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July 1

Add Maths Project Work 2007

How do you combine an ARITHMETIC PROGRESSION & a GEOMETRIC PROGRESSION?

Formula for A.P. : T_n=a+(n-1)d

                  S_n=n/2[2a+(n-1)d]

Formula for G.P. : T_n=ar^n-1

                  S_n=a(rⁿ-1)/(r-1); for r>1
                  S_n=a(1-rⁿ)/(1-r); for r<1

Do I just combine the formulae by substituting?

What do you mean by "combine"? -- Meni Rosenfeld (talk) 00:36, 1 July 2007 (UTC)[reply]

Are you quite sure you don't mean the arithmetic-geometric mean? Otherwise, I can see no good answer. --KSmrq^T 00:45, 1 July 2007 (UTC)[reply]

I believe what he's asking is, given the sequence for instance T_n+1 = 3 + 5T_n, what is its sequence of partial sums? If each term doesn't differ from the previous by a constant amount, or a constant factor, but rather by both, how do you sum the series? Black Carrot 06:26, 1 July 2007 (UTC)[reply]

Nobody else is answering, so I'll do the best I can. The answer is actually pretty easy to find, if you're willing to do the algebra for it. Have pencil and paper ready. Let's say T₀ = c, and T_n+1 = aT_n+b. If you write out the full polynomial formulas for T₁, T₂, T₃, T₄, and so on, you should spot a pattern. In fact, each looks almost exactly like a geometric progression, with a slight hiccup at the beginning. The same trick that gives the formula for a geometric series will give you a closed form solution for T_n. You multiply the polynomial by (a-1), simplify, then divide by (a-1) and leave it as a fraction. Once you've done that for several choices of n, you should spot the pattern. Then, you want S_n, where S₀ = T₀ and S_n+1 = S_n + T_n+1. In other words,

S_{n}=\sum _{i=0}^{n}T_{i}

. When you get the formula for T_n, substitute that into the formula for S_n. You wind up with a formula that can be easily simplified. Keep in mind that a constant factor can be distributed into and out of a sum, that sums can be broken into two pieces across a plus sign, and that you already know how to solve a geometric series. Let me know if you need more help, or if I answered the wrong question. Black Carrot 08:01, 2 July 2007 (UTC)[reply]

arithmetic-geometric mean

Could you give an example of the arithmetic-geometric mean? I don't understand the concept after reading the page on arithmetic-geometric mean.

Jevous aimeclsm Jevous aime

Consider Arithmetic-geometric mean and suppose x=6 and y=24. We want the arithmetic-geometric mean of x and y.

a1 = (x+y)/2 = 15. b1 = √(x*y) = 12.

a2 = (a1 + b1)/2 = 13.5. b2 = √(a1 * b1) = √(180) = 13.41640786...

And so on.

n  a_n               b_n
0  6                 24
1  15                12
2  13.5              13.41640786500..
3  13.45820393250..  13.45813903099..
4  13.45817148175..  13.45817148171..
...

The arithmetic-geometric mean is the limit of these 2 sequences. This is approximately 13.45817148173. PrimeHunter 01:46, 1 July 2007 (UTC)[reply]

To confuse matters, the arithmetic/geometric mean inequality is often referred to as the AGM inequality or sometimes just AGM. iames 04:37, 1 July 2007 (UTC)[reply]

I have added PrimeHunter's example to the arithmetic-geometric mean article. Gandalf61 12:26, 1 July 2007 (UTC)[reply]

Was the 0th entry excluded from the table on purpose? In my opinion it helps see the big picture. -- Meni Rosenfeld (talk) 12:29, 1 July 2007 (UTC)[reply]

As it is written, the article starts the series with a₁ and g₁ and does not define a₀ and g₀, so it would have beeen inconsistent to start the table at 0. I didn't bother to rewrite the article to define a₀ and g₀, but I have no objection if someone else wants to do that. Gandalf61 18:22, 1 July 2007 (UTC)[reply]

Nice work. I see Meni Rosenfeld found a way to include the 0th entry without changing the definition. PrimeHunter 21:51, 1 July 2007 (UTC)[reply]

Cycloid paradox

A spot on the tire of a bicycle wheel would describe a cycloid and touch the earth once every four diameters of the wheel, in other words 4d. But if you removed the tire, cut it across, and laid it flat it would only be (pie)d long. In laypersons terms, why is there a difference between the 4d and the (pie)d? Where does the extra distance come from? 80.0.126.168 18:40, 1 July 2007 (UTC)[reply]

Oops! I see now that while the length of the cycloid curve is 4d, the distance between the points at which the spot would touch the earth is still (pie)d. So no paradox. 80.0.126.168 19:16, 1 July 2007 (UTC)[reply]