Good ol' Veeky Forums humor thread

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

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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.

The border collie looks like Einstein.

no such function exists LIAR

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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.

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

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

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

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

Where's the joke?

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

Correct!

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So this is /int/ humour?

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

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Is it funny because it's scientifically illiterate?

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

of they could both be arbitrary constant functions

/pol/ get out of here please

f=Id and g in C1

i smell jew

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.

Nice

wrong

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

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.

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