Possible short webpage title: How to Prove a+bx+cx² = d+ex+fx² for all x?

[Philcorp]

given
a+bx+cx² = d+ex+fx², for all x, you know that a=d, b=e and c=f...
how does one go about proving this?
Any pointers would be appreciated

Philcorp

 

[jcsd]

Well the co-efficent of x^2 is given solely by a on the left and d on the right, for the two equations to be equal this co-efficent must be the same therefore a=d again the same with b and e for the co-efficent of x and also c and f must be equal as they give the units (or the co-efficent of x^0 if you like).

 

[Hurkyl]

Form three equations by plugging in -1, 0, and 1 for x. Solve those equations for a, b, and c in terms of d, e, and f.

 

[Hmmmm . . .]

hmmm . . .

have you ever seen a proof such as if ax + by = cx + dy then a = c, and b = d for all x,y in the reals?


same principle

 

[STAii]

Originally posted by Hurkyl
Form three equations by plugging in -1, 0, and 1 for x. Solve those equations for a, b, and c in terms of d, e, and f.

How do you know that the numbers -1, 0, 1 will not give a special case result ?
i mean, i know if you plug in -1, 0, 1 you will get a=d, b=e, c=f, but how you would explain that this is not a special case (you know, you chose three special numbers, so the result can be special case result).

 

[Hurkyl]

We're told that the equation holds for all x, so it must be true for any choices of x!

By solving the equations, we've proven that, if the equation is true for all x, then a = d, b = e, and c = f. It's trivial to go the other way and prove that if a = d, b = e, and c = f then the equation is true for all x.

 

[suffian]

A rather simple solution can be obtained by calculus:

a + bx + cx² = d + ex + fx²
b + 2cx = e + 2fx (taking derivatives, think of each side as a function of x)
2c = 2f
c = f

Now plug in c = f, cancel out the x² terms and repeat.

In this way, you could extend the applicability to a polynomial of any degree.

 

[teddy]

another(?) way is to just use the substitution x=0 again and again as follows:

if p(x)= ax^n + bx^(n-1)...+ z= 0 for all x

then @ x=0
p(0)=0 => z=0

then take x common and write
x(ax^(n-1) + bx^(n-2)...+ y) = 0
(y is the coefficient of x in p(x))

since this is of the form g(x)*h(x)=0 and g is not identically zero we have h(x)=0.
now substitute x=0 again and we get y=0.similarly all other coefficients turn out to be zero.

 

[AndersHermansson]

Sounds like it's not even necessary to prove. Aren't they equal by definition. If d = a, e = b, f = c

 

[HallsofIvy]

No, they are not "equal by definition". The definition of
"f(x)= g(x)" is that for any value x= a, the values f(a) and g(a) are equal. While the value, of course, depends on the coefficients, it is incorrect to say that the definition of equality of two polynomials is that they have the same coefficients.

I would also note that jcsd's suggestion (which is essentially the same thing) works, it probably would not be accepted as a proof since I suspect that the purpose of this problem is to see WHY the fact that two polynomials have the same values for all x implies that they have the same coefficients.

 

[KLscilevothma]

In general

Lemma: If the polynomial f(x)=a_nxⁿ+a_n-1x^n-1...+a₁x + a₀ = 0 for more than n distinct values of x, then a₀ = a₀ =...=a_n = 0

Proof: Supose f(a₁)=f(a₂)=...=f(a_n+1)=0 where a_i are distinct.

f(a_n+1) = a_n(a_n+1-a₁)(a_n+1-a₂)...(a_n+1-a_n) = 0
==> a_n=0
==>f(x)=a_n-1x^n-1...+a₁x + a₀

Repeat the above process and the result follows.

Thm: Let
f(x) = a_nxⁿ + a_n-1x^n-1...+ a₁x + a₀
g(x) = b_nxⁿ + b_n-1x^n-1...+ b₁x + b₀
be 2 polynomials
If f(x) = g(x), then a_i = b_i for all i=1,2,3,...,n

Pf:
f(x)=g(x)
a_nxⁿ + a_n-1x^n-1...+ a₁x + a₀ = b_nxⁿ + b_n-1x^n-1...+ b₁x + b₀ "x belong to complex numbers (more than n distinct values of x)

(a_n-b_n)xⁿ + (a_n-1-b_n-1)x^n-1 + ... + (a₀-b₀) = 0
==> a_i = b_i

QED

Edit: typo

 

[suffian]

Originally posted by KL Kam
Lemma: If the polynomial f(x)=a_nxⁿ+a_n-1x^n-1...+a₁x + a₀ = 0 for more than n distinct values of x, then a₀ = a₀ =...=a_n = 0

Proof: Supose f(a₁)=f(a₂)=...=f(a_n+1)=0 where a_i are distinct.

f(a_n+1) = a_n(a_n+1-a₁)(a_n+1-a₂)...(a_n+1-a_n) = 0
==> a_n=0
==>f(x)=a_n-1x^n-1...+a₁x + a₀

The proof of the lemma seems a little sketchy to me; or, very confusing. It seems your using the same notation (a_i) to refer to both the coeffecients and the roots of the polynomial.

Let me rewrite what I understand:

Suppose x₁, x₂, x₃, ..., x_n, x_n+1 are roots of the polynomial.

Then you rewrite f(x) as follows:
f(x) = a_n(x - x₁)(x - x₂)...(x - x_n)

My problem is this, how do you know you can rewrite f(x) as such (and knowing you can do so whether or not a_n != 0)?

 

[Hurkyl]

Choose n to be the highest exponent of x with a nonzero coefficient.


f(x) can be factored as such because we know f(x) has n different roots, which means (x - x_j) must be a factor of x for each root x_j. Multiplying out all n of these factors makes it clear we must also have an additional factor a_n.

 

[suffian]

Originally posted by Hurkyl
Choose n to be the highest exponent of x with a nonzero coefficient.


f(x) can be factored as such because we know f(x) has n different roots, which means (x - x_j) must be a factor of x for each root x_j. Multiplying out all n of these factors makes it clear we must also have an additional factor a_n.

Okay, here is my point. You stated that if the highest coefficient is non-zero, then there are precisely n roots of f(x) (counting "double", "triple" roots their respective number of time). If we assume f(x) has n + 1 distinct roots, then we immediately know the highest coefficient must be zero for this situation to even possibly exist. Unfortunantly, because the highest coefficient is now zero, we don't know whether we can rewrite the polynomial as desired. Therefore, we must avoid taking that route in a proof.

Although from here the proof is relatively easy to complete. Now that the highest coefficient is zero we have a new equivalent polynomial of degree n-1 which equals f(x) (but we don't know whether this new polynomial's highest coefficient is zero). Again, since we have n+1 distinct roots to f(x), and our law states that a polynomial of degree n-1 can have at most n-1 distinct roots if it's highest coefficient is non-zero, then we know (since this is not the case) that the new equivalent polynomial of degree n-1 has a highest coefficient of zero. Repeating this for all coefficients completes the proof.

 

[Hurkyl]

There are lots of ways to actually prove this. Though I think you and Kam have essentially the same proof, just Kam left out a detail or two. My mental proof is slightly different, and I would employ the "let m be the highest integer so that a_m is inequal to 0)" approach.

 

[suffian]

Originally posted by Hurkyl
There are lots of ways to actually prove this. Though I think you and Kam have essentially the same proof, just Kam left out a detail or two. My mental proof is slightly different, and I would employ the "let m be the highest integer so that a_m is inequal to 0)" approach.

I think there is a bit of confusion. I was only referring to KL Kam's lemma, where the goal was to prove that all the coefficients are zero (i.e. there is no highest non-zero coefficient). Also, I must disagree that his lemma only requires a quick fix as that transformation (which he cannot preform) is the key step. This is certainly not a crusade against KL Kam's work. I respect his response and only feel it justified to critique considering the time he put in. Anyways, enough of that, we all know that it is provable and there is no need to spend more time on it.