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5h
comment Evaluate the line integral $\int_C \ x^2 dx+(x+y)dy \ $
Very nice, elegant solution.+1
5h
comment Antiderivative of $e^{au}$
@Ethan, chain rule: put $\;f(u)=e^u\;,\;\;g(u):=au\;$ , so $$e^{au}=f(g(u))\implies (e^{au})'=f'(g(u))\cdot g'(u)=e^{g(u)}\cdot a=ae^{au}$$and from here the solution...
5h
answered Evaluate the line integral $\int_C \ x^2 dx+(x+y)dy \ $
5h
comment Solving $\phi(n)=84$
Not only will naslundx be sad if you break math: you will have to pay for it!
6h
comment Antiderivative of $e^{au}$
@Lemur...No me acostumbro a este nick...
6h
answered Antiderivative of $e^{au}$
11h
comment proving determinant of lower triangular matrix from definition
Well @Warz , in general I meant. The application to this problem is another matter: $\;\sigma\neq Id.\iff \exists\,i\;\;s.t.\;\;\sigma(i)\neq i\;$
11h
comment Find the sum of the following series
Do you mean the series $$\sum_{n=1}^\infty-\frac9{n6^n}\;\;?$$
11h
comment Calculated by multiplying the arithmetic progression terms
Do you expect a closed, cute and short formula? Have you done some examples with small $\;n\;$ ? What do you get?
11h
answered Normalizer of a subgroup of $GL_2(\mathbb{R})$
12h
answered Prove that $\sin(\frac{\pi}{3}+x)=\cos(\frac{\pi}{6}-x)$
12h
comment $\exists a\in G-H$ such that $aHa^{-1}=H$
Looks fine, @abe...and about knowing: experience and practice.
13h
comment Help defining the statistical number that pi posses a philosophical question in the first 15 characters.
Hmmm...numbers, infinite, codes, gods, mumbo-jumbo...this ain't going to end well.
13h
comment $\tan(x)=\cot(90^\circ-x)$??
@NateEldredge, are you sure USA is an english speaking country? Just kidding...but not much. :)
13h
comment $\tan(x)=\cot(90^\circ-x)$??
@rah,I meant in that past comment that right angled is the correct term to what I called "straight angled" ...
13h
comment $\exists a\in G-H$ such that $aHa^{-1}=H$
No @abe, it is not that at all...since in any finite group we have that $\;x^{|G|}=e\;\;\forall\,x\in G\;$ ...
13h
answered $\exists a\in G-H$ such that $aHa^{-1}=H$
13h
comment $\exists a\in G-H$ such that $aHa^{-1}=H$
@Abe, have you already studied nilpotent groups? Do you know a finite $\;p$-group is nilpotent?
13h
comment $\exists a\in G-H$ such that $aHa^{-1}=H$
No @abe, wikipedia doesn't have anything of the like. I think you're misunderstanding something in that link you wrote...
13h
comment $\exists a\in G-H$ such that $aHa^{-1}=H$
how come $\;Z(H)\le Z(G)\;$ ? How did you prove that?