Video není dostupné.
Omlouváme se.
DESTROYING
Vložit
- čas přidán 22. 11. 2023
- Advanced MathWear:
my-store-ef6c0...
Complex analysis lectures:
• Complex Analysis Lectures
If you like the videos and would like to support the channel:
/ maths505
You can follow me on Instagram for write ups that come in handy for my videos:
...
My LinkedIn:
/ kamaal-mirza-86b380252
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!
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.
Damn bro you spittin🔥 lemme see what I can do
@@maths_505 yes... haha YAAAAA-
Graduate mechanics texts are a great place to start!
@@PopPhyzzle I have studied them. Just need some revision before posting some problems here. Thanks for the recommendation bro.
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
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
Thanks.....I'll give it a shot
Wolfram alpha can’t solve it
That's cuz I haven't told Wolfie how to solve it yet 😎😂🔥
Lobachevsky integral formula will do it here
@@maths_505😂😂beautiful comment
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
Does anyone have the link to MichaelPenn's video on this?
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
I solved it like this too. Nice to see this highlighted.
Smart Idea for solving integral. So, thank you for your fruitful effort.
THE BEST WAY:
Integrate by parts like you did
2:53 Frulani integral
Like you said
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.
The thumbnail for this one was like "Can you guess the trick??" and I was like, looks like Feynman to me. You too 😂
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
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
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?
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?
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
Excellent as always my friend
@@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
@@maths_505 speaking about Michael Penn's videos
I like this video czcams.com/video/g3cc1OgU_Kg/video.html
because this idea helped me to expand generating function for Hermite polynomials
also channel mathmajor looks promising
In which program you made this video?
Samsung notes
Yayyy feynmann's again
❤
Genial
0:32
The weirdest integration by parts notation I have seen in my life
I like my udv+vdu=d(uv) al dente
I do it that way too, it's much cleaner