Discrete mathematics and its application 2.4 problem 26

[GoGoDancer12]

Homework Statement

Find a formula for when m \sum k=0 the flooring function of[k^1/3 ] ,m is a positive integer.

Homework Equations

n\prod j=m a_j

The Attempt at a Solution

the flooring function of[k^1/3] = 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.

 

[GoGoDancer12]

omit the second part

2. Homework Equations

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

 

[GoGoDancer12]

omit the second part

2. Homework Equations

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

 

[Mark44]

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

 

[GoGoDancer12]

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

 

[Mark44]

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.

 

[farleyknight]

Homework Statement

Find a formula for when m \sum k=0 the flooring function of[k^1/3 ] ,m is a positive integer.

Homework Equations

n\prod j=m a_j

The Attempt at a Solution

the flooring function of[k^1/3] = 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?

 

[Mark44]

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.

 

[GoGoDancer12]

exactly

 

[Mark44]

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]

after expanding the summation I got this :

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

 

[Mark44]

And that works if m is, say, 64?

 

[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.

 

[Mark44]

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]

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.