## How to prove this statement of the geometric series

If we have $$sum_ {n = 0} ^ { infty} r ^ {i}$$, that converges towards $$S = frac {1} {1-r}$$, shows CA $$S – s_n = frac {r ^ {n + 1}} {1-r}$$, or $$s_n$$ is the sum of the first n terms.

My approach,

$$S – s_n = frac {1} {1-r} – frac {1-r ^ n} {1-r} = frac {r ^ n} {1-r}$$

So, there is no $$r ^ {n + 1}$$ in my answer as in the question. If we have $$s_n$$, that means the sum of the first $$n$$ terms, so there should not be any $$n + 1$$.

Where am I wrong?

## geometric measurement theory – Hausdorff dimension of the set of levels of the basic function of Conway 13

Yesterday I discussed the function of Conway's base 13 (https://en.wikipedia.org/wiki/Conway_base_13_function) (and the fancy properties it has). During this discussion, the other person explained that he represented the sets of levels as fractals. This made me curious and motivated the following question:

Do we know what is the Hausdorff dimension of the level sets of Conway's base 13 function?

If you're worried that it's not measurable (that is), see is Conway's base 13 function measurable?

## sequences and series – In a geometric progression, \$ S_2 = \$ 7 and \$ S_6 = \$ 91. Estimate \$ S_4 \$.

In a geometric progression, $$S_2 = 7$$ and $$S_6 = 91$$. Assess $$S_4$$. Alternatives: 28, 32, 35, 49, 84.

Here is what I have tried until now:

$$S_2 = frac {a_1 (1-r ^ 2)} {1-r} implies 1-r = frac {a_1 (1-r ^ 2)} {7} \ S_6 = frac {a_1 (1-r ^ 6)} {1-r} implies 1-r = frac {a_1 (1-r ^ 6)} {91}$$

Then:
$$frac {1-r ^ 2} {1} = frac {1-r ^ 6} {13} \ r ^ 6 – 13r ^ 2 + 12 = 0$$

Now, I can not solve this equation, there may be an easier way …

## Ag.algebraic geometry – Reading geometric properties of an appropriate scheme from its refinement

Let $$k$$ to be a field, $$X rightarrow mathrm {Spec} , k$$ to be an appropriate morphism.

• Yes $$X$$ is geometrically reduced, so $$mathcal {O} _X (X)$$ is the direct product of finely separable finite extensions of $$k$$.
• Yes $$X$$ is geometrically connected, then $$mathcal {O} _X (X)$$ is an irreducible geometrically $$k$$-algebra.

Are the reverse statements true?

## UMVUE geometric distribution

What is the UMVUE of $$p ^ 2$$ for geometric distribution?

## geometry – Law of parallelogram Geometric proof

I've therefore approached the parallelogram law in many ways over and over again now. And algebraically (or maybe I mean arithmetically) it seems perfectly logical to me – I can prove it, understand it and accept it. However, most authors use a representation of a parallelogram with diagonals or drawn vectors, like that of Wikipedia:

This makes sense to me, but in the end, the parallelogram law connects the squares of these segments, not the segments themselves (directly), so I have no intuition to draw from this diagram. Can any one extend a geometric light on this subject without resorting to algebra (or with a minimal call)? If this is the standard proof (that is, similar to the Wikipedia proof), that does not interest me. Alternatively, if someone can show why my request is not possible, it would cool too.

Thank you.

## SQL server – Calculate the geometric mean in a SQL query

Sample table

``````Field - Value
A = 30.21
B-15.03
C = 5.06
B - 10.20
B - 12.89
C = 1.12
``````

Calculation of the geometric mean

``````A = (30.21) ^ 1/1
B = (15.03 * 10.20 * 12.89) 1/3
C = (5.06 * 1.12) ^ 1/2
``````

The results

``````A = 30.21
B = 12.23
C = 2.37
``````

I have tried something like this.

``````POWER (10.00, SUM (LOGT (CAST) (REPONSEVALVALUE DECIMAL (30,2)))), 1 / (CAST (COUNT (ANSDET.CAMPO) AS DECIMAL (10,2)))
``````

An invalid floating point operation has occurred.

I need the SQL formula to return these results averages … any ideas?

## geometry – Triangular geometric construction from 2 medians and from one side

So here's the problem: using only a ruler and a compass, geometrically build a triangle from the three given segments representing the base (side $$c$$) and the medians of the other two sides ($$m_ {a}$$ and $$m_ {b}$$).

The 3 given segments

I've looked at many websites that were trying to get help for this build, but I found only two-sided triangles and a median or two medians builds and an altitude. I know I'm supposed to start with the side $$c$$ and draw circles using the medians, but that's about it all. How can I do this?

## Geometric probability in square unit

On the opposite sides of a unit square, there are two randomly selected points that are joined to a line. This line cuts the square into two parts. What is the probability that the area of ​​the remaining part is greater than 0.3?

All I know is that this problem solving should probably include integrals, but I do not know where. How should I start with this?

## groups of links – Geometric Rubik's Cube (Smooth)

My friend and I think of a smooth analog of Rubik's cube. An idea is:

Consider the 2-dimensional sphere $$S ^ {2}$$. We choose three parameters: $$(L, H, theta)$$. Right here $$L$$ is a ray that crosses the origin, $$H$$ is a plan that crosses $$L$$, the sphere and orthogonal to $$L$$. $$theta in [0, 2pi]$$ is an angle that we will rotate the part of the sphere. Here is the photo for action:

Now, I have some questions about a group $$G$$ generated by this action on $$S ^ {2}$$.

1. is $$G$$ has a finite dimensional Lie group? If so, what is the size of the group?

2. Is there an interesting and non-trivial relationship between the elements? For example, in $$mathrm {SO} (3)$$the composition of two rotations is also a rotation. But our group clearly does not satisfy such things.

edit. As Elkies said, it is an infinite dimensional group, because it contains an infinite dimensional abelian group (generated by elements to $$L$$ and variant $$H & theta$$. But we can think of certain subgroups that seem to be finite or even finite.

Before doing this, there is a minor problem with the definition of $$G$$ that Jim Conant mentioned. I want to ignore the action that moves only measure 0 parts, like an equator.
To do this, for any two functions $$g_ {1}, g_ {2}: S ^ {2} to S ^ {2}$$ we can define an equivalence relation as
$$g_ {1} sim g_ {2} text {iff} {x in S ^ {2} ,: , g_ {1} (x) neq g {2} (x) } text {has a measure 0}$$
Now consider the group quotient by this equivalence relation, we will then have a more reasonable group. There is a reason why I think this is reasonable – there are certain subgroups I want to think about.

1. Suppose we allow only "half-spheres". We therefore consider the subgroup of $$G$$ where the plane $$H$$ should pass the origin.
Is it a finite dimensional group? I think $$mathrm {SO} (3)$$ is a codimension 1 subgroup of this half-sphere group, but I'm not sure.

2. What will be the finite subgroups of this $$G$$? There is a natural but not trivial way to build a finite subgroup of $$G$$: think real Riddles of the Rubik cube type. For example, place the Rubik's cube in the center of the sphere and bend it along the planes that correspond to the sliced ​​sides of the Rubik's cube (I hope you understand what I mean), then turn them around. pieces along the axis of the cube. In other words, it is an isomorphic subgroup to the group of Rubik's cube. We can do the same with Pyraminx, Metaminx, Dogic, Skewb or any ordinary polytope. I want to know if they cover all possible finite subgroups of $$G$$.

3. What about higher-dimensional spheres, or other highly symmetrical varieties (like a torus or hyperbolic spaces)?