guess I'm not going crazy. that's the 3rd typo I've found in this book tonight

 

[osnarf]

wait wait... then the rest of the proof doesn't make sense. next it says:

...so we can write:

f(x) - a_k+1(x - a) = (x - a)g(x) + b

or

f(x) = (x - a)[g(x) + a_k+1] + b

which is the required form.

so...? Can't combine the (x - a)'s if one is to the k+1th power

 

[Mark44]

I'm not following that step either. f(x) is a degree k + 1 polynomial, so subtracting a_k+1(x - a) doesn't get you to a degree k polynomial.

Can you include the full text of what you're looking at, consisting of the induction hypothesis and what follows?