Algebra: If P(x), degree n, shares n points with x^n, then it is x^n

  • #1

Homework Statement


Prove that if a Polynomial P(x) of degree n, shares n points with xn, then P(x)=xn.

(a more general proof would be almost the same but for two polynomials, but I think I've proved this if this is proved)


Homework Equations


(FTA) Fundamental Theorem of Algebra - Every P(x) of degree n has at most n roots.

The Attempt at a Solution


An idea was to make an analogy between the proof of FTA and this probem. To use for example the factorization: P(x)= ∏(x-ri) Which when proved valid, proves the FTA too. But I don't understand the proofs given online too well.

But also if we accept FTA, we could also possibly analyze P(x)-xn which shall be called D(x).
D(x) has degree (n-1) since xn-xn=0.
Also since P(x) and xn have n equal points, then D(x) has n zeroes.
But FTA says that the D(x) of degree (n-1) has at most n-1 zeroes. But it has n zeroes. Therefore it must equal 0 itself. P(x)-xn=0 ⇔ P(x)=xn

Is this correct? Is there any simpler way to prove or address this or the more general problem?
 
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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
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Homework Statement


Prove that if a Polynomial P(x) of degree n, shares n points with xn, then P(x)=xn.

(a more general proof would be almost the same but for two polynomials, but I think I've proved this if this is proved)


Homework Equations


(FTA) Fundamental Theorem of Algebra - Every P(x) of degree n has at most n roots.

The Attempt at a Solution


An idea was to make an analogy between the proof of FTA and this probem. To use for example the factorization: P(x)= ∏(x-ri) Which when proved valid, proves the FTA too. But I don't understand the proofs given online too well.

But also if we accept FTA, we could also possibly analyze P(x)-xn which shall be called D(x).
D(x) has degree (n-1) since xn-xn=0.
Also since P(x) and xn have n equal points, then D(x) has n zeroes.
But FTA says that the D(x) of degree (n-1) has at most n-1 zeroes. But it has n zeroes. Therefore it must equal 0 itself. P(x)-xn=0 ⇔ P(x)=xn

Is this correct? Is there any simpler way to prove or address this or the more general problem?
Looking at D(x) is the right way to do it. BTW D(x) doesn't necessarily have degree (n-1). The largest degree it could have is n-1, but it might have less.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
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Homework Statement


Prove that if a Polynomial P(x) of degree n, shares n points with xn, then P(x)=xn.

(a more general proof would be almost the same but for two polynomials, but I think I've proved this if this is proved)


Homework Equations


(FTA) Fundamental Theorem of Algebra - Every P(x) of degree n has at most n roots.

The Attempt at a Solution


An idea was to make an analogy between the proof of FTA and this probem. To use for example the factorization: P(x)= ∏(x-ri) Which when proved valid, proves the FTA too. But I don't understand the proofs given online too well.

But also if we accept FTA, we could also possibly analyze P(x)-xn which shall be called D(x).
D(x) has degree (n-1) since xn-xn=0.
Also since P(x) and xn have n equal points, then D(x) has n zeroes.
But FTA says that the D(x) of degree (n-1) has at most n-1 zeroes. But it has n zeroes. Therefore it must equal 0 itself. P(x)-xn=0 ⇔ P(x)=xn

Is this correct? Is there any simpler way to prove or address this or the more general problem?
You can prove the result without knowing the FTA. If you have
[tex] x_i^n + a_{n-1} x_i^{n-1} + \cdots + a_0 = x_i^n, i = 1,2,\ldots,n\\
\text{where } x_1 < x_2 < \cdots < x_n. [/tex]
you need to show that this gives you
[tex] a_{n-1} = a_{n-2} = \cdots = a_0 = 0.[/tex]
Note that you have a set of n equations in the n unknowns aj, and the coefficient matrix is a Vandermonde matrix.

RGV
 
  • #4
You can prove the result without knowing the FTA. If you have
[tex] x_i^n + a_{n-1} x_i^{n-1} + \cdots + a_0 = x_i^n, i = 1,2,\ldots,n\\
\text{where } x_1 < x_2 < \cdots < x_n. [/tex]
you need to show that this gives you
[tex] a_{n-1} = a_{n-2} = \cdots = a_0 = 0.[/tex]
Note that you have a set of n equations in the n unknowns aj, and the coefficient matrix is a Vandermonde matrix. RGV
I've looked the Vandermonde matrix up, but I had never seen it before.
I've set a system which is equivalent to say P(aj)=ajn
If we subtract xn from both sides. We get p(aj)=0 , p of degree n-1
If we set all the equations we get a square matrix. Whose determinant can be taken.
If VC (Vandermonde times coefficients) is a matrix that represents the system of all equations p(aj)=0
Then we get VC=[0] ⇔ |VC|=0
V has no null lines and they're not linearly dependent.(actually one a could be zero)
=> |V|≠0, therefore C ought to be 0 since all it does is make a linear combination of the columns of V.
Or possibly another is to argue that either V is not invertible and |V|=0 which is false, or |V| is invertible and V-1VC=0⇔C=0

PS:I'm a bit tired today, sorry for the trouble with clearly formalizing my thoughts.
Thank you for the help so far.
 

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