Can Lower Bound of Summation Be Any Real Number?

  • Context: Undergrad 
  • Thread starter Thread starter smslca
  • Start date Start date
  • Tags Tags
    Bounds Summation
Click For Summary
SUMMARY

The discussion clarifies that in standard mathematical practice, the lower bound of a summation (Σ) must be an integer. While variations exist, such as the Mobius function, which sums over divisors, the conventional rule dictates that bounds are typically integers. Exceptions arise in specific cases, such as summing the reciprocals of prime integers or using non-integer increments, but these are not the norm. For instance, a summation starting at π with increments of √2 can be expressed as Σ_{j=1}^n (π + (j-1)√2).

PREREQUISITES
  • Understanding of summation notation (Σ)
  • Familiarity with integer and non-integer sequences
  • Knowledge of mathematical functions like the Mobius function
  • Basic concepts of prime numbers and their properties
NEXT STEPS
  • Explore the properties of the Mobius function in number theory
  • Learn about summation techniques involving non-integer increments
  • Investigate the implications of summing over prime integers
  • Study advanced summation notation and its applications in mathematics
USEFUL FOR

Mathematicians, educators, and students interested in advanced summation techniques and the properties of integer sequences will benefit from this discussion.

smslca
Messages
63
Reaction score
0
can the lower bound of a summation(sigma) be any real number ?
i.e ex: sigma(LB:sqrt(2) or (9/2) etc )
Even a lower bound be a real number is possible or not can upper bound be any real number or is it a strict rule that '1' should be added to lower bound to get the consecutive number.?
i.e. ex: LB + sqrt(2) or (9/2) etc.
 
Physics news on Phys.org
You mean something like

\Sigma_{i=\pi}^{\pi^2}i

?

If so, no. In standard usage, the bounds are always integers.

Of course, you could always choose a different definition. Many variations on Σ exist, like the Mobius function, which sums over the divisors of an integer.
 
"Is it a strict rule that '1' should be added to lower bound to get the consecutive number.?"

There are some exceptions: for instance, if we wanted to write an expression for the sum of the reciprocals of the prime integers (note the primes do not change by differences of 1) we write

<br /> \sum_{j \text{ prime}} {\frac 1 j }<br />

(or something similar). If you wanted the "jump" between successive terms in a sum to be \sqrt 2, and start at \pi, you might do something like this:

<br /> \sum_{j = 1}^n {(\pi + (j-1)\sqrt{2})}<br />
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 24 ·
Replies
24
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 85 ·
3
Replies
85
Views
9K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K