Proving that a solution to a recurrence relation is true

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    Recurrence Relation
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Discussion Overview

The discussion revolves around the methods for proving the correctness of a proposed solution to a recurrence relation, specifically the relation ##H_n = 2H_{n-1} + 1##. Participants explore the validity of using both substitution and mathematical induction as proof techniques, along with the implications of having an initial condition.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the necessity of both plugging the proposed solution into the recurrence relation and using induction to prove its correctness.
  • Another participant argues that plugging in the solution is equivalent to performing the induction step, suggesting that both methods yield the same result if no initial value is provided.
  • A later reply questions the validity of the approach if a starting value, ##H_1 = 1##, is given, implying that induction may then be necessary.
  • It is noted that ##H_n = 2^n - 1## is not the only solution to the recurrence relation.
  • One participant clarifies that if a starting value is present, the proposed solution must also satisfy this initial condition, aligning the approach with full induction.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether both methods are necessary in all cases, particularly when an initial condition is involved. There are competing views on the sufficiency of each proof method.

Contextual Notes

The discussion highlights the dependence on initial conditions and the potential for multiple solutions to the recurrence relation, which may affect the proof strategy.

Mr Davis 97
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I am a little confused about how we prove that a solution for a recurrence relation is correct. For example, say that I have the recurrence relation ##H_n = 2H_{n-1} +1##. Using an iterative process, we guess that the solution is ##2^n - 1##. Now, to prove that this is correct, it seems that it would be sufficient to just plug it into the original recurrence relation to see if everything checks out. However I have seen some sources that claim we also need to use induction on ##H_n = 2^n - 1## to verify that it is true. Why do I have to do both to prove that it is true? Why can't I just do the former or just do the latter?
 
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Plugging it in is equivalent to the induction step.
If you have no starting value given, there is no need for the start of the induction, so both approaches are identical.

Hn = 2n-1 is not the only solution by the way.
 
mfb said:
Plugging it in is equivalent to the induction step.
If you have no starting value given, there is no need for the start of the induction, so both approaches are identical.

Hn = 2n-1 is not the only solution by the way.
What if I do have a starting value, ##H_1 = 1##? Would proof by induction be the only valid way in this case?
 
You would have to show that your solution also satisfies this starting value. And then your approach is identical to full induction.
 
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