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camjohn

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- Thread starter camjohn
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camjohn

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Vorde

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The Integral Sign is a very special case of the sigma summation, an Infinite Riemann Sum to be specific. You can have all sorts of sums without the 'f(x) d(x)' part which is required of Integrals. Summations only mean you are taking the total of a bunch of different things, it has nothing to do with Integrals if you don't want it to.

For example: It wouldn't make much sense to take the integral of your balloon and my balloon to find out how many balloons we have, but it would make sense to take the sum.

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Bacle2

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If I understood you well, if your function was locally-constant, e.g., a collection of

rectangles put together, as a step function , then a sum would

make more sense. Or if your function was well-approximated by rectangles.

Think, too, on finding the area of an equilateral triangle. You can break down

this triangle into two triangles by bisecting using the height. Then you have two

triangles , whose areas are simple as 1/2bh . Using the equation of two lines

to desscribe the triangle and then using a Riemann sum would seem overkill.

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Vorde

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Bacle2

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help clarify? Maybe by sigma summation s/he is referring to a dimensionless sum,

or,at most a 1-dim sum ( making me hungry for Chinese!) where it would make

sense to add numbers?

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camjohn

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- #7

Vorde

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The only problem with your logic is that I think you are under the impression that sums are just the discrete version of the integral.

When trying to figure out area under a graph (or any other interpretation of the same value), integrals of one form or another will

However, as I tried to explain before, sums are a much broader topic that just riemann sums. It might be worthwhile to google both 'summation' and 'riemann sums' to understand the difference.

I love analogies, so I'll try another one:

We all have been taught in grade school that multiplication is just repeated addition, and after one solves 5 x 4 by adding 5 four times, and solves 16 x 2 by adding sixteen twice, he might be tempted to ask why we bother defining a multiplication operator at all when in fact all we ever do is repeated addition.

The reason he asks this is because he has never been exposed to a problem like 16 x 1.56, where it is obvious that the unique methods of multiplication are needed rather than simple repeated addition. Likewise, I think your question stems from the fact that you have only been exposed to summation in the form of riemann sums leading up to integrals and therefore have no exposure to the broader use of sums.

It is a shame that the more rigorous foundation of riemann sums is not taught when developing basic calculus, but I totally understand why. I was lucky enough to learn calculus out of a very old textbook which looked at integration with just enough mathematics to make me appreciate Riemann Sums and not enough to dissuade me.

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camjohn

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Vorde

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