- #1

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Can this be expressed through some function? Like the sum of 1 + 2 .. +n is n(n+1) /2

Or as a Riemann sum?

I'm trying to add only the odd numbers from a random set of N integers to a sum.

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- Thread starter caljuice
- Start date

- #1

- 70

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Can this be expressed through some function? Like the sum of 1 + 2 .. +n is n(n+1) /2

Or as a Riemann sum?

I'm trying to add only the odd numbers from a random set of N integers to a sum.

- #2

- 795

- 7

Can this be expressed through some function? Like the sum of 1 + 2 .. +n is n(n+1) /2

Or as a Riemann sum?

I'm trying to add only the odd numbers from a random set of N integers to a sum.

Do you mean in some programming language? If '/' is integer division, then x is even if and only if

x = 2 * (x/2). [using '=' as equality, not assignment]

But if you already have a mod operator like % then you can just use that.

- #3

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2*n+1 is always an odd number, for any value of n.

- #4

- 38

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yes the modulus function works well.

if(x%2 == 1){

...

}

alternatively, you can test the last bit to see if it's 1.

if(x&1 == 1){

...

}

this is slightly faster.

- #5

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For example

Let's say the set of integers is from some function( not random anymore)

[itex]\sum \limits_{i=1}^n A

- #6

pwsnafu

Science Advisor

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If finding the number was odd or mod or % 2 was rewritable as a function or riemann sum, then I could reduce or solve the problem.

I fail to see how Riemann sums would be relevant in what appears to be a discrete math question.

- #7

coolul007

Gold Member

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For example

Let's say the set of integers is from some function( not random anymore)

[itex]\sum \limits_{i=1}^n A(A\mod{2})[/itex] is only adding odd terms but the equation is not reducible, so I was hoping the % could be expressed differently so I could solve or reduce the problem. If finding the number was odd or mod or % 2 was rewritable as a function or riemann sum, then I could reduce or solve the problem.

I think he wants do something like the trick with minus signs alternating by using n for an exponent on -1. Just a guess here.

- #8

pwsnafu

Science Advisor

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I think he wants do something like the trick with minus signs alternating by using n for an exponent on -1. Just a guess here.

So something like

##\frac{1 - (-1)^n}{2}##

- #9

coolul007

Gold Member

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So something like

##\frac{1 - (-1)^n}{2}##

adding all the odds would look like this:

[itex]\sum n(\frac{1 - (-1)^n}{2})[/itex]

where n is a positive integer

- #10

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- #11

- 841

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For example

Let's say the set of integers is from some function( not random anymore)

[itex]\sum \limits_{i=1}^n A(A\mod{2})[/itex] is only adding odd terms but the equation is not reducible, so I was hoping the % could be expressed differently so I could solve or reduce the problem. If finding the number was odd or mod or % 2 was rewritable as a function or riemann sum, then I could reduce or solve the problem.

I don't understand the problem here. All odd numbers equal 1 mod 2, so adding n odd numbers will give you the result in mod 2 the following:

[itex]\sum \limits_{i=1}^n A

- #12

pwsnafu

Science Advisor

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I don't understand the problem here. All odd numbers equal 1 mod 2, so adding n odd numbers will give you the result in mod 2 the following:

[itex]\sum \limits_{i=1}^n A\mod{2}) = n [/itex] so n odd numbers will give an even number if n is even and an odd number if n is odd. There is no need to use more complex functions.

He doesn't want to

##A = (1,2,3,5,7,8,10)##, he wants an expression which evaluates 1+3+5+7.

- #13

- 795

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He doesn't want tocountthe number of odd A, but to sum only the odds. So if

##A = (1,2,3,5,7,8,10)##, he wants an expression which evaluates 1+3+5+7.

Oh I see. Well then how about

x = [itex]\displaystyle \sum_{k=1}^n a_ksin^2(a_k\pi) [/itex]

If a

- #14

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Hmm, this looks neat, so you are going to add all the odd terms in f(x)? So you need, as a function, [itex]\displaystyle \sum_{x=a}^n f(x)\frac{1-(-1)^{f(x)}}{2}[/itex]. That looks like a really cool and useful function! I have literally filled notebooks finding sums like that, so I have a huge collection, but I've never used a power of f(x) for anything other than f(x)=c or f(x)=x. Do you mind if I play with this? The most that I can simplify this too, without any extra exploration is:

[itex]\displaystyle \sum_{x=a}^n f(x)\frac{1-(-1)^{f(x)}}{2} = \sum_{x=a}^n \frac{f(x)}{2}-\sum_{x=a}^n f(x)\frac{(-1)^{f(x)}}{2}[/itex]

For my own purposes, I will want to look at x=0, specifically. If I want to look at specific parts of (f(x), I will use f(x-c).

**EDIT:** I realised after working through this a bit that I forgot to multiply by f(x). The formulas above are corrected (I hope). However, I do have a list of solutions that relate to the correct version (where f(x)=x^{a} for some non-negative integer a). Typically, the first part of that sum would be calculated with the Hurwitz Zeta function and the Riemann Zeta function (though I have a different way that I prefer). The second part, I believe, can be similarly computed. For cases where f(x) is a polynomial, this should not be difficult to compute, especially as it can be expressed as a linear combination of c_{n}x^{n}

[itex]\displaystyle \sum_{x=a}^n f(x)\frac{1-(-1)^{f(x)}}{2} = \sum_{x=a}^n \frac{f(x)}{2}-\sum_{x=a}^n f(x)\frac{(-1)^{f(x)}}{2}[/itex]

For my own purposes, I will want to look at x=0, specifically. If I want to look at specific parts of (f(x), I will use f(x-c).

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