# Complete Induction Proof

• NATURE.M
In summary, the conversation discusses a mathematical proof by induction, specifically showing that the sequence r is O (log2 (log2 n)). The conversation delves into the definition of big oh and the use of induction to prove the inductive step. It is suggested to assume the inductive hypothesis to be true and use the monotonicity of the log function to simplify the expression. The conversation ends with a question about how to get from 1 + rfloor(√(n+1)) to 1 + rfloor(√n).

## Homework Statement

We are given the sequence r defined by: r1 = 1, and rn = 1 + rfloor(√n)
, n≥2
We need to show by induction that rn is O (log2 (log2 n)).

## The Attempt at a Solution

Definition of big oh: ∃c∈ℝ+, ∃B∈ℕ, ∀n∈ℕ, n≥B => f(n) ≤ cg(n)
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So the basic proof format is fairly simple. My issue is with the inductive step. We let for n≥3, P(n): rn ≤ 4log2(log2n)). Now this comes from the defn. of big oh and is what we have to use induction to prove. In addition, from the definition of big oh, we defined c = 4, since it seems to work.

My issue is in the inductive step I have to show P(n+1) but have no idea how to proceed. How do I get from rn+1 = 1 + rfloor(√(n+1)) to 4log2(log2n+1) ? I'm trying to invoke the IH but don't know how to simplify either expression. Any hints would be appreciated.

Also maybe this thread should belong to the computer science section instead.

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I would suggest not assuming c=4 just yet. Leave it as c.
What do you get when you substitute for rfloor(√(n+1)) using the IH?

haruspex said:
I would suggest not assuming c=4 just yet. Leave it as c.
What do you get when you substitute for rfloor(√(n+1)) using the IH?

Ok I'll assume c is arbitrary for now. But i can't just use the IH directly on rfloor(√(n+1)) since i have to simplify it first. My issue is I have no idea how to go about simplifying it.
And as a note: We let n∈ℕ, then the IH is P(n) or for complete induction it would be for 3≤k<n, P(k).

NATURE.M said:
Ok I'll assume c is arbitrary for now. But i can't just use the IH directly on rfloor(√(n+1)) since i have to simplify it first. My issue is I have no idea how to go about simplifying it.
And as a note: We let n∈ℕ, then the IH is P(n) or for complete induction it would be for 3≤k<n, P(k).
Like haruspex suggested, assume the IH to be true:
$$r_n = 1 + r_{\lfloor\sqrt{n}\rfloor} \leq 1 + c\log(\log\sqrt{n})$$
Recall then that you'd like a ##c\log(\log n)## term on the RHS of the inequality.

Edit: I did the problem using different notation, and it came out as nonsense here. Removed that part.

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milesyoung said:
Like haruspex suggested, assume the IH to be true:
$$r_n = 1 + r_{\lfloor\sqrt{n}\rfloor} \leq 1 + c\log(\log\sqrt{n})$$
Recall then that you'd like a ##c\log(\log n)## term on the RHS of the inequality.

Edit: I did the problem using different notation, and it came out as nonsense here. Removed that part.
So I'm kinda confused where did the square root come from in the expression log2(log2√n)

NATURE.M said:
But i can't just use the IH directly on rfloor(√(n+1)) .
You want to show rn+1 <= something, call it X. So you want to show rfloor(√(n+1)) <= X. The simplification is to find some Y such that rfloor(√(n+1)) is clearly <= Y, then show Y <= X. Is the r sequence monotonic?

haruspex said:
You want to show rn+1 <= something, call it X. So you want to show rfloor(√(n+1)) <= X. The simplification is to find some Y such that rfloor(√(n+1)) is clearly <= Y, then show Y <= X. Is the r sequence monotonic?
We aren't given whether the sequence is monotonic or not. So I understand that I need to find Y. My issue is I don't know how to reduce rfloor(√(n+1)) into some Y, from which I can employ the IH.

NATURE.M said:
We aren't given whether the sequence is monotonic or not. So I understand that I need to find Y. My issue is I don't know how to reduce rfloor(√(n+1)) into some Y, from which I can employ the IH.
Sorry, I asked the wrong question. I should have asked whether the log function is monotonic.

NATURE.M said:
We aren't given whether the sequence is monotonic or not. So I understand that I need to find Y. My issue is I don't know how to reduce rfloor(√(n+1)) into some Y, from which I can employ the IH.
Can you show that the sequence is monotonic?

Try writing out the first part of the sequence. It increases very slowly, which is not a surprise, considering what you're asked to prove.

For what values of n is ##\ r_{n+1}>r_n \ ## ?

SammyS said:
Can you show that the sequence is monotonic?

Try writing out the first part of the sequence. It increases very slowly, which is not a surprise, considering what you're asked to prove.

For what values of n is ##\ r_{n+1}>r_n \ ## ?
But let's say its monotonic, in what way does that help us ? Maybe if it was monotonically decreasing, then it would help us in the proof as follows:
rfloor(√(n+1)) ≤ rfloor(√(n)) , and then we can invoke the inductive hypothesis. But if its increasing which it is than how does it help us ?

NATURE.M said:
But let's say its monotonic, in what way does that help us ?
As I wrote in post #8, what's useful is to think about monotonicity of the log function.
Apply the IH to 1 + rfloor(√n). What do you get? What's an upper bound for that?

haruspex said:
As I wrote in post #8, what's useful is to think about monotonicity of the log function.
Apply the IH to 1 + rfloor(√n). What do you get? What's an upper bound for that?
Okay that makes more sense since the log function is increasing monotonically.

Applying the induction hypothesis: 1 + rfloor(√n) ≤ clog2(log2n)., and from here we can use the monotonicity of the logarithmic function to get to clog2(log2n+1). Although, we will have to prove the monotonicity of the log function separately I guess.

And I'm still unsure how to get from 1 + rfloor(√(n+1)) to 1 + rfloor(√n) ?

NATURE.M said:
Applying the induction hypothesis: 1 + rfloor(√n) ≤ clog2(log2n)
No, not just 'n' inside the logs on the right. What does n represent in the IH?

haruspex said:
No, not just 'n' inside the logs on the right. What does n represent in the IH?
Just to clarify which IH are you referring to, for simple induction of for complete induction. If the latter, than n is the upper bound. This is the case we want to show is true. Not sure if that is what you mean.

NATURE.M said:
Just to clarify which IH are you referring to, for simple induction of for complete induction. If the latter, than n is the upper bound. This is the case we want to show is true. Not sure if that is what you mean.
I mean in rn ≤ c log2(log2n)). How do you apply that to rfloor(√(n+1))?

haruspex said:
I mean in rn ≤ c log2(log2n)). How do you apply that to rfloor(√(n+1))?
But that was my initial question ? I'm not sure how to apply it to rfloor(√(n+1)).

NATURE.M said:
But that was my initial question ? I'm not sure how to apply it to rfloor(√(n+1)).
Maybe it would help if we used two different indexes. The inductive hypothesis is, say, rk≤ c log2(log2k)). What do you have to set k to so that it applies to rfloor(√(n+1))?

haruspex said:
Maybe it would help if we used two different indexes. The inductive hypothesis is, say, rk≤ c log2(log2k)). What do you have to set k to so that it applies to rfloor(√(n+1))?
We would have to set k = n + 1. So then we have 1+ rfloor(√(n+1) = 1+ rfloor(√(k) but we can't conclude that its ≤ c log2(log2k))
since our induction hypothesis only assume rn to be true.

NATURE.M said:
We would have to set k = n + 1.
No. You need to set k to something which will make the left hand side look like rfloor(√(n+1). I'm not asking you to do anything difficult here.

haruspex said:
No. You need to set k to something which will make the left hand side look like rfloor(√(n+1). I'm not asking you to do anything difficult here.
Oh just set it equal to k = floor(√(n+1)). Then rk is ≤ the log term by the IH.

NATURE.M said:
Oh just set it equal to k = floor(√(n+1)).
Right. What inequality does the IH give you? Can you see how to get rid of the floor function by exploiting monotonicity of log?

So here is my logic (by complete induction)
Induction Hypothesis: rk ≡ 1 + rfloor(√(k)) ≤ clog2(log2k) for 3≤k<n.
Then, rn = 1+ r floor(√(n)). By the IH, we know rfloor(√(n)) ≤ clog2(log2floor(√(n))) since floor(√(n)) < n.
So then we have 1+ r floor(√(n)) ≤ 1 + clog2(log2floor(√(n))). And the log function is increasing, so 1 + clog2(log2floor(√(n))) ≤ 1 + clog2(log2n).

So I think the above should be right for the most part, I'm a bit uncertain what to do with the 1 however.

NATURE.M said:
since floor(√(n)) < n
No, you've thrown away too much. You're only trying to get rid of the floor function.

haruspex said:
No, you've thrown away too much. You're only trying to get rid of the floor function.
Isn't that what I did ? I got rid of the rfloor(√n) function, since rfloor(√n) = 1 + rfloor(√(floor(√n)) ≤ clog2(log2floor(√(n))) by IH.

As an aside, thanks for being so patient.

In doing this:
NATURE.M said:
1 + c log2(log2floor(√(n))) ≤ 1 + c log2(log2n).
you've thrown away the √ as well as the floor. Only get rid of the floor.

haruspex said:
In doing this:

you've thrown away the √ as well as the floor. Only get rid of the floor.
Oh I figured it out, cause then you use that to break the log term and to cancel out the + 1 term.

And just two last questions then. Should i just leave c as is for the entire proof, as an arbitrary positive real ?
And since I'm using complete induction in my IH, I said P(k) for 2 ≤ k < n. Now is it true that floor(√(n)) < √n only if n > 1. So its fair if I prove for ≥2.

NATURE.M said:
Should i just leave c as is for the entire proof, as an arbitrary positive real ?
It isn't arbitrary. You should find it has to be at least some value for the result to work. Please post your working from where it left off.
NATURE.M said:
Now is it true that floor(√(n)) < √n only if n > 1
Putting n = x2, x > 0, (and correcting from '<' to '<=') you are asking whether floor(x) <= x only if x > 1. Is it?

haruspex said:
It isn't arbitrary. You should find it has to be at least some value for the result to work. Please post your working from where it left off.

Putting n = x2, x > 0, (and correcting from '<' to '<=') you are asking whether floor(x) <= x only if x > 1. Is it?

Okay, so 1+ r floor(√(n)) ≤ 1 + clog2(log2floor(√(n))) ≤ 1 + clog2(log2(√(n))) = 1 + clog2(1/2)(log2(n))
= 1 + clog21/2 + clog2(log2(n)) = clog2(log2(n)), as desired.

Now I'm not entirely sure how to solve for c.

And well floor(0) <= 0, so no. However, since the sequence is only defined for n>=1 that case wouldn't be important. The case where n = 1, we can prove as a separate case (or directly) since the sequence is defined differently. However, this would require us knowing what c is. And for n>=2 we proceed as above.

NATURE.M said:
1 + c log21/2 + c log2(log2(n)) = c log2(log2(n)),
There's a mistake there. You have effectively assumed a value for c.

haruspex said:
There's a mistake there. You have effectively assumed a value for c.
Oops I missed that. In that case, can't we just let c = 1, at the beginning of the proof since the definition of big oh asks for you to choose a c. (∃c∈ℝ+).

NATURE.M said:
Oops I missed that. In that case, can't we just let c = 1, at the beginning of the proof since the definition of big oh asks for you to choose a c. (∃c∈ℝ+).
Sure. The choice for c is 1 or greater. Since you'd generally look for the tightest bound, choose 1.

haruspex said:
Sure. The choice for c is 1 or greater. Since you'd generally look for the tightest bound, choose 1.

Ok. And also just realized that n has to be ≥3 otherwise the proof fails. Namely, in the definition of big oh our choice of B should be 3. Since n = 1 is undefined for the double log term and n=2 doesn't satisfy the predicate with c = 1.

NATURE.M said:
Ok. And also just realized that n has to be ≥3 otherwise the proof fails. Namely, in the definition of big oh our choice of B should be 3. Since n = 1 is undefined for the double log term and n=2 doesn't satisfy the predicate with c = 1.
OK.
All done?

haruspex said:
OK.
All done?
Yeah I believe that is it. I can't thank you enough for your efforts and being patient with me throughout.

NATURE.M said:
Yeah I believe that is it. I can't thank you enough for your efforts and being patient with me throughout.
A pleasure.