# Proof by induction

## Homework Statement

http://img62.imageshack.us/img62/3433/questionl.png [Broken]

## The Attempt at a Solution

Can anyone help me with this induction?

If $P(z)=Q(z)$ for all $z$ then setting $z=0$ shows that $a_0 = b_0$ .

We must also have $P'(z)=Q'(z)$ and again setting $z=0$ shows that $a_1 = b_1$

Now I want to show, by induction that each successive derivative, at $z= 0$, gives the equality of the next coefficients.

Do I say that $P^n(0) = Q^m(0) \Rightarrow a_ n = b_m$

and then $P^{n+1}(z) = Q^{m+1}(z) \Rightarrow 0 = 0$ ??? In particular I need to show that $m=n$ .

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Assume without loss of generality that $$n \leq m$$, and then compare $$P^{(n)}$$ and $$Q^{(n)}$$.

HallsofIvy
Homework Helper
You can, for example, take z= 0 to show that $a_0= b_0$, then subtract that common value and factor out a "z":
$$z(a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1})= z(b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1})$$
For z not equal to 0, we can cancel and have
$$a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1}= b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1}$$

But polynomials are continuous so we have
$$a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1}= b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1}$$
for all z. Now take z= 0 again.

You can, for example, take z= 0 to show that $a_0= b_0$, then subtract that common value and factor out a "z":
$$z(a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1})= z(b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1})$$
For z not equal to 0, we can cancel and have
$$a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1}= b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1}$$

But polynomials are continuous so we have
$$a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1}= b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1}$$
for all z. Now take z= 0 again.
So how do I set up my induction in this situation? Doing this again and again will give $a_n=b_m$ eventually but does this prove that $m=n$ ?