# Math Challenge - October 2019

• Challenge
• fresh_42
• Featured
In summary, this conversation covers various questions and solutions related to mathematics, including proving inequalities, determining the maximum value of a sum of distances, calculating probabilities, and finding the center points of chords on a sphere. It also discusses Jensen's inequalities and their usefulness in functional analysis and non-equilibrium statistics.
fresh_42 said:
Yes and no. Yes, as this is part of the question, and no, since this procedure has to be done with any second point. We have a fixed single point ##P## and a set ##\mathbb{S}^2##. Then we consider all lines ##\overline{PQ}## from ##P## to ##Q## to another point ##Q## on the sphere. This line has a center ##M_Q##. The problem asks for ##\{\,M_Q\in \overline{PQ}\,|\,Q\in \mathbb{S}^2\,\}\,.##
Like this?
Supposing that ##P=(x,y,z)##, the set of all midpoints between ##P## and ##Q## by varying ##Q## would be ##\{(\frac{x+x_Q}{2},\frac{y+y_Q}{2},\frac{z+z_Q}{2})|(x_Q,y_Q,z_Q)\in\mathbb{S}^2\}##.

archaic said:
Like this?
Supposing that ##P=(x,y,z)##, the set of all midpoints between ##P## and ##Q## by varying ##Q## would be ##\{(\frac{x+x_Q}{2},\frac{y+y_Q}{2},\frac{z+z_Q}{2})|(x_Q,y_Q,z_Q)\in\mathbb{S}^2\}##.
Yes, but which equation describes those points? What is it what you have written down?

Question 14
Isn't this just a scaling of the original sphere? So this would be a sphere at point P whose slope at P is tangent to the original sphere and has radius half of the original sphere.

Okay so I thought about an equation but for the love of me I cannot seem to translate the sphere into the correct position to be a tangent to P.

Last edited:
lekh2003 said:
Question 14
Isn't this just a scaling of the original sphere? So this would be a sphere at point P whose slope at P is tangent to the original sphere and has radius half of the original sphere.

Okay so I thought about an equation but for the love of me I cannot seem to translate the sphere into the correct position to be a tangent to P.
Well, it's not a scaling anymore if the center is shifted. Your description is right, as is the set which @archaic defined. What I wanted to see is a simple transformation from one description to the other one. As you both have basically solved it, let me give the solution I had in mind. (Please note the two directions! The centers are on the smaller ball, and all points of the smaller ball are indeed a center of some chord!)

The variable endpoint ##X## of the chord is on the sphere, so for its position vector we have ##\vec{x}^2=r^2##. The position vector of the center of the chord ##\overline{PX}## is thus
$$\vec{c}=\dfrac{\vec{p}+\vec{x}}{2} \;\Longleftrightarrow \; \vec{x}=2\vec{c}-\vec{p}$$
hence ##r^2=(2\vec{c}-\vec{p})^2## or ##\left( \vec{c}-\dfrac{\vec{p}}{2} \right)^2=\dfrac{r^2}{4}##.

So the set of points we were looking for are all on a sphere with center ##\overline{OP}/2=\vec{p}/2## and radius ##r/2##. All points of this sphere are on the other hand a center of some chord of the original sphere with endpoint ##P##, since we can go back. The point ##P## itself is the center of the chord ##\overline{PP}##.

@fresh_42 ahh, thanks for the explanation, that makes sense.

4 challenge problems left, who can solve them?

New ones on Friday!

Greg Bernhardt said:
4 challenge problems left, who can solve them?

New ones on Friday!
I would try, but i don't fancy speed learning topology right now :P

fresh_42 said:
5. Let ##A=\sum_{k=0}^\infty a_k\, , \,B=\sum_{k=0}^\infty b_k## be two convergent series one of which absolutely.
Prove: The Cauchy-product ##C=\sum_{k=0}^\infty c_k## with ##c_k=\sum_{j=0}^ka_jb_{k-j}## converges to ##AB##.
Give an example that absolute convergence of at least one factor is necessary.
This is The Theorem of Mertens, Knopp, Theory and Applications of Infinite Series, pg 321, 188.
An example when absolute convergence of at least one factor is necessary is the other challenge problem I just did, the Cauchy product ##\left(\sum_{n=0}^\infty \frac{(-1)^n}{\sqrt{n+1}}\right) ^2##.

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