1
$\begingroup$

I saw this in a proof for the Central Limit Theorem:

$\ln(1-x) \approx -x$ when $x$ is small

It seems to be true when I plug in small values of $x$. But why does it work?

$\endgroup$
  • 4
    $\begingroup$ Compute $$\lim_{x \to 0} \frac{\ln(1-x)}{-x}$$ $\endgroup$ – user384138 Jun 12 '17 at 18:02
  • 3
    $\begingroup$ You can look at the Taylor expansion to obtain this result. $\endgroup$ – Severin Schraven Jun 12 '17 at 18:05
  • 2
    $\begingroup$ Dang I blinked after I clicked post your answer and now there's 5 answers lol $\endgroup$ – mrnovice Jun 12 '17 at 18:07
  • 2
    $\begingroup$ @mrnovice The "fastest gun in the west" problem $\endgroup$ – user223391 Jun 12 '17 at 18:11
5
$\begingroup$

We can write $$\ln(1-x)$$ as the infinite series:

$$\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots$$

so for $x$ is small, we have that $x^2, x^3, x^4, \dots$ are even smaller, neglectible amounts so we can say:

$$\ln(1-x) \approx -x$$

| cite | improve this answer | |
$\endgroup$
  • 5
    $\begingroup$ It's a good intuitive argument if one knows the sum, but.... formalizing it to argue that the sum of infinitely many negligible terms is still negligible is not that trivial. $\endgroup$ – Clement C. Jun 12 '17 at 18:09
  • $\begingroup$ You are absolutely right. If the OP wants me to give further information, I will do that. $\endgroup$ – user370967 Jun 12 '17 at 18:11
  • $\begingroup$ I agree with @ClementC. Taylor's theorem is extremely relevant in proving this, and at that point, you're over-killing it since it uses the MVT to prove in the first place and you may as well stop after once step and appeal to the more basic theorem. $\endgroup$ – Adam Hughes Jun 12 '17 at 18:27
  • 1
    $\begingroup$ Thank you, Taylor series is a good enough argument for me. Also Open Ball's suggestion in the comments is also good. $\endgroup$ – foobar Jun 12 '17 at 22:59
2
$\begingroup$

Assuming $x$ is real, the Taylor series of $\ln (1-x)$ about zero is $$ \ln (1-x) = \ln(1) + \frac{d}{dx}\ln(1-x)|_{(x = 0)}x + \mathcal{O}(x^2) $$ or $$ \ln (1-x) = 0 - x + \mathcal{O}(x^2) = -x + \mathcal{O}(x^2) $$ For small $x$ (that is, much less than one) all terms of order $x^2$ are negligible so we have $$ \ln(1-x) \approx -x. $$ Note that the statement as written cannot be true, for if $y = \ln(1-x)$ then $e^y = 1-x$, which is $< 1$ when $x$ is positive. Thus $y$ must be negative when $x$ is positive and positive when $x$ is negative in the linear limit.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The easiest way to see this is to use the Mean Value Theorem. Note that since $\log(1-x)$ is twice differentiable (in particular its derivative is continuous) that means that

$$f(x)\approx f'(0)(x-0)+f(0)={-1\over 1-0}(x-0)+0=-x$$

for $x$ near $0$. (i.e. small $x$)

You may also know this process by another name: "linearization."

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Because $\ln(1)=0$ and $\left.\frac{d}{dt}\ln(1-x)\right|_{t=0}=-1$. Therefore, when $x$ is small, $\ln(1-x)$ is approximately $0+(-1)\times x=-x$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

This can also be seen using Taylor expansions. Let $f(x) = \ln(1-x)$. We have $f(0) = \ln(1) = 0$. And since $f'(x) = \frac{-1}{1-x}$, $f'(0) = -1$. This gives a linear Taylor approximation of

$$ f(x) = \ln(1-x) \approx 0 + (-1)\cdot x = -x.$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I am surprised by the earlier answers, all of which agree that the statement

$\ln(1 - x) \approx -x $ when $x$ is small

was true. Actually, it is not, and the problem lies in the notion of smallness.

Smallness

All earlier answers seem to assume that the closer a number $x$ is to $0$, the smaller it is. However, if we assume the usual order of the real numbers, i.e., $$ \ldots < -2 < -1 < 0 < 1 < 2 < \mathrm{e} < 3 < \pi < \ldots \;, $$ then a number $x$ is the smaller, the closer it is to $-\infty$.

It is true that the closer the absolute value $|x|$ is to $0$, the smaller this absolute value. But the absolute value $|x|$ is not the same as $x$ itself.

Now let us investigate two approaches suggested among the answers.

Suggested approach 1

The first approach suggests to look at the series $$ \ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \ldots $$ @Math_QED claims that

"for $x$ is small, we have that $x^2, x^3, x^4, \dots$ are even smaller".

This is not true. If $x$ is small, i.e., a negative number with large absolute value, then the odd powers $x^1,x^3,x^5,\ldots$ are indeed smaller, but the even powers $x^2,x^4,x^6,\ldots$ are greater. So, this approach fails.

Suggested approach 2

Another approach suggests to compute the limit of $$ \dfrac{\ln(1-x)}{-x} $$ as $x$ becomes smaller. Actually, @Open Ball suggests to compute $$ \lim\limits_{x \to 0} \dfrac{\ln(1-x)}{-x} = 1 \;. $$ However, instead of letting $x \to 0$, we have to let $x \to -\infty$, because any negative number is still smaller than $0$, and we want that $x$ becomes as small as possible. The result of the limit is $$ \lim\limits_{x \to -\infty} \dfrac{\ln(1-x)}{-x} = 0 \;, $$ and this disproves the statement.

smallness

A visual depiction is given in the figure. The smaller $x$ becomes, the more do $\ln(1 - x)$ and $-x$ diverge from each other.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ lol, take x = 0,01, then x^2 = 0,0001 is not smaller? $\endgroup$ – user370967 Jun 12 '17 at 19:28
  • $\begingroup$ It is, but your example is inappropriate, because $0.01$ is not small! A number $x$ is small if it is negative and its magnitude is large. But the magnitude of $0.01$ is not large! By the way, if you pick this out to claim that I was wrong, then you really miss the point. $\endgroup$ – Björn Friedrich Jun 12 '17 at 19:31
  • $\begingroup$ To the downvoter: Next time, you can just hint to the error, so that I can fix it. :-) $\endgroup$ – Björn Friedrich Jun 12 '17 at 19:35
  • 3
    $\begingroup$ Small means close to 0 $\endgroup$ – Flame Trap Jun 12 '17 at 19:38
  • 1
    $\begingroup$ @FlameTrap: Assuming the usual order of real numbers, $-5$ is smaller than $1$ even though $1$ is closer to $0$. $\endgroup$ – Björn Friedrich Jun 12 '17 at 19:45

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.