TITLE: Let $U=\{p \in \mathcal{P}_4(\mathbb{R}):p''(6)=0\}$. Find a basis of $U$
QUESTION [0 upvotes]: Let $U=\{p \in \mathcal{P}_4(\mathbb{R}):p''(6)=0\}$. Find a basis of $U$. I just have a question about how to show that the list that I found is a basis.
Here is what I found:
Let $p \in U$, then $p(x)=a+bx+cx^2+dx^3+ex^4$ with $a,b,c,d,e \in \mathbb{R}$. We determine now $a,b,c,d,e$ using second derivative of $p$ at $6$.
We have that $p''(6)=2c+36d+432e=0 \iff c=-18d-216e$. Thus, each polynomial $p$ of $U$ is of the form: $p(x)=a+bx+d(x^3-18x^2)+e(x^4-216x^2)$. So, the list $(1,x,x^3-18x^2,x^4-216x^2)$ might be a basis of $U$. $(1,x,x^3-18x^2,x^4-216x^2)$ is clearly linearly independent, since they all have different degrees.
But, I don't really see how to show that each $p \in U$ is linear combination of $(1,x,x^3-18x^2,x^4-216x^2)$. Does it follow directly after I found $a,b,c,d,e$ using second derivative?

REPLY [1 votes]: Your argument is fine: Form $p(x)=a+bx+d(x^3-18x^2)+e(x^4-216x^2)$ you can deduce that ever polynomial in $U$ is a linear combination of your basis; and the linar independence follows form the degree argument.
Another approach is to identificate every polynomial of degree less or equal than 4, with a vector in $\Bbb R^5$ though its coefficientes:
$$a+bx+cx^2+dx^3+ex^4+fx^5\in{\cal P}_4(\Bbb R) \longleftrightarrow (a,b,c,d,e,f)\in\Bbb R^5.$$
So you can take your basis of $U$, convert into $\Bbb R^5$ vectors and try to show that the range of the matrix formed by these vectors has range 4.


TITLE: Can grounding wrist prevent electrical shock?
QUESTION [0 upvotes]: Today while repairing some electronics and wearing anti-static wristband like on picture below I got thinking - in theory, would it be possible to avoid shock while working on mains by grounding yourself using such antistatic wristband / some copper wire wrapped around your wrist?
In theory I think such band / wire should give less resistance than body to ground and it should make electrical current go through it to ground rather than through whole body to feet, but maybe I'm wrong? Also I know that anti-static wristband probably wouldn't be suitable for higher voltages / currents but it's used as proof of concept, below I also sketched what I'm talking about to clarify

yellow = mains voltage e.g. 230v AC
brown = ground
green = grounding wire running from wrist to ground
black = test specimen
We assume that in both cases on the left and right test specimen is touching ground with his both feet even though on picture it doesn't look like it and that wire is making perfect contact with his wrist and ground, and "prevent shock" means that electrical shock effects would be slightly reduced and current flowing through whole body would be much smaller

REPLY [3 votes]: No, it will not protect you and might even put you in greater danger.   A grounding strap or wire provides a path from the high voltage source to ground that passes through your body. That's the opposite of what you want for safety.  The grounding wire in an outlet or appliance provides a path to ground from the metal housing, directing any current AWAY from the user.
The reason grounding straps are used is to siphon away any charge or voltage that builds up on the person, so they do not become the high voltage source that ruins an electronic part.  It's a totally different situation and hazard being prevented.
I echo the stern safety warning of the previous reply – if you have to ask this question, you are NOT even remotely qualified to work on electrical equipment.  Hopefully this was just an academic question.


TITLE: $\iiint_V \ x^{2n} + y^{2n} + z^{2n} \,dx\,dy\,dz$
QUESTION [6 upvotes]: $$\iiint_V \ x^{2n} + y^{2n} + z^{2n} \,dx\,dy\,dz$$
where V is the unit sphere.
No information is given about n but I assume it is an integral. All I could think to do was to convert to spherical co-ordinates and use reduction formulae, but I ended up with really messy answers. Any help would be brilliant.

REPLY [2 votes]: Following runaround's excellent hint, the required integral is
$$3\int_{-1}^1 z^{2n}\hskip-1.4cm \underbrace{\int\int dx\,dy}_{\mbox{area of disk of radius }\sqrt{1-z^2}} \hskip-1.3cm dz=3\pi \int_{-1}^1 z^{2n}(1-z^2)\,dz={12\pi\over(2n+3)(2n+1)}.$$