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Advanced MathWear:
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Пікірлер: 44

@renzo711 9 ай бұрын

You can also do that “Nice integral trick” that you did from a past video here! Let f(x)=(1-e^-x)^2=1-2e^-x+e^-2x, and g(x)=1/x^2 and apply the laplace stuff!

@BadlyOrganisedGenius 9 ай бұрын

Nice solve! If you want some more Feynman integrals, here's a tricky one: Integral from (-infinity to +infinity) of tan(x)*ln[1+cos(x)] / x

@maths_505 9 ай бұрын

Thanks.....I'll give it a shot

@anupamamehra6068 9 ай бұрын

Wolfram alpha can’t solve it

@maths_505 9 ай бұрын

That's cuz I haven't told Wolfie how to solve it yet 😎😂🔥

@violintegral 9 ай бұрын

Lobachevsky integral formula will do it here

@Ghostwriter_zone 9 ай бұрын

@@maths_505😂😂beautiful comment

@PopPhyzzle 9 ай бұрын

Oh dude I've got a new content curveball for ya. How familiar are you with Jacobi Elliptical coordinates or elliptical differential equations? I never see anyone cover elliptic trig functions. They're analogous to the regular unit circle trig functions in the context of an ellipse. They even obey almost identical identities.

@maths_505 9 ай бұрын

Damn bro you spittin🔥 lemme see what I can do

@PopPhyzzle 9 ай бұрын

@@maths_505 yes... haha YAAAAA-

@PopPhyzzle 9 ай бұрын

Graduate mechanics texts are a great place to start!

@maths_505 9 ай бұрын

@@PopPhyzzle I have studied them. Just need some revision before posting some problems here. Thanks for the recommendation bro.

@raonimesquitadossantos7175 9 ай бұрын

Bro, I thought it stopped at hyperbolic trig but apparently not. I'm actually wondering how many trig funcrion can you make by combining inverse trig, hyperbolic/elliptical and whatever sinc(x) is

@ElliotUnbound 9 ай бұрын

Found some cool integrals: sqrt(1-(lnx)^2) from 1/e to e sinx/coshx from 0 to infinity cosx/sqrt(x^2+1) from -infinity to infinity sqrt(1-(sinhx)^2) from -1 to 1

@matteofiorillo9599 9 ай бұрын

I've also solved this integral after watching Michael Penn's video with Feynman's trick, good work man😎. We can apply the trick directly from the beginning putting e^(-αx) and resolving as usual

@MrWael1970 9 ай бұрын

Smart Idea for solving integral. So, thank you for your fruitful effort.

@magicianwizard4294 9 ай бұрын

I solved it like this too. Nice to see this highlighted.

@nightmareintegral5593 9 ай бұрын

THE BEST WAY: Integrate by parts like you did 2:53 Frulani integral Like you said

@DendrocnideMoroides 7 ай бұрын

Does anyone have the link to MichaelPenn's video on this?

@manstuckinabox3679 9 ай бұрын

2:16, wouldn't be using hospital's rule be redundant since by definition, f'(0) of (1-e^-x)^2 = lim x-->0 of (1-e^-x)^2-f(0)/x which is the desired limit? then by computing the derivative it would be the value of 2(1-e^-x)e^-x at 0? I know it's literally the same thing, but it's basically the same issue with sin(x)/x, right? 3:17 can't we immediately define e^-ax/x and integrate the derivative from 1 to 2? Anyways awesome Vid Mah Man! I remember this integral being posted a while back.

@holyshit922 9 ай бұрын

If you like differential equations you can solve (1-x^2)y''-xy'+n^2y=0 two ways with change of independent variable and with power series And with this equation y''(t) + 2(t - x)y'(t)+2y(t)=0 y(0) = 1 y'(0) = 2x particular solution is easier to guess after reduction to Riccati

@holyshit922 9 ай бұрын

My proposition how to reduce to Riccati and then to Bernoulli when particular solution is easy to guess Lets start from equation y''(t)+p(t)y'(t)+q(t)y(t) = 0 Let's try to get quotient rule on the LHS y''(t)=-p(t)y'(t)-q(t)y(t)| : y(t) y''(t)/y(t) = -p(t)(y'(t)/y(t)) -q(t)| - (y'(t)/y(t))^2 y''(t)/y(t) - (y'(t)y'(t))/y(t)^2 = -(y'(t)/y(t))^2 -p(t)(y'(t)/y(t)) -q(t) (y''(t)y(t) - y'(t)y'(t))/y(t)^2 = -(y'(t)/y(t))^2 -p(t)(y'(t)/y(t)) -q(t) Now we heve quotient rule on the LHS so (y'(t)/y(t))' = -(y'(t)/y(t))^2 -p(t)(y'(t)/y(t)) -q(t) Let z(t) = y'(t)/y(t) then we have z'(t) = - z(t)^2 - p(t)z(t)-q(t) y'(t) = y(t)z(t) Let'a assume that we somehow found particular solution of this Riccati equation We can reduce it to Bernoulli following way Let z1(t) be the particular solution of Riccati z'(t) = - z(t)^2 - p(t)z(t)-q(t) (Eq 1) z1'(t) = -z1(t)^2-p(t)z1(t)-q(t) (Eq 2) Lets subtract Eq2 from Eq 1 z'(t) - z1'(t) = -(z(t)^2 - z1(t)^2) -p(t)(z(t)-z1(t)) (z(t) - z1(t))' = -(z(t) - z1(t))(z(t) + z1(t))-p(t)(z(t)-z1(t)) (z(t) - z1(t))' = -(z(t) - z1(t))((z(t) - z1(t)) + 2z1(t))-p(t)(z(t)-z1(t)) (z(t) - z1(t))' = -(z(t) - z1(t))^2 -2z1(t)(z(t) - z1(t))-p(t)(z(t)-z1(t)) (z(t) - z1(t))' = -(z(t) - z1(t))^2 - (p(t) + 2z1(t))(z(t) - z1(t)) (z(t) - z1(t))' + (p(t) + 2z1(t))(z(t) - z1(t)) = -(z(t) - z1(t))^2 Let w(t) = z(t) - z1(t) and we have following Bernoulli equation w'(t) + (p(t) + 2z1(t))w(t) = -w(t)^2

@jkid1134 9 ай бұрын

The thumbnail for this one was like "Can you guess the trick??" and I was like, looks like Feynman to me. You too 😂

@holyshit922 9 ай бұрын

My way was Rewrite integrand as (1-exp(-x))^2/x^2 then integration by parts with u = (1-exp(-x))^2 and dv = 1/x^2 dx Substitution u = exp(-x) Leibniz rule for differentiating under integral sign

@maths_505 9 ай бұрын

Excellent as always my friend

@holyshit922 9 ай бұрын

@@maths_505 Your way remids me calculating Laplace transform of (1-exp(-t))/t which is ln(1+1/s) and if we plug in s = 1 we will have the result

@holyshit922 9 ай бұрын

@@maths_505 speaking about Michael Penn's videos I like this video kzbin.info/www/bejne/nWTGlGSFnLqSgck because this idea helped me to expand generating function for Hermite polynomials also channel mathmajor looks promising

@giuseppemalaguti435 9 ай бұрын

Uso feyman con e^(-ax)...I(a=1)=I...derivò due volte,ma risulta I"=0,I'=0..I=0,..boh,non riesco a trovare l'errore...ok,con la trasformata di Laplace e la proprietà risulta 2ln2...ma con feyman risulta I=0...why?

@VibesStudy 9 ай бұрын

Hi, its been quite a while watching your channel, and i love it, the way you compete and present solution, like its so fun. I'm a calc 1 student, and watching your vids, to improve my skills. I was struggling in a limit problem, I know the standard limit lim h->0 ((x^h-1)/h) = ln(x) But why is, lim h->0 (((x+h)^h-1)/h) = ln(x) Do we use partial limit, or is there any similar way to prove this, as before?