Question

the value of z transform of (1/n!) is​

Answer 1

Answer:

hope this helps u

Step-by-step explanation:

Answer 2

Answer:   $Z(\frac{1}{n} )= log(\frac{Z}{Z-1} )$

Step-by-step explanation:

                    $Z(\frac{1}{n} )=$∑$Z^{-n} (\frac{1}{n})$ (from, $n=1$ to $n= infinite$)

                             $=Z^{-1} +\frac{Z^{-2} }{2} +\frac{Z^{-3} }{3} +.............$

• We know that the  expansion of $log(1-x)$ is

         $=-x-\frac{x^{2} }{2} -\frac{x^{3} }{3} -......................$

∴we can write as,

             $log(1-x)=-(x+\frac{x^{2} }{2} +\frac{x^{3} }{3} +......................)$

• Now replace $x$ with $\frac{1}{2}$

           $log(1-\frac{1}{2} )=-(Z^{-1} +\frac{Z^{-2} }{2} +\frac{Z^{-3} }{3} +.....)$

          $-log(\frac{Z-1}{Z} )=(Z^{-1} +\frac{Z^{-2} }{2} +\frac{Z^{-3} }{3} +...)$

            $log(\frac{Z}{Z-1} ) = Z(\frac{1}{n} )$