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Proof by induction

  • Thread starter Ted123
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  • #1
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Homework Statement



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

The Attempt at a Solution



Can anyone help me with this induction?

If [itex]P(z)=Q(z)[/itex] for all [itex]z[/itex] then setting [itex]z=0[/itex] shows that [itex]a_0 = b_0[/itex] .

We must also have [itex]P'(z)=Q'(z)[/itex] and again setting [itex]z=0[/itex] shows that [itex]a_1 = b_1[/itex]

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

Do I say that [itex]P^n(0) = Q^m(0) \Rightarrow a_ n = b_m[/itex]

and then [itex]P^{n+1}(z) = Q^{m+1}(z) \Rightarrow 0 = 0[/itex] ??? In particular I need to show that [itex]m=n[/itex] .
 
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Answers and Replies

  • #2
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Assume without loss of generality that [tex]n \leq m[/tex], and then compare [tex]P^{(n)}[/tex] and [tex]Q^{(n)}[/tex].
 
  • #3
HallsofIvy
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You can, for example, take z= 0 to show that [itex]a_0= b_0[/itex], then subtract that common value and factor out a "z":
[tex]z(a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1})= z(b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1})[/tex]
For z not equal to 0, we can cancel and have
[tex]a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1}= b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1}[/tex]

But polynomials are continuous so we have
[tex]a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1}= b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1}[/tex]
for all z. Now take z= 0 again.
 
  • #4
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You can, for example, take z= 0 to show that [itex]a_0= b_0[/itex], then subtract that common value and factor out a "z":
[tex]z(a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1})= z(b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1})[/tex]
For z not equal to 0, we can cancel and have
[tex]a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1}= b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1}[/tex]

But polynomials are continuous so we have
[tex]a_1+ a_2z+ \cdot\cdot\cdot+ a_nz^{n-1}= b_1+ b_2z+ \cdot\cdot\cdot+ b_mz^{m-1}[/tex]
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 [itex]a_n=b_m[/itex] eventually but does this prove that [itex]m=n[/itex] ?
 

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