Proving that $\sqrt{1 + x^2}$ is not a polynomial function

Solution 1:

Suppose it is a polynomial, then $$\sqrt{1+x^2}= ax^n+...+b$$

so $$1+x^2= (ax^n+...+b)^2 = a^2x^{2n}+...$$

Since polynomial are equal iff they have same degree we have $2=2n$ so $n=1$ and now we have $$ 1+x^2 = (ax+b)^2 = a^2x^2+2abx+b^2 \implies ab = 0$$ a contradiction, since $|a|=|b|= 1$.

Solution 2:

Alt. hint: following up on OP's idea to use derivatives, note that $\,f(x) \cdot f'(x) = x\,$, but the only polynomials which satisfy that identity are $\,f(x)=\pm x\,$ (why?).

Alternatively, show that there exists no polynomial $\,f\,$ such that $\,f^2(x)=x^2+1\,$.

Solution 3:

Suppose $f$ is a polynomial,namely $$f(x)=\sqrt{1+x^2}={a_0+a_1x+\cdots+a_nx^n}=p(x)$$ for some $n$. Then $$\frac{f(x)}{x}=\frac{p(x)}{x}$$ for all non zero $x$. Hence $$\lim_{x \rightarrow \infty} \frac{p(x)}{x}=\lim_{x \rightarrow \infty}\frac{f(x)}{x}=\lim_{x \rightarrow \infty} \frac{\sqrt{1+x^2}}{x}=1$$

which means $p(x)$ and $x$ are polynomials of same degree, a contradiction!

Solution 4:

Another demonstration:

A polynomial of degree one or higher when squared, should result in a polynomial that has all its roots double or more than doubles.

$f ^ 2 = x ^ 2 + 1 = (x + i) (x-i)$

That has two roots but they are not equal. And by the fundamental theorem of algebra, that factorization is unique.

Therefore f can not be a polynomial.