What's going wrong here? (slight repost)

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Homework Statement



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Homework Equations



The way to solve is basically

Ax = b ---> QRx = b, where Q is an orthonormal basis for A and R is an upper triangular basis ---> Rx = QTb ----> x = R-1QTb

The Attempt at a Solution



No bugs, but for some reason I'm getting the same answer for x and y ... which doesn't make sense, since x is solving for alpha and beta, while y is solving for alpha, beta, and gamma.


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In problems of this type (even when using software) it is always a good idea to re-scale the data to avoid big numbers, because solution procedures may involve squares, or cubes, or fourth powers of those big numbers, leading to matrix manipulation problems that are unstable due to roundoff errors. You should re-scale to something like s = (t - 1900)/10, so your independent variable data are s = 0, 1, 2, 3,..., 11 instead of t = 1900, 1910 1920,..., 2010. Subtracting 1900 is more important than dividing by 10, but since doing both is so easy, why not go all the way? Personally, throughout my professional life I have encountered many examples where using even sophisticated software and double-precision computation gives horribly wrong answers in an un-scaled problem but gives great accuracy, even using single-precision computation, in the scaled version.

RGV
 
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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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