Derivation of cosh and sinh
ฝัง
- เผยแพร่เมื่อ 16 ก.ย. 2024
- In this video, I derive the formulas for cosh and sinh from scratch, and show that they are indeed the hyperbolic versions of sin and cos. I also explain what the input x of cosh(x) means. Included is a calculation of the integral of sqrt(x^2-1)
Note: A big thanks to Alex Zorba, who came up with the idea and the proof, thank you 🙏
![Shareปลาวาฬ ส่งท้าย SS1 พาเปียโน ขึ้น Ferrari ก่อนไปขาย แลก Nissan l [Nickynachat]](http://i.ytimg.com/vi/KAknGkTNIvg/mqdefault.jpg)
Damn!
HAHAHA BLACKPENREDPEN YEAHHH
hello blackpenredpen, I love your videos!
Seriously needed this, one of the reasons I love doing maths is being able to derive the content from the foundations of my own knowledge and you have done just that sir, thank you!
triton62674 in my humble opinion, i think these insights about the foundations are often the most valuable!
I am a math major. I never learned hyperbolic trig functions in high school or university at least as far as their application is concerned.
When Dr. P. drew the bottom half to say S = theta, the entire thing clicked in, in an instant. Brilliant. And thank you.
Wonderful!
I never thought about hyperbolic functions in terms of area parameters.
Now not only I know how hyperbolic functions come into existence from scratch but also got the firm understanding of circular trig functions.
Really appreciated sir.
This fills in the gaps of all textbooks which start hyperbolic functions by saying we let cosh x =(e^x +e^-x)/2 and sinh x = (e^x - e^-x)/2 like it is the most natural thing in the world to do.
That's so right! I searched everywhere for a derivation but could never find it, only some hints. This bothered me for years as I don't like to assume anything, hence this video
Even a boring lecture can be turned into an interesting one when an lecturer explains you with a smiling face and soothing voice like this ❣️❣️
Amazing! The explanation i have been searching for a long time, and couldn’t find in any book have been staring me in the face all along! Thank you very much Dr Peyam!
I can tell how much you love what you're talking about. Nothing is better than a teacher that is passionate about what they teach. Thank you so much and please don't stop making videos.
Wow; I just loved this: was doing complex numbers in my first year course and suddenly had sing and cosh with no idea where they even came from; this has really helped! Keep up the great work
Thank you so much!!
An all American genius, no doubt. Always smiling, short trousers, left handed. Well, no baseball cap but he simply lost it on his way into the studio. These guys prove everything within two minutes.
Incrediblely well explained. Now I feel I have a stronger understanding of the HB fns. The algebra/calculus used is truly "magical".
I've always had trouble wrapping my head around the hyperbolic trig functions, but this really helped me understand them. Thank you for a great presentation!
Just stunning how elegant geometry can be. I love it. Thank you very much.
The way you explain things is really amazing. I didn't even realise it was 30 minutes long before the video ended. Really simple and easy to understand. Thanks for this.
Great lecture! I might've seen this derivation, but forgot it, and you explained it perfectly. Thanks!
YES THANKS FOR THIS VIDEO!!
I really needed this
I wish I could half an hour of calculus and algebra with a smile on my face. I love math, but this man is something to aspire to.
superb. Finally someone has derived these odd looking expo functions from scratch. many thanks.
Love you, sir......I really thought about the idea for 5-6 hours but i was unsuccessful.. I couldn't find any perfect answers on internet explaining the reason of the values of coshx, sinhx...
Then i found this.
Amazing.... I'm very much pleased now...keep up good works, sir...
Again, lot's of love and respect who found the idea to derive the formulas.. ❤️❤️❤️❤️❤️❤️🙏🙏🙏🙏🙏
Thank you Dr. Peyam!! My doubt has been solved
Searching for while making accounts in websites to find but know i finally found this thank you sir☺️
Wow i was searching for this proof thumbs up man!!
1:55 I almost chocked from laughing too hard :D
At least my thought on sinh and cosh is: mathematicians came across the function (e^x+e^-x)/2 and (e^x-e^-x)/2 for whatever reason --> then they realize they have lots of identities similar to sin and cos --> then they realize they are exactly the parametric form of points on hyperbola --> let's called them sinh and cosh
Yeah this is what would have happened.
Thank you so much for posting this video! You're a legend!
No, thank YOU for the idea, you’re awesome 👍👍👍
Yes Alex, thank you very very much.
Finally found some solid proof. Thx Sir.
I always wanted to know. There's a costant like π but in the hyperbole? Maybe it has a relation with π (cause it's everywhere)
Didn't you notice 'e'?
If you want a relation with π:
e^(π*i) = -1
Best I could find is the limit of difference in length between the asymptotic straight line and the hyperbolic curve as alpha goes to infinity which is around 0.40092988 according to Matlab, and I'm a bit lazy to compute it but it's can be calculated by [LaTeX code]
\int_0^{+\infty} \frac{{\left( \cosh t - \sinh t
ight)}^2}{\sqrt{\cosh^2 t + \sinh^2 t}} \,\mathrm{d} t
All the other things I found goes to 1/2 or 1.
Change of basis between euclidean and hyperbolic space should give you that constant?
@@lukedavis6711 Idk
That was truly inspiring! Thank you Dr P.
Great work! You focus on very cool stuff..and explain it well.
Love from India !! & I still didn't get who are those 6 Students who had disliked the video ( might be they are having some integral issues 😂).....
I hope you got 1000M subscribers in near Future... #Believe
I have waited so long for this moment!!
Legit I need this because I have been using these functions in my courses but we never really looked much into the mathematics of the hyperbolic functions. We just kinda suddenly used them in my Enginerring courses. Thanks for this haha.
Mind blowing dude ❤❤ Can you please explain why the derivative of coshx is sinhx. Like why there isn't any negative sign. Kindly, make a video on that one as well. Loved this video btw ❤❤
You just use the definition of cosh in terms of exponentials
@@drpeyam that was very helpful. Thanks a lot once again ❤️❤️
Interesting how to find the exponential definition of cosh and sinh, you first have to find their inverse function (i.e. you search for α = ln(x + sqrt(x² - 1)) ) and then solve for x. Either way, nice video!
If you integrate with respect to y you might not need to do this. Haven't checked this though.
Amazing! It's great to see videos like this fill in the 'motivation' of math.
Earned a subscriber, fabulous! ❤
That is really cool,thanks.
No body realizes how smart this guy really is , defo phd level
Cool, loving it! More of this stuff please :) Always nice to watch a reminder :)
More to come!
This was absolutely AMAZING!!!! :-O Thank you Professor!!!
oooh my cosh !!!! great video Dr tigre Peyam!! Thanks
From now on I will answer the question from one of my friends, "Man, do you wanna come to the party this evening?" with " Yeah, sin(theta)". xD
curious, is there a similar "derivation" you could do for the complex definitions of cosx and sinx?
Possibly, but you could more easily take Euler's Formula and apply the formulas for Re(z) and Im(z)
Gold161803 yeah I know about that derivation
I also saw that if you do the even and odd function deconpositions of exp(x), you get exactly cosh and sinh
That was awesome!
Unique
Ace Peyam - the Cosh Xo near the end is a 'rabbit out of a hat' trick. Magic without magic.
You are my favorite
how does sec theta = 1 ime confused? If youre finding theta you do inverse sec on both sides then 1/cos (1). Cos (1) is 0? and 1/0 is undefined...?
love your shirt here! so funky!
So so good!
this was awesome man! such a tour! tho i think a u sub would have been way faster for the integral of sqrtx^2-1. Good one to practice at home for me! Thanks a lot dr pi*m!!
i love this guy
Maybe you could do a video solving that integral with complex numbers. sqrt(x^2-1)=i*sqrt(1-x^2). The latter can be integrated using the substitution x=sin(t). Then, you'll end up with some arcsin(x). Since x0 is greater than 1, you need complex numbers to find it's arcsin and, hopefully, all imaginary terms will be canceled and you'll get everything real and correct
Very beautiful derivation....!!!
i was trying to do this today for the hell of it. i knew that cosh and sinh were just the "unit hyperbola" equivalents of the circular trig functions. i came up with doing everything in terms of half an area. but i went way off course in my derivation. even going so far as trying to integrate sqrt(x^2 - 1) by substituting x = cos(t) and deciding that cos^2(t) - 1 = sin^2(t) in a moment of absolute brain fartness.
i ended up with a horrific expression that was utterly unsolvable. after watching your video i went back, wondering why cos/sin identities didn't work as well as sec/tan and realized my ridiculous error.
Excellent. Thank-you very much.
From the conic sections, now that i wonder, if there's a circular sine and cosine function, and a hyperbolic one, is there a parabolic definition for those trigonometric ones?
Well, what really connects the hyperbolic and circular trig functions isn't that they measure areas under conic sections, but that they find areas of sections of quadratic forms(this means that the x and y variables are both squared in equations for hyperbola and circles). Another difficulty is in deciding what the "center" of a parabola is, although I suppose you could choose its focus.
@@tracyh5751 i was thinking on how to generalize a definition for these functions by generalizing the exponential form of the "classic" trigonometrics ones, If, for example, sin(x)=(e^ix-e^-ix)/2i and sinh(x)=(e^x-e^-x)/2, the parabolic sine, say sinp(x) would be sinp(x)=(e^(1+i)x/√2-e^-(1+i)x/√2)/2i, but it seems very arbitrary
excellent video :)
Thank you.
amazing video
Beautiful beautiful beautiful😍💓😍💓😍💓😍💓.
I love you for this
Hi Dr. Peyam. I wonder if the argument of hyperbolic functions is dimensional, cos it is an area. Isn't it?. How you measure that argument given certain hyperbola and the point (x,y) over the hyperbole. And why expositors of hyperbolic function they use always a unitary hyperbole, and never use a general hyperbola (x^2 - y^2 = a)?. I know it takes a little more algebra but it is better to see how they compare with trigonometric functions
int 1/(x^n+a^n), V n>2,a>1
i like your short and your brain
that was very nice :)
I liked the video half way in, but at the end I forgot I already liked it so I ended up deleting my like xd (I liked it again btw)
Whenever I get those feelings of grandeur I come here to be reminded of just how stupid I really am. Thanks professor.
Good doc 👌🍺
Why we integrate √(x^2-1) separately
Please give reason why not entire Ah
I needed this video
You could say that the derivation is so easy, it's a sinh
(if 'Oh My Cosh' is taken)
As always, super ;-)
Thank you!
hey, I really Love your Videos! But I think they would be even better if you work on you blackboard arrangement XP
Question:
To convert a function f(x) to be even function you put f(|x|) or f(-|x|)
So how you convert a function g(x) into an odd function?
In general 1/2 (g(-x)-g(x)) works
@Dr. Peyam's Show
However, if g(x) is even, this is NOT the way since then the function will always be 0, and this function isn't odd, I want a method works for every function, not only if it's not even
For even functions 1/2 (f(x)+f(-x))
Hmmm, with that I might be able to continue my attempt at inventing "parabolic trigonometry" hahaha
Although I have already realised that if this does make sense, it's going to be weird because the fundamental equation is gonna be something like cosp (x) - sinp^2 (x) = 1, which obviously does not define any "regular" or "nice" geometry since the cosp term is not squared, thus no quadratic form, but hey, should still be fun
Isnt it easier doing the integral with sub of t=sin(theta)? atleast thats how I would do it
Martin Björn you should get the relation between sin and sinh (and cos and cosh), since you’d get an imaginary answer
This is the derivation of the derivation .....!!!!
I would have calculated the integral by integrating the sqrt(x)*derivative of x^2 -1
101st like in base 13.
but the triangle line isn't straight
Cooooool!
What is the jacobian of hyperbolic coordinates.
Still 1 since cosh^2 - sinh^2 = 1
@@drpeyam the jacobian of polar coordinates is r, if we let x = ucoshv and y = usinhv then wouldn’t the jacobian be u the “radial part” of a standard hyperbola?
Nice
Thanks for the inclusion of the unnecessary spoiler "alert"!
Belum
Why we calculate sector area
Dr peyam give answer
I am eating salt and vinegar potato chips right now
cute af
19:34 - Were you drunk out of a sudden or just farting? Hehe, just kidding, Dr. Peyam! As always ... a great video :)
dp121273 HAHAHAHAHA 😂😂😂😂 I was trying to avoid tripping on my microphone cable, but I see what you mean 😂😂😂😂
that intro is touching my spivak nerves
27:37 wot
I always love this derivation:
math.stackexchange.com/questions/757091/alternative-definition-of-hyperbolic-cosine-without-relying-on-exponential-funct/757241#757241
Math ..... 😎
28:00 "Turns out this is always positive" ....literally unwatchable
;)
Stupid
Are you in the southern hemisphere?
No, why?
@@drpeyam
because we are in January and you wear summer clothes!
Hahaha, that’s why 😂 I recorded this video in October, lol
@@m.a.4794 Lol Southern California has no winter tho
Doctor, please help! My dog ate ham, now has diarrhea and vomit. :(
Kokote!