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Anonymous Thu 15 Apr 11:18:30 2021 No.12978237 View Reply Original Report

Quoted By: >>12978271 >>12978309 >>12982968 >>12984586 >>12986075

Why does the erf function show up in the solution to this 2nd order differential equation?

(y'')^-1 = y

Anonymous

Anonymous Thu 15 Apr 2021 11:34:11 No.12978271 Report

Quoted By: >>12978302 >>12979209

>>12978237
>the erf function
Perhaps you should submit your inquiry to the Department of Redundancy Department instead.

Anonymous

Anonymous Thu 15 Apr 2021 11:51:26 No.12978302 Report

Quoted By: >>12978312 >>12979209

>>12978271
You are an idiot. "erf" here refers to THE NAME OF THE FUNCTION, i.e. a mathematical symbol, not an abbreviation. If this is too abstract for you to understand... then fuck off, NORMIE.

Anonymous

Anonymous Thu 15 Apr 2021 11:56:09 No.12978309 Report

Quoted By: >>12979209

>>12978237
It may have something to do with how differentiation is defined and what erf means.
But I am just guessing here.

Anonymous

Anonymous Thu 15 Apr 2021 11:57:02 No.12978312 Report

Quoted By: >>12979095 >>12979209

>>12978302
>"erf" here refers to THE NAME OF THE FUNCTION
Correct. It refers to the error function, Erf. It does not refer to the "error function function," or the "Erf function."
Sounds like you have a bad case of the HIV virus in your brain. Maybe you should stop fucking monkeys.

Anonymous

Anonymous Thu 15 Apr 2021 15:45:14 No.12979095 Report

Quoted By: >>12979186 >>12979209

>>12978312
A "name of a function is a LOGICAL concept. "erf" is the LABEL of the erf-function, just as "exp" is the label of the exp-function and "log" of the log-function (and "f" of the f-function etc.) Please learn how FORMAL SYSTEMS work before you opine on mathematics.

Anonymous

Anonymous Thu 15 Apr 2021 16:05:33 No.12979186 Report

Quoted By: >>12979209 >>12979725

>>12979095
>"erf" is the LABEL of the erf-function
Listen, monkey fucker: erf is the """LABEL""" of the error function, not the error function function. It's fine to write exp function, because there's no redundant f. If you wrote "expf function" for exponential function function, then that would be equally stupid.

Anonymous

Anonymous Thu 15 Apr 2021 16:10:31 No.12979209 Report

Quoted By: >>12979259 >>12981259

>>12979186
>>12979095
>>12978312
>>12978309
>>12978302
>>12978271
autism

Anonymous

Anonymous Thu 15 Apr 2021 16:23:08 No.12979259 Report

Quoted By:

>>12979209
Sorry anon, we all hope that you may recover.

Anonymous

Anonymous Thu 15 Apr 2021 17:42:03 No.12979595 Report

Quoted By: >>12979914 >>12984426

Why the fuck do you guys argue about semantics????
Clearly everyone gets what op is talking about. Now answer the fucking question. Stop bombing the thread will useless shit. Otherwise don't comment.

Anonymous

Anonymous Thu 15 Apr 2021 18:14:41 No.12979725 Report

Quoted By: >>12979914

>>12979186
You seem to not understand the language of MATHEMATICS, because you apply the semantics of ENGLISH to it. "erf" is just a SYMBOL in the LEXICAL ENVIRONMENT of the formal system we are working in. The "redundant f" is a phenomenon of ENGLISH, not of the ACTUAL FORMAL LANGUAGE used... You would not write er(x) instead of erf(x) "because it is clear from the context that er is a FUNCTION and therefore 'er' and 'erf' have the same ``informational content''!" (or would you?? you never know with RETARDS!)

Anonymous

Anonymous Thu 15 Apr 2021 18:54:35 No.12979914 Report

Quoted By:

>>12979725
>You would not write er(x) instead of erf(x)
Correct. You would never use "er" in an equation, because er is not used an abbreviation for the error function.
And you would never use "erf function" in an equation, either, idiot.
And if you write either of those things in an English sentence, you are either redundant, retarded, or (in your case) both.
>>12979595
>Now answer
nypa, faggot

Anonymous

Anonymous Thu 15 Apr 2021 23:58:11 No.12981259 Report

Quoted By:

>>12979209
no u

Anonymous

Anonymous Fri 16 Apr 2021 10:57:49 No.12982968 Report

Quoted By: >>12986067 >>12986209

>>12978237
1/y''=y
y''=1/y
y''y'=y'/y
(y'^2)/2=log(y)+C
y' = sqrt(2log(y) + 2C)
replace x by g(t) where g is the inverse of y.
From the identity y(g(t))=t we get y'(g)*g' = 1.
So 1/g' = y'(g).
The d.e. becomes
1/g' = sqrt(2log(t) + 2C)
g' = 1/sqrt(2log(t)+2C)
Integrate
let t = e^(z^2/2 -C), dt = z*e^(z^2/2 -C)dz
The integrand becomes e^(z^2/2 -C)dz
you get an erf here.

Anonymous

Anonymous Fri 16 Apr 2021 17:22:28 No.12984426 Report

Quoted By:

>>12979595
brainlets rather argue about what they understand enough to argue about rather than what they don't. this includes you too btw. also me.

Anonymous

Anonymous Fri 16 Apr 2021 17:47:54 No.12984586 Report

Quoted By:

>>12978237
dunno, maybe because the convexity of the solution at point x should be inversely proportional to the value at that point and -efr(x) supposedly satisfies that (I mean look at its graph).

Anonymous

Anonymous Fri 16 Apr 2021 21:04:45 No.12986067 Report

Quoted By:

>>12982968
What is ur IQ?

El Cabrón

El Cabrón Fri 16 Apr 2021 21:07:23 No.12986075 Report

Quoted By: >>12986155

>>12978237
Because you’re integrating an e-like thing

Anonymous

Anonymous Fri 16 Apr 2021 21:20:52 No.12986155 Report

Quoted By: >>12986209

>>12986075
whats e?

El Cabrón

El Cabrón Fri 16 Apr 2021 21:28:46 No.12986209 Report

Quoted By:

>>12986155
>>12982968

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