Proof by Induction Homework: Can Anyone Help?

  • Thread starter Thread starter Ted123
  • Start date Start date
  • Tags Tags
    Induction Proof
Click For Summary

Homework Help Overview

The discussion revolves around a proof by induction related to polynomial coefficients and their equality at specific points, particularly focusing on the derivatives of two polynomials P(z) and Q(z). The original poster seeks assistance in establishing a formal induction argument to demonstrate that the coefficients of these polynomials are equal.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to show that if the polynomials are equal at z=0 and their derivatives are also equal, then the coefficients must be equal. They question how to properly set up the induction process and whether their reasoning about the relationship between n and m is valid.

Discussion Status

Participants have provided various approaches to the problem, including suggestions to compare derivatives and manipulate polynomial expressions. There is an ongoing exploration of how to establish the induction step and whether the assumptions made about the coefficients are sufficient to conclude equality.

Contextual Notes

Some participants assume without loss of generality that n is less than or equal to m, which introduces a potential constraint on the induction argument. The original poster expresses uncertainty about how to proceed with the induction setup and the implications of their findings.

Ted123
Messages
428
Reaction score
0

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 .
 
Last edited by a moderator:
Physics news on Phys.org
Assume without loss of generality that n \leq m, and then compare P^{(n)} and Q^{(n)}.
 
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.
 
HallsofIvy said:
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 ?
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K