Undergrad Proving that a solution to a recurrence relation is true

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To prove the correctness of a solution to a recurrence relation, such as H_n = 2H_{n-1} + 1, both substitution and induction methods can be used. Substituting the guessed solution, 2^n - 1, into the original relation checks its validity, but induction is necessary to confirm it holds for all integers, especially when a starting value like H_1 = 1 is provided. Without a starting value, substitution alone suffices, as both methods yield the same result. However, with a specified starting point, induction is essential to demonstrate that the solution satisfies this condition. Thus, both approaches are important for a comprehensive proof of the solution's correctness.
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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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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