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14h
comment Why do we need to assume continuity in the proof of the chain-rule?
@Nick I’m not sure I see how it fails. What’s written there is pretty much the definition of being continuous at $x$, no continuity anywhere else is needed.
16h
answered Finding Tangent Line to Graph
1d
awarded  Nice Answer
Dec
23
comment Easy example why complex numbers are cool
Most? I’d say most integrals cannot be solved in closed form, but we may simply be looking at different distributions over the space of possible integrands…
Dec
23
answered Easy example why complex numbers are cool
Dec
18
comment How to determine if the sequence $a_n= (30+12\arctan(n!))/6^n$ is divergent or convergent
Wasn't the question about the sequence itself, not the series?
Nov
6
revised An element of a group has the same order as its inverse
an exponent was missing its ^. Took out a few {} because I need 6 characters
Nov
6
suggested approved edit on An element of a group has the same order as its inverse
Oct
25
comment Why doesn't $\arccos x = -\tfrac12\sqrt{3}$ have any solutions?
No need to limit yourself to real $x$. There's no complex $x$ with that property either. (In both real and complex numbers, that is true if you take the standard primary branch.)
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?