Reputation
10,393
Top tag
Next privilege 15,000 Rep.
Protect questions
Badges
2 15 61
Impact
~200k people reached

33m
answered If $V$ is a vector space, then, proving that…
2h
comment If $f(x)\ll1$ is it safe to assume that $f^{\prime}(x)\ll1$?
It depends on what you mean by "behaves like $e^{-x}$". If you mean it monotonically decreases and does not pass zero, then you are actually imposing conditions on $f'$. "Monotonically decreases" == $f'(x) < 0$, and "doesn't cross zero" implies $f'(x) \to 0$ if $\lim f'(x)$ exists. Even in this situation though, it is possible to make $f'$ have spikes arbitrarily far out to make $\lim f'(x)$ not exist.
3h
comment If $f(x)\ll1$ is it safe to assume that $f^{\prime}(x)\ll1$?
In general, there is no bounding condition you can put on $f$ (besides demanding $f=0$) in order to bound $f'$. What you can say is that $f'(x)$ is small, then $f(x)$ is small, but you cannot go the other way around.
3h
comment If $f(x)\ll1$ is it safe to assume that $f^{\prime}(x)\ll1$?
My example does have $f(x)\to 0$ as $x\to \infty$, but $f'(x)$ oscillates.
3h
answered For what values of $a$, $b$, and $c$ the above system has: One solution. Infinitely many solutions. No solutions.
4h
answered If $f(x)\ll1$ is it safe to assume that $f^{\prime}(x)\ll1$?
4h
comment If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$?
@user103816, you are thinking about it too hard. I did not show that first setting $u=v$, then setting $v=w$, then setting $w=x$ will somehow increase the product three times and be maximal. Instead, start with the maximum $(u,v,w,x)$. Is it possible that $u\neq v$? No, then I could increase the product. Is it possible that $v \neq w$? No, then I could increase the product. Is it possible that $w \neq x$? No, then I could increase the product. Thus they must all be equal.
19h
answered Numerical Integration by Undetermined coefficients
23h
comment the probability density function (PDF) of concatenation of two Gaussian variables
This only holds if $x,y$ are jointly normal, which essentially by definition implies $z$ is normal.
23h
comment What maths courses are needed for Machine Learning
class.coursera.org/ml-005/lecture May be of interest to you.
23h
comment Let $(X,Σ,μ)$ be a measure space and $f$ and $g$ are positive integrable functions and $h=f-g$
@Upgrade, the statement that $f,g$ are integrable includes the fact that $f,g$ must be measurable.
1d
comment If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$?
@user103816 That is not what my proof says. My proof shows that for $(u,v,w,x)$ if some two of them are not equal, then $uvwx$ is not maximized under the given constraint. The statement "$uvwx$ is maximized" implies that the hypothesis of my implication is false, i.e. "not (some two of $(u,v,w,x)$ are not equal))" holds. Expanding this we see it is equivalent to "every two of $(u,v,w,x)$ are equal", i.e. $u=v=w=x$.
Jul
1
comment If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$?
@user103816. By compactness, the maximum must exist. Now suppose that the max occurs with $(u,v,w,x)$. If $u\neq v$, then my proof shows that I could do strictly better, which would contradict that $uvwx$ is a maximum. Thus $u=v$. Similarly, if $v \neq w$, the proof would show $uvwx$ is not a max, so it must be that $v=w$. Finally, if $w \neq x$, again $uvwx$ wouldn't be a max, so it must be that $w=x$. Thus $u=v=w=x$.
Jun
30
comment If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$?
@GEdgar, my symmetry comment is meant to show the reader where I got the proof idea from. It is intuition, not a proof. Thankfully I followed it up with an actual proof. Notice that if my intuition had said "minimum" instead of "maximum" the $>$ would have to be a $<$ in the string of iffs, and I would not have been able to complete the proof.
Jun
30
comment If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$?
@user103816, the variables go from $(u,v,w,x)$ to $(\frac{u+v}{2},\frac{u+v}{2},w,x)$. The variables $w$ and $x$ stay exactly the same, they do not decrease. You can check that $5=u+v+w+x = \frac{u+v}{2}+\frac{u+v}{2}+w+x$ so there is no need to change $w,x$. I'm not claiming that this new set of vars is the maximum, I'm just showing that unless $u=v=w=x$, then the vars you chose are not a maximum.
Jun
29
answered If $a+b+c+d=1$ then why is the maximum value of $(a+1)(b+1)(c+1)(d+1)$ is ${\left(\frac{5}{4}\right)}^4$?
Jun
29
revised definition of limsup of a function
deleted 2 characters in body
Jun
25
answered Joint PDF of two random variables with Uniform Distribution
Jun
16
comment What does it mean that a sine wave is unchanged when added to another sine wave?
@AnonSubmitter85 Yes that is correct. This only applies for functions $f:\mathbb{R}\to\mathbb{R}$, otherwise the "change the phase" condition wouldn't make any sense.
Jun
14
awarded  Popular Question