Reputation
Next tag badge:
score
answers
Badges
3 33 98
Newest
 Enlightened
Impact
~1.4m people reached

23m
comment Find the width of a rectangle with an area of $x^2 -4x -12$ and the length of $x-2$
@user3491648: Give the new user a chance. This is hardly a welcoming stance, even if they want spoon feeding.
27m
comment Find the width of a rectangle with an area of $x^2 -4x -12$ and the length of $x-2$
By any chance did you mean $x^2+4x-12$?
1h
comment Is convex or non convex function?$J(u,c)=\int K(x).u.(f(x)-c)^2dx$
In the paper, the $c_1,c_2$ are fixed. In the problem above, $c$ is a parameter to $J$. If $c$ is fixed, then for each fixed $x$, the function $u \mapsto K(x) u (f(x)-c)^2$ is linear, hence convex and it follows that the integral is also convex (as a function of $u$). So, when you ask is $J$ convex, you need to specify which parameters that you are considering from a convexity standpoint. The function $x \mapsto xy$ is convex, the function $(x,y) \mapsto xy$ is not.
1h
comment Is convex or non convex function?$J(u,c)=\int K(x).u.(f(x)-c)^2dx$
I believe the Hessian of $(x,y) \mapsto x(f-y)^2$ always has a negative eigenvalue (for $y \neq f$).
2h
comment Is convex or non convex function?$J(u,c)=\int K(x).u.(f(x)-c)^2dx$
The term $(u,c) \mapsto u (f(x)-c)^2$ is not convex.
2h
comment Linear functional and Hessian
I'm not sure, I imagine any multivariable calculus text deals with partial derivatives. I found some related material in on old text I have; Marsden's "Elementary Classical Analysis".
2h
comment Is this condition on continuity extraneous or troublesome?
The function $f$ is not necessarily surjective, so, in general, you won't have $f(N) = M$. Also, you need open sets.
9h
comment What are the values of $p$ so that equation $x^3+(p-2)x^2+(5-2p)x-10=0$ has exactly $2$ real roots…
Why the downvote?
10h
comment 2 restaurants located randomly
Same idea, but you need to draw the $|x-y|\le L$ lines and compute the areas.
10h
comment 2 restaurants located randomly
Are they chosen independently? (I presume so, just confirming.)
10h
comment Little confusion about connectedness
Being not connected is more than disjoint, otherwise all sets with at least two elements would be not connected. Any open set containing $B$ must contain an open set around $(0,0)$ and hence intersect $X$.
10h
comment Little confusion about connectedness
The closure doesn't contain $C$.
10h
comment Solve for real value of $x$: $|x^2 -2x -3| > |x^2 +7x -13|$
Well, I used brute force to get the zeros, the above approach gets them 'for free'. The above is a cute trick.
11h
comment If the surface area of a box is 32 and its volume is doubled what is the new surface area?
Whoa. I have written an alternative approach to a possibly ill posed question. There were already two answers which added an additional assumption, adding yet another seems truly pointless. But downvoting because my interpretation is different to yours seems harsh, not to mention the implication of bad faith.
11h
comment If the surface area of a box is 32 and its volume is doubled what is the new surface area?
The question asks what the surface area and mentions nothing about proportions. Rather than downvoting, you might see the answer as an interesting take on the question. I am sure that you understand that 'obviously has a well-defined shape' and 'hints that the dilation is isotropic' are a little at odds? Perhaps the question was intended exactly as written?
12h
comment Solve for real value of $x$: $|x^2 -2x -3| > |x^2 +7x -13|$
+1: Smoother than mine.
12h
comment Solve for real value of $x$: $|x^2 -2x -3| > |x^2 +7x -13|$
Thanks! ${}{}{}$
12h
comment How to prove this has no real solution?
Here is a brute force approach: Find an upper and lower bound on the values of $x$. Then look at $x \in {1 \over 32} \mathbb{Z}$ in this range (hint: there are none!).
12h
comment How to prove this has no real solution?
There are only finitely many points to consider :-).
12h
comment How do you prove the commutativity of multiplication of all real numbers?
It starts with the integers, then the rationals,...