I need help on this question Thanks
As we know recursion refers to function calling itself.
Example x+y
f(x,0)=x
f(x,y+1)=f(x,y)+1
f(1,0)=1
f(1,1)=f(1,0)+1=1+1=2
f(1,2)=f(1,1)+1=(f(1,0)+1)+1=1+1+1=3
f(5,0)=5
f(5,1)=f(5,0)+1=5+1=6
f(5,2)=f(5,1)+1=(f(5,0)+1)+1=5+1+1=7
Example g(x)=x^2
Compare x^2 with x^y
x^0=1
x^(y+1)=x^y*x
f(1,0)=1
f(1,1)=x^(0+1)=x^0*x=1*1=1
f(1,2)=x^(1+1)=(x^(0+1))*x=x^0*x*x=1*1*1=1
f(5,0)=1
f(5,1)=x^(0+1)=x^0*x=1*5=5
f(5,2)=x^(1+1)=(x^(0+1))*x=x^0*x*x=1*5*5=25
I need help on this question Thanks 1. Let g(x) = x2 and h(x, y, z) =x+ y + z, and let f(x, y) be the function defined...