7
$\begingroup$

What is gap $\log(n+1)-\log(n)$ between log of consecutive integers? That is what precision of logarithms determines integers correctly?

$\endgroup$
1
  • 4
    $\begingroup$ You should specify the base. All answers are assuming base e, because if you mean base 10 most answers are wrong (by a fixed constant) $\endgroup$
    – user
    Jul 22, 2018 at 18:52

5 Answers 5

20
$\begingroup$

$\log(n+1)-\log n=\log(1+\frac1n)$. Using the Taylor series for $\log(1+x)$, this is $$\frac1n-\frac1{2n^2}+\frac1{3n^3}-\cdots\approx\frac1n.$$

$\endgroup$
17
$\begingroup$

Especially Lime's answer is absolutely correct and a very good approach. Let me show a conceptually somewhat simpler one, that only uses the derivative, namely $(\log x)'=1/x$. This is a strictly decreasing function, so $\log x$ is concave. In particular, its graph is below the tangent line at any point of the graph.

We obtain $\log (n+1)< \log n + 1/n$ and $\log n< \log (n+1) - 1/(n+1)$. To sum up:

$$\frac{1}{n+1} < \log (n+1)-\log n < \frac{1}{n}$$

$\endgroup$
10
$\begingroup$

This is rather similar to previous answers, but I think it's still worth pointing out. You're asking about the slope of a chord of the graph of $\log x$, the chord joining $(n,\log n)$ to $(n+1,\log(n+1))$. By the mean value theorem, this equals the slope of the tangent line, $1/x$, at some $x$ between $n$ and $n+1$.

$\endgroup$
8
$\begingroup$

Just added for your curiosity.

In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as $$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{2}{2 n+1}$$ $$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{6 n+3}{6 n^2+6n+1}$$ $$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{60 n^2+60 n+11 } {60 n^3+90 n^2+36 n+3 }$$ $$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{420 n^3+630 n^2+260 n+25 }{420 n^4+840 n^3+540 n^2+120 n+6 }$$

These are respectively equivalent to Taylor series to $O\left(\frac{1}{n^3}\right)$, $O\left(\frac{1}{n^5}\right)$, $O\left(\frac{1}{n^7}\right)$ and $O\left(\frac{1}{n^9}\right)$.

$\endgroup$
5
$\begingroup$

Here's a geometric way to get a good approximation for $\ln(n+1)-\ln(n)$:

Use the fact that $\ln(x)=\int_1^x \frac1t dt$. (For $x>0$.)

Then $\ln(n+1)-\ln(n)$ is the area under the curve $f(x)=\frac1x$ from $n$ to $n+1$.

We are looking for the area over an interval of length $1$. So numerically the area should be equal to the 'average' height. Because the $\frac1x$ function is strictly decreasing, we get a good approximation to that 'average' height by evaluating the $\frac1x$ function a the midpoint of the interval (namely at $n+\frac12$).

So $\ln(n+1)-\ln(n)\approx \frac{1}{n+\frac12}$, with the approximation getting better and better as $n$ gets large.

$\endgroup$
1
  • 1
    $\begingroup$ Notice that $\frac{1}{n+1/2}$ is the same as the Padé approximant given in @ClaudeL's answer of $\frac{2}{2n+1}$! $\endgroup$
    – Silverfish
    Jul 22, 2018 at 23:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .