Assuming it is a cubic function

i=(i)^(0.5)

f(0)=1=-(i^2)^0

f(1)=-i^2=-(-1)=1=-(i^2)^1

f(2)=-1=-(i^2)^2

f(3)=-1(i^2)^3

b=1

let f(x)=a(x)^3+b(x)^2+c(x)+d

f(0)=1=d

f(1)=1=a+b+c+1=>a+b+c=0

f(2)=-1=8a+4b+2c+1

6a+2b=-2

3a+b=-1=> 24a+8b=-8

f(3)=27a+9b+3c+1=-i

24a+6b=-i-1

2b=i+7

b=(i+7)/2

a=-(1+b)/3=-(i+9)/6

c=-(a+b)=(-i-9+3i+21)/6=(2i+12)/16=(i+6)/8

d=1

f(x)=-((i+9)/6)x^3+((i+7)/2)x^2+((i+6)/8)x+1

 

Categories: math