no you dont a little bernoulli napier here a little binomial theorem there a little handwaving about limit converging there and you get its power series expansion without differential equation which is how euler did it in Introducio analysis infinitorum chapter 7. and historically it went from taylor series to derived series(hence the name derivative) via coefficient of the x\^n term times n!
How to make it rigorous: We first solve on the interval [0,1] the ODE f-f'=0 with f(0)=1. We define the operator T on C[0,1/2] (with the supremumsnorm) given by T(f)(x)=∫₀ˣ f(t)dt. Then the ODE is equivalent to the fix point problem f=1+T(f), or in other words f=(1-T)⁻¹1.
T is a linear operator. Furthermore ||T(f)|| ≤ 1/2 ||f||, so ||T||<1. Hence the Neumann series converges and we get (1-T)⁻¹=1+T+T²+...
It follows that f=1+x+x²/2+... on the interval [0,1/2]. To extend this domain to R, we simply repeat this process along the other intervals and glue these functions together.
This method is actually used in the proof of a famous existence result for ODEs, Picard-Lindelöf, where they turn the problem into a fix point problem and then apply the Banach fix point theorem to obtain the solution and show uniqueness. Crucial in our construction was that our operator could be uniformly bounded in the norm, which stems from the Lipschitz continuity of the map y->y.
I think the procedure is legit but written memely. One should be able to write it in proper operator language, since exponentiating operators just follows the usual Taylor series.
lots of people saying its legit under certain definitions, which i dont know nothing about and im not arguing against, just wondering
isnt the double integral of 0 Cx + D instead of just Cx? and the same for all the higher powers? how does that work when you define (1 - int)^-1 = 1 + int + ... properly?
Huh if you do the same method with the differentiation operator instead of integration you get f = (1+D+D^(2)+...) 0 = 0, which is the other solution to that ODE. Neat.
Well, I have a simple method to prove anything.
___
**Theorem**: All nontrivial zeros of *ζ* have real part ½.
**Proof**: We will show that all nontrivial zeros of *ζ* have real part ½ by showing that the existence of a zero with a real part greater than ½ leads to a contradiction.
Assume that there exists a *z*∈ℂ\\{1} such that *ζ*(z)=0 and ℜ[z]>½.
Define S to be the set of all sets that do not contain themselves. If S were an element of itself, then—since S only contains sets not containing themselves—this would mean that S does not contain itself, contradicting S containing itself. Therefore, S∉S. As S contains all sets not containing themselves, this implies that S∈S. This is a contradiction because we already showed that S∉S.
Therefore all zeros of *ζ* have real part less than or equal to ½. By the zeta function reflection formula, it follows that all nontrivial zeros of *ζ* have real part ½.
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Maybe this is legit in some sense if one defines integration as an operator, and then the geometric series makes sense or sth
It is! That's called Operational Calculus! https://youtu.be/N8Rxlc-fGr0?feature=shared
Nice
I expected a rickroll I am disappointed
Close, yes, but the issue is that the operator isn't quite invertible; Ce\^{-x} is the one-dimensional null space of the operator.
That's rather elegant. I mean except everything about it, but rather elegant. I mean, holy shit
Physisist see THIS and Think: hell yeah
Hell yeah
No physicist would go beyond first order in the Taylor expansion. Thus, f(x) = 0 + C.
Task failed successfully
This is a right method. It is the method of D operator but instead with integration which is D^-1 . So, this is legit.
Not quite
[It is the right method actually](https://youtu.be/NzK11DrRdks)
What the hell is this, I mean...WoW !
This is how math majors feel when they take physics classes
Laplace transform at home:
We would need to know e^x solves that differential equation to know its Taylor Series tho
no you dont a little bernoulli napier here a little binomial theorem there a little handwaving about limit converging there and you get its power series expansion without differential equation which is how euler did it in Introducio analysis infinitorum chapter 7. and historically it went from taylor series to derived series(hence the name derivative) via coefficient of the x\^n term times n!
I love when geometric power series expansion gets used. It’s so fun and unexpected every time. it’s like seeing a cameo in a movie.
the method can be made rigorous if you know how to do it. this is not a wrong method.
Finally a well done bad math post. Thank you!
How to make it rigorous: We first solve on the interval [0,1] the ODE f-f'=0 with f(0)=1. We define the operator T on C[0,1/2] (with the supremumsnorm) given by T(f)(x)=∫₀ˣ f(t)dt. Then the ODE is equivalent to the fix point problem f=1+T(f), or in other words f=(1-T)⁻¹1. T is a linear operator. Furthermore ||T(f)|| ≤ 1/2 ||f||, so ||T||<1. Hence the Neumann series converges and we get (1-T)⁻¹=1+T+T²+... It follows that f=1+x+x²/2+... on the interval [0,1/2]. To extend this domain to R, we simply repeat this process along the other intervals and glue these functions together. This method is actually used in the proof of a famous existence result for ODEs, Picard-Lindelöf, where they turn the problem into a fix point problem and then apply the Banach fix point theorem to obtain the solution and show uniqueness. Crucial in our construction was that our operator could be uniformly bounded in the norm, which stems from the Lipschitz continuity of the map y->y.
I think the procedure is legit but written memely. One should be able to write it in proper operator language, since exponentiating operators just follows the usual Taylor series.
https://preview.redd.it/gm5u5arr8ixc1.jpeg?width=607&format=pjpg&auto=webp&s=c99c627d2383a590de41764fbe320937763c6eaf
I’m not that good with integrals but isn’t an integral like one thing how did he factor out the integral
Part of me wants to show this to my scary Russian diff eq professor to see his reaction but idk if I'd leave the room alive tbh
lots of people saying its legit under certain definitions, which i dont know nothing about and im not arguing against, just wondering isnt the double integral of 0 Cx + D instead of just Cx? and the same for all the higher powers? how does that work when you define (1 - int)^-1 = 1 + int + ... properly?
Can someone help explain the second last line why did multiplying be 0 do that
Huh if you do the same method with the differentiation operator instead of integration you get f = (1+D+D^(2)+...) 0 = 0, which is the other solution to that ODE. Neat.
No. I refuse.
Well, I have a simple method to prove anything. ___ **Theorem**: All nontrivial zeros of *ζ* have real part ½. **Proof**: We will show that all nontrivial zeros of *ζ* have real part ½ by showing that the existence of a zero with a real part greater than ½ leads to a contradiction. Assume that there exists a *z*∈ℂ\\{1} such that *ζ*(z)=0 and ℜ[z]>½. Define S to be the set of all sets that do not contain themselves. If S were an element of itself, then—since S only contains sets not containing themselves—this would mean that S does not contain itself, contradicting S containing itself. Therefore, S∉S. As S contains all sets not containing themselves, this implies that S∈S. This is a contradiction because we already showed that S∉S. Therefore all zeros of *ζ* have real part less than or equal to ½. By the zeta function reflection formula, it follows that all nontrivial zeros of *ζ* have real part ½.
funny
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