Studying Mathematics Thread

[GrandizerII]

Ok, now I'm confused about the following:

The improper integral of 1/x from -e to e is said to diverge yet can be assigned a CPV value of 0.

Intuitively, the 0 answer makes sense (since symmetrically opposite areas, even infinite, should completely cancel each other out), but it's considered to be a problematic answer. Can anyone ELI5 why 0 is not always the right answer here? What does definite integral really mean then?

[GrandizerII]

Ok, now I'm confused about the following:

The improper integral of 1/x from -e to e is said to diverge yet can be assigned a CPV value of 0.

Intuitively, the 0 answer makes sense (since symmetrically opposite areas, even infinite, should completely cancel each other out), but it's considered to be a problematic answer. Can anyone ELI5 why 0 is not always the right answer here? What does definite integral really mean then?

After some reading, I think I have my answer:

Intuitively speaking, there is nothing wrong with the thinking that, even if infinite, symmetrical and opposite arithmetic areas cancel each other out. And this accords well with the CPV. However, mathematically speaking, the way definite integrals are defined, symmetry isn't assumed by the definition. So when dealing with improper integrals (like the one above), one can arbitrarily choose whatever values they want close to 0 (one to the left of 0 and the other to its right), and so the values need not be opposites of each other. Since that's the case, when calculating the limit, it could be anything depending on the arbitrary values chosen as the bounds closest to 0 (e.g., 0 in the case of a and -a, ln(2) in the case of a and -2a, ln(3) in the case of a and -3a).

[robvalue]

Right, it depends on how you define the replacement integrals. Since you're originally trying to integrate across a singularity, the integral is undefined. You're splitting it into two integrals, and it depends how fast you have them approach the singularity relative to each other. The CPV assumes they approach at the same speed.

[polymath257]

Ok, now I'm confused about the following:

The improper integral of 1/x from -e to e is said to diverge yet can be assigned a CPV value of 0.

Intuitively, the 0 answer makes sense (since symmetrically opposite areas, even infinite, should completely cancel each other out), but it's considered to be a problematic answer. Can anyone ELI5 why 0 is not always the right answer here? What does definite integral really mean then?

After some reading, I think I have my answer:

Intuitively speaking, there is nothing wrong with the thinking that, even if infinite, symmetrical and opposite arithmetic areas cancel each other out. And this accords well with the CPV. However, mathematically speaking, the way definite integrals are defined, symmetry isn't assumed by the definition. So when dealing with improper integrals (like the one above), one can arbitrarily choose whatever values they want close to 0 (one to the left of 0 and the other to its right), and so the values need not be opposites of each other. Since that's the case, when calculating the limit, it could be anything depending on the arbitrary values chosen as the bounds closest to 0 (e.g., 0 in the case of a and -a, ln(2) in the case of a and -2a, ln(3) in the case of a and -3a).

A roughly similar situation happens with integrals from -infinity to infinity.

So, what is the integral from -infinity to infinity of x dx?

If you do it symmetrically, from -b to b, and take the limit, the answer is 0.

If you do it from -b to b+1/b, and take the limit as b goes to infinity, you will get 2.

If you do it from -b to b+1, you will get infinity.

Play around with the function 2x/(x^2 +1). This function goes to 0 at infinity, but the doubly infinite integral is still undefined since the positive and negative are both infinite.

The symmetric integral is 0.

The integral going from -b to b^2 gives 2.

The integral from -b to b^3 gives 3.

Such is the issue with divergent integrals. They are like conditionally convergent series: you can get any value out of them if you play enough.

[Edwardo Piet]

Actual infinites that refer to something in external reality can't be possible if the universe itself is finite... or the infinities wouldn't fit in the universe lol.

Of course, mathematical infinities is a whole different matter. We can lay out what an infinity would be in principle, even if it doesn't exist.

Personally I think the universe is eternal but finite.

[GrandizerII]

So what does a definite integral actually mean?

Say we have a continuous function f(x), with a domain interval [a,b]:

The definite integral of f(x) is the sum of all the values of f(x) within the domain [a,b]. Or, in other words, the net sum area of the region(s) bounded by the function graph and the x-axis.

As manually calculating such a sum is virtually impossible for most functions, we need to make use of the concept of the limit itself. Just like with derivatives, definite integrals can be defined in terms of the limit of some expression. But what expression should it be?

Well, let's approximate the total region area bounded by the graph (whether from the top or the bottom) using adjacent rectangles of equal widths and varying heights, each extending from the x-axis and having its top/bottom edge touching f(x) at some point, with the leftmost rectangle touching the x-axis at x=a with its bottom left corner and the rightmost rectangle touching the x-axis at x=b with its bottom right corner. We need equal widths in order to make easy use of the limit eventually.

Once the rectangles are drawn, we can approximate the area of interest by simply adding up the positive/negative areas of all the rectangles. This is an approximation of course, as there will be gaps in the region of interest that are not covered by the rectangles themselves and rectangle parts extending beyond the curve.

To get to the exact value of the area, you need to visualize the width of each rectangle equally shrinking at the same rate. As the rectangles become narrower and narrower, more equally wide rectangles are able to fit into the region, better approximating the area. As the width approaches 0, each rectangle in the region gradually evolves into a vertical straight line, with each line stopping at the graph at exactly one point. That's when you get the exact area value (since the height of each "fully shrunken rectangle" would exactly match the value of f(x) at the point it touches).

0 widths means infinite rectangles, and remember that the area of a rectangle is width * height. Therefore, the standard definition formula for the definite integral should now make sense.

As the number of rectangles approaches infinity, the sum of the net areas of the rectangles approaches the exact value of the total net area, or the definite integral.

For reference, here's the standard formula for the definite integral:
https://pasteboard.co/Hi0l1le.png

[Divinity]

I'd like to understand math better. My granddaughter is struggling with Algebra this year, and my son realized something startling: All of our family sucks at math.

She actually did pretty good at the beginning. (It's 8th grade algebra). It got a bit more complicated once they got into graphing and complex numbers.

Any suggestions on where to start? Last time I took Algebra we Reagan was President. And contrary to what my 10th grade algebra teacher told me, I never had to use it once since then.

[Fireball]

I have a degree in physics. I learned to memorize the material. It takes a lot of work to get it memorized. Maybe that's just called "studying", but in the end, I always got that AHA! moment. Don't give up too soon, that's the key. If you and your daughter are reasonably smart (and I can tell you are more than that, reading your posts), you should be able to work through it. Also, unless the teacher forbids use of a particular method for problem solving, any means available are fair game for use. Except cheating, of course. 

"There is no royal road to geometry, sire" is said to be a response of Euclid to King Ptolemy I when he asked for a short cut to the Elements (of Geometry).

One thing to know is that every mathematician has had to struggle with it. Mathematics is a harsh Mistress who tolerates no errors.

[GrandizerII]

I'd like to understand math better. My granddaughter is struggling with Algebra this year, and my son realized something startling: All of our family sucks at math.

She actually did pretty good at the beginning. (It's 8th grade algebra). It got a bit more complicated once they got into graphing and complex numbers.

Any suggestions on where to start? Last time I took Algebra we Reagan was President. And contrary to what my 10th grade algebra teacher told me, I never had to use it once since then.

Have you heard of Khan Academy? I would strongly suggest this for both you and your granddaughter. It's all free and accessible online on their site, and lots of topics are covered in video and article formats, with lots of quizzes and exercises and progress tracking to boot.

[polymath257]

So what does a definite integral actually mean?

Say we have a continuous function f(x), with a domain interval [a,b]:

The definite integral of f(x) is the sum of all the values of f(x) within the domain [a,b]. Or, in other words, the net sum area of the region(s) bounded by the function graph and the x-axis.

As manually calculating such a sum is virtually impossible for most functions, we need to make use of the concept of the limit itself. Just like with derivatives, definite integrals can be defined in terms of the limit of some expression. But what expression should it be?

Well, let's approximate the total region area bounded by the graph (whether from the top or the bottom) using adjacent rectangles of equal widths and varying heights, each extending from the x-axis and having its top/bottom edge touching f(x) at some point, with the leftmost rectangle touching the x-axis at x=a with its bottom left corner and the rightmost rectangle touching the x-axis at x=b with its bottom right corner. We need equal widths in order to make easy use of the limit eventually.

Once the rectangles are drawn, we can approximate the area of interest by simply adding up the positive/negative areas of all the rectangles. This is an approximation of course, as there will be gaps in the region of interest that are not covered by the rectangles themselves and rectangle parts extending beyond the curve.

To get to the exact value of the area, you need to visualize the width of each rectangle equally shrinking at the same rate. As the rectangles become narrower and narrower, more equally wide rectangles are able to fit into the region, better approximating the area. As the width approaches 0, each rectangle in the region gradually evolves into a vertical straight line, with each line stopping at the graph at exactly one point. That's when you get the exact area value (since the height of each "fully shrunken rectangle" would exactly match the value of f(x) at the point it touches).

0 widths means infinite rectangles, and remember that the area of a rectangle is width * height. Therefore, the standard definition formula for the definite integral should now make sense.

As the number of rectangles approaches infinity, the sum of the net areas of the rectangles approaches the exact value of the total net area, or the definite integral.

For reference, here's the standard formula for the definite integral:
https://pasteboard.co/Hi0l1le.png

This is known as the Riemann integral. In it, we divide up the x-axis, the form the product of the length of the interval on the x-axis and the height of the function. Then we add up tttttt products. Finally we take a limit. All is well and good.

But there is another approach due to Lebesgue. Instead of dividing up the x-axis, divide up the y-axis. For each interval on the y-axis, look at the size of the set on the x-axis where the function values are in that interval (for nice functions, this is the sum of the lengths of a bunch of intervals). Again, multiply this size on the x-axis with the height on the y-axis, sum up, and take a limit.

The two methods give the same end result for 'nice' functions (read: continuous), but it turns out that the Lebesgue formulation works in far more generality and is much better when dealing with limit properties of functions. This is the start of the subject of 'Measure Theory'.