I Bet You've Never Seen THIS Before
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- Опубліковано 9 гру 2023
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Today we derive the maclaurin series expansion of the exponential function using functional analysis and integral operator magic. Enjoy! =D
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To be clear, the manipulation from 1 = (1 - I)(exp) to exp = (1/(1 - I))(1) requires that exp be a function not in the kernel of the operator (1 - I), since otherwise, the inverse of the operator does not exist. Hence, it is convenient to know what the kernel is before hand: we want to know all functions f such that (1 - I)(f) = 0, which is equivalent to I(f) = f. We can turn this into a differential equation, so we have f' = f, which implies f(x) = Aexp(x), and so I(f)(x) = Aexp(x) - A = f(x) = Aexp(x), which holds for all x if and only if A = 0, which implies f = 0. This means that, in fact, (1 - I) is an injective operator, so it is meaningful to apply this with any function.
Now, for the derivative operator, we have D(exp) = exp, which means (1 - D)(exp) = 0, and notice the implication here: exp is in the kernel of 1 - D, so there is no meaning in applying the inverse here. Additionally, there is the issue that D is an unbounded operator.
Hey! Found you again. It's been a while
Wow! I understood absolutely nothing of that.
what the FUCK is a kernel i signed up for MATH not POPCORN
Yoo thanks
Yay more weird operator manipulation videos
What you are using is actually a functional analytic argument. You are implicitly considering the Volterra operator on a space of continuous functions. Then one can observe that this operator, although it has an operator norm equal to one, has a spectral radius equal to 0. In particular, the spectral radius is strictly less than one, which allows using the Neumann series (which is basically just the geometric series for operators). I used to show it to physics students in my exercise classes :D
Acutally very happy to see a new Video every day.
I though you were gonna do something like turn the integral operator into an equation and then do some broken stuff to it.
That's a great derivation for the taylor series expansion, never seen that before. Thanks for sharing
This is eye opening, thank you for this content papa flammy ❤
Yo it's President McKinley
That's just brilliant! Thank you for sharing this!
i want papa Flammy to read out the discount code to his merch shop accurately at least once
No, you don't. It's the name of a Lovecraftian Old One.
that's the point: witnessing something beyond conscious comprehensibility is truly a spiritual experience
Love waking up to a new video of PAPPA FLAM!
I fan since 2018!
Our boy is starting to shoot unnecessary shots at physics 😂 EACH VIDEO
I saw it on instagram once but did in an horrible way
I tried it after and saw that it did work out after all, but yeah with more precautions
good thing you make a video of your own about this !!
He is bad ass !!... I love it... but I would use I for the identity operator and S for the integral operator.. it just makes the notation harmonize better with the intent.
This is what my professor would call dirty math
The bit with geometric series expansion on an def integral operator...how is that even legit?
Von Neumann Series
Something something Neumann series
Fantastic result.
That is crazy, I love it
I thought the title should be: How REAL men find Taylor series
Love your wonky shit. You’re one of a kind and I appreciate you sharing this wicked video!
Is this useful for differential equations?
Awesome, but how can one apply the classical operations to operators and still getting the right answer? One could have also exponentiation or integration over a operator? This kind of topic make me wish study math instead of physics
Pretty sure it has to be a bounded, linear operator. Exponentiation as an operator isn't linear. An operator is linear if for constants a,b and functions f,g the operator O fulfills:
O(a*f+b*g)=a*O(f)+b*O(g)
You can exponentiate the derivative and get the shift operator because that's essentially the Taylor expansion. Explicitly, exp(d/dx) f(x) = f(x) + f'(x) + ½f"(x) + ⅙f'"(x) + ... = f(x+1).
that was really nice video, never heard about something like this, exiting
That is sick!
I've never seen the series representation for exp() derived this way, but it makes perfect sense!
(and yes, it has to be a bounded operator Papa Flammy)
I don't know much about functional analysis. But isn't the answer (exp(x)) * (x+1). because if we apply (1- INTGRAL from 0 to x) operator on that we get exp(x) . so if we apply its inverse on exp(x) we have to get that answer.
THIS IS BEAUTIFUL! Now I can understand why we can manipulate the backshift operator in time series like a fucking geometric series :D
Can this be considered as proof for the taylor series of e^x?
YOU ABSOLUTE GIGACHAD THIS IS PEAK
Thank you for the amazing content +1
I find this immensely funny
Can anyone explain how we can approach this with derivative operator?
So cursed, yet it works out in the end
Should have been a halloween video because this shit is spooky
Thats so freaking nice
holy shit, wow.
Really cool
That was fuckin cool dude. I didn’t get a chance to learn the Taylor series in college, and to see one proved like that was awesome.
Operational calculus. 1/(1 - D) f(x) = f(x + 1) . There operation form of Taylor's theorem. D = d/dx
wake up babe, new cursed math video just dropped
i’m not at all qualified to say this, but wouldn’t factoring out the e^x not be allowed, as it’s kind of like factoring out the x out of sinx just leaving sin?
the integral is not a function
Was babbelst du
the fuck just happened
I wanna learn this
alright wait that was actually so cool >:0
🔥
challenge: integral from 0 to infinity of ((x)^(-x))dx
Laplace is typing…
Where can I get a better understanding of this apparent nonsense that puts the same variable in both the integral and in the limits of the integral? I remember that being forbidden and/or confusing in calculus classes. And I don't think this whole thing would work if you had to start from "I e^t = e^x - 1", or if I was defined as the functional operator taking an integral in x from 0 to t of the integrand in x (giving us "I e^x = e^t - 1"). I haven't really studied functional analysis, so maybe I need a place to get a lot more comfortable with that in general.
I'm smart enough to know that the end doesn't justify the means. Just because the final result is a correct expression doesn't mean that the reasoning along the way isn't flawed. So, like, what exactly makes this massive repeated use of the variable x valid?
Nice
Wow that’s dope ,,, but actually how can we proof that this steps is valid mathematically
this is done in functional analysis
I think this is what's going on. A linear map over a vector space induces a polynomial where multiplication is now composition. In this case, we're looking at the integral operator on the space of infinitely differentiable functions, and the "polynomials" here can also be uniquely factored because it's over the complex numbers.
Geek
Is this legal
this is cracked
Trying to decide whether this is legit, or whether it happens to work for e^x but falls apart for anything else and this is just a big troll.
It is legitimate if and only if the function you start with is not in the kernel of the operator 1 - I.
@@angelmendez-rivera351Do you have a reference for that fact?
1h ago
Where is the dx in the thumbnail?
this is the most ridiculous thing ive ever seen
asnwer=1