Why hyperbolic functions are actually really nice
HTML-код
- Опубликовано: 3 июн 2024
- Keep exploring at ► brilliant.org/TreforBazett. Get started for free for 30 days - and the first 200 people get 20% off an annual premium subscription!
Today we unpack everything to do with hyperbolic functions. In calculus we often see an analytic definition of hyperbolic cosine and hyperbolic sin expressed as exponential functions. But why? Analytically, this is just asking for the even and odd components of e^x. However, we can connect this geometrically to hyperbolas, and specifically to the area enclosed by a hyperbola.
0:00 Even and Odd Functions
2:24 Analytic Definition of cosh and sinh
3:19 Graphic cosh and sinh
3:53 Taylor series and derivatives
5:12 Hyperbolas
6:42 Defining trig functions geometrically
9:22 Defining hyperbolic trig functions geometrically
10:45 The geometric and analytic definitions are the same
14:03 Euler's Equation
15:06 Brilliant.org/TreforBazett
Check out my MATH MERCH line in collaboration with Beautiful Equations
►beautifulequations.net/pages/...
COURSE PLAYLISTS:
►DISCRETE MATH: • Discrete Math (Full Co...
►LINEAR ALGEBRA: • Linear Algebra (Full C...
►CALCULUS I: • Calculus I (Limits, De...
► CALCULUS II: • Calculus II (Integrati...
►MULTIVARIABLE CALCULUS (Calc III): • Calculus III: Multivar...
►VECTOR CALCULUS (Calc IV) • Calculus IV: Vector Ca...
►DIFFERENTIAL EQUATIONS: • Ordinary Differential ...
►LAPLACE TRANSFORM: • Laplace Transforms and...
►GAME THEORY: • Game Theory
OTHER PLAYLISTS:
► Learning Math Series
• 5 Tips To Make Math Pr...
►Cool Math Series:
• Cool Math Series
BECOME A MEMBER:
►Join: / @drtrefor
MATH BOOKS I LOVE (affilliate link):
► www.amazon.com/shop/treforbazett
SOCIALS:
►Twitter (math based): / treforbazett
►Instagram (photography based): / treforphotography
In school we are rarely ever taught the connection between the hyperbola and sinh(x), etc...Very interesting.
It’s so weird to be omitted!
@@DrTrefor I first stumbled on the term, when my textbook used the Gateway Arch in St Louis as an "example" of a parabola, with a fine print note that it's really a hyperbolic cosine. Playing with my graphing calculator, I attempted every combination of hyperbolas and cosines I could think of, like 1/cos(x) and cos(1/x), and couldn't find anything resembling it. Eventually learning it for real, I figured out on my own what properties hyperbolics have in common with standard trig, and could connect the dots on at least that part of its namesake.
@@DrTrefor "It’s so weird to be omitted!" It seems they do it on purpose to make math boring...
@@DrTrefor I think it's the fact that we never learn the area approach you showed in terms of relating cos and sin to the unit circle, which makes it unnatural to think of the hyperbolic parametrizations cosh and sinh in the same way.
In my case, in school we are [never] ever taught the hyperbola and sinh(x), etc...
I always thought hyperbolic functions were just some weird made up versions of regular trig functions. I didn't realize how intuitive and natural they are.
In some regards, the hyperbolic functions are more natural than the circular functions ('circular' is a more appropriate adjective to use than 'trigonometric').
Cool integration trick somebody taught me: if you’re integrating some gnarly function over some interval that symmetrically straddles zero (say, between -1 and +1), split the integrand into even and odd functions and see if the even function is more amenable to analysis. This is because the contributions of the odd function will cancel out and can be ignored. EDITED TO CORRECT ERROR THAT WAS POINTED OUT.
Love that trick
It is the other way. The integral of an odd function over a symmetric domain is 0, not an even function.
@@ethanbottomley-mason8447 You’re right, my bad, I corrected it. Thanks for pointing it out.
You still have an error.
@@oqardZ Can you rephrase it appropriately please?
Love it. I'm currently working on my dissertation, which heavily involves the complex exponential function, and cosh seemed to appear out of nowhere. This helps make sense of it, especially how cosh and sinh come from the real part in the same way cos and sin are in the imaginary part.
This really is a superb introduction to hyperbolic functions. All of the key ideas in 15 minutes explained perfectly!
Glad you think so!
I also made a video on the subject but with detailed computations of the integral. Check the video on my channel for more details:
video title: "he Geometric Definition of the Hyperbolic Functions, and Derivation of their Formulas"
True. Best intro to the topic I've ever seen.
I've somehow managed to never have a class on hyperbolic functions even though they show up occasionally. This video is mind blowing and really puts together so many disparate puzzle pieces for me. Truly incredible work!
Glad it was helpful!
It's these kinds of videos that make mathematics actually interesting.
Wow I'm in my first engineering year and even the professors never explained it like that
Really appreciate the amount of work you've put in this video!
Thank you for this video. Hyperbolic trigo is not even taught in schools where I live so most people don’t even know they exist even until they graduate from high school.
The complex relation between the regular and hyperbolic trigo functions also explains the similarity between their derivative properties, and their taylor series.
The taylor series for sinx and cosx have that alternating factor. The hyperbolic functions have the exact same terms just without the alternating.
That's really, really awesome. I was wondering recently why these functions were called "hyperbolic". The analogy with circle and sin and cos is great!
This is crazy! We were never taught that in school. It makes so much sense
It’s weird how hyperbolic functions are taught imo!
If you want a further generalization, look at geometric algebra. It explains how you can interpret i as an oriented area, and generalize the exponential to operate on oriented planes in 3d space. This provides a nice encoding of rotations (quaternions).
Like that approach starting with odd and even function, easily one of the best video on hyperbolic functions
The way you speak every topic is really heartwarming.😊
The way i first found hyperbolics was when i was curious on what cos(ix) was, so i used the maclaurin expansion and found it wasthis cool, and surprisingly real valued, mix of e^x and e^-x. It was only much later when i realised that was infact cosh(x). I love hyperbolic functions man
Note: this video starts with the analytic definition and proves that it works with the geometric one. But it's possible to go the other way around!
Let's start with cosh(a) and sinh(a). We know nothing about them other than these:
- The point (cosh(a), sinh(a)) is on the hyperbola x² - y² = 1.
- The area traced out by this certain region is a/2.
Notice that just through the geometric definition it's already possible to deduce a few identities.
First: that cosh(a) is even and sinh(a) is odd. Just flip the area upside down. The x-coordinate stays put, and the y-coordinate is negated.
And the other important one: cosh²(a) - sinh²(a) = 1.
(cosh(a), sinh(a)) is a point on the hyperbola, so it should satisfy x² - y² = 1 by definition.
The next step is to verify the integral stuff. It's the same process in the video, except we get stuck here:
(1/2)cosh(a)sqrt(cosh²(a)-1) + (1/2)ln|cosh(a) + sqrt(cosh²(a)-1)| - (1/2)cosh(a)sinh(a)
If there is a god then this better be equal to a/2.
Here we can use an identity from earlier, just rewritten a little:
cosh²(a) - 1 = sinh²(a)
Then the above result simplifies and cancels into
(1/2)ln|cosh(a) + sinh(a)| = a/2.
A little more algebra and we get
cosh(a) + sinh(a) = e^a.
Here we can use our other identities: cosh is even and sinh is odd. We're forced straight into the analytic definition:
cosh(a) = (1/2)(e^a + e^(-a))
sinh(a) = (1/2)(e^a - e^(-a))
Oh. And before you get suspicious about the whole cosh(ix) = cos(x) thing, plug in ix into the definition of cosh.
Then cos(x) can be written as
(1/2)(e^ix + e^(-ix)), and sin(x) as
(1/2)(e^ix - e^(-ix)).
Euler's identity makes things work out nicely in the end.
Which means cos(x) = 2 has a solution, and it's i*arcosh(2).
And also means that sinh(i*2pi) = 0.
Not sure if it's possible to take the derivative of cosh(x) and sinh(x) without first finding the analytic formulas but considering it's possible with cos(x) and sin(x) I assume it requires some squeeze theorem
Love this!
This is a flawed analysis. In actuality, it is not possible to derive analytic formulae for cos and sin from the geometric definition alone, which is why formal proofs involving cos and sin use their analytical definitions and not geometric definitions. You can derive the geometric definition from the analytical definition, but not the other way around. This is not a coincidence: if you start with the axioms of Euclidean geometry, deriving the axioms of real analysis is impossible, but you can derive all of the axioms of Euclidean geometry from the axioms of real analysis. Geometry is grounded in analysis.
Very nice presentation, thank you.
This is such a clear explanation of hyperbolic functions! What a perfect timing too, since I was wondering about them after my multivariable calculus professor briefly mentioned them in lecture a few days back, but never bothered to go over them in detail since they were irrelevant.
Glad it was helpful!
That was amazing connection. Thank you
4:40 in to the video. So cool to see the Taylor series expansion with sinh and cosh pointed out. I realized that if you take the derivative of any term in the expansion, you get the term to the left of it. It makes the derivative obvious. Blew me away. Thanks!
Glad it helped!
Great video! I want more videos continuing explaining this now!
Thanks! Will do!
Great presentation. You may want to expand this presentation to include RF transmission line theory, and the associated hyperbolic function utilization to solve those equations.
Hi Dr. Bazett!
So cool!
That's by far the best explanation of hyperbolic functions I have ever seen.
All the others seemed ad hoc. The properties were proved, but it was never explained why the functions were considered in the first place.
Everything in your video was very well motivated, thank you.
That''s exactly what I was going for, thank you!
Most beautiful math explanation
Very nice indeed! I wasn't aware of the geometric definition of the hyperbolic functions. Whilst the use of areas to define the trig functions is not quite so natural as using angles, the analogous result for the hyperbolic functions is really quite satisfying.
It's worth noting that angles can't work to parameterise the hyperbolic functions as they aren't periodic, so we need a parameter than can go off to infinity without repeating points on the curve. Angles don't work for this, but areas fit the bill perfectly.
Angles are not a useful quantity, but arclength still is, and the two are equal when it comes to circles.
At 2:18, i think it would have been relevant to mention that this decomposition is unique, especially for the part with taylor series. Aside of that, good video, like always
Great point!
Less known is that lχ| is alctually the arc length of the hyperbola from (1,0) to (cosh χ, sinh χ) when the ty-plane has the geometry of special relativity wherein, given t > y, the time elapsed along the line segment from (0, 0) to (t, y) is the square root of t^2 - y^2 (with time unit and light speed both set to 1 for simplicity). Hyperbolic angles are largely analogous in this context to circular angles in Euclidean geometry.
I like them because they're like circular trig functions, but stretchy!
In UK further maths A level we learn Osborn's rule where the hyperbolic trig functions act the same way as normal trig functions in terms of identities etc. In the ultimate part you basically justified why this is so. It even now makes sense why you have yo flip the sign of the product of two sines as it is i^2.
Calculating the area of A directly is relatively easy as well. Just parametrize the points in the area as r(cosh(t), sinh(t)) with r in [0, 1] and t in [0, a], then the jacobian is r(cosh(t) * sinh'(t) - sinh(t) * cosh'(t)) which happens to cancel to just r, so integrating f(r, t) = r in the rectangle [0, 1] x [0, a] we get a/2
U could also have talked about the explicit formula for the inverses
of Cosh and Sinh !
Amazing and the best introduction to Hyperbolic I was thinking about hyperbolic functions a day before or so and this video came out! Coincidence ?
Oh cool!
excellent video❤
Thanks! Good explanation …
lovely video!
I mean hyperbolic functions are called sinh/cosh has to have a reason and there has to be a connection between sin/cos and sinh/cosh. This video helped me understand it. 👍
great vid!
Love you Prof!⭐
Sir please come up with a series on Numerical methods for ODE & PDE.
What’s really interesting is that you can change the hyperbolic version of Euler’s formula into a two-dimensional analogue to the formula by using j, so that e^jx = coshx + jsinhx, where j^2 = 1 (instead of -1). It results in the split-complex plane, which has some weird geometry, like distance being equal to the square root of x^2 *minus* y^2
Superb !! 👍
Hello sir, been subscriber to your channel since sometime..love the content..thanks for uploading.. Lots of love and appreciation from india 🇮🇳
Thanks so much!
You are a great math sorcerer.
Also sir, considering that youre a maths professor.. Could you please make a video on statistics for machine learning
That’s not really my area but I’m interested…
@@DrTrefor me too sir, and I'm sure with your style of teaching,with such clear explanation and beautiful animations.. People would love it!
Excellent.
I wish I could give 2 thumbs up to that great video. Such great content!
Thank you!!
Enjoyed the math, but also - what a great T-shirt!
Thank you! Merch link in description lol
Thanks!
hey thanks so much!
14:28. One little thing I want to point out is that we don’t know yet that this is necessarily true, for example, cosh(ix) could have an imaginary component which would make this comparison faulty. The statement made is in fact true and you can figure that out by representing sin and cos in terms of the exponential or by looking at the tailor series of the functions. Point is, statement is right but the reasoning given is faulty. Otherwise the video is great and gives a good intro to hyperbolic trig
where was this when I was in my first year of astrophysics TvT still I really enjoyed this video
Thanks
We finished Hyperbolic functions in A Level Further Math and this video is exactly what I needed. The visuals are so much help as well as the plethora of analogies with other topics, thank you so much🥹🥹🥹
Glad it helped!
As someone with A levels 55 years in my past I remember the words Cosh and Sinh and hyperbolic functions. Perhaps (no defiinitely) had I watched this video back then the dust would not have settled so thickly on my memory.
So my congratulations on being young enough to have timely access to this resource.
Subscribed 👍
Make a playlist about complux analysis please😢
I can't believe I finally got an explanation about what the actual fuck a sinh(x) is months after I was supposed to write an exam over it at uni by it randomly stumbling into my youtube feed
Cool, like this video already and i haven' t even see all of it:)
What is the aplication to create the video?
It’s really worth mentioning that e^(ix) = cos x + i sin x is not a definition. I honestly dislike e^x notation in complex numbers because #PowersAreComplicated in complex numbers (for exponents that aren’t natural numbers). Fact of the matter is, what is meant is the application of the exponential function exp, defined as exp(x) = 1 + x + x²/2! + x³/3! + …; this definition works fine on complex inputs as well. The powers in this series are not complicated, it’s just repeated multiplication. In my Analysis I class, we have exp(ix) = cos x + i sin x by definition, because sin and cos were defined by this equation: cos x = Re(exp(ix)) and sin x = Im(exp(ix)).
Actually, that is not correct. Open up just about any calculus text, and it will state that any "proof" of Euler's identity is not accurate. Rather, it is a definition motivated by the series expansions of sinx, cosx, and e^x with x = i*theta.
I strongly agree! I wish more mathematics educators on RUclips took note of this.
Great!
12:17 variable “t” is introduced out of nowhere and gets substituted for as though it was x later, is that a mistake, should that be x instead?
Oh yes, thank you! I forgot whether my dummy variable of integration was t or x, it doesn’t matter as long as it is all x or all t.
If we habitually moved close to light speed, this would be so intuitive
Alright. The correspondence between hyperbolic and trigonometric functions when multiplying x by i was very cool. I was not aware of that fact, but your explanation makes it seem so trivial.
The problem is... I watched this video at 1am and should be going to bed. Now Im sitting here with a notebook playing around with these functions.
Why do you do this to us mathematics!?
At 6:15, x=tan(theta) and y = sec(theta) gives x²-y²=-1. You said it the other way around.
Oh right, thank you!
That's the most beautiful thing about mathematics, isn't it?
Your new Subscriber
For special relativity this is a God sent
12:32 t should be cosh(a). From-to notation was a little bit messy. x = 1 to cosh(a) applies to both terms.
Professor , in 6:14 , x = sec(theeta) and y = tan(theeta) isn't it ?
what kinds of tools do you use to generate such plots/videos?
This is all just Desmos
sec^2(θ)-tan^2(θ)=1
That is cooler than what you have. It leads to the cone you see in my thumbnail. Which can probably be used to do physics. I don't know if it would be better, but I'm pretty sure it can measure change in energy levels.
The functions cosh and sinh are much more convenient to work with than sec and tan.
@@angelmendez-rivera351 have you noticed that the sum of sec(arctan(x)) and tan(arctan(x)) is quadratic? If x is a complex number, you're in C^2 territory. If you're working in 3d, that's C^3. I've seen people do RUclips videos on C^2 with magnetism and relativity stuff.
I wonder about their practical application, one is free hanging wire or rope (not a chain bridge), other is pursuit curve. Third is Mercator projection I think. But I never used these function in my live and they seem to be on every scientific calculator and even some advanced slide rules. All I know is they are solution to some differential equations where second derivate is the same as function. For sine it's 4th derivative and for e^x first one. But all youtube videos are about abstract concepts or identities. There must be some motivation why they exists.
The fact that you have never used these functions in your life says absolutely nothing about how many applications they have. Also, this video explains precisely why these functions exist.
Thanks for the interesting video! Is there any way to visually verify the equation "cosh(i x) = cos(x)" in desmos? I know desmos doesn't have complex numbers, but you can just augment desmos by adding any required operation, like multiplying complex numbers m(P,Q) = (P.x Q.x - P.y Q.y, P.x Q.y + P.y Q.x).
Intuitively, we might notice that when we compare the equations of the circle and the hyperbola, x^2 + y^2 = 1 and x^2 - y^2 = 1, that changing the y^2 from positive to negative is the same as multiplying y by i. Think of what this means if we consider a "circle" on C^2, where x^2 + y^2 = 1 for a pair of *complex* numbers x and y. We can see that the cross section across the real components of x and y of this hypercircle is a circle, and the cross section with real x and imaginary y makes a hyperbola. Thus, if our trigonometric functions extend from normal circles over R^2 to complex circles on C^2 in an intuitive fashion, we should expect this sort of identity to fall out.
@@Keldor314 Hmmm, I think it becomes rather complicated to visualize functions from C^2 to C. But I've noticed that I can verify the equation visually by plotting both cos(x+iy) and cosh(x+iy) at wolfram alpha, restricting attention to the real part of both plots and verifying that the cross section with the X=0 plane for cosh(x+iy) matches up with the cross section with the Y=0 plane for cos(x+iy), which seems to confirm the identity cos(x)=cosh(ix).
Let me just ask one question ?
Why don't we take the other part of the hyperbola to define the hyperbolic sine and hyperbolic cosine function ? Is the results the exact same as you obtained by considering the other part of the function?
Hyperbolic cosine comes from the x-position of the point on the unit hyperbola. Hyperbolic sine comes from the y-position of the same point. This is for a hyperbola that opens to the left and right, on the standard x-y plane.
at 9:35 what does a in cosh(a) sinh(a) geometrically represent? i still dont get it. a is not an angle, but an area?
examples of conversation
A. in circle, i found an enemy, rotate your gun 30 degrees, then you get the enemy
B. in hyperbola, i found the enemy, change your WEIRD area to 3.7/2 then you find an enemy? before you measure the required area, you will be probably already dead by the enemy or you will be 100 years old. is it possible to assess "a" as a function of vertices or foci and etc
started sounding like Fourier series
Fourier series is the end boss:D
Great
And the "Hyper-Fourier" transform for all this. Maybe another video?
Dear Dr could you please proof lambert w function formula w(xe^x)=x
There is nothing to derive: this formula is the definition of W.
I NEVER learned this in school. I had to research this independently... I don't know why they omit the derivations smh
Is there also something like parabolic trig functions?
Why _YES! INDEED!_
While ultimately unnecessary, I'm going to quickly define a symbol j such that j² = 1 (1 satisfies this equation, but much like how i isn't ℝeal, j doesn't have to be either.) Much like Euler's formula,
exp(jϕ) = cosh(jϕ) + sinh(jϕ) = cosh(ϕ) + jsinh(ϕ)
And as a refresher:
exp(iϕ) = cosh(iϕ) + sinh(iϕ) = cos(ϕ) + isin(ϕ)
Now that we're refreshed on the similarities of their structures, lets now define ε such that ε² = 0 (again, it need not be ℝeal even if their is a ℝeal number that satisfies it).
Here, it's easiest to use the Taylor series exp(εϕ) = 1 + εϕ + ε²ϕ²/2 + ... Since ε² = 0, every term past that is 0 leaving us with
exp(εϕ) = cosh(εϕ) + sinh(εϕ) = 1 + εϕ
Yup. Parabolic cosine is the constant _1,_ and parabolic sine is the identity function. Wrap your head around _that._ Then again, if they should make sense if you consider the small-angle approximation. ε can be thought of as an angle (hyperbolic or elliptic) that's infinitesimally small.
The main thing that makes it questionably "parabolic" is that its "unit circle" isn't actually a parabola, but rather a pair of vertical lines as x² = 1. This can be found using the conjugate formula for the magnitude of a complex number, generalized to a hyperbolic or dual number (the official names for multiples of j and ε added to ℝeal numbers).
|x + yi|² = (x + yi)(x - yi) = x² - i²y² = x² + y²
|x + yj|² = (x + yj)(x - yj) = x² - j²y² = x² - y²
|x + yε|² = (x + yε)(x - yε) = x² - ε²y² = x²
@@angeldude101 Wow, I can't believe I've never thought of this before. That's amazing.
@@angeldude101 I would argue that the equation not being a parabola makes those functions not parabolic at all.
@@angelmendez-rivera351 They're more "flat trig" than "parabolic trig", but some might argue they could be called the latter.
@@angeldude101 Calling it flat trigonometry makes far more sense. I cannot get behind calling it parabolic trigonometry.
This is where I think we need slow down , really slow down just like Fourier and Laplace transformer
Something feels so circular about this and many other video that does similar things with Hyperbolic trig. Is it circular logic?
I love your shirt.
Thanks! Merch link in description lol
Hi, I do not get 14:30 argument. like how 2+3 == 4+1 can result to 2=4 and 3=1 !!!
12:30 i think there is a t instead of a x, else i dont understand :D
6:12
Tried plugging it in and it's -1.
So you had it backwards. x should be sec theta and y should be tan theta.
I just changed my major from Finance to Maths!!
Cool!
Holy moly that was some sexy af math
So that’s where that function from this week’s tutorial comes from 😅
I should rename this channel “stuff I wanted to do in tutorial but didn’t want a revolt”:D
Muito bom. Tô pensando em fazer meu TCC em funções trigonométricas de números complexos e suas relações com funções trigonométricas hiperbólicas.
I also made a video on the subject but with detailed computations of the integral. Check the video on my channel for more details:
video title: "he Geometric Definition of the Hyperbolic Functions, and Derivation of their Formulas"
Can we call y=x a parabolic sine, and y=x² a parabolic cosine? 🤔
I believe the parabolic cosine is the constant function 1.
Remember folks, radians, not degrees.
Could you teach me where you bought that fucking beautiful T-shirt?
Haha it’s my merch link in description!
Thanks for your video! You slow down your speech. You are running now! Just speak in a normal speed as a lecturer would do in his classroom.
Leibniz's calculus > Newton's calculus
The complex part seemed like the biggest deal
🤠
Why do you only deal with the real numbers?
This gets a lot more general when one proceeds to complex numbers, and above.