Using basic calculus, I can prove that $f(t)=t-1-\log t \geq 0$ for $t > 0$ by setting the first derivative to zero \begin{align} \frac{df}{dt} = 1 - 1/t = 0 \end{align} And so I have a critical point at $t=1$ and $f(1)=0$. Then I calculate the second derivative $\frac{d^2f}{dt^2} = 1/t^2 \geq 0$ meaning that $f$ is a convex function with a minimum value of 0 so $f \geq 0$ for $t > 0$.

However, something in my gut tells me there's a way to prove this without even using the first or any derivative of $f$. I've been thinking about this for a while and I haven't been able to do this.

Question is: can you prove $f\geq 0$ without relying on any derivatives of $f$?

  • 3
    $\begingroup$ What is your definition of $\log$ without derivatives? If I'm allowed to use $\log(t) =\int_1^t dx\, x^{-1}$ then your question is easy (but if feels like cheating). $\endgroup$
    – Fabian
    Apr 7, 2011 at 19:27
  • $\begingroup$ @Shai Covo: Why did you remove your answer earlier? I was going to vote it up. $\endgroup$ Apr 7, 2011 at 20:02
  • $\begingroup$ @Eric Naslund: Because it corresponded to $t-1$ being nonnegative. $\endgroup$
    – Shai Covo
    Apr 7, 2011 at 20:15
  • $\begingroup$ @Shai Covo: It's obvious when $t-1\geq 0$, but it also holds if $0\lt t\lt 1$ (using the Taylor series as well). The "tail" of the series starting with the quadratic term is an alternating series whose terms are strictly decreasing in absolute value, and start with a positive term, so the "tail" is positive and the inequality holds. See my (corrected) answer. $\endgroup$ Apr 7, 2011 at 20:17
  • $\begingroup$ @Arturo Magidin: Thanks, I just explained why I decided to remove that answer. $\endgroup$
    – Shai Covo
    Apr 7, 2011 at 20:30

2 Answers 2

  1. With the definition of $\log t$ as an integral. We can define $$\log t = \int_1^t \frac{1}{x}\,dx.$$ The function $\frac{1}{x}$ is decreasing, so a left hand sum approximation is always an over estimate, and a right hand sum approximation is always an underestimate. Dividing the interval $[1,t]$ into a single interval of length $t-1$ and evaluating at the left endpoint, we get $$\ln(t) = \int_1^t\frac{1}{x}dx \leq f(1)(t-1) = t-1$$ giving the desired inequality.

    If $0\lt t\lt 1$, then we first switch limits and use a right hand sum with one interval we get: $$\ln(t) = \int_1^t\frac{1}{x}dx = -\int_t^1\frac{1}{x}dx \leq -f(1)(1-t) = t-1$$ (we have $\int_t^1\frac{1}{x}dx \geq f(1)(1-t)$ since the right hand sum is an underestimate), so multiplying by $-1$ gives the inequality above), giving the desired inequality again.

    If $t=1$, the inequality reduces to $0\geq \log(1)$, which is of course true.

  2. With exponentials. $t-1-\log t\geq 0$ if and only if $\log t\leq t-1$, if and only if $t \leq e^{t-1}$.

    • With the Taylor series definition of $e^t$. Since $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ then $$e^{t-1} = 1 + (t-1) + \frac{(t-1)^2}{2!} + \frac{(t-1)^3}{3!} + \cdots.$$ If $t\geq 1$, then $e^{t-1}\geq 1+(t-1) = t$, giving the desired inequality. If $0\lt t\lt 1$, then we have an alternating series $$ \frac{(t-1)^2}{2!} + \frac{(t-1)^3}{3!} + \frac{(t-1)^4}{4!}+\cdots$$ with ever decreasing terms: $$\frac{|t-1|^{n+1}}{(n+1)!} \lt \frac{|t-1|^n}{n!} \Longleftrightarrow |t-1|\lt n+1,$$ which holds because $n\geq 2$ and $|t-1|\lt 1$. Thus, the "tail" (starting in the quadratic term) of the series is positive, so $e^{t-1} \geq 1+(t-1) = t$ still holds, giving the desired inequality as well.

    • With the definition of $e^t$ as a limit. We have $$e^x = \lim_{n\to\infty}\left(1 + \frac{x}{n}\right)^n$$ so $$e^{t-1} = \lim_{n\to\infty}\left(1 + \frac{t-1}{n}\right)^n.$$ If $t\geq 1$, the sequence is nondecreasing (we are compounding interest, so the more often we compound the bigger the payoff). In particular, $e^{t-1}\geq 1 +\frac{t-1}{1} = t$, giving the desired inequality. If $0\lt t \lt 1$, then we have $$e^{t-1} = \lim_{n\to\infty}\left(1 - \frac{1-t}{n}\right)^n.$$ Again, the sequence is increasing (this is like paying off a debt with fixed interest; if you pay down the capital more often, your total interest will be smaller in the end). So again we have $e^{t-1} \geq 1 - \frac{1-t}{1} = t$.

  • $\begingroup$ Thanks! I should have thought of exponentials. So obvious, now that I see it. $\endgroup$
    – JasonMond
    Apr 7, 2011 at 23:57

With the definition of $\log$ as an integral $$t-1 -\log t = \int_1^t \frac{x-1}{x} dx.$$

Because the integrand is positive for $x>1$, we have $t-1 -\log t \geq 0$ for $t>1$.

Because the integrand is negative for $x<1$, we have $t-1 -\log t \geq 0$ for $0<t<1$.


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