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GROUPS:

# Formula for computing sqrt(2) of binary numbers

Posted 8 years ago
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Posted 9 months ago
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Posted 9 months ago
 Hello, here is one I have found recently: The n'th binary digit of sqrt(2) is given by floor(Sum_{n>=1} A005875(n)/exp(Pi*n/(2^((2/3)*k+(1/3))))) mod 2 Will give the k-th binary digit of sqrt(2), where A005875 is the number of ways to write n as sum of 3 squares. See image of the formula : r_3(n) is A005875.NOTE : there is one for sqrt(10) also which is NOTE 2: The series converge quite slowly but there is a marker for the decimal point which enables to get the n'th. The actual precision is very high when k >> 1.
Posted 3 years ago
 I found this very interesting so I tried to find a similar formula for the n-th digit of $\sqrt2$ in decimal. For that I used the following, already existing formula for the decimal expansion of $\sqrt2$ (from http://oeis.org/A002193): $a(n)=-10 \cdot \lfloor 2^{-\frac{3}{2}+n} \cdot 5^{-2+n}\rfloor + \lfloor 2^{-\frac{1}{2}+n} \cdot 5^{-1+n} \rfloor$,applying this on it: $\lfloor x\rfloor =x-\frac{1}{2}+\frac{\sum _{k=1}^{\infty } \frac{\sin (2 k \pi x)}{k}}{\pi }$and then following the same steps you did for base 2. This way I got the following formula: $B_{10}(n) = \frac{9}{2}-\frac{10 \tan ^{-1}\left(\cot \left(\sqrt{2} \pi10^{n-2} \right)\right)}{\pi }+\frac{\tan ^{-1}\left(\cot \left(\sqrt{2} \pi 10^{n-1} \right)\right)}{\pi }$, which was working as expected. At this point I had a suspicion for a generalized formula for the n-th digit in any base $b \geq 2$ : $B_{b}(n) = \frac{b-1}{2}-\frac{b \tan ^{-1}\left(\cot \left(\sqrt{2} \pi \cdot b^{n-2} \right)\right)}{\pi }+\frac{\tan ^{-1}\left(\cot \left(\sqrt{2} \pi \cdot b^{n-1} \right)\right)}{\pi }$which is the formula you found for $b=2$ and the one I found for $b=10$ and I also tried $b=16$ for Hexadecimal and it returned correct values. But at this point I had another suspicion, namely that you could generalize it so that you get the b-th base expansion for any positive real number $0 < u \in \mathbb{R}$: $B_{b}(n) = \frac{b-1}{2}-\frac{b \tan ^{-1}\left(\cot \left( u \cdot \pi \cdot b^{n-2} \right)\right)}{\pi }+\frac{\tan ^{-1}\left(\cot \left(u \cdot \pi \cdot b^{n-1} \right)\right)}{\pi }$, and indeed, this formula could return for example the binary and decimal expansions of $u=\pi$ or $u=\frac{1}{9}$. I am not sure how useful these formulae would be though because you need the value of $u$ to calculate the n-th digit with it.
Posted 9 months ago
 Try, where u is Pi: F[n_, b_, u_] := Floor[b^(-1 + n) *u] - b*Floor[b^(-2 + n) *u] ; {Table[F[n, 10, ((2^(2 + 4 n) (1 - 3^(-1 - 4 n)) (4 n)!)/ EulerE[4 n])^(1/(1 + 4 n))], {n, 1, 50}], mm = N[Pi, 50]; RealDigits[mm, 10][[1]]} // MatrixForm (*Pi for base = 10 *) {Table[F[n, 2, ((2^(2 + 4 n) (1 - 3^(-1 - 4 n)) (4 n)!)/ EulerE[4 n])^(1/(1 + 4 n))], {n, 0, 55}], mm = N[Pi, 17]; RealDigits[mm, 2][[1]]} // MatrixForm (*Pi for base = 2 *) Formula the n-th digit off Pi for base=b: $B_b(n)=\left\lfloor b^{-1+n} \left(\frac{2^{2+4 n} \left(1-3^{-1-4 n}\right) (4 n)!}{E_{4 n}}\right){}^{\frac{1}{1+4 n}}\right\rfloor -b \left\lfloor b^{-2+n} \left(\frac{2^{2+4 n} \left(1-3^{-1-4 n}\right) (4 n)!}{E_{4 n}}\right){}^{\frac{1}{1+4 n}}\right\rfloor$Where: EulerE[4 n] are Euler numbers.  f[n_] := I Sum[Binomial[k, l] ((-1)^l*(k - 2 l)^(4 n + 1))/(2^k*I^k*k), {k, 1,4 n + 1}, {l, 0, k}] Table[{EulerE[4*n], f[n]}, {n, 1, 5}] // MatrixForm Regards M.I
Posted 5 years ago
 I am interested in proving that the proportion of binary digits of $\sqrt{2}$ is exactly $\frac{1}{2}$, and I have worked on this for a long time. You wrote a great article using the Fourier series expansion for the fractional part function. I am looking at a simpler formula (I have yet to find one!) but have asked my question on StackExchange, here. For your convenience, my question is posted below:Is it possible to find an expression of the form $$\lfloor 2^n \sqrt{2}\rfloor = \sum_{k=1}^r (\alpha_k + \beta_ki)\cdot\Big(a_k+b_k i\Big)^n,$$ where $i^2 = -1$ and $\alpha_k, \beta_k, a_k, b_k$ real numbers? Maybe with $r< 5$ if possible. I am trying to compute the mean value (average) of $\{ 2^n \sqrt{2}\}= 2^n \sqrt{2} - \lfloor 2^n \sqrt{2}\rfloor$ over $n=0, 1,2, \cdots$. Let us denote as $p$ this mean value: $p$ is the proportion of binary digits of $\sqrt{2}$ equal to one, and $p$ is widely believed to be equal to $\frac{1}{2}$. Note that $$p=\lim_{n\rightarrow\infty} p_n, \mbox{ with } p_n = \frac{1}{n}\sum_{k=1}^n \Big( 2^k \sqrt{2} - \lfloor 2^k \sqrt{2}\rfloor\Big).$$ Assuming we have a simple closed form for $\lfloor 2^k \sqrt{2}\rfloor$ as suggested in the first formula, then we can have a closed form for $p_n$, involving a finite number of terms. I don't expect this to be true if you replace $\sqrt{2}$ by (say) $\pi$ or $\log 2$, but I would expect this to work with any quadratic irrational.Of course you can always use the Fourrier series to represent the fractional part function, but I don't think it would be easy to handle: it involves an infinite number of terms ( $r=\infty$.)
Posted 8 years ago
 Finding formula for the n-th digit in the binary representation of Sqrt[2] using Floor function.  sol = RSolve[{b[n + 2] == a[n + 1] - 2*a[n], a[n] == Floor[Sqrt[2]*2^n], b[0] == 1, b[1] == 0}, {b[n], a[n]}, n] sol2 = FullSimplify[b[n] /. sol[[1]], Assumptions -> {n > 0, n \[Element] Integers}]  $$B_n=-2 \left\lfloor 2^{-\frac{3}{2}+n}\right\rfloor +\left\lfloor 2^{-\frac{1}{2}+n}\right\rfloor$$This is formula for the n-th digit in the binary representation of $\sqrt{2}$ for $n > 0$ and $n\in \mathbb{Z}$ using Floor function.  Table[sol22 /. n -> m, {m, 1, 50}]  ${1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, \ 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, \ 1, 0, 0, 1}$
Posted 8 years ago
 Just checking: is this, then, some kind of spigot algorithm for sqrt(2)?https://en.wikipedia.org/wiki/Spigot_algorithm
Posted 8 years ago
 No is not algorithm, it's better than that. It simple, it's a formula, or function.An example:To start calculating binary digits from, say, the 8th place only need substitute for n=8,9,10,11,12,13,14,15....That it's all. formula =1/2 - (2 ArcTan[Cot[2^(-(3/2) + n) \[Pi]]])/\[Pi] + ArcTan[Cot[2^(-(1/2) + n) \[Pi]]]/\[Pi] sol = Table[formula, {n, 8, 15}] // N // Round  ${1, 0, 0, 0, 0, 0, 1, 0}$and say from n=1....15:  sol2 = Table[formula, {n, 1, 15}] // N // Round  ${1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0}$
Posted 8 years ago
 I think it is then a spigot algorithm, because you have extensions, like the famous one for Pi in base 16 which calculates the nth digit directly... Algorithm here can be read as function:> A further refinement is an algorithm which can compute a single arbitrary digit, without first computing the preceding digits: an example is the Bailey-Borwein-Plouffe formula, a digit extraction algorithm for ? which produces hexadecimal digits.https://en.wikipedia.org/wiki/Spigot_algorithm
Posted 8 years ago
 You're right.So let it be.Have a nice day. :)Mariusz.
Posted 8 years ago
 Appendix: To simplify $B_n$ formula.We have: $$B_n=\frac{i \left(-i \pi +2 \log \left(1-e^{-i 2^{-\frac{1}{2}+n} \pi }\right)-2 \log \left(1-e^{i 2^{-\frac{1}{2}+n} \pi }\right)-\log \left(1-e^{-i 2^{\frac{1}{2}+n} \pi }\right)+\log \left(1-e^{i 2^{\frac{1}{2}+n} \pi }\right)\right)}{2 \pi }$$  sol2 = (I (-I \[Pi] + 2 Log[1 - E^(-I 2^(-(1/2) + n) \[Pi])] - 2 Log[1 - E^(I 2^(-(1/2) + n) \[Pi])] - Log[1 - E^(-I 2^(1/2 + n) \[Pi])] + Log[1 - E^(I 2^(1/2 + n) \[Pi])]))/(2 \[Pi]) sol5 = FullSimplify[Re[sol2] // ComplexExpand, Assumptions -> n \[Element] Integers] Table[sol5, {n, 1, 5}]  $\left\{\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(\sqrt{2} \pi \right)}{1-\cos \left(\sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(2 \sqrt{2} \pi \right)}{1-\cos \left(2 \sqrt{2} \pi \right)}\right)}{2 \pi },\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(2 \sqrt{2} \pi \right)}{1-\cos \left(2 \sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(4 \sqrt{2} \pi \right)}{1-\cos \left(4 \sqrt{2} \pi \right)}\right)}{2 \pi },\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(4 \sqrt{2} \pi \right)}{1-\cos \left(4 \sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(8 \sqrt{2} \pi \right)}{1-\cos \left(8 \sqrt{2} \pi \right)}\right)}{2 \pi },\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(8 \sqrt{2} \pi \right)}{1-\cos \left(8 \sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(16 \sqrt{2} \pi \right)}{1-\cos \left(16 \sqrt{2} \pi \right)}\right)}{2 \pi },\frac{\pi -4 \tan ^{-1}\left(\frac{\sin \left(16 \sqrt{2} \pi \right)}{1-\cos \left(16 \sqrt{2} \pi \right)}\right)+2 \tan ^{-1}\left(\frac{\sin \left(32 \sqrt{2} \pi \right)}{1-\cos \left(32 \sqrt{2} \pi \right)}\right)}{2 \pi }\right\}$Now we need find a pattern.The first ingredient in the formula: We have: $$\left\{4 \tan ^{-1}\left(\frac{\sin \left(\sqrt{2} \pi \right)}{1-\cos \left(\sqrt{2} \pi \right)}\right),4 \tan ^{-1}\left(\frac{\sin \left(2 \sqrt{2} \pi \right)}{1-\cos \left(2 \sqrt{2} \pi \right)}\right),4 \tan ^{-1}\left(\frac{\sin \left(4 \sqrt{2} \pi \right)}{1-\cos \left(4 \sqrt{2} \pi \right)}\right),4 \tan ^{-1}\left(\frac{\sin \left(8 \sqrt{2} \pi \right)}{1-\cos \left(8 \sqrt{2} \pi \right)}\right),\tan ^{-1}\left(\frac{\sin \left(16 \sqrt{2} \pi \right)}{1-\cos \left(16 \sqrt{2} \pi \right)}\right)\right\}$$  FindSequenceFunction[{1, 2, 4, 8, 16}, n] (*2^(-1 + n)*) Finding the second ingredient in the formula is the same how I found at first:  FindSequenceFunction[{2, 4, 8, 16, 32}, n] (*2^n*) Substitution: $$\frac{\pi -\left(4 \tan ^{-1}\left(\frac{\sin \left(n \sqrt{2} \pi \right)}{1-\cos \left(n \sqrt{2} \pi \right)}\right)\text{/.}\, n\to 2^{n-1}\right)+\left(2 \tan ^{-1}\left(\frac{\sin \left(n \sqrt{2} \pi \right)}{1-\cos \left(n \sqrt{2} \pi \right)}\right)\text{/.}\, n\to 2^n\right)}{2 \pi }$$  FullSimplify[(\[Pi] - (4 ArcTan[Sin[n Sqrt[2] \[Pi]]/(1 - Cos[n Sqrt[2] \[Pi]])] /. n -> 2^(n - 1)) + (2 ArcTan[Sin[n Sqrt[2] \[Pi]]/( 1 - Cos[n Sqrt[2] \[Pi]])] /. n -> 2^n))/(2 \[Pi])] // Expand  $$B_n = \frac{1}{2}-\frac{2 \tan ^{-1}\left(\cot \left(2^{-\frac{3}{2}+n} \pi \right)\right)}{\pi }+\frac{\tan ^{-1}\left(\cot \left(2^{-\frac{1}{2}+n} \pi \right)\right)}{\pi }$$
Posted 8 years ago
 Neat! Is it your result and is it new? Do you think expression can be simplified? This does not work $$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$but maybe similar trigonometry.
Posted 8 years ago
 Yes, I found this formula.Is it new? Probably so. I searched in the web, but did not find a similar formula.I tried to simplify, but I have not yet succeeded.