How Are g1, g2, g3 Derived and What Does L Represent in Numerical Analysis?

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To get g_1, they simply added x to both sides of x^3- x- 5= 0 to get x^3- 5= x.

To get g_2, they added x+ 5 to both sides to get x^3= x+ 5 and took the cube root of both sides.

To get g_3, they subtracted 5 from both sides to get x^3- x= 5, then factored, x(x^2- 1)= 5 and, finally, divided both sides by x^2- 1 to get x= 5/(x^2- 1).
 
ok i solved a similar equation:
x^3+2x^2+4-x=0
g1=x^3+2x^2+4

why for g1 L=0.5
how to find L for other g
what is the meaning of L?
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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