Stellar Number Series- Mathematical Series

[Dr. Oza]

Hi guys, I had some problem with my Math IB work. I can't find a general statement for the stellar series. I tried using a series of quadratic equations and trying to discern a pattern but it didn't work out. Can anyone help me?

I specifically need help with number 4 (circled)

[IMG][/IMG]

[Sphere]

121 , 181


Solution If you want it.

Spoiler Alert, click show to read: 

S(n) = 6n^2 - 6n + 1

[John Doe]

Try to decompose your problem.

1) find a patern for how many dots there are in the new star only
2) S(n) is the addition of all the dot in this and the previous stars.

little bit more help

Spoiler Alert, click show to read: 

New star,
U₁=1
U_n= function of n

S_n= U₁+U₂+....+U_n
and somehow you should find the n(n+1)/2=1+2+3+...+n pattern hidden within S_n and find the solution given by Sphere.

[Dr. Oza]

Okay, thanks for the input.. i actually found all this.. Using recursive triangular numbers and quadratic expressions.

I've written a 18-page report on this. But i still can't seem to come up with an explanation of "why" it works. My teacher said that aside from an algebraic explanation and the general statements, there should be a geometric reason. Is he crazy?

Also, is anyone willing to look at my 18-page discussion/project ?

[Sphere]

Absolutely not.

Just share the general stellar number series in terms of "p" and "n". I am curious as what it actually is.

I can not but assume you butchered it's beauty if you went on for 18 pages about it. Math's elegance is in its simplicity, and so I am weary of anything that takes over a page.

[Dr. Oza]

Well, it could have been 8-10 (which most peoples' are) but i went out of my way to make my mathematical work clear, maybe too much.

I also basically did the main problem twice. Once via system of equations and once using this:

The following general statement seemed to hold true, with T_(n-1) indicating one triangular term less than n.

However, tS_n=1+(T_(n-1)*12)

I recalled the general statement for the n^(th) triangular number I discovered earlier:

S_n=(n^2+n)/2


To avoid confusion, I used the term T_n instead:

T_n=(n^2+n)/2

However, the value I need is the T_(n-1) value, thus:

T_(n-1)=(〖(n-1)〗^2+(n-1))/2

T_(n-1)=(n^2-2n+1+n-1)/2

T_(n-1)=(n^2-n)/2

I substituted T_n into the S_n for the stellar numbers:

S_n=1+(T_(n-1)*12)

S_n=1+(((n^2-n)/2)*12)

S_n=1+((12n^2-12n)/2)

S_n=1+(6n^2-6n)

S_n=6n^2-6n+1

[Sphere]

Excellent.

This pleases me.

[Agent Miles]

Don’t be too pleased yet.

If n has a value of zero (still part of geometry the last time I checked), then the solution should be no points. However, the equation gives a result of 1, which is invalid. So you should also stipulate that this is for all values for n greater than zero.

[Sphere]

I am a chocoholic, only with booze.

You are a grammer Nazi, only with math.

[Gaidin]

You are a grammer Nazi, only with math.

Welcome to real math. Stipulations are a part of the game.

[John Doe]

The following general statement seemed to hold true, with T_(n-1) indicating one triangular term less than n.

That's the part I'm most concerned with.... I would have got a F^- for this.

[Mathias]

That's the part I'm most concerned with.... I would have got a F^- for this.

Yeah, it's not the Riemann hypothesis you are talking about. The only times you are allowed to express yourself in that way in mathematics are in two cases:
i) "Testing it for a couple of numbers, it seems to hold true. Lets try to prove it.".
ii) "This seems to hold true, but it is ridiculously hard to prove/I don't room for a proof in the small space I have alotted for proofs, so I'll just leave it as conjecture so people can bang their heads against the wall trying to prove it for some hundred years"

[Nikitn]

You see that the difference added between each star = d*(n) = 12(n) ? You could make an arithmetic series out of that, and then make an expression of its sum. Then you can add that sum to S1=1, and you're all set.

Edit: I'll solve it. Use the formula for the sum of an arithmetic series, (S1+Sn)*n/2 = [S1+(S1+d*(n-1))]*n/2

[12+(12+12*(k-1))]*K/2 + S1 = [24-12+12k]*K/2 + S1 = 6k +6k^2 +1= 6(k+k^2)+1. We see that this expression is one step too high, ie when K=0 the expression equals S1, when K=1 the expression equals S2, when K=2 the expression equals S3 etc. So we set K=(n-1).

6(n-1+(n-1)^2)+1 = 6(n-1+1-2n+n^2)+1 = 6(-n+n^2)+1. <-- Here is the correct formula. You _REALLY_ don't need 10 pages to explain this.

The geometric reason as to why it works is also added in my formula. Each new star which gets added 2 the figure is 12 points bigger than the last one.

[Agent Miles]

Great, we all agree that this is the correct equation (most of the time). In square and triangular numbers the solutions also work for n=0, but this solution does not. I wonder why?

We have not found a solution for:

“(8) Hence, produce the general statement, in terms of p and n, that generates the sequence of p-stellar numbers for any value of p …”

Thus for (9) it doesn’t work for “black holes” (n=0).

(10) Why not?

Something like, the solution favors odd numbers (the star-shaped geometries), which is a problem with zero being technically an “even” number.

[Nikitn]

Why isn't the formula correct for n=0? ie why does S0 have to equal 0, and not 1?

Heck, shouldn't the formula be correct for all whole numbers (.....-3,-2,-1,0,1,2,3....), anyway?

[Agent Miles]

It’s the numerical solution for a “stellar” geometry. A geometrical representation of 0 is not one point. The solutions for squares and triangles in the OP validate this (when n=0 the solution is also 0), but the stellar solution does not.

[John Doe]

Yeah, it's not the Riemann hypothesis you are talking about. The only times you are allowed to express yourself in that way in mathematics are in two cases:
i) "Testing it for a couple of numbers, it seems to hold true. Lets try to prove it.".
ii) "This seems to hold true, but it is ridiculously hard to prove/I don't room for a proof in the small space I have alotted for proofs, so I'll just leave it as conjecture so people can bang their heads against the wall trying to prove it for some hundred years"

i) like using recurrence , or just stating that adding a spot on 12 segments would add 12 spots to the new star compared to the previous one. True, it's easy enough but it must be done. May be the teaching of math changed in the last few years, but "it's so because it looks so" didn't go down well with any of my teachers.
ii) like an axiom? If you are doing a Phd and your professor agrees with you, then yes.

Why isn't the formula correct for n=0? ie why does S0 have to equal 0, and not 1?

Heck, shouldn't the formula be correct for all whole numbers (.....-3,-2,-1,0,1,2,3....), anyway?

You have to solve the problem that is given to you, S is only defined from index "1", so there should be a reminder in the formula about the starting point (n>=1 or n>0) as pointed earlier. You only solve the equation for all the index of the members of the serie, not for the value of any integer n. Instead, if the problem said that S₀ was the single dot, or you study U_n=S_n+1, that wouldn't be much of an issue with n=0 (it would be a different formula in the solution obviously), but you still have to define n as representing the value of the index (n>=0 for U).

Usually series start either at 0 or 1 to simplify the equation, but you can start anywhere you want and you'll even study series with S_f(n) (and later S_{(f(n),g(p),h(q),...)})

[Agent Miles]

http://wiki.answers.com/Q/What_is_the_general_statement_of_6-stellar_numbers

Apparently this is the correct solution, so I guess I’m the only one who sees a problem with it.

As the saying goes, “Genius is when you look at something that everyone else has looked at, and you see something that no one else has seen.”

[John Doe]

http://wiki.answers.com/Q/What_is_the_general_statement_of_6-stellar_numbers

Apparently this is the correct solution, so I guess I’m the only one who sees a problem with it.

As the saying goes, “Genius is when you look at something that everyone else has looked at, and you see something that no one else has seen.”

Nikitn is right,

In S_n, n isn't really a number, it's an index, or the position of an element within a family (like the n^th position in an array). It just happens that in most math problem, the value of that element is a fonction of its position. In this problem, S₁ is defined as the first element of the family, and not S₀ (S₀ does not even exist). And the n^th element of this family will have the value 6n²-6n+1.

If you start at S₀, then S_n will be the (n+1)^th element of the family (since S₀ is the first). and S_n=6(n+1)²-6(n+1)+1

[Agent Miles]

“In this task you will consider Geometric shapes that lead to special numbers.”

I submit that in the first two examples, squares are represented by n^2 and triangles by (n^2+n)/2. These are valid generalizations that actually work “for all values of n” to include n=0 when no shape/dot is present. These would not be generalizations if they did not work for all values. (You cannot have a negative number of dots, but you can have no dot.) S is merely the function in each case that renders the number of dots. It is not described as a function that creates these shapes via a matrix of elements.

The stellar shape is a contrivance. The statement “…the first four representations for a star with six vertices…” is just wrong. Since when does one single dot have six vertices? A moment ago, a single dot was a square or a triangle. It also does not work for n=0, when logically no dot should be present. This is the limitation of the general statement about stellar shapes. This is some kind of universal problem that people are fed verbatim and apparently are required to answer with blinders on.