Wikipedia:Reference desk/Archives/Mathematics/2010 October 12

= October 12 =

Rings with infinite height
Is there an example of ring (commutative, with unit) in which every prime ideal has infinite height? The closest example I came up with is
 * $$R=\{\sum_{i=1}^n a_ix^i|i\in \mathbb{R}, i\geq 0,a_i\in \mathbb{Q}\}$$

but I'm not sure it is right.--151.71.87.31 (talk) 11:33, 12 October 2010 (UTC)


 * Take a look at this paper. It looks at polynomial extension rings of commutative unitary rings, and proves that prime ideals have finite height if and only if they are finitely generated. So to find your example, you need to find one where none of the prime ideals are finitely generated. — Fly by Night  (  talk  )  12:21, 12 October 2010 (UTC)

A simple exponentiation math problem
I'm not sure how to solve this: '''2n+3n=4. Solve for n.''' I'm pretty sure it's nothing I had to learn in high school or college, but I could be wrong. I'm trying to simplify some mortgage amortizations and I'm not sure the best way to solve them. (I could probably use Newton's method, but I'm trying to look for an exact or direct approach.) Thanks in advance! ~a (user • talk • contribs) 19:45, 12 October 2010 (UTC)
 * n=0.7604913577476414 by Newton's method. I don't think there is an elementary solution. 72.89.116.142 (talk) 20:02, 12 October 2010 (UTC)
 * Awww, that's too bad. Thanks for your time!  ~a (user • talk • contribs) 20:37, 12 October 2010 (UTC)

Here's a link: Newton's method. Michael Hardy (talk) 20:21, 12 October 2010 (UTC)
 * Yeah I know about Newton's method (I mentioned it above), but I was hoping for (what anon calls) an "elementary solution". Regardless, thanks for your help!  ~a (user • talk • contribs) 20:37, 12 October 2010 (UTC)


 * I just tried Newton's method and the iterations seem to be periodic, i.e. they don't settle down to a limit and just jump back and forth between two values. Which initial value did you use 72.89.116.142? — Fly by Night  (  talk  )  21:39, 12 October 2010 (UTC)
 * I started with n=0 and got the same answer. "for(int i=0;i<10;i++)n-=f(n)/fp(n);"  and f(n) is "2n+3n-4" and fp(n) is "2n*ln(2)+3n*ln(3)".  ~a (user • talk • contribs) 22:03, 12 October 2010 (UTC)


 * I've just run it again, with x0 = 0 and the answer oscillates between 0.7604913576 and 0.7604913579. I ran the iteration 500 times. Asking Maple to solve the equation for me gives x = 0.7604913576. Doing some algebra I get:
 * $$ x_{n+1} = x_n - \frac{2^{x_n}+3^{x_n}-4}{2^{x_n}\ln 2 + 3^{x_n} \ln 3} \ . $$
 * Putting xn = 0.7604913576 into this gives xn+1 = 0.7604913579. Maybe it's a rounding problem in Maple. Could you do that last substitution at your end and let me know what the output is, please? — Fly by Night  (  talk  )  22:49, 12 October 2010 (UTC)
 * It's obviously a rounding error. -- Meni Rosenfeld (talk) 08:32, 13 October 2010 (UTC)
 * Thanks for that Meni... — Fly by Night  (  talk  )  10:47, 13 October 2010 (UTC)

Using the J (programming language), expand the function as a truncated (i.13) taylor series (t.) and use the polynomial equation solver (p.). {:{:>p.((2&^)+(3&^)-4:)t.i.13 0.76049135775 Bo Jacoby (talk) 23:35, 12 October 2010 (UTC).
 * Ah taylor series? So $${\sum_{i=0}^{\infty}\frac{ln(2)^i+ln(3)^i}{i!}n^i}=4$$ ?  That's kind of hard to find an exact solution for n too, right?  ~a (user • talk • contribs) 13:53, 13 October 2010 (UTC)
 * Right, but there are some special-purpose numerical algorithms for polynomials, and apparently J only has a numerical polynomial solver, not a general root-finder. Anyway, it should be $$i!\;\!$$ in the denominator. -- Meni Rosenfeld (talk) 14:44, 13 October 2010 (UTC)
 * Truncating the taylor expansion to 13 terms gives a polynomial of degree 12 having 12 roots, and J computes them all: 4.44295j5.75916 4.44295j_5.75916 _5.91012 _5.18925j2.80887 _5.18925j_2.80887 _3.22387j4.90441 _3.22387j_4.90441 _0.567437j5.75809 _0.567437j_5.75809 1.63828j5.44476 1.63828j_5.44476 0.760491. The untruncated series has an infinite number of roots. No programming language has a root-finder computing that infinitely long row of complex numbers. Bo Jacoby (talk) 18:14, 13 October 2010 (UTC).
 * Oops! Thanks, fixed.  ~a (user • talk • contribs) 15:55, 13 October 2010 (UTC)
 * I think the problem is that 2x + 3x = 4 does not have an exact solution in terms of elementary functions. The best you can hope for is a numerical solution. — Fly by Night  (  talk  )  15:02, 13 October 2010 (UTC)
 * For this to make sense, you'd have to exclude constant functions from being elementary. The solution is trivially elementary according to the definition in the linked article.—Emil J. 15:10, 13 October 2010 (UTC)
 * You're right. I had an inverse of y(x) = 2x + 3x in mind. Thanks. — Fly by Night  (  talk  )  16:40, 13 October 2010 (UTC)
 * Ah, ok. Thanks.  So, then, solving the generalized problem an+bn=c is impossible?  ~a (user • talk • contribs) 15:55, 13 October 2010 (UTC)
 * Very much so. Just look at Fermat's Last Theorem. Looks really easy on the face of it. Number theory is full of problems that are easy to state and seemingly impossible to solve. — Fly by Night  (  talk  )  16:40, 13 October 2010 (UTC)


 * For those without a knowledge of Newton's Method or Taylor series expansions, the "Goal Seek" function in Excel (or any similar spreadsheet) gives a very easy way to solve a large number of such equations, but Excel achieves only five significant figures, even with a thousand iterations. (Excel actually uses the Newton-Raphson method)   D b f i r s   15:23, 13 October 2010 (UTC)
 * Does Excel really use Newton-Raphson? I find it hard to believe that Excel can symbolically differentiate, and if it evaluates the derivatives numerically then it may as well just use the secant method. -- Meni Rosenfeld (talk) 15:36, 13 October 2010 (UTC)
 * Agreed. According to this MS support page Excel Goal Seek uses "a simple linear search". From the description, it could be the secant method or possibly the false position method. Gandalf61 (talk) 16:15, 13 October 2010 (UTC)
 * Apologies for my error. I jumped to a false conclusion based on inadequate research, and if I'd thought about it (as Meni did), I'd have realised that it was unlikely!  Sorry!    D b f i r s   01:53, 14 October 2010 (UTC)
 * Wolfram Alpha can also solve this, to arbitrary precision. -- Meni Rosenfeld (talk) 15:42, 13 October 2010 (UTC)
 * Oh wow, that's cool. I'll have to keep that URL.  ~a (user • talk • contribs) 15:55, 13 October 2010 (UTC)
 * There is no need for symbolic differentiation with Newton's method. Numerical differentiation works just fine and AFAIK it's usually done that way.  72.229.127.238 (talk) 05:32, 14 October 2010 (UTC)
 * As I said, I am fairly sure that the secant method is superior to Newton's method with numerical derivatives. Personally I have used Newton's method many times with symbolic derivatives, never with numerical. -- Meni Rosenfeld (talk) 07:06, 14 October 2010 (UTC)

If the n's don't have to be the same, on the theory that "you can't step into the same river twice", an exact solution is 2^0 + 3^1 = 4. also the average of those exponents (0 and 1) is 0.5, or 0.75 if you'd rather have something than nothing, which is pretty close to the above answers by more complicated or less whole/precise methods. 92.230.70.59 (talk) 23:25, 14 October 2010 (UTC)
 * Someone please look the numerical solution up in Plouffe's Inverter in case it has an explicit form we're missing. The website doesn't load for me right now.  &#x2013; b_jonas 22:40, 16 October 2010 (UTC)

The equation $$ 2^x+3^x=4$$ can also be solved without using differentiation (such as Newton's method and Taylor's formula do). Rewrite the equation
 * $$ 3^x=4-2^x$$

Take the logarithm
 * $$ x\log 3=\log(4-2^x)$$

Divide by log 3
 * $$ x=\frac{\log(4-2^x)}{\log 3}$$

Now the unknown is nicely isolated on the left hand side, but alas it also occurs on the right hand side. So try the iteration
 * $$ x_0=0$$
 * $$ x_{n+1}=\frac{\log(4-2^{x_n})}{\log 3}$$

for n= 0,1,2,3,... The J expression is simply 3&^.@(4-2&^)^:_]0 0.76049135775 Bo Jacoby (talk) 16:52, 17 October 2010 (UTC).