Let f(z) be an entire function such that fn+1(z) equivalent 0, Prove that f is a polynomial of degree at most n? Solution We can prove the above formula using mathematical induction. So lets us take multiple cases. For f(x) = x, f(x)\'=1, f(x)\'\'=0. so for f(x)^2 the polynomial power is 1, for f(x)=x^2, f(x)\'=2x, f(x)\'\'=2, f(x)\'\'\'=0, so for f(x)\'\'\' the polynomial is 2. So we can deduce that for a polynomial with n power it should be differentiated n+1 times to get fn+1(z)=0, So it could be stated otherwise if fn+1(z)=0 then f(z) has a power of atmost x^n..

1. Let f(z) be an entire function such that fn+1(z) equivalent 0, Prove that f is a polynomial of
degree at most n?
Solution
We can prove the above formula using mathematical induction.
So lets us take multiple cases.
For f(x) = x, f(x)'=1, f(x)''=0. so for f(x)^2 the polynomial power is 1,
for f(x)=x^2, f(x)'=2x, f(x)''=2, f(x)'''=0, so for f(x)''' the polynomial is 2.
So we can deduce that for a polynomial with n power it should be differentiated n+1 times to get
fn+1(z)=0,
So it could be stated otherwise if fn+1(z)=0 then f(z) has a power of atmost x^n.