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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 suggested by James Cranch, let postulate for simplicity that the both functions are discrete-analytic, that is equal to their Newton expansions.

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

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

Or in the weaker case, bi-directional Newton expansion

$$f(x) = \lim_{n\to\infty}\sum_{k=-n}^n \binom{x}k \Delta^k f\left (n\right)$$

$$g(x) = \lim_{n\to\infty} \sum_{k=-n}^n \binom{x}k \Delta^k g\left (n\right)$$

Am I right that hyperbolic sine and cosine are the most close even and odd functions to each other?

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

enter image description here

enter image description here

Note that since we can expand the definition of discrete-analytic functions to those having poles in integer points, using this formula

$$f(x)=\lim_{n\to\infty}\frac{\sum _{k=-n}^n \frac{(-1)^k \lim_{t->k}\left( f(t)\prod_{j=0}^\infty(t-x_j)\right)}{(x-k) (k+n)! (n-k)!}}{\sum _{k=-n}^n \frac{(-1)^k\prod_{j=0}^\infty(k-x_j)}{(x-k) (k+n)! (n-k)!}}$$

where $x_j$ are the poles, we can similarly talk about such most close even and odd functions even if they have poles:

enter image description here

The motivation for this question is as follows: it seems that this relation connects elementary functions to zeta function and differentiated gamma functions. Particularly, the most close odd function to the even elementary function $f(x)=\csc^2 x - \frac1{x^2}$ is not elementary.

On the graphic above, the blue function is elementary while similarly looking red one is not!

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    $\begingroup$ Of course, for the problem as stated, the best solution is to take $g(x) = f(x)$ for $x>0$, $g(0) = 0$ and $g(x) = f(-x)$ for $x<0$. I cannot think of any set of extra conditions that makes the problem interesting. $\endgroup$ Commented Mar 4, 2014 at 17:33
  • $\begingroup$ I suspected you might say that. But then there is no best answer. Choose some positive real $\epsilon$, and let $g(x) = f(x)$ for $x>\epsilon$ and hence $g(x) = -f(x)$ for $x<\epsilon$ and let $g(x)$ be linear on the interval $[-\epsilon,\epsilon]$. Then as you take $\epsilon$ to be very small $g$ gets increasingly close to $f$. $\endgroup$ Commented Mar 4, 2014 at 17:40
  • $\begingroup$ @James Cranch I have added the condition that f(x) and g(x) are equal to their Newton series expansions so to avoid the trivial solutions. Another way to avoid the trivial solutions would be use of the distance measure involving higher derivatives. $\endgroup$
    – Anixx
    Commented Mar 4, 2014 at 17:55
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    $\begingroup$ Meta thread: meta.mathoverflow.net/questions/1530/… $\endgroup$ Commented Mar 4, 2014 at 22:12
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    $\begingroup$ @Anixx, could you please avoid deleting your own comments when they are required to understand others' comments? $\endgroup$ Commented Mar 4, 2014 at 23:26

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