# Discrete mathematics and its application 2.4 problem 26

1. Feb 15, 2010

### GoGoDancer12

1. The problem statement, all variables and given/known data

Find a formula for when m $$\sum$$ k=0 the flooring function of[k1/3 ] ,m is a positive integer.

2. Relevant equations

n$$\prod$$ j=m aj

3. The attempt at a solution

the flooring function of[k1/3] = K

the summation of K is $$\frac{m(m+1)}{2}$$

There's a table of useful summation formulas in the Discrete Mathematics and Its Application sixth edition textbook pg.157 and that's where I got the summation formula for K. Just plug in m in the formula for the summation of K.

2. Feb 15, 2010

### GoGoDancer12

omit the second part

2. Relevant equations

nLaTeX Code: \\prod j=m aj
....not part of the problem

3. Feb 15, 2010

### GoGoDancer12

omit the second part

2. Relevant equations

nLaTeX Code: \\prod j=m aj
....not part of the problem

4. Feb 15, 2010

### Staff: Mentor

It's not clear to me what you're asking. When $$m\sum_{k = 0}^? floor(k^{1/3})$$ does what?

5. Feb 15, 2010

### GoGoDancer12

I have to find the summation formula for floor(K1/3):; and m is the on top of the summation symbol.

6. Feb 15, 2010

### Staff: Mentor

Start by expanding the summation:
floor(1) + floor(21/3) + floor(31/3) + ... + floor(m1/3).

Use enough terms so that you can find out how many of the terms will be equal to 1, to 2, to 3, and so on. That's what I would start with.

7. Feb 15, 2010

### farleyknight

I'd be willing to offer some help but I can't quite read you equations! Could you try and format them a bit better?

8. Feb 15, 2010

### Staff: Mentor

GogoDancer12 wants to find a closed for expression for
$$\sum_{k = 0}^m \lfloor k^{1/3}\rfloor$$

The $\lfloor$ and $\rfloor$ symbols are for the "floor" function, the greatest integer less than or equal to the specified argument.

9. Feb 15, 2010

### GoGoDancer12

exactly

10. Feb 15, 2010

### Staff: Mentor

Again, try expanding as I suggested in post #6. You're going to get a bunch of terms that are 1, a bunch that are 2, and so forth. See if you can figure a way to count how many 1s, 2s, and so forth.

11. Feb 15, 2010

### GoGoDancer12

after expanding the summation I got this :

1+1+1+1+1+1+m

12. Feb 15, 2010

### Staff: Mentor

And that works if m is, say, 64?

13. Feb 15, 2010

### GoGoDancer12

after expanding the summation some more I got this:
1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+...+m

and I'm still lost.

Last edited: Feb 15, 2010
14. Feb 15, 2010

### Staff: Mentor

How many 1s are there? How many 2s? 3s? Can you find any pattern? At what value of k do the 1s turn to 2s? Do the 2s turn to 3s? Are you sure the last number will be m?

15. Feb 16, 2010

### JSuarez

Here's a tip: look at the inverse function of $x^{1/3}$, which is $x^{3}$. Then you'll see this that:

$$\left\lfloor k^{1/3}\right\rfloor=1 \Rightarrow k \in \left[1,2^{3}\left[=\left[1,8\left[$$

$$\left\lfloor k^{1/3}\right\rfloor=2 \Rightarrow k \in \left[2^{3},3^{3}\left[=\left[8,27\left[$$

and so on...

From this you should be able to count the number of 1,2,etc.