Wikipedia:Reference desk/Archives/Mathematics/2011 December 6

= December 6 =

Population increase percentage
I was hoping that someone would be able to calculate the population increase percentage for the following figures:

2000 Census - 18,278,559

2010 Census - 21,813,334

Thanks. --Ghostexorcist (talk) 01:19, 6 December 2011 (UTC)


 * (21,813,334 - 18,278,559) / 18,278,559 ≈ 0.1933, so the population of Xīnjiāng Wéiwú'ěr Zìzhìqū grew 19% during that decade. -- 203.82.66.205 (talk) 01:35, 6 December 2011 (UTC)


 * Thank you. --Ghostexorcist (talk) 01:36, 6 December 2011 (UTC)

Obtaining an absolute sum.
If I have a set of $$N$$ real numbers $$x_i$$ whose values are unknown, but I do know the values of a corresponding set of numbers$$s_i$$ given by $$s_i = \sum_{j=0}^N x_j^i$$ for all natural numbers $$i$$, is there a way of using the knowns $$s_i$$ to evaluate the sum $$\sum_{j=0}^N |x_j|$$. Or indeed is there any information that can be gained on the possible range of values of this sum.

Thank you. — Preceding unsigned comment added by 92.14.38.60 (talk) 10:52, 6 December 2011 (UTC)
 * If it is useful it is also known that all $$s_i>0$$. — Preceding unsigned comment added by 92.14.38.60 (talk) 10:54, 6 December 2011 (UTC)


 * The $$s_i$$ are the power sums of the $$x_i$$. Given the values of the first N $$s_i$$, you can use Newton's identities to find the values of the elementary symmetric polynomials in the N $$x_i$$. From these values, you can construct an N-degree polynomial whose roots are the $$x_i$$. If N is greater than 4 then there is no guarantee that you can find the roots of this polynomial using algebraic methods. However, you can use numerical methods to approximate the roots, and if, for example, you also know that the $$x_i$$ are all integers then numerical methods can give you an exact solution. Gandalf61 (talk) 11:29, 6 December 2011 (UTC)


 * For example, if N is 3, and we are given
 * $$s_1 = 8 \quad s_2 = 30 \quad s_3 = 134$$
 * we use Newton's identities to find
 * $$e_1 = x_1 + x_2 + x_3 = s_1 = 8$$
 * $$e_2 = x_1x_2 + x_1x_3 +x_2x_3 = \frac{s_1e_1 - s_2}{2} = 17$$
 * $$e_1 = x_1x_2x_3 = \frac{s_1e_2-s_2e_1+s_3}{3}= 10$$
 * and from these we construct the cubic
 * $$x^3 - 8x^2 + 17x - 10$$
 * By inspection, x = 1 is a root of this cubic, so we have
 * $$x^3 - 8x^2 + 17x - 10 = (x-1)(x^2 - 7x + 10)$$
 * and solving the quadratic tells us that the other two $$x_i$$ are 2 and 5. Gandalf61 (talk) 11:37, 6 December 2011 (UTC)


 * And another method if you really do ave the values for all natural numbers you can get the billionth root of the billionth $$s_i$$ and that will very closely approximate the largest $$x_i$$, remove this from the sums and work backwards using smaller powers. This may not be quite so practical as the previous method depending on how easy it is to find the $$s_i$$ but lower values than a billion might lead to good approximations which one can iterate from. ;-) Dmcq (talk) 11:41, 6 December 2011 (UTC)


 * e.g. with 1 2 and 5 we have 1^10+2^10+5^10=9766650 and its tenth root is 5.00005248 at least so googles calculator assures me.
 * 1^5+2^5+5^5 - 5.00005248^5 = 32.8359966 and its fifth power is 2.01034244
 * 1+2+5 - 5.00005248 - 2.01034244 = 0.98960508
 * So there we have a fairly reasonable approximation without going anywhere near billionth powers. I'd be interested what is the best powers to choose. Dmcq (talk) 11:51, 6 December 2011 (UTC)
 * Use http://www.wolframalpha.com/input/?i=x^3+%E2%88%92+8x^2+%2B+17x+%E2%88%92+10 to find the roots of your polynomial. Bo Jacoby (talk) 11:17, 7 December 2011 (UTC).