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seen Jul 19 at 22:24

Jun
5
comment 'Obvious' theorems that are actually false
@Asaf, maybe you could clarify what exactly you mean by “height” here. Each element of the set you defined clearly has order $\omega$, and while I believe we mean the same thing for “the $n$-the level” when $n$ is finite, if I read $f\upharpoonright n$ as $f$ limited to the preimage set $\{0,\dots,n\}$. Now, none of these actually is in your set. Each element of your set is the union of a countable number of such finite sequences, which you can choose to view as finite paths. But what would, with your definition of “height,” be such a tree of height $\omega$?
Jun
5
comment 'Obvious' theorems that are actually false
@Asaf, I fail to see how you get order type $\omega+1$ entries in elements of the set $\{0,\dots,9\}^\mathbb{N}$. E.g., I think we agree there is exactly one element in this set corresponding to the decimal expansion of $\pi-3$. If that element, as you say, has order type $\omega+1$, then it has a last digit. Which one is that and why? And how does it come to be in a function from $\mathbb{N}$ to $\{0,\dots,9\}$?
Jun
5
comment 'Obvious' theorems that are actually false
Since the decimal expansion of, say, $1/3$, does not have a last digit, the “tree” does not have a final level in this sense.
Jun
4
comment 'Obvious' theorems that are actually false
This is a bit of a stab in the dark, but can there even be such a g if f is a space-filling curve?
Jun
4
comment 'Obvious' theorems that are actually false
In which sense does your tree have a “final level” and what does that term even mean here?
Jun
1
comment Why do we drop the abolute value bars when doing indefinite integration?
@GitGud $\ln(x)$ is an antiderivative, just like $\ln(x)+H(x)$ (with $H$ the Heaviside step function) or $\ln(|x|)$ or an uncountably infinite number of other expressions are. Sure, you can ask that all of these include a “+ C” term, which I regard as pointless boilerplate. But there’s no reason to regard $\ln(|x|)$ as inherently better. Sure, it’s one of the real-valued antiderivatives. So what, big deal. It’s also one that makes differential-algebraic manipulations harder than they need be. Last discussion tidbit on here from me, I promise.
Jun
1
comment Why do we drop the abolute value bars when doing indefinite integration?
@GitGud: What is wrong? Listing $\int\frac 1x\,\mathrm{d}x = \ln(x)$ is perfectly correct, just like $\int\frac 1x\,\mathrm{d}x = \ln(|x|)$ is – just take the derivative to check. The difference between the two is a piecewise constant – for real $x$. The form without the absolute values is even valid for complex $x$.
May
18
comment Interesting “real life” applications of serious theorems
I’m sure some drivers would call that a mean application of some value theorem.
May
18
comment Interesting “real life” applications of serious theorems
For a simple proof of the first equation, note that $\sum_{j=0}^n\binomial nj (-1)^j$ is just the multinomial expansion of $(x+1)^n$ evaluated at $x=-1$.
Mar
24
comment Integrating factor in linear differential equations
Just starting with $\log y$ would also work in the complex plane. :-)
Mar
24
comment Integrating factor in linear differential equations
Link-only answers are not really a good fit for SE, could you summarize the most relevant points here, as far as they are not well covered by other answers yet?
Mar
24
comment Integrating factor in linear differential equations
Nice answer. But why introduce the absolute value first, only to drop it (without an explanation why we can)? The formula remains right if you start by saying that $y'/y = \mathrm{d}/\mathrm{d}x \log(y(x))$.
Mar
13
comment Interval around a root of a function
@Ale A function that is nowhere constant will not have the property that it is zero at every point in $[x_0-\epsilon, x_0+\epsilon]$, since then it would be constant in this interval.
Mar
12
comment Interval around a root of a function
@Ale You did ask for a function that is constant in an interval.
Mar
12
comment Interval around a root of a function
Nitpick: 180 is not a zero of sin(x). 180°, which equals π, is. (This is a math site, after all.)
Mar
1
awarded  Yearling
Feb
21
answered Factoring a hard polynomial
Feb
4
comment Understanding non-solvable algebraic numbers
Also, there are algebraic dependencies between the different zeroes of a polynomial. In particular, things like $\sum_{r, p(r)=0}q(r)$ can be expressed in the base field for polynomial (or even most rational) functions $q$.
Feb
4
answered Find the approximate center of a circle passing through more than three points
Jan
23
revised Convergence of integral of $-\ln(x)^{a}$
formula typesetting