TITLE: Standard deviation sig figs
QUESTION [0 upvotes]: Let's say I calculated a mean to be 2.475, but the data values had the least significant figure in the tenths place (i.e. 2.6, 2.8 etc.) so I round the mean value to 2.5 for correct sig figs. However, my standard deviation is 0.0835, and choosing my 1 sig fig for SD, I round to 0.08 So my SD is uncertain in the hundredths place, but my mean is in the tenths place....should I round my SD to 0.1 to match the mean? Or is it okay to have uncertainty at a decimal place further right than your mean?

REPLY [0 votes]: There are many rules about how many decimal places to show, some formulated by statisticians (seeking useful answers), others by editors (seeking pretty typeset pages).
You're on the right track, seeking to show only as many decimal places as convey useful information.
As a guide to the precision of parameter estimates, you can look at the widths of 95% confidence intervals. Crucially missing from the information about your example, as given in your question, are the sample size $n,$ the sample mean $\bar X$ and the sample SD $S.$
So I can only speculate.
Suppose $n = 100, \bar X = 50.0, S = 4.0,$ then a 95% CI for $\mu$ would be $50 \pm 2(4)/10$ or $50 \pm 0.8.$ Also a 95% CI for $\sigma$ would use the relationship
$\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = n-1).$ This leads to the 95% CI $(3.51,4.65)$ for $\sigma.$
sqrt((99*(16)/qchisq(c(.975,.025), 99)))
[1] 3.512027 4.646701

Both CIs seem to suggest that one-place accuracy is
enough. However, if original data were rounded to two
or three places, I might show the mean and SD (and if appropriate, their CIs) to two places. This would be
more a matter of tact than of fact, but civility is a
worthwhile goal.

Note: Without regret, I gave two-place accuracy in my 95% CI
$(3.51,4.65)$ for $\sigma.$ But if anyone thinks the second decimal place is crucial, then ponder the following 96% and 94% CIs. (As a matter of practical decision making, does anyone
really consider the three intervals to be crucially
different?)
sqrt((99*(16)/qchisq(c(.98,.02), 99)))
[1] 3.490703 4.681064
sqrt((99*(16)/qchisq(c(.97,.03), 99)))
[1] 3.530201 4.618012

But $(3.512, 4.647)$ for the 95% CI would
pretend a degree of precision that is only
a fantasy.


TITLE: In what medium do Gravitational waves propagate?
QUESTION [4 upvotes]: I am having a hard time understanding how gravitational waves move.
The only reasonable thing that comes to my mind would be the aether but that idea was scrapped.

REPLY [7 votes]: As you are an undergraduate, have you any knowledge of General Relativity?
Gravitational waves do not exist in classical Newtonian Mechanics. They are waves in the fabric of space and time itself, according to General Relativity. This is a simplified explanation of what gravitational waves are. As you will see, since it is space time itself that displays wave like behavior there is no need for a medium.


TITLE: Isomorphic polynomial rings
QUESTION [0 upvotes]: Let $\mathbb K$ be a finite field and let $x, y$ be two indeterminates.
My intuition tells me that
$$\mathbb K[x, y]\cong\mathbb K[x, y+1],$$
but $\mathbb K[x, y]\ncong\mathbb K[x, xy],$ or $\mathbb K[x, y]\ncong\mathbb K[x, x+y]$.
So, my question is:

Let $\mathbb K$ be a finite field and let $x_1, x_2, ..., x_n$ indeterminates. If $f_1, f_2, ..., f_n$ are polynomials in $n$ variables, what conditions must satisfy the polynomials so that we have an isomorphism
  $$\mathbb K[x_1, ..., x_n]\cong \mathbb K[f_1(x_1, ..., x_n),\dots, f_n(x_1,..., x_n)]?$$

REPLY [1 votes]: Your intuition is in error. For any ring $R$, the homomorphism
$$ R[x,y] \to R[x,xy] : f \mapsto f(x, xy) $$
is an isomorphism. It's clearly a homomorphism since the domain is a polynomial ring and the map is given by evaluation. One way to see that it's an isomorphism is to observe

$R[x,y]$ is a free $R$-module with basis $\{ x^i y^j \mid i,j \in \mathbb{N} \}$
$R[x,xy]$ is a free $R$-module with basis $\{ x^i y^{i+j} \mid i,j \in \mathbb{N} \} = $

and the map $f(x,y) \mapsto f(x, xy)$ is a bijection between the two bases.
For the other example you can even say something stronger: you actually have equality $R[x,y] = R[x,x+y]$, because $y = (x+y) - x$, and so $y \in R[x,x+y]$.

Regarding your actual question, $\mathbb{K}$ being finite is irrelevant. Allowing $\mathbb{K}$ now to be any field, the condition you are looking for is

There is an isomorphism fixing $\mathbb{K}$ if and only if the $f_i$ are algebraically independent — that is, there does not exist a nonzero polynomial $g$ of $n$-variables over $\mathbb{K}$ with the property that
  $$ g(f_1, \ldots, f_n) = 0 $$

You always have a surjective homomorphism given by
$$\mathbb{K}[x_1, \ldots, x_n] \to \mathbb{K}[f_1, \ldots, f_n] : h \mapsto h(f_1,  \ldots, f_n)$$
So the above condition is clearly sufficient, because it is equivalent to saying that the kernel of this homomorphism is the zero ideal.
As for being necessary, if such a $g$ existed, then the field extension $\mathbb{K}(f_1, \ldots, f_n) / \mathbb{K}$ would have transcendence degree strictly less than $n$. However, $\mathbb{K}(x_1, \ldots, x_n) / \mathbb{K}$ has transcendence degree exactly $n$.
Any isomorphism $\mathbb{K}[x_1, \ldots, x_n] \cong \mathbb{K}[f_1, \ldots, f_n]$ that fixes $\mathbb{K}$ would extend to an isomorphism of the above field extensions, which would contradict the fact they have different transcendence degree.