4

I read this related thread but it doesn't give me a satisfactory answer to the following question:

Must it be true that the order of indices in a sum is relevant? A finite sum is essentially adding up all elements in a set. Who cares it that set is ordered forwards or backwards? Consider the following examples.

Common convention tells us that

$$\sum_{n=5}^0n=0$$

even though we are essentially trying to add the elements of {$n\in[0,5]$} backwards. The validity of this becomes paramount in the following case,

$$\sum_{m=0}^n f(n-m)\ne0$$

when we make a simplifying change of variable $m=n-i$. Then, the sum becomes

$$\sum_{i=n}^0 f(i)=0, \text{by convention.}$$

Obviously nothing significant has changed with the sum when we just change a variable definition, Why, then, does the convention dictate that the value of the sum must change? It seems this particular sum convention is nonsense and should be abandoned.

avikarto
  • 270
  • 1
  • 7

1 Answers1

3

If you write the summation as $$ \sum_{0\le m\le n}f(n-m) $$ and do the substitution $m=n-i$, then the condition in the sum becomes $$ 0\le n-i\le n $$ which is equivalent to $$ n\ge i\ge 0 $$ which in turn can be written as $0\le i\le n$.

Therefore $$ \sum_{0\le m\le n}f(n-m)=\sum_{0\le i\le n}f(i) $$


A summation such as $$ \sum_{n=5}^0 f(n) $$ can (should?) be written as $$ \sum_{5\le n\le0}f(n) $$ and no index $n$ satisfies the condition, so the summation is by definition $0$.

egreg
  • 238,574