A strange differential equation (this is awesome!)
HTML-код
- Опубликовано: 17 фев 2024
- Functional differential equations have interesting solution developments because they almost always require us to be creative.
My 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 and DM me in case you need math help:
maths.505?igshi...
My LinkedIn:
/ kamaal-mirza-86b380252
Advanced MathWear:
my-store-ef6c0f.creator-sprin...
I was wondering regarding the last possible solution, when you multiply g(x) and g’(x) won’t you get 1/2 * g(2x)? since sin(βx)*cos(βx) = 1/2 sin(2βx)
Yes indeed. I should've written it as sin(2βx)/β and the same adjustment for the hyperbolic function.
What about f(x) = 2x? My first thought was that it would be a linear function
Nice one 👍🏻
e^x immediately comes to mind lol. Now let's watch.
yep thought of the same
Came here to say this
Or f(x)=2x
I started by trying with f(x) = Ax^B.
So we have:
f'(x) = BAx^(B-1)
f(x)f'(x) = A^2B x^(2B-1)
f(2x) = A2^B x^B
Solving the equation system, we get:
B=1
A^2 =2A-> A=0 or A=2
Therefore f(x) = 2x is a possible solution
f(x)=2x is a possible solution
The video claims at 10:30 - 11:15 to have shown that the only solutions with non-zero f(0) are of the exponential form, but it did no such thing. Showing that the exponential functions do satisfy the constraints is not the same as showing that nothing else can satisfy the constraints. You would have to actually show that the recursion relation on the coefficients of the power series has only one solution for a given non-zero a_0.
On a related point, the video is also wrong at 9:47 when it says that given a_0 we can determine a_k for k≥1. That is true if a_0 ≠ 0, but when a_0 = 0 you have two choices for a_1. If a_1 = 0 then all the coefficients of the series must be 0 and you get the zero solution (f(x)=0), which is perfectly valid even though the video seems dismissive of it. The other choice is a_1 = 2, and then you get two possible families of real-valued solutions which have both been essentially found by other commenters: sin(2bx)/b and sinh(2bx)/b for a constant b. (You can get the f(x)=2x solution by letting b->0 in either case.) Along with the previous cases, these are all of the possible analytic solutions.
The video didn't even show that the simplest case, i.e. all a_n =1 is a solution of the recurrence relation. (Or did I miss something?) It wouldn't actually be very hard, at least if one multiplies the recurrence relation with k!, identifies the binomial coefficients (k choose n) and recalls that the sum of the k-th row in Pascal's triangle is 2^k.
@@Notthatkindofdr "That is true if a_0 ≠ 0"
Well, he started the proof by assuming that f(0) is not equal to zero, and hence a0 can't be zero...?
3:10, never subscribed faster in my life
The last function lacks one detail to make it work, which is making the argument of sine 2ßx instead of just ßx. This will make our derivative (sin(2ßx)/ß)'=2cos(ßx), then on multiplication sin(2ßx)*2cos(2ßx)/ß=sin(4ßx)/ß which does satisfy the initial differential equation, by the way this beta is a little sus
change \alpha to i\beta in \alpha exp(x/\alpha), chill and don't use those trigonometric formulae ever again
I immediately spotted a particular sol f= sin2x
but another fun solution that is very similar is f=sinh(2x)
f’ = 2cosh(2x)
f’(x)f(x) = 2 sinh2x cosh2x = sinh4x
f(x) = 2 sin(x) is also a solution, since 4 sin(x) cos(x) = 2 sin(2x).
8:05 Why is it, when something happens, it is always you three?
Ok i got that chest hair but where are my inches?
those are extra rewards that come after buying the merch
My cal3 professor gave me this one a year and a half ago! I never would’ve thought to see it here, wild how these things happen. ❤
From the equation at 9:54, we can prove by induction a_k = a_0^(-k+1). Hence f(x)=alpha*e^(x/alpha)
3:08 oh my GAWD
Pretty Analysis for this DE. Thank you.
2:40 only fans solutions 🤣🤣🤣
If exp and sin are both solutions, this suggests the general solution is some linear combination of hyperbolic functions.
Can you elaborate? Is that a hyperbolic geometry thing?
This was given to me by my teacher after he taught us derivatives, and only cared about the polynomial solution
disclaimer: the narrator has no chest hair, since there is a solution of form f(x) = 2x + ax^3+ 3a^2/20x^5 +3a^3/280x^7+...
2:45 Maths 505 you never fail to make me smile!
😂😂 math is itself funny and math 505 makes it immensely funnier
Awesome! Is there any chance you could prove the in the limit as x goes to zero from the right of any even tetration of x is one and odd tetration is zero? I can’t find a general proof of this anywhere.
It is like Frobenius method!
I was working with this method for Bessel and Hypergeometric functions. It is beautiful!
But I am wonedring from where Bessel ODE came from or what this model presents...
Anyway, nice video!
Btw Have you seen Hazbin Hotel?
Nah bro haven't seen that one
I've seen hazbin hotel lmao
Incredible solution development. How about simple functions like sin or sinh was my thought cause it leads back to the double angle formulas.
Yes you're right.....I mentioned sin(ax)/a at the end of the video and sinh(ax)/ax is also a valid solution.
@@maths_505 ah right sorry I didn't watch all the way to the end.
If all the solutions are of the form
f(x) = a exp (x/a)
Couldn’t any function be turned into a Fourier integral and also plug in?
Wait, let me check for linearity…
Let g(x) == [a exp (x/a)] + [b exp (x/b)]
g(x) g’(x) =? g(2x)
[a(exp(x/a)) + b(exp(x/b)] • [exp(x/a) + exp(x/b] =
a exp 2(x/a) +
b exp 2(x/b) +
2ab exp (x/a+x/b)
Never mind. The cross terms mean the solutions are not linear.
Since in the differential equation, f is mulitplied by its derivative, this is not a linear differential equation, and hence linear combinations of solutions in general won't be solutions again.
Thank You So Much Sir ❤ I'am student in applied mathematics and i like tour proof's videos
50k soon:)
This equation reminds me sine and i found particular solution
f(x) = 2sin(x)
but
f(x) = e^{x} is also sulution
hlo sir can u please bring question in combinatorics they r quite absurd i think ur teaching will help me
(1/a)e^(ax)
As a subscriber for over a year now, I can confirm that my package has grown a couple inches.
I went for something way more simplistic and somehow dumb and got just f(x) = 2x
Junior High School smuttiness seems out of place here . . . then again, maybe not.
Junior high is probably when female contact ended for you then lmao.
Chest hair I am not worried about but the extra length would be great.
I grew few inches of hair down there and lost my chest🤣🔥love this channel
🤣🤣🤣🤣🤣
"There is nothing more masculine than solving cool math problems"?? WTAF? RIP Emmy Noether, Maryam Mirzakhani, Maryna Viazovska...
Nice job finding 3 masculine women in an entire field. 🤭
Why only fans 😂
All the extra hair on my chest interferes with typing and my offering other possible solutions. ;-) Cool problem.
😂😂😂
i got my inches but where is my chest hair?
onlyfans?