Agreed. The BoS has been a little sneaky in the past and added questions involving the hyperbolic functions, but, of course, they had to give definitions and explain them. Nevertheless, still good to see them there!
These are definitions, no derivation for definitions. You may ask that how the points on the graph of x²-y²=1 {x ≥ 1} -which is the unit hyperbola- are given by the ordered pair ((e^x+e^(-x))/2, (e^x-e^(-x))/2). This is the interesting part for me
He didn’t even touch on the relation to the unit hyperbola, as a parallel to the unit circle. Or the cosh^2 - sinh^2 = 1 analogy to Pythagorean. And, fwiw, I learned and have always used the pronunciation as “sinch” not ‘shine”. Never heard of “shine”.
12 y/o me first known about them when playing with a Calculator. Thats where all joy of maths begin. All hatred comes from a result slip with low grades.
I always found the hyperbolic functions somewhat superfluous, since they were just combinations of exponentials. They only really ever came in handy for notational convenience.
That is true for basically everything which isn't + an -. Using words and those two things you describe everything. Multiplication, division, therefore surds and indices, and all the trig functions....
It is still Pythagorean Theorem. Instead of x try y/x or v/c î:= (x,y) -> cos(arctan(y/x))=x/sqrt(x^2+y^2) ĵ:= (x,y) -> sin(arctan(y/x))=y/sqrt(x^2+y^2) [î(x,y), ĵ(x,y)] Those are unit vectors in R^2. You'll find that î looks a lot like gamma 'Γ' (the 'γ' looks like 'y', I don't want to use it). That's because that is what it is if you use an imaginary value. So, hyperbolic math is trig with 'i'. You'll probably want to keep it on the y-axis. This takes place on the complex plane. î(x,iy)+ĵ(x,iy) î(c,iv)+ĵ(c,iv) It is used in physics. Momentum is imaginary. The vector that you do angle adition to get to 1 is -(i*y-x)/(i*y+x).Which is Pythagorean Triple unit vectors. Something cool going on there. So, that tangent is different from the other one. i`:= (x,y) -> cos(arctan(-(I*y-x)/(I*y+x)))=sqrt(2)*(I*y+x)/(2*sqrt(x^2-y^2)) j`:= (x,y) -> sin(arctan(-(I*y-x)/(I*y+x)))=-sqrt(2)*(I*y-x)/(2*sqrt(x^2-y^2)) [i`(x,y),j`(x,y)] is four dimensional. It probably has more to do with this sphere [cos(θ)*cos(φ), cos(θ)*sin(φ), sin(θ)*cos(φ), sin(θ)*sin(φ)] but this isn't spherical, it's a 4d hyperbolic beast. Good luck.
on the positive side of the graph yes, they approach eachother think of it as this, 2 graphs, f(x) = a+b, and g(x) = a-b in this case, a = e^x, and b = e^-x since e^-x approaches 0, b approaches 0, hence a+b approaches a, likewise a-b approaches a. thus f(x) approaches g(x)
If you include complex numbers, then you can relate them, yes. If z = a + bi, in other words a complex number, then: cosh (iz) = cos (z), and sinh (iz) = i*sin (z)
According to whom? There are several ways to pronounce this word: ˈsɪŋ, ˈsɪntʃ, and ˈʃaɪn [1]. [1] From Wikipedia, from (1999) Collins Concise Dictionary, 4th edition, HarperCollins, Glasgow, ISBN 0 00 472257 4, p. 1386
You are a great teacher Eddie
Another great video Eddie!
I remember when I first discovered hyperbolic functions. Still amazes me today :P
Agreed.
The BoS has been a little sneaky in the past and added questions involving the hyperbolic functions, but, of course, they had to give definitions and explain them. Nevertheless, still good to see them there!
Eddie! Eddie who? Exactly
So where is the derivation or intuition behind these functions and the relationship to e^x?
Visit my channel! You are going to understand the basic of the functions.
I guess you will get your answer in these two videos:
ua-cam.com/video/zd3RyRk6wYI/v-deo.html
ua-cam.com/video/Wfpb-fniSSk/v-deo.html
@@namasthapa1277 do you have the complete playlist's link from zero?
@@NoName-cm9tx I'm not sure what you are asking. Said so I believe I don't have the link you are asking.
These are definitions, no derivation for definitions. You may ask that how the points on the graph of x²-y²=1 {x ≥ 1} -which is the unit hyperbola- are given by the ordered pair ((e^x+e^(-x))/2, (e^x-e^(-x))/2). This is the interesting part for me
He really did it, explained so beautifully
i had no clue what the pho fucking fuck that "h" meant. Now, I do.
He didn’t even touch on the relation to the unit hyperbola, as a parallel to the unit circle. Or the cosh^2 - sinh^2 = 1 analogy to Pythagorean. And, fwiw, I learned and have always used the pronunciation as “sinch” not ‘shine”. Never heard of “shine”.
You have saved me from failing a test
12 y/o me first known about them when playing with a Calculator. Thats where all joy of maths begin. All hatred comes from a result slip with low grades.
amazing
Man that was great
I always found the hyperbolic functions somewhat superfluous, since they were just combinations of exponentials. They only really ever came in handy for notational convenience.
That is true for basically everything which isn't + an -. Using words and those two things you describe everything. Multiplication, division, therefore surds and indices, and all the trig functions....
great
thankyou!
I wish this was my teacher
It is still Pythagorean Theorem. Instead of x try y/x or v/c
î:= (x,y) -> cos(arctan(y/x))=x/sqrt(x^2+y^2)
ĵ:= (x,y) -> sin(arctan(y/x))=y/sqrt(x^2+y^2)
[î(x,y), ĵ(x,y)]
Those are unit vectors in R^2. You'll find that î looks a lot like gamma 'Γ' (the 'γ' looks like 'y', I don't want to use it). That's because that is what it is if you use an imaginary value. So, hyperbolic math is trig with 'i'. You'll probably want to keep it on the y-axis. This takes place on the complex plane.
î(x,iy)+ĵ(x,iy)
î(c,iv)+ĵ(c,iv)
It is used in physics. Momentum is imaginary. The vector that you do angle adition to get to 1 is -(i*y-x)/(i*y+x).Which is Pythagorean Triple unit vectors. Something cool going on there. So, that tangent is different from the other one.
i`:= (x,y) -> cos(arctan(-(I*y-x)/(I*y+x)))=sqrt(2)*(I*y+x)/(2*sqrt(x^2-y^2))
j`:= (x,y) -> sin(arctan(-(I*y-x)/(I*y+x)))=-sqrt(2)*(I*y-x)/(2*sqrt(x^2-y^2))
[i`(x,y),j`(x,y)] is four dimensional. It probably has more to do with this sphere [cos(θ)*cos(φ), cos(θ)*sin(φ), sin(θ)*cos(φ), sin(θ)*sin(φ)] but this isn't spherical, it's a 4d hyperbolic beast. Good luck.
Wait shouldn't sinh x and cosh x be approximately the same curve?
on the positive side of the graph yes, they approach eachother
think of it as this, 2 graphs, f(x) = a+b, and g(x) = a-b
in this case, a = e^x, and b = e^-x
since e^-x approaches 0, b approaches 0, hence a+b approaches a, likewise a-b approaches a. thus f(x) approaches g(x)
sin means a division of particular 2 foots of triangle. Likewise how to represent sinh ???
omg dood i had the same doubt but i got it cleared
@@battlewing221 then ducking explain it😂
@@banana6108 well, we just define sinh and cosh in a different way. It need not be related to the way we define sin and cos.
the main bring away is
they have no relationship with your best friend sine and cosine
Hello
Sir how to evaluate sinhz=coshz
d/dx(sinhz) = coshz
What is the level at which you are teaching these concepts? High school? Graduation? And where(country)?
I wondered the same.. 🤔
@@TheBrickagon high school in NSW, Australia
Eddie, can you express sinh(x) interms of sin(x) and vice versa?
If you include complex numbers, then you can relate them, yes.
If z = a + bi, in other words a complex number, then:
cosh (iz) = cos (z), and sinh (iz) = i*sin (z)
@ 0:22 There are more ways to pronounce "sinh":
ua-cam.com/video/er_tQOBgo-I/v-deo.html
According to whom? There are several ways to pronounce this word: ˈsɪŋ, ˈsɪntʃ, and ˈʃaɪn [1].
[1] From Wikipedia, from (1999) Collins Concise Dictionary, 4th edition, HarperCollins, Glasgow, ISBN 0 00 472257 4, p. 1386
@@CaptchaSamurai Thanks! I obviously didn't know that. I adjusted my reaction.
@@jacobvandijk6525 You re welcome &
Happy New Year!
@@CaptchaSamurai The same to you, Kamil.
I was pronouncing sinsh until i read some textbook
I love u Sir
When Mathematicians have no work to do then they come up with these stuff.
I think the definition of Hyperbolic Sine & Cosine on this lecture is wrong. They should be defined on S-Plane.
What do you mean
Even tho he explained it brilliantly.. i still understood nothing lol zero.. not a thing
You’ll get there soon
Bored