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January 26

Coefficient of determination

https://www.tylervigen.com/spurious/correlation/2733_the-distance-between-jupiter-and-the-sun_correlates-with_the-number-of-secretaries-in-alaska says:

r² = 0.9017589 (Coefficient of determination)

This means 90.2% of the change in the one variable (i.e., The number of secretaries in Alaska) is predictable based on the change in the other (i.e., The distance between Jupiter and the Sun) over the 13 years from 2010 through 2022.

Forget that it is a fake paper, that's besides the point.

Is its judgment valid? Does r squared tell that?

Does it apply to meta-analysis, like in the following papers?

• https://www.tandfonline.com/doi/full/10.1080/10550887.2024.2401680

• https://academic.oup.com/hcr/article-pdf/43/3/315/22776525/jhumcom0315.pdf

Previously asked at Wikipedia:Reference desk/Science#Percentage of explained variation, but I didn't get much wiser.

Why do I ask? If the reported correlations getting squared show the amount of explained variance, that means the predictive validity of the tested theory is ridiculously weak. My hunch is that the theory is completely bunk. But I want to know this for sure. tgeorgescu (talk) 20:23, 26 January 2026 (UTC)

Here is the basic mathematics.
For simplicity I only consider the case of two variables. So suppose we have two numerically-valued quantities $X$ and $Y$, whose values vary with spacetime, and a data set containing a number of observations of the pair. For whatever reason, some data scientist views $X$ as driving $Y.$ (This may sometimes be a priori perfectly reasonable and at other times a priori manifestly ludicrous, depending on what we know of what these quantities represent and how they interact in systems involving both quantities.) For the sake of convenience, our scientist only considers a linear relationship of the form $Y=\lambda X+{\text{“stuff”}},$ in which the last term is thrown in because the relationship will not be perfect. This type of relationship is especially convenient because the usual suspects common statistics packages allow their users to explore such linear relationships with one click of a button without ever needing to think about what anything means. Now, if the data scientist also does this and asks for the value of the linear coefficient $\lambda$ that gives the best "fit", meaning that it minimizes the variance of $Y-\lambda X,$ being the extra "stuff", we get the following:
    ${\text{Var}}(Y)=\lambda ^{2}{\text{Var}}(X)+{\text{Var}}(Y-\lambda X),$
which holds since the best-fit $\lambda$ ensures that ${\text{cov}}(X,Y-\lambda X)=0.$ So this partitions the variance of quantity $Y$ into a part that can be considered "explained" by the variance of quantity $Y$ assuming that it is legitimate to view $X$ as driving $Y,$ and a residual variance "explained away" by the adage "s✶✶✶ stuff happens".
If the assumption of directional causation is not only justified but also correct, a change $\Delta X$ will on the average correspond to a change $\Delta Y=\lambda \Delta X.$
Now Pearson's correlation coefficient $\rho _{X,Y}$ – commonly called the correlation coefficient because people are unaware it's not the only one in town – satisfies
    $\rho _{X,Y}^{2}={\frac {{\text{cov}}(X,Y)}{{\text{Var}}(X){\text{Var}}(Y)}}={\frac {\lambda ^{2}{\text{Var}}(X)}{{\text{Var}}(Y)}}.$
The multivariate case requires more symbols but is essentially the same, except for the requirement that if more explaining quantities $X_{1},X_{2},...,X_{n}$ are assumed and these are not independent, the maths gets more complicated and there may be no basis for apportioning culpability among these quantities as being responsible for the explainable variance.  ​‑‑Lambiam 22:54, 26 January 2026 (UTC)

So, am I right that one paper explains about 1% of the variance, and the other even less than that?

So: squaring those coefficients is an exact / approximate / no measure of the explained variance? tgeorgescu (talk) 09:41, 27 January 2026 (UTC)

When, for example, $\rho _{X,Y}=-0.06,$ we have that its square $\rho _{X,Y}^{2}=0.0036.$ This means that, if the assumption of causality is justified, about 0.36% of the observed variance in the dependent quantity is driven by the variance in the causating quantity. If we then further see that the 95% confidence interval for the estimated correlation is given by $-0.09\leq \rho _{X,Y}\leq -0.02,$ we see that we cannot conclude with confidence that more than 0.04% of the variance is explained under the assumption of causation. But note that variance itself is the mean of a squared quantity ${(X-{\text{E}}[X])}^{2},$ so, like objects in the mirror, the typical deviation of the mean is larger than it appears. Rewriting a formula given above as
    ${\frac {\Delta Y}{\sigma _{Y}}}=\rho _{X,Y}{\frac {\Delta X}{\sigma _{X}}},$
we see that one might say, imprecisely but with some justification, that about $|\rho _{X,Y}|$ × 100% of the variation of $Y$ around the mean $\mu _{Y}$ is due to the variation of $X.$
The observations of a data set are always just samples from a population. Any statistic computed from these observations is no more than an estimate of a population parameter. Adding to the unavoidable uncertainty resulting from even the most unbiased sampling, there are also systematic errors resulting from sampling bias, observational errors, and uncertainty possibly caused by changes in a dynamic population while sampling is in progress.  ​‑‑Lambiam 12:18, 27 January 2026 (UTC)

It's not our task to judge if the assumption of causality is justified. All we can say: according to these papers, the theory is a good/lousy explanation, assuming that the assumption of causality is justified. tgeorgescu (talk) 21:53, 27 January 2026 (UTC)

The essence of the theory is that X causes Y. If we assume without questioning that this is indeed the case, what is it that needs to be explained? This is not a matter of mathematics but of science, but shouldn't we be testing the theory, and in doing so, can't we challenge the assumption of causality, which is an essential aspect of the theory? A dampening effect of binge watching pornography on the desire to engage in carnal engagement with live partners can IMO not be discounted out of hand. Furthermore, it is IMO also justified to question the theory that the fluctuating gravitational pull on Earth (due to Jupiter's motion) has an effect on the influx of secretarial energy in Alaska. Isn't it about requally likely that the rise and fall of Alaskan secretarial energy drives Jupiter's motion?  ​‑‑Lambiam 01:07, 28 January 2026 (UTC)

January 29

Attempting to prove Lindemann's Theorem

While trying to write a neat rigorous proof for the Lindemann–Weierstrass theorem, I am trying to prove this weaker statement:

For all $0\neq z\in {\overline {\mathbb {Q} }}$ we have ${\text{e}}^{z}\!\notin \mathbb {Q}$.

The method is similar to the transcendence proof for $\pi$:
Let $z_{1},\ldots ,z_{n}$ be a complete set of algebraic conjugates.
Let us assume falsely that ${\text{e}}^{z_{1}}\!\in \mathbb {Q}$, meaning $a+b{\text{e}}^{z_{1}}\!=0$ for some $a,b\in \mathbb {Z} \setminus \{0\}$. We get:

${\begin{aligned}0&=(a+b{\text{e}}^{z_{1}}\!)\cdots (a+b{\text{e}}^{z_{n}}\!)\\[3pt]&=a^{n}+a^{n-1}b^{1}\sum _{i\,=\,1}^{n}{\text{e}}^{z_{i}}+a^{n-2}b^{2}\!\!\!\!\!\!\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!\!{\text{e}}^{z_{i_{1}}}{\text{e}}^{z_{i_{2}}}+a^{n-3}b^{3}\!\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<i_{2}<i_{3}\leq n}\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}}{\text{e}}^{z_{i_{2}}}{\text{e}}^{z_{i_{3}}}\!+\cdots +b^{n}\!\!\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<\cdots <i_{n}\leq n}\!\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}}\!\cdots {\text{e}}^{z_{i_{n}}}\\[3pt]&=a^{n}{\text{e}}^{0}+a^{n-1}b^{1}\sum _{i\,=\,1}^{n}{\text{e}}^{z_{i}}+a^{n-2}b^{2}\!\!\!\!\!\!\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!\!{\text{e}}^{z_{i_{1}}+\,z_{i_{2}}}+a^{n-3}b^{3}\!\!\!\!\!\!\!\!\!\sum _{1\leq i_{1}<i_{2}<i_{3}\leq n}\!\!\!\!\!\!\!\!{\text{e}}^{z_{i_{1}}+\,z_{i_{2}}+\,z_{i_{3}}}\!+\cdots +b^{n}{\text{e}}^{z_{1}+\,\cdots \,+\,z_{n}}\\[3pt]&=\sum _{i\,=\,1}^{2^{n}}c_{i}{\text{e}}^{\beta _{i}}\end{aligned}}$

The exponents $\beta _{i}$ are symmetric polynomial in $z_{1},\ldots ,z_{n}$, and among them are $n\leq m\leq 2^{n}-1$ non-zero sums. That is:

$c_{0}\!+c_{1}{\text{e}}^{\beta _{1}}\!+\cdots +c_{m}{\text{e}}^{\beta _{m}}=0$.

From here we continue by constructing a polynomial in $\mathbb {Z} [z]$ with the $\beta _{i}$'s as its roots, and reaching a contradiction.

Unfortunately, I believe this can be done only if we can show that $c_{0}\!\neq 0$. Can we actually do that? יהודה שמחה ולדמן (talk) 19:06, 29 January 2026 (UTC)

What is c₀? Is it aⁿ? JRSpriggs (talk) 02:17, 1 February 2026 (UTC)

$c_{0}$ is the sum of all the coefficients $c_{i}$ of which $\beta _{i}=0$.

I am not able to show that $c_{0}\!\neq 0$.

But the truth is, the Lindemann–Weierstrass theorem is proved by constructing symmetric polynomials of which their roots are algebraic integers.

Unfortunately, the theorem's proof is written quite poorly, and I barely follow any details of it. יהודה שמחה ולדמן (talk) 07:57, 1 February 2026 (UTC)

To aid our understanding: taking, for a simple example, $n=2$ and expanding the product $(a+be^{z_{1}})(a+be^{z_{2}})$ into the sum $a^{2}+ab(e^{z_{1}}+e^{z_{2}})+b^{2}e^{z_{1}+z_{2}},$ what would the coefficients and exponents in $c_{0}+c_{1}e^{\beta _{1}}+\cdots +c_{m}e^{\beta _{m}}$ become, concretely?  ​‑‑Lambiam 17:55, 1 February 2026 (UTC)

Assuming that $a,b$ are non-zero integers and that $z_{1},z_{2}$ are algebraic conjugates, we get:

${\begin{aligned}&c_{0}\!=a^{2}\!+b^{2},\,c_{1}\!=c_{2}\!=ab\\&\beta _{1}\!=z_{1},\,\beta _{2}\!=z_{2},\,\beta _{3}=0\end{aligned}}$

But for $n>2$ it might be that $c_{0}\!=0$. Can we actually prove otherwise? יהודה שמחה ולדמן (talk) 20:10, 1 February 2026 (UTC)

I made a mistake: $\beta _{3}$ is not necessarily 0. יהודה שמחה ולדמן (talk) 20:24, 3 February 2026 (UTC)

February 3

Was the Golden Ratio used by Egyptians to build the pyramids of Giza?

I'm not sure if I should post this in mathematics or history, but I will start here. When I was reading a math book a couple of years ago, I read somewhere that the pyramids of Giza were built using the Golden Ratio. Does this mean that the Egyptians knew of the ratio at the time? Or was it just an extremely aesthetically pleasing ratio that they happened to use it and the math came later? Ilovezucchinii (talk) 19:03, 3 February 2026 (UTC)

If you cannot remember more about it than that, then we should just dismiss it. Which book? Which page number? Which pyramid or pyramids? What aspect of the pyramid is in that ratio? And so forth. I find it highly implausible that an engineering feat as difficult as building a pyramid in those days would be influenced by someone's mere esthetic feeling for the golden ratio. At best, it is a coincidence. More likely, a mistake or a hoax. JRSpriggs (talk) 19:41, 3 February 2026 (UTC)

The original measurements of the Great Pyramid, before millenia of erosion took their toll, are thought to have been a height of 280 cubits and a side length of the square base of 440 cubits. So the slides slope up with a tangent of 280/(1⁄2×440); the slope, expressed as an angle, is the arctangent of 280/220, which is about 51°50'40". The secant of the slope angle equals 1.61859..., which is close to the golden ratio, φ = 1.61893... . This is widely reported as establishing that these measurements were based on the golden ratio. But how solid is this conclusion? Were the measurements really chosen to achieve this, or is this perhaps a coincidence? If you compute a large number of quantities, you increase the likelihood of hitting on one that is close to a well-known mathematical constant. For example, the ratio of the circumference of the base to the height of the Great Pyramid is equal to 4×440/280 = 6.28571..., close to 2π ≈ 6.28319. There are no known further indications that the secants of slopes played a particular role in Egyptian architecture, and those of the other two major pyramids of Giza are not nearly as close to φ.  ​‑‑Lambiam 21:10, 3 February 2026 (UTC)

There have been as many "false positive" reports of golden ratio sightings as instances where it was actually used. It was used by Euclid, who called it the "extreme and mean ratio", to construct a regular pentagon. But reports of it being used in the design of the Pyramids or Parthenon are highly doubtful as there is no direct evidence. Unfortunately, math has a number of such myths, repeated uncritically by "popularizers" who don't bother to check for original sources or direct evidence. See Golden ratio#Disputed observations for more information. --RDBury (talk) 16:09, 4 February 2026 (UTC)

February 4

Paradox Logic

Can you help me understand the difference between Graham Priest's logic of paradox and Newton da Costa's approach? ~2026-78268-0 (talk) 21:45, 4 February 2026 (UTC)

Priest's LP (logic of paradox) is described in: Priest, Graham. "The Logic of Paradox." Journal of Philosophical Logic 8, no. 1 (1979): 219–41. JSTOR 30227165. It is basically a 3-valued logic, in which the third value is not something "between true and false" but more like "kind of both true and false". It can be formally derived from CL (classical logic) by giving it a model, as follows.

Denote the classical truth values by $\top$ representing truth and $\bot$ representing falsehood. The truth values of LP in the model are the three non-empty subsets of the set of classical truth values: $t=\{\top \},f=\{\bot \},p=\{\top ,\bot \}.$ The logical operators are derived from the classical logical operators by applying the latter to all possible combinations of classical truth values of the operands. For negation, we get $\neg _{\text{LP}}A=\{\neg _{\text{CL}}a~|~a\in A\},$ so

$\neg _{\text{LP}}t=\neg _{\text{LP}}\{\top \}=\{\neg _{\text{CL}}\top \}=\{\bot \}=f;$

$\neg _{\text{LP}}f=\neg _{\text{LP}}\{\bot \}=\{\neg _{\text{CL}}\bot \}=\{\top \}=t;$

$\neg _{\text{LP}}p=\neg _{\text{LP}}\{\top ,\bot \}=\{\neg _{\text{CL}}\top ,\neg _{\text{CL}}\bot \}=\{\bot ,\top \}=p.$

Likewise, $A\land _{\text{LP}}B=\{a\land _{\text{CL}}b~|~a\in A,b\in B\}.$ Now the principle of explosion allowing the derivation of any statement $B$ from the premise $A\land \neg A$ does not hold, since $p\land _{\text{LP}}\!(\neg _{\text{LP}}p)=p,$ and $p\to _{\text{L}}f=(\neg _{\text{LP}}p)\lor _{\text{L}}f=p\lor _{\text{LP}}f=p.$ This differs from the logic described in the section Paraconsistent logic § An ideal three-valued paraconsistent logic, in which ${\mathsf {b}}\to {\mathsf {f}}={\mathsf {f}},$ but I am not sure the latter is not an error in the table given there.

I did not find a complete exposition of da Costa's system, but a sketch is given in Section 3.5: Logics of Formal Inconsistency of the article "Paraconsistent Logic" in the online Stanford Encyclopedia of Philosophy.  ​‑‑Lambiam 18:58, 5 February 2026 (UTC)

The table is correct. We must have (b→f)=f or else there will be no modus ponens and no tautologies. Logic would cease to exist. See Paraconsistent logic#Strategy. JRSpriggs (talk) 21:06, 5 February 2026 (UTC)

Priest's LP also has tautologies, such as $A\to A.$ It has contraposition but not modus ponens, while the "ideal three-valued paraconsistent logic" has modus ponens but not contraposition.  ​‑‑Lambiam 03:23, 6 February 2026 (UTC)

I should have said, no non-trivial tautologies. The point is that you cannot deduce anything in his system. What good does it do to supposedly save contrapositive if you throw out the whole system? JRSpriggs (talk) 15:10, 6 February 2026 (UTC)

I'm not defending it, although I do not know where to draw the line between trivial and non-trivial tautologies. If it is possible to make "non-trivial" deductions in da Costa's system (using, in an essential way, premises that are paraconsistent in the sense that $A\land \neg A$ is not explosive), this would appear to be a significant difference between the systems.  ​‑‑Lambiam 15:47, 6 February 2026 (UTC)