Solving Sum(1,x) = [Ax+b] for x: A & b Constants

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In summary, the conversation discussed the concept of summing a series from 1 to x using different functions. The first part explored the use of basic number theory and the definition of the sum. The second part delved into the problem of finding a formula for a sum when the terms are defined by [ax+b]. Different approaches were discussed, including the use of Euler's sum identity and Abel's summation formula. The conversation concluded with a suggestion to define a(n) = [ax+b]/[x] in order to obtain the desired sum. However, it was noted that using clear and precise notation is important in mathematics.
  • #1
eljose79
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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.
 
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  • #2
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?
 
  • #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.
 
  • #4
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
 
  • #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]
 
  • #6
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
 

Related to Solving Sum(1,x) = [Ax+b] for x: A & b Constants

1. How do I solve for x in this equation?

The first step in solving this equation is to simplify the left side by using the formula for the sum of consecutive integers, which is n(n+1)/2. This will give you (x+1)/2. Then, you can use the distributive property to expand the right side to Ax + b. From there, you can rearrange the equation to get x by itself on one side.

2. What do A and b represent in this equation?

A and b are constants in this equation, which means that they have fixed values. A represents the coefficient of x, while b represents a constant term. These values will be used in the process of solving for x.

3. Can this equation be solved algebraically?

Yes, this equation can be solved algebraically. As mentioned before, the first step is to simplify the left side and expand the right side. Then, you can use algebraic properties to rearrange the equation and isolate x. Remember to perform the same operations on both sides of the equation to maintain its balance.

4. Are there any other methods for solving this equation?

Yes, there are other methods for solving this equation. One method is to graph both sides of the equation and find the point(s) of intersection. Another method is to use a calculator or computer program to solve the equation numerically.

5. How do I know if my solution for x is correct?

To check if your solution for x is correct, you can substitute it back into the original equation and see if both sides are equal. If they are not equal, then your solution is incorrect. You can also use a graphing calculator or computer program to graph the equation and see if your solution falls on the point of intersection.

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