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Ted123

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## Homework Statement

## Homework Equations

## The Attempt at a Solution

How do I go about Q1 and showing the coefficients are unique and then Q2?

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- Thread starter Ted123
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- #1

Ted123

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How do I go about Q1 and showing the coefficients are unique and then Q2?

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What is your attempt at the solution?

- #3

Ted123

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What is your attempt at the solution?

I was trying some sort of induction argument.

[itex]P(0)=Q(0) \implies a_0 = b_0[/itex] so the result is true for [itex]z=0[/itex]

Now suppose for induction that [itex]P(k)=Q(k)[/itex] ...

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- #5

Ted123

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Bearing in mind z is complex does normal differentiation still hold?

[itex]P'(0)=Q'(0) \implies a_1=b_1[/itex]

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- #6

Ted123

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Could I say without loss of generality that [itex]n \geq m[/itex] .

Then if [itex]P(z)=Q(z)[/itex] then [itex]P(z)-Q(z)=0[/itex]

So collecting up like terms we see that

[itex]P(z) - Q(z) = (a_0 - b_0) + (a_1 - b_1)z + (a_2 - b_2)z^2 + ... + (a_m - b_m)z^m + a_{m+1}z^{m+1} + ... + a_nz^n[/itex]

This implies that [itex]a_j=b_j[/itex] for all [itex] 0 \leq j \leq m[/itex] (by equating coefficients).

So The first [itex]m[/itex] terms are all 0.

So we have [itex]a_{m+1}z^{m+1} + ... + a_nz^n=0[/itex] .

Can you help me finish?

- #7

Ted123

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Instead, if [itex]P(z) = Q(z)[/itex] for all z and [itex]F(z) = P(z)-Q(z)[/itex], then [itex]1, 2, 3, ..., m+n+1[/itex] are all roots of [itex]F(z)[/itex], and so [itex](z-1)(z-2)...(z-(m+n+1))[/itex] divides [itex]F(z)[/itex] (factor theorem). Then what does this implies about the degree of F?

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