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Basic difference between summation and integration?

  1. Jun 19, 2012 #1
    can you plz tell me what is the basic difference between summation and integration..? i was going through the Poisson distribution function and in one case it was discrete and we had to make summation to get the result and other cases for continuous function we integrated it...now what is meant by discrete and continuous in probability and why do we integrate or sum up in different cases?
     
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  3. Jun 19, 2012 #2

    chiro

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    Hey cooper607 and welcome to the forums.

    Integration is a special kind of summation. For the Riemann integrals (the one you study in first year calculus and usually problems all throughout undergraduate years), the Riemann is a sum where the length of the boxes get smaller and approach zero but never get there. In other words, think of it as an infinite sum of an infinite number of boxes in n-dimensions (for area, we have a 2D box, for volume 3D and so on).

    Now the area of the boxes both in discrete and continuous both correspond to probabilities.

    The difference with continuous is that the probability at 1 point for a continuous distribution is zero. In order to get a non-zero probability, you have to have a region of integration that has a non-zero integral (i.e. from a to b where b > a not b = a).

    But in terms of what is going on, the area under the curve for both discrete and continuous distributions is the probability related to the events that are being summed (discrete) or integrated (continuous).
     
  4. Jun 19, 2012 #3
    thanks a lot chiro for this basic explanation, but can you go a little deep to help me get hold of the concept with real life circumstances? like i am asking if we can practically impose these summation or integration to differentiate them?

    regards
     
  5. Jun 19, 2012 #4

    micromass

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    I struggled with this question a long time. The formula's looked the same in both cases, except that in one case we sum and in the other case we integrate. There didn't seem a particular reason for this.

    My questions were finally answered when I learned about measure-theoretic probability. This unites the two cases very nicely and you can easily see there that both integration and summation are two special instances of something more general.

    Sadly enough, this is not easy theory, so it's not something I can explain here. I think for now it is best to accept that it works this way. Later you can always find out why.
     
  6. Jun 19, 2012 #5

    chiro

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    The easiest physical example I can think of is finding total displacement from a definition of velocity over some region or total velocity given acceleration.

    Remember that when you add up all the velocities over time, you get a total displacement. If we assume velocity doesn't change, then it means if we start at A and head in a straight line with the velocity towards some point B, then the displacement is given by dv/dt x 1 assuming t is in seconds and we want to find the displacement from point A after 1 second.

    In general as long the requirements are met for the integral to be solved using these techniques, the velocity can be instantaneously changing all the time from start to finish so instead of dv/dt = c we can have dv/dt = f(t) or even dv/dt = f(v,t) which is common in physics.
     
  7. Jun 20, 2012 #6
    umm...struggling with it still :(.......better not to try to get the hang until i am done with the undergraduate maths....

    anyway let me ask something more basic...why do we have a integrational constant in calculus? if we sum up the thing do we need any constant like this??
     
  8. Jun 20, 2012 #7

    chiro

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    The basic reason has to do with an initial value.

    Let's say you want to find y(x) and you are given dy/dx = f(x). Then you get y(x) = F(x) + C where F(x) is your anti-derivative.

    Now let's say you have been given an initial condition y(0) = 2 and we know F(0) = 0. Then we know y(0) = F(0) + C = 2 which implies C = 2 and we have solved our equation y(x).

    The reason why you don't need this in definite integrals is because the C's cancel out. Lets say we integrate f(x) with respect to x from a to b b > a. Then we end up getting the result [F(b) - C] - [F(a) - C] = F(b) - F(a) which means the C's always cancel out for an integral that is definite (as opposed to indefinite).

    Visually the C value is the starting value given some initial condition and you can think of it in the graph as some point.
     
  9. Jun 20, 2012 #8
    hmm got the c part...thanks a lot dear
     
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