If `e^(A)` is defined as `e^(A)=I+A+A^(2)/(2!)+A^(3)/(3!)+...=1/2 [(f(x),g(x)),(g(x),f(x))]`, where `A=[(x,x),(x,x)], 0 lt x lt 1` and I is identity matrix, then find the functions f(x) and g(x).

Explanation

`A=[(x,x),(x,x)]`
`implies A^(2)=[(2x^(2),2x^(2)),(2x^(2),2x^(2))], A^(3)=[(2^(2)x^(3),2^(2)x^(3)),(2^(2)x^(3),2^(2)x^(3))]` and so on
Then `e^(A)=I+A+A^(2)/(2!)+A^(3)/(3!)+...`
`=[(1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+...,x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+...),(x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+...,1+x+ (2x^(2))/(2!)+(2^(2)x^(3))/(3!)+...)]`
`=[(1/2(1+2x+(2^(2)x^(2))/(2!)+...)+1/2,1/2(1+2x+(2^(2)x^(2))/(2!)+...)-1/2),(1/2(1+2x+(2^(2)x^(2))/(2!)+...)-1/2,1/2(1+2x+(2^(2)x^(2))/(2!)+...)+1/2)]`
`=1/2 [(e^(2x)+1,e^(2x)-1),(e^(2x)-1,e^(2x)+1)]`
`implies f(x)=e^(2x)+1` and `g(x)=e^(2x)-1`