Recall the expansions of the square of a binomial and the cube of a binomial.
(u+v)^2&=u^2+2uv+v^2
(u+v)^3&=u^3+3u^2v+3uv^2+v^3
Let's use these with u=z and v=1,2,3,4, and5. We will need these in finding f(z+1), f(z+2), f(z+3), f(z+4), and f(z+5). First, let's see the quadratic expansions.
Expression
Expansion
Simplification
(z+ 1)^2
z^2+2z( 1)+ 1^2
z^2+2z+1
(z+ 2)^2
z^2+2z( 2)+ 2^2
z^2+4z+4
(z+ 3)^2
z^2+2z( 3)+ 3^2
z^2+6z+9
(z+ 4)^2
z^2+2z( 4)+ 4^2
z^2+8z+16
(z+ 5)^2
z^2+2z( 5)+ 5^2
z^2+10z+25
Let's also write the cubic expansions.
Expression
Expansion
Simplification
(z+ 1)^3
z^3+3z^2( 1)+3z( 1^2)+ 1^3
z^3+3z^2+3z+1
(z+ 2)^3
z^3+3z^2( 2)+3z( 2^2)+ 2^3
z^3+6z^2+12z+8
(z+ 3)^3
z^3+3z^2( 3)+3z( 3^2)+ 3^3
z^3+9z^2+27z+27
(z+ 4)^3
z^3+3z^2( 4)+3z( 4^2)+ 4^3
z^3+12z^2+48z+64
(z+ 5)^3
z^3+3z^2( 5)+3z( 5^2)+ 5^3
z^3+15z^2+75z+125
Let's now find and simplify f(z), f(z+1), f(z+2), f(z+3), f(z+4), and f(z+5), as asked. To find f(z) we replace x with z in the given f(x).
f(z)=az^3+bz^2+cz+d
To find f(z+1), let's use the expansions of (z+1)^2 and (z+1)^3 we found above.
We can find the other expressions similarly.
f(z)&=az^3+bz^2+cz+d
f(z+1)&=az^3+(3a+b)z^2+(3a+2b+c)z+(a+b+c+d)
f(z+2)&=az^3+(6a+b)z^2+(12a+4b+c)z+(8a+4b+2c+d)
f(z+3)&=az^3+(9a+b)z^2+(27a+6b+c)z+(27a+9b+3c+d)
f(z+4)&=az^3+(12a+b)z^2+(48a+8b+c)z+(64a+16b+4c+d)
f(z+5)&=az^3+(15a+b)z^2+(75a+10b+c)z+(125a+25b+5c+d)
In the following table we list these expressions and their differences. Remember that these differences are found by subtracting each term from the next. For example, the first difference in the table is found by simplifying f(z+1)-f(z).
First-Order Differences
az^3+bz^2+cz+d
3az^2+(3a+2b)z+(a+b+c)
az^3+(3a+b)z^2+(3a+2b+c)z+(a+b+c+d)
3az^2+(9a+2b)z+(7a+3b+c)
az^3+(6a+b)z^2+(12a+4b+c)z+(8a+4b+2c+d)
3az^2+(15a+2b)z+(19a+5b+c)
az^3+(9a+b)z^2+(27a+6b+c)z+(27a+9b+3c+d)
3az^2+(19a+2b)z+(37a+7b+c)
az^3+(12a+b)z^2+(48a+8b+c)z+(64a+16b+4c+d)
3az^2+(27a+2b)z+(61a+9b+c)
az^3+(15a+b)z^2+(75a+10b+c)z+(125a+25b+5c+d)
Let's now look for the second-order differences; the differences of the first-order differences.
Second-Order Differences
3az^2+(3a+2b)z+(a+b+c)
6az+(6a+2b)
3az^2+(9a+2b)z+(7a+3b+c)
6az+(12a+2b)
3az^2+(15a+2b)z+(19a+5b+c)
6az+(18a+2b)
3az^2+(19a+2b)z+(37a+7b+c)
6az+(24a+2b)
3az^2+(27a+2b)z+(61a+9b+c)
Finally, let's find the third-order differences, the differences of the second-order differences.
Third-Order Differences
6az+(6a+2b)
6a
6az+(12a+2b)
6a
6az+(18a+2b)
6a
6az+(24a+2b)
We can see that the third-order differences are indeed constant.
