More astounding to me is that "t" = ln(cosh(t)+sinh(t)) where cosh(t) = x and sinh(t) = y on the curve: x^2-y^2=1 Hence, take ln(x+y) of any (x,y) on that curve and you have the hyperbolic angle associated with that point. Hyperbolae are SO MUCH MORE INTERESTING than circles. :)
When you work on a long maths problem and you and up with a good result you feel highly elated. There is an endorphin rush. You are literally getting high via your brain's natural chemicals.
Yeah, because this entire video was cool although I don't understand what he is talking about. I gave it a like because it is wonderful to see shiny happy people all around.
Since your first video on the hyperbolic trig functions, I've been wondering. We've used the unit circle and hyperbola. How about the other conic sections? Are there analogous elliptical trig functions and parabolic trig functions as well? If so, are they useful like circular and hyperbolic ones or just mathematical curiosities? I have a science background (chemistry) which included a fair amount of math but can't say I recall learning them but it seems like there's no reason for them to not exist.
An ellipsis has got two radii and have a wide range of shapes. A unit ellipsis is, as you'd guessed, an ellipsis with rx and ry both being 1 i.e. a unit circle. You can say that a elliptical function exist although they're derived after the trigonometric functions by multiplying them with a constant. On the other hand, parabolic functions are a bit more complicated. A unit parabola's exist as it's graph is defined as 0 = x² - y, more commonly known as y = x². However, I don't think you can derive parabolic functions from that or if they do, they're unnecessary unlike the trigonometric and hyperbolic functions (seriously, I can't come up with the logic of the parabolic functions if you can create them).
Archimedes figured out the area inside a parabolic arc 2 centuries BC and he didn’t need no stinking calculus or exponential function to do it. The parabola is simple because there’s really only one of them, y equals x squared. Elliptical paths and the areas inside ellipses, on the other hand, became important to figure out when Kepler deduced the laws of planetary orbits. Squashing a circle was no longer good enough and Bessel had to compute the Bessel functions. So the answer to the question is no, there is no table of parabolic sines or elliptical cosines, but the mathematics of the curves are very important and practical.
Soo cool! I saw Dr payems video where he derived tye definition of sinh and cosh, but he started with the assumption of the outer area being t/2 (he used alpha instead of t...Doesn’t matter) and I was curious as to why. This video nicely answers that! I am just not able to take things for granted in math, I must see why!!!
Actually, he has published a short and simple "proof", but he relies on the so-called Todd function, which is a function that he has formulated and has not published the proof of yet, except for his paper that is under peer review. So, we wait.
There will be no need for that. It's quite a complicated situation... see meta.mathoverflow.net/questions/3894/is-there-a-way-to-discuss-the-correctness-of-the-proof-of-the-rh-by-atiyah-in-mo for some background.
@@michel_dutch He did claim that he had come up with a simple proof, which sounds kind of suspicious. But I guess that time will tell, since his work needs to be sufficiently examined first.
How does t relate to the “normal” angle θ? I haven’t been able to find information on that relationship. All I know is that t must approach infinity as θ approaches 45 degrees.
First of all : i enjoy your videos very much. How you play with math is a joy for the brains :-) But now a question. cosh(t) is defined as 1/2(e^t + e^-t). From the picture the x coordinate of any point on the hyperbola is defined as cosh(2A) where a is the drawn area. What i don't get immediate is that this x coordinate also equals the definition of cosh.
Don't matter how good you're on something, there will always be an asian better than you, specially in maths! Great explanation! Thank you so much and greetings from Brazil!
Dear...I'm not totally convinced of what you've done. Please explain why you changed the variable t by u. Can explain also why you use sometimes t to talk about time and also to talk about 2 times the area of the figure? Why you mixed both variables if both are not the same?
1/(tan(x)+cot(x)+sec(x)+cosec(x))= 1/(sinx/cosx+cosx/sinx+1/cosx+1/sinx)= sinxcosx/(sin^2x+cos^2x+sin(x)+cos(x)= sinxcosx/(1+sinx+cosx)= sinxcosx(1-sinx-cosx)/((1+sinx+cosx)*(1-sinx-cosx))= sinxcosx(1-sinx-cosx)/((1-(sinx+cosx)^2)= sinxcosx(1-sinx-cosx)/((1-1-2sinxcosx)= sinxcosx(1-sinx-cosx)/(-2sinxcosx)= (sinx+cosx-1)/2 so the integral becomes (cosx-sinx-x)/2
I like hyperbolic cosine theta equals the quantity e to the i theta plus e to the minus i theta the quantity divided by two. I think cosh and sinh are wrong. I can get the right answer with normal trig functions.
Question why can you Just say 0 to t at the integral because if i translate the x to the t i would geht the Formula x=cosh(t) and so the First x is 1 you could Just See that arccosh(1)=t=0 so okay but If i would do the same with b then i would get arccosh (b)=t so the integral should Go from 0 to arccosh(b) isn't it? Or am i allowed to say b= cosh(t) so i would get t again?
Just to clarify: The integral goes from a to b for x. a is defined as the left edge of the parabola i.e. 1 and b where the area ends which is cosh(t). Now let's follow the substitution: In the substitution, X is defined as cosh(u). In order to get the values for u, we solve for it so u = arccosh(X). Next, we change the integral borders due to the changed variable: a = 1 so the new bottom border is arccosh(1) = 0 and b = cosh(t) so the top border becomes t. So yes, b = cosh(t).
Well done but your starting off with the premise that x (t) and y (t) equal the hyperbolic functions and going from there. Try to prove that if the area in question is t/2 then x (t) & y (t) are the hyperbolic functions. That'll take you a while.
@@blackpenredpen That is soooooo cool, I moved to the US from Taiwan about 2 years ago. I am now enrolling in AP Calc AB class while self studying BC content, your video does help a lot and really fun to watch! Ignore those critics and KEPP IT UP!
@@blackpenredpen Noooo don't say that xp You're such an amazing teacher though; i understand if you won't do it, but I can guarantee every single one of your subscribers believe you can do it! Perhaps a certain... like goal on the next video to persuade you? ;)
I wish all instructors were as excited as you when they teach. More people would learn faster when you show such enthusiasm. Plus, it's infectious.
It's amazing that the only difference between the equations for a circle and a hyperbola is a minus sign but they produce very different functions
The similarity is due to them both being conic sections :)
@@carmangreenway I have a question: Why do we study conic sections?
@@createyourownfuture5410 they're very useful for orbits in particular. Zach Star did a great video on that and other uses :)
@@carmangreenway oh, thanks
@@createyourownfuture5410 i think they're useful for 3d things in general (quadric surfaces and that kind of things)
He gets so excited at the end it’s the sweetest thing I’ve ever seen🥺
More astounding to me is that "t" = ln(cosh(t)+sinh(t)) where cosh(t) = x and sinh(t) = y on the curve: x^2-y^2=1 Hence, take ln(x+y) of any (x,y) on that curve and you have the hyperbolic angle associated with that point. Hyperbolae are SO MUCH MORE INTERESTING than circles. :)
that's so cool, indeed the hyperbola is really interesting
In this video blackpenredpen has become blackpenredpenbluepengreenpen
Harris Idham Onepentwopenredpenbluepen
Don’t mind the purple pen
Thanks for taking the time to produce this video, I didn’t think the proof would be so simple. It’s long-winded but quite straight forward.
He got very excited at the end of the video lmao
*cancelling intensifies*
When you work on a long maths problem and you and up with a good result you feel highly elated. There is an endorphin rush. You are literally getting high via your brain's natural chemicals.
@@WitchidWitchid now i cant view math the same thanks
Honestly this is exactly why I adore maths so much. Great video, great solution. Keep it up!
The most interesting thing is that: THIS IS THE SAME FORMULA OF THE AREA OF A SLICE OF A UNIT CIRCLE! (Just change t -> theta)
Excellent! You tend to do the work that other channels don't or are unable to do. Keep up the good work 👍👍👍
you are the very most accurate teacher
I was just as excited as him in the end. this was really cool
The isn't its are back! I'm so happy!!
Another great video! Keep it up
Master JH thanks!!!!
Awesome video, blackpen!
Pranay Venkatesh thanks!!
He is truely passionate about math
He's the man!
This was so cool! I love proofs from calculations. Formal logic and proofs are my Achilles heel.
So cool, great video, great result, thanks for the video!!!
New subscriber , dude ur videos are awesome and unique keep doing MATHS love from INDIA.👌👌
Just what I needed. Thank you!
Thank you! I appreciate your hard job!!!
well done, teacher!
if we can definite the t with the length of arc? instead of t/2 with the area. like a circle
He was so excited at the end! 😂😁
Yeah, because this entire video was cool although I don't understand what he is talking about. I gave it a like because it is wonderful to see shiny happy people all around.
11:28 you were really excited to conclude the viedo
"oh, look at that!"
Since your first video on the hyperbolic trig functions, I've been wondering. We've used the unit circle and hyperbola. How about the other conic sections? Are there analogous elliptical trig functions and parabolic trig functions as well? If so, are they useful like circular and hyperbolic ones or just mathematical curiosities? I have a science background (chemistry) which included a fair amount of math but can't say I recall learning them but it seems like there's no reason for them to not exist.
Interesting idea.
An ellipsis has got two radii and have a wide range of shapes. A unit ellipsis is, as you'd guessed, an ellipsis with rx and ry both being 1 i.e. a unit circle. You can say that a elliptical function exist although they're derived after the trigonometric functions by multiplying them with a constant.
On the other hand, parabolic functions are a bit more complicated. A unit parabola's exist as it's graph is defined as 0 = x² - y, more commonly known as y = x². However, I don't think you can derive parabolic functions from that or if they do, they're unnecessary unlike the trigonometric and hyperbolic functions (seriously, I can't come up with the logic of the parabolic functions if you can create them).
Archimedes figured out the area inside a parabolic arc 2 centuries BC and he didn’t need no stinking calculus or exponential function to do it. The parabola is simple because there’s really only one of them, y equals x squared. Elliptical paths and the areas inside ellipses, on the other hand, became important to figure out when Kepler deduced the laws of planetary orbits. Squashing a circle was no longer good enough and Bessel had to compute the Bessel functions. So the answer to the question is no, there is no table of parabolic sines or elliptical cosines, but the mathematics of the curves are very important and practical.
Cool, great explanation thanks
Great !!! Love it
Very cool 👍🏻
Soo cool! I saw Dr payems video where he derived tye definition of sinh and cosh, but he started with the assumption of the outer area being t/2 (he used alpha instead of t...Doesn’t matter) and I was curious as to why. This video nicely answers that! I am just not able to take things for granted in math, I must see why!!!
Can you do video on new riemanns hypothesis ,,proof"?
It hasn't been published yet, it is currently under peer review.
Actually, he has published a short and simple "proof", but he relies on the so-called Todd function, which is a function that he has formulated and has not published the proof of yet, except for his paper that is under peer review. So, we wait.
Oh ok, thank you
There will be no need for that. It's quite a complicated situation... see meta.mathoverflow.net/questions/3894/is-there-a-way-to-discuss-the-correctness-of-the-proof-of-the-rh-by-atiyah-in-mo for some background.
@@michel_dutch
He did claim that he had come up with a simple proof, which sounds kind of suspicious. But I guess that time will tell, since his work needs to be sufficiently examined first.
Can u pls explain the Riemann Hypothesis proof?
Thank u.
Excellent
How does t relate to the “normal” angle θ? I haven’t been able to find information on that relationship. All I know is that t must approach infinity as θ approaches 45 degrees.
Very good!
But what is the relation of 't' with the slope of the point (x.y)?
First of all : i enjoy your videos very much. How you play with math is a joy for the brains :-) But now a question. cosh(t) is defined as 1/2(e^t + e^-t). From the picture the x coordinate of any point on the hyperbola is defined as cosh(2A) where a is the drawn area. What i don't get immediate is that this x coordinate also equals the definition of cosh.
OK, I figured it out myself :-) 2 times the area A = ln(x+sqrt(x^2-1). And yes, 1/2(e^2A+e^-2A) = x ! QED
just tea over two and two beneath tea
Amazing!
wonderful
so good
daniel : )
At the end the area is equal to t/2 it's great
Afaf Salem yup!!
Don't matter how good you're on something, there will always be an asian better than you, specially in maths! Great explanation! Thank you so much and greetings from Brazil!
It's wonderful... 😍😍
the tea world and the yew world :D
How about turning it 45° and getting the hyperbola xy=1/2?
Pierre Abbat
trying to be funny?
Pierre Abbat
雙曲線方程式
y^2-x^2=1
如何旋轉pi/4成
xy=1/2
如下
原座標
e^ia=x+iy=cos(a)+isin(a)
新坐標b=a-pi/4
e^ib= x1+iy1 = cos(b)+isin(b)
則a=b+pi/4
e^ia= e^i(b+pi /4 )
=cos(b+pi /4)+isin(b +pi /4)
=((cos(b)*cos (pi /4 ) -sin (b)* sin (pi /4 ))+
i(cos (b)* sin (pi /4 )+sin (b) *cos (pi /4 )))
= ((cos(b)/sqrt(2) -sin (b) /sqrt(2) )+
i(cos (b) /sqrt(2) +sin (b) /sqrt(2) ))
其中x1= cos(b) ，y1= sin(b)
原式為
x=x1 /sqrt(2)-y1 /sqrt(2)=(x1-y1) /sqrt(2)
y=x1 /sqrt(2)+y1 /sqrt(2)=(x1+y1) /sqrt(2)
雙曲線方程式
y^2-x^2=1
(x1+y1) ^2/sqrt(2)^2-(x1-y1)^2 /sqrt(2)^2=1
1/2*((x1^2+2x1y1+y1^2)-(x1^2-2x1y1+y1^2))=1
1/2(4x1y1)=1
則
x1y1=1/2
此即為新座標雙曲方程式
@@blackpenredpen how can I message you privately?
@@umar-ot6mi u canot XD
very nice
Dear...I'm not totally convinced of what you've done. Please explain why you changed the variable t by u. Can explain also why you use sometimes t to talk about time and also to talk about 2 times the area of the figure? Why you mixed both variables if both are not the same?
Man I've never seen you so excited before..
: )
Could you make a video relating the angle it makes with the curve to T
what is t? is it the 4th dimension plane variables?
could you do a video on hyperbolic angles
great video! can you plz solve this integral dx/(tanx+cotx+secx+cosecx)
What is sec(x)?
@@franzschubert4480 sec(x) is the secant function. sec(x)=1/cos(x)
1/(tan(x)+cot(x)+sec(x)+cosec(x))=
1/(sinx/cosx+cosx/sinx+1/cosx+1/sinx)=
sinxcosx/(sin^2x+cos^2x+sin(x)+cos(x)=
sinxcosx/(1+sinx+cosx)=
sinxcosx(1-sinx-cosx)/((1+sinx+cosx)*(1-sinx-cosx))=
sinxcosx(1-sinx-cosx)/((1-(sinx+cosx)^2)=
sinxcosx(1-sinx-cosx)/((1-1-2sinxcosx)=
sinxcosx(1-sinx-cosx)/(-2sinxcosx)=
(sinx+cosx-1)/2
so the integral becomes (cosx-sinx-x)/2
Why can you ignore the negative part of the hyperbola when you integrated?
5:13 x isn't a function of time, but just a function of t.
IMO area = (1+exp(-2t))/4 + t/2
What is that small print on your shirt?
I like hyperbolic cosine theta equals the quantity e to the i theta plus e to the minus i theta the quantity divided by two. I think cosh and sinh are wrong. I can get the right answer with normal trig functions.
please make video on e^x power series
Too good
and that's pretty much it.
Why not when you integrate y dx use y=x^2-1 and continue subs.
M. Shebl
Bc I wanted to show how cool that parameterization is! : )
He kinda got overexcited in the end...😂😂
Question why can you Just say 0 to t at the integral because if i translate the x to the t i would geht the Formula x=cosh(t) and so the First x is 1 you could Just See that arccosh(1)=t=0 so okay but If i would do the same with b then i would get arccosh (b)=t so the integral should Go from 0 to arccosh(b) isn't it? Or am i allowed to say b= cosh(t) so i would get t again?
Oh i think it does not Matter at the end but in reality you should have to write arccosh(b) but you need to substitute Back in anyway soooo😂
Just to clarify: The integral goes from a to b for x. a is defined as the left edge of the parabola i.e. 1 and b where the area ends which is cosh(t).
Now let's follow the substitution: In the substitution, X is defined as cosh(u). In order to get the values for u, we solve for it so u = arccosh(X). Next, we change the integral borders due to the changed variable: a = 1 so the new bottom border is arccosh(1) = 0 and b = cosh(t) so the top border becomes t.
So yes, b = cosh(t).
MarioFanGamer t and u are basically the same so yeah xD i've understood everything got just a little stuck for a moment no problem but thanks :)
live long sir
please differential equation!!
What is t ??
5:12 "x is a function of time" we're on the hyperloop? lol
came from twitter
Dr Peyam mode lol
Well done but your starting off with the premise that x (t) and y (t) equal the hyperbolic functions and going from there. Try to prove that if the area in question is t/2 then x (t) & y (t) are the hyperbolic functions. That'll take you a while.
老師好， 請問您是從台灣來的還是中國來的呢～?
Taiwan!! : )
@@blackpenredpen That is soooooo cool, I moved to the US from Taiwan about 2 years ago. I am now enrolling in AP Calc AB class while self studying BC content, your video does help a lot and really fun to watch! Ignore those critics and KEPP IT UP!
謝謝, 你也很棒喔!老師為你加油!!
Why 480p
Jota oh god, I forgot to change back the setting
Cara faz vídeos legendados em português.
OMG! The guy is writing numbers and other stuff without computers or keyboards. How does he do that? Also his remote controller is very small.
Reijo P. ?
blackpenredpen It was a joke. People don't usually use pens anymore, at least in Finland. Not even teachers, so this is suddenly very cool.
Reijo P. Oh I see!!
Why are your videos suddenly low-res? What happened to your camera?
1:27 "Let's just go ahead do the meth Now!"
xD sounds so wrong out of context (we know its a math channel so thats context)
Wouldnt it be nicer to include the area under the x-axis so that all together, the area is t?
Sure. But in that case it wouldn't be similar to the unit circle situation.
Just for fun, do you think you can do that integral I sent you?
It would be awesome to see someone actually do it lol
#YAY
Sir Rahmed which one is it??
@@blackpenredpen It was the indefinite intregral of:
(x^2)/( cos(x) + sin(x) ) dx
Sir Rahmed oh! But I don't think I can do it tho..
@@blackpenredpen Noooo don't say that xp
You're such an amazing teacher though; i understand if you won't do it, but I can guarantee every single one of your subscribers believe you can do it!
Perhaps a certain... like goal on the next video to persuade you? ;)
Sir Rahmed Multiply and divide by (cos x - sin x). Then you get x^2 (cos x - sin x) / cos2x. From here, it's pretty easy.
thank u sir = please name book =thank u sir
Change the name! You're not blackpenredpen, you're blackpenredpengreenpen.
: )
Long proofs taste much better.
But for t=0 area = 1/2, not 0
Hyper-balla? Never heard anyone say it like that. hyPERboLUH
Este chinito tiende a complicar artificialmente el asunto. Supongo para que el vídeo dure un poco más.
#YAY!