What are rational approximations to $e$ and Euler-Mascheroni constant $\gamma$ ?

(Like $ 22/7 , 355/113, ...$ are for $\pi.$)


(Number theory is not my cup of tea..). I wish to be able to learn if for every irrational number there exists an infinite set of progressively reducing error rational numbers as approximation and a procedure by which it is obtained.


Are continued fraction the fastest or the only method to obtain rational successive approximations to irrationals?

  • $\begingroup$ An answer to the title of your question can be found here. $\endgroup$ – barak manos Sep 12 '16 at 11:10
  • $\begingroup$ With regards to your bottom-line question, I believe that the answer is subjective, i.e., depends on the irrational value that you want to approximate. For example, Newton-Raphson method for integral roots converges pretty quickly. $\endgroup$ – barak manos Sep 12 '16 at 11:15
  • $\begingroup$ Depends on what measure? Is there in mathematics a * degree* of irrationality defined for irrational numbers? $\endgroup$ – Narasimham Sep 12 '16 at 11:25
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    $\begingroup$ :) so the woods are deep. Thank you sir, for above patient comments, $\endgroup$ – Narasimham Sep 12 '16 at 11:56
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    $\begingroup$ @Théophile: That's very interesting. I was not aware of this. Thank you. $\endgroup$ – barak manos Nov 25 '16 at 4:39

You can calculate continued fraction representations, where you can choose to terminate after you've reached some level of accuracy:


Here are some of the first results:

$$e\approx\frac{19}{7}, \frac{87}{32}, \frac{106}{39 },\dots$$

Another (beautiful, but non-simple) continued fraction representation of $e$ is $$e=2+\frac{1}{1+\frac{1}{2+\frac{2}{3+\frac{3}{3+\cdots}}}}.$$

Same for $\gamma$:

$$\gamma=[0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40,\dots],$$

where the first few (useful) convergents are $$\gamma \approx \frac{11}{19}, \frac{15}{26},\frac{71}{123},\dots$$

Edit as a response to OPs edit:

From the Wikipedia-page: every irrational number $\alpha$ is the value of a unique infinite continued fraction, whose coefficients can be found using the non-terminating version of the Euclidean algorithm applied to the incommensurable values $\alpha$ and 1.

  • $\begingroup$ OK,thanks. Is it the * fastest * with respect to successive approximation number?Because the target value ( of convergence) is known? $\endgroup$ – Narasimham Sep 12 '16 at 10:46
  • $\begingroup$ @Narasimham I'm not sure that measure is well-defined, as the steps in the algorithm are somewhat arbitrarily chosen (through convention)... for instance, I could just define that one step covered, say, two steps in the continued fraction, and then my algorithm would converge faster (even though it is doing exactly the same).. But I'm not sure. However the continued fraction-approach is surely faster converging than a series representation. $\endgroup$ – Bobson Dugnutt Sep 12 '16 at 10:56

For $e$, the first few are $2, 3, 8/3, 11/4, 19/7, 87/32, 106/39$. You get these by grinding out the continued fraction for $e$ and then computing the convergents. See http://mathworld.wolfram.com/eContinuedFraction.html.

Also see http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfCALC.html where you can input a decimal approximation for the Euler constant and get the continued fraction and the convergents.


Euler's approximation (with function in link): 19/7, 87/32, 106/39, etc... http://www.johndcook.com/blog/2013/01/30/rational-approximations-to-e/

Euler-Mascheroni constants can be found here with references to the original findings: http://mathworld.wolfram.com/Euler-MascheroniConstantApproximations.html


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