a geometric approach to a famous integral
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- Опубліковано 4 тра 2024
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9:32 -> Ridiculous place to stop
20:03 -> Good place to stop
This is hilarious
Love this 😆
Exactly my thoughts
👍👍👍
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.
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
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.
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
Who would have thought that A1=B1 and A2=B2? Very interesting video!
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.
👍 You are correct!
It seems that the mistake he made cancels out. ✔
This is beautiful. Keep up the good work.
Now we need a geometric argument why A1=B1 and A2=B2.
9:31 Not a good place to stop
20:03
This was great fun.
Thank you, professor.
Cool video! I like when you put together geometry and calculus
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).
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! 😃👍
oi
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
Nice explanation!
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:)
Wow. Really nice approach.
well done - i liked the visual representation of what was going on -
This integral proof is so complicated that I just don't see it as having merit. The standard trigonometric substitution is so simple.
13:47 Mistake here. Should be x=1/√(t²+1)
Wow increible
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.
Any hyperbolic equivalent?
good job 🙂
most enjoyable derivation!
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?
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 👍🏻
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). ✔
Probably because he can then say area A1=area B2 and area A2=area B1 near the end.
At 14:00, tx-1=t/sqrt(t^2+1)-1
Nice topic.
I just saw another video about that this week! ua-cam.com/video/ZcbWjD6HA_s/v-deo.html 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!
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.
13:48 He means x^2=
You had all the coordinates of the intersections. Why did you use similar triangles? :)))
15:40 He wrote the similarity in the wrong order.
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.
Because the whole point is to avoid calculating any integrals.
Why introduce the factor of 1/2?
Because it appears at the end when finding the area of the sector.
@@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?
Nice
Like everyone else commented, the slope is 1/t
I only get abouyt 1/50 of your problems out before the video ends. Behold! This was one of them.
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.
🙏😺
Where is +c?
Great video, enjoyed it a lot.
I have wondered it for a long time if he would rename us channel to michael pen²
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
:D
Original publication by "A Insel". Like most mathematics nerds.
16:52 He says “C times D” when he means “the length of CD.”
So? It happens...