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1d
comment Why is the derivative of a vector orthogonal to the vector itself?
@YagnaPatel If you imagine the two radial vectors getting closer and closer, the difference of the two velocity vectors will be along the length of the curve.
1d
comment Evaluating $\int{ \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}}dx$
This is a very, very good answer. I never would have imagined the solution would be so nice. $+1$!
2d
comment Linear operator, image, transformation
Okay so let's see what happens when we evaluate $A(p(x)+q(x))$. $A(p(x)+q(x)) = A((p+q)(x)) = (p+q)(x+3)$. Can you take it from there?
2d
comment Linear operator, image, transformation
Do you know what it means for an operator to be linear?
2d
comment Does there exists an entire function $f: \mathbb C \to \mathbb C$ which is bounded on real line and imaginary line?
Use \, in $\rm\LaTeX$ to put in artificial spaces.
2d
comment Does there exists an entire function $f: \mathbb C \to \mathbb C$ which is bounded on real line and imaginary line?
@Learner For sure.
2d
comment Does there exists an entire function $f: \mathbb C \to \mathbb C$ which is bounded on real line and imaginary line?
You should remove some of the ellipses.
2d
comment Does there exists an entire function $f: \mathbb C \to \mathbb C$ which is bounded on real line and imaginary line?
See this Math Overflow page for some really good answers towards this direction: mathoverflow.net/questions/190837/…
2d
comment How to define a group ring when the group is infinte?
The group ring has different characteristics depending on whether or not your group is discrete. If it is not discrete, the group ring is defined as the compactly supported continuous functions on $G$. In the discrete case, the group ring is similar. It's the sets of formal sums with finitely many nonzero terms.
Jun
26
revised How to obtain a presentation for each group of order $64$
rolled back to a previous revision
Jun
26
revised How to obtain a presentation for each group of order $64$
edited tags
Jun
26
comment order isomorphism maps $f < g$ to $Tf < Tg$?
What is meant here by $f \le g$ and $f < g$? Is it that $f(x)\le g(x)$ and $f(x) < g(x)$ for all $x$, resp.? Or are you imposing some other order on $A(X)$? Also why introduce $A(X)$ in the first place? It seems like you're just restating $C(X)$..
Jun
25
comment On the Fourier transform of $f(x)=e^{-x^2+2x}$
Yes you most definitely are.
Jun
25
comment On the Fourier transform of $f(x)=e^{-x^2+2x}$
Mathematica does not use the same definition of the Fourier transform as the one you are using as default. It uses the opposite convention where there the sign is positive in the Fourier kernel. You have to use FourierParameters->{0,-1} to get the convention you are using.
Jun
25
comment On the Fourier transform of $f(x)=e^{-x^2+2x}$
Yes I know. That's why I said I am pretty sure it's a typo in the book. It should have a negative sign, not a positive sign. Typos like this are common in papers which are much shorter than books. Books are rife with minor typos like this.
Jun
25
comment On the Fourier transform of $f(x)=e^{-x^2+2x}$
I find it interesting that you used the acronym EDO instead of ODE. Are you from a romantic language speaking country?
Jun
25
comment On the Fourier transform of $f(x)=e^{-x^2+2x}$
You made a typo in your post then. The RHS of the translation property should be $e^{-iak}\hat{g}(k)$. In which case, I think this is a typo in the text. I've worked it out myself and double checked with Mathematica.
Jun
25
comment On the Fourier transform of $f(x)=e^{-x^2+2x}$
What is your definition for the Fourier transform? The answer hinges on that. Most commonly the Fourier kernel has a negative sign in the exponent which changes the translation property.
Jun
25
comment $\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$
That's exactly what I was going for. I've seen multiple people do this and it gets under my skin a little. It's great for mature mathematicians like you and I but doesn't do much for students who are struggling with the basic concepts. Thanks for adding an explanation.
Jun
25
comment $\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$
Yes I agree but you didn't compute the value for the gamma function, instead you just stated it.