The notation of the norm of polynomials

td21
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what is x_i? is it the coefficient of x or simply add up 1-5?
i found the notation different from http://mathworld.wolfram.com/PolynomialNorm.html
so i am confused. Thx!
 

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Basically, that middle part is equivalent to saying:

\sqrt{(x_1-x_1^2)^2 + (x_2-x_2^2)^2 + (x_3-x_3^2)^2 + (x_4-x_4^2)^2 + (x_5-x_5^2)^2}

Is that what you were asking?
 
Char. Limit said:
Basically, that middle part is equivalent to saying:

\sqrt{(x_1-x_1^2)^2 + (x_2-x_2^2)^2 + (x_3-x_3^2)^2 + (x_4-x_4^2)^2 + (x_5-x_5^2)^2}

Is that what you were asking?

sorry, i know the notation. But i don't know what x_i means? coefficient of x?
 
No, it actually means the ith dimension of x, I believe.
 
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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