I’m making an attempt to know the process undergone to find out a perform of three unbiased variables that satisfies orthogonality relations for roots and powers larger than two.

Firstly I’ll write an instance of what I’m making an attempt to enhance on:

If we now have a foundation of oscillations $ e_{i} = {x_{0}^{0}, x_{0}^{1}, x_{0}^{2}, .., x_{0}^{i}} $ for the $i^{th} $ oscillation, below a remodel $ a_{i} = (a_{1}, a_{2}, a_{3}, .., a_{i}) $ for $ f(x) $

Equally for $ g(y) $: $ e_{j} = {y_{0}^{0}, y_{0}^{1}, y_{0}^{2}, .., y_{0}^{j}} $ and $ b_{j} = (b_{1}, b_{2}, b_{3}, .., b_{j}) $

In order that $ {e_{i}, e_{j} } in B $

Then the transformations $ T(e_{i}) = sum_{i=1}^{n} a_{i}e_{i} $ and $ T(e_{j}) = sum_{j=1}^{m} b_{j}e_{j} $ will be mixed below that foundation utilizing a sq. norm.

It’s understood that for coefficients $ (a,b) $ two capabilities are orthogonal by their interior product $ langle x,y rangle = sum_{i=1}^{infty} x_{i}y_{i} $ if:

$$ langle f,g rangle = int_{a}^{b} f(x) g(x) .dx = 0 $$

I can see that is true through the use of $ y(x) $ in order that $g(y(x)) = g(x) $ in order that we will say $ { f(g(x)), g(f(y)) }leqslant B $

Nonetheless I’m caught when contemplating a 3rd perform $ h(z) $, as a result of earlier than we might use a sq. norm of each transformations:

$$ sqrt = sqrt{a_{i}b^{j}} sqrt{ e_{i} cdot e_{j}} $$

for a circle centered round $ (a,b) $ for $ sup(x + y) leqslant sup(a + b) $

$$ (x+a)^{2} + (y-b)^{2} = r^{2} $$

that could be a convergent for when

when $ sqrt{a_{i}b^{j}} = sqrt{ e_{i} cdot e_{j}} $ which create statistics it approaches $|x||y| $

What do you do for $ f(x), g(y), h(z) $ with $ y(x), z(x) $?

It could appear apparent to up the norms index by one, nevertheless we’re nonetheless constrained to $ r^{2} $, at the least with out utilizing $ S_{3} $. How do you heart a sphere round a cubic norm?

Is there such a factor are a triple interior product? $ langle f | h | g rangle $ with some notion of separation $ ok $ that I’m overlooking?