Chapter 1 - Limits and Their Properties - 1.4 Exercises - Page 82: 126

Let the polynomial be $P(x)=a_0+a_1x+a_2x^2...$ We know, $P(0)=0$ $\implies P(0)=a_0=0$ We know if we substitute $x^2+1$ in place of $x$. We will get constant value too. Which are the sum of the coefficients. But we proved above that the polynomial $P(x)$ should not have a constant term. hence, the coefficients should be equal to one. Thus, the polynomial $P(x)=x+x^2+...$ We have given $P(x^2+1)=(P(x))^2+1$ On the right-hand side, there is only one constant. But, if we substitute $x^2+1$ in the place of $x$. There will be as many constants as there is $x$. Hence, polynomial $P(x)=x$ is the only polynomial with satisfy the conditions $P(x^2+1)=(P(x))^2+1$ and $P(0)=0$.

Let the polynomial be $P(x)=a_0+a_1x+a_2x^2...$ We know, $P(0)=0$ $\implies P(0)=a_0=0$ We know if we substitute $x^2+1$ in place of $x$. We will get constant value too. Which are the sum of the coefficients. But we proved above that the polynomial $P(x)$ should not have a constant term. hence, the coefficients should be equal to one. Thus, the polynomial $P(x)=x+x^2+...$ We have given $P(x^2+1)=(P(x))^2+1$ On the right-hand side, there is only one constant. But, if we substitute $x^2+1$ in the place of $x$. There will be as many constants as there is $x$. Hence, polynomial $P(x)=x$ is the only polynomial with satisfy the conditions $P(x^2+1)=(P(x))^2+1$ and $P(0)=0$.