Wikipedia:Reference desk/Archives/Mathematics/2024 January 20

= January 20 =

Sumudu transform v Laplace transform?
Hi all,

Back in January 2015 "Sumudu transform" was deleted via an uncontested WP:PROD as "merely the Laplace transform with a change of variables"

The article has been recreated as "Sumudu transform" and "Sumudu Transform".

I would appreciate your assistance with this Shirt58 (talk) 🦘 11:06, 20 January 2024 (UTC)

Here is the link: Sumudu transform. Michael Hardy (talk) 18:58, 21 January 2024 (UTC)


 * I have no opinion on this, I don't know anything about the subject. Suntooooth, it/he (talk/contribs) 11:13, 20 January 2024 (UTC)


 * I'm pretty sure this discussion should be on WT:WPM. It sounds, without researching the subject, like it would have been better to add Sumudu transform as a new section in the Laplace transform article. Inclusion in another article is usually the best compromise for topics with questionable notability. --RDBury (talk) 13:48, 20 January 2024 (UTC)


 * PS. The article isn't really readable, so perhaps a PROD is in order for the time being. That doesn't imply that the subject is not notable though. From this Hindawi article, the Sumudo transform is defined by
 * $$F(u) = \int_0^\infty f(ut)e^{-t} \, dt.$$
 * Meanwhile the Laplace transform is defined by
 * $$F(s) = \int_0^\infty f(t)e^{-st} \, dt.$$
 * So it certainly seems like one can be obtained from the other by a change of variable. (I make no judgements as to whether Hindawi is considered a reliable source.) The closest thing to a secondary source I saw in my Google search was this JETIR article. (Again, I make no judgements as to whether jetir.org is a reliable source.). --RDBury (talk) 14:30, 20 January 2024 (UTC)
 * For reference if the issue gets discussed at WT:WPM:
 * Deakin, Michael A. B. (1997). The “Sumudu transform” and the Laplace transform. International Journal of Mathematical Education in Science and Technology, 28(1), 159–160.
 * While it appears notable, it seems not to deserve more attention than a note in our article Laplace–Carson transform. The article under discussion does not explain the concept in a reasonable way and contains, even in its current stubby form, multiple errors. --Lambiam 14:49, 20 January 2024 (UTC)
 * I'm having a hard time seeing why Laplace–Carson is notable, seeing as it's the Laplace transform with an extra variable and a factor thrown in. Maybe the Laplace transform article should have a section called "Variations" where these tweaks can be listed. --RDBury (talk) 16:24, 20 January 2024 (UTC)

At this page I've linked to this present discussion. That page is the proper place for it. Michael Hardy (talk) 18:57, 21 January 2024 (UTC)

Approximating with Pythagorean triples
Can any positive real number be approximated by a fraction p/q as close as desired, with p,q,r being a Pythagorean triple? Thank you. Hevesli (talk) 21:30, 20 January 2024 (UTC)


 * Yes. This StackExchange post shows that you can reduce the problem to finding approximations by arbitrary fractions. GalacticShoe (talk) 21:47, 20 January 2024 (UTC)
 * Notice that the inverse stereographic projection of the rational number line onto the unit circle gives the set of all rational coordinates of form $$(p/r, q/r)$$ on the circle. So one way you can find an approximation is to take your real number slope $$x$$ and project it onto the circle $\bigl(1/\!\sqrt{1 + x^2}, x/\!\sqrt{1 + x^2}\bigr),$ then take the stereographic projection (half-tangent) $x \big/ \bigl(1 + \sqrt{1 + x^2}\bigr).$  Approximate this as closely as you like by a rational number, then take the inverse stereographic projection back onto the circle and you will have rational coordinates which you can multiply by the denominator to recover the legs of an integral Pythagorean triangle. –jacobolus (t) 22:10, 20 January 2024 (UTC)
 * Thank you both. Using the second method I computed the triple (394303317,125510644,413797085) to approximate pi. I didn't quite understand how it works but I used the formulas. Maybe Stack exchange is the same method? I don't know. Hevesli (talk) 00:12, 21 January 2024 (UTC)
 * The example you got is pretty good. The continued fraction for $\pi \big/ \bigl(1 + \sqrt{1 + \pi^2}\bigr)$ is [0; 1, 2, 1, 2, 1, 1, 3, 1, 1, 5, 1, 7, 1, 1, 23, 2, 2, 4, 3, 1, 11, 158, 1, 1, 1, 1, 4, 2, 1, ...], and your example triple is the result of cutting this just before the 23 to get a pretty close rational approximation 16979/23223, then taking the inverse stereographic projection, yielding the triple $$23223\cdot16979,\ $$$$ \tfrac12(23223^2 - 16979^2),\ $$$$\tfrac12(23223^2 + 16979^2)$$ or 394303317, 125510644, 413797085. –jacobolus (t) 01:26, 21 January 2024 (UTC)
 * A marginally simpler procedure is to find a rational approximation $$s/t$$ of $$x+\sqrt{x^2+1}.$$ Then $$(p,q,r)=(s^2-t^2,2st,s^2+t^2)$$ is a Pythagorean triple for which $$p/q$$ approximates $$x.$$ The continued fraction for $$\pi+\sqrt{\pi^2+1}$$ is $$[6,2,3,1,1,3,2,1,16,47,...].$$ The approximant cutting this of just before the $$47$$ is $$20101/3122,$$ giving again the Pythagorean triple $$(20101^2-3122^2,2\cdot20101\cdot3122,20101^2+3122^2)=$$ $$(394303317,125510644,413797085).$$ --Lambiam 09:06, 21 January 2024 (UTC)
 * It is a bit of a coincidence that both procedures share the approximation $$394303317/125510644$$ of $$\pi.$$ The following table has four columns. The second column shows shared approximations, while the first and third show those approximations that appear only with one of the procedures. The fourth column is the errror, rounded to 15 decimals.
 *         0/1                            -3.141592653589793
 * 12/5                           -0.741592653589793
 * 24/7                           +0.286978774981635
 * 35/12       -0.224925986923127
 * 176/57                          -0.053873355344179
 * 165/52               +0.031484269487130
 * 988/315              -0.005084717081857
 * 3283/1044     +0.003043361735878
 * 12648/4025                        +0.000767594857412
 * 10353/3296             -0.000512556502414
 * 65720/20919            +0.000048916274923
 * 683669/217620   -0.000020187823779
 * 2032596/646997                      -0.000007762160621
 * 1414647/450296           +0.000000875999632
 * 174241192/55462695                    -0.000000056980487
 * 221476772/70498245                    +0.000000048469670
 * 394303317/125510644        -0.000000001097308
 * 436618757080/138980066871     +0.000000000019055
 * --Lambiam 20:45, 21 January 2024 (UTC)