Let *n* stand for a non negative integer and let A_{n} denote the number of solutions of the Diophantine equation: x + 5y + 10z + 25u + 50v = n in non-negative integers then the series A_{0}+ A_{1}ζ +A_{2}ζ^{2} +A_{3}ζ^{3} +... + A_{n}ζ^{n} + ... represents a rational function of ζ. Find it. Before to solve this problem (originally posted Wed 2008-08-27, 14:00), I want you get confidence with Diophantine equations.
**Example 1. **
Suppose, you go to your favorite SuperMarket and purchase the following food: - Hamburger Bread - Euro 1.50
- Pringles Rice Paprika - Euro 1.99
- Baked Salmon - Euro 6.85
- Blue de Bresse French Cheese - Euro 3.79
- Toblerone Swiss Chocolate - Euro 1.28
Diophantine Equations regards integer, but we can multiply by 100, and move the numbers to integers. Now, suppose that you purchase, each time, exactly 1, 5, 10, 25 and 50 quantities of this food, each time, you go to the SuperMarket. (It is a mere example). So, each time, you will spend: 150 + 5 • 199 + 10 • 685 + 25 • 379 + 50 • 128 = n How much is n? N (we multiply by 100, to hold numbers between integers), N = 23,870. This simply algebraic connection get clear, in this case. **Example 2. **
The second example may be used to know the number of products to sale to touch a specific target. Suppose you have five products. One cost 1 US$, the others in sequence: 5, 10, 25 and 50 US$. How many product you need to sold to produce 100,000 US$ ? Now, we will have, x + 5y + 10z + 25u + 50v =100000 The posted problem ask for the number of solutions to this equation, and in this case, what is A_{100000} ? Well, let us find ... just one solution. We can assign a number to each variable. If you choose 1000 for v, we will have, x + 5y + 10z + 25u = 50000 Again, like for the solution of (In how many ways we can change a dollar), we see that the problem is recursive. If you sold 2000 products for those that cost, 25 US$, we will arrive at the point. Because, 25 • 2000 + 50 • 1000 = 100000. A_{100000} <= A_{100} • 1000 = 296,000. Now, we arrive to the point to solve the problem. But this is easy. Consider, s_{n} = 1 + x + x^{2}+ ... x^{n-1} = (1 - x^{n}) / 1 - x. Now, if |x| < 1, then this sum tends to 1/1-x. And, so Σ_{n=0} A_{n}ζ^{n }= A_{(x+5y+10z+25u+50v)}ζ^{(x+5y+10z+25u+50v)} =1/(1-ζ)(1-ζ^{5})(1-ζ^{10})(1-ζ^{25})(1-ζ^{50}) Please read and solve exercises from 3 to 31on the Book: George Polya and Gabor Szego - Problems and Theorems in Analysis I. Лобановскийсправка нетрудоспособностинастенные электрические котлыsms imobisвыгодно ли печь торты на заказлазанью как готовитьскребки для языкакарта sdhcкотлы для отопления частного домарасширительный бак 100 лноутбук акцияраскрутка интернет магазина |