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Anonymous Sat 10 Oct 22:27:40 2020 No.12216841 View Reply Original Report

Quoted By: >>12217125 >>12217217 >>12217292

y' = y as the y'---> infinity in other words y''''''''...... What behavior can you find from this differential equation as you increase the order?

Anonymous

Anonymous Sat 10 Oct 2020 22:30:34 No.12216856 Report

Quoted By: >>12216869

e^x

Anonymous

Anonymous Sat 10 Oct 2020 22:32:46 No.12216869 Report

Quoted By:

>>12216856
that's for y' = y but I'm asking as y^n th derivative as n tend to infinity

Anonymous

Anonymous Sat 10 Oct 2020 22:33:38 No.12216870 Report

Quoted By:

idk fuckin nerd i just come here to pretend that my IQ score from an online test that i took 7 times to score 132 on guarantees my success despite my general apathy towards life, in an effort to numb the pain from a lifetime of repeated failure

Anonymous

Anonymous Sat 10 Oct 2020 23:53:42 No.12217125 Report

Quoted By: >>12217155 >>12217164 >>12217247

>>12216841
Let $\mathcal{A}:=\{ y \in C_b^\infty (\Omega) \mid \Vert y - y^{(n)} \Vert \rightarrow 0 \}$ .
Then clearly, for any $y \in \mathcal{A}$ it holds that $y^{(k)} \in \mathcal{A}$ for any k.
Thus: $\Vert y'-y \Vert \leq \Vert y' - y^{(n+1)} \Vert +\Vert y^{(n)} - y^{(n+1)} \Vert +\Vert y - y^{(n)} \Vert \rightarrow 0$ .
In other words, $y'=y$ for any $y \in \mathcal{A}$ and hence $\mathcal{A}$ is a linear space of dimension one, spanned by the exponential on $\Omega$ .

Anonymous

Anonymous Sun 11 Oct 2020 00:05:07 No.12217155 Report

Quoted By:

>>12217125
Math major?

Anonymous

Anonymous Sun 11 Oct 2020 00:07:10 No.12217164 Report

Quoted By: >>12217174

>>12217125
wanted an answer from quora but it's full of retards that have 20-year-old degrees LOL

Anonymous

Anonymous Sun 11 Oct 2020 00:10:18 No.12217174 Report

Quoted By: >>12217179

>>12217164
What makes you think their 20-year-old degrees are invalid?

Anonymous

Anonymous Sun 11 Oct 2020 00:12:11 No.12217179 Report

Quoted By: >>12217185

>>12217174
bc I told you that they didn't give the response I was looking for

Anonymous

Anonymous Sun 11 Oct 2020 00:14:06 No.12217185 Report

Quoted By: >>12217189

>>12217179
could you link me the thread?

Anonymous

Anonymous Sun 11 Oct 2020 00:15:24 No.12217189 Report

Quoted By: >>12217194 >>12217468 >>12219393

>>12217185
https://www.quora.com/Y-y-as-the-y-infinity-in-other-words-y-What-behavior-can-you-find-from-this-differential-equation-as-you-increase-the-order

Anonymous

Anonymous Sun 11 Oct 2020 00:17:23 No.12217194 Report

Quoted By:

>>12217189
lol

Anonymous

Anonymous Sun 11 Oct 2020 00:28:00 No.12217217 Report

Quoted By:

>>12216841

I'm not sure you can say a general solution to a 4th order or higher differential equation exists, let alone one equal to itself, that wasn't trivial.

To solve even 3rd order equations requires constraints on what kinds of functions could even be considered.

Anonymous

Anonymous Sun 11 Oct 2020 00:28:46 No.12217219 Report

Quoted By:

it's very interesting, the hidden relationships between e^x and the hyperbolic trig functions
the only difference between the differential equations y'=y and y''=y is the degree of the DE yet the solutions yield entirely different functions

Anonymous

Anonymous Sun 11 Oct 2020 00:37:46 No.12217247 Report

Quoted By: >>12217451

>>12217125
There is a mistake in my argument here: it is not at all clear that $y^{(k)} \in \mathcal{A}$ .

Anonymous

Anonymous Sun 11 Oct 2020 00:54:12 No.12217292 Report

Quoted By:

>>12216841
maybe this is a problem in the domain of Godel's incompleteness theorem LOL

Anonymous

Anonymous Sun 11 Oct 2020 01:50:56 No.12217451 Report

Quoted By:

>>12217247
unless n tends to inf :)

Anonymous

Anonymous Sun 11 Oct 2020 01:55:07 No.12217468 Report

Quoted By:

>>12217189
lol

Anonymous

Anonymous Sun 11 Oct 2020 16:03:44 No.12219393 Report

Quoted By:

>>12217189
i mean i didn't get your question until someone else answered, maybe write it like y^(n) n->\infty

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