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]

[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]