221 reputation
18
bio website tedyin.com
location Chengdu, China
age 20
visits member for 1 year, 11 months
seen Aug 1 at 6:12

An IT enthusiast, GNU/Linux-lover and arty idiot.


Aug
9
awarded  Popular Question
Apr
19
awarded  Popular Question
Sep
25
awarded  Yearling
Feb
5
comment What symbol should I invoke in order to present the angle between two vectors?
It's just a matter of notation, since we don't like to write the words "the angel between vector a and b" over and over again in our claims.
Feb
5
accepted Calculate $\lim\limits_{x \to 0}{\frac{||a + xb|| - ||a||}{x}}$.
Feb
5
asked Calculate $\lim\limits_{x \to 0}{\frac{||a + xb|| - ||a||}{x}}$.
Feb
5
comment What symbol should I invoke in order to present the angle between two vectors?
I know that symbol. But it seems to be used in the occasion where an angle is made by three points or just to use an alphabet to represent an angle without indicating form of the angle.
Feb
5
asked What symbol should I invoke in order to present the angle between two vectors?
Jan
27
revised Prove that $\lim\limits_{x\rightarrow+\infty}\frac{x^k}{a^x} = 0\ (a>1,k>0)$
added 123 characters in body
Jan
27
comment Prove that $\lim\limits_{x\rightarrow+\infty}\frac{x^k}{a^x} = 0\ (a>1,k>0)$
You're right. But this problem expects some kind of way that use basic properties and definition of limits.
Jan
27
comment Prove that $\lim\limits_{x\rightarrow+\infty}\frac{x^k}{a^x} = 0\ (a>1,k>0)$
Note that x is not necessarily a integer. You've just proved it in sequence situation.
Jan
27
comment Prove that $\lim\limits_{x\rightarrow+\infty}\frac{x^k}{a^x} = 0\ (a>1,k>0)$
@Clayton Sorry, but that rule is not allowed here. > <
Jan
27
asked Prove that $\lim\limits_{x\rightarrow+\infty}\frac{x^k}{a^x} = 0\ (a>1,k>0)$
Oct
14
comment Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$
I'm trying to do this. But how can I bound $(f(x)A)^{\frac{1}{3}}$?
Oct
14
comment Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$
@EuYu Yes. Because this problems comes from the exercises of my textbook, I suppose that I should prove it without proving more general situation (or this specific question seems to be useless). Thanks for the material. :)
Oct
14
comment Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$
Well, I'd like to use epsilon-delta definition to prove that. (Requirement for homework)
Oct
14
revised Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$
added 93 characters in body
Oct
14
comment Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$
@AustinMohr Sorry, I forgot to add a crucial condition. That's my fault.
Oct
14
revised Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$
added 93 characters in body
Oct
14
comment Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$
Excuse me, but I wonder why you downvote?