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July 10

Definition of proof

How does Gödel defined a proof in his version of Intermediate logic? 2A02:8071:60A0:92E0:78B6:4D3A:774B:E50C (talk) 18:34, 10 July 2024 (UTC)

Can you give us a pointer to a source defining "Gödel's version" of intermediate logic? Also, is there evidence that Gödel defined the notion of proof for this logic?  --Lambiam 18:46, 10 July 2024 (UTC)

July 13

Given the results from powers of tau in the trusted setup ceremony ; the verifying and the proving key, how can I find the point [f] resulting from the trusted setup in Groth16 ?

Moved to the Computing section of the Reference desk —  --Lambiam 13:36, 16 July 2024 (UTC)

July 15

Next Julian period

Moved to the Science section of the Reference desk —  --Lambiam 13:24, 16 July 2024 (UTC).

July 16

In SageMath, how to use GF() on a very large finite field ?

Moved to the Computing section of the Reference desk —  --Lambiam 13:37, 16 July 2024 (UTC)

July 18

Functions whose every derivative is positive growing slower than exponential

Is there any smooth function with the following two properties:

$f^{(n)}(x)>0$, i.e. the nth derivative of f is strictly positive for every x and n.

$\lim _{x\rightarrow \infty }{\frac {f(x)}{b^{x}}}=0$ for every b > 1. The hard case is when b is small.

Functions like $a^{x}$ (for a > 1) are the only ones I can think of with the first property, but none of them has the second property because you can always choose b < a. So I am asking whether there is any function with the first property that grows slower than exponential.

120.21.218.123 (talk) 10:09, 18 July 2024 (UTC)

Wouldn't any power series with positive coefficients that decrease compared to the coefficients of the exponential do? The exponential is $1+\sum _{k=1}^{\infty }{\frac {1}{k!}}x^{k}$, so e.g. $f(x)=1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}$ should do the trick. The next question is whether you can find a closed-form expression for this or a similar power series. --Wrongfilter (talk) 13:02, 18 July 2024 (UTC)

Good thinking. It is of course the case that the first property holds for any power series where all coefficients are positive. Plotting on a graph, I think your specific example doesn't satisfy the second property, but others where the coefficients decrease more rapidly do. 120.21.218.123 (talk) 13:26, 18 July 2024 (UTC)

$1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}>1+\sum _{k=1}^{\infty }{\frac {1}{k{+}1}}{\frac {1}{k!}}x^{k}=$ $1+\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k}=$ $1+{\frac {1}{x}}\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k{+}1}=$ $1+{\frac {1}{x}}\sum _{k=2}^{\infty }{\frac {1}{k}}x^{k}={\frac {\exp x-1}{x}}=$ $\Omega (2^{x}).$  --Lambiam 13:45, 18 July 2024 (UTC)

A half-exponential function will satisfy your requirements. Hellmuth Kneser famously defined an analytic function that is the functional square root of the exponential function.  --Lambiam 14:04, 18 July 2024 (UTC)

References

A variant of Wrongfilter's idea that I think does work:

$f(x)=\sum _{k=0}^{\infty }{\frac {x^{k}}{(k!)^{2}}}$

(taking $0^{0}$ to be $1$).

Numerical evidence suggests that $\log f(x)\sim 2{\sqrt {x}}.$ One might therefore hope that $e^{2{\sqrt {x}}}$ would also work. However, its second derivative is negative for $x<{\tfrac {1}{4}}.$  --Lambiam 22:20, 20 July 2024 (UTC)

And some higher derivatives are negative for even larger values of x. The eighth derivative is negative for $x<6.17$, for instance. 120.21.79.62 (talk) 06:48, 21 July 2024 (UTC)

The fourteenth derivative is negative for $x<20$. That's as high as WolframAlpha will let me go. 120.21.79.62 (talk) 06:54, 21 July 2024 (UTC)

Yes, shifting the graph along the x-axis by using $e^{2{\sqrt {x+c}}}$ won't help.  --Lambiam 11:46, 21 July 2024 (UTC)

July 21

If the sum of the first m factorial numbers is equal to the sum of the first n positive integers

If the sum of the first m factorial numbers is equal to the sum of the first n positive integers, i.e. 1! + 2! + 3! + … + m! = 1 + 2 + 3 + … + n, then (m,n) = (0,0), (1,1), (2,2), (5,17), right? 220.132.216.52 (talk) 20:30, 21 July 2024 (UTC)

The triangular numbers, modulo 19, are reduced to one of 10 possibilities: 0, 1, 2, 3, 6, 7, 9, 10, 15 and 17. The sum 1! + 2! + 3! + ... + m!, for m > 17, modulo 19, is reduced to 8. Therefore no further factorial sums are triangular.  --Lambiam 22:19, 21 July 2024 (UTC)

I looked at mod 7 with about the same result. The left hand side is 5 for m≥6 and the right hand side can never be 5. So you only have to check m from 1 to 5. (Btw, I would count 0! = 1 as a factorial number, so the sums of factorials would be 1, 2, 4, 10, 34, ... . (sequence A003422 in the OEIS)) --RDBury (talk) 22:28, 21 July 2024 (UTC)

Wrote up a quick MATLAB script to find numbers which can be used as modulos to show that the list is finite, it starts: $7,13,14,17,19,21,25,26,28,34,35,38,\ldots$. Obviously if a number appears in the list then all its positive multiples do too. The list of nontrivial numbers starts $7,13,17,19,25,59,67,73,79,83,94,97,\ldots$ GalacticShoe (talk) 03:01, 22 July 2024 (UTC)

July 22

Computing the centre of a triangle...

(On math section as it's a geometric/trig problem essentially)

In CSS, color-mix() takes 2 Params.. However I have colors to mix that take 3 or more params.

Whilst with 2 params you can do a simple linear interpolation, based on the weights of the 2 params, I wasn't sure how it could be done for 3.

One approach I had considered was (at least for an RGB blend) is to compute the centrepoint of a triangle in 3D space, where the 3 points of the triangle are the three colors. However that would assume equal weights of each color, I figured

So for a given "triangle" defined by (r1,g1,b1),(r2,g2,g2), (r3,b3,g3) and a mix ratio of w1:w2:w3  compute the centroid(?) of the triangle representing the blended color. ?

Alternatively is there a different math/geometrical technique that is used in actual computer graphics work?
ShakespeareFan00 (talk) 17:18, 22 July 2024 (UTC)

I don't have an actual computer graphics answer, but the interpolation method still works for three points, simply take the weighted sum of points assuming $w_{1}+w_{2}+w_{3}=1$ (if not, then just define new values ${\frac {w_{1}}{w_{1}+w_{2}+w_{3}}},{\frac {w_{2}}{w_{1}+w_{2}+w_{3}}},{\frac {w_{3}}{w_{1}+w_{2}+w_{3}}}$ which do add to $1$.) In other words, you can just take $(w_{1}r_{1}+w_{2}r_{2}+w_{3}r_{3},w_{1}g_{1}+w_{2}g_{2}+w_{3}g_{3},w_{1}b_{1}+w_{2}b_{2}+w_{3}b_{3})$ (or, more concisely, $w_{1}(r_{1},g_{1},b_{1})+w_{2}(r_{2},g_{2},b_{2})+w_{3}(r_{3},g_{3},b_{3})$.) GalacticShoe (talk) 17:45, 22 July 2024 (UTC)

Thanks. I thought I was thinking along the right lines..
In case you are wondering why I asked -s:Page:The_color_printer_(1892).djvu/55 ShakespeareFan00 (talk) 17:52, 22 July 2024 (UTC)

The weighted average makes sense for additive colour mixing, but the colour resulting from pigment mixing is not so easily determined. For example, the colours   and   are complementary. Their sum in RGB colour models is   and their average is  , as grey as it gets. However, mixing red and green paint gives more of a brown colour.^[1] A colour model that is more likely close to that of The Color Printer is the RYB colour model. If the pigments are fully opaque, the subtractive model is adequate, but generally pigments are not fully opaque.  --Lambiam 21:22, 22 July 2024 (UTC)