Відео

Thank you for this video. If you would like to calculate the x value corresponding to a 95% confidence interval with df > 2, how can you apply the approximate function? It will be exciting if I can watch an example in your new video!

🌲🌳 log

freaky sonic video next

ur bald fans (me) were anxiously awaiting this video

One can make life simpler by choosing b as the Feynman parameter. dI/db = 2b \int^{\infty}_{-\infty} dx i/(x^2+a^2)(x^2+b^2) = 2\pi /a 1/(b+a) by contour integration. Then I =( 2\pi/a )ln(b+a) + c. Setting b=0, we get J=\int^{\infty}_{-\infty} dx ln(x^2)/(x^2+a^2) = (2/pi /a )ln a . Thus, c = 0. So, I = ( 2\pi/a )ln(b+a).

zesty

can you explain eulers identity

integrating for funsies?

ginger math biology takeover when?

this is so neat i love it when people do math neat

I think I’m in the wrong classroom

You ate and left no crumbs

ginger what about cultural ecosystem services :(

deforestation bad :( tree good :)

love this man

Imagine taking APES

I can barely hear you

hi! i'm so so confused, do you mind explaining the concept of squircles for a beginner in math to then explain deriving the volume formula. please😭

ceil(x)-x=1-{x} 0<={x}<1, so the geometric series for 1/(1-{x}) converges. The problem, then, is finding the ceiling of the sum, because if {x} is something like .999 the ceiling will be much higher than when it's something smaller like .6

ISABELLA GOT A LOLLIPOP!!!!!!

I understand every bit of this!

I always thought those floors and ceilings were so hard.

Nice video! Are you not forgetting a (-1)^m in the final solution due to the i^2m?

Hi 🙂Maybe at 16:41 you should add (-1)^m

4:45 Missed a t multiplying the second log.

wow the mic quality is so good

Looks interesting, but I can't figure out what steps you're skipping when you claim an bounds of zero and two for the integral wrt t.

specific heat of potato is about 3.4 kJ/kg

Here’s to hoping you continue this series

Nice one

Very interesting

WOAH!!! 🤯🤯🤯🤯

I've used f(z) = ln(z + ib) / (z² + a²) , b>0 And then contour integration by using upper side rectangular contour with residue at z = ia , and then equating real parts and we got it.

Integration by parts twice and substitution x = tan(t) leads us to the integral -4\int_{0}^{\frac{\pi}{2}}\ln{(cos{(t)})}dt

I'm curious to know how we can the pole at z=i because it is of infinite order and the power series can't be used as well due to the 1/(z+i) being there...any ideas on how to calculate it...?

I=int[-♾️,♾️]((1+x^2)^-(n+1))dx x=tan(y) dx=sec^2(y)dy I=int[-pi/2,pi/2](sec^(-2n)(y))dy I=2•int[0,pi/2](sin^0(y)cos^2n(y))dy beta(u,v)=2•int[0,pi/2](sin^(2u-1)(y)cos^(2v-1)(y))dy I=beta(1/2,n+1/2) I=sqrt(pi)gamma(n+1/2)/gamma(n+1)

y=x^t x=y^(1/t) dx=1/t•y^(1/t-1)•dy I=1/t^2•int[0,1](ln(y)ln(1-y)/y)dy u=ln(y) dv=ln(1-y)/y du=dy/y v=-Li_2(y) I=1/t^2•(-ln(y)Li_2(y)|[0,1]+int[0,1](Li_2(y)/y)dy) I=Li_3(1)/t^2 Li_3(1)=sum[k=1,♾️](1^k/k^3)=zeta(3) I=zeta(3)/t^2

what about the arcs tho?

I didn't use complex analysis but simple integration techniques and I got the result Integral(-½ to +½) Γ(1+x) Γ(1-x) dx = (4/π) β(2) = 4G/π Here β is dirichlet beta function.

Please dont say lohopitals)

Choise of contours always seems completely arbitrary to me ://

bro forgot to edit out the first take

L'Hopital)

Take the derivative of both sides to give y’ = y’’ + y’’’ + … Then plug into the original equation to get y = 2y’ Giving the solution y = Ae^ (x/2)

cool method before i watch the video i think that you will use the box contour but i suprise by using the semicircle contour really wonderful 🥰🥰

really cool🥰🥰 i love this contour integration🤩 but sir can you explain why the second integral =- pi res(0) because i couldn't really understand. thank you !! and keep going! more contour integration😊

Another way you could do it: find a power series for ln(1-x^t)/x and then you pretty much all that's left in the integral is ln(x) ⋅ power function which is easily handled by integration by parts

try and get rid of the annoying background noise in future videos