Відео

I completely agree. Rebecca did a very fantastic job on this video.

Glad you liked it!

I am glad to have helped!

I am glad you liked it. The book LADO, and this accompanying lecture video series, is a byproduct our exactly what you mentioned. The standard text didn't quite fit the introductory level of linear algebra. I tried lots of different texts, most of which were too light or too heavy on theory. Students need a conceptual understanding to thrive in linear algebra, but most students, much like my own, are not ready to be proving or abstracting anything yet about linear algebra as they are learning it for the first time. I really tried to strike a better balance, such as doing linear algebra over a finite field. The whole point to introduce at the beginning is just to help abstract linear algebra without lengthy proofs that are lost in the students. Thanks also for reporting some errors on a few of the other videos.

For certain! It's this exact same algorithm! The process to find the RREF only requires the elementary row operations, which requires add and multiplication of scalars, which can be done in any field, including finite ones. Once you have the RREF, you can use it to solve the reduced homogeneous system, which will give you a basis for the null space. This does not depend on the field whatsoever.

I am glad to have helped.

Glad it was helpful!

You're very welcome!

In many problems in calculus, we take the derivative with respect to x, that is d/dx, but in this one we are taking the derivative with respect to time, that is d/dt. Similarly, we abbreviate dy/dx as y', but here y' stands for dy/dt. Likewise, x' = dx/dt, the rate that x varies over time.

Best of luck!

I'm glad it helped.

Thanks

🐻🐻🐻🐻🐻🐻🐻🐻🐻

Glad you liked it.

Proof by delegation. It's a valid proof technique, but only if you have a PhD 😉

We can tell that it is proper since the numerator has smaller degree than the denominator. If it is improper, then we do polynomial division to make it proper.

@@Misseldine thanks

You're welcome

I am glad you liked it. This is the very last video, so I hope this means you made it all the way through. I will be honest, UA-cam does not provide an easy way to post a lecture series, as playlists have some limitations. At some point, I hope to organize them better on my website, but that hasn't happened as of this comment. This is a great question. We have to return to the definition of a lune. Intuitively, it is the area enclosed between two lines, not line segments. It might be easier to see in on the projective real plane (double point elliptic geometry). Using the hemispherical model, imagine one of the boundary lines to the lune is the equator of the hemispherical and the other line is any other line in the geometry. The two angles of the line are vertical and congruent and less than 180°. Given the "third" angle of the line is flat and 180°, the same of these three angles is strictly less than 3*180° = 540°. Allow this non-equatorial to converge toward the equatorial line. When they touch, the two angles will become flat, that is, 180° each. Thus the angle sum is exactly 540°, but we do not have a lune anymore, as the area is now that which is enclosed by a single line, not two. In the real projective plane, our adopted model of elliptic geometry (since we chose line determination over plane separation), a line does not separate the plane, that is, the area enclosed by a single line is the whole geometry. In spherical geometry, aka one-point elliptic geometry, where we do have plane separation and not line determination, two intersecting lines bound four distinct regions (as opposed to only two in projective geometry). Each lune comes with a partner on its antipode that is congruent. Allowing these two lines to converge toward each other with have the effects that two of the congruent lines will squish to nothing and the other congruent lunar pair will converge to complementary hemispheres. The three angles of the line are now 180° each. Thus, a hemisphere is the limit of lunes and represents the maximum a lune could be. If the lines continued past their meeting the small lines would return and the hemispheres would lose some area. I imagine you were imagining two longitude lines on the earth diverging until they meet and enclose the whole goal. It is reasonable to say that figure has two 360° angles, giving it a 720°, but such a figure is not a lune, as it cannot be enclosed by to spherical lines even though it is enclosed by to spherical line segments.

​@@Misseldine Thank you for the detailed answer. Yes, I have studied each and every lecture in this series (meaning several of them, I have watched multiple times even :), to get every nuance you added as an informal help to get closer to the idea). Nevertheless, I still have some plot holes here. I think I follow your argument above step by step (I get the problem with segment- or line-based lunes also), but I still don't get why you speak about a "third" angel in the case of a "biangle", which has only two angles by definition. I mean, if you can reach 540° in the limit, then there should exist an ordinary biangle having an angle sum of 539.9, or, let's just say 361°. Which one is that? I also get that in the case of the hemisphere model, we have only a single vertex but with two angles still, the sum of which cannot exceed 360° either. What do I miss? Thank you!

Glad it was helpful!

Glad it was helpful!

You certainly can compute the characteristic polynomial for a 3x3 matrix using the methods illustrated in this video, but calculations get exponentially harder for determinants as the matrix grows in size. The complexity of a determinant using Laplace cofactor expansion is O(n!), which is a beast. Using Gaussian elimination to find the determinant is much better, requiring O(n^3), which is still quite difficult though. Of course, this gets a bit messy though as that methods will lead to rational functions in the matrix entries eventually. One could compute the characteristic polynomial without determinants, say through iterated powers of the matrix. We compute the powers through A^n, which must have a dependency relation. But this still leads to a O(n^3) complexity. More sophisticated algorithms are necessary in order to improve the complexity. For this reason in practice people don't usually compute the characteristic polynomial, instead accepting numerical approximations of the eigenvalues.

If you click on my UA-cam homepage, you can find a playlist for all my classes, including Math 4220/4230. I need to find a better method to share the whole course, but you can at least find it there, lecture by lecture.

@@Misseldine Your videos helped with my midterm this week. Much appreciated.

My lecture notes, which you can find linked in the description, are originally based upon Tom Judson's Abstract Algebra Theory and Applications, although it does not follow the ToC perfectly. These videos are based upon my lecture notes, which are fairly detailed, but you can find more details in Judson's book. AATA is an open source textbook which you can find here: abstract.ups.edu/aata/aata.html

Thank you

It is (50-30)/2 = 10. The difference between the top and bottom is 50-30 = 20. By the symmetry of the shape, half the excess is on the left and right, hence 10 on both sides.

Glad it helped!

Glad it helped!

Most welcome

Wow, thank you! It is good to hear from you again. I do hope graduate school is going well for you. It can be rough, but very satisfying.

To explain the four, we note that the side length of the squares are twice the y-coordinate of the cross section inscribed in the circle. When we squared this side to find the area of a single square cross section, we get the area as 4y^2. This is where the 4 came from, but this does not seem to be your concern from the comment. There is an extra factor of 2 that comes from symmetry. There is two ways to see this. First, the integrand is an even function. Likewise the interval is symmetric about the origin. This the area under the even function is twice the area of just the region to the right of the y-axis. You will notice that an extra factor is 2 appears (giving a total scalar of 8) at the same time as the lower limit of the integral is replaced with a 0. This is the symmetry play so common in integral applications. The second way to see the symmetry is just looking at the solid itself. By the nature of the cross sections and the symmetry of the base circle, the y-axis is a line of symmetry for the solid. Hence, doubling half the volume gives the whole. Hope that helps.

@@Misseldine Hi, Thank you for your detailed explanation. Your video's helped me pass my midterm earlier. I appreciate your insight into the symmetry at work.

That's fantastic to hear. Congrats on the passing score on your midterm. Do it again on your final!

Three point geometry is self-dual, that is, it is isomorphic to is dual. Hence a graph of the geometry is also a graph of the dual. This is true for all fan geometries, the smallest of which is three point geometry.

Thanks

It's my pleasure

The link to my lecture notes can be found here: drive.google.com/file/d/1UjEQOEXBmgEE3xg6-M_34ETqC1U7_Acz/view?usp=drivesdk The lecture series loosely follows the Boof of Proof by Hammack: www.people.vcu.edu/~rhammack/BookOfProof/. The videos follow my lecture notes perfectly, but the numbering of sections follows the Book of Proof so my students can go there for further readings, if necessary. Another good reference is Transition to Advanced Mathematics by Doud and Nielsen: mathdept.byu.edu/~pace/Transition_v104.pdf. I would occasionally borrow an example from them here or there when I wanted to supplement the narrative from Hammack.

You are very welcome

I'm glad it was helpful.

I'm glad it was helpful.

Glad it was helpful!

Excellent question. Your cross sections would still be right triangles. Set up your integral using the same variables. The height of the triangle is parameterized by its location along the spectrum and would be unaltered. The length along the base side is what is changed now. Since the plane intersects the base circle as a line segment, the amount of the base side changed by a constant value, an amount either added to or subtracted from depending on where the plane intersects the base circle's center. It should only involve a very minor change to the integral. The area of a specific cross section should be V = 1/2*b*h*dx = 1/2*(r^2-x^2-a)^(1/2)*(r^2-x^2)^(1/2)dxx. But note that small change dramatically complicates the calculation of the antiderivative. I would recommend a numerical estimate. The original setup using isosceles triangle is a great simplifier.

Thank you

You must use the Hermitian inner product, which is like the standard dot product but you take the conjugate of the first complex vector. Hence the length of [1,0,i] would be 1(1)+0(0)+(-i)(I)=1+0+1=2.

Glad it was helpful!

What are you developing exactly?

Glad I could help!