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1d
comment Help with function $f_r(x^q)=q^rx^{q-1}$
@user58533 The division by $X$ should not carry any problem, as $h$ contains multiplication by $X$. If you want, you can instead replace the first $h$ by just derivation: $$ h_r : D \circ h \circ h \circ ... \circ h$$ where $D$ is derivation.
1d
comment Checking proof that $f(x)=x^2+1$ is continuous
$|x^2-c^2| \neq |x-c|^2$. Same way $|x-c|^2 \neq |x-c|$.
1d
answered Help with function $f_r(x^q)=q^rx^{q-1}$
1d
comment Help with function $f_r(x^q)=q^rx^{q-1}$
What is the domain of $f_r$? Note that in order to use the derivative, $f_r$ needs to be defined on functions not numbers.
2d
awarded  Nice Answer
Jul
23
answered How to find a pointset with unique distances
Jul
23
answered How can I evaluate this indefinite integral? $\int\frac{dx}{1+x^8}$
Jul
23
comment How do I find this integral $\int\frac{dx}{2x^4+2x^2-1}$
@5xum yes, ty..
Jul
23
answered How do I find this integral $\int\frac{dx}{2x^4+2x^2-1}$
Jul
21
comment Removing cycle from the complete graph.
Or equivalent hint: the complement of $K_{m,n}$ is disconnected.
Jul
21
comment Equality of Sums
No it doesn't. As you said, for each $k$ there exists a $k'$ so that $\left\{ \frac{ak}{n} \right\}= \left\{ \frac{bk'}{n} \right\}$. Which means that $k\left\{ \frac{ak}{n} \right\}$ becomes $k\left\{ \frac{bk'}{n} \right\}$....
Jul
21
comment Equality of Sums
But doesn't the first sum become $\sum_{k=1}^n k \left\{ \frac{bk'}{n}\right\}$ instead of $\sum_{k=1}^n k \left\{ \frac{bk}{n}\right\}$?
Jul
20
comment Equation $a^{n}+b^{n}=2008$ has no integers solutions.
@MariusDamian Was a mistake, I think I fixed now that c ase.
Jul
20
revised Equation $a^{n}+b^{n}=2008$ has no integers solutions.
added 280 characters in body
Jul
20
comment Equation $a^{n}+b^{n}=2008$ has no integers solutions.
The point is that you can have $n \geq 11$ odd...So considering $ 2 \leq n \leq 10$ is not enough.
Jul
20
revised Equation $a^{n}+b^{n}=2008$ has no integers solutions.
added 280 characters in body
Jul
20
answered Equation $a^{n}+b^{n}=2008$ has no integers solutions.
Jul
20
comment Equation $a^{n}+b^{n}=2008$ has no integers solutions.
What if $b$ is negative?
Jul
17
comment Prove $f(x) < 0 \forall x$
I am also puzzled by the hint, as I don't see why $f'' < 0$, which is the key to your hint....
Jul
17
comment A strange puzzle having two possible solutions
@AlexR If I remember right, the proper way to define limits of sets is $\cap_{n=1}^\infty \cup_{m=n}^\infty A_m$. And with this definition, the limit of sets is the emptyset, which has 0 elements.