Leibniz quest for Pi pt1

[daubabylon]

bad audio, great class

[alifeofreason]

Leibniz = laibnits

[alifeofreason]

It's pronounced Laibnits, not Libnis...

[thefifthlord1]

pi/4 = 180/4 = 45 degrees not 90, pi/4 = 1/8th a circle.

[gregg4]

@thefifthlord1

surface area circle is PI * r * r

if r = 1, surface area of quarter circle is PI / 4

You mix up with circumference of circle. Circumference circle is 2 * r * PI. If r = 1, circumference is 2 * PI and

half-circle (180 degrees) would be PI.

Point of video is the find surface area (integration -> area under curve) under quarter circle, i.e. PI/4

Gregg

[GelandnaleG]

2:00

[chirrrs]

gooooooooooood tiiiiiiiiiiiiiiimesssssss!

11:47 PM

[DoubleDutchBust]

3:06AM

[wepf2000]

You actually get (1/2)y - int(xdy) ( both from zero to 1). Which gives you:

(1/2)y(1) - integral (xdy, from 0 to 1). Now you can figure out y(1) by using the proportion x/2=y^2/(1+y^2) in part 3 . That's how you get y(1)=1.

[sourasteroid]

that was really some clever thinking by the man...

[supermegaburne]

so easy. let's talk about complex numbers!

[x24isis]

wow! thank you for this fantastic tour of mathematical genius from the mind of Liebniz! amazing!

[Patsan120]

I still dont get how ds = OS and dx = OR. I cant understand the congruency between the two trinagles.

[davis1337]

lets call the angle onto PQ; N (90°). angle RS0 = angle PQN -> these are F-corners. SR0 = PQN = 90°. that leaves us with one corner which is 180°-90°-PQN = 180°-90°-RS0. so there is an AAA-congruency (three of the same angles).

[gregg4]

@Patsan120

ds need not be equal to OS or dx to OR; what is equal is their ratio

(OS / OR ) = ( ds / dx )

Gregg

[mazemaster225]

9:00-10:35: nice move

[BernardoDW]

very very nice !

[davis1337]

i don't get that part...for example: what is d in d/dx?

what rules of partial integration does he use?

[OJReadMore]

d in (d÷dx) means 'take derivative of'.

But basically all he did was split the value of 1 into x and (d÷dx) because x times d÷dx) is equal to 1.

∫v×du = u×v - ∫v×du

(solving for right side) where:

u = y  du = (dy÷dx) dx

v = x dv = x (d÷dx) dx

Special: notice [ x (d÷dx) ] = 1

So if you just plug in the above values, that gets you to where he is at 10:06. Then he evalutes from y(1) - y(0).

(Integration by Parts is taught in second semester Calculus.)

[OJReadMore]

(sorry, the above formula is actually)

∫u×dv = u×v - ∫v×du

[HomerWells007]

I'm scratching my head. Something seems to escape me. Maybe I'm not understanding it clearly but how did you jump from the integrand of y from [0,1] to a nonlinear expression d/dx xy?

[Onyxyte]

Interesting. Thanks for sharing your knowledge.

[mexaguil]

Really good thank for that

[sanchezdot]

:::)