1 - When you are using the symbol of the sum with the lower and upper index defined [;\sum_{i=1}^{n}a_i;], It is implicit that you are computing the summation of the terms of a sequence a_1, a_2, .... which has a natural order.

Generally you don't need specify these limits. You can compute the sum under a set (not a n-upla) of index "I" finite or not [;\sum_{I}a_i;]. When it is countable, It can be seen as an integral, just using the counting measure.

2 - Supose that we have defined [;0^0=a\neq 0;]. So, [;a=0^0=0^{1-1}=(0^1)*(0^{-1})=0/0;]. Thus we also have defined [;a=0/0;], which is know undefined. But why?

Consider the question: What is the solution for the equation [;0*a=0;]?? Of couse [;a=1;] is a solution, but also [;a\in\R;] is solution for this. So it is undefined, like [;0^0;].

The answer to your first question is zero. There are no summands. It's as simple as that.

Douglas Duarte Novaes has explained you both answers very well. In addition, if you want a simpler way to look at it here it goes:

1) Sums which are defined the way you state, assume a consecutive, natural, order, hence, quite often the index of the sum belongs to the natural numbers. The result of that operation is the addition of all the terms defined with a *generic* term, which is after the Sum symbol, and indexed from the lower to the upper index. For example, Sum (lower limit, index "i" = 1) to (upper limit, index, say, "N =3") of "i" (itself) is 1+2+3. If it is a more complicated function, then one usually knows what it is. Say (the same sum) but now of the function "x (sub-index "i"). Here, one is just stating a formal addition, but to actually perform it, one must know what the values of "x_1", "x_2", and "x_3" are; so assume x_1 = 5, x_2 = 8, and x_3 = 11, their sum will be 24.

2) Most operations on zero are either "zero" or "undetermined" (in computers NAN = Not A Number) by definition. This is because is zero acts upon arithmetical operations mostly like either a "place-holder" or a "conservation balance" OPERATOR (e.g. "ten" has 10 by the way we have defined decimal notation, meaning that are "zero" units, and "one" decade, but that's it as concerns of zero; also 1-1 = 0, but here is "zero" balances out a conservation principle.) Whenever one wants to use some definitions with itself both as OPERATOR and OPERAND, indeterminations usually appear. Philosophically speaking, it falls in the realm of statements like "What happens when an unstoppable force meets an unmovable object?"

It is called as empty sum..so the result is zero.

The category of sets has a terminal object which consists of one element.

It follows that if we define a sum of nonegative integers $\sum\limits_{i\in I}n_i$ as the cardinality of the coproduct $\bigsqcup\limits_{i\in I}X_i$ where $X_i$ are finite sets with $|X_i|=n_i$, then we obtain $\sum\limits_{i\in\emptyset}n_i= 1$.

$lim_{x \to 0} (x/x) = (d(x)/dx)/(d(x)/dx) = 1/1= 1$,

$0/0$ is not necessarily undefined it depends on the context.

This fact coupled with the breakdown given above by Douglas Duarte Novaes:

$(x^0 = x^1 \times x^{-1} = x/x)$ therefore implies that $0^0 = 1$.

Also consider a polynomial $y = x^2+x^1+x^0$ this is not suddenly undefined or $0$ at $x=0$. $y(0)=1$

Stuart,

[;lim_{x \to 0} (x/x)=1;] does not imply that [;0/0=1;], neither in this context. It is a limit process that studies the behavior of the function near x=0 but never to x=0.

If you consider the function [;f(x)=x/x;], it is indeed undefined for [;x=0;]. However it can be continuously extended for x=0 as

[;f(x)=x/x;] for [;x\neq0;] and [;f(x)=1;] for $[;x=0;]. But, note that we do not evaluate 0/0.

The polinomial function of degree n is a identification with the polinomial ring P^n[x] which is given by the n-uplas (a0,a1,...,an), we do not need to evaluate x^0. Thus even in this case, 0^0 undefined.

Sometimes it's appropiate to define o^0=1, when you study power series, but it's only for certain situations

Q1: I sort of agree with Ridouan Rani, namely summations are similar to integarls, but discrete. Namely, you want [sum: x_i (i=a to b)] + [sum: x_i (i=b+1 to c)] to equal sum:x_i(i=a to c). Replacing b+1 by b, solving for sum:x_i(i=b to i=c) gives the answer of: sum:-x_i(i= c to b+1). Namely, sum:x_i(i=7 to 4) =-( x_7 + x_6 + x_5).

Dear ASRAT

Some things are defined and others are considered by convention

if you need to start the summationin on a way that the lower bound is greater than the upper one,,go a head but you need to state and define this new form..

Also concerning 0^0 this is equivalent to 0^(n-n) =0^n/0^n=0/0 for any n not 0 and this is equivalent 1/0.0 equivalent to infinity.0 and so on ... all 7 indeterminate forms are

0/0 , inf./inf, 0.inf , inf - inf , 0^0 , inf^0 and 1^inf and all of them are equivalent

to test the value of convergence just use the L'hopital rule as in the seondary school calculus.

I thank you all for your brief explanations to my questions!!!

For 1. context should be provided.

For 2. also, but at at first glance the answer is "undefined".

1- If we define a sum of nonegative integers $\sum\limits_{i\in I}n_i$ as the cardinality of the coproduct $\bigsqcup\limits_{i\in I}X_i$ where $X_i$ are finite sets with $|X_i|=n_i$, then for empty $I$, the coproduct is initial object and we obtain $\sum\limits_{i\in\emptyset}n_i= 0$. But the product of empty set of numbers equals 1.

The order of summation is irrelevent therefore it does not matter in what order the indices of the sum apear or in what possibly completely random order the elements of the set of indices over which the sum is computed are arranged, the resulting sum will have the same value (unless of course we are talking about quaternions or something other algebra where the addition operation is not both comutative and associative).

This is different from the Integral which is the limit of a sum of products of $$f(x)$$ with $$\delta x$$ as $$\delta x \to 0$$. Since the order of the indices in the integral implies a direction/sign (+/-) for $$\delta x$$ reversing the indices reverses the direction/sign of $$\delta x$$ and hence the sign of it's product with $$f(x)$$ and therefore of the resulting integral (ie. $$\int_a^b f(x)dx = - \int_b^a f(x)dx $$).

Q1 I think how the limits of summation are treated is a convention e.g. the limits are integers and the lower index is smaller than the lower.

Q2: It is good to use logs for such limits

ln 0^0=ln(0) *0 = inf*0 = undefined

Asrat, could you please tell us how did you arrive at such an unconventional summation?

Essentially I agree with Stuart's thinking, i.e. the order of summation doesn't matter (at least for the number of summands being finite), but I can think of still another interpretation: the sum runs over all integers (i.e. from -\infty to +\infty) EXCEPT the indexes explicitly shown (and all those between them, of course).

So, is your question motivated by pure curiosity, or is it an unexpected result of some previous calculations?

!0^5 means multiplication of 10 by itself five times. So 0^0 means multiplication of 0 by itself 0 times and any value multiplied by zero becomes zero. Therefore, claim that not defined is not valid from my primary understanding of mathematics, I cannot say what advance mathematics makes of it.

The expression 0/0 is indeterminate, i.e. does not have a specific value. The reason is very easy to explain: any number multiplied by 0 gives 0 as result. For this reason you have an indefined number of solutions.

1)It is a matter of sorting and you can afford negative step, all software languages have this ability, don't worry about formal definitions of summation etc.

2)it depends on the situation: limit(sin(x)/x,x-->0)=1, limit(x/x^3,x-->0)= +infinity, other limits are undefined etc. Where did you find your 0/0 case?

Both are entirely context-dependent, and any author should point out the convention that they intend to use. I find this situation to be quite unsatisfactory, but I learned long ago that I'm stuck with that.

@Mohammad Firoz Khan:

Following your line of reasoning 5^0=5. So, where's the catch? Simple: multiplying 5 by itself zero times means NO MULTIPLICATION takes place. If there is no multiplication then how do you know its result?? More precisely: your reasoning is only good for powers which are natural numbers. Unfortunately, zero is not the natural number. Besides, how would you evaluate 5^(1/2)?

The latter means that anything raised to a power is multiplied by itself that many TIMES, or instances; 0 multiplied by itself zero times, still leaves that ONE instance of 0, or 1.  In the same vein, if you were to multiply yourself zero times, you still would be left with yourself or 1.