here is one proof that I know but I am not totally sure if it is acceptable-

exponential functions are exponential: no matter how many times you differentiate them e.g-

first derivative f`(x)= e^x
2nd derivative f``(x)= e^x
3rd derivative f```(x)=e^x

and so on.

now if you differentiate a polynomial function- let's say,

f(x)= x^5
1st derivative f`(x)= 5x^4
2nd de3rivative f``(x)= 20x^3
3rd derivatives f```(x)=60x^2
4th derivative f````(x)=120x
5th derivative f`````(x)= 0

like this every polynomial finally gets differentiated to zero or a constant . this proves that the polynomials are not exponential.

                        **is my proof ok**  

I want more alternate proofs and a brief explanation about this one.

  • $\begingroup$ Your proof is fine if we just want to prove that $p(x)$ is not $e^x$. But how do we know that $p(x)=e^{f(x)}$ has no solution whatever function $f$ is? For example, if $p(x)=x^2+1$, $e^{\ln(p(x))}=p(x)$ which is a polynomial. $\endgroup$ – Bernard Massé May 23 '20 at 18:17

Your proof is correct. You can also say that $\lim_{x\to-\infty}e^x=0$, whereas you have$$\lim_{x\to-\infty}P(x)=\pm\infty$$if $P$ is a non-constant polynomial function. And, clearly, the exponential function is not constant.


Suppose $e^x=P(x)$, where $P$ is a polynomial of degree $n$. Note first that $n\gt0$, since $e^x$ is nonconstant. It follows that $P(2x)$ and $(P(x))^2$ are polynomials of different degrees, namely $n$ and $2n$. But $P(2x)=e^{2x}=(e^x)^2=(P(x))^2$ says they are of the same degree, which is a contradiction. So $e^x$ is not equal to any polynomial.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.