I posted this on PF but it didn't get very far:

 

Suppose {a[n]} is an increasing sequence and whenever m divides n, then

 

a[n] = a[n/m] + d

 

where m is a positive integer and d is a real number. Show that a[n] = Θ(lg n).

 

I can show that a[n] = O(lg n). All I need to do is show that a[n] = Ω(lg n). This is where I'm stuck. Any hints?

sorry dude, way out my league. can u care to explain what

 

Ω(lg n).

Θ(lg n).

O(lg n).

 

those functions are?

Actually those are sets. They are used in order analysis to determine how a particular function grows.

 

We say that g(x) = O(f(x)) iff for any constanct c > 0 their exists a k >= 0 such that g(x) <= cf(x) for all x >= k. In other words, g(x) = O(f(x)) iff g(x) grows no more than f(x), i.e. O(f(x)) is the set of functions that grow no more that f(x). Example: x^2 + x + 3 = O(x^2). Also x = O(x^2), but x^3 != O(x^2).

 

Similarly, we say that g(x) = Ω(f(x)) iff for any constant c > 0 their exists a k >= 0 such that g(x) >= cf(x) for all x >= k. In other words, Ω(f(x)) is the set of functions that grow no less than f(x), practically the opposite of O(f(x)).

 

If g(x) = Ω(f(x)) and g(x) = O(f(x)) then g(x) = Θ(f(x)) and vice versa.

 

So basically what I need to show is that a[n] grown no more and no less than lg n.

x^2 + x + 3 = O(x^2). dont understand how those two can be equal. one is a function and the other is a set of functions

 

did u mean

[math]x^2+x+3 \in O(x^2)[/math]?

 

its a new concept to me. and finding a lil bit difficult to visualise. I will have another look at it later

Oh yeah. I forgot to mention that it's a common convention to say

 

[math]x^2 + x + 3 = O(x^2)[/math]

 

instead of

 

[math]x^2 + x + 3 \in O(x^2)[/math]

 

They are both equivalent. It's just convention.

Where you have n/m are you meaning the floor/ceiling of n/m.

If so, I think you can use the "Master Theorem". This gives asymptotic bounds.

Where you have n/m are you meaning the floor/ceiling of n/m.

If so' date=' I think you can use the "Master Theorem". This gives asymptotic bounds.[/quote']

No, I don't mean the floor or the ceiling of n/m. Read the post carefully. Anyways, I would like to solve this without the use of the Master theorem.

So do you mean it doesn't matter what the values of a[n] are when m doens't divide n, as long as the property of it being an increasing sequence is maintained?

 

If so, then isn't a[n] lower bounded by the sequence {b[n]} with b[n] defined as:

 

b[n] = b[Floor[n/m]] + d

 

In that case you could still use the Master Theorem

Even if you don't want to use the master theorem, does that help at all?

Let b[1] = c

b[m] = c + d

b[m^2] = c + 2d

 

b[n] = c + Floor[Log_m[n]]*d

b[n] = ?(1) + ?(lg n)

So do you mean it doesn't matter what the values of a[n'] are when m doens't divide n, as long as the property of it being an increasing sequence is maintained?

Yes. The recurrence is defined ONLY when m divides n.

 

If so, then isn't a[n] lower bounded by the sequence {b[n]} with b[n] defined as:

 

b[n] = b[Floor[n/m]] + d

Interesting, but first I would need to show that b[n] is less than or equal to a[n] for all n. I haven't attempted this yet so bear with me.

 

In that case you could still use the Master Theorem

I would but the book I got this problem from does not mention the Master Theorem anywhere.

 

Even if you don't want to use the master theorem, does that help at all?

We'll see.

 

Let b[1] = c

b[m] = c + d

b[m^2] = c + 2d

 

b[n] = c + Floor[Log_m[n]]*d

b[n] = ?(1) + ?(lg n)

Note that the "d" in the recurrence of a[n] is not constant (in wouldn't make sense otherwise). I'll examine your suggestion more carefully later (I'm working and I shouldn't be doing this). Thanks.

I forgot to mention two important details: m > 1 and d > 0.

 

OK. I've managed to show that b[n] ≤ a[n] for all n assumming b[0] = a[0].

 

All I need to do now is show the following: Let f and g be functions of x such that f(x) ≤ g(x) for all x. For some function h, if f = O(h) then g = Ω(h). Proof: Define h such that f(x) ≤ h(x) ≤ g(x).

 

In order to prove that a[n] = Ω(lg n), let f(n) = b[n] - 1 so f(n) < b[n] ≤ a[n] and since f(n) = O(lg n), then a[n] = Ω(lg n).

 

Somehow I have a feeling that I've made an error somewhere. Can anybody check this for me please?

I revive this old post in order to resolve the matter in question once and for all. I have conferred this problem with a colleague of mine. He suggested that m and d are both constants, which I certaintly agree on. The misunderstandings came from the wording of the problem (i.e. the problem does not explicitly state that d and m are constants). Here is a similar, albeit harder, ambigous problem:

 

Suppose {a[n]} is an increasing sequence and that whenever m divides n, a[n] = ca[n/m] + d, where c and d are real numbers satisfying c > 1 and d > 0, and m is an integer satisfying m > 1. Show that a[n] = [math]\textstyle \Theta (n^{\log _m c})[/math].

 

Nowhere does the problem state that c, d, and m are constants, though one must assume this to solve the problem. Here is the solution:

 

Let n = m^k so a[n] = a[m^k]. Define b[k] = a[m^k]. The given recurrence can be written as b[k] = cb[k - 1] + d. "Expanding" the recurrence yields b[k] = c^kb[0] + c^{k - 1}d + c^{k - 2}d + ... + cd + d. Simplifing this expression yields

 

a[m^k] = b[k] = c^ka[1] + d(c^k - 1)/(c - 1)

 

where a[1] = b[0] (by definition). Since [math]\textstyle c^k = m^{\log _m c^k} = m^{k\log _mc} = n^{\log _m c}[/math] then [math]\textstyle a_{m^k} = \Theta (n^{\log _m c})[/math]. But what happens if n is not a power of m? Algebra says m^{k - 1} < n ≤ m^k so a[m^{k - 1}] < a[n] ≤ a[m^k] or

 

c^{k - 1}a[1] + d(c^{k - 1} - 1)/(c - 1) < a[n] ≤ c^ka[1] + d(c^k - 1)/(c - 1)

 

Concentrating on the left-hand side, one can see that

 

c^{k - 1}a[1] + d(c^{k - 1} - 1)/(c - 1) ≥ c^{k - 1} = [math]\frac{1}{c} m^{k \log _m c} \ge \frac{1}{c}n^{\log _m c} = \Omega(n^{\log _m c})[/math]

 

On the right-hand side, one can use algebra and derive

 

c^ka[1] + d(c^k - 1)/(c - 1) = [math]m^{k \log _m c}a_1 +\dots < cn^{\log _m c}a_1 + \dots = O(n^{\log _m c})[/math]

 

Therefore [math]\textdisplay a_n = \Theta(n^{\log _m c})[/math]. What I have essentially proven here is a simpler version of the Master Theorem. The original problem (see first post) is an even simpler version of the Master Theorem. How interesting!