Big-O Solving Recurrence Relation by iteration with fractions

Question

I was trying to solve the recurrence relation in order to get a some big-O bound $$ B(n) = B(n-4) + \frac{1}{n} + \frac{5}{n^{2} + 6} + \frac{7n^{2}}{3n^{3} + 8}$$ by following the accepted answer here. I expanded which resulted in $$ B(n) = B(n-8) + \frac{1}{n-1} + \frac{5}{(n-1)^{2} + 6} + \frac{7(n-1)^{2}}{3(n-1)^{3} + 8} + \frac{1}{n} + \frac{5}{n^{2} + 6} + \frac{7n^{2}}{3n^{3} + 8}$$

This lead to $$ B(n) = B(n-4k) + \frac{1}{n-k} + \frac{5}{(n-k)^{2} + 6} + \frac{7(n-k)^{2}}{3(n-k)^{3} + 8} + \frac{1}{n-k} + \frac{5}{(n-k)^{2} + 6} + \frac{7(n-k)^{2}}{3(n-k)^{3} + 8}$$ However, I get stuck when replacing $k$ with $n$ since this results in dividing by zero. How do I proceed? Can it solved by the iteration?

The first calculation is wrong. It should be $B(n) = B(n-8) + \frac{1}{n-4} + \frac{5}{(n-4)^{2} + 6} + \frac{7(n-4)^{2}}{3(n-4)^{3} + 8} + \frac{1}{n} + \frac{5}{n^{2} + 6} + \frac{7n^{2}}{3n^{3} + 8}$ — John L., Oct 17 '18 at 18:09
Welcome to [cs.SE]! Your question looks nice. If this problem comes from an online source such as a programming contest or coding camp, please provide a URL. If it comes from a textbook or a paper, a reference. If it comes from some real task in your life, some background? All those information motivate and help people answer the question faster and better. Please add those information in the question since people and search engine are not expected of looking at comments. — John L., Oct 17 '18 at 18:13
Can you replace n by (n-4) everywhere in $B(n) = B(n-4) + \frac{1}{n} + \frac{5}{n^{2} + 6} + \frac{7n^{2}}{3n^{3} + 8}$? The only place that can be simplified is $(n-4)-4=n-8$. — John L., Oct 17 '18 at 18:16
@Apass.Jack thanks. This problem is from my own lecture notes — dreamin, Oct 17 '18 at 18:18
@Apass.Jack I made edits to the problem but I still don't see how to come by a solution — dreamin, Oct 17 '18 at 18:25
Can you answer @RickDecker's question above? I do not think there is closed formula for $B(n)$. — John L., Oct 17 '18 at 18:28
@Apass.Jack I'm not sure. I guess. I just saw this recurrence and I took the above steps in an attempt to solve it. How do you determine that there is no closed formula? — dreamin, Oct 17 '18 at 18:31
Many years of experience on math, if I have to say. Or I could define what is a closed formula in some way such as a quotient of two integer polynomials in $n$. Then I can prove it. — John L., Oct 17 '18 at 18:51
What was the course for which your lecture notes was taken? Then we can make a educated guess what was the initial goal. — John L., Oct 17 '18 at 18:54
@Apass.Jack The course is a graduate level analysis of algorithms — dreamin, Oct 17 '18 at 18:57
By the way, your first calculation is still wrong. You may take a look at my first comment. (Also please note I am planning to remove most of my comments here.) — John L., Oct 17 '18 at 19:01
@RickDecker would it possible to walk me through the steps of coming up with a solution by setting $B(n) \approx \frac{n}{4}$ ? — dreamin, Oct 17 '18 at 19:05
There is no way the last expression for $B(n)$ is right. The Wolfram Cloud gives me a much worse expression that is not even comparable. You have to sum up the stuff on the right from the first expression $k$ times, lowering $n$ by 4 each time — HackerBoss, Oct 17 '18 at 19:35
@HackerBoss that was my attempt at solving which I see now is not right. — dreamin, Oct 17 '18 at 19:59

score 1 · Answer 1 · answered Oct 18 '18 at 04:28

We can rewrite your recursion as $$ B(n) = B(n-4) + \Theta\left(\frac{1}{n}\right). $$ It follows that $$ \begin{align*} B(n) &= \Theta\left(\frac{1}{n} + \frac{1}{n-4} + \frac{1}{n-8} + \cdots\right) \\ &= \Theta\left(\frac{1}{n/4} + \frac{1}{n/4-1} + \frac{1}{n/4-2} + \cdots \right) \\ &= \Theta(\log (n/4)) = \Theta(\log n), \end{align*} $$ using the formula for the sum of a harmonic series.

score 0 · Answer 2 · answered Oct 17 '18 at 20:29

The base case will depend on $n$, but I will assume that $B(0)$, $B(1)$, $B(2)$, and $B(3)$ are known. I used the Wolfram Cloud to crunch some stuff. The following summarizes the 'simplified' results when $n=4k,\,4k+1,\,4k+2,\,\text{and}\,4k+3$ (these are given in Mathematica input form for later use in the Wolfram Cloud, etc):

B[n_]=Piecewise[{{(144*EulerGamma - (60*I)*Sqrt[6]*PolyGamma[0, (-I/2)*Sqrt[3/2]] + (60*I)*Sqrt[6]*PolyGamma[0, (I/2)*Sqrt[3/2]] + (60*I)*Sqrt[6]*PolyGamma[0, (-I/4)*(Sqrt[6] - I*n)] -
(60*I)*Sqrt[6]*PolyGamma[0, (I/4)*(Sqrt[6] + I*n)] + 144*PolyGamma[0, 1 + n/4] + 112*n^2*RootSum[8 + 3*n^3 - 36*n^2*#1 + 144*n*#1^2 - 192*#1^3 & , PolyGamma[0, -#1]/(n^2 - 8*n*#1 +
16*#1^2) & ] - 896*n*RootSum[-184 + 144*n - 36*n^2 + 3*n^3 - 576*#1 + 288*n*#1 - 36*n^2*#1 - 576*#1^2 + 144*n*#1^2 - 192*#1^3 & , (PolyGamma[0, -#1] + PolyGamma[0, -#1]*#1)/(16
- 8*n + n^2 + 32*#1 - 8*n*#1 + 16*#1^2) & ] + 1792*RootSum[-184 + 144*n - 36*n^2 + 3*n^3 - 576*#1 + 288*n*#1 - 36*n^2*#1 - 576*#1^2 + 144*n*#1^2 - 192*#1^3 & , (PolyGamma[0,
-#1] + 2*PolyGamma[0, -#1]*#1 + PolyGamma[0, -#1]*#1^2)/(16 - 8*n + n^2 + 32*#1 - 8*n*#1 + 16*#1^2) & ] - 7*n^2*RootSum[-1 + 24*#1^3 & , PolyGamma[0, -#1]/#1^2 & ] +
14*n*RootSum[-1 + 24*#1^3 & , (n*PolyGamma[0, -#1] + 4*PolyGamma[0, -#1]*#1)/#1^2 & ] - 7*RootSum[-1 + 24*#1^3 & , (n^2*PolyGamma[0, -#1] + 8*n*PolyGamma[0, -#1]*#1 +
16*PolyGamma[0, -#1]*#1^2)/#1^2 & ])/576, Mod[n, 4] == 0}, {((-15*I)*Sqrt[6]*PolyGamma[0, (-I/4)*(-3*I + Sqrt[6])] + (15*I)*Sqrt[6]*PolyGamma[0, (I/4)*(3*I + Sqrt[6])] +
(15*I)*Sqrt[6]*PolyGamma[0, (-I/4)*(Sqrt[6] - I*n)] - (15*I)*Sqrt[6]*PolyGamma[0, (I/4)*(Sqrt[6] + I*n)] + 28*n^2*RootSum[8 + 3*n^3 - 36*n^2*#1 + 144*n*#1^2 - 192*#1^3 & , PolyGamma[0,
-#1]/(n^2 - 8*n*#1 + 16*#1^2) & ] - 224*n*RootSum[-184 + 144*n - 36*n^2 + 3*n^3 - 576*#1 + 288*n*#1 - 36*n^2*#1 - 576*#1^2 + 144*n*#1^2 - 192*#1^3 & , (PolyGamma[0, -#1] +
PolyGamma[0, -#1]*#1)/(16 - 8*n + n^2 + 32*#1 - 8*n*#1 + 16*#1^2) & ] + 448*RootSum[-184 + 144*n - 36*n^2 + 3*n^3 - 576*#1 + 288*n*#1 - 36*n^2*#1 - 576*#1^2 + 144*n*#1^2 -
192*#1^3 & , (PolyGamma[0, -#1] + 2*PolyGamma[0, -#1]*#1 + PolyGamma[0, -#1]*#1^2)/(16 - 8*n + n^2 + 32*#1 - 8*n*#1 + 16*#1^2) & ] - 28*n^2*RootSum[-89 + 324*#1 - 432*#1^2 +
192*#1^3 & , PolyGamma[0, -#1]/(9 - 24*#1 + 16*#1^2) & ] + 56*n*RootSum[-89 + 324*#1 - 432*#1^2 + 192*#1^3 & , (-3*PolyGamma[0, -#1] + n*PolyGamma[0, -#1] + 4*PolyGamma[0,
-#1]*#1)/(9 - 24*#1 + 16*#1^2) & ] - 28*RootSum[-89 + 324*#1 - 432*#1^2 + 192*#1^3 & , (9*PolyGamma[0, -#1] - 6*n*PolyGamma[0, -#1] + n^2*PolyGamma[0, -#1] - 24*PolyGamma[0,
-#1]*#1 + 8*n*PolyGamma[0, -#1]*#1 + 16*PolyGamma[0, -#1]*#1^2)/(9 - 24*#1 + 16*#1^2) & ] + 144*DifferenceRoot[Function[{y, n}, {(4*n - n)*y[n] + 2*(-2 - 4*n + n)*y[1 + n] + (4 + 4*n
- n)*y[2 + n] == 0, y[0] == 0, y[1] == n^(-1)}]][(-3 + n)/4])/144, Mod[n, 4] > 2}, {((-15*I)*Sqrt[6]*PolyGamma[0, (-I/4)*(-I + Sqrt[6])] + (15*I)*Sqrt[6]*PolyGamma[0, (I/4)*(I + Sqrt[6])] +
(15*I)*Sqrt[6]*PolyGamma[0, (-I/4)*(Sqrt[6] - I*n)] - (15*I)*Sqrt[6]*PolyGamma[0, (I/4)*(Sqrt[6] + I*n)] + 28*n^2*RootSum[8 + 3*n^3 - 36*n^2*#1 + 144*n*#1^2 - 192*#1^3 & , PolyGamma[0,
-#1]/(n^2 - 8*n*#1 + 16*#1^2) & ] - 224*n*RootSum[-184 + 144*n - 36*n^2 + 3*n^3 - 576*#1 + 288*n*#1 - 36*n^2*#1 - 576*#1^2 + 144*n*#1^2 - 192*#1^3 & , (PolyGamma[0, -#1] +
PolyGamma[0, -#1]*#1)/(16 - 8*n + n^2 + 32*#1 - 8*n*#1 + 16*#1^2) & ] + 448*RootSum[-184 + 144*n - 36*n^2 + 3*n^3 - 576*#1 + 288*n*#1 - 36*n^2*#1 - 576*#1^2 + 144*n*#1^2 -
192*#1^3 & , (PolyGamma[0, -#1] + 2*PolyGamma[0, -#1]*#1 + PolyGamma[0, -#1]*#1^2)/(16 - 8*n + n^2 + 32*#1 - 8*n*#1 + 16*#1^2) & ] - 28*n^2*RootSum[-11 + 36*#1 - 144*#1^2 +
192*#1^3 & , PolyGamma[0, -#1]/(1 - 8*#1 + 16*#1^2) & ] + 56*n*RootSum[-11 + 36*#1 - 144*#1^2 + 192*#1^3 & , (-PolyGamma[0, -#1] + n*PolyGamma[0, -#1] + 4*PolyGamma[0,
-#1]*#1)/(1 - 8*#1 + 16*#1^2) & ] - 28*RootSum[-11 + 36*#1 - 144*#1^2 + 192*#1^3 & , (PolyGamma[0, -#1] - 2*n*PolyGamma[0, -#1] + n^2*PolyGamma[0, -#1] - 8*PolyGamma[0, -#1]*#1
+ 8*n*PolyGamma[0, -#1]*#1 + 16*PolyGamma[0, -#1]*#1^2)/(1 - 8*#1 + 16*#1^2) & ] + 144*DifferenceRoot[Function[{y, n}, {(4*n - n)*y[n] + 2*(-2 - 4*n + n)*y[1 + n] + (4 + 4*n - n)*y[2
+ n] == 0, y[0] == 0, y[1] == n^(-1)}]][(-1 + n)/4])/144, Mod[n, 4] == 1}}, ((-15*I)*Sqrt[6]*PolyGamma[0, (-I/4)*(-2*I + Sqrt[6])] + (15*I)*Sqrt[6]*PolyGamma[0, (I/4)*(2*I + Sqrt[6])] +
(15*I)*Sqrt[6]*PolyGamma[0, (-I/4)*(Sqrt[6] - I*n)] - (15*I)*Sqrt[6]*PolyGamma[0, (I/4)*(Sqrt[6] + I*n)] + 28*n^2*RootSum[8 + 3*n^3 - 36*n^2*#1 + 144*n*#1^2 - 192*#1^3 & , PolyGamma[0,
-#1]/(n^2 - 8*n*#1 + 16*#1^2) & ] - 224*n*RootSum[-184 + 144*n - 36*n^2 + 3*n^3 - 576*#1 + 288*n*#1 - 36*n^2*#1 - 576*#1^2 + 144*n*#1^2 - 192*#1^3 & , (PolyGamma[0, -#1] +
PolyGamma[0, -#1]*#1)/(16 - 8*n + n^2 + 32*#1 - 8*n*#1 + 16*#1^2) & ] + 448*RootSum[-184 + 144*n - 36*n^2 + 3*n^3 - 576*#1 + 288*n*#1 - 36*n^2*#1 - 576*#1^2 + 144*n*#1^2 -
192*#1^3 & , (PolyGamma[0, -#1] + 2*PolyGamma[0, -#1]*#1 + PolyGamma[0, -#1]*#1^2)/(16 - 8*n + n^2 + 32*#1 - 8*n*#1 + 16*#1^2) & ] - 7*n^2*RootSum[-2 + 9*#1 - 18*#1^2 + 12*#1^3 &
, PolyGamma[0, -#1]/(1 - 4*#1 + 4*#1^2) & ] + 14*n*RootSum[-2 + 9*#1 - 18*#1^2 + 12*#1^3 & , (-2*PolyGamma[0, -#1] + n*PolyGamma[0, -#1] + 4*PolyGamma[0, -#1]*#1)/(1 - 4*#1 +
4*#1^2) & ] - 7*RootSum[-2 + 9*#1 - 18*#1^2 + 12*#1^3 & , (4*PolyGamma[0, -#1] - 4*n*PolyGamma[0, -#1] + n^2*PolyGamma[0, -#1] - 16*PolyGamma[0, -#1]*#1 + 8*n*PolyGamma[0, -#1]*#1
+ 16*PolyGamma[0, -#1]*#1^2)/(1 - 4*#1 + 4*#1^2) & ] + 144*DifferenceRoot[Function[{y, n}, {(4*n - n)*y[n] + (-4 - 8*n + 2*n)*y[1 + n] + (4 + 4*n - n)*y[2 + n] == 0, y[0] == 0, y[1] ==
n^(-1)}]][(-2 + n)/4])/144]

This line assigns a function B[n_] in Mathematica which gives $B(n)$ given $n$, using the best explicit formula I could get, which was still really ugly.

I'm not sure I was clear in what I was trying to do and also how this relates to the question so I provided more clarity to the post. — dreamin, Oct 18 '18 at 02:52

Big-O Solving Recurrence Relation by iteration with fractions

2 Answers2