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2d
comment About the differentiability of $|x|$?
Exactly. I was a little bit lazy when typing, sorry.
2d
comment About the differentiability of $|x|$?
I made a silly mistake, my impression was actually: I am starting to guess that for a function to be differentiable at a point, all it's derivatives should also be continuous at the same point.
2d
comment About the differentiability of $|x|$?
Sorry, I made a mistake, it's actually: I am starting to guess that for a function to be differentiable at a point, all it's derivatives should also be continuous at the same point.
2d
comment About the differentiability of $|x|$?
@Rahul Isn't $|x|$ sometimes defined as $\sqrt{x^2}$?
Apr
4
comment Equations of the tangent lines of $y=x^4$ at the point $(2,0)$?
@DanielEscudero Yes! There are lines in both directions. That should explain the lines.
Apr
4
comment Equations of the tangent lines of $y=x^4$ at the point $(2,0)$?
@DanielEscudero Yes. I also think it's weird to ask about lines, In my mind, I also think it's only one line. But he used the plural retas in here.
Apr
4
comment Equations of the tangent lines of $y=x^4$ at the point $(2,0)$?
@DanielEscudero Oh, I get it. What is being asked is the equation of the line that passes through $(2,0)$ and touches the curve. I guess it's this.
Apr
3
comment What is the meaning of division for quasigroups in here?
@vadim123 Oh yes. Sorry. I was too lazy that I didn't even think about googling quasigroups. I blame the lack of coffee.
Apr
2
comment How to show that $\frac{-1}{x^2}=0$ has no solutions?
Good. But multiplying both sides by $x^2$ is the only possible operation to solve that, isn't it?
Apr
2
comment Are there two meanings to induction?
Very good. Your answer is the first thing that made the proof tick in my mind. You've made all steps really clear, I am satisfied. Thanks a lot!
Apr
1
comment Are there two meanings to induction?
I don't know, but I feel that part a little artificial. But perhaps it's only my own stupidity.
Apr
1
comment Are there two meanings to induction?
Exactly! But the hypothesis is made out of nowhere. It is not a consequence of what was written before. Or if it is, then I really do not see how that could be.
Mar
26
comment Identify the subset of the plane composed by $R\cos \theta=5$ by means of polar coordinates $R,\theta$?
@N.F.Taussig I figured it now! Thanks.
Mar
25
comment Identify the subset of the plane composed by $R\cos \theta=5$ by means of polar coordinates $R,\theta$?
I don't know, I'm a bit confused with the book instructions. It says that the polar coordinates are $(R, \theta)$ and the cartesian coordinates of this point are $(R \cos \theta, R \sin \theta)$. I'm a bit confused.
Mar
16
comment Help on Apostol's explanation of Archimedes method?
@neuguy For example, why start with $(k+1)^3$? I'm not really sure of what is happening in there, then I asked it vaguely: I was hoping that someone could rephrase it adding a little bit more detail perhaps?
Feb
23
comment How to solve this distribution problem with generating functions?
@Taussig Sorry,I meant 3 boxes. I forgot to add it.
Feb
19
comment Is it possible to represent any sequence as the coefficients of arbitrary generating functions?
@whacka What's the meaning of sufficiently radical? I googled it and most of the results are about radical christianism(?!) and suffixing "math", yields something that also seems completely unrelated.
Feb
18
comment Is it possible to represent any sequence as the coefficients of arbitrary generating functions?
@whacka Got it. I got curious on a generating function which can't be expressed in a nice closed-form. Could you give me an example?
Feb
18
comment Is it possible to represent any sequence as the coefficients of arbitrary generating functions?
@whacka The problem is here: "you can form a corresponding generating function". I've looked (in my lectures) at few methods for finding generating functions. They are basically the expansion $(1-x)^n$ and $\left[ \frac{1}{1-x} \right]^n$, derivating and integrating one series and performing the sum/multiplication of two series. With this, it's not really clear that every sequence could be produced in the coefficients of the generating function.
Feb
17
comment On the calculus of recurrence relations using generating functions?
Oh, I got it. I was being stupid of not checking it by expanding a few terms. I did it now and I noticed they are the same. $$a_0+\sum_{k\ge 1}(ba_{k-1}x^k+cx^k)=a_0+(ba_0x^1+cx^1+\cdots)+(ba_1x^2+cx^2+\dots)$$ $$a_0+bx\sum_{k\ge 0}a_kx^k+cx\sum_{k\ge 0}x^k=a_0+(ba_0x^1+ba_1x^2+\dots)+(cx^1+cx^2+ \dots ) $$