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Given an even function $f(x)$, how to obtain the most close to it continuous odd function $g(x)$?

By most close I mean that $\int_0^\infty |f(x)-g(x)| dx$ be the minimum possible and the difference $|f(x)-g(x)|$ be monotonously decreasing for x>0.

To avoid trivial solutions, let postulate that the both functions are discrete-analytic, that is equal to their Newton expansions.

$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)$$

$$g(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k g\left (0\right)$$

Am I right that hyperbolic sine and cosine satisfy these conditions?

I would prefer a plain and simple expression for this transformation operator.

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Particularly I am interested in the odd counterpart to $\csc ^2(x)-\frac{1}{x^2}$.

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    $\begingroup$ Your problem is not currently well-posed: If $\phi:(0, \infty) \to \mathbf{R}$ is arbitrary, there exists a unique odd function extending $\phi$, and a continuum of even functions extending $\phi$ (the value at $0$ is free). That is, unless you impose additional conditions, you can arrange that $f(x) \equiv g(x)$ for $x>0$. Even imposing smoothness, there is no minimum of $\int_0^\infty |f(x) - g(x)|\, dx$.... $\endgroup$ – Andrew D. Hwang Mar 4 '14 at 17:40
  • $\begingroup$ @user86418 thanks, I have addressed this. $\endgroup$ – Anixx Mar 4 '14 at 17:58
  • $\begingroup$ I don't have a feel for discrete-analytic functions, but if $\tanh$ is an example, then the second part of my earlier comment seems still to apply: If $f$ is an even function that doesn't grow too rapidly (e.g., $f = \cosh$), then $g_n(x) = f(x) \tanh(nx)$ is an odd function for which $\int_0^\infty |f(x) - g_n(x)|\, dx$ can be made as small as you like. Assuming the space of discrete-analytic functions is closed under multiplication, division, and composition, I'm not optimistic that your goal can be met. $\endgroup$ – Andrew D. Hwang Mar 4 '14 at 21:37
  • $\begingroup$ @user86418 tanh(nx) is not discrete-analytic for large enough n, see here for a similar case: mathoverflow.net/questions/99166/… $\endgroup$ – Anixx Mar 4 '14 at 21:50
  • $\begingroup$ Ah. In that case, I retract my skepticism. Unfortunately, I don't have any promising ideas toward your question. $\endgroup$ – Andrew D. Hwang Mar 4 '14 at 22:43

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