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Discrete mathematics and its application 2.4 problem 26 
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#1
Feb1510, 12:56 PM

P: 14

1. The problem statement, all variables and given/known data
Find a formula for when m [tex]\sum[/tex] k=0 the flooring function of[k^{1/3} ] ,m is a positive integer. 2. Relevant equations n[tex]\prod[/tex] j=m a_{j} 3. The attempt at a solution the flooring function of[k^{1/3}] = K the summation of K is [tex]\frac{m(m+1)}{2}[/tex] 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
Feb1510, 03:01 PM

P: 14

omit the second part
2. Relevant equations nLaTeX Code: \\prod j=m aj ....not part of the problem 


#3
Feb1510, 03:02 PM

P: 14

omit the second part
2. Relevant equations nLaTeX Code: \\prod j=m aj ....not part of the problem 


#4
Feb1510, 03:31 PM

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P: 21,216

Discrete mathematics and its application 2.4 problem 26
It's not clear to me what you're asking. When [tex]m\sum_{k = 0}^? floor(k^{1/3})[/tex] does what?



#5
Feb1510, 05:09 PM

P: 14

I have to find the summation formula for floor(K^{1/3}):; and m is the on top of the summation symbol.



#6
Feb1510, 05:24 PM

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P: 21,216

Start by expanding the summation:
floor(1) + floor(2^{1/3}) + floor(3^{1/3}) + ... + floor(m^{1/3}). 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
Feb1510, 06:12 PM

P: 128




#8
Feb1510, 06:55 PM

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P: 21,216

GogoDancer12 wants to find a closed for expression for
[tex]\sum_{k = 0}^m \lfloor k^{1/3}\rfloor[/tex] The [itex]\lfloor[/itex] and [itex]\rfloor[/itex] symbols are for the "floor" function, the greatest integer less than or equal to the specified argument. 


#9
Feb1510, 07:11 PM

P: 14

exactly



#10
Feb1510, 07:46 PM

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P: 21,216

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
Feb1510, 08:05 PM

P: 14

after expanding the summation I got this :
1+1+1+1+1+1+m 


#12
Feb1510, 08:28 PM

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P: 21,216

And that works if m is, say, 64?



#13
Feb1510, 09:03 PM

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. 


#14
Feb1510, 10:30 PM

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P: 21,216

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
Feb1610, 11:14 AM

P: 403

Here's a tip: look at the inverse function of [itex]x^{1/3}[/itex], which is [itex]x^{3}[/itex]. Then you'll see this that:
[tex]\left\lfloor k^{1/3}\right\rfloor=1 \Rightarrow k \in \left[1,2^{3}\left[=\left[1,8\left[[/tex] [tex]\left\lfloor k^{1/3}\right\rfloor=2 \Rightarrow k \in \left[2^{3},3^{3}\left[=\left[8,27\left[[/tex] and so on... From this you should be able to count the number of 1,2,etc. 


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