Wikipedia:Reference desk/Archives/Mathematics/2012 October 14

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October 14

Solving Polynomials

I"ve got a problem I'm trying to solve (curiosity, not hw) that boils down to solving this, where a,b,c,d are constants and I want to solve for x,y,z: ${\frac {d^{2}}{x^{2}+y^{2}+z^{2}}}={\frac {(a-x)^{2}}{y^{2}+z^{2}}}={\frac {(b-y)^{2}}{x^{2}+z^{2}}}={\frac {(c-z)^{2}}{x^{2}+y^{2}}}$ How would you go about solving this problem? Black Carrot (talk) 20:28, 13 October 2012 (UTC)

I don't think you'll be able to find any sort of closed form expression for x, y, z in terms of a, b, c, d. With Macaulay2 I found a Gröbner basis for the system, and eliminated x and y. The result was a square-free polynomial in z, a, b, c, d of degree 22, with 1003 terms. If you have values of a, b, c, d you could plug them in and get a degree 15 polynomial in z, which you could solve numerically. Of course in that case you could just solve the original system numerically so I'm not sure that's helpful. Rckrone (talk) 00:14, 14 October 2012 (UTC)

Sometimes the symmetry of a problem helps simplifying. If you can find power sums $s_{n}=x^{n}+y^{n}+z^{n}$ for, say, n=−1 and n=1 and n=2, then each of x and y and z satisfies a cubic equation like

$x^{3}-s_{1}x^{2}+{\frac {s_{1}^{2}-s_{2}}{2}}x-{\frac {s_{1}^{2}-s_{2}}{2s_{-1}}}=0.$

So your problem is reduced to symmetrizing the equations for finding equations for the power sums. Bo Jacoby (talk) 04:53, 14 October 2012 (UTC).

Ya, that's pretty much what I thought would happen. Thanks for the reference to Macaulay2, I'll check that out. Black Carrot(talk) 04:34, 15 October 2012 (UTC)

Expressing equation

I have the following

g = 1/a * (sqrt(b^2 + 2*a*n) - sqrt(b^2 + 2*a*N - 2*a*n)) + T

h = 1/((b^2 + 2*a*n)^(-1/2) + (b^2 + 2*a*N - 2*a*n)^(-1/2))

and I want to eliminate n and express this as a relationship between g and h, preferably in the form "h = some function of g". Any ideas? Anyone got some super-dooper software that could possibly solve this (if it can be done)? Note that the variables n and N are distinct, in case that messes up anything. 86.160.222.148 (talk) 03:19, 14 October 2012 (UTC)

So you have g as a function of a, b, N, n, and T, while h is a function of a, b, N, and n ? Yikes ! StuRat (talk) 03:29, 14 October 2012 (UTC)

For this purpose, a, b, N and T can be considered constants. They do not make the problem excessively more complicated in the way you seem to be suggesting. 86.160.222.148 (talk) 03:38, 14 October 2012 (UTC)

By multiplying a(g-T) and 1/h, you can solve for n as $n=N/2+(g-T)/4h$. Substituting that in to one of your equations you will get an expression in terms of h and g which doesn't include n, although it won't have the form h = f(g). Rckrone (talk) 05:55, 14 October 2012 (UTC)

[ec]$g-T$ is of the form ${\frac {1}{a}}({\sqrt {x}}-{\sqrt {y}})$ and h is of the form ${\frac {1}{1/{\sqrt {x}}+1/{\sqrt {y}}}}={\frac {{\sqrt {x}}{\sqrt {y}}}{{\sqrt {x}}+{\sqrt {y}}}}$. We can see that ${\sqrt {x}}+{\sqrt {y}}={\frac {x-y}{a(g-T)}}$ and ${\sqrt {xy}}={\frac {x+y-a^{2}(g-T)^{2}}{2}}$. We have $x+y=2(b^{2}+aN),\ x-y=4an-2aN$ giving $h={\frac {(g-T)(2b^{2}+2aN-a^{2}(g-T)^{2})}{8n-4N}}$. You still need to solve for n to get $n={\frac {1}{4}}(2N\pm {\sqrt {(4b^{2}+a(4N-a(g-T)^{2}))(g-T)^{2}}})$; I don't think it can be simplified further. -- Meni Rosenfeld (talk) 06:08, 14 October 2012 (UTC)

Ah, I misread the exponents in the second equation. Fixing that mistake I also get a quadratic in n which makes things a lot messier.Rckrone (talk) 06:13, 14 October 2012 (UTC)

Awesome, thanks. 86.128.6.37 (talk) 00:57, 15 October 2012 (UTC)

What's the functional square root of sin(x)?

--128.42.153.38 (talk) 22:04, 14 October 2012 (UTC)

Have you looked at our functional square root article? It shows a plot of the function square root of the sine function, and gives a reference to the source. (Like the great majority of functional roots, it can't be written in explicit form.) Looie496 (talk) 22:50, 14 October 2012 (UTC)

Also, the coefficients of the power series expansion are given at OEIS 048602 and 048603. --Meni Rosenfeld (talk) 06:34, 16 October 2012 (UTC)

Infinite sum

Hi. Today I did a math contest and one infinite sum problem kinda messed me up. The question was to compute the sum of the nth Fibonacci numbers over 5^n, viz: $S=\sum _{n=1}^{\infty }{\frac {F_{n}}{5^{n}}}$ where F_n is of course the nth Fibonacci number (so F_1=1, F_2=1, F_3=2, etc., etc.). I bounded the sum and determined that 1/4<S<1/3 but that's about it. I knew the identity $F_{n}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{n-1}-({\frac {1-{\sqrt {5}}}{2}})^{n-1}}{\sqrt {5}}}$ for n>1 but I did not... ahem, avail myself of it was suspicious because it should not have been expected, I might have misstated it, and I suspected there was a more elegant solution. Any thoughts?24.92.74.238 (talk) 23:46, 14 October 2012 (UTC)

What was the question? To evaluate the series? --Tango (talk) 23:55, 14 October 2012 (UTC)

Oh yes, I should have been more explicit. Corrected above. 24.92.74.238 (talk)

See geometric series. Sławomir Biały (talk) 01:29, 15 October 2012 (UTC)

Or divide the Fibonacci recurrence relation by ${5^{n}}$ and sum.John Z (talk) 01:36, 15 October 2012 (UTC)

Let $S_{2}=\sum _{n=2}^{\infty }{\frac {F_{n}}{5^{n}}}=S-{\frac {1}{5}}$ and $S_{3}=\sum _{n=3}^{\infty }{\frac {F_{n}}{5^{n}}}=S-{\frac {6}{25}}$. You fave $F_{n+2}=F_{n+1}+F_{n}$ and so $25S_{3}=5S_{2}+S$, solve to get $S={\frac {5}{19}}$. -- Meni Rosenfeld (talk) 07:41, 15 October 2012 (UTC)

That's brilliant, thank you!

Resolved

Hi, everybody! Are you guys sure it's correct to iterate sums with variable k when the terms are indexed with n...? --CiaPan(talk) 11:54, 16 October 2012 (UTC)

Since $\sum _{k=1}^{\infty }{\frac {F_{n}}{5^{n}}}$ doesn't make much sense, I assume it ws just a typo. The OP probably meant$\sum _{n=1}^{\infty }{\frac {F_{n}}{5^{n}}}$, and Meni simply copied the typo through cut-and-paste - an easy mistake to make.Gandalf61 (talk) 12:20, 16 October 2012 (UTC)

$\sum _{k=1}^{\infty }{\frac {F_{n}}{5^{n}}}$ sure makes sense, it just diverges for every $n\neq 0$ :). But yeah, I fixed my comment and took the liberty to fix the OP as well.

Speaking of mistakes, the formula given in the OP for $F_{n}$ is not correct for the offset given by $F_{1}=F_{2}=1$. Should be$F_{n}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{n}-({\frac {1-{\sqrt {5}}}{2}})^{n}}{\sqrt {5}}}$. -- Meni Rosenfeld (talk) 20:16, 16 October 2012 (UTC)