a geometric approach to a famous integral

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[@tommasoantonelli7176]

9:32 -> Ridiculous place to stop 20:03 -> Good place to stop

[@edwardfyodorov8268]

This is hilarious

[@Happy_Abe]

Love this 😆

[@redpepper74]

Exactly my thoughts

[@Ahmed-Youcef1959]

👍👍👍

[@williamperez-hernandez3968]

The slope of the line is 1/t, so we get x=t/sqrt(t^2 +1). This does give y=1/sqrt(t^2+1) -1 as given in the vid.

[@nahuelcaruso]

Yes, the misleading slope t leads to a lower bound equal to (t/sqrt(t^2+1) -1). However, with this little correction the proof runs without problem

[@JeanYvesBouguet]

The best part of this video is when showing the equality of areas between the As and the Bs respectively. This is the most interesting and non intuitive part of the method in my opinion.

[@hydropage2855]

Several mistakes starting at around 13:00. The slope was supposed to be 1/t. You forgot the square root when solving for x, and even though you remembered it later you forgot the t when plugging into y = tx - 1. Should’ve been t/sqrt(t^2 + 1) - 1. You got lucky and your mistakes canceled out

[@djttv]

Who would have thought that A1=B1 and A2=B2? Very interesting video!

[@yutaj5296]

13:05 The equation of the line should be y=x/t-1. The result that y=1/sqrt(t^2+1)-1 is correct.

[@Jack_Callcott_AU]

👍 You are correct!

[@Jack_Callcott_AU]

It seems that the mistake he made cancels out. ✔

[@goodplacetostop2973]

20:03

[@BrianDominy]

9:31 Not a good place to stop

[@MathTutor1]

This is beautiful. Keep up the good work.

[@Milan_Openfeint]

Now we need a geometric argument why A1=B1 and A2=B2.

[@bsmith6276]

When introducing the substitution u=sqrt(2y)-1 I think if you broke up the integrand into sqrt(1-2y) / sqrt(2y) then the substitution may seem a bit more motivated since dy/sqrt(2y) is the differential of sqrt(2y).

[@gp-ht7ug]

Cool video! I like when you put together geometry and calculus

[@manucitomx]

This was great fun. Thank you, professor.

[@coreymonsta7505]

That’s the bigger picture of FTC1,2! Anti derivatives are like cumulative area functions, since they’re of that form up to a constant

[@user-ow2ll9jf8z]

and 1/(1+x2)=(i/2(x+i))-(i/2(x-i)),so tanx=(i/2)ln((x+i)/(x-i))+C.You can find the value of C by taking the derivative of sine x or cosine x.This is how most civilizations in the universe connect the real and complex domains:)

[@bellfoozwell]

Nice explanation!

[@ryoikitokuiten]

Wow. Really nice approach.

[@The_Green_Man_OAP]

13:47 Mistake here. Should be x=1/√(t²+1)

[@estudematematica]

Yet another great video, but I have a feeling that it got unnecessarily rushed from minute 15 or so… we’ll be around if it takes an extra minute or two, Mr. Penn! 😃👍

[@Bruno-yg9lu]

oi

[@maurobraunstein9497]

I just saw another video about that this week! kzbin.info/www/bejne/kJTFiJ16a610ldU Another Roof's geometric interpretation is quite different, drawing a circle under the Witch and comparing the area of a slice of the Witch to the area of a slice of the circle. In the more than 20 years I've known that the integral ∫dx/(1 + x^2) is arctan(x) + C, I had never, until a few days ago, even considered that there might be a relatively simple geometric interpretation, and now I know two!

[@ingiford175]

It is seems to be the same general proof, but the location of the circle is in a different place. Penn has his circle under the x axis, while the other proof is a circle half the radius within the the curve and the x axis. Both are interesting and forgot I saw the other version earlier this month.

[@randomuser-xc2wr]

I was thinking the other day that all these transcendental functions should really be defined by the integral definition of their inverse, so we have inverse of exponential = int( 1 / sqrt(x^2) ) inverse of elliptical sine = int( 1 / sqrt(1 - x^2) ) inverse of hyperbolic sine = int( 1 / sqrt(1 + x^2) ) inverse of parabolic sine = int( 2-x / 2sqrt(1 - x) ) -Well, I wasn't sure about the last one which sent me down the rabbit hole of parabolic trig functions, but it all looked so familiar and of course this channel had a video about parabolic trig functions some time ago- -I re-watched the video but sadly no inverse functions and the net was no help....so I gave up, but I still think- I found the formula for the inverse of parabolic sine from the margin of: *Parabolic Trigonometry* by G. Dattoli · E. Di Palma · J. Gielis · S.Licciardi. It's the same paper (or at least same authors) used by prof. Penn in his video on parabolic sine and cosine. These integrals should be the starting point of the whole silly transcendentals....start with the integrals, define them as numeric functions to be calculated by approximation. Then define the exponential & trig functions as their inverses...makes much more sense, derive their relation to arc length of an ellipse, hyperbola, parabola with no need of angle nonsense, just a simple application of line integrals and Bobs your uncle. It would be interesting to find a relation between the exponential and the parabolic sine cosine functions, such that we can write: exp( i theta ) = cos(theta) + i sin(theta) exp( theta ) = cosh(theta) + sinh(theta) exp( ??) = cosp(theta) ?? sinp(theta) There are helpful hints in the papers by Dattoli et al. and I might find something in the margin (again!), if anyone has an idea let know.

[@randomuser-xc2wr]

should the exponential function be renamed to *Linear Sine* function?! The Line is not considered a conic section (at least according to Wikipedia) but you can easily see if the plane is parallel with the vertical axis of the cone and the tip point of the cone is in the plane then the section will be two lines that meet in the tip point, which can be represented with abs(x) or sqrt(x^2) as a function. And what is the integral definition of the inverse exp.? well it's int(1/sqrt(x^2))!

[@randomuser-xc2wr]

Well I looked into this a little bit further and indeed the lines are a conic section with equation y^2 - x^2 = 0 arc length for a given x = int( sqrt(2) , t=0...x) = sqrt(2)*x arc length for a given y = int( sqrt(2) , t=0...y) = sqrt(2)*y the sqrt(2) comes from the equation above and the definition of line integral, because: sqrt(2) = sqrt( 1 + (dy/dx)^2 ) sqrt(2) = sqrt( 1 + (dx/dy)^2 ) but what about the integral of 1/sqrt(x^2) ?? where does that come from? is this the inverse of Linear sine & cosine and not the exponential? Is the exponential something else but somehow connected to the conic sections? is there a fifth section? What I am having trouble with understanding is how the angle theta & the tangent be defined? It almost makes more sense to take the equation: y^2 - (x-1)^2 = 0 instead....hmm.

[@randomuser-xc2wr]

I am quite convinced that *Arcometric Function* (i.e. Arc-length measuring functions for a given point on the curve) are the correct starting points while the traditional Trigonometric Functions (i.e. Line-length measuring functions for a given angle) are not well-defined and are better viewed as the inverse of the Arcometric Functions (which is the opposite of the current formulation). There are FOUR Conic Sections (notice that it is the section between a single plane and a single conic) even when the ellipse and circle are considered one conic section: The Linear Conic Section (looks like a \/) given by the equation: y^2 - x^2 =0 The Hyperbolic Conic Section (single branch) given by the equation: y^2 - x^2 = 1 The Elliptical Conic Section (includes the circle) given by the equation: y^2 + x^2 = 1 The Parabolic Conic Section given by the equation: y^2 + x = 1 Here I am using the standard unit "conic section" similar to the unit circle used in Trigonometry, but by using more general forms we can show that the Limit of of the Hyperbolic CS goes to the Linear CS and The limit of the Ellpitical goes to the Parabolic CS. Then for a given x or y you can differentiate the equation and use the line integral to define an arc-measuring function, i.e. an Arcometric Function, for each CS (see comments above) That's how we get the familiar arcsin(x) and arcsinh(x) but also two new others to give the length of a linear and a parabolic arc (see the comments above). All these functions must be calculated *numerically* this includes calculating Pi (for the circle) and sqrt(2) (for the line) and this fact should be front and center (which, again, is the opposite of the current teaching methods that emphasis finding rational and exact values for sine & cosine, when we know that only a handful of angles satisfy that: the famous 30, 45, 60, 90 in the first quarter and even here the sine of 45 is 1/sqrt(2) which has to be calculated numerically eventually) I am not sure how ln(x) comes up (maybe the linear arctangent, but I am not sure) but what is known is that the arcsin(x) and arcsinh(x) both have a logarithmic forms as well as their integral definitions: en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms en.wikipedia.org/wiki/Inverse_hyperbolic_functions#Definitions_in_terms_of_logarithms These forms follows from the formula of Exp(i*x) and Exp(x) in terms of, respectively, sin(x) & cos(x) and sinh(x) & cosh(x). Finding similar formulas for Linear and Parabolic Arcometric functions or logarithmic forms would be an interesting path of research. I think I reached the limit of what I can do here, I'll leave these comments as long as possible for someone maybe will be inspired to write it all up more thoroughly and completely but I will eventually delete it so please copy it if you're interested :)

[@randomuser-xc2wr]

ln(x) = int( 1/t , t=1...x) arctan(x) = int( 1 / (1 - t^2), t=0...x) arctanh(x) = int( 1 / (1 + t^2), t=0...x) arctanp(x) = int( 1 / (1- t), t=0...x) which gives theta = -ln( 1 - tanp(theta)) 1 - tanp(theta) = exp(-theta) so we can write: exp(i^0 x ) = cosh(x) + sinh(x) exp(i^1 x) = cos(x) + i sin(x) exp(i^2 x ) = 1 - sinp(x)/cosp(x) The integral definitions of the natural log, the arc tangent and the hyperbolic arc tangent and the similar defined parabolic arc tangent are all very easy to compute numerically and to understand visually and they all should be the starting point for studying the exponential and trigonometric functions.

[@CTJ2619]

well done - i liked the visual representation of what was going on -

[@backyard282]

only thing i didn't quite understand is why you you were using the function (1/2) * 1/(x^2+1). Why the 1/2?

[@Aditya_196]

For the last part the way we calculated the arcs area we got the 1/2(theta)*1² ..if u won't do that u will still get same sort of stuff but I believe the people who come up with the proof already had the basis of proof as using that arcs area so they just did it with 1/2 factor in the beginning to avoid some extra computational efforts 👍🏻

[@Jack_Callcott_AU]

Hey @backyard282, I think the reason for that is that 1/( X ^2 + 1) is an even function, so the integral from -t to t on (-t, t) is 2 times the integral from 0 to t on (0, t). ✔

[@bsmith6276]

Probably because he can then say area A1=area B2 and area A2=area B1 near the end.

[@DeJay7]

Never explained the reasoning behind the 1/2 factor in the initial function, very strange. I think it just happens to make Area(A1) = Area(B1) and Area(A2) = Area(B2) instead of having a factor of 2.

[@nicolascamargo8339]

Wow increible

[@pieters286]

most enjoyable derivation!

[@byronwatkins2565]

At 14:00, tx-1=t/sqrt(t^2+1)-1

[@bethhentges]

13:48 He means x^2=

[@marc-andredesrosiers523]

good job 🙂

[@levprotter1231]

Any hyperbolic equivalent?

[@easymathematik]

Nice topic.

[@erfanmohagheghian707]

You had all the coordinates of the intersections. Why did you use similar triangles? :)))

[@udic01]

Like everyone else commented, the slope is 1/t

[@bethhentges]

15:40 He wrote the similarity in the wrong order.

[@jounik]

Why was it necessary to even split the area below the x-axis? It's a triangle with base t and height 1, so its area is t/2. It would've been easier - and less error-prone - to just complete the integral for A1 and show that A1 and A2 sum to t/2.

[@Detka48]

Because the whole point is to avoid calculating any integrals.

[@kruksog]

Why introduce the factor of 1/2?

[@bethhentges]

Because it appears at the end when finding the area of the sector.

[@kruksog]

@@bethhentges don't you think it might be a good idea to explain or say something about that, rather than allowing it to be an utter surprise at the end of a nearly half hour derivation?

[@klofat]

Where is +c? Great video, enjoyed it a lot.

[@EqSlay]

Nice

[@letitiabeausoleil4025]

I only get abouyt 1/50 of your problems out before the video ends. Behold! This was one of them.

[@zh84]

Fascinating. I never would have thought of this, but you led me right through it. Thank you Back in 1991 I tried expanding exp(-x²) with Mathematica in powers of 1/(1+x²) because I thought the two functions looked similar (both equal to 1 when x = 0, both asymptotic to the x-axis) but it turned out to be horribly messy.

[@PillarArt]

🙏😺

[@Aditya_196]

I have wondered it for a long time if he would rename us channel to michael pen²

[@MacHooolahan]

Is there a (different) proof here that involves d-theta-ing some angle round from the origin to intersect with that pythag-y function line? Feels like there is but it's probably not going to come from me right now, having had several beers! :O

[@fonzi102]

:D

[@jimskea224]

Original publication by "A Insel". Like most mathematics nerds.

[@bethhentges]

16:52 He says “C times D” when he means “the length of CD.”

[@leiv4642]

So? It happens...