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July 6th, 2018, 02:11 AM   #1
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Inverse Logic

From A we solve B or $\displaystyle A\rightarrow B$
By inverse we get $\displaystyle B\rightarrow A \rightarrow A\equiv B$
Let's use it to solve equations below:
Given equation $\displaystyle a(x)=b(x)$we transform to $\displaystyle a(x)-b(x)=0$ or $\displaystyle z(t)=0$
From $\displaystyle \frac{dz}{dt} =0 $ we find extremum of $\displaystyle z(t)$ (if it exist ofc)
Now by inverse theory we find value $\displaystyle t_0$ for which $\displaystyle \int z(t)dt=E[\int z(t)dt]$ so now we have an
equivalence $\displaystyle \frac{dz}{dt}=0 \equiv \min\{ \int z(t)dt \}=\int z(t)dt$
Let's take a great example below:
Given equation $\displaystyle x=\sin x$ we transform to $\displaystyle x-\sin x =0$ now if value $\displaystyle x_0 $ satisfies the equation $\displaystyle \frac{1}{2} x^2 +\cos x = \min( \frac{1}{2} x^2+\cos x )$
then $\displaystyle x_0 $ holds for $\displaystyle x_0=\sin x_0$
For those who further info, I have solved equations with this method like below:
$\displaystyle e^x=x+1$ , $\displaystyle e^x=ex$ , $\displaystyle a^x=x+1 ...$ etc

Last edited by skipjack; July 6th, 2018 at 06:07 AM.
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July 6th, 2018, 02:36 AM   #2
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Yes, root finding problems and minimization problems are closely related. So what is your point?
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