Integrals: Unveiling the Logic Behind "Norm of the Partition

[gikiian]

And why are the partitions not equal to one value? Why x1, x2, ... , xk, ... , xn-1, xn ?
And why |the norm| -> 0 ?

I was just curious if there is some specific logic behind it or if it is just there to discuss things in general.

Thanks a lot.


P.S.: Norm is the partition having the greatest magnitude of all.

 

[Mark44]

And why are the partitions not equal to one value? Why x1, x2, ... , xk, ... , xn-1, xn ?
And why |the norm| -> 0 ?

You're confused on some of the terms. A partition isn't a number, so it makes no sense to say that it is equal to one value. A partition is a collection of points in the interval of concern, [a, b]. You can divide the interval into subintervals of equal length, or you can divide it into subintervals that are different in size. If you have a function that is relatively constant or linear in one part of the interval, but increases or decreases a significant amount in another part of the interval, an approximation will be better if the subintervals are shorter where the function's slope is steepest.

So the answer to your question is that for the sake of generality, a partition is not assumed to define subintervals of equal lengths.

I was just curious if there is some specific logic behind it or if it is just there to discuss things in general.

Thanks a lot.


P.S.: Norm is the partition having the greatest magnitude of all.

No, the norm of a partition is the length of the longest subinterval. For example, let's take the interval to be [0, 2]. Here is one partition: P = {0, .5, 1.0, 1.25, 1.5, 1.6, 1.7, 1.8, 2.0}. The norm of this partition, norm(P), is .5.

 

[lurflurf]

And why are the partitions not equal to one value?
The partition does not matter so we can chose any partition we like.
Often certain partitions have certain advantages.

Why x1, x2, ... , xk, ... , xn-1, xn ?
It is a dummy variable take
y1, y2, ... , yk, ... , yn-1, yn
if you like it does not matter.
And why |the norm| -> 0 ?
An integral is the limit of a sum
The sum depends upon the partition
In order for the integral exist the function must be nice in the opinion of the integral.
When the function is nice the sum depends less upon the partition as the norm becomes smaller, as the norm becomes small the partition does not matter.

For example we might have
|sum-integral|<C*norm(Partition)
for some C>0
So would we know if the norm is small the integral and the sum are close together.

 

[gikiian]

Thanks :)

 

[Stephen Tashi]

I suspect we have not given a complete answer to this question. There are probably mathematical technicalities that would cause defining integration over partitions having equal length subintervals to produce a definition of integral that is different than the standard one. (By "different", I mean that functions that are integrable by the standard definition would turn out not to be integrable by the alternate definition - or vice versa.) It's more than a question of being able to do convenient numerical approximations. For example, if you define integration of f(x) on [0,1] by taking subintervals of equal length, the length of the interval is always a rational number. If your definition evaluates the function f(x) at the endpoints of the intervals, you leave yourself open to someone who wants to make up a tricky example of a function that obeys one formula on the rational numbers and another forumula on the irrational nunbers.