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Mar
26
revised Prove not a violation of Stokes theorem
added 861 characters in body
Mar
26
comment Prove not a violation of Stokes theorem
@Faust7: You mean you're having trouble actually doing the integral? I'll add a bit about that.
Mar
26
comment Prove not a violation of Stokes theorem
@Faust7: Added explanation.
Mar
26
revised Prove not a violation of Stokes theorem
full answer
Mar
26
answered Prove not a violation of Stokes theorem
Mar
25
comment Prove not a violation of Stokes theorem
Hint: review the precise hypotheses assumed by the theorem. Does it hold in this situation?
Mar
23
comment Looking at the numerator first or denominator?
That's not what the question is asking.
Mar
17
comment Point of Inflection or Turning Point?
I don't think it's so clear. If the rate of change at a point is $-2$, is that a higher or lower rate of change than where it's $0$. Maybe it was meant to be interpreted as the absolute value of the rate of change.
Mar
17
comment Calculating $\int_0^\infty(\log t)^n e^{-t}\ dt$
Actually, sorry, but I don't.
Mar
17
comment Calculating $\int_0^\infty(\log t)^n e^{-t}\ dt$
Sorry, I should have been clearer. Expanding $t^{s-1}e^{-t}$ as a power series in $s-1$, you get this integral where the nth derivative goes. But in my question I asked how to calculate it, so this is besides the point.
Mar
17
comment Calculating $\int_0^\infty(\log t)^n e^{-t}\ dt$
Well, yes, that's where I got it from.
Mar
17
asked Calculating $\int_0^\infty(\log t)^n e^{-t}\ dt$
Mar
16
comment Looking for help with a proof that n-th derivative of $e^\frac{-1}{x^2} = 0$ for $x=0$.
Hint: show that all derivatives of $f$ are of the form $R(x)e^{-1/x^2}$ for some rational function $R$.
Mar
13
awarded  Notable Question
Mar
12
comment Order of calculation in all math equations
@AJMansfield: That could very well be interpreted as $6/(2/(1+2))$. The only unambiguous way is to use parentheses.
Mar
12
comment $\operatorname{im} A = \ker A$ for a $2 \times 2$ matrix $A$?
What makes you think they have to be different?
Mar
12
comment Is this proof about functions correct?
Try to format your question a bit better, it's really hard to read like this. Also, this has nothing to do with number theory.
Mar
12
revised Dividing matrices
edited tags
Mar
12
comment Dividing matrices
Short answer: You can't. Maybe there's a subset of the set of all matrices that allows division, but I'm not sure.
Mar
12
comment Evaluate each of the numeric expressions: $\sqrt{(-3)}$, $\sqrt{3}$, $\sqrt{-3}$
As far as I know, while the equation $x^2 = a$ has two solutions for $a > 0$, $\sqrt{a}$ is defined as the positive square root, not both. This only applies when working in $\mathbb{R}$, since in $\mathbb{C}$ you can't distinguish positive and negative.