- #1

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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

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- #1

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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

- #2

jcsd

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- #3

Hurkyl

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- #4

Hmmmm . . .

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

- #5

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How do you know that the numbers -1, 0, 1 will not give a special case result ?Originally posted by Hurkyl

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).

- #6

Hurkyl

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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.

- #7

suffian

a + bx + cx

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 x

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

- #8

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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.

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HallsofIvy

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"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.

- #11

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Lemma: If the polynomial f(x)=a

Proof: Supose f(a

f(a

==> a

==>f(x)=a

Repeat the above process and the result follows.

Thm: Let

f(x) = a

g(x) = b

be 2 polynomials

If f(x) = g(x), then a

Pf:

f(x)=g(x)

a

(a

==> a

QED

Edit: typo

Last edited:

- #12

suffian

Originally posted by KL Kam

Lemma: If the polynomial f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}...+a_{1}x + a_{0}= 0 for more than n distinct values of x, then a_{0}= a_{0}=...=a_{n}= 0

Proof: Supose f(a_{1})=f(a_{2})=...=f(a_{n+1})=0 where a_{i}are distinct.

f(a_{n+1}) = a_{n}(a_{n+1}-a_{1})(a_{n+1}-a_{2})...(a_{n+1}-a_{n}) = 0

==> a_{n}=0

==>f(x)=a_{n-1}x^{n-1}...+a_{1}x + a_{0}

The proof of the lemma seems a little sketchy to me; or, very confusing. It seems your using the same notation (a

Let me rewrite what I understand:

Suppose x

Then you rewrite f(x) as follows:

f(x) = a

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

- #13

Hurkyl

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f(x) can be factored as such because we know f(x) has n different roots, which means (x - x

- #14

suffian

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, weOriginally posted by Hurkyl

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}.

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.

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Hurkyl

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- #16

suffian

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 itOriginally posted by Hurkyl

wouldemploy the "let m be the highest integer so that a_{m}is inequal to 0)" approach.

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