I thought if the lower and upper bound of a summation are equal, that from that would follow that th

dokezwa17

dokezwa17

Answered question

2022-05-23

I thought if the lower and upper bound of a summation are equal, that from that would follow that the sum is always equl to 0. But when trying this in Wolfram Alpha the result is just one iteration of the summation term. Can someone please explain if my hypothesis therefore is wrong?
<math xmlns="http://www.w3.org/1998/Math/MathML" "> j = 0 0 1 = 1

Answer & Explanation

soymmernenx

soymmernenx

Beginner2022-05-24Added 10 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Intuitively:

The sum
i = 0 n a i
denotes the sum of all a i where i goes from 0 (inclusive) to n (inclusive). In other words,
i = 0 n a i = a 0 + a 1 + + a n
so, if n=0, you would have the sum of all a i for i from 0 (inclusive) to 0 (inclusive). The index i=0 is the only index that satisfies this condition, so the sum would be a 0 .

From definitions: by most definitions,
i = 0 o a i
is equal to a 0 .
In fact, the standard way of defining the sum in general would be a recursive definition, i.e.
i = 0 0 a i = a 0 i = 0 n a i = a n + i = 0 n 1 a i
This definition, for example, says that
i = 0 1 a i = a 1 + i = 0 0 a i .
dokezwa17

dokezwa17

Beginner2022-05-25Added 1 answers

Alternatively, at that point, you could also define
<math xmlns="http://www.w3.org/1998/Math/MathML" "> i = m n a i = i = 0 n a i i = 0 m 1 a i

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