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osnarf
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Spivak's "Calculus," chapter 3 - problem 7 - a
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.
<here spivak divides x3 - 3x + 1 by x - 1 using long division>
<he arrives at x3 -3x +1 = (x - 1)(x2 + x -2) - 1>
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) = ak + 1xk+1 + ... + a1 + a 0.
agreed so far...
then he writes, immediately following the previous quote:
Now the polynomial function h(x) = f(x) - ak+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
ak+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?
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) = ak + 1xk+1 + ... + a1 + a 0.
agreed so far...
then he writes, immediately following the previous quote:
Now the polynomial function h(x) = f(x) - ak+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
ak+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?