Undergrad Is there a name for this approximation?

[hilbert2]

Because it holds that

$\displaystyle\int_{1}^{x}\frac{dt}{t} = \log x$, and

$\displaystyle\int_{1}^{x}\frac{dt}{t^a} = \frac{1}{a-1}\left(1-\frac{1}{x^{a-1}}\right)\hspace{20pt}$when $a>1$

it could be expected that

$\displaystyle\frac{1}{a-1}\left(1-\frac{1}{x^{a-1}}\right) \rightarrow \log x$ when $a\rightarrow 1$.

Trying this with Wolfram Alpha for $a = 1 + 10^{-8}$ or similar, it seems that the graphs of the natural logarithm and the other result overlap quite well.

I just realized this a while ago, and have never seen this approximation before... Is there any name for it?

 

[fresh_42]

Because it holds that

$\displaystyle\int_{1}^{x}\frac{dt}{t} = \log x$, and

$\displaystyle\int_{1}^{x}\frac{dt}{t^a} = \frac{1}{a-1}\left(1-\frac{1}{x^{a-1}}\right)\hspace{20pt}$when $a>1$

it could be expected that

$\displaystyle\frac{1}{a-1}\left(1-\frac{1}{x^{a-1}}\right) \rightarrow \log x$ when $a\rightarrow 1$.

Trying this with Wolfram Alpha for $a = 1 + 10^{-8}$ or similar, it seems that the graphs of the natural logarithm and the other result overlap quite well.

I just realized this a while ago, and have never seen this approximation before... Is there any name for it?

I doubt it has a name. The (German) Wikipedia page (https://de.wikipedia.org/wiki/Logarithmus#Als_Potenzreihe) has it (with some minor adjustments) as a direct consequence of the power series expansion of $\log (1+x)$.

 

[Matt Benesi]

I doubt it has a name. The (German) Wikipedia page (https://de.wikipedia.org/wiki/Logarithmus#Als_Potenzreihe) has it (with some minor adjustments) as a direct consequence of the power series expansion of $\log (1+x)$.

I never understood why they do the 1+x power series instead of the log(x/y) power series.

 

[hilbert2]

Thanks. I was quite surprised that there is a short approximative formula with only algebraic numbers in it.

 

[mfb]

It is not a nice formula, numerically. You get a huge term multiplied by a small term, where the small term is the nearly perfect cancellation of two numbers, and you have to evaluate x^a-1 with a high precision to get a reasonable approximation.

 

[hilbert2]

It is not a nice formula, numerically. You get a huge term multiplied by a small term, where the small term is the nearly perfect cancellation of two numbers, and you have to evaluate x^a-1 with a high precision to get a reasonable approximation.

Yes, I noticed that too.

Taking the inverse function it's quite obvious that $\displaystyle e^x = \lim_{n\to\infty}\left(\frac{n}{n-x}\right)^n$, and the approximation with large but finite $n$ blows up to infinity when $x\to n$.

 

[Svein]

Another fact: \int_{1}^{x}\frac{1}{t}dt=\ln(x). Doing a Riemann approximation of the integral: \sum_{n=2}^{N}\frac{1}{n}<\int_{1}^{N}\frac{1}{t}dt<\sum_{n=1}^{N-1}\frac{1}{n}. It can be shown that \sum_{n=1}^{N}\frac{1}{n}\approx \ln(N)+\gamma for large N, where γ is the Euler-Mascheroni constant (0.5772156649...).