Extending Sigma Sums to Real Numbers

[TylerH]

I mean an extension analogous to the extension of factorial to gamma(ignoring the shift).

I want to extend the iterative number, usually written below the sigma, to be a real number, rather than an exclusively natural number.

The easiest way I can think of is to convert the sum into a definite integral, since the bounds of an integral can be any real number(or infinity). Is there any known way to convert sigma sums into definite integrals? It seems as if it should be possible, seeing as the integral is just an infinite sum of infinitesimals.

 

[dimitri151]

Intgerals are usually defined as limits, or infimums or supremums of "sigma-type" expressions. You can use expressions for the sum of squares, cubes of integers to derive some integrals, but you have to take the limit of those expressions.

 

[TylerH]

I'm having trouble grasping what you mean. Could you show me an example, please?

 

[Stephen Tashi]

You have to define what you mean by "extending". When we "extend" something we usually want it to be the way it was on where it already existed. The only place we want it to do something different is in the new territory where we extend it.

The relation between $$\sum_{i=1}^n f(i)$$ and $$\int_{x=1}^{x=n} f(x) dx$$ is that you can visualize the sum as a very coarse Riemann sum with bases of the rectangles having length = 1 (pretty big for a $$\triangle x$$) that approximates the integral. I've seen this idea used in deriving Sterlings approximation to the n!. Look at log n! = log 1 + log 2 + ... + log n and compare it to the integral of log n.

To me it is more interesting to look at the analogy between anti-differentiation to solve definite integrals and anti-differencing to solve summations. To get a closed for formula for $$\sum_{i=1}^n f(i)$$ all you need to do is to discover a function $$F(i)$$ such that $$F(i+1) - F(i) = f(i)$$ because:

$$\sum_{i=1}^n f(i) = \sum_{i=1}^n (F(i+1) - F(i))$$

$$= (F(2)-F(1)) + (F(3) - F(2)) + (F(4) - F(3) ) + ... (F(n+1) - F(n))$$

$$= F(n+1) - F(1)$$ due to the "telescoping" sum.

That field of study is called The Calculus Of Finite Differences. Learning to find "anti-differences" is like a weird version of learning to find anti-derivatives.

 

[dimitri151]

I don't know if you're thinking is going to work. You're trying to iterate a summation over some interval or subset of the real numbers? I think the index of summation plays a different role than the boundaries of integration.

I don't know if this will help. Take y=x² on the interval [0,1]. Divide that interval into n equal parts. At the end point of each interval (on the x-axis) erect the perpendicular to y=x² . Connect the points of intersection of the parabola and the perpendiculars. You'll have a bunch of trapezoids. Form a sigma expression for the area of the sums of those trapezoids. This will be some expression in n. If you take the limit of that expression as n approaches infinity it should equal what you know of as the integral of x² from [0,1]. The way the sigma notation is converted to an integral is by taking the limit (in this case).

 

[TylerH]

You have to define what you mean by "extending". When we "extend" something we usually want it to be the way it was on where it already existed. The only place we want it to do something different is in the new territory where we extend it.

By "extending" I mean extending the set the index belongs to from the integers to the reals.

 

[Stephen Tashi]

By "extending" I mean extending the set the index belongs to from the integers to the reals.

.

You could regard $$\sum_{i=0}^n f(i)$$ as $$f(0) 1 + f(0+ 1))1 + f(0 +1+1)1 +$$ etc. and generalize this to
$$\sum_{x=a, h}^n f(x) = f(a) h + f(a+h) h + f(a+h+ h) h +$$ etc., which is just a Riemann sum -- not very exciting.

 

[TylerH]

not very exciting.

Yes... :zzz: Oh well...

 

[Hurkyl]

I want to extend the iterative number, usually written below the sigma, to be a real number, rather than an exclusively natural number.

What are you really trying to do, that you think doing something like this will help?

 

[TylerH]

What are you really trying to do, that you think doing something like this will help?

I really didn't have a purpose in this. There are a few places I think it would be interesting. Like the sum: $$\sum^b_{a=c} a$$ for a real a,b,c. For example, I wondered what this would be for b=2.5 and c=0. Since it should be additive, you could break it apart into $$\sum^2_{a=0} a + \sum^{2.5}_{a=2}a$$, but wtf is $$\sum^{2.5}_{a=2}a$$?

 

[TylerH]

The solution to $$\sum_{a=0}^x a$$ is $$\frac{x^2}{2}+\frac{x}{2}$$. How would I get the latter from the former? I found it using Eureqa(fitness of R^2 = 1), but I would prefer a mathematic way.

 

[Stephen Tashi]

TylerH,

I have no idea what you're saying in that last post. I'll put the thought of your original post this way:

Given a function $$f(i)$$ defined on the non-negative integers, find an integrable function $$g(x)$$ on the real numbers such that for all (or as many as possible!) non-negative integers n

$$\sum_{i=0}^n f(x) = \int_0^n g(x) dx$$

And you may be wanting $$g(x)$$ to be a smooth function instead of a collection of interpolations between the partial sums of the series.

 

[TylerH]

TylerH,

I have no idea what you're saying in that last post.

Sorry, I didn't really explain what I meant. I meant $$\sum_{a=0}^x a=\frac{x^2}{2}+\frac{x}{2}$$. How would I find this, for a general $$\sum_{a=c}^x f(a)$$, without brute forcing(which is how I found it, with Eureqa)?

I'll put the thought of your original post this way:

Given a function $$f(i)$$ defined on the non-negative integers, find an integrable function $$g(x)$$ on the real numbers such that for all (or as many as possible!) non-negative integers n

$$\sum_{i=0}^n f(x) = \int_0^n g(x) dx$$

And you may be wanting $$g(x)$$ to be a smooth function instead of a collection of interpolations between the partial sums of the series.

I think this is getting too focused on integration. It doesn't have to be done through integration. The goal, to clear any question, is to find a continuous(and hopefully differentiable) extension to sigma.

 

[Stephen Tashi]

Sorry, I didn't really explain what I meant. I meant $$\sum_{a=0}^x a=\frac{x^2}{2}+\frac{x}{2}$$. How would I find this, for a general $$\sum_{a=c}^x f(a)$$, without brute forcing(which is how I found it, with Eureqa)?

The way I look at it, you simply wrote the summation formula for n consecutive integers using an $$x$$ instead of an $$n$$

$$\sum_{i=0}^n i = \frac {(n)(n+1)} {2} = \frac {n^2}{2} + \frac {n}{2}$$

 

[dimitri151]

How about introducing a parameter into the sigma notation like the "step" specification in some programing languages. You have for some sigma expression k=1 to 10. There's an implied step of one, so you evaluate at k=1,2,...,10. Then you specify a step of 1/2, you evaluate at k=1,3/2,2,5/2,...,10. I think you'll need to multiply the whole expression by your step though otherwise it's going to go to infinity (typically).

 

[Stephen Tashi]

How about introducing a parameter into the sigma notation like the "step" specification in some programing languages. You have for some sigma expression k=1 to 10. There's an implied step of one, so you evaluate at k=1,2,...,10. Then you specify a step of 1/2, you evaluate at k=1,3/2,2,5/2,...,10. I think you'll need to multiply the whole expression by your step though otherwise it's going to go to infinity (typically).

That would be the suggestion of post #7. It gives Riemann sums.

Tyler_H's example is to take the closed form formula for a sum and changes the variable in the formula to a real number.
$$\sum_{i=0}^n i = \frac{ i (i+1) }{2}$$ generalizes to
$$\sum_{i=0}^{i=x} i = \frac{ x(x+1)}{2}$$

Suppose we have a series $$\sum_{i=0}^n f(i)$$ and we don't know a closed form formula for it. How would we define the extension? Do the examples of extending closed form formulas suggest any properties that the extension should have besides producing the right answer at integer values?

 

[TylerH]

I can't think of any properties, other than being equal at integer values, worth defining. The goal is to have a continuous, and hopefully differentiable, sigma.

On second thought, to prevent arbitrary max/mins in the extension: For an extension, g(x): If, for any a < b, $$\sum_{i=0}^a f(i) < \sum_{i=0}^b f(i)$$ then the g'(x) > 0 for all x in (a,b). And, vice versa.

 

[Stephen Tashi]

I suppose we are assuming $$f()$$ is a function defined on the reals and
$$g(n) = \sum_{i=0}^n f(i)$$.
I don't think we should insist $$g'(x) > 0$$. We don't want a definition that only applies for f(i) that are non-negative. For example, it would nice if the definition worked for $$f(i) = (-1)^i$$

Does a shift operation suggest anything?

$$\sum_{i=0}^n f(i + 1) = \sum_{i=1}^{n+1} f(i) = g(n+1) - g(0)$$

We can look at smaller shifts.

For $$0 < h < 1$$ the expression $$\sum_{i=0}^n f(i+h)$$ makes sense.