Infinite polygons

[Mobius]

Finding limits of infinite series is possible but I don't know how to calculate the limit of a multiplicative series.

 

I put the problem I have on a website including the background to the problem, would be interested in any insights or alternative ways of doing the sum....

 

http://www.geocities.com/mobiusmaths/infpoly.htm

[matt grime]

the product

 

(1+x_1)(1+x_2)(1+x_3)...

 

converges if and only if the series sum x_i converges.

 

or in this case the product (y_1)(y_2)(y_3)...

 

converges if and only if the series sum(y_i - 1) converges so translate it to a sum, what is the series?

[Mobius]

The series is attached.

series.doc

[Mobius]

Ooops sorry in the attachement it has from n=1 to inf, it should be from n=3 to inf...

 

[math]R=\prod \limits _{3}^{\infty }{1\over sin({(n-2)\times 180\over 2n})}[/math]

[matt grime]

that isn;t a series, that is product. you were supposed to work out the series to sum and see if it converged thus telling you if the product converged.

[Mobius]

Not so good at my limits it seems...

 

On your logic is this the series my product changes to?:-

 

[math]\lim \limits {1\over sin({(n-3)\times 180\over 2(n-1)})}[/math]

 

Surely this is an infinite sequence as the angle is always less then 180 degrees therefore the sign of the angle is always positive and between 0 and 1 (it would be quickly approaching 1) and therefore 1/this number will be greater than 1, so the infinte series diverges!

 

The computer seems to agree with this also.

 

Is there no way to work out the original limit of?:-

[math]\prod \limits_{3}^{\infty}{1\over sin({(n-2)\times 180\over 2n})}[/math]

 

The computer says it does diverge to 8.7

 

As this problem is based on increasing regular polygons it stands to reason that the distance between consecutive circles gets shorter and shorter and should converge.

[matt grime]

your first piece of latex is once more not a series, and you even then call it a sequence, and no that isn't what i said the terms of the series ought to be. secondly things can't diverge to 8.7 that would be a typo, tough.

 

 

read it agina carefulyl:

 

prod(1+x_i)

 

converges if

 

sum x_i

 

converges

 

notice that the terms in the product are different from the ones in the sum.

 

i#ve no idea if the associated sereis converge.

[Mobius]

Yeah typo: I meant the second eqn. converges to 8.7.

 

Not to worry I will put some thought into your points over the next few days as I am not sure how to convert it to a series (I just took 1 from the two n's in the equation).

 

The first equation is the limit (from n =4 to infinity) of the sum of the equations (just didn't come out that way).

 

Cheers for the help...

[matt grime]

why did you take 1 off the n's? I told you to take 1 from each term to go from y_i to Y_i-1. is the link between the terms of the series and the product not clear frmo the typesetting? we go from the product (1+x_1)(1+x_2).... to the series x_1+x_2+x_3... so going back subtract one off the terms, not the indices.

[Mobius]

Ok here we go...

 

Subtracting 1 from the original terms gives...

 

[math]\sum_{3}^{\infty}({1\over sin({(n-2)\times 180)\over 2n})}-1)[/math]

 

Not sure of how to get the limit of the series but I can find the limit of the expression as n approaches [math]\infty[/math]

 

multiplying out the sin expression:-

 

[math]\lim (n\rightarrow \infty)({1\over sin({180n-360\over 2n})}-1)[/math]

 

[math]\Rightarrow \lim (n\rightarrow \infty)({1\over sin({180-{360\over n} \over 2})}-1)[/math]

 

Now as n approaches infinity the 360/n term approaches 0.

 

[math]\Rightarrow ({1\over sin(90)}-1)[/math]

 

[math]\Rightarrow 1-1=0[/math]

 

Therefore the limit series does indeed converge.

 

Is this the idea are have I made another mistake?