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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!

Ben, thanks for your comment. Your frustration with the navigation is mine as well. CZcams itself does not offer the right features to organize a lecture series the size of this one. For this reason, I am currently on a creation hiatus so that I can better organize my website to make the series easier to search, although that is not available yet. I hopefully can offer an update soon. It is well organized in my students' Canvas page, but those are not publicly available. I need to transfer these pages into my webpage.

Thank you for your kind words.

You're welcome

Admittedly, the first three geometries on your list are explicitly considered in this video. They are illustrated on the video preview. For these examples, is the answer to your question already provided? If not, maybe I do not understand the request, for which please clarify. Pappus geometry, like Fano geometry, is self-dual. Thus the original diagram is also a diagram for the dual. Technically speaking, this applies to any geometry, not because they are self-dual because most are not. It is because the terms "point" and "line" are undefined terms and could be interpreted as anything. Draw a diagram for Young geometry. That is also a diagram for the dual of Young geometry if we call the circles "lines" and the links "points." Maybe that is not a satisfying diagram. Instead maybe we want a diagram where "points" look like vertices and "lines" look like edges. To draw the dual of Young geometry, we start by drawing 12 vertices since Young geometry has 12 lines. Next we will draw 9 lines, representing the 9 points of Young geometry. Each of these lines will be incident to exactly four points, since in Young geometry each point is on 4 lines. In order to get the incidence right, label Young geometry and use these same labels for the dual. This process will get you the diagram you seek. Will it be a pretty diagram? Probably not. You might need to adjust the diagram to make it look symmetrical, but that is a separate question. Repeat this process for the last geometry.

@@Misseldine Ohhh thank you very much is the dual axiom of those geometry are the same as the original axiom?

Interchange the word "point" and the word "line" and you get your dual axioms.

@@Misseldine Thank you very much, so all of the listed geometry are applicable for duality also their axioms?

Exactly, the dual axioms give the dual geometry.

You're welcome!

You're welcome 😊

Thanks for the comment. Let's be clear on which function we are talking about. If you are talking about the square wave function, then it has a jump discontinuity each time it goes up or down. If we are talking about the Fourier approximations, that is, the partial sums, then those are all continuous functions. Now, as the partial sun becomes larger and larger, then the part of the graph near where the square wave has its discontinuities becomes steeper and steeper, but never vertical. I agree with you that these partial sums are continuous. Finally, there is The Fourier series, which is the whole infinite sum. This function does have discontinuities, jump discontinuities at the exact same places that the square wave function does. The square wave function and its Fourier series agree on all points in their domains, except at the jump discontinuities. Both functions have jumps, but the Fourier series assigns y=1/2 to each jump, as that is the average of the left and right limits at the jumps. This is all guaranteed by the Fourier Convergence Theorem. This example shows that a sequence of continuous functions can produce a limit which does have discontinuities.

@@Misseldine Thanks for the information!!!

Glad it helped!

Very good and true point! In my class, I present the product rule and its proof much earlier in the class, like day 2 of derivative rules. The proof follows from difference quotients and looks like, "I see why it works, but I wouldn't have ever thought of it myself." This proof is much slicker, the one I would select in a more advanced calculus course than this one is intended to be.

Yes, of course. In this very video, you see the proof of the power rule using logs. In another video, we handle the power-exponential function x^y using log differentiation. You likewise can handle exponentials too, although there is a small sticky point. In this lecture series, we found the derivative of a^x using the chain rule on e^x since a^x = e^(x ln a). But still requires e^x. You can easily prove the derivative of e^x using log differentiation, but how do you prove log differentiation works? One school of thought is to define ln as the antiderivative of 1/x and then you can derive everything else about it. That works just fine logically, but is very backward thinking for most college students, the log is defined abstractly and its inverse is proven to be an exponential function, implying that it is the inverse of an exponential, aka a logarithm. Stewart has a very good appendix in his Calculus book that carries out this whole development. Conversely, I find that students find the exponential the more natural function and hence develop it first, leaving logarithms for later in the course. We define e as a limit and later prove it's derivative using that limit and the squeeze theorem. It's a very nice argument very approachable to early calculus students without any logical gaps or delays. As transcendental functions, I delay them to the near end of the derivative rule chapter because they are more foreign to my typical student. It does mean that many arguments are more complicated than necessary but allows for better student learning in the long run. Great comment!

@@Misseldine I don't know but I have learnt in my school that the derivatives of (cosx)^sinx , x^y can be found using log differentiation. Then I understood that for exponential functions taking log on both sides and using power rule of log, we can easily get their derivatives. Then I thought to prove the derivatives of x^n and a^x using log differentiation as they are exponentials too. Then I got proofs of their derivatives using log differentiation. Literally I came here to search if the proofs I have done are correct. I wanted to find some video in which the derivative of x^n is solved by log differentiation. And then I found your video. I then believed that the proof that I have done is correct. Thanks to you for making this video sir.

@@Misseldine y = a^x Taking log on both sides we get log y = x(log a) Differentiating wrt to x (1/y) (dy/dx) = log a dy/dx = y(log a) = a^x loga Thanks a lot sir for replying to my comment. I am not a Mathematics expert like you sir. I just wanted to share my thought. Thank you sir.

@@Misseldine So, we can also prove the derivative of x^n same as your proof of derivative of a^x as follows: x^n = e^(logx^n) = e^(nlogx) Then differentiating wrt x We get d/dx (x^n) = [e^(nlogx)]×(n/x) = [e^(log x^n) ] × (n/x) = (x^n) × (n/x) = n[x^(n-1)] If we can find the derivative of x^n using log differentiation, we can also find the derivative of a^x using log differentiation. They both are exponential functions only. Thank you sir.

You're welcome. Best of luck on your exam!

You are so welcome!

You're welcome.

I am glad to hear it. Best of luck on your final.

You're welcome.

Happy to help!

Once explained it makes sense. I think i'm in the wrong but self learning is pretty hard.

It is pretty hard. My opinion is that most math textbooks skip many steps that students find very illuminating. It is my goal to try and share those skipped steps, but I am not always perfect about it myself.

Glad you think so!

Glad you liked it.

Thank you.

Glad you liked it!

Glad it could help.

Well said.

Thank you very much!

You're welcome. Happy π day!

The exercise was to prove the bisector theorem. To do so, the student would likely use the midpoint of a crossbar of the angle such that the angle and crossbar form an isosceles triangle. Then using SSS you can argue that the two triangles formed are congruent. As corresponding parts of congruent triangles are congruent, the angle has been bisected. The median of this triangle then is a bisector of the angle. On the other hand, the median of a generic triangle is not necessarily a bisector. It was necessary that the considered triangle is isosceles. As a last note, the perpendicular bisector of a segment requires a separate argument since a flat angle is exceptional and the triangle formed in the argument here would have been degenerate on a flat angle. The new argument is even easier as you can erect a perpendicular out of the midpoint of the segment.

@@Misseldine Thanks! Didn’t know if “crossbar of an angle” implied that the triangle formed would be isosceles in general- I had assumed not. So at 20:09, if and only if it passes through the midpoint of every crossbar (that forms an isosceles triangle with the angle).

@@ericsonw13Yes, every midpoint of a crossbar associated to an isosceles triangle. That isosceles triangle bit is critical to the proof, and I ought to have mentioned that in the video. Thanks for pointing it out.

In earlier math courses involving functions, such as precalculus or calculus, that is the way one thinks about the domain of a function. One thinks of the default domain as all real numbers except you throw out of the domain all those values for which the function is undefined. As the function is given by some kind of formula, one discards those real numbers for which the formula is undefined (as a real number), e.g. division by zero, square roots of negatives, etc. This approach, called the Domain Function, works for an elementary coverage of functions , such as in calculus, but it is insufficient when in settings of advanced mathematics. That is not to say one cannot ever define a function via this convention. Of course you can still do that, but what is necessary is an explicit domain set A, even if the elements of the specific domain set are not yet explicitly determined. So, you could define the set A to be the subset of real numbers x such that ln(x) is itself a real number. This defines A as a subset of real numbers, even if a priori we didn't know that A is likewise the set of positive reals. So, the domain set is explicitly set as part of the definition of the function even if we the operator have to determine the exact elements of the domain after defining the function itself. I hope that answers your question.

@@Misseldine yes, as I can comprehend is a matter of convention, important thing is that one clarifies in advance what is meant in the heading of the function "f: A -> B". Thank you very much.

Any time

Subtraction has all the problems of being a group. First, subtraction is not associative. For example, 2 - (3 - 4) = 2 - (-1) = 3 but (2 - 3) - 4 = -1 - 4 = -5. To your point about identities, subtraction does not have an identity element, there is no element e such that for all elements x we have x - e = e - x = x. There is a universal element e such that x - e = x. This is e = 0. This gives us a "right identity." But 0 is not a left identity, since 0 - 7 = -7, not 7. In an associative operation the presence of a one-sided identity (left or right) implies it is actually a two-sided identity (that is, the standard definition of identity). To the issue of inverses, because the operation has no identity it can have no inverses. The inverse axiom is conditional to having an identity. After all, the law says there for all x there is a y such that x - y = y - x = e, where here e is the (two-sided) identity. As there is a right identity, there could be inverses (one-sided or two-sided) associated to it. So while it seems like there are two-sided inverses here (an element is it's own inverse), the existence is associated to a half identity and hence not real inverses. On the other hand, the operation of subtraction is cancellation, meaning that for all x if a - x = b - x or x - a = x - b then a = b. Cancellation (or the stronger latin square axiom) is a property that essentially as the same thing the inverse axiom but is not dependent on the existence of an identity. This is the approach of a quasigroup, a set with binary operation which satisfies the Latin square axiom, namely for all a and b there exists a unique x and y such that a*x = b and y*a = b. A quasigroup has the cancellation property as a consequence of the LSA. Likewise, if a quasigroup has an identity, such a structure is called a loop, then it will automatically have inverses too. Admittedly, without associativity the benefit is very, very limited compared to what we see in group theory, but they exist nonetheless. The structure of subtraction is that if a quasigroup.

@@Misseldine thank you Andrew 🙏

Good question. While we know there exists a set of three non-collinear points, such as the set {A, B, D}, we do not have that every set of three points is non-collinear. The statement is existentially qualified, not universally. Some sets of three points may be collinear, but we are guaranteed at least one set of three non-collinear points. I hope that helps.

@@Misseldine That helps a lot, thank you!

Norms of complex vectors are computed using conjugates. Hence, |(1+i, 2)|^2 = (1+i)(1-i) + 2(2) = 1 - i + i - i^2 + 4 = 1 + 1 + 4. In general, (a+bi)(a-bi) = a^2 + b^2. So the (square of) norm of a vector is a sum of squares, real or complex.

Thanks

Honestly, I often lecture too fast for my own students, but the advantage of the video is pause, rewind, slow down. It can make challenging math concepts more digestible.

@@Misseldine exactly! Thanks again

Glad it helped!

I'm glad it helped.

Thank you

Yes, that is correct. For a general radius r and angle θ, the method shown here would generalize to V = 2/3*r^3*tan θ.

Glad you think so!

You're welcome.

Good thing that's not the temperature, right?

Thank you for your generous comments. My research background is in algebraic combinatorics, which, among other things, means we get obsessed with classification problems of finite, diverse things. Finite groups is one important such problem. Finite geometries are another. If incidence geometries are to groups, then affine geometries are to cyclic groups, as they are completely characterized up to their order. That's at least true for finite ones. Infinite affine geometries require a greater knowledge of field theory. Every field gives rise to a unique affine geometry using the field as a set of coordinates for the geometry, that is, points are ordered pairs of coordinates and lines are linear equations (up to equivalent equations). This is always an affine geometry if the coordinates are a field (an abstract analytics geometry, you can say). The other direction is only conjectured, that is, affine geometries can always be coordinated. Every known affine geometry can be. An interesting exercise for any student is to try to classify the incidence geometries of small order. One of order 3, two of order 4, four or order 5, and I think 17 or 18 of order 6. I would have to look up the last one as I don't remember off the top of my head, but it explodes after that and 6 is a good place to stop for most students. Too challenging to just guess but easy enough that it can be solved.

Thank you for that excellent info! These videos have been great for filling in a lot of details for me, but that bigger picture of geometry as nested structures really elevates the entire subject for me to a whole new level of appreciation which is golden!