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December 12[edit]

Smallest power with a given property[edit]

A recent question was on how to determine the last two digits of 3^1033 and similar expressions. I'm wondering how the smallest power having a specified pattern can be determined analytically. In particular, what is the smallest power of 3 which ends in 000001? Since φ(1000000)=400000, 3^400000 must be 1 mod 1000000, but some experimentation with a program showed that 3^50000, but no lower power, also had the property. Can the value of 50000 be calculated?→194.72.9.25 (talk) 14:10, 12 December 2008 (UTC)[reply]

Unless I miss my guess, you're describing the discrete logarithm problem, which, in full generality, is a sufficiently hard problem that it forms the basis for several cryptographic systems. Our article on the problem contains links to several slightly "better" algorithms (for certain definitions of better), but no computationally easy ones. Ray (talk) 15:47, 12 December 2008 (UTC)[reply]

Actually, the smallest power of 3 which ends in 000001 is 3⁰ = 1. If you want the smallest positive k such that 3^k ends in 000001, i.e., the order of 3 in (Z/1000000Z)^×, you can use the fact that it has to divide every other l satisfying 3^l = 1 (mod 1000000), and in particular, it divides φ(1000000). In general, if you want the order of a modulo n, you can first compute k = φ(n), and factorize k = p₁·…·p_r. Then you keep dividing the p_is out of k as much as possible while maintaining a^k = 1 (mod n). This is an exponential-time algorithm because of the two factorizations (one implicit in the computation of φ), but I guess that it's still faster than trying all a¹, a², a³, … until reaching 1 (mod n). In your particular example, you have φ(1000000) = 400000 = 2⁷·5⁵, and then you compute modulo 1000000: 3²⁰⁰⁰⁰⁰ = 1, 3¹⁰⁰⁰⁰⁰ = 1, 3⁵⁰⁰⁰⁰ = 1, 3²⁵⁰⁰⁰ = 500001 ≠ 1, 3¹⁰⁰⁰⁰ = 200001 ≠ 1, hence the order of 3 modulo 1000000 is 50000. — Emil J. 16:48, 12 December 2008 (UTC)[reply]

(OP, using the secure server to bypass the current problem of BT putting all users on the same IP address) Thanks for replies.→217.43.211.175 (talk) 12:13, 14 December 2008 (UTC)[reply]

Solution sought[edit]

please give me solution for the set of equations (way of solving the equations)

     a^2 + b = 28 and a + b^2 = 14

and also let me know what type of equations are called simultaneous equations. 117.200.0.16 (talk) 17:11, 12 December 2008 (UTC)[reply]

See Simultaneous equation and Diophantine equation.

Note that, it depends on whether you want the solutions to be integers or not. If not, then the solution set is simply the points of intersection between the two curves:

y = 28-(x^2)

y^2 = 14-x

There will be precisely four solutions (you know this how?). One solution is a=5 and b=3; all the other solutions are non-integer solutions. Generally, the basic way (I know that there are others) to tackle such problems would be to 'guess and check' (if you are looking for integer solutions). For a start, b must be between -3 and -3. Also, b can't be 0 (because 28 is not a perfect square), b can't be 1, etc... so you can eliminate these possibilities.

Topology Expert (talk) 17:29, 12 December 2008 (UTC)[reply]

The standard way of solving two equations with two unknowns

a² − 28 + b = 0 (eq1)

a − 14 + b² = 0 (eq2)

is to eliminate one of the variables. Multiply equation 1 by b and subtract equation 2. This eliminates b² and gives

(a² − 28)b − a + 14 = 0 (eq3)

Multiply equation 1 by (a² − 28) and subtract equation 3. This eliminates b and gives

a⁴ − 56a² + a + 770 = 0 (eq4)

Equation 4 is solved numerically. The four solutions are approximately:

a = −5.69543 , a = −4.86382 , a = +5 , a = +5.55925 .

This can be substituted into eq1 to give the corresponding values for b. You may also eliminate a from the original equations to obtain an exact equation for b.

Bo Jacoby (talk) 19:34, 12 December 2008 (UTC).[reply]

If you're happy to solve a quartic you could just substitute b=28-a^2 into the second equation and expand out. --Tango (talk) 19:58, 12 December 2008 (UTC)[reply]

It is possible to solve equation 4 algebraically using Galois theory. Topology Expert (talk) 19:42, 12 December 2008 (UTC)[reply]

It's possible, but it's a complete mess. I doubt anyone actually does so unless they are taking a course in Galois theory (that's certainly the only time I've ever done so). --Tango (talk) 19:58, 12 December 2008 (UTC)[reply]

To OP: See this for a method of finding the exact roots of your equation. To Tango: It is a complete mess but it is certainly very interesting to solve it in the general case (the first time but after that it gets boring!). Topology Expert (talk) 20:29, 12 December 2008 (UTC)[reply]

Oh, yes, it's a great bit of maths, but I would never actually use it. --Tango (talk) 20:34, 12 December 2008 (UTC)[reply]

Notice that in this case, the OP actually asked for the method so it is fine to give it to him. Even if he would have asked for only the answer, it would have made no difference (we would have given him both). The point is that sometimes it is impossible to give the answer without actually giving the method. Anyway, this section was short and sweet and hopefully all sections will be the same from now on. Topology Expert (talk) 21:21, 12 December 2008 (UTC)[reply]

The Ferrari formula for quartics does not really solve the equation, but merely 'reduces' it to a complicated expression involving Nth roots, which are computed by solving equations which are not easier than the original one. The method of substituting b=28-a^2 into the second equation works in this case but not in the general case where b cannot be written explicitely as a function of a, while the method of elimination works generally. The observation that the integer +5 is a solution can be used to find the simpler equation

a³ + 5a² − 31a − 154 = 0

for the other three solutions, by computing the quotient

(a⁴ − 56a² + a + 770) / (a − 5)

Bo Jacoby (talk) 10:04, 13 December 2008 (UTC).[reply]

If you only want to know the integer solutions, here is an easy way.

1. you subtract the second equation from the first.

$a^{2}-a-b^{2}+b=14\,\!$

2. multiply it by 4.

$(4a^{2}-4a+1)-(4b^{2}-4b+1)=2^{3}*7\,\!$

3. put A=2a-1, B=2b-1

$A^{2}-B^{2}=2^{3}*7\,\!$

$(A+B)(A-B)=2^{3}*7\,\!$

4. It's fairly straightforward from here. If you consider all 8 possibilities, and cross reference it with the original equations, the answer would be (a,b)=(5,3)Johnnyboi7 (talk) 06:00, 15 December 2008 (UTC)[reply]

π = –i log(–1) ?[edit]

Starting with Euler's identity and solving for π, I get

e^{i\pi }=-1

i\pi =log(-1)

(principal branch)

\pi ={\frac {1}{i}}\,log(-1)

Thus

\pi =-i\,log(-1)

(principal branch)

Is this correct? If so, I think it would make a reasonable addition to the List of formulae involving π#Miscellaneous, since it's such a beautiful identity. — Loadmaster (talk) 23:06, 12 December 2008 (UTC)[reply]

Well, the principal branch of log is only defined on the complex plane minus the negative real line, so it isn't defined for -1. You'll have to pick another branch, but if you do I can't see why it isn't entirely correct. I guess that's why people don't talk about that identity much, though - it's too confusing to use strange branches of log! If you want to, though, you could get an even weirder identity -

\pi =\log((-1)^{-i})

. Bonus marks if you can give it a geometric interpretation. --Tango (talk) 23:17, 12 December 2008 (UTC)[reply]

If you look at that article you'll see most of the formulae have names. This is because they are notable which is a prime criterion for inclusion in Wikipedia. In WP this means that people have noticed them and written things about them. What you have wouldn't be counted as notable and moreover since you did it yourself would be counted as original research which is a no-no for inclusion in WP. Sorry WP just isn't the place for cool new things you've found out, but if you get enough people elsewhere interested then it does become eligible. Dmcq (talk) 10:59, 13 December 2008 (UTC)[reply]

I find the definition of original research in terms of maths and science articles very fuzzy, as if something is directly deduced from known formula, it is hardly original research (I.e. if I take coulombs law, and newtons law III, and cancel out the F's thats not anything new) as it has always been known as long as the 2 laws have, its just the application of things we know, however that is also exactly how discoveries in maths are made, by just discovering things that have always been there but not been discovered yet. So if the equation above were defined, it would not be original research, it would just be a rearrangement of eulers identity. However this does create a very wishy washy boundry as to where you stop deducing things that we already know, to where you start deducing things that could fall under the somewhat umbrella term of 'original research'.

I wouldn't give a separate entry to something so transparently equivalent to another. —Tamfang (talk) 02:49, 15 December 2008 (UTC)[reply]

First of all: it is $ln$ not $log$ , secondary: it is not new information - every student knows that $\ln(-r)=\ln(r)+i\pi \$ , and every student at least one time in his studies has got the given formula for π: $\pi ={\frac {1}{i}}\,log(-1)$ 194.99.216.134 (talk) 13:07, 20 December 2008 (UTC)[reply]

Mathematics Question[edit]

Please can any one tell me if the Boubaker polynomials have any Exponential Generating Function and if their roots have any analytical expression ? (and please where can one find the Exponential Generating Function for Chebyshev Polynomials ?? please answer in my talk page. Duvvuri.kapur (talk) 23:37, 12 December 2008 (UTC)[reply]

I have no knowledge of this topic, but I see we have Boubaker polynomials. -- SGBailey (talk) 16:38, 13 December 2008 (UTC)[reply]