Interval Bisection: General Formula?

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Is there any general formula for interval bisection, rather than consider two similar triangles each time?

Thanks
 
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I must be misunderstanding what you mean by "interval bisection". The point bisecting the interval from (x_0,y_0,z_0) to (x_1, y_1, z_1), in 3 dimensions, is ((x_0+ x_1)/2, (y_0+ y_1)/2, (z_0+ z_1)/2), "averaging" the coordinates. That can be derived by using 'similar triangles' but you surely don't have to rederive it for each application.
 
Thanks for the reply. Sorry, I meant linear interpolation, not interval bisection - is there a general formula for this?

Thanks
 

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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