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visits member for 3 years, 5 months
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Aug
30
comment Question about proof of Cauchy-Schwarz inequality.
I see, the usage of the unit vector is kind of like justification for normalization, ok, ok. Sorry, I didn't stare at it hard enough I guess.
Aug
30
comment Question about proof of Cauchy-Schwarz inequality.
O wait, I think I understand...
Aug
30
comment Question about proof of Cauchy-Schwarz inequality.
When I read this proof, it makes sense. But, the original proof doesn't use the properties of linearity to prove the inequality. On one hand, you have confirmed beyond doubt that the proof has confirmed beyond doubt that it must work in the interval $(0,1)$, but by doing something (that feels) different altogether. Is there way to justify the current outline of proof? Sorry if what I am saying doesn't make sense.
Aug
30
asked Question about proof of Cauchy-Schwarz inequality.
Jul
7
comment Calculate $\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$
I like how the top two answers, at least at the moment, are virtually identical...
Jul
6
suggested rejected edit on Limits of $s(x)$ and $H(x)$
Jul
6
comment Mathematical Induction Problem with Fraction
@Kekker: So since $16 \neq 17$ does this equation hold for $n=2$? And what can you say about all $n$?
Jul
2
awarded  Curious
Jun
30
awarded  Popular Question
Jun
30
comment Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.
I don't know if editing this would change the intent of this question... but, $\lim_{x \to 0} \frac{\sinh x}{x} \neq 0$
Jun
30
comment Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.
\sin hx should be \sinh x... :)
Jun
29
comment Draw a function $g$ such that $g$ is defined on the interval $[-9,9]$
Are you sure the question said: $lim_{x\rightarrow 5} g(x)$ does not exist? Are you sure it isn't supposed to be $-5$?
Jun
28
comment Find $G'\left( x\right)$.
You don't want G^', you just need G'... That will stop the latex error.
Jun
28
comment How to interpret integration of a discontinous function
I believe it has to do with (informally speaking): Rationals are countably infinite, irrationals are uncountably infinite... Hence, the point discontinuities of the rationals are insignificant in the end. Which is why the integral is zero. Someone correct me if I'm wrong.
Jun
28
comment When are there no critical points?
@Gahawar: Wouldn't that function have a critical point everywhere? Since there exists no derivative?
Jun
28
revised Why is the function integrable?
added 2 characters in body
Jun
28
revised Why is the function integrable?
added 6 characters in body
Jun
28
answered Why is the function integrable?
Jun
27
comment What do I not understand about one-to-one functions?
@JBKing: Sorry I meant -1... Typed it too quickly.
Jun
27
accepted What do I not understand about one-to-one functions?