4

I need to prove this approximation, but I am unable to conclude $$\log \left(1+\frac{1}{n}\right) \approx \frac{1}{n}$$

Zev Chonoles
  • 129,973
linux mint
  • 145
  • 5
    I'm confused. how does one prove an approximation ? How would I prove that $5$ is approximately equal to $17$ ? – mercio Jul 14 '15 at 15:26
  • 2
    What exactly does '$\approx$' mean? How close should LHS and RHS be? – BCLC Jul 14 '15 at 16:04
  • @mercio I suspect '$LHS \ \approx \ RHS$' is a heuristic for $|LHS - RHS| < \epsilon$ for some $\epsilon \in (0,1)$ – BCLC Jul 14 '15 at 16:04
  • 3
    8 people think this question is well researched? – Alec Teal Jul 14 '15 at 20:31
  • possible duplicate of Showing y≈x for small x if y=log(x+1) –  Jul 14 '15 at 21:12
  • @AlecTeal 9 :( In fact, 3 people think the accepted answer is useful. Is usefulness relative to the reader? I think a necessary condition of usefulness is well-definedness of notation. Oh anyway, it was put on hold as off-topic \m/ – BCLC Jul 15 '15 at 11:22
  • 1
    I rarely degree on " off-topic ". To me any math question is on topic. Duplicate or such is more appropriate. How can an answer be off topic and have a few good answers ? – mick Aug 20 '16 at 18:56
  • @mick 2 years later, do you still disagree/do not understand? (Or understand now but later disagree if you didn't understand before) – BCLC May 26 '18 at 19:10

6 Answers6

21

We have $$\log\left(1+\frac{1}{n}\right)=\log\left(n+1\right)-\log\left(n\right)=\int_{n}^{n+1}\frac{1}{t}dt $$ then by first mean value theorem for integration we have that exists some $c_{n}\in\left[n,n+1\right] $ such that $$=\frac{1}{c_{n}}\left(n+1-n\right)=\frac{1}{c_{n}}\approx\frac{1}{n}. $$

Marco Cantarini
  • 33,062
  • 2
  • 47
  • 93
  • 2
    using integral test , wonderful proof – linux mint Jul 14 '15 at 06:22
  • In what sense do you mean ≈? How do you know OP means the same thing as you? – BCLC Jul 14 '15 at 17:26
  • @BCLC Normally $\approx$ means approximately. I don't know if you use it with another sense, but I always see it with this meaning. However I think it's quite obvious what I mean in this answer, and in fact OP doesn't ask more clarifications. – Marco Cantarini Jul 14 '15 at 17:44
  • Why the downvote? What's the problem? – Marco Cantarini Jul 15 '15 at 06:37
  • @MarcoCantarini When do you say two numbers are approximately equal? http://math.stackexchange.com/questions/1360393/how-to-prove-this-approximation-for-a-logarithm/1360440?noredirect=1#comment2768067_1360440 – BCLC Jul 16 '15 at 19:58
  • @BCLC I really don't understand your”battle”. If your goal is to show that what OP asked is not precise, well, you're right. But I think that also a mathematician can use a “common sense” to understand a problem. And note that you can find a lot of question in this site with $\approx $. So I suggest you to do the same question in every single thread. By the way, what I intend in this situation is that we have an error $$\left|\log\left(1+\frac{1}{n}\right)-\frac{1}{n}\right|=\left|\frac{1}{c_{n}}-\frac{1}{n}\right|\leq\frac{1}{n\left(n+1\right)}\leq\frac{1}{n^{2}}$$ hence, smaller than $1/n$. – Marco Cantarini Jul 17 '15 at 09:59
  • Marco, maybe I'll flag instead. >:P Also this question was put on hold...due to imprecision? – BCLC Jul 17 '15 at 10:30
  • @BCLC Yes and because the OP didn't write his attempts to prove that. – Marco Cantarini Jul 18 '15 at 07:26
  • @MarcoCantarini How do you know it is due to that? :P – BCLC Jul 22 '15 at 11:35
  • @BCLC "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." And this is my last comment. – Marco Cantarini Jul 22 '15 at 12:39
15

The Taylor series for the function $\log(1+x)$ about the point $x=0$ is $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$ so when $|x|\ll 1$ we have $\log(1+x)\approx x$. See more at the Wikipedia article on Taylor's theorem.

Zev Chonoles
  • 129,973
  • In what sense do you mean $\approx$? How do you know OP means the same thing as you? – BCLC Jul 14 '15 at 16:08
10

You may apply the fact that, as $x \to a$, for any differentiable function around $a$, we have

$$ \frac{f(x)-f(a)}{x-a} \to f'(a). $$

Then take $f(x)=\log (1+x)$, with $f'(x)=\dfrac1{1+x}$, giving as $x \to 0$, $$ \frac{\log(1+x)-\log(1+0)}{x-0} \to \frac1{1+0}=1 $$ $$ \frac{\log(1+x)}x \to 1, $$ that is, as $n \to \infty$,

$$ \log\left(1+\frac1n\right) \sim \frac1n. $$

Olivier Oloa
  • 120,989
5

By definition,

$$e^{\ln(1+1/n)}=1+\frac1n.$$

Then, using the crude approximation* $e^x\approx1+x$,

$$1+\ln\left(1+\frac1n\right)\approx1+\frac1n.$$

*By the definition of $e$ and the binomial theorem,

$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n=\lim_{n\to\infty}1+x+\frac{n-1}{2n}x^2+\frac{(n-1)(n-2)}{3!n^2}x^3\cdots\approx1+x.$$

Using the next term will yield

$$1+\ln\left(1+\frac1n\right)+\frac12\ln^2\left(1+\frac1n\right)=1+\frac1n,$$ which can be solved for the logarithm.

$$\left(\ln\left(1+\frac1n\right)+1\right)^2=1+\frac2n.$$

A pretty contrived method.

  • In what sense do you mean $\approx$? How do you know OP means the same thing as you? – BCLC Jul 14 '15 at 16:06
  • 3
    @BCLC - How do you know we both see the same color red? – Nathaniel Bubis Jul 14 '15 at 16:29
  • @nbubis We don't, conspiracy keanu :)) To be precise, what I mean to ask is: maybe OP meant $LHS \approx RHS$ as $|LHS - RHS| < \epsilon = 0.5$ while Yves Daoust meant $|LHS - RHS| < \epsilon = 0.0000000000001$? Furthermore, what is the domain? I think Yves Daoust's $e^x \approx 1 + x$ is for values of $x \in (-1,1)$. Consider $e^5$. Unless of course Yves Daoust means something different than me by '$\approx$'. – BCLC Jul 14 '15 at 16:32
  • 3
    @BCLC the notion $\approx$ is perfectly standard for anyone who has ever ventured outside the realm of pure mathematics. – Nathaniel Bubis Jul 14 '15 at 16:47
  • 1
    @nbubis Enlighten me, please? – BCLC Jul 14 '15 at 17:26
  • 3
    I wonder if I could use that crude approximation while building a bridge. It would save time. – Christopher King Jul 14 '15 at 17:41
  • 2
    The approximation $e^x\approx1+x$ clearly doesn't hold. One is linear the other is exponential. – kasperd Jul 14 '15 at 18:47
  • 1
    Guys, the context of the question is $x=1/n$, where the exponential behavior has not developed and the goal is precisely to explain the linear approximation! –  Jul 14 '15 at 21:13
  • 1
    @BCLC Generally, $f(x)\approx g(x)$ means that the difference is little o of something as $x$ approaches a limit. Here, the something is "the smallest order term", and the limit is 0. So $e^x\approx 1+x$ means that $e^x-1-x\in o(x)$ as $x\to 0$ and $\log(1+\frac{1}{n})\approx\frac{1}{n}$ means that $\log(1+\frac{1}{n})-\frac{1}{n}\in o(n^{-1})$ as $n\to\infty$. – Teepeemm Jul 15 '15 at 02:17
  • 1
    An alternative is $f(x)/g(x)\to1$ as $x$ approaches its limit. The usage should usually be specified by the author, but often isn't. But it doesn't mean $|f(x)-g(x)|<\epsilon$ for a specific $\epsilon$, unless you're looking at an $\epsilon(x)$ that approaches 0. – Teepeemm Jul 15 '15 at 02:24
  • @Teepeemm Are those interpretations of $\approx$ two equivalent (the one involving little o notation (only knew big O before) and the one involving asymptotic or whatever you call it)? – BCLC Jul 15 '15 at 11:19
  • 1
    They aren't equivalent (for example, $e^x\approx 1+5x$ satisfies the second but not the first). In the first definition, you could replace $o(x)$ with $o(1)$ to get an equivalent (but weaker) definition. – Teepeemm Jul 15 '15 at 14:02
  • @Teepeemm So you disagree with nbubis ('perfectly standard') ? – BCLC Jul 16 '15 at 19:55
  • @PyRulez How about finance? – BCLC Jul 16 '15 at 20:00
  • The short answer is I up-voted his comment, so yes. The longer answer is that I think it’s standard enough that the meaning could be inferred from context (especially if that context involves Taylor series). If the context isn’t Taylor series (for example, yours of $\pi(x)\sim x/\log x$), then it may require a definition (but notice that we have a different symbol). I would also classify myself a pure mathematician, so his comment is vacuously true in my case. – Teepeemm Jul 16 '15 at 21:58
  • @Teepeemm: this isn't the place to chat. –  Jul 17 '15 at 06:41
  • @Teepeemm I don't understand. You disagree yet upvoted? – BCLC Jul 17 '15 at 10:31
  • @bclc: nor for you. –  Jul 17 '15 at 10:32
4

Use the inequality $e^x\ge x+1$. Substitute in $x=\ln(1+\frac1n)$ and $x=-\ln(1+\frac1n)$ to obtain an upper and lower bound for $\ln(1+\frac1n)$.

Akiva Weinberger
  • 23,062
2

We have $\ln(x)=\lim_{m\to\infty}m(\sqrt[m]x-1)$.

By the generalized binomial theorem,

$$\lim_{m\to\infty}m\left(\sqrt[m]{1+\frac1n}-1\right)=\\ \lim_{m\to\infty}m\left(\frac1m\frac1n-\frac{m-1}{2m^2}\frac1{n^2}+\frac{(m-1)(2m-1)}{3!m^3}\frac1{n^3}-\frac{(m-1)(2m-1)(3m-1)}{4!m^4}\frac1{n^4}\cdots\right)\\ =\frac1n-\frac1{2n^2}+\frac1{3n^3}-\frac1{4n^4}\cdots$$