# Show that, for $t>0$, $\log t$ is not a polynomial.

How can I show that? I've tried to reverse the logarithm to it's exponential form in a trial to show that but I got no success. Can you help me?

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I think it is not even a rational function. – Shubhodip Mondal Nov 26 '12 at 19:58
A polynomial is a polynomial for all of $t$. So the very fact that log isn't defined for $\le 0$ means it is not a polynomial. – Kaz Nov 26 '12 at 20:21
I don't know Kaz, $g(t)=\exp(\log(t))$ could be thought of as a polynomial function even though it is only defined when $t>0$. – Jonas Meyer Nov 26 '12 at 20:23
You can differentiate log an arbitrary number of times and never get a function that is identically zero. – tomcuchta Nov 29 '12 at 15:57
@tomcuchta: Yes, that is in beauby's answer. – Jonas Meyer Nov 30 '12 at 4:43

If the logarithm were a polynomial, we could write $$\log(x) = a_n x^n +a_{n-1}x^{n-1}+\cdots +a_0,$$ for constants $a_0,a_1,\ldots,a_n$, where the formula holds for all $x>0$. Assuming this is the case, a contradiction will be reached. Using $\log(x^2)=2\log(x)$, we have $$a_n x^{2n}+a_{n-1}x^{2n-2}+\cdots + a_0 = 2a_n x^n+2a_{n-1}x^{n-1}+\cdots+2a_0.$$ Using the fact that equal polynomials have equal coefficients, this implies that $\log(x)=0$ for all $x$. That isn't true, and thus the assumption that $\log$ is a polynomial leads to a contradiction.

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I can't figure out how you went from my question to this answer. – Voyska Nov 27 '12 at 6:22
Gustavo: I was inclined to use a familiar property of logarithms that is foreign to polynomials. Each answer uses such a property. Another: $\log(t)\to\infty$ as $t\to \infty$ but $\dfrac{\log(t)}{t}\to 0$ as $t\to\infty$. No polynomial (or rational function) has this property. (But the answer above doesn't require limits.) – Jonas Meyer Nov 27 '12 at 6:27
Gustavo: $\log(t)$ is a function of $t$. (More precisely it might be better to say that $\log$ is the function.) This is a standard proof by contradiction. A polynomial function on the positive real numbers has the form $p(t) = a_nt^n +\cdots +a_0$ for some constants $a_i$. We want to show: The function $\log(t)$ is not equal to any polynomial function on the positive reals. Proof by contradiction: Suppose $\log(t)$ is equal to a polynomial function on the positive reals. (Explicitly, that means there is a polynomial function $p(t)=a_nt^n+\cdots+a_0$ such that $\log(t)=p(t)$ for all... – Jonas Meyer Nov 27 '12 at 7:19
...$t>0$.) Next, show that this leads to a contradiction, as the answer above does in sketch form. (Uniqueness of coefficients needs justification. Another exercise in that book deals with this.) – Jonas Meyer Nov 27 '12 at 7:21
Great! I got it! Both are functions, how couldn't I see this? Really thanks for your answer, coments, time and for helping me to become less dumb. – Voyska Nov 27 '12 at 7:22

Do you mean $t \mapsto \log t$ ? If this is the case, consider the derivatives and the fact that a polynomial's derivatives are eventually null.

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This is a good answer. Another formulation: polynomial is eventually equal to its Taylor series. This is never true for $\log$ since it's Taylor series are infinite. – yo' Nov 26 '12 at 20:10
tehecz: A technicality on your formulation: The logarithm is equal to its Taylor series in an interval about each point in its domain, because the Taylor series allows infinitely many terms. However, polynomials are characterized by being equal to a finite Taylor series. – Jonas Meyer Nov 26 '12 at 20:15

You can also use the fact that $\lim_{t \to \infty} \frac{ \log t}{t} = 0 \,.$

For a polynomial of degree at least 2, the corresponding limit is infinity, while for a linear polynomial, the limit is zero if and only if the polynomial is constant....

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To be clear, the limit for polynomials is $\pm \infty$. – Austin Mohr Nov 26 '12 at 20:00

What do you mean? Even if the domain is restricted to $(0,\infty)$, a polynomial function will have a finite right hand limit as $t\rightarrow0^+$, while the logarithmic function has not.

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I don't know, I just copied the question from the Book. – Voyska Nov 26 '12 at 19:55
Why not also read the answer in the book? – GEdgar Nov 26 '12 at 20:52
Maybe the answer in the book cannot satisfy Gustavo's intellectual needs. At least I feel more content after reading Jonas Meyer's proof. – user1551 Nov 27 '12 at 9:37

The series $\sum \frac{1}{n \log n}$ diverges and the series $\sum \frac{1}{n (\log n)^2}$ converges, and this isn't true with $\log n$ replaced by any polynomial (with finitely many terms removed if the polynomial has integer zeroes) or indeed with any rational function (and most of the other arguments given so far are not enough to prove this). Actually this argument probably shows that $\log n$ is not even algebraic.

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The function $\log(t)$ can be expressed via Taylor series as the infinite summation $$\log(t) = (t-1)-\frac{1}{2}(t-1)^2+\frac{1}{3}(t-1)^3-\frac{1}{4}(t-1)^4+\cdots,$$ which involves terms of $t^n$ for unbounded $t$.

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That series is valid when $0<t<2$, but the point is still valid because for example the coefficient of $(t-1)^n$ is $\frac{1}{n!}\log^{(n)}(1)$, and therefore this shows that the derivatives of $\log$ aren't eventually $0$. (It is related to beauby's answer, but they can also be thought of independently, and one doesn't need to bring in derivatives to elaborate your answer.) – Jonas Meyer Nov 26 '12 at 20:21

It does not match in 0: $\lim_{x \to 0^+} \log x = -\infty$ but for any polynomial $\lim_{x \to 0^+} (a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0) = a_0$. Hence there is no $f(x) = (a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0)$ such that $\lim_{x \to 0} (\log x - f(x)) = 0$ so as we get closer to 0 we get divergence from any polynomial.

The $\lim_{x \to 0} \log x$ can be obtained as $\log a = b \Leftrightarrow a = e^b$ and $\lim_{x \to -\infty} e^x = 0$ (assuming logarithm as reverse of exponentiation is well defined).

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The function $f:t\mapsto \log t$ has a vertical asymptote at $0$ (i.e. $\lim_{t\to0}f'(t)=\infty$), so it has a non-removable discontinuity. Polynomials are continuous everywhere.

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If $\log(t)$ is a polynomial, its derivative $\frac{1}{t}$ also is. But this is clearly impossible (the only invertible polynomials are constants because of the degree).

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