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/sci/ - Science & Math


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8932789 No.8932789 [Reply] [Original]

Good ol' /sci/ humor thread
more like pic related pls.

>> No.8932792
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>> No.8932793
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>> No.8932800
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>> No.8932804
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>> No.8932846
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>> No.8932851
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>> No.8932867
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>> No.8932898
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EZ AF

>> No.8932902
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>> No.8932904

[eqn](f\cdot g)'(x) = f'(x)g'(x) + C [/eqn]

>> No.8932929
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>>8932902

>> No.8932934
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>> No.8932938

>>8932904
QUICK EXERCISE FOR THE ENTIRE BOARD

Find all real functions such that the equality
[math] (f\cdot g)'(x) = f'(x)g'(x) [/math] holds.

If you can't do this then leave this board immediately.

>> No.8932944

>>8932934
The border collie looks like Einstein.

>> No.8932948

>>8932938
no such function exists LIAR

>> No.8932954
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8932954

>> No.8932956

>>8932948
And that's why we have the empty set. That said, you have not given any proof so what you are saying is worth less than dog shit.

Anyway, let me re-state my problem formally.

[math] \text{Characterize the set } A = \{ (f,g) \in (C^1 ( \mathbb{R} ))^2 : (f\cdot g)'(x) = f'(x)g'(x) \} [/math]

In other words, find all pairs of functions such that they have that property.

>> No.8932958
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8932958

>>8932954

>> No.8932966
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>>8932956

>> No.8932969

>>8932948
f = g = R->R
x->0

>> No.8932971

>>8932938
f=0, g arbitrary C1 function or the other way around

>> No.8932989

>>8932956
[math]A=\mathbb{R}[/math]
t: wildberger

>> No.8933002

>>8932971
That's the trivial solution. If you think it is the only one, prove that no other element can be in the set.

>> No.8933010

>>8932958
>>8932966
Where's the joke?

>> No.8933016

>>8932956
f(x) = e^x
g(x) = x

>> No.8933020

>>8933016
Correct!

>> No.8933024
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8933024

>>8933010

>> No.8933026

>>8933024
So this is /int/ humour?

I'm guessing you didn't see what board we're on?

>> No.8933028
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>>8933026

>> No.8933357

>>8933028
Is it funny because it's scientifically illiterate?

>> No.8933366

>>8933357
Are you seriously responding to obvious /pol/ b8 newfag?

>> No.8933371

>>8932971
of they could both be arbitrary constant functions

>> No.8933379

>>8932934
>>8932954
>>8932958
>>8932966
>>8933024
/pol/ get out of here please

>> No.8933400

>>8932938
f=Id and g in C1

>> No.8933417
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8933417

>>8933379
i smell jew

>> No.8933456

>>8932938
Right.
Suppose [math] f'\cdot g + f \cdot g' = f' \cdot g' [/math]. Divide through by [math] f \cdot g [/math]. You're left with [math] \ln(f)' + \ln(g)' = ln(f)' \ln(g)' [/math]. This is easily solved for ln(g)' : [math] \ln(g)' = 1 - \frac{1}{\ln(f)' - 1}. [/math].
Choose an f and you can solve this for g. Probably need to take care about the interval where they're defined but this should give the non-trivial solutions.

>> No.8933461

>>8933456
Nice

>> No.8933467

>>8933456
wrong
>>8933016

>> No.8933470 [DELETED] 

>>8933456
Should've been 1+ instead of 1-, whoops.

>> No.8933474 [DELETED] 

>>8933456
>>8933467
Fucked up with my babby algebra: you end up with [math] \ln(g)' = \frac{\ln(f)'}{\ln(f)' - 1}. Simplify that as you wish, or not, plug in an f, and solve for g.

>> No.8933476

>>8933456
>>8933467
Fucked up with my babby algebra: you end up with [math] \ln(g)' = \frac{\ln(f)'}{\ln(f)' - 1} [/math]. Simplify that as you wish, or not, plug in an f, and solve for g.

>> No.8933477
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