Proof by Induction Homework: Can Anyone Help?

[Ted123]

Homework Statement

http://img62.imageshack.us/img62/3433/questionl.png

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 .

 

[ystael]

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

 

[HallsofIvy]

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.

 

[Ted123]

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 ?