\(Fc \{x^{(\frac{-1}{2})}\} = \int_0^? x^{(\frac{-1}{2})} cos(px)dx = constant * p^{(\frac{-1}{2})} \)
Inverse fourier transform of \( p^{(\frac{-1}{2})} = constant * x^{(\frac{-1}{2})} \)
Hence function \(x^{(\frac{-1}{2})} \)is self reciprocal.

\(e^{-2|x|}= e^{-|2x|}= F(2x) \)
\(F\{F(2x)\} = \frac{1}{2} f(\frac{p}{2}) \)
\( = \frac{1}{2} \frac{2}{(1+\frac{p^2}{4})} \)
\( = \frac{4}{(4+p^2 )}.\)

?(n) exists only at n=0 and ?(n+1) exists only at n=1. Therefore while substituting this function, Z – Transform at every or place becomes zero except at n=1. Therefore Z-Transform of ?(n+1) is \(\frac{1}{z}. \)

\(-a \, e^{-ax} = \frac{d}{dx}(e^{-ax}) = F?(x) \)
\(Fs\{F?(x)\} = -pfc(p) \)
\( = \frac{-p^2}{a^2+p^2} \).

This can be solved using determinant.
\(\begin{bmatrix}
y_n & 1 & 1\\
y_{n+1} & 3 & 5\\
y_{n+2} & 9 & 25\\
\end{bmatrix} \) = 0. Now, by solving determinant, we get required difference equation.

Taking lim_z?? U(z), we get 0 which is u₀. Now using shifting property and again using limit we get u₁ which is 0. Again, by using shifting property we get u₂ which is 3. Now, by using shifting by 3 properties, we get value of u₃ which is 14.

\(F\{e^{-|x|}\} = \frac{2}{1+p^2} \)
\(F\{t^2 e^{-|x|}\} = (-i)^2 \frac{2}{1+p^2} = \frac{-2}{1+p^2} \).

First, substitute z as \((\frac{1}{y})\) and n expand got result. This is in format of Z-Transform expansion. Thus we get required results.

Fourier transform of \( e^{-a|x|} = \frac{2a}{a^2+p^2} \)
Fourier transform of \(e^{-b|x|} = \frac{2b}{b^2+p^2} \)
fourier transform of \(e^{-a|x|}* e^{-b|x|}= \frac{2a}{a^2+p^2}.\frac{2b}{b^2+p^2} \)
\( = \frac{4ab}{(a^2+p^2)(b^2+p^2)} \).

The first step is to expand given equation by replacing every ?y_n by (y_n+1– y_n). Order of a difference equation is given by, \(\frac{n+3-n}{1}\) which is actually 3.