Show that not exists any polynomial function such that $f(x) = \log (1+x)$.

Question

Does anyone have any idea on that problem?

Let $f : \mathbb{R} \to \mathbb{R}$ be a polynomial function. Show that not exists any $f$ such that $f(x) = \log (1+x)$.

It's easy to show that $a_0 = 0$ and $a_1 = 1$. But after i don't have any idea. Any point? Thanks!

This is a bit of a dumb answer, but polynomials exist and are continuous $\forall x \in \mathbb{R}$, whereas $\log(1+x)$ is not. — K.defaoite, May 31 '20 at 14:13
More generally logarithm is not an algebraic function. The current question is very easily handled by noting that derivative of a polynomial is also a polynomial. — Paramanand Singh, May 31 '20 at 15:38

score 5 · Accepted Answer · answered May 31 '20 at 14:11

Maybe there are many ways to show this; but what I have in mind is this: If $f$ is a polynomial of degree $k$, then its $(k+1)$-th derivative is zero. But this is not the case for $\log(1+x)$; the $(k+1)$-th derivative of $\log(1+x)$ does not vanish.

score 1 · Answer 2 · answered May 31 '20 at 14:12

1

If there was, then it would not be constant and we would have$$\lim_{x\to\infty}\frac{\log(x+1)}{f(x)}=1.$$But, for every non-constant polynomial function, we have$$\lim_{x\to\infty}\frac{\log(x+1)}{f(x)}=0.$$

answered May 31 '20 at 14:12

José Carlos Santos

427,504

score -1 · Answer 3 · answered May 31 '20 at 15:04

-1

If $\log(1+x)$ is polynomial, $f$ is a transcendental polynom function which doesn't exist.

answered May 31 '20 at 15:04

EDX

1,797

Show that not exists any polynomial function such that $f(x) = \log (1+x)$.

3 Answers3