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2KiB, 456x147, z.png

converting sigma notation to integral notation

Anonymous Tue 27 Oct 2015 17:58:55 No.7619816 View Reply Report Delete

Quoted By: >>7619845 >>7619861 >>7619912

Could anyone please explain why the first equation can simply be integrated to find the answer, but the second one can not? what is it about sigma notation that differentiates it from integration to disallow this sort of conversion?

apologies in advance if my notation is wrong (i suspect that 1 should be 0), i'm absolute garbage at maths.

mtp !EKFQOBEsKo

mtp !EKFQOBEsKo Tue 27 Oct 2015 18:03:38 No.7619834 Report Delete

Quoted By: >>7619860

but theyre not equal. LHS=30n, RHS=30(n-1)

Anonymous

Anonymous Tue 27 Oct 2015 18:04:01 No.7619835 Report Delete

Quoted By: >>7619852

The first one works just by fluke.

When you turn a sigma thing into an integral, you need a differential element. A delta x. You need to be multiplying by something that you can make arbitrarily small.

sum (5x delta x) = integral (5x dx), as the limit is taken for delta x -> 0.

Anonymous

Anonymous Tue 27 Oct 2015 18:07:42 No.7619845 Report Delete

Quoted By: >>7619854 >>7619860

>>7619816
The first equation is wrong, the LHS is 30n and the RHS is 29n.
Why are you integrating x^2/2 and not just x?

Anonymous

Anonymous Tue 27 Oct 2015 18:10:31 No.7619852 Report Delete

Quoted By:

>>7619835
so is it even possible to convert between the two or am i better off memorizing sigma shortcuts without understanding them?

Anonymous

Anonymous Tue 27 Oct 2015 18:12:15 No.7619854 Report Delete

Quoted By:

>>7619845
oof didn't even see i did that shit. you're right it should be x, i got ahead of myself.

Anonymous

Anonymous Tue 27 Oct 2015 18:14:53 No.7619860 Report Delete

Quoted By:

>>7619834
>>7619845
that's the mistake i suspected in the original post, should be 0 to n, not 1 to n

Anonymous

Anonymous Tue 27 Oct 2015 18:15:10 No.7619861 Report Delete

Quoted By:

>>7619816
Simga is just notation for an integral over \N wrt. the counting measure. Naturally, such an integral does, in general, not agree with an integral over \R wrt. the Lebesgue measure.

Anonymous

View Same Google ImgOps iqdb SauceNAO z.png, 2KiB, 456x147

Anonymous Tue 27 Oct 2015 18:16:18 No.7619866 Report Delete

Quoted By:

here's what (i think) i meant to say

Anonymous

Anonymous Tue 27 Oct 2015 18:29:53 No.7619890 Report Delete

Quoted By:

I think i see what Anon meant
Sigma seems to be the RRAM delta x = 1 of an f(x), while the integral has one of them infinitely small deltas.

Anonymous

Anonymous Tue 27 Oct 2015 18:43:04 No.7619912 Report Delete

Quoted By:

>>7619816
The two are not comparable in general.
What you can say in general, however, is that $\int_0^n f(x)dx = \sum_{k=1}^n \int_{k-1}^k f(x)dx$

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