Discrete mathematics and its application 2.4 problem 26


by GoGoDancer12
Tags: application, discrete, mathematics
GoGoDancer12
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#1
Feb15-10, 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[k1/3 ] ,m is a positive integer.

2. Relevant equations

n[tex]\prod[/tex] j=m aj



3. The attempt at a solution

the flooring function of[k1/3] = 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.
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GoGoDancer12
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#2
Feb15-10, 03:01 PM
P: 14
omit the second part

2. Relevant equations

nLaTeX Code: \\prod j=m aj
....not part of the problem
GoGoDancer12
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#3
Feb15-10, 03:02 PM
P: 14
omit the second part

2. Relevant equations

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

Mark44
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#4
Feb15-10, 03:31 PM
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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?
GoGoDancer12
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#5
Feb15-10, 05:09 PM
P: 14
I have to find the summation formula for floor(K1/3):; and m is the on top of the summation symbol.
Mark44
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#6
Feb15-10, 05:24 PM
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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.
farleyknight
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#7
Feb15-10, 06:12 PM
P: 128
Quote Quote by GoGoDancer12 View Post
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[k1/3 ] ,m is a positive integer.

2. Relevant equations

n[tex]\prod[/tex] j=m aj



3. The attempt at a solution

the flooring function of[k1/3] = 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.
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?
Mark44
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#8
Feb15-10, 06:55 PM
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P: 20,937
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.
GoGoDancer12
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#9
Feb15-10, 07:11 PM
P: 14
exactly
Mark44
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#10
Feb15-10, 07:46 PM
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P: 20,937
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.
GoGoDancer12
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#11
Feb15-10, 08:05 PM
P: 14
after expanding the summation I got this :

1+1+1+1+1+1+m
Mark44
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#12
Feb15-10, 08:28 PM
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And that works if m is, say, 64?
GoGoDancer12
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#13
Feb15-10, 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.
Mark44
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#14
Feb15-10, 10:30 PM
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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?
JSuarez
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#15
Feb16-10, 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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