My Favourite Golden Ratio Fact
Вставка
- Опубліковано 11 чер 2024
- The gradient of y = tan(x) is the golden ratio, at the point where the graph of tan(x) intersects cos(x). We explore why the golden ratio has arisen, and construct another similar example using hyperbolic trig functions.
00:00 Intro
00:41 Verifying
03:34 Second example
fun fact: the bisector of y=x and y=3x is y=(phi)x
Your two examples seem related by a correspondence I noticed some time ago between the circular and hyperbolic trig functions, viz. sin ~ tanh, cos ~ sech, tan ~ sinh, cot ~ csch, sec ~ cosh, csc ~ coth. The latter are all "stretched out" versions of the former, satisfying relations such as sin² + cos² = 1 = tanh² + sech² and artanh(sin x) = arsinh(tan x). Many years later, I found out about the gudermannian function, which neatly encapsulates all of this.
I didn't understand. Cosine is a stretched out version of sine? One is a translation of the other one. What do you mean by "stretched out"?
@samueldeandrade8535 Draw a graph of sine. Then, take the (-pi/2, pi/2) section and 'stretch' it (by varying amounts) so it covers the real line. It will now resemble the graph of hyperbolic tangent. Likewise, cosine will become hyperbolic secant, etc. This is all made much more precise by the (somewhat obscure) gudermannian function. Alas, I can only go so far in a UA-cam comment.
@@tomkerruish2982 thank you for this great insight! I hadn't noticed this stretching out relationship before. Thanks also for the reference to the gudermannian function.
This is gd(x)=arctan(sinh x)=∫(0 to x) sech t dt, mapping ℝ to ]-π/2,π/2[.
Your stretching out function is its inverse gd⁻¹(x)=arsinh(tan x)=∫(0 to x) sec t dt, mapping ]-π/2,π/2[ to ℝ.
(from the Wikipedia article on the gudermannian function)
From your comment and knowledge of an antiderivative of sec x from calculus, we get three ways of expressing the same function gd⁻¹(x)=∫(0 to x) sec t dt defined on ]-π/2,π/2[:
ln(sec x+tan x)
arsinh(tan x)
arctanh(sin x)
We can check this by finding the derivative of each function on ]-π/2,π/2[.
[ln(sec x+tan x)]'=(sec x+tan x)'/(sec x+tan x)=(sec x tan x+sec²x)/(sec x+tan x)=sec x
[arsinh(tan x)]'=sec²x/√(1+tan²x)
(as (arsinh x)'=1/√(1+x²))
=sec²x/√(sec²x)=sec²x/sec x (as sec x>0 for x∈]-π/2,π/2[)
=sec x
[arctanh(sin x)]'=cos x/(1-sin²x)
(as (artanh x)'=1/(1-x²))
=cos x/cos²x=1/cos x=sec x
This confirms each function has derivative sec x. Since all three functions equal 0 at x=0, the functions are equal on ]-π/2,π/2[, and in fact equal ∫(0 to x) sec t dt.
That's really cool.
5:43 Another way to rule out the negative solution for cosh(x) is to note cosh is never negative ;)
Good point!
Very cool... The two curves are also orthogonal at that intersection so the slope of the cosine curve is -1/phi which is technically also the golden ratio (just negative).
Very elegant! Thanks for sharing!
That's a nice fact about the Golden Ratio, but personally, I prefer the appearance of the Golden Ratio when one considers the inflection points of a quartic polynomial function. :)
Your videos are always interesting
"Ok, so if you have a graph of y equals cossacks-"
Really threw me off there for a moment.
That's crazy how I got a bit annoyed of math vulgarization because I don't often find new curious subjects, but I don't know how you are the only one who still interests me with cool and new stuff
Amazing 👍
Math video poster at 2am
Me at 2:10 : 👁👁
I think he uploaded it at a more reasonable time where he lives.
..the brilliant Dr Barker
It's crazy how often you see the golden ratio, or the golden quadratic, as I like to call that equation, in math...
It pops up almost as much as pi.
Interesting!
two curves are also orthogonal? Is that connected?
coshx is never negative (or null) anyway...
Didn't see someone also mentioned it...
How unique is this property of the golden ratio? Do the other metallic ratios have similar properties?
That’s awesome
Hi Dr. Barker!
Please, i am new to existence and I want to ask you why 1/sinx = 1/cos^2(x) is gradient? Sorry, I am a new human.
The derivative/gradient of tan(x) is sec^2(x) or 1/cos^2(x). The condition which satisfies the intersection of the two graphs is tan(x) = cos(x), which is equivalent to sin(x) = cos^2(x).
Substituting cos^2(x) = sin(x) in the equation for gradient leads to gradient = 1/sin(x).
Hardly surprising since the point of intersection of x*cot( (pi*x)/2) intersect one of the axis at 2/pi:
ua-cam.com/video/bNd1gqoyvdc/v-deo.html
😮
👍
cos x = tan x
cos^2 x = sin x
sin^2 x + sin x - 1 = 0
4 sin^2 x + 4 sin x = 4
(2 sin x + 1)^2 = 5
sin x = (-1 ± √5)/2
(tan x)' = sec^2 x = 1/cos^2 x
cos^2 x = 1 - sin^2 x = sin x since sin^2 x + sin x - 1 = 0
1/cos^2 x = 2/(-1 + √5) = 2(√5 + 1)/4 = (√5 + 1)/2 - the golden ratio.
Please say ‘tangent x’ and ‘cosine x’. Students who are just learning trig are confused enough without adding tan/tangent and cos/cosine (but not sin/sine) to the mix.
I don't think the video is intended for students just learning trigonometry
Surely students just learning trig functions encounter their abbreviations right from the outset anyway. That's how these functions are denoted on their calculators. In 37 years of teaching, I never met one who found the use of these abbreviations (whether orally or in writing) to cause confusion.
In Germany, tan is called "tangens", not "tangent". I think calling it "tangent" leads to even more confusion, because a tangent actually is a line, not a ratio of two lengths in a right triangle.
Using the short names is closer to being universal
the denotions can be different, but they're always understandable for the trig things, down here (where i am from) we use "tg" for the tangent and "ctg" for cotangent