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Apr
5
answered Suppose that $f$ is a real valued function such that its second derivative is discontinuous.Can you give some example?
Apr
5
comment Does $a^n \mid b^n$ imply $a\mid b$?
@awllower: you're right that it's a duplicate, but I think this one has better answers.
Apr
5
accepted Does $a^n \mid b^n$ imply $a\mid b$?
Apr
5
asked Does $a^n \mid b^n$ imply $a\mid b$?
Mar
28
comment Show $f(z)=2u\left( \frac z2, \frac{-iz}{2}\right ) + \text{constant} $ if $f(z)=u(x,y)+iv(x,y)$ is an analytic function
My complaint still stands. $u$ and $v$ are functions from $\mathbb{R}^2$ to $\mathbb{R}$, you can't give them complex numbers as arguments.
Mar
28
comment Show $f(z)=2u\left( \frac z2, \frac{-iz}{2}\right ) + \text{constant} $ if $f(z)=u(x,y)+iv(x,y)$ is an analytic function
The question as stated makes no sense. $u$ and $v$ are functions of two real variables, you can't feed them complex numbers.
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