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2h
revised $x^2+y^2+9=3(x+y)+xy$ Find all pairs of real $x,y$ that meet this equation
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2h
comment $x^2+y^2+9=3(x+y)+xy$ Find all pairs of real $x,y$ that meet this equation
You didn't consider if $y=3$, which puts $\sqrt{\Delta}\in\mathbb{R}$.
2h
answered $x^2+y^2+9=3(x+y)+xy$ Find all pairs of real $x,y$ that meet this equation
3h
answered Proving $ab(a+b)+ac(a+c)+bc(b+c)$ is even
13h
comment How come $32.5 = 31.5$?
It's not exactly a coincidence that these are ratios of Fibonacci numbers. That's a good way to end up with ratios that are quite close to each other. In this case, both are approximants to $\frac{3-\sqrt{5}}{2}$, the limit of $\frac{F_n}{F_{n+2}}$.
17h
revised How (the graphic of) a $\mathcal C^1$ but not $\mathcal C^2$ function looks like
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19h
answered How (the graphic of) a $\mathcal C^1$ but not $\mathcal C^2$ function looks like
1d
comment Is it true that we can get zero for all $(x,y,z)\in\mathbb{N}^3$?
@AhmedHussein Ah OK, I read you in reverse.
1d
comment Is it true that we can get zero for all $(x,y,z)\in\mathbb{N}^3$?
@AhmedHussein An explanation why this is false with only $2$-tuples: The transformation leaves the sum invariant. In order to reach $0$, the previous step requires having two equal values, and so the sum would have to be even. Thus if you start with any pair having an odd sum, for instance $\{2,3\}$, it would be impossible to reach $\{0,5\}$ or $\{5,0\}$. Note that this example $\{2,3\}$ shows there is no requirement that $y=nx$ for the $2$-tuple to be incapable of leading to $0$.
1d
answered What does it mean to attach a cell to a space by a map?
1d
revised How to caluclate the integral of $\int \frac{1}{\sqrt{4x^{2}+1}}dx$ using a trig substitution?
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1d
answered How to caluclate the integral of $\int \frac{1}{\sqrt{4x^{2}+1}}dx$ using a trig substitution?
1d
comment Can someone explain why $x^{\log(a)} = a^{\log(x)}$?
@goblin Oh yeah, of course. Just writing too fast without thinking.
1d
comment Can someone explain why $x^{\log(a)} = a^{\log(x)}$?
Incidentally, you can use this to put a field structure on $\mathbb{R}_{\geq 0}$, where the "addition" operation is given by $x\cdot y$ with identity $1$, and the "multiplication" operation is given by $x^{\ln(y)}$ with identity $e$. This fact is what makes the "multiplication" commutative.
2d
comment $\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}$ is Irrational
@AvZ Sure about that?
Jan
28
comment Show$\:\frac{1}{\left|x^2+x+1\right|}\:\ge \:\frac{1}{x^2-\left|x\right|-1}$
@kobe See my answer. It's not correct if $x^2-|x|-1$ is negative either.
Jan
28
answered Show$\:\frac{1}{\left|x^2+x+1\right|}\:\ge \:\frac{1}{x^2-\left|x\right|-1}$
Jan
28
comment Show that $\operatorname{span}(\operatorname{span}\{\vec{x},\vec{y}\}) = \operatorname{span}\{\vec{x},\vec{y}\}$
Nothing in the statement of this question suggests $\vec{x}$ and $\vec{y}$ are linearly independent. They could be equal even, and the claim is still true.
Jan
28
answered Show that $\operatorname{span}(\operatorname{span}\{\vec{x},\vec{y}\}) = \operatorname{span}\{\vec{x},\vec{y}\}$
Jan
26
comment Prove that $\log_{36} 30 $ is irrational number.
Note $\log_{36}(30)=\frac12\log_6(30)=\frac12\log_6(6)+\frac12\log_6(5)=\frac12+ \frac{1}{2}\log_6(5)$. So if it streamlines the argument at all, you could just focus on whether $\log_6(5)$ is irrational.