Let's Solve A Homemade Exponential
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- Опубликовано: 8 сен 2024
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Very nice and not easy. I tried to use logs but I lost it after a few moves. I did not know that you could manipulate the polynomial as the exponent to get the Lambert. In this case it was adding 1 to (x^2+2x) in e. But what was done to the RHS to equate that? Multiplying 1 by e? If that's the case then it was kind of a constriction on numbers. If it does not fit, force it.
We should have expected that after squaring both sides we could have introduced extraneous solutions. That's what happened with x=0!
He mentioned it.
Clearly -2 is the only solution. The range values set tne value of x to be -2. Archimides decomposition wiill work as well.Equations everyday.yeah..
exp (( ( x + 1) ^2 - 1) /2) = - 1/( x + 1)
exp ( (z^2 - 1) /2) = - 1/z
Has a trivial solution at x + 1 = z = -1
This implies x = - 2
Very nice problem and solution.
Thank you 🙂
👍
Well done
Cool
Loved it
It's nice 👍👍👍
Nice!
Can you multiply by (x+1) and integrate both sides? LHS is f'e^f
??? Non si può fare
@@giuseppemalaguti435
(x+1)e^(x+x^2/2)=-1
Integrating from -1 to some X we get:
e^(x+x^2/2)= -(x+1)
But we know that LHS is also equal to -1/(x+1), thus:
-(x+1) = -(1/x+1)
(x+1)^2 = 1
=> x = -2 or x = 0
I wonder if it's a coincidence
x=-√W(e)-1=-2
I am always confused how solving for solutions and then you check to find some 'don't work'
Any approach in these "Lambert's" problems to only find working solutions?
It's wrong from about 4 minutes
what’s wrong
Sorry I was wrong