WoW! I haven't been this blown away since when I was shown Euler's identity!
@fernandogaray16816 жыл бұрын
I love this kind of videos. I love all the proof videos! Thanks!
@blackpenredpen6 жыл бұрын
Fernando Garay thank you
@andresxj16 жыл бұрын
I've been a Brilliant member for a year and a half now, and it all began with your special offer. I'm delighted with the app! I've learnt a lot and I've enjoyed it so much! So thank you for introducing me to Brilliant and thank Brilliant for sponsoring you!
@blackpenredpen6 жыл бұрын
Thank you Andy! Glad to hear that you like it!!!
@nomadr13493 жыл бұрын
this is by far the best take on hyperbolic functions I found on youtube so far. And I looked far and wide too!
@plaustrarius6 жыл бұрын
looking at the series expansions for exp(x) cosh(x) and sinh(x) is what really drove this point home for me.
@retired55486 жыл бұрын
complex relationship well played, good sir, well played
@idolgin776 Жыл бұрын
I've been fascinated by these patterns for a while, and yours is an excellent explanation. Thanks!
@axelreispereiravaz16996 жыл бұрын
I always asked myself why the hyperbolic trigs functions and the complex trigs functions looked so similar. Even my teacher didn't showed this relation. Now i have my answer ! Thanks BPRP !
@angeldude1013 жыл бұрын
I actually discovered this when I noticed that cosh and sinh had a structure similar to the inner and outer products of geometric algebra, which are defined as the symetric and antisymetric components of the full product. But the inner and outer products are usually defined with sin and cos... along with i. This is what led me to realize the relation between them and the real and imaginary / even and odd parts of the exponential. The only reason one relation is x^2 + y^2 = 1 and the other is x^2 - y^2 = 1 is because the imaginary factor of y flips the sign when squared. It felt so awesome to find that on my own. Now I kind of want to make a visualization of the complex exponential's even and odd parts to try and get the hyperbolic and spherical trig functions to appear on different axes of the same graph.
@theomegaspec7923 Жыл бұрын
Very interesting. I was wondering about the relation between the hyperbolic trig functions and the complex definitions of the trig functions after seeing one of your videos, and you explained these concepts so clearly.
@aidan88586 жыл бұрын
cos(it) + cosh(t) = cosh(it)
@cringy7-year-old5 Жыл бұрын
that implies cosh(t) = cos(t)/2
@penguincute3564 Жыл бұрын
So 2cosh(t) = cosh(it)?
@penguincute3564 Жыл бұрын
@@cringy7-year-old5that is seriously wrong…
@Questiala124 Жыл бұрын
Cos(it)=0?
@kennethx78013 жыл бұрын
An easy way to remember this is that, e^(ix)=cosx+isinx on one hand, on the other, e^ix=cosh(ix)+sinh(ix). Match the even part of one side with the even part of the other side, and do the same with the odd part. You get that cosh(ix)=cosx and sinh(ix)=isinx. Now evaluate these functions at x=it and you get the rest ;)
@Riiisuu6 жыл бұрын
Give this problem a try and when you’re ready, continue the video. Did *You* figure it out?
@markuswilliams44756 жыл бұрын
Reece 5..4..3..2..1
@markuswilliams44756 жыл бұрын
Math meanies 😡
@davidadegboye7735 жыл бұрын
Hey guys it's presh talwaker making sure you mind your decisions
@rafaellisboa84936 жыл бұрын
It's like you can read my mind comrade! second time I was studying some maths and you made a vid exactly about what I was studying, great vid!
@kostantinos22976 жыл бұрын
Is there a geometrical representation of tanh(t), coth(t) etc, just like cosh(t) and sinh(t) are the x and y values of the points of the hyperbola?
@cuzeverynameistaken12836 жыл бұрын
Putting this comment just so if someone else finds it. Right now its late where Im from so I'll try and see if there is one in the morning
@filyb5 жыл бұрын
@@cuzeverynameistaken1283 did you find one?
@joea-497kviews24 жыл бұрын
@@filyb he’s still working on it
@filyb4 жыл бұрын
@@joea-497kviews2 lmao
@paulniziolek92003 жыл бұрын
@@joea-497kviews2 eta perhaps?
@ramkrishnapandey77374 жыл бұрын
You solve mathematics like you are hanging out with Ur friends😜😜 And Ur excitement after solving is just awesome. Just because of teacher like u I'm happy of being a mathematic student. Thank you🙏
@zralok6 жыл бұрын
Did you saw the joke isn't it? So the similarity of "i sin(it)" to isn't it.
@blackpenredpen6 жыл бұрын
der Ultrahero Nice catch!!!
@mike4ty46 жыл бұрын
Or "I sign it"?
@IrateUngulate6 жыл бұрын
I just discovered your channel. Your videos are brilliant! Good thing they're your sponsor :D
@_DD_156 жыл бұрын
Omg.. The biggest problem of my life.. Finally solved 😱😱😱😱😱impressed
@blackpenredpen6 жыл бұрын
DD Yup!!!! : )
@_DD_156 жыл бұрын
@@blackpenredpen I have plenty of calculus books and have never seen that one around, weird :)
@6root913 жыл бұрын
I was searching for these formulae (didn't need the proofs, but they were cool too) for about an hour until I found it here and was able to answer my question.
@hemanthkotagiri88656 жыл бұрын
Your videos are pretty amazing man. Keep going. 👌
@ffggddss6 жыл бұрын
Before 1 min: There's a 3rd way to interpret the angle - the arc length subtended on a unit circle, whose equation you've written: x² + y² = 1. This may or may not work for the unit rectangular hyperbola; I'm checking into that. It does have the right behavior near 0, and it does go to ∞, but those are no guarantee... Fred
@rockapedra11302 жыл бұрын
Excellent video! This is super interesting! Thanks for making these videos!
@koltonjones8666 жыл бұрын
Your videos should be required viewing for most math classes. Do you do anything for dicrette algebra?
@irrelevantgaymer61955 жыл бұрын
What I think is cool is if you were to somehow create a 4D graph and declare your x, y, z, and t axis, and call the z axis the imaginary input and call the t axis the imaginary output, the function x^2+y^2=1 on the t axis looks like x^2-y^2=1 and vice versa. So I kind of think of the hyperbolic function as a complex version of the circle function and vice versa
@leeluu9985 жыл бұрын
I hope you're gonna be a math teacher because yours videos are so clear and precises
@m_riatik6 жыл бұрын
please continue this series!
@luuksemmekrot450923 күн бұрын
Cool again! Love your content!
@carlosraventosprieto2065 Жыл бұрын
Wow!! Thank you for the video!
@nishasharma-gk5bo3 жыл бұрын
Look at this cute face he is blushing while playing with Maths 😍 ,maths must be his love.
@borg9726 жыл бұрын
Great one, thanks! If you could do more parameterization videos it would be great since finding them is always so confusing.. also integrations along a curve with parameterization
@tm896812 жыл бұрын
Nice lecture👍
@RichardCorongiu8 ай бұрын
Throw in a bit of an explanation of Eulers formula in terms of the Taylor series of e^x polynomial... love your passion...😊
@Anonymous-rr5cx5 жыл бұрын
Sir at 6:50 u said imaginary looking Theta = it So t is also imaginary so that they can be real Sin theta = real function Sin h x= imaginary function ?????? Sir please please clear this doubt Thank you
@spelunkerd6 жыл бұрын
I'm headed back to your channel to find the link to "even" and "odd" parts of e^t, described at 15:18. Not sure where to look....
@spelunkerd6 жыл бұрын
Ah, found it here. kzbin.info/www/bejne/pX29oHp7mK9lj6c
@blackpenredpen6 жыл бұрын
It's here: kzbin.info/www/bejne/pX29oHp7mK9lj6c
@ian-ht1nf6 жыл бұрын
9: 26 "isin(it)"?
@crappypoopycrap98006 жыл бұрын
nice one :)
@alejrandom65927 ай бұрын
Easy way: exp(it)=cos(t)+i*sin(t) But also exp(it)=cosh(it)+sinh(it) Pairing up the odd part with odd part, and even with even we get: cosh(it) = cos(t) sinh(it) = i*sin(t)
@Lucky102792 жыл бұрын
0:57 Shouldn't the area = t? The area 0f a unit circle is A = 2π and the area of a sector of a circle is A*(corresponding angle of sector/2π), assuming the angle is in radians. Hence, the area in the diagram should be 2π•t/2π = t.
@ma7-s8j2 жыл бұрын
Area of circle with r = 1: pi * r ^ 2 = pi. Not 2pi.
@thomasolson7447 Жыл бұрын
I noticed that to. But I went the arctan route. cos(i*arctan(3/4))-i*sin(i*arctan(3/4))=1.9031323020709 cos(arctan(i*(3/4)))+sin(arctan(i*(3/4))) =sqrt(7)(4/7+3/7i) for tan= i*3/4 Works just fine this way. You can do geometry with it. It's just pythagorous' theorem with an 'i' in it. [x/sqrt(x^2+y^2), y/sqrt(x^2+y^2)]
@tunneloflight3 жыл бұрын
Plot them! The hyperbolic sin and cos “jump” off the tops and bottoms of the sin and cos at right angles in the y-i plane. Likewise, when sinh and cosh are real, sin and cos are at right angles in the y-i plane. Etc…. It is beautiful. Next extend to tan and tanh, sec and sech …. Then extend to Bessel and J functions!
@holyshit9226 жыл бұрын
Rational paramerization of hyperbola is based on observation (1-t^2)^2+(2t)^2=(1+t^2)^2 (2t)^2=(1+t^2)^2-(1-t^2)^2 1=\left(\frac{1+t^2}{2t} ight)^{2} -\left(\frac{1-t^2}{2t} ight)^{2}
@nicholaslau31946 жыл бұрын
Damn clickbait title! I wish professors can use clickbait to make lectures more interesting
@thalesbastos39156 жыл бұрын
Thank you sooooo much!!!
@billazz91764 жыл бұрын
RIGHT HERE, RIGHT HERE, RIGHT HERE
@edgardojaviercanu47403 жыл бұрын
Beautiful!
@leoarzeno4 жыл бұрын
great video
@siddharthsengar88596 жыл бұрын
after all i've been through in last year , "Imaginary" is a inappropriate title.
@artey66716 жыл бұрын
You don't even need Euler's formula to show that cos(it) = cosh(t). You can also show that their power series are the same.
@koenth23596 жыл бұрын
Yeah, Tibees just did that Bob Ross style!
@artey66716 жыл бұрын
You mean her newest video? I don't see any cosh in there.
@koenth23596 жыл бұрын
@@artey6671 yeah guess you are right. May have misremembered.
@gwalla Жыл бұрын
Are there other conic section analogues of the trigonometric functions? Parabolic sine? Elliptical cosine?
@debaprasadparui47575 жыл бұрын
Sir you are awesome....!!!!
6 жыл бұрын
Cool!!!! So one can get the derivative of sinh and cosh using chain+product rule from the equal sin/cos statement, never thought on that :-O I always did that from the definition of sinh/cosh only ("e-stuff").
@TheNerd4846 жыл бұрын
We just went over sinh and cosh in my calc class today. What are the chances? This is a much more complete explination than we got.
@abdonecbishop Жыл бұрын
makes me want to say.....this is wonderful short video.... beautiful work...so ...so ..excellent...but i think you need to add a quick physic conclusion to your video.........this certainly is one of the slickest short video in circulation....why?....because you connects non-Euclidean equilateral triangle's surface area(excess/deficit) change to a Euclidean triangle's total energy change and the triangle's inertial mass change dependent on (a function off) the average of the total number of summed ''-' , '+' and '0' Gaussian curved triangle edges counted ......
@rishinandha_vanchi5 жыл бұрын
ellipse eqn in complex extended x-plane-y-axis will be a hyperbola in the Im-x-side. This so parametric forms cos and cosh are complex and real counterparts
@rishinandha_vanchi5 жыл бұрын
Oh You mentioned it? Just now say it.
@mehwishbhatti62072 жыл бұрын
Can you please make a video relating tan and tanh
@Thoalfeqargamer4 жыл бұрын
i love you man 💕💕💕💕
@armchairtin-kicker503 Жыл бұрын
Then there is Osborn's Rule, a very useful relationship between trigonometric and hyperbolic functions and identities.
@h4c_186 жыл бұрын
What about x=sec(t) and y=tan(t) for 0
@ugursoydan81874 жыл бұрын
thanks
@alaba50856 жыл бұрын
¡¡Lo máximo!!
@bernardfinucane20616 жыл бұрын
Very cool
@blackpenredpen6 жыл бұрын
Yay!
@RichardCorongiu8 ай бұрын
How's your stock of whiteboard pens ? 😊
@canaDavid14 жыл бұрын
Wait... Are the trig functions C -> N? Or can some input a+bi give imaginary output?
@justacutepotato29454 жыл бұрын
they're C->C. Also, you put C->N, pretty sure you meant C->R or C-> [-1,1].
@canaDavid14 жыл бұрын
@@justacutepotato2945 yeah, you're right. And I meant C->R.
@decay2__6 жыл бұрын
You probably don't know this but you made a pun at 5:49
@helloitsme75536 жыл бұрын
Tbh I've always felt like this is true because I can say integral of 1/1-x^2 dx = integral of 1/1+(ix)^2 and then use u-sub. But at the same time, the integral is tanh(x)
@davidmorochnick4985 жыл бұрын
QUESTION: With -i out in front of sin(it), [-isin(it)], doesn't the proof fail?
@a.a.sunasara92026 жыл бұрын
Bruh😍awesome.... Love ot
@nicolasinostrozamoreno42486 жыл бұрын
Why you don't have spanish subtitles ?? Its so interesting
@Koisheep6 жыл бұрын
Well to some extent you can also use x(t)=sqrt(1-t) and y(t)=sqrt(t) I mean (?)
@bullinmd4 жыл бұрын
Ever heard of gd(x), the Gudermannian function?
@comingshoon27174 жыл бұрын
Tengo sueño ... pero igual veo estos videos aunque hayan sido contenidos que vi hace muchos años!....
@drshamajain41494 жыл бұрын
What we do with the part of the hyperbola on the left side
@KwongBaby6 жыл бұрын
What's the usage of sinh and cosh?
@afafsalem7396 жыл бұрын
Yes it's very cool
@khaled014z6 жыл бұрын
hey bprp, there was an integral video involving cos's and sin's I think and you solved it with a creative way of adding 2 solutions of 2 integrals together and I can't find that video, any ideas? thank you :D
@azmath20596 жыл бұрын
great video. but try starting from first principles and proving that for a hyperbola x=cosht and y=sinht and see how long that takes you!!
@shoobadoo1233 жыл бұрын
What about cosh(it)
@snejpu25086 жыл бұрын
What do you need hyperbolic functions for in math? Of course, except defining them and solving equations with them?
@nicolastroncoso17916 жыл бұрын
to simplify the work or notation of multiple real life problems, instead of putting an enormous amount of digits you simply use hyperbolic functions, same as trigonometry in general
@geoffstrickler3 жыл бұрын
3^2 - 2^2 = 1 too. 😎
@musicandmathematics78974 жыл бұрын
We have theta be real and t be real also. So, how can we put theta = it ??
@justacutepotato29454 жыл бұрын
We're extending the theta and t to complex world
@kbotter39556 жыл бұрын
What does E equal?
@lewisbulled67646 жыл бұрын
2.71828182... it is transcendental.
@markorezic31316 жыл бұрын
Approximately 2.7182818 Its a irrational and transcendental number like pi, its related to the exponential function
@andi_tafel6 жыл бұрын
(1+1/n)^n if n goes to infinity
@dranxelaa67706 жыл бұрын
e=3=pi *isn't* *it* *?*
@omarifady6 жыл бұрын
Sum from 0 to infinity of 1/n! 😃
@mathteacher26515 жыл бұрын
Another great video - kid!
@rafaellisboa84936 жыл бұрын
Could you make a vid about Lobachevsky space pleaseee? don't make me beg
@zralok6 жыл бұрын
That's in my textbook xD
@jimallysonnevado39736 жыл бұрын
Still unsatisfied how can you say that this equation can make the area t/2 and is there a way to come up with this formula using calculus like what you can do with sin and cosine by knowing the derivatives first then using taylor then coming up with a formula like eulers identity
@blackpenredpen6 жыл бұрын
Jim Allyson Nevado as I said in the video. I will do a proof for that. So stay tuned!
@jimallysonnevado39736 жыл бұрын
blackpenredpen waiting for that
@jimallysonnevado39736 жыл бұрын
blackpenredpen oops i apologize for not listening carefully
@menjolno5 жыл бұрын
Notice how close they are to co shit.
@AmogUwUs4 жыл бұрын
*slaps theta* TAG YOU'RE (it)
@rot60156 жыл бұрын
OMG!!!! @&@&&@&@&@; THIS IS SO EXTREME LIKE THE TITLE!!!
Can u Integrate xtan(x)? 😛 Help me if u can. I really love ur Videos. I learn a lot from them. Thanks
@_DD_156 жыл бұрын
Btw, is this a newly discovered relation?
@tupperwallace90486 жыл бұрын
Yes, considering that the history of mathematics goes back millennia. Wikipedia dates them to the 1760s.
@aswinibanerjee62616 жыл бұрын
Why you don't advatise for patrion
@quahntasy6 жыл бұрын
ALGEBRAIC EXPRESSIONS HATE HIM.
@MuthuKumar-mk13206 жыл бұрын
lim x → 0 sin2x^(tan2x) ²
@duggydo6 жыл бұрын
cos(it)=cosh(t) is interesting. cosh(it)=? is a more interesting question though.
@MarioFanGamer6596 жыл бұрын
cosh(it) = cos(t) Like, it's simply inserting i*t into x for cosh(x) and get cos(t) as the result just like as if you have inserted i*t for cos(x) and get cosh(t) as a result.
@kingbeauregard6 жыл бұрын
Is it possible to actually understand complex numbers, or are they simply an abstract tool in the world of mathematics? Like, I can understand integers as "counting things", real numbers as "measuring things", and sines and cosines as "the vertical and horizontal weights of a diagonal line". But imaginary numbers and complex numbers ... ? I don't see them in the real world anywhere. Maybe I'm not looking hard enough?
@camilincamilero6 жыл бұрын
I think of them as vectors that rotate at a certain frequency, or phasors. They are used all the time in electrical engineerig, in the field of power systems.
@kingbeauregard6 жыл бұрын
I don't doubt the utility of complex numbers in real-world calculations, but they seem to be a way to arrive at a real-world result rather than a representation of the real-world result: for example, if two electrical signals are out of phase with one another, the result is the superposition of sine curves, with no visible evidence of any imaginary components. So I'm wondering if they appear visibly in the real world anywhere. If someone asked where I can see Fibonacci Numbers I could point to a sunflower. Is there anywhere I could see complex numbers?
@Harkmagic6 жыл бұрын
Complex numbers are as real as the rotation you are doing in your problem. For most engineering and physics applications those rotations are very real. Hell, 3d rotation can't even be mathematically described properly without quaternions, which are basically more complex imaginary numbers.
@kingbeauregard6 жыл бұрын
Show me an imaginary number. Don't just tell me that they factor into the arithmetic of sine waves; show me an imaginary number in the real world. I'm not sure it can be done, but if it can, I'd like to see it.
@Harkmagic6 жыл бұрын
Showing an imaginary number is no more possible than showing a real number. Numbers are not real physical things that you can point to or reach out and touch, even our real numbers are just abstractions used to represent reality. So if I tell you to stand up and turn and face to your left and tell you that is the number 'i' it is just as valid as saying telling you to count the number of fingers on your hand and telling you that it is the number '5'.
@urluberlu27574 жыл бұрын
you have 99% of like and 1% of dislike... Not too bad... I like ;-)