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- เผยแพร่เมื่อ 5 ก.ย. 2024
- Analysis of an interesting case study in differential equations leading to some delightful complex valued functions
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0:11 a very elegant way of putting it
Mathematical bars😂
@@maths_505 lol you should make a math rap
it would be so funny
Yes, this is A particular solution. However the more interesting question is what is THE general solution what we don't know (it might be not equal to this particular solution).
For instance we can see that the equation is not defined for x=0. An important question should be asked whether f is a function from C to C or if it can be from R to R (probably not).
We can also see that the limit of f' at infinity is f^(-1)(0). One can then discuss what is the behaviour of f depending on the number of solutions of f(x)=0. If there are none, f' would not be defined at infinity.
One can also differentiate the equation and get information about the convexity depending on the sign of f''(x) if f(x) is real. Etc.
Hi,
I give you the stupid thought that I probably adopted when young to remember that sin 30° = 1/2 : the number thirty : thirty minutes is .... half an hour.
I say "probably", I am not sure, but I guess it's that, as you know, mnemonic processes have these secrets.
I love your videos.
"ok, cool" : 0:19 ,
"terribly sorry about that" : 3:30 , 5:10 .
When writing out beta and doing calculations involving raising it to its own power, I think it is nicer to write beta in exponential/polar form in the base, and rectangular form in the exponent so that some stuff will cancel out and separate nicely.
Oh yeah that would've been cool...I think Euler would write it that way😂
YES MIDDLE EARTH IS BACK
@@kingzenoiii FOR GONDOR!!!!
@@maths_505 lmaoo
Sin(pi/6) =1/2. Also second beta maybe beta = -exp(-ipi/6), but it should be complex conj so i dont understand then
Oh yeah but that only means I'm off by a factor of 1/2😂 and yes the other beta is the conjugate of the first one.
@@maths_505 oh no you are right nvmnd
Holy crap. I did this problem before watching the video and thought, no way, I'm overthinking it. But apparently not!
Next video: solve the basel problem and aprey's constant PLEASE.
Search Basel problem maths 505 and you'll get my video
@@maths_505 I see but do you think you could solve it like a similar way to Apostol's method, ie: using multivariable calculus since that feels more intuitive (I have no idea what residue theorem is). Also, I think then you could expand it to solving the zeta function of 3 and perhaps some complex inputs as well.
Yes it can be generalised to an integral in N space for ζ(N).
how do you just assume that the function f(x) = ax^b and that you can just take away the x's and set a^(-1/b) = ab?
I had the same confusion, didn't see how ab=a^(-1/b) was obtained
Me: My world isn't real anymore.
Him: what's your problem?
Me: it's complex.
Почему решение ищется сразу в виде степенной функции?
I explained why at the beginning of the solution.
Типа производные и интегралы полиномиалов тоже полиномиалы. Я таки думаю есть и другие функции которые подходят.
@Jalina69 вот я тоже думаю, что есть и другие функции. Например, можно продифференцировать обе части уравнения, и воспользоваться тем, что производная обратной функции равна обычной производной этой же функции в минус первой степени. Или попробовать взять интеграл от обеих частей уравнения. Или поменять переменную.
Are those all solutions?
The alpha parameter is defined as a complex power function so we have to be careful that this the principle value we're talking about. In that case, indeed these are the only 2 solutions (or maybe just the only ones I've found 😂).
Dr. Michael penn a similar one before
Michael Penn also did this one I think
@@dang-x3n0t1ct nope
@@maths_505 https:/ /th-cam.com/video/rO8z0wUdu_A/w-d-xo.html it's this one
@@maths_505 it's from the "A beautiful differential equation" video
@@dang-x3n0t1ct still nope. I searched the video and the RHS is f(1/x) whereas my solution pertains to f^{-1}(1/x)
@@maths_505 i must have gotten confused, apologies
I think I remember something like this uploaded earlier if im not wrong
Nah that was for just the inverse
Didn't Michael Penn make a video on this exact problem? :O
Nah this one's a different case
And even the case he discuss was solved by Dr Peyam 5 years before Penn's video.
👍
Yo dawg I heard you like inverse functions so I chose an inverse function at an inverse argument
Why you supose that y=a x ^b?????
I explained that in the video
Hi. Should I spell your name as Kamal, Kemal or Camal?
Kamaal
3rd
sure but what if f(1) = 1
f’(1) = f^-1(1/1) = 1
f’’(1) = 1/f’(f^-1(1/1))*-1/1^2 = -1
dunno the solution but it seems like you get a nicer result
f'''(1) = 3
f''''(1) = -12
f'''''(1) = 57
-303, 1761
OK Kool
The least satisfying solution ever
Now split all that into real and imaginary parts.
I was a commerce student, why the heck I'm watching this 😂
u started this , u can't go back to simple math now , u will feel bored 😂
is it possible to solve these equations from the ground up (without assuming the form of f(x)) by integrating the inverse? i'm sure i've seen some sort of formula for the integral of an inverse function
That would give us the function in terms of an integral which is much less clear of a solution.
it's hard to integrate either function at 1/x
@@maths_505 Might other basic forms be plausible candidates for alternative solutions, for example exponentials of x? Could there be, in principle, classes of solutions that differ in such a profound way? I should just try it, I guess...
I am the first view
let f(x)=ax^b
f'(x)=abx^(b-1)
(x/a)^(1/b)=f^-1(x)
f^-1(1/x)=(ax)^-(1/b)
abx^(b-1)=(ax)^-(1/b)
b-1=-1/b
b^2-b+1=0
b=(1+-isqrt(3))/2=e^(+-ipi/3)
ab=a^-(1/b)
b=a^-(1+1/b)
b^-(b/(b+1))=a
b+1=(3+-isqrt(3))/2
=sqrt(3)e^(+-ipi/6)
a=(e^(+-ipi/3))^-(e^(+-ipi/6)/sqrt(3))
a=e^(-+ipi/(3sqrt(3))•e^(+-ipi/6))
a=e^(-+ipi/(3sqrt(3))(cos(pi/6)+-isin(pi/6)))
a=e^(pi/(6sqrt(3))-+ipi/6)
a=e^(pi/(6sqrt(3)))•(sqrt(3)-+i)/2
f(x)=e^(pi/(6sqrt(3)))•(sqrt(3)-+i)/2•x^((1+-sqrt(3)i)/2)
f'(x)=e^(pi/(6sqrt(3)))•(sqrt(3)+-i)/2•x^((-1+-sqrt(3)i)/2)
f^-1(1/x)=e^(-pi/(3sqrt(3))+-ipi/6)•x^(-1+-isqrt(3))/2)
smth aint right
This is one hell of an ugly result
bro I think I need to have an appointment with an ophthalmologist as soon as possible. if u can't see and enjoy the beauty in results like this simple one's ,one day u won't be able to enjoy the beauty of elegant results.
one more thing kaamal might have had a busy day . so he just didn't want to take more strain today .
Im the second comment
Could you tell me the app used for note handwriting?
@@jakybel884 Samsung notes
@@maths_505 ❤️❤️ thanks
what if i smash my laptop? will you buy me a new one?
Don't worry bro them $1 patrons be floodin' in all we need is a few more😭😭😭
bro we ourselves are poor bro
we just don't talk about it publicly
so please don't smash ur laptop and if u bought new also don't smash it just recycle it or if it's working condition donate it to someone who u know is in need of it .
@@aravindakannank.s. but make sure to subscribe before donating so that the new owners have cool math they can watch 😂
@@maths_505 and cycle starts once again 😂