Why don't they teach simple visual logarithms (and hyperbolic trig)?

[@johnpearcey]

It seems that you are able to provide a geometric interpretation for any mathematical concept that exist. Simply amazing.

[@digitig]

Nicolas Bourbaki will be furious. 😉

[@WillToWinvlog]

Yes, even reality itself is interpreted geometrically from our perspective!

[@blvckbytes7329]

Because there *is* nothing else than geometry! All other concepts are abstractions, but everything is still rooted in reality which in essence represents geometry. Teaching math nowadays is just hugely disconnected from what's actually going on. Students should always be presented with this kind of visualization. Sadly, we prefer recipes for robots over deep understanding.

[@johnpearcey]

@@blvckbytes7329 Well, I'm not so sure. Give me a geometric representation of a discrete metric space.

[@devalapar7878]

@@blvckbytes7329 You can't represent everything geometric. And geometric proofs are sometimes very complicated. Here he didn't really prove anything. He just said how it is. And algebra and analysis is often simpler. We usually take the approach that's simple to understand. If you thought there is a conspiracy against geometry, you are wrong. The real problem is math is one of the fastest growing disciplines and there is too much to teach. We learn actually a lot of geometry but it's advanced geometry like differential or algebraic geometry. The only thing I don't like is that geometry isn't really taught in schools anymore. I was very lucky. I was taught Euclidean and affine geometry in 3d before I entered university.

[@lambdaprog]

Here's the thing. I just got home and am extremely tired and definitely not ready to get my brain working on math, or anything demanding that is. So, I got hooked in the first few seconds and my brain started working again. Your skills, sir, are of another world.

[@Mathologer]

Never knew I had to power to bring people back from the dead. Great :)

[@AvoidsPikes-]

Same thing happened to me, but I was already at home.

[@philipsamways562]

I agree

[@romain.guillaume]

For the demonstration of A(X) + A(Y) = A(XY) Definitions : - we define A’(x,y) as the area under the curve between x and y (we notice that A’(1,y) = A(y) and that A’(a,b) + A’(b,c) = A’(a,c)) - we define f_p(z) the transformation of an area z by a factor p Let’s find an expression for f_p: Given z=A’(a,b), the top left hand corner of the area is mapped to: 1/p * 1/a = 1/(ap) (squeezing) which corresponds to the inverse function for a value ap (shifting). Same goes for the top right hand corner. We can conclude that: f_p(A’(a,b)) = A’(pa,pb) = A’(a,b) Let now prove the theorem : A(x) + A(y) = A’(1,x) + A’(1,y) = A’(1,x) + f_x(A’(1,y)) = A’(1,x) + A’(x,xy) = A’(1,xy) = A(xy)

[@TodosDuo85]

I'm almost 40 and back in school to become an engineer, not having touched math since my teens. This type of visualization is so helpful getting my old brain back on track! I have a feeling I'm gonna be replaying this video a lot 🤓

[@Mathologer]

That's great. Glad you like this video so much :)

[@mtb095]

Congrats and good luck on your degree!

[@peppybocan]

I think your visual derivations of the fundamental mathematical concepts is amazing. It helps the people who struggle with visualising the algebraic operations with something visually tangible and making connections which were forgotten by the maths teachers.

[@v2ike6udik]

Forgotten? Gooby, pls. World Wide Chewmmunist Degradation Agenda. 2023 should be the time to undersand why they did it, whitout being called a paranoid and crazy.

[@patrickirwin3662]

Exactly.

[@AvoidsPikes-]

Yes!

[@abramcz]

The fact that amyone on earth can connect to the web and have their mind expanded in this manner, for free, is wonderful. I am very grateful for this and other maths content on KZbin. Thank you!

[@andrewharrison8436]

Degree level teaching but with good graphics - agree, it's fabulous.

[@francimcdonald2721]

Exhilarating indeed

[@walterkipferl6729]

When I saw the 1+1/2 case for figuring out the base (and mentally extrapolated the rest of the proof), I was shocked- such beauty!

[@diribigal]

I thought I'd know all of this already, but was pleasantly surprised when you put the circle and hyperbola together and at the geometric proof of the e^α identity. Great video!

[@learn123]

Es hat wie immer sehr viel Spaß gemacht. Die Zusammenhänge der Funktionen geometrisch in dieser Form zu erleben ist wirklich der Hammer.

[@orterves]

This is definitely going on my rewatch list

[@Mathologer]

Great, and thank you :)

[@NittyLittyNiturzion]

I cant believe after all these years of calculus ive finally seen an intuitive argument for how lnx appears when integrating 1/x 😮

[@Mathologer]

Glad you enjoyed this one :)

[@vibaj16]

yeah, all the normal proofs just randomly throw lnx into the equations and it ends up working, but I'm just like "but why that particular logarithm?"

[@dennis5130]

I just discovered this channel a couple hours ago, and I have spent over 3 hours in a row watching your videos about sequence, derivatives, and of course trig and log .. it’s funny because I’m not really a math person because I feel like knowing something by just memorizing its efficient enough to apply it on a test but not into real life so that really doesn’t motivates me into doing maths. So it’s nice seeing a deep explanation of these concepts and not just something superficial.

[@idrisbalavakos]

I put on the captions for my deaf mother and we both appreciate not only the effort you put into them but also the smilies :)

[@Mathologer]

That's great, glad the captions are being put to good use. Always takes forever to put those together :)

[@nice3294]

I love all the visual intuition for why logs and e fall out of 1/x with all the squeeze maps. This video gave me so many aha! moments.

[@stingrayx2182]

A great way to look at familiar stuff. Top 5 Mathologer video of all time.

[@henridelagardere264]

The Professor's workroom looks like a boy's playroom, and vice versa; an anti-shapeshifting space, home to a brilliantly playful and playfully brilliant mind.

[@pauselab5569]

for the 1/x is a anti-shapeshifter if you squish the graph by 2 along the y axis, then each value of y gets mapped to y/2, while values of x stay where they are. the function that describes this graph is f(x)=(1/x)/2 = 1/(2x) now let's stretch this graph by 2 along the x axis each value of x gets mapped to 2x while y doesn't change so f(2x) = 1/(2x) and so f(x)=1/x the function of this graph is 1/x; same as before

[@3Max]

This is a good proof and it shows there is a bijection between the points under the y=1/x curve, but is any more care needed to show that this bijection is area preserving? Not all bijections are area preserving, but at the same time, it looks like the same shape is still covered. Although we are "pulling points from infinity", so maybe a limiting argument is required?

[@jonadams8841]

Thanks! Great little vid. I think I was trained that the hyperbolic sin is pronounced “sinch”. Of course, for me little use in daily life for hyperbolics except for catenary curves and there are tables for that 😂

[@AshishRanjan-jn7re]

I feel it's easy enough to understand that equation is xy=1. So if you scale x up by any factor, y must be scaled down by same factor and equation remains same.

[@patrickmcginty632]

Could you help me follow your logic near the end, and correct my logic where I get yours wrong? You say "each value of x gets mapped to 2x while y doesn't change so f(2x) = 1/(2x)" but if you start with the new function after the y-axis squish -- call it g(x) = 1/(2x) -- then stretch g(x) along the x axis, which would be g(2x) = 1/[2(2x)], don't we end up with 1/4x? Thanks for the help!

[@alexanderdaum8053]

​@@patrickmcginty632 When stretching g(x) along the x axis, you need to replace x by x/2 and not by 2x. I like to think about it like this: We want to get a new function g'(x) that is equal to g(x) stretched by a factor of 2. So the new function g'(x) should have the same value at twice the value of x, so g'(2x) = g(x). From this follows, that g'(x) = g(x/2)

[@grumpyparsnip]

Thanks for taking us on a stimulating mathematical journey! I love the animated proof at the end. It's almost like a poem.

[@johnchessant3012]

I paused right after you defined anti-shapeshifters and I tried to find one. I was only looking for finite-area ones though, so here's what I found: the ellipse x^2 + y^2/4 = 1, when stretched horizontally by 2 and squished vertically by 2, becomes x^2/4 + y^2 = 1, which is the same ellipse just rotated by 90 degrees. Every ellipse has one stretch/squish factor for which this works. Also, as a jaded know-it-all calculus demon, I was completely enamored by 1:50, 14:06, and 28:51. Really excellent!

[@Mathologer]

Good idea :) And you definitely picked the right bits and pieces to like.

[@santerisatama5409]

How about this point-free approach? Start from top of the exponent tower written as < > and/or > < > < > etc. Concatenating mediants is the familiar Stern-Brocot algorithm, check out yourself if and how you can find the Stern-Brocot connection in the generated strings. :) Any case, the character count of the rows, including the blanks, this pretty thing: 1^n+2^n+3^n The problem is, the claim that the area is infinite is mereologically inconsistent with the finite volume of Gabriel's horn. The problem seems deeply related with incomparable metrics, it seems. On the other hand coordinate system is by definition point-reductionism (in which Zeno-machine ontology nothing could move, hence everything would be anti-shapeshifting pure entropy :P), on the other hand proof-attempt utilizes geometric intuition of mereologically continuous intervals with metric of at least semantically mereological fractions. The point-free construction of Stern-Brocot type fractions hinted above gives a natural mereological metric of computationally increasing finite resolution, instead of assuming non-computational instantly infinite resolution. Intuition suggests that the (minimum mereological) logarithm 1^n+2^n+3^n from the shortest possible chiral symbol pair seed for string generation of Stern-Brocot type metric is somehow deeply related to the Gabriel's horn problem, as well as the question about logarithmic constant.

[@wyattstevens8574]

What would the general factor be for ax^2 + cy^2 + dx + ey +f = 0?

[@dominicellis1867]

@@wyattstevens8574it would be the gradient vector of f(x,y) = E(x,y,a,b). When we integrate a differential we get the average value over the region. This is the general constant factor on a manifold M denoted E*M(x,y).

[@wyattstevens8574]

@@dominicellis1867 I was just wondering: in the expanded form, is there a closed-form formula for that in A, C, D, and E?

[@DitDede]

What a pleasure! So many new perspectives that seem to have been just waiting to be noticed: 1. Looking at y=1/x as a shape invariant to stretch&compress 2. Defining ln(x) and e^x using the curve 3. deriving cosh+sinh identity from their geometric definitions (as opposed to the exponential identities definitions)

[@LilyOliver-r6m]

This is my boyfriend's favorite video. He watches it at least once a week.

[@WolfdogMusic]

"I can't be bothered to do the hard cases" in my day = "Left as an exercise." "I can't be bothered to do the hard cases" today = "Leave your thoughts in the comments."

[@Anytus2007]

Apparently the names of the hyperbolic functions are pronounced differently in different places! I had no idea. Perhaps this is a US vs. UK thing? In my schooling in the US, I have never heard anything other than COSH for cosh(x) and SINCH (homophonous with cinch) for sinh(x). The other functions I've heard more variation. For tanh(x) some people say TANJ like 'tangerine' and some TANCH. The most difficult one by far to incorporate, I think, is csch(x) wherein you can try to say COSEECH or instead bail out to '1 over SINCH.'

[@iabervon]

Likewise, but I think it's a bit ambiguous whether sinh ends with a "ch" sound or an "sh" sound. I think I press my tongue against the roof of my mouth a bit harder saying "cinch", but I don't know that it sounds any different.

[@excentrisitet7922]

Surprisingly even in Russian the same trick is used. Sinh is called shinus (шинус) and cosh - coshinus (кошинус). Letter «и» is stressed one in both cases.

[@MarkRawling]

In Australia, it's cosh and shine.

[@WarmongerGandhi]

Same. In the US, I've only heard it pronounced "sinch".

[@misterlau5246]

Yup. English is not my first language and I got into quantum grad at UCLA 😳 😟 ANTIMATTER or anti anything, USA teachers also had the different pronunciation of the "A" 😳😟😭 It was like cosh and sine withe ea ch or sinch Cosh was easier 🧐 Just like arcos 🤔 src ar.. 🤔 😬 🤣 What the.. Is a four vector? 🤔🧐 Tetra vector, amigo. Another.. Greek letters. Pie, psi, etc, now you pronounce the "" i" like in the alphabet 🥺? Oh, my SEENUS not sinus 🤓😅😅😅😅🖖🖖🖖 Zero was "O" "Z"

[@jakobthomsen1595]

The visualization for finding the base (14:30) to (16:30) is marvelous :D

[@Mathologer]

Yes, I am also pretty pleased with how this one came out :)

[@shortfatboy]

I am curious if this visualization can be attributed to @Mathologer. The limit is well-known, but I never saw the limit presented in this way before.

[@Mathologer]

@@shortfatboy Yes, I think I may very well be the first to present the derivation of this limit like this :)

[@tiusic]

15:50 The missing chunks look approx triangular, so I calculated the size of the missing triangles, and used that to find a correction factor for the (1+1/n)^n approximation of e. If you multiply that approximation by (2n+2)/(2n+1), you get a much more accurate approximation. For n=100, (1+1/n)^n has an error of 1.35%, but ((1+1/n)^n)*(2n+2)/(2n+1) has an error of just 0.001%

[@learn123]

(2n+1)/(2n+1)=1

[@tiusic]

@@learn123 oops, typo. Fixed.

[@Mathologer]

Good idea :)

[@JCOpUntukIndonesia]

20:58 All this time, I never knew that ln stands for _logarithmus naturalis_ always thought that it's just **natural** to use ln (pun intended) Also, another beautiful visual explanation from Prof. Polster Thank you.

[@jursamaj]

Yes, much of the famous terminology in math derives from Latin or German terms.

[@namanjain989]

As year 10 students were explicitly mentioned:- First the horizontal strech by a factor of 2, means that the input x was changed to x/2, hence plugging that in gives 2/x. The vertical squish divides that by 2, leaving us with 1/x.

[@vaanivijay6552]

Wait bro I’m a year 11 student can u explain that to me pls?

[@the_moist]

Vertical : multiplying the whole function by something, like 2f(x) Horizontal : multiplying x and only x by something, like f(2x) For a function like y=x you can't really make that distinction tho, so if you turn it into y=2x, that can be considered either a vertical stretch or a horizontal squish. If you have trouble remembering which is which, i usually imagine the graph is play doh, so vertical stretch is pulling it apart up and down and horizontal compression is pushing it in, but notice vertical stretching and horizontal compressing does the same thing Also notice that horizontal stuff works the opposite as vertical because as my teacher told me "x's (exes) always lie"

[@mikemcculley]

Great ways to conceptualize and visualize properties of logarithms. One of my favorite videos.

[@kwgm8578]

Always, fascinating! I followed, surprisingly, until the last example. That's when my brain went flat and would inflate no more, until patched with a second cup of tea. Why, at the end of each video, is there a smile on my face, and a chuckle in my heart? Perhaps they have much to do with your own gentle encouragement along the way with a chuckle here and a giggle there, however, I'm also quite fond of that sweet tune, pleasingly harmonized on a collection of folksy strings, that serves as a segway and ending to your videos. As always, thank you and I'll look forward to the next time. 😊

[@Mathologer]

All this makes you an ideal Mathologer viewer :)

[@Gunbudder]

i have never seen the bridge from conic to catenary explained this way! now i understand why so much effort went into trying to find the catenary as a conic section in early days of math

[@accountname1047]

This is how I was taught the logarithm definition in school, great video

[@ignacioperezlarrain4632]

Solid, it gave life to hiperbolic trignonometric that felt allways pretty blury. Thank you!

[@conrad5342]

It is nice to see this geometric approach to the topic. The result is not new to me, but the path 😊

[@gamerpedia1535]

For people in Calc, this will make sense given integration work with logarithms. The integral of 1/x is ln(x), and the integral is the area under a curve.

[@Mathologer]

Yep, I am coming from the opposite direction in this video. From basic geometry to calculus :)

[@ectoplasm12345]

When you showed the equal area segments on the hyperbola I finally felt like I "got" why it was a rotation. I was aware of the various taylor expansions and exponential related similarities but this was a really good thing to comfort my gut about it.

[@DrDeuteron]

It’s a boost

[@yinq5384]

Great video as always! 2:28 "1/x was hiding in a spinning cube" because a hyperboloid of two sheets is a ruled surface. 9:00 (Wish we could use A(X)=ln(X) here :P) Consider the region bounded by {x=1, x=Y, y=0, y=1/x}, after squishing and stretching it becomes the region bounded by {x=X, x=XY, y=0, y=1/x}, thus A(Y)=A(XY)-A(X). 25:40 We have the parametrization (cosh(a), sinh(a)) for x^2-y^2=1. But I never realized the parameter a is indeed the area.

[@JohnSmall314]

Excellent explanation. I especially liked the explanation of the hyperbolic trig functions. It's always puzzled me why they're called hyperbolic. I appreciate the work you've put into this. It must have taken hours and hours to get right. Well done.

[@Mathologer]

Yep, took quite a while to put this one together:)

[@RhythmnOfThought]

Your excitement is absolutely contagious! What an enjoyable watch!

[@Mathologer]

Glad you enjoyed it!

[@enbyarchmage]

This was one of the clearest mathologer videos yet. Everything worked. I loved it!

[@Mathologer]

Glad you think so. Maybe also check out part 2 (the next video :)

[@RobinHillyard]

OMG. I wrote my first (FORTRAN) program 55 years ago to find the solution for x = cosh(x). And yet I didn't know that cosh and the catenary were the same thing until just now! Thank you, Mathologer :)

[@OlivierMIEL]

Love you mathologer. Great video. I was particularly excited by the proof of the value for the base, it was my first time seeing it.

[@Mathologer]

Actually, you won't see this done the way I did it anywhere else :)

[@poutineausyropderable7108]

You have a nice argument for let g(n) = (1+1/n) And f(n) = g(n)^n You show a nice visual argument for the squeeze: f(n-1) < f(n) < e < g(n)^(n+1) So since you know e

[@vrixphillips]

you were right, that animated proof at the end was absolutely worth the wait! and thank you SO MUCH for explaining natlogs. I took trig in high school and in college and never understood where e came from or what a natlog was :/ this made sense! thank you!

[@nape6625]

I have discovered the "expressing angle as area inside the curve" idea myself when I first encountered hyperbolic sin and cos, but this video makes an incredible job of it and another AMAZING way to find e inside 1/x. Thank you!

[@howwitty]

Excellent introduction to logarithms, I liked the video! Thanks.

[@WikTerrOdoc]

Great video! I recently had a project related to hyperbolic trigonometry and it was really nice to see a completely different approach.

[@nourshagruni3027]

hi, i am student in grade 11 looking for math idea for my Math Extended Essay in IB , can you help me in this?

[@WikTerrOdoc]

@@nourshagruni3027 haha, that project I mentioned was actually a Math Extended Essay for IB which I wrote. I got a C so I'm not sure if I can help much but I can tell you my research question was How can complex numbers be used to derive trigonometric identities. One of my friends also wrote an EE on limits and their applications in math and physics, he got a B iirc

[@Gunbudder]

the length of this one is perfect, and i love the geometric proofs. the logical proofs get so boring! and its always helpful to confirm what you know by coming from a different starting place and taking a different path

[@antoniomonteiro1203]

I don't agree... The mathologer videos are always too short!... :D

[@beatoriche7301]

Nice work! Gives me some flashbacks to Paul Lockhart's lovely book _Measurement,_ which is like the one textbook that I, as a soon-to-be math teacher, actually really like; of course, I can't base my teaching on it because of these soul-crushing standardized requirements and tests, though (I guess I can at least use it for inspiration). One of my favorite related proofs when it comes to understanding the logarithm based on the graph of the function defined by f(x)=1/x that could have been mentioned, though, is a lovely visual argument that the alternating harmonic series sums to the natural logarithm of two. You restrict yourself to the little region from 1 to 2 and ponder the area under the graph - it's the natural logarithm of two, of course, but you can also approximate it with rectangles. If you use a 1x1 square, you're overshooting a lot, but you can remedy that by taking a rectangle out horizontally to remove the second half, where the overcounting is most egregious. You then add the smallest rectangle you can that still covers the desired area in that region, which has a height of 1/(3/2)=2/3 and a length of 1/2, so its area is a third. I think most of you will know where this is going, right? Our estimation, which still overcounts but is much more accurate, is 1 - 1/2 + 1/3. Now, we remove the second half of the first rectangle (area 1/4) and add a better fit back in that has an area of 1/5, and we also remove the second half of the second big rectangle we have (area 1/6) to get a better approximation with an area of 1/7. And so on and so forth - it's not too hard to convince yourself that this approximation strategy does indeed amount amount to evaluating partial sums of the alternating harmonic series, and the argument shows that they approach the logarithm of 2. Isn't that beautiful, even if it's tricky to describe everything just with words? This is exactly the kind of thing I want to teach eventually.

[@Mathologer]

Actually something like this will be in the next video, but even nicer than what you are describing there. Something to look forward to, promise :)

[@beatoriche7301]

@@Mathologer Lovely! Looking forward to it, then.

[@KingstonCzajkowski]

Ugh, soul-crushing standardized requirements... Paul Lockhart's "A Mathematician's Lament" is worth a read whenever you feel like no one understands you.

[@beatoriche7301]

@@KingstonCzajkowski It really is! Lockhart's Lament is such a beautifully eloquent treatise of what is fundamentally wrong with present-day mathematics education, and it hasn't lost a modicum of relevance in the past 20 years - if anything, it's become even more accurate. I really do intend to be transparent about this with my students - I might actually read and discuss Lockhart's essay with my older students for the first week of class. The relation of our culture to mathematics is a mess - education authorities don't know the first thing about math but insist on its profound relevance, so they inflict their bastardized idea of what mathematics is on unsuspecting students, making things really hard for well-meaning math teachers in the process; most adults in students' lives don't know a lot about math outside of painful school memories, so they basically treat it as a test of endurance; and finally, math teachers themselves generally don't know enough about math to be able to actually take a step back and question this soul-crushing regimen, as well as our culture's misunderstanding of the mathematical enterprise more broadly. And I include myself in that, by the way; I'm certainly not a math prodigy, and I feel like I know just enough to see that what we're doing clearly doesn't capture the mathematical spirit or work for the vast majority of students.

[@KingstonCzajkowski]

@@beatoriche7301 I'm so happy you're going to be a math teacher. We need more people like you.

[@Brightgalrs]

I liked the motivation and the path it took us down, but now I want to know if there are more of these anti-shapeshifters, besides the class we explored!

[@TheEternalVortex42]

If we have y = f(x) then to have the "anti-shapeshifter" property, we need that to equal f(x/c)/c. Thus f(x) = f(x / c) / c for arbitrary c. I'm not sure how to solve this for absolutely general functions, but let's say we assume our function has a Laurent series, then: Σ a_n x^n = Σ (a_n / c) (x / c) ^n = Σ a_n x^n (1/c)^(n+1) Comparing coefficients we get a_n = a_n (1/c)^(n + 1) for arbitrary c. This implies that a_n = 0 except possibly for n = -1. Thus we see that f(x) = 0 is a solution, as well as f(x) = A/x for any constant A, and those are the only solutions for "nice" functions.

[@Mathologer]

@@TheEternalVortex42 Very good :)

[@landsgevaer]

If it is about shapes, like that cat, not just functions, then any area bounded between two (aligned) hyperbolas works too, and so does any combination of such (aligned) areas. If the factors by which you stretch and shrink are limited to e.g. rationals, then all kinds of fractal-like dust qualifies too.

[@DrDeuteron]

@@landsgevaerso d3p/E from QFT, maybe

[@Christopherus3]

A wonderful video! By the way: because alpha is the arclength in sin/cos and the area in sinh/cosh the inverse functions are called arcsine/arccos and ar(ea)sinh/ar(ea)cosh.

[@maxfern]

I have had a wonderful math professor, he never watered down the theory but didn't get in more depth than we engineering students needed. So he called the inverse hyp functions Ar(gument). Per the comment above it is clear that they could and should be called Ar(ea).

[@InverseHackermann]

10:46 For negative integers, A(x^n x^-n) = A(x^n) + A(x^-n) = nA(x) + A(x^-n), so A(x^-n) = -nA(x). For rationals, qA(x^(p/q)) = A(x^p) = pA(x), so A(x^(p/q)) = (p/q)A(x). For reals, I'm pretty sure we have to assume A is continuous? For any real number c, we get a sequence of rationals q_1, q_2, ... approaching it. By continuity, the limit of A(x^(q_i)) is A(x^c). To compute this, we want the limit of q_iA(x), which is cA(x). If we don't assume A is continuous, I'm not 100% certain. My real analysis class has traumatized me to the point that I'm pretty sure there's a weird discontinuous counterexample lurking around the corner, but I can't find it.

[@MuffinsAPlenty]

Excellent work! And yes, you should be wary of discontinuous "logarithms" lurking in the dark :) The axiom of choice guarantees their existence (but that also explains why you can't find any of them - because they're non-constructive). Essentially, the multiplicative group of positive real numbers forms a vector space over Q (taking vector addition to be multiplication and scalar multiplication to be exponentiation). So the axiom of choice guarantees that there is a Hamel basis for the multiplicative group of positive real numbers over Q, and we can assume e is part of that basis. Similarly, the additive group of all real numbers is a vector space over Q, so the axiom of choice guarantees us a Hamel basis here, too, which we can assume contains 1. (Note that we can prove the dimension of both of these spaces over Q is |R|.) Then we can create a function which maps the Hamel basis from the multiplicative positive real numbers to the Hamel basis of the additive real numbers, choosing e to get mapped to 1, and choosing it so that some basis element d of the multiplicative positive reals is mapped to something other than ln(d). But a function defined from a basis of one vector space to elements of another vector space defines a unique linear transformation between the two vector spaces. Let's call this linear transformation L. We can then check by the definition of these vector spaces and the definition of linear transformations that it has all the algebraic properties of the natural logarithm. If a and b are two positive real numbers, then L(ab) = L(a)+L(b) because vector addition in the domain is actual multiplication but vector addition in the codomain is actual addition. For any positive real number a and any rational number c, we have L(a^c) = cL(a) since a^c is scalar multiplication of a by c in the domain but scalar multiplication in the codomain is actual multiplication. We also have L(1) = 0 since 1 and 0 are the identities respectively, and L(e) = 1 by choice. So we have the "properties" of a logarithm that you were able to pin down without continuity. But we have L(d) = something other than ln(d), so L is not the same thing as the natural log. (As an added bonus, I made sure L is bijective too!) But your argument pretty much shows that ln(x) is completely pinned down by ln(e) = 1, ln(ab) = ln(a)+ln(b) for all real positive numbers a and b, and continuity. So the only conclusion here is that this nasty L function cannot be continuous. And since it has the other properties of logarithms, it is in some sense a discontinuous "natural log".

[@Mathologer]

​@@MuffinsAPlenty Excellent work yourself :) Would be nice if one could somehow automatically collect the best comment at the top of comment sections.

[@davidquirk8097]

You've taken me back 39 years (!) to my days at Technical College and the seemingly endless amounts of Calculus theory that we had to learn and then had to apply to physical laboratory works it was great to predict the answers mathematically and then prove them on a test rig (we were taught to think in terms of 'analogues' so we could apply the theory equally happily to mechanical or electrical systems) Today my work involves less mathematics but it still stretches my brain and memory. I'm working on a problem at the moment that involves a lot of electrical relays and, after having spent three days going through electrical wiring diagrams, it has occurred to me that I need to recreate the original logic diagram for the system in order to identify the cause of the problem. I wonder if I still have a good chart template kicking around...

[@jamesmstern]

This video was a joy to watch.

[@chrisladd5855]

Nothing short of brilliant visual representation. Thank you for this!

[@allanjmcpherson]

That visual proof that e^a = cosh(a) + sinh(a) is really satisfying! I've always wondered how the exponential related to hyperbolas, i.e. why we call them hyperbolic trig functions. I'd seen the construction in terms of hyperbolas, and I'd seen the representation in terms of the exponential, but whenever I tried to find an explanation how these two were connected, all I got was, "That's just the definition." (Read: I don't know, and I've never thought about it.)

[@_unkown8652]

for the demo at 3:21: formally, "squishing" by k is really dividing the graph by k. Then, "stretching" is to set remplace x by x/k. Intuitively, you can understand this because of the fact that by dividing all the x by k, you basically slow down the growth of x, so it is just like stretching the x axis. Finally, you get 1/[k(x/k)] = 1/x

[@nathanbarajas9174]

I can’t get over how big the dude’s forehead in the thumbnail is.

[@abysmal7000]

find the indefinite integral of that forehead

[@bilkishchowdhury8318]

It's to fit his big brain

[@smotala11]

Absolutely deserves a follow! This was amazing. There are too many amazing channels on KZbin. It really is a golden age ot maths learning! How will I ever have time to watch them all!

[@Myrus_MBG]

One at a time. Never lose your curiosity :)

[@smotala11]

@@Myrus_MBG finished all of 3Blue1Brown's videos for now at least

[@wulfifi]

ngl, when i saw where the visual proof that the base of the "area function" is e was going i had the biggest grin on my face

[@T0NYD1CK]

Thank you so much for providing the "missing link" at long last! Many years ago I had to present some information about Fourier Transforms and signal analysis, so, obviously, sine waves were super important. With a friend of mine we pieced together everything we could have been taught about trig functions but never were. To our surprise, they had nothing to do with triangles but everything to do with circles. You can relate the meaning of the words sine, tangent and secant to lines on a diagram based on a circle. The worrying thing was that in a previous existence my colleague had been a maths teacher and I was an engineer so both of us had been familiar with trig functions for years at that point but we never saw the "big picture." Subsequently, I decided there ought to be a link to the hyperbolic functions. I discovered the area link but could never see a reason for it - until today! The "turning circle" did it for me. Thanks again.

[@Luizabf]

Simplesmente maravilhoso

[@georgykrikhan2503]

One of the very few channels I have subscribed to! This video saved to favorites. Keep up the fantastic work!

[@arikgr]

Great video, as always! Fun fact: in French, ln is usually referred to as "logorithme népérien" or "Neperian logarithm", from Scottish mathematician John Napier who created the first log tables.

[@greatquux]

Wow great video on a new way of thinking about what I thought were old and familiar formulas and identities!

[@Mathologer]

Glad you liked it!

[@Adityarm.08]

Amazing content, as always :)

[@polettix]

25:25 not sure why the relation to the area remains the same when moving to the hyperbolic functions. Additionally, using the same symbol "a" for the area before and after the animation is confusing, it seems like we're talking about the same quantity but it's not.

[@ClementinesmWTF]

I know that cosh has a pretty obvious pronunciation in English and that sinh doesn’t, but I’ve often heard that there is no one way of pronouncing sinh because of that. I was taught at least 3 different ways of pronouncing it, but none of them were “shine”. I’ve since opted to use “sinch” since it at least expresses the “h” being at the end and sounds similar enough to cosh to be obviously the related function cosh.

[@Mathologer]

Maybe time to switch to shine my personal favourite (as you can probably tell :)

[@ClementinesmWTF]

@@Mathologer boo!! I think it may be time to start the shine vs sinch vs shin vs sink vs singe vs skine wars! 😉

[@mehill00]

I’ve always heard “sinsh” or “sinch” for sinh in physics circles. I’m US based. While “shine” is nice, it seems that if we’re gonna use “shine” for sinh then we should follow the pattern and use “chose” for cosh but “chose” is only _n choose n_ letter off from another famous spoken version of an expression, which, I add parenthetically, exclaims its own importance each time it is written.

[@timharig]

There is a really big problem with that. It sounds far too much like the pronunciation for the sinc() = sin(x)/x function. As an electrical engineer, I see sinc() far more than I see sinh(). The difference between the pronunciation of the words sink and cinch can be subtle -- especially when heard from somebody with a non-English accent. So I have heard sinc() pronounced pronounced very similarly to cinch.

[@tobybartels8426]

​@@mehill00: You can use ‘shine’ and ‘coshine’ (‘co-shine’), taking ‘cosh’ as just an abbreviation of the latter.

[@skipper144]

Sometime before Christmas 2023, i started revisiting Trig functions then went down a math rabbit hole. I have been binging videos like this. In high school, we never truly covered log, ln, e. Merely memorized formulae. You have opened up doors for me to understand log/ln/e. Similarly in high school, hyperbolic Trig was avoided. I had no idea until watching this video that trig and logarithms were connected like this. Love the visuals and the in-depth explanations. Keep up the amazing work.

[@OlivierMIEL]

20:57 As a French, I was taught "ln" stands for "logarithme népérien" after John Napier, discoverer of logarithms (John Napier's name is often francizied as "Jean Neper").

[@Bushviking]

In Napier's time they publicized in Latin, so Napier was called Neperus. Hence "logarithme népérien".

[@TheOneThreeSeven]

omg when you said "even the jaded know it all calculus demons will like it, guaranteed" i was like "oh reallly?" and then you wrote down (1+1/2) and I was like OOOMMGGG. you were right

[@59de44955ebd]

concerning the one-liner: f(x) = 1/x => 1/c * f(x * c) = 1/c * c/x = 1/x Q.E.D.

[@Halopend]

I feel like 3:50 starts with the assumption of infinity to eventually prove it's infinite (when you say the 2 areas are the same plus a difference and the only way that can be true is if the areas are infinite). You have shown that the area can only be the same if the areas is infinite given the containment factor and the clear visual difference between the 2 states though, which is not a small thing...... I'm not following the next jump though as it seems to require either the infinite assumption or pre-existing knowledge that the areas are the same (neither of which can be determined visually as far as I can see since it requires an arbitrary segmentation to produce the tailfin meaning knowledge of the squish and stretch being unchanged no longer applies). It's possible that the squish and stretch concept though requires infinite area in order to produce a shape that remains unchanged (as otherwise how can one dimension not change when divided). I suppose if you start there than it does make sense. As you have the missing piece of info that allows you to assume an infinite area no matter how segmented. Think I get it now.

[@Mathologer]

It's mostly the flow of the story that I am telling and my target audience that dictates what I am saying here. I could also have said that because the shape squishes and stretches into itself it is "obviously" infinite :)

[@logiciananimal]

Galileo was one of the first to notice the "part equals the whole" weirdness with infinities. I find it interesting to contemplate how this then plays out in another puzzle he wrestled with - the shape of cables or ropes under gravity. (He got it wrong- he guessed that they were parabolic.)

[@carultch]

Before people knew the analytic solution to the shape of hanging cables under gravity, the architects would build models upside down, with hanging strings and fabric, and then study the shape to specify how to build arches and domes right-side-up. They recognized that since a string hangs with a purely axial load, and so should an arch be built in a shape so it also carries a purely compressive axial load.

[@Mathologer]

@@carultch I really have to do that video on the catenary I've been meaning to do for a long time :)

[@landsgevaer]

Fun fact: for a suspension bridge, if the weight of the bridge dominates that of the cable, you *do* get a -hyperbola- parabola.

[@carultch]

@@landsgevaerI think you mean parabola.

[@landsgevaer]

@@carultch Yeah, -typo- brainfart. Jeez. Edited... Thnx

[@DanielKRui]

Wow the proof at 15:00 is such a nice way of visualizing this limit; intuitively the sequence is increasing and always below e! Super memorable!

[@rgfs71]

This is a really nice visualization. Here’s another you may appreciate… draw a 45 [deg] line through the origin intersecting with f(x)=1/x and label these points B and B’ , with B being the intersection in the positive quadrant. Now draw a circle with diameter BB’ (i.e. centred about the origin with radius |B|). Pick a point, C, on the circle. Take the tangent to the circle at C and intersect it with the line BB’ to get D. From D, draw a perpendicular line and intersect it with the line B’C to get E. The point of intersection, E, will lie on f(x)=1/x and |CD|==|DE|. As a bonus, reflect f(x)=1/x about the circle to generate a very special shape 😉

[@Mathologer]

Very nice :)

[@misterlau5246]

The transformations of your cube are glorious 🥰🥰🥰🥰👏👏👏

[@erikstanton3908]

I've always pronounced sinh as sinch.

[@damiandassen7763]

3:20 if we consider f(x) = 1/x, then squishing by a factor of "n" is like dividing f(x) by n yielding f(x)/n, stretching on the other hand is dividing x by n, leaving us with f(x/n)/n. Filling this in gives us f(x/n)/n = 1/n * 1/(1/n *x) = n/n(n*x) = 1/x

[@JatinSanghvi1]

At 9:00 - For adding A(X) and A(Y), we widened the shape with width (Y-1) by a factor X to have a new width of X(Y-1). We then moved the shape's left side to X units on the horizontal axis. The right side hence touched X + X(Y-1) or "XY" units on the horizontal axis. Hence A(X) + A(Y) gave A(XY).

[@MarioFanGamer659]

How the introduction i.e. spinning Rubix Cube features _a lot_ of foreshadowing; the link between rotation (i.e. circles), hyperboles (which essentially are perpendicular to circles) and 1/x (which is a hyperbole rotated by 45°).

[@kappasphere]

3:09 Proof that the shape defined as "the curve y=1/x stays the same curve when you both stretch vertically and squish horizontally by a factor s": A curve y=f(x) stretched vertically by a factor v is just y=v f(x), and squished horizontally by a factor h is just y=f(h x). With that, we can show that in our case, y=s f(s x) still lies on the curve defined by y=f(x), because with f(x)=1/x, y=s 1/(s x) = 1/x = f(x). So the transformed curve is the same as the original curve, making it resistant to shape-shifting, and any section of the curve will turn into another section still on the curve when applying the transformation. 9:00 So you have A(X) and A(Y), and want to stretch A(Y) so that it fits on the right edge of A(X). You know that A(Y) has a height of 1 on the left and A(X) has a height of 1/x on the right, so for them to fit together, you need to vertically stretch A(Y) by a factor of 1/X (aka vertically squish by a factor of X). The bottom edge of A(Y) originally had the length Y-1 (because the edge goes from 1 to Y), but because we want the new shape to fit the 1/x curve, we also need to stretch it horizontally by X after having squished it by that amount. So the bottom length of the new A(Y) will be X(Y-1) = XY - X. The resulting total shape after adding A(X) and the transformed A(Y) together will fill the curve until the point where A(X) ends plus the length of the transformed A(Y), so in total, X + XY - X = XY. Per definition of A, this area is A(XY). But we also know that this area is just A(X) + A(Y), because the transformed A(Y) has the same area as the original. 10:47 Proof that it works for all integers: If log(XY)=log(X)+log(Y) for all (X, Y), then 0=log(1)=log(X/X)=log(X) + log(1/X), and therefore, log(1/X)=-log(X), and by extension, log(X^-N)=-log(X^N)= -N log(X). For the exponent 0, you can just use log(X^0)=log(1)=0=0 log(X). For rational powers of X, so for the rule p/q log(X)=log(X^(p/q)), we can take log(X^p)=(q log(X^p))/q =(pq log(X))/q=p/q log(X^q). Now in that equation, substitute Y=X^q (which implies X=Y^1/q), and you get log(Y^p/q) = p/q log(Y). This equation must hold for all Y, because any Y can be expressed in terms of X using the earlier substitution rule. For irrational exponents, you can argue that the logarithm must be continuous on any compact interval, as 1/x is continuous as well, and because the rationals are dense in the reals, that shows that the rule must also be true for any real number.

[@doyourclanwars]

This is incredible! You deserve much more subscribers

[@Pan_Tarhei]

Amazaing visualisation and explanation! You do great job!

[@phyarth8082]

In physics e number is unique sclalarity non-dimensional constant. For example x*y = x + y solution of equality expressed from independent t variable is: x = t + 1 y= t+1/t=1+1/t where (1+t - square hyperbola) if t=1 we get most famous math idiom easy like 2+2=4 or 2*2=4 or 2^2=4. It makes only sense if and only if x*y = x + y are non-dimensional unit, because x*y is [Area unit], x+y is [Length unit], thus definition or identity of exponents multiplication is e^x*e^y ≡ e^(x+y). And x and y must be non-dimensional units not to distort physicals reality. e - number is hidden scalarity unit or shapeshifting constant of other units, we so preoccupied with Euclid straightforward proportionality and straight linear symmetry but our World is complex and non-linear.

[@sdsa007]

very good entertainment! thank you, for the excellent visuals and calm music selection! this is one of my favorite math videos!

[@Mathologer]

Glad you enjoyed it!

[@rnoro]

When I saw your fascinating animation, it reminded me of a monograph I read in the Little Mathematics Library when I was young. Those were some very nice little math books! I am moved a lot by your work! One of my friends once said that people are moved by art and music, but why not science and math? Now I understand what he meant!

[@Mathologer]

This makes me very happy :)

[@soyuzssr]

Brilliant !! This was precisely the video I had been waiting for all my life !!!

[@زكريا_حسناوي]

الفكرة أنّه إذا كان لدينا قطعين زائدين يحصران بين مقاربيهما نفس الزاوية فسيكونان قطعين متشابهين، تحويل السحق والتمديد لم يغير الزاوية بين المقاربين ولذلك صنع قطعاً زائداً مشابهاً، لكن إذا كان هناك شكلين متشابهين يمتلكان نفس المساحة فسيكونان متطابقين، هذا يثبت الخاصية في بداية الفيديو

[@koendos3]

My brain exploded out of excitement at 14:44 when I realised your argument might be the most beautiful one in existence!😂

[@Mathologer]

Glad that someone agrees with me in this point :)

[@weatherinme6882]

Your laugh though. Its too contagious! Also I greatly appreciate this explanation!

[@Mathologer]

No, because THEY don't understand themselves :)

[@HarryFortyTwo]

Perfect pedagogical presentantion of mathematics! Showing it a the wonder it is, showing the magic, lighting that spark of fascination - thats how schools should do it. i enjoy your videos massively even if i have a pretty solid maths background from physics; but many of your approaches via geometrical concepts give me a fresh perspective. Fun and enlightening!

[@Mathologer]

Glad you enjoyed it!

[@spacelem]

14:30 Wow... I remember during my maths degree we went through the proof that lim{n->inf} (1+x/n)^n = e^x, during some long tutorial session, but that visual argument was like a mind blown moment!

[@ttd972]

The animated proof at the end was really nice

[@snowob]

this is the first one of your videos i've been able to follow in a while

[@Mathologer]

Maybe not the last one, but the one before that "Powell's Paradox" should be very accessible. Maybe give that one a go :)

[@wegwang7795]

Each time I open a mathologer video,its maginficence hits me.

[@MattMcIrvin]

9:03 : To get A(a) + A(b), we will have to squash the A(b) shape by 1/a, which means we stretch it out by a factor of a to retain the same area. But the width of this shape is actually b-1. So we get A(a) + A(b) = A(a + a*(b-1)) = A(a + a*b - a) = A(a*b).

[@aradhyajain5355]

3:23 equation is xy=1. Therefore when you stretch and squish by a factor k (really just stretching one of these by a factor k and the other by 1/k) the product remains the same.

[@frankied.2828]

Finally, just what we needed. AW, I wish I had this in highschool, it really updates your mindset and philosophy.

[@vinod.tanwani]

Too beautiful. Keep making such videos

[@deinauge7894]

28:14 but what if you optimize the chain in a way, that at every point it is just strong enough to not break under its weight. This way, it could be thin and light in the middle, but gets stronger and heavier the further out you go. Interestingly the result involves the normal cos... it has the form -ln(cos(x)) (insert some constants for a more general result)

[@Mathologer]

I didn't check but definitely an interesting idea. I like it :)

[@deinauge7894]

i thought about that on a bike tour a few days ago. Derivation and solving of the differential equation without writing anything down, so no guarantees ;) and i set the strength proportional to thickness^2 and therefore to (mass/length).

[@deinauge7894]

@@Mathologer cosh makes a comeback even here: the chain strength at distance d (along the chain) from the midpoint is cosh(d). And this makes sense, as a dangling chain with the same optimization has an exponential profile.