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
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.
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:)
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).
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.
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). ✔
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! 😃👍
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.
I just saw another video about that this week! th-cam.com/video/ZcbWjD6HA_s/w-d-xo.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.
@@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?
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.
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
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
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
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.
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. ✔
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:)
20:03
Who would have thought that A1=B1 and A2=B2? Very interesting video!
This is beautiful. Keep up the good work.
9:31 Not a good place to stop
13:48 He means x^2=
13:47 Mistake here. Should be x=1/√(t²+1)
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).
Now we need a geometric argument why A1=B1 and A2=B2.
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.
Cool video! I like when you put together geometry and calculus
At 14:00, tx-1=t/sqrt(t^2+1)-1
This was great fun.
Thank you, professor.
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
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.
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
Wow. Really nice approach.
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.
well done - i liked the visual representation of what was going on -
Nice explanation!
Personally I think it'd save some time and effort at 12:40 by spotting a pair of similar triangles rather than solving simultaneous equations
I just saw another video about that this week! th-cam.com/video/ZcbWjD6HA_s/w-d-xo.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.
This integral proof is so complicated that I just don't see it as having merit. The standard trigonometric substitution is so simple.
Any hyperbolic equivalent?
Wow increible
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 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?
most enjoyable derivation!
Where is +c?
Great video, enjoyed it a lot.
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.
good job 🙂
I only get abouyt 1/50 of your problems out before the video ends. Behold! This was one of them.
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
Like everyone else commented, the slope is 1/t
Nice topic.
I have wondered it for a long time if he would rename us channel to michael pen²
Nice
🙏😺
16:52 He says “C times D” when he means “the length of CD.”
So? It happens...
Original publication by "A Insel". Like most mathematics nerds.
:D