# Are there various other analytic features with this building of sinc function?

This inquiry is encouraged by my previous post concerning sinc function.

Confirm or refute that $\frac{\sin x}{x}$ is the only nonzero whole (i.e. analytic almost everywhere) function $f(x)$ on $\mathbb{R}$ such that $$\int_0^\infty f(x) dx=\int_0^\infty f(x)^2 dx$$ or $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx.$$

If $f$ is called for just to be continual, after that various other instances are feasible, as an example the also expansion of the list below function : $$ f(x)=\left\{\begin{array}{ll} -2(5+\sqrt{65})x^2+(7+\sqrt{65})x-1 & 0\le x\le \frac{1}{2}\\ 2(5+\sqrt{65})x^2-(13+3\sqrt{65})x+4+\sqrt{65} & \frac{1}{2}\le x\le 1\\ \frac{1}{x^2} & x\ge 1 \end{array}\right. $$

As commented listed below, it ends up that there are very easy response to the above inquiry. ADVERTISEMENT additionally revealed a function listed below that additionally pleases $$ \sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty f(n)^2=0. $$ In sight of these solutions, my inquiry is currently changed to :

Prove or refute that $\frac{\sin x}{x}$ is the only nonzero whole function, $f(x)$ on $\mathbb{R}$ such that $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx=\sum_{-\infty}^\infty f(n) =\sum_{-\infty}^\infty f(n)^2 $$

Similar to the pointer of Zaricuse, that is take a whole function $f$ such that $\int_{-\infty}^\infty f(x) dx \ne0$ after that address for $$\int_{-\infty}^\infty af(x)dx = \int_{-\infty}^\infty (af(x))^2dx$$ Then $g(z)=af(z)$ addresses fifty percent of the trouble. To get to $$\sum g(n)=\sum g(n)^2$$ we might as an example start with $f(z)=\sin (\pi z) \cdot h(z)$ where $h$ is a various other integrable whole function.

In the spirit of Zaricuse is remedy without the amount need, take any kind of completely well acted features f and also g. After that you need to have the ability to locate a straight mix af+bg that satisfies both formulas. If you allow

$$if=\int_{-\infty}^\infty f(x) dx$$ $$if2=\int_{-\infty}^\infty f(x)^2 dx$$ $$sf=\sum_{n=1}^\infty f(n)$$ $$sf2=\sum_{n=1}^\infty f(n)^2$$

and also in a similar way for fg and also g, we have

$$a*if + b*ig=a^2*if2+2ab*ifg+b^2ig2$$ and also

$$a*sf + b*sg=a^2*sf2+2ab*sfg+b^2sg2$$

which can be addressed for an and also b most of the times.

**Included ** In feedback to the new demand that the values of both integrals and also 2 amounts all suit, I simply require adequate handles to transform. Specify $g(k,x)=\exp(-kx^2)$ and also take $f(x)=g(1,x)+ag(2,x)+bg(3,x)+cg(4,x)$ The wonderful feature of this $f$ is that $f^2$ is created in regards to $g(k,x)$, though k rises to 8. The indispensable of $g(k,x)$ is simply $\sqrt{\frac{\pi}{k}}$ and also the amount is computed by Wolfram Alpha as $\vartheta_3(0,\exp(-k))$. We can make a table :

$$\begin{array}{ccc}k&\int g(k,x)&\sum g(k,x)\\1&1.772453851&1.77264\\2&1.253314137&1.27134\\3&1.023326708&1.09959\\4&0.886226925&1.03663\\5&0.79266546&1.01348\\6&0.723601255&1.00496\\7&0.669924586&1.00182\\8&0.626657069&1.00067\end{array}$$

So the indispensable of f is $\sqrt{\pi}(1+a/\sqrt{2}+b/\sqrt{3}+c/\sqrt{4})$ The indispensable of f ^ 2 is $\sqrt{\pi}(1/\sqrt{2}+2a/\sqrt{3}+(a^2+2b)/\sqrt{4}+(2c+2ab)/\sqrt{5}+(b^2+2ac)/\sqrt{6}+2bc/\sqrt{7}+c^2/\sqrt{8})$ with comparable expressions for the amount in regards to theta3. We intend to locate a, b, c to make sure that the integrals and also amounts all suit. Unless my matrix of coefficients has a really not likely dependancy it will certainly be readily available. Excel asserts $a=-3.782590725, b=4.503400057, c=-1.83137936$ is really near a remedy, and also there need to be extra.

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