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I am learning mathematics to help further mathematical development in my country.


Jun
11
comment $\frac{|| \overline{AM}||}{|| \overline{AB}||}=\frac{|| \overline{AN}||}{|| \overline{AC}||}=\frac{|| \overline{MN}||}{|| \overline{BC}||}$
Can you expand a little more on how did you get these steps? I'm a little lost.
Jun
11
comment $\frac{|| \overline{AM}||}{|| \overline{AB}||}=\frac{|| \overline{AN}||}{|| \overline{AC}||}=\frac{|| \overline{MN}||}{|| \overline{BC}||}$
I didn't get why you assumed that $\overrightarrow{AM}+\overrightarrow{AN}+\overrightarrow{MN}=0$.
Jun
11
revised $\frac{|| \overline{AM}||}{|| \overline{AB}||}=\frac{|| \overline{AN}||}{|| \overline{AC}||}=\frac{|| \overline{MN}||}{|| \overline{BC}||}$
added 9 characters in body
Jun
11
comment $\frac{|| \overline{AM}||}{|| \overline{AB}||}=\frac{|| \overline{AN}||}{|| \overline{AC}||}=\frac{|| \overline{MN}||}{|| \overline{BC}||}$
@Duncan Show that $\frac{|| \overrightarrow{AM}||}{|| \overrightarrow{AB}||}=\frac{|| \overrightarrow{AN}||}{|| \overrightarrow{AC}||}=\frac{|| \overrightarrow{MN}||}{|| \overrightarrow{BC}||}$.
Jun
10
revised $\frac{|| \overline{AM}||}{|| \overline{AB}||}=\frac{|| \overline{AN}||}{|| \overline{AC}||}=\frac{|| \overline{MN}||}{|| \overline{BC}||}$
added 41 characters in body
Jun
10
asked $\frac{|| \overline{AM}||}{|| \overline{AB}||}=\frac{|| \overline{AN}||}{|| \overline{AC}||}=\frac{|| \overline{MN}||}{|| \overline{BC}||}$
Jun
10
revised Is Courant's Introduction to Calculus and Analysis still up-to-date?
added 4 characters in body
Jun
10
asked Is Courant's Introduction to Calculus and Analysis still up-to-date?
Jun
9
awarded  Nice Question
May
31
accepted Why to see that $\overline{B}(x;r)$ is closed if it was just defined?
May
30
awarded  Popular Question
May
28
comment On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$?
What about the choice for $a$ in $|a_n-a|<\epsilon$? In some sequences, I can find an $a$ intuitively, I'm not sure if there is an algorithm to find it. I've tried to do the following: $$|a_n-a|<\epsilon\\-\epsilon < a_n-a< \epsilon\\-\epsilon -a_n<-a<-a_n +\epsilon\\\epsilon +a_n>a>a_n -\epsilon$$ But I don't know how to proceed from here - presuming that this is a way to do it.
May
28
accepted On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$?
May
27
awarded  Popular Question
May
27
asked Why to see that $\overline{B}(x;r)$ is closed if it was just defined?
May
26
revised On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$?
edited title
May
26
comment On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$?
So basically I must look for some $\epsilon$ that $\forall\epsilon\gt 0,\exists N\in\Bbb N :\forall n\ge N:|a_n-1|\lt\epsilon$ does not hold? The question seems to be utterly trivial, but I was thinking that some algorithmic process should be applied in order to automatically obtain those $\epsilon$'s.
May
26
revised On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$?
edited body
May
26
asked Are there functions that converge to $P$ when $f:\mathbb{N}\to\mathbb{R}$ and to $Q$ when $f:\mathbb{R}\to\mathbb{R}$ with $Q\neq P$?
May
26
asked On the thought process for choosing $\epsilon$'s to check the convergence of $(-1)^n$?