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Problem with a sum

  1. Mar 20, 2004 #1
    we know from basic number theory that

    Sum(1,x)1=[x] being [x] the integer part of x

    my problem is when we have the formula

    Sum(1,x)?=[Ax+b] being A and b arbitrary constant .. then

    ?=[ax+b]/[x] or not?..thanks.
     
  2. jcsd
  3. Mar 20, 2004 #2

    matt grime

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    Erm, when you sum(1,x)1

    do you mean:

    [tex]\sum_1^x1[/tex]

    cos if you do then it is assumed that x is an integer. It doesn't make sense to talk of the sum from 1 to pi, say, of a series - what is the indexing set that you're summing over?
     
  4. Mar 20, 2004 #3
    Yes Matt this is the definition i make.
    I took this definition from a math book but let,s suppose x is an integer so fgor general a nad b [ax+b] is not but my problem is to get the function so

    Sum(1,x)g(n)=[ax+b] x is an integer now i think g(x) is g(x,n)=[ax+b] but i am not sure.
     
  5. Mar 20, 2004 #4

    matt grime

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    Stating g(n)=[ax+b] is saying that the n'th term in the series is independent of n, so you might just be adding up a number [ax+b], some number of times, possible [x] but i'm not sure. You should cleary state what the index set of the sum is and what the terms of the sum are.


    It could be read as saying that you want to know if there is nice formula for

    [tex]\sum_{x=1}^{n}[ax+b][/tex]

    for a and b some real numbers.

    Off the top of my head I know of no nice way of doing it, other than working it out. Floor is not a great function: it would be nice if [u+v]=+[v] but it doesn't (let u=v=1/2). It would be nice if [rs]=[r] but that isnt' true either (r=2 s=1/2)

    Or it could be you want

    [tex]\sum_{n=1}^{[x]}[ax+b][/tex]

    which would be [x][ax+b]

    The latter is closer to what you wrote I think
     
  6. Mar 20, 2004 #5
    My problemdeals with euler,s sum identity:

    Sum(1,x)f(n)=f(x)[x]-f(1)-Int(1,x)[x]f`(x)dx

    where the comma means differentiation..now i would like to have a similar identity but with the function [ax+b], in fact using Abel,s summ formula

    Sum(1,x)a(n)f(n)=[ax+b]f(x)-[a+b]f(1)-Int(1,x)[ax+b]f`(x)dx now the problem is to get the a(n) so
    Sum(1,x)a(n)=[ax+b]
     
  7. Mar 20, 2004 #6

    matt grime

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    for future reference and clarity, always state what the index set is, ie sum(n=1,[x])

    and you want the sum(n=1,[x])a(n) to be equal to [ax+b] ?


    Your short hand confused me, I thought you wanted a(n)=[ax+b] which didn't make sense to me, but that part was me being dense.

    Note: you really ought to have [x] in the limits if x is not an integer, which is not me being dense.

    In that case defining a(n) = [ax+b]/[x] will give you the required sum, cos you're just adding up a constant [x] times, where the a's on either side are unrelated, of course.

    Bad notation is the bane of mathematics
     
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