Series integral (1 Viewer)

[henry08]

This is not part of the HSC course I just wondered how it is done.

How do you represent the series 1 + 2 + 3 + 4 + ... + n counting up to 1,000,000 where 2 = 1,000,000 as an integral? i know how to d it using Sigma notation, but how as an integral?

 

[3unitz]

1 + 2 + 3 + ... + (n-2) + (n-1) + n = n(n+1)/2

= (n^2/2) + (n/2)

take the derivative:

d/dn [(n^2/2) + (n/2)] = n + (1/2)

thus 1 + 2 + 3 + ... + (n-2) + (n-1) + n is equal to:

S{0->n} x + (1/2) dx

eg. 1 + 2 + 3 + ... + 1 000 000 = S{0-> 1 000 000} x + (1/2) dx

 

[Iruka]

This is not part of the HSC course I just wondered how it is done.

How do you represent the series 1 + 2 + 3 + 4 + ... + n counting up to 1,000,000 where 2 = 1,000,000 as an integral? i know how to d it using Sigma notation, but how as an integral?

That function is only defined over a set of discrete values (i.e., the natural numbers), so if you represent it as an integral, where you are assuming that the function is defined over a set of continuous numbers, you will only get an approximation.

I mean, you can do what 3unitz just did, but since you need to know how to calculate the discrete sum to do that anyway, I don't quite know what the point of doing that would be. (Note that he got that interesting little correction term that you would lose if you just treated it as a function over x rather than n.)

There is a branch of microeconomics called marginal analysis where they substitute continuous valued functions for discrete valued ones so that they can use basic calculus. Of course, being economists, they have to give everything weird names. (For example, "marginal blah-blah-blah function" just means the derivative of the function blah-blah-blah.)