$$ f(x)+\frac{y.f'(x)}{1!}+\frac{y^2 f^{''}(x)}{2!}+\cdots=e^{yS}(f(x))=f(x+y) \text{ where }\frac{d}{dx}=S$$ is a operator
$$ f(x)+\frac{y.f''(x)}{1!}+\frac{y^2 f^{(4)}(x)}{2!}+\cdots=e^{yS^2}(f(x))=?$$ how can it be expressed as f(x)?
The most nearly expression I had:
$$ f(x)+\frac{y.f''(x)}{2!}+\frac{y^2 f^{(4)}(x)}{4!}+\cdots=\frac{1}{2} (e^{\sqrt{y}S}(f(x))+e^{-\sqrt{y}S}(f(x)))=\frac{1}{2}(f(x+\sqrt{y})+f(x-\sqrt{y}))$$
Is there any way to express $e^{yS^2}(f(x))$ as closed form of $f(x)$?
Thanks for answers
EDIT: I thought fourier transform may help to get progress, What I tried:
Fourier transform of $f(x)$ can be written $$F(x)= \int_{-\infty}^{+\infty} f(t)e^{-2πixt} \mathrm{d}t$$
inverse Fourier transform of $f(x)$ can be written $$f(x)= \int_{-\infty}^{+\infty} F(t)e^{2πixt} \mathrm{d}t$$
Thus
$$f^{(2n)}(x)= \int_{-\infty}^{+\infty} F(t) (2πit)^{(2n)}e^{2πixt} \mathrm{d}t$$
$$\frac{y^nf^{(2n)}(x)}{n!}= \int_{-\infty}^{+\infty} F(t) \frac{y^n(2πit)^{(2n)}}{n!}e^{2πixt} \mathrm{d}t$$
$$\sum_{n=0}^\infty\frac{y^nf^{(2n)}(x)}{n!}= \int_{-\infty}^{+\infty} F(t) (\sum_{n=0}^\infty\frac{y^n(2πit)^{(2n)}}{n!})e^{2πixt} \mathrm{d}t$$
$$\sum_{n=0}^\infty\frac{y^nf^{(2n)}(x)}{n!}= \int_{-\infty}^{+\infty} F(t) (\sum_{n=0}^\infty\frac{((2πit)^2y)^{(n)}}{n!})e^{2πixt} \mathrm{d}t$$
$$\sum_{n=0}^\infty\frac{y^nf^{(2n)}(x)}{n!}= \int_{-\infty}^{+\infty} F(t) e^{(2πit)^2y} e^{2πixt} \mathrm{d}t$$
$$\sum_{n=0}^\infty\frac{y^nf^{(2n)}(x)}{n!}=e^{yS^2}(f(x))= \int_{-\infty}^{+\infty} F(t) e^{-4π^2yt^2+2πixt} \mathrm{d}t$$
$$e^{yS^2}(f(x))= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(u)e^{-2πiut} e^{-4π^2yt^2+2πixt} \mathrm{d}u \mathrm{d}t$$
$$e^{yS^2}(f(x))= \int_{-\infty}^{+\infty} f(u) \int_{-\infty}^{+\infty} e^{-2πiut} e^{-4π^2yt^2+2πixt} \mathrm{d}t \mathrm{d}u$$
Now Need to find
$$\int_{-\infty}^{+\infty} e^{-2πiut} e^{-4π^2yt^2+2πixt} \mathrm{d}t=\int_{-\infty}^{+\infty} e^{-4π^2yt^2+2πi(x-u)t} \mathrm{d}t $$
EDIT:(after @Winther comment)
$$\int_{-\infty}^{+\infty} e^{-4π^2yt^2+2πi(x-u)t} \mathrm{d}t =e^{4π^2y\alpha^2}\int_{-\infty}^{+\infty} e^{-4π^2y(t-\alpha)^2} \mathrm{d}t$$
Where $$2\alpha.(-4\pi^2y)=2\pi i(x-u)$$
$$\alpha=\frac{-i(x-u)}{4\pi y}$$
$$\int_{-\infty}^{+\infty} e^{-4π^2yt^2+2πi(x-u)t} \mathrm{d}t =e^{\frac{-(x-u)^2}{4y}}\int_{-\infty}^{+\infty} e^{-4π^2y(t-\alpha)^2} \mathrm{d}t$$
$$z=t-\alpha$$
$$\int_{-\infty}^{+\infty} e^{-4π^2yt^2+2πi(x-u)t} \mathrm{d}t =e^{\frac{-(x-u)^2}{4y}}\int_{-\infty}^{+\infty} e^{-4π^2yz^2} \mathrm{d}z=\frac{e^{\frac{-(x-u)^2}{4y}}}{2\sqrt{\pi y}}$$
$$e^{yS^2}(f(x))= \int_{-\infty}^{+\infty} f(u) \int_{-\infty}^{+\infty} e^{-2πiut} e^{-4π^2yt^2+2πixt} \mathrm{d}t \mathrm{d}u=\int_{-\infty}^{+\infty} f(u) \frac{e^{\frac{-(x-u)^2}{4y}}}{2\sqrt{\pi y}} \mathrm{d}u$$
Finally the result can be written as
$$e^{yS^2}(f(x))=\frac{1}{2\sqrt{\pi y}}\int_{-\infty}^{+\infty} f(u) e^{\frac{-(x-u)^2}{4y}} \mathrm{d}u$$
Please check my result if I made a mistake during calculations. Thanks for comments and advice.