Ginger Math

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@RashmiRay-c1y Күн бұрын

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).

@mini_eiger3493 15 күн бұрын

zesty

@mini_eiger3493 21 күн бұрын

can you explain eulers identity

@mini_eiger3493 21 күн бұрын

integrating for funsies?

@GingerMath 21 күн бұрын

Always for funsies :) But that would be a great title - Fourier for Funsies... ok yeah that's going to be an upcoming video (and an excuse to do Fourier things (if there needs to be a reason in the first place :) ))

@mini_eiger3493 21 күн бұрын

ginger math biology takeover when?

@GingerMath 21 күн бұрын

You tell me >:|

@mini_eiger3493 27 күн бұрын

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

@teodagreat 28 күн бұрын

I think I’m in the wrong classroom

@sebas31415 28 күн бұрын

You ate and left no crumbs

@mini_eiger3493 28 күн бұрын

ginger what about cultural ecosystem services :(

@GingerMath 28 күн бұрын

I'm sure there will be a Part 2 at some point :)

@mini_eiger3493 28 күн бұрын

deforestation bad :( tree good :)

@jakebroxterman640 29 күн бұрын

love this man

@jw4215 29 күн бұрын

Imagine taking APES

@RanBlakePiano Ай бұрын

I can barely hear you

@lincolnabena770 Ай бұрын

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😭

@maxvangulik1988 Ай бұрын

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

@mini_eiger3493 Ай бұрын

ISABELLA GOT A LOLLIPOP!!!!!!

@GingerMath Ай бұрын

YES SHE DID! I should include bloopers like that more often

@teodagreat Ай бұрын

I understand every bit of this!

@GingerMath Ай бұрын

AWESOME - wanna share it with some friends then? :D

@teodagreat Ай бұрын

@@GingerMath Already did😂

@Jalina69 Ай бұрын

I always thought those floors and ceilings were so hard.

@GingerMath Ай бұрын

Ditto! But if you split them up like this they're (hopefully) manageable

@angusclark6170 Ай бұрын

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

@GingerMath Ай бұрын

Yes I am... whoops - thanks for pointing that out lol

@Mario_Altare Ай бұрын

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

@Mephisto707 Ай бұрын

4:45 Missed a t multiplying the second log.

@mini_eiger3493 2 ай бұрын

wow the mic quality is so good

@worldnotworld 2 ай бұрын

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.

@TungstenCarbideProjectile 2 ай бұрын

specific heat of potato is about 3.4 kJ/kg

@SaintC0bain 2 ай бұрын

Here’s to hoping you continue this series

@Jalina69 2 ай бұрын

Nice one

@carbazone619 2 ай бұрын

Very interesting

@mini_eiger3493 2 ай бұрын

WOAH!!! 🤯🤯🤯🤯

@Anonymous-Indian..2003 2 ай бұрын

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.

@holyshit922 2 ай бұрын

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

@DihinAmarasigha-up5hf 2 ай бұрын

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...?

@carbazone619 2 ай бұрын

The singularity at z=i is no longer a pole in this case, but an essential singularity because the Laurent series doesn’t terminate for some finite negative power.

@maxvangulik1988 3 ай бұрын

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)

@maxvangulik1988 3 ай бұрын

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

@maxvangulik1988 3 ай бұрын

what about the arcs tho?

@Anonymous-Indian..2003 3 ай бұрын

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.

@Anonymous-Indian..2003 3 ай бұрын

But ginger sir , your solution is also really cool.

@Jalina69 3 ай бұрын

Please dont say lohopitals)

@Jalina69 3 ай бұрын

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

@thedefenestrator 3 ай бұрын

bro forgot to edit out the first take

@GingerMath 3 ай бұрын

Well that's embarrassing... but hey it's fixed now :D

@Jalina69 3 ай бұрын

L'Hopital)

@henrydaley9686 3 ай бұрын

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)

@GingerMath 3 ай бұрын

@@henrydaley9686 I meannn you could… but Laplace transforms are fun…

@shivx3295 2 ай бұрын

@@GingerMathi mean why would we waste our time

@carlgauss1702 Ай бұрын

Marvelous solution. Not the most original idea, but it never gets old.

@bahiihab-y2r 3 ай бұрын

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 🥰🥰

@bahiihab-y2r 3 ай бұрын

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😊

@GingerMath 3 ай бұрын

Glad you liked it!! Whenever we have a pole completely inside a contour (going counterclockwise), we would have 2pi*i*Res, but since the pole at zero lies ON the contour, we integrate around it with a semicircle (taking the limit as its radius goes to zero). The result is that instead of 2pi*i*Res is -pi*i*Res, as it is only a semicircle (so pi radians) and we are going around it clockwise, hence the minus sign. I'll be sure to add a bit on this for the next complex analysis video!

@Samir-zb3xk 3 ай бұрын

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

@WonderfulDeath 3 ай бұрын

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

@Potatos2980 2 ай бұрын

try keeping that to yourself in the future

@WonderfulDeath 2 ай бұрын

@@Potatos2980 lol be quiet child

@Jalina69 3 ай бұрын

Really)

@GingerMath 3 ай бұрын

Yes really 😁 - thoughts?

@Jalina69 3 ай бұрын

@@GingerMathmy wild guess is you never were taught cursive)

@GingerMath 3 ай бұрын

@@Jalina69 Man what? 🤣 I know it, my handwriting is just awful

@jw4215 3 ай бұрын

That other dude seems pretty cool

@乔峯 3 ай бұрын

Genius

@holyshit922 2 ай бұрын

Solution too complicated

@jw4215 4 ай бұрын

Academic WEAPON

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