also at 9:26 just assuming the answer meant to be written at the blurred part is - 2/3 or we’ll say negative two thirds :)
@Kate-tc5uu2 күн бұрын
7:37 just for nit-picky clarity since the audio is a little muffled here and the auto yt transcript is a little off, what’s being said is: when the denominator is irrational, you must rationalize it
@arghyadas87339 күн бұрын
Hi, Can You Please Suggest Me A Textbook That I Can Use To Follow This Lecture Series?
@Misseldine9 күн бұрын
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.
@Tee031314 күн бұрын
thank you very much for this wonderful video
@Misseldine14 күн бұрын
You are very welcome
@Tejas-s3u19 күн бұрын
Super Sir
@Misseldine19 күн бұрын
I'm glad it was helpful.
@Tejas-s3u19 күн бұрын
Super Sir
@Misseldine19 күн бұрын
I'm glad it was helpful.
@shinluching803926 күн бұрын
Thanks, this is helpful :)
@Misseldine25 күн бұрын
Glad it was helpful!
@Pr0fBruceАй бұрын
what if the angles slice did not pass through the midpoint of the cylinder? XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXX XXXXXX XXXX XX
@MisseldineАй бұрын
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.
@user-kg3mx8oy7oАй бұрын
Nice explanation
@MisseldineАй бұрын
Thank you
@nulmonas9951Ай бұрын
what if the vector goes [1 0 i] which norm is 0? what unit vector will it be?
@MisseldineАй бұрын
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.
@Blueberry_GSDАй бұрын
Thank you sir This illustration video is such a clear and high-quality one!
@MisseldineАй бұрын
Glad it was helpful!
@amiraslkhalili5638Ай бұрын
close but not what i am trying to develop , well i am actually developing
@MisseldineАй бұрын
What are you developing exactly?
@RB_Universe_TVАй бұрын
Oh! Now I understand At first I thought why are the answers different Now, I understand
@MisseldineАй бұрын
Glad I could help!
@BenHutchisonАй бұрын
Hi Andrew, First up, thank for for putting your lectures out freely. I find it some of the clearest and most articulate math content available online. You clearly take pride in your work.. it is appreciated 🙏 However, I do find navigation of the courses difficult (and frustrating tbh), for several reasons. One is that the playlists do not seem to align with the linear order of the lectures. For example, this lecture finishes with you saying "next up, well look at the properties of relations". But the next video in the list is "combining proof techniques", which is from a different topic. The subsequent lecture is actually 3 lectures along in the playlist. The second reason is KZbins search functionality seems "deliberately blunt", and quite unsuitable for research and technical use. I'll be searching using maths terminology and it will offer a yoga video, because I watched one last week! I wish there was a way to turn recommendation mode off when searching and get results strictly related to the search terms. Anyway, no easy answers to that problem... 😖
@MisseldineАй бұрын
Ben, thanks for your comment. Your frustration with the navigation is mine as well. KZbin 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.
@tylerbakeman2 ай бұрын
This is the best lecture on semilattices on KZbin. I have watched so many examples, and read off of 5 sites for an intro… This one is the clearest, and simplifies the topic Without context, a semilattice seems like a special poset with a infimum or supremum; contextualizing, union and intersect, makes it easier to understand the upcoming topics: Relations as products of sets, Topological spaces as locales, Net theory (the research is hard to find for Moore-Smith sequences tbh) into Frames (which has to do with Nets and Semilattices somehow), Fibre (Fiber) theory (which is also hard to study) So. Thank you!
@MisseldineАй бұрын
Thank you for your kind words.
@omardaoud-z3n2 ай бұрын
thnx
@MisseldineАй бұрын
You're welcome
@DanmarkGonzalez2 ай бұрын
Can you help me with this I'm having a hard time understanding the duality Contruct a diagram for the dual of the following finite geometries : 1. Four point Geometry 2. Four Line Geometry 3. Fano's Geometry 4, Young's Geometry 5. Pappus` Geometry 6. Desargues Geometry Thank you in advance :)
@Misseldine2 ай бұрын
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.
@DanmarkGonzalez2 ай бұрын
@@Misseldine Ohhh thank you very much is the dual axiom of those geometry are the same as the original axiom?
@Misseldine2 ай бұрын
Interchange the word "point" and the word "line" and you get your dual axioms.
@DanmarkGonzalez2 ай бұрын
@@Misseldine Thank you very much, so all of the listed geometry are applicable for duality also their axioms?
@Misseldine2 ай бұрын
Exactly, the dual axioms give the dual geometry.
@mariac32782 ай бұрын
Great explanation! Is the result f(x) in a trigonometric form? How can we transform it to a complex exponential form? Does it require a different computation to get that? Or we always come out to a same result?
@tobiathan2 ай бұрын
Thank you!!!!
@Misseldine2 ай бұрын
You're welcome!
@yusufkareem84152 ай бұрын
Thank you so so so much❤
@Misseldine2 ай бұрын
You're welcome 😊
@samandersen12282 ай бұрын
Just out of curiosity... where did you see that notorious single word proof? Also, hello from Los Alamos prof Misseldine!
@panchovilla54003 ай бұрын
NIce explanation, but BRO, It does not have discontinuities. Even though is a vertical line it is a VERITABLE continuous line. The line is continous as far as I´m concerned from the time it is a rough squiggly line to a more and more and more polished square form. I am telling you this from observation, not because of a text definition. I hope you come from the same place. Show me where it is discontinious!!!!!
@Misseldine3 ай бұрын
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.
@panchovilla54003 ай бұрын
@@Misseldine Thanks for the information!!!
@user-cr4yh9yw2s3 ай бұрын
thanks for your clear explanation! it helps.
@Misseldine3 ай бұрын
Glad it helped!
@ankithsai13 ай бұрын
Also the differentiation of two functions in multiplication rule comes from logarithmic differentiation. Let y = f(x)•g(x) then apply logarithm log y = log f(x) + log g(x) Then differentiating wrt. to x we get dy/dx = f(x)•g'(x) + f'(x)•g(x)
@Misseldine3 ай бұрын
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.
@ankithsai13 ай бұрын
Logarithms are most important in differentiation. Even the derivatives of a^x, e^x, x^n, x^y and any other exponential function are found using logarithmic differentiation.
@Misseldine3 ай бұрын
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!
@ankithsai13 ай бұрын
@@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.
@ankithsai13 ай бұрын
@@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.
@ankithsai13 ай бұрын
@@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.
@philosan3383 ай бұрын
hey ; have a university exam and you are one of the few good people that explain this topic . thank you 🥰🥰🥰
@Misseldine3 ай бұрын
You're welcome. Best of luck on your exam!
@chesleywayne19394 ай бұрын
That was a very elegant and easy to follow tutorial! Thanks much.
@Misseldine4 ай бұрын
You are so welcome!
@neshirst-ashuach18814 ай бұрын
Thanks, super useful.
@Misseldine4 ай бұрын
You're welcome.
@glennxhose72174 ай бұрын
This is one of the best lectures I ever came across 🫡specifically proof that a finite p-group implies it's center is non-trivial .. you saved my semester 😂😂
@Misseldine4 ай бұрын
I am glad to hear it. Best of luck on your final.
@MaxwellBird-m8v4 ай бұрын
thank you!
@Misseldine4 ай бұрын
You're welcome.
@Steven-ek4ny4 ай бұрын
Thank you for these videos! I think I have the hang of it now. Onto Surface revolution!
@Misseldine4 ай бұрын
Happy to help!
@Steven-ek4ny4 ай бұрын
Self teaching myself calc 2 and thank you for the explanation. Sometimes the textbook assumes knowledge which isn't helpful. Thanks!
@Steven-ek4ny4 ай бұрын
Once explained it makes sense. I think i'm in the wrong but self learning is pretty hard.
@Misseldine4 ай бұрын
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.
@wyattcrowell19854 ай бұрын
Are you able to explain how you derived the domain with a little more detail? I just cannot understand it.
@the-boss-984 ай бұрын
great explanation
@Misseldine4 ай бұрын
Glad you think so!
@koik90824 ай бұрын
hehehe
@Misseldine4 ай бұрын
Glad you liked it.
@BililgnSamson4 ай бұрын
That's so nice ❤
@Misseldine4 ай бұрын
Thank you.
@lollipopjoe55304 ай бұрын
such a great video, thank you!
@Misseldine4 ай бұрын
Glad you liked it!
@shubhammehra55275 ай бұрын
Smj Gaya sir
@Misseldine5 ай бұрын
Glad it could help.
@MOTM12345 ай бұрын
Thank you The tense you're referring to is known as the "pluralis modestiae" or the "royal we." It's a stylistic device where speakers or writers refer to themselves using the plural pronoun "we" instead of the singular "I" to express modesty, formality, or inclusiveness. It's often used in academic or formal writing to convey a sense of collective responsibility or to imply that the speaker is speaking on behalf of a group, even when they are acting alone. using "we" to refer to yourself indicates that you're not speaking solely on your own behalf, but rather in a broader sense, which could include your colleagues, the scientific community, or the general population.
@Misseldine5 ай бұрын
Well said.
@UhrBushaltestelle5 ай бұрын
Great video especially the order of operations
@Misseldine5 ай бұрын
Thank you very much!
@josehenriquetarginodiasgoi74245 ай бұрын
I liked it very much. May you share further references to this class?
@zianiera6 ай бұрын
Very well explained.Thanks
@Misseldine6 ай бұрын
You're welcome. Happy π day!
@ericsonw136 ай бұрын
Was the exercise for the students at 19:48 to prove that a median of any triangle is also an angle bisector?
@Misseldine6 ай бұрын
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.
@ericsonw136 ай бұрын
@@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).
@Misseldine6 ай бұрын
@@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.
@giack62356 ай бұрын
Is it correct to define A as the domain of the function? I mean, couldn't we have: f := ln(x) : A = R -> R and then compute the domain as the subset of A = R such as x > 0 ? Anyway, thank you so much for the very nice video.
@Misseldine6 ай бұрын
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.
@giack62356 ай бұрын
@@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.
@diegotorres63196 ай бұрын
Thanks man
@Misseldine6 ай бұрын
Any time
@BenHutchison6 ай бұрын
If I'm not mistaken,the problem with subtraction relates to the Identity law not the inverse law. 7 - 0 is not equal to 0 - 7. It seems like subtraction can satisfy the inverse law with every element it's own inverse
@Misseldine6 ай бұрын
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.
@BenHutchison6 ай бұрын
@@Misseldine thank you Andrew 🙏
@allylukens456 ай бұрын
I'm sort of confused on how this model satisfies Incidence Axiom #3, which states "there exists three distinct points with the property that no line is incident with all three". Here in our model, it seems like there does exist three distinct points with the property that a line is incident with all three (line DBE and line ABC), and if this is true, then wouldn't this model not satisfy IA3 and thus not be a model of incidence geometry? Please let me know where I'm mistaken - thanks!
@Misseldine6 ай бұрын
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.
@allylukens456 ай бұрын
@@Misseldine That helps a lot, thank you!
@capturedart06 ай бұрын
in |v1| and |v2|, i^2 = -1 so why are you still adding? like (1+1+4)=6 but shouldn't it be (1-1+4)=4?
@Misseldine6 ай бұрын
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.
@henryguo81656 ай бұрын
Good Example!
@Misseldine6 ай бұрын
Thanks
@intuitiveclass64016 ай бұрын
im sure its not my prof's fault, but my brain struggles to be sharp and quick like the others so when she just speeds through a lecture I'm always lost on concepts that aren't even all that difficult. thank goodness for this video!
@Misseldine6 ай бұрын
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.