does the sequence 
defined by the initial condition 
and the recursion 
have 
for all 
,
which contains 
such that 
?
is the quaternion group with 8 elements 
,
of the ring 
of all Hamiltonian quaternions.
Y by
Let 
be a real-valued function that is twice continuously differentiable throughout 
, and define
![\[
\rho (x,y)=yh_x -xh_y .
\]](//latex.artofproblemsolving.com/7/3/6/736993597c4b45cf77b9013185d2a4d931a7d231.png)
Prove or disprove: For any positive constants 
and 
with 
, there is a circle 
of radius whose center is a distance away from the origin such that the integral of 
over the interior of is zero.
be a real-valued function that is twice continuously differentiable throughout 
, and define![\[
\rho (x,y)=yh_x -xh_y .
\]](http://latex.artofproblemsolving.com/7/3/6/736993597c4b45cf77b9013185d2a4d931a7d231.png)
Prove or disprove: For any positive constants 
and 
with 
, there is a circle 
of radius whose center is a distance away from the origin such that the integral of 
over the interior of is zero.
and let 
,
and radius 
Parameterizing 
by 
with 
,
Now consider a function 
with 
,
traverses continuously along 
.
is continuous, and since 
(easy to check from above), by the Intermediate Value theorem there is an ![$\alpha_0\in[0,1]$](http://latex.artofproblemsolving.com/8/6/b/86bd1abc03ef8428b49e0008ce013c5f0cd01e40.png)
with 
.
!
is the circle of radius 
,
,
with period 
.
.
.
.
as 
.
.
is continuous in 
,
.
.
,
.
and since 
over the circle, 
is useful in the first place and isn't just a lucky guess.
.![\[S=\sum f(a_i)f(a_{i+1}) < 0,\]](http://latex.artofproblemsolving.com/2/3/1/231adf870f651a3ddbb5be6dba2d9e63b3d6655e.png)
or whatever" type. In these cases it's natural to take the expected value over all such permutations and hope 
.
,
.
![\[\int\!\!\!\!\int_S\rho(x,y)\,\text dA=\int\!\!\!\!\int_S(yh_x(x,y)-xh_y(x,y))\,\text dA.\]](http://latex.artofproblemsolving.com/1/f/2/1f257daf0498b010bc0ddb8bf3cf48beaa820526.png)
If possible, we want some vector field such that ![\[\mathrm{curl}\,\mathbf F(x,y)=\rho(x,y)=yh_x(x,y)-xh_y(x,y),\]](http://latex.artofproblemsolving.com/3/5/c/35c2eaa7893e06dfa2a32c1166611b5ccee9bc3c.png)
where 
refers to 2-dimensional (scalar) curl so that we can apply Green's Theorem. A candidate for 
is 
which in fact does satisfy ![\[\mathrm{curl}\,\mathbf F(x,y)=yh_x(x,y)-xh_y(x,y),\]](http://latex.artofproblemsolving.com/2/e/3/2e3334d1e554023e49d5363023fa75c241652481.png)
and so ![\[\int\!\!\!\!\int_S\rho(x,y)\,\text dA=\oint_C\mathbf F(x,y)\cdot\text d\mathbf r=\oint_Cxh(x,y)\,\text dx+\oint_Cyh(x,y)\,\text dy,\]](http://latex.artofproblemsolving.com/a/e/1/ae1ac7d0a60f16ae9dbb856bffa3e63970b7c535.png)
where 
is positively oriented. One parameterization of 
is 
with 
(I will abbreviate 
and 
)![\[\oint_Cxh(x,y)\,\text dx=-\int_0^{2\pi}rh(\mathbf c(t))(d_0+r\cos t)\sin t\,\text dt,\]](http://latex.artofproblemsolving.com/1/0/f/10f92a5d506ff6c426514d5108090f840f97f64b.png)
and the second integral converts to ![\[\oint_Cyh(x,y)\,\text dy=\int_0^{2\pi}rh(\mathbf c(t))(d_1+r\sin t)\cos t\,\text dt.\]](http://latex.artofproblemsolving.com/2/6/5/265ef66b9a0f0c43ff67de0f95d8577166a2e87c.png)
As it turns out, many things simplify; adding these two integrals gives ![\[{\oint_Cxh(x,y)\,\text dx+\oint_Cyh(x,y)\,\text dy=\int_0^{2\pi}rh(\mathbf c(t))(d_1\cos t-d_0\sin t)\,\text dt=\int_0^{2\pi}rdh(\mathbf c(t))(\sin t\cos t-\cos t\sin t)\,\text dt=\int_0^{2\pi}0\,\text dt=\boxed0}.\]](http://latex.artofproblemsolving.com/5/4/f/54f69a3a1b200c0e8ce4f81c1dda294d34831241.png)
So, the integral of ![\[\int\!\!\!\!\int_S\rho(x,y)\,\text dA=0,\]](http://latex.artofproblemsolving.com/0/0/9/009817ae0b5aca8576736a2c3cf621717d0cbd3e.png)
and the statement is true. (I believe I only used the fact that 
has first-order continuous partial derivatives and proved this for any circle on the plane, but I may have used some hidden assumption.)
