Studying Mathematics Thread

[Divinity]

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.

No, I hadn't.  Thanks for that.  My son had been using IXL, which has good questions, but limits to 10 questions a day for non-members (and he doesn't want to pay $10/month for something that won't get a LOT of use).  

I think my biggest problem with math is that it all jumbles together.  A lot of times I just have no idea what they're asking me.  It's like I'm looking at a goddamn foreign language that nobody's bothered to teach me.  You can tell me how to get the answer, and I'll be able to repeat it.  But I have no understanding of it, and will forget it by tomorrow.

[GrandizerII]

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.

No, I hadn't.  Thanks for that.  My son had been using IXL, which has good questions, but limits to 10 questions a day for non-members (and he doesn't want to pay $10/month for something that won't get a LOT of use).  

I think my biggest problem with math is that it all jumbles together.  A lot of times I just have no idea what they're asking me.  It's like I'm looking at a goddamn foreign language that nobody's bothered to teach me.  You can tell me how to get the answer, and I'll be able to repeat it.  But I have no understanding of it, and will forget it by tomorrow.

See how you go with KA. Start one level at a time, and build up from there. It's important to first gain mastery of the basics before moving on to the next level in difficulty, if you wish to improve your intuition about the various concepts in mathematics. That, and you need to be interested and motivated enough to keep going, despite hurdles along the way.

[GrandizerII]

So ... arc length formula derivation, here we go:

Say we have a continuous and differentiable function f(x). What is the general formula for the length of the curve of the function within the domain interval [a,b]?

Let's call the length L.

Like before, let's come up with a valid approximation of the answer, and then use the limit to arrive at the exact value.

To do this, divide the interval into smaller intervals of equal size, with a segment drawn in each interval from the first point of the interval to the last. All segments should have the same delta-x value, but varying delta-y values. Note that the length of each segment can be derived via the Pythagorean theorem. Adding up the segment lengths should yield a fair approximation of the total length of the arc. If you visualize the intervals getting smaller and smaller, each ever-shrinking segment gets closer and closer to a point on the graph of the function f(x). As each segment progressively gets closer and closer to zero-length segments, the sum of the segment lengths gets closer and closer to the exact total length. So now the first line in the image below should make sense.

To get to the second line in the image, one must keep in mind the Mean Value Theorem: there is a point in each of the small intervals, whereby the derivative of the function at that point is equal to the slope of the corresponding segment.

Interestingly enough, the second line is a good example of the definition formula for the definite integral (discussed in an earlier and recent post), and so we have the definite integral line as the third line in the image.

https://pasteboard.co/Hi3LTHc.png

And so there we have it: the general formula for the arc length L of f(x) within the specified domain interval [a,b].

[robvalue]

That's amazing I've probably seen it before but I've forgotten it if so. The integral function makes more intuitive sense, I think, when viewed as a "summing" of a continuous series. This is shown here by the discrete summing function becoming an integral as the intervals between the terms tends to zero.

It's again hilarious to imagine getting an exact answer any other way. Draw the graph, then get a piece of string...

Of course, you could use a computer program to get more and more exact approximations. But it's astonishing to me how you can cut right through that to get an exact answer very quickly, assuming you can complete the integral of course! Otherwise, you're back to approximate methods anyway.

[polymath257]

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.

No, I hadn't.  Thanks for that.  My son had been using IXL, which has good questions, but limits to 10 questions a day for non-members (and he doesn't want to pay $10/month for something that won't get a LOT of use).  

I think my biggest problem with math is that it all jumbles together.  A lot of times I just have no idea what they're asking me.  It's like I'm looking at a goddamn foreign language that nobody's bothered to teach me.  You can tell me how to get the answer, and I'll be able to repeat it.  But I have no understanding of it, and will forget it by tomorrow.

And it should be approached as a foreign language. First, get the basic concepts and words. Then gradually expand outward, acquiring fluency by degrees and with lots of practice.

[GrandizerII]

That's amazing I've probably seen it before but I've forgotten it if so. The integral function makes more intuitive sense, I think, when viewed as a "summing" of a continuous series. This is shown here by the discrete summing function becoming an integral as the intervals between the terms tends to zero.

It's again hilarious to imagine getting an exact answer any other way. Draw the graph, then get a piece of string...

Of course, you could use a computer program to get more and more exact approximations. But it's astonishing to me how you can cut right through that to get an exact answer very quickly, assuming you can complete the integral of course! Otherwise, you're back to approximate methods anyway.

It's amazing to me how useful the limit has been as a concept in deriving formulas for certain variables that otherwise would've probably been impossible to attain without it. Despite everything I've posted here, I'm still at the very beginning of the journey of maths, and only recently have I started to realize just how amazing some of these theorems and concepts and tools really are.

[polymath257]

So ... arc length formula derivation, here we go:

Say we have a continuous and differentiable function f(x). What is the general formula for the length of the curve of the function within the domain interval [a,b]?

Let's call the length L.

Like before, let's come up with a valid approximation of the answer, and then use the limit to arrive at the exact value.

To do this, divide the interval into smaller intervals of equal size, with a segment drawn in each interval from the first point of the interval to the last. All segments should have the same delta-x value, but varying delta-y values. Note that the length of each segment can be derived via the Pythagorean theorem. Adding up the segment lengths should yield a fair approximation of the total length of the arc. If you visualize the intervals getting smaller and smaller, each ever-shrinking segment gets closer and closer to a point on the graph of the function f(x). As each segment progressively gets closer and closer to zero-length segments, the sum of the segment lengths gets closer and closer to the exact total length. So now the first line in the image below should make sense.

To get to the second line in the image, one must keep in mind the Mean Value Theorem: there is a point in each of the small intervals, whereby the derivative of the function at that point is equal to the slope of the corresponding segment.

Interestingly enough, the second line is a good example of the definition formula for the definite integral (discussed in an earlier and recent post), and so we have the definite integral line as the third line in the image.

https://pasteboard.co/Hi3LTHc.png

And so there we have it: the general formula for the arc length L of f(x) within the specified domain interval [a,b].

OK, so the basic formula for this derivation is the Pythagorean result, (ds)^2 =(dx)^2 +(dy)^2.

We can 'divide by (dx)^2) to then get (ds/dx)^2 = 1 +(dy/dx)^2, take a square root to get

ds/dx = sqr{{1+ (dy/dx)^2 }

and then integrate to get the correct formula.

So, what happens if we use polar coordinates? For such, x=r cos(A) and y=r sin(A) and we use (r,A) as our coordinates as opposed to (x,y).

Well, take differentials:

dx = cos(A) dr -r sin(A) dA
dy= sin(A) dr + r cos(A) dA

Now, square these

(dx)^2 = cos^2 (A) (dr)^2 -2r sin(A) cos(A) dr dA + r^2 sin^2 (A) (dA)^2
(dy)^2 = sin^2 (A) (dr)^2 +2r sin(A) cos(A) dr dA + r^2 cos^2 (A) (dA)^2.

Adding, we get

(ds)^2 = (dr)^2 + r^2 (dA)^2

Now, we can divide by (dA)^2, take a square root, and integrate to get the arc length formula for polar coordinates:

s= int sqrt{ r^2 + (dr/dA)^2 } dA

This is useful when the radius is given as a function of the angle A.

More later.

[polymath257]

OK, so the formulas
(ds)^2 = (dx)^2 + (dy)^2
= (dr)^2 + r^2 (dA)^2

Give us arc length formulas for the plane in rectangular and polar coordinates, respectively.

These very naturally generalize to three (or more) dimensions:

(ds)^2 = (dx)^2 + (dy)^2 + (dz)^2 <--- rectangular
= (dr)^2 + r^2 (dA)^2 + (dz)^2 <----cylindrical
= (dr)^2 + r^2 (dA)^2 + r^2 cos^2 (A) (dB)^2 <---spherical

In the last, A represents latitude and B represents longitude. This is slightly different than typical (usually, spherical uses the angle down from the z-axis instead of latitude, which is up from the equator)

But, suppose we live on a sphere? (well, we do, at least approximately). Just fix r in the spherical coordinates and so dr=0. So, on a sphere we have

(ds)^2 = r^2 (dA)^2 + r^2 cos^2 (A) (dB)^2.

Now, these formulas for the arc length are called 'line elements' and each surface has one that describes the arc length on that surface in whatever coordinates you want to use.

It turns out that these line elements contain a LOT of information about the surface. In particular, we can determine how to find areas on the surface from them. We can determine curvature of the surface.

For example, on the sphere, we can determine area by doing a *double integral* over a region of r^2 cos(A) dA dB. This is the square root of the product of the two terms in the line element.

For spherical coordinates in three dimensions, we can compute *volume* by doing something similar and getting a *triple integral* of r^2 cos(A) dr dA dB.

More later.

[polymath257]

In previous posts, I discussed the concept of a line element. It is a description in terms of differentials of a distance equation.

The nice thing about these line elements is that they exist for every surface and easily generalize to higher dimensions (which for cylindrical and spherical coordinates in three dimensions).

So, at this point, we know how to find the arc length of a curve on a surface by integrating ds over that curve.

Now, a new concept: a geodesic is a curve between two points that has the smallest possible arc length for curves between those two points.

As an easy example, a straight line is a geodesic in the plane and a great circle is a geodesic on a sphere.

So, how do we find the equations for a geodesic?

This will involve finding the minimum value of an integral over all possible curves between two points. Following the intuition from calculus, we want to take a derivative *with respect to the curves* and set that derivative equal to zero.

The procedure is called the calculus of variations and is a very general and very powerful way of finding *functions* that minimize certain integrals. Just like we find ordinary algebraic or trigonometric equations when minimizing functions in calculus, in the calculus of variations, we get a *differential* equation whose solution is the function we seek. This differential equation is called the Euler-Lagrange equation for the integral we are trying to minimize.

Next, we can determine that our surface (or, better, higher dimensional manifold) is *curved* if two geodesics that start out parallel start to accelerate either towards or away from each other. Imagine the great circles on a sphere that we have two of them through close points that are parallel (in the sense of having the same direction). Those two great circles will inevitably get closer together as we move away from our initial points. This is because a sphere is curved.

What this ultimately says is that the line element (now also called a metric) tells us a great deal about the geometry of the surface (or manifold) we are investigating. It tells us which curves are geodesics and those tell us about the curvature of the manifold.

next step: General Relativity.

[GrandizerII]

General relativity, eh? I'll eventually catch up to what you're saying, but for now, I'm personally going to address a confusion some have with the concept of the constant of integration C. In the derivation of the formula for integration by parts, C just isn't there anywhere in the final formula. Where did it magically disappear?

Lo and behold, here's my derivation of the formula that puts this in perspective.

https://pasteboard.co/HiCvx8S.png

Fucking math nerds, the bunch here.