# Discrete mathematics and its application 2.4 problem 26

by GoGoDancer12
Tags: application, discrete, mathematics
 P: 14 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.
 P: 14 omit the second part 2. Relevant equations nLaTeX Code: \\prod j=m aj ....not part of the problem
 P: 14 omit the second part 2. Relevant equations nLaTeX Code: \\prod j=m aj ....not part of the problem
 Mentor P: 21,215 Discrete mathematics and its application 2.4 problem 26 It's not clear to me what you're asking. When $$m\sum_{k = 0}^? floor(k^{1/3})$$ does what?
 P: 14 I have to find the summation formula for floor(K1/3):; and m is the on top of the summation symbol.
 Mentor P: 21,215 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.
P: 128
 Quote by 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.
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?
 Mentor P: 21,215 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.
 P: 14 exactly
 Mentor P: 21,215 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.
 P: 14 after expanding the summation I got this : 1+1+1+1+1+1+m
 Mentor P: 21,215 And that works if m is, say, 64?
 P: 14 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.
 Mentor P: 21,215 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?
 P: 403 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.

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