Vtyjkyuk

Question:

For the given linear homogeneous difference equation, find the general solution:

$$y_n+2 + y_n+1 + y_n = 0$$

With the initial conditions of:

$$y(0)=sqrt3, y(1) = 0$$

Attempted Answer:

I approached the problem normally as one would except by solving the auxiliary equation which yields:

$$m^2 + m + 1 = 0$$
$$implies m = frac-12 pm fracisqrt32$$

Now the general solution is of form:

$$y_n = A left(frac-12 - fracisqrt32right)^n
+ B left(frac-12 + fracisqrt32right)^n $$

Here is the part where I get stuck when I substitute the initial conditions to form a system of equations:

$$A + B = sqrt3$$
$$A left(frac-12 - fracisqrt32right)
+ B left(frac-12 + fracisqrt32right) = 0$$

Now to make my life easier (which it really didn't), I decided to take a 'shortcut' and rewrite the 2nd equation as the sum of an imaginary and real part:

$$frac-12(A + B) + ifracsqrt32(-A + B) = 0 + 0i$$

Since the real part on the LHS must equal the real part on the RHS, and the same applies for the imaginary part, we obtain:

$$frac-12(A + B) = 0$$
$$fracsqrt32(-A + B) = 0$$

So the above implies that $A = B = -B$ which can only be true if $A = B = 0$. However if that is the case, then the first equation in the system of equations ($A + B = sqrt3$) implies that 0 = $sqrt3$.

I know that there is a mistake somewhere in the logic of my reasoning (I think perhaps when I equated each real and imaginary part of the LHS to the RHS) but I don't know why and where exactly. Please point out my mistake.

What happens
when the roots of the
characteristic polynomial are complex
is that the solutions
have a periodic component.

Your case is
$y_n
= a left(frac-12 - fracisqrt32right)^n
+ b left(frac-12 + fracisqrt32right)^n
=a r^n + bs^n
$.
Note that
$r+s = -1$
and
$r-s = -isqrt3$.

If
$y_0 = u$
and
$y_1 = v$,
then
$a+b = u$
and
$ar+bs = v$.
Since
$b = u-a$,
$v
= ar+(u-a)s
=a(r-s)+us
$,
so
$a
=fracv-usr-s
$
and
$b
=u-a
=u-fracv-usr-s
=fracu(r-s)-v+usr-s
=fracur-vr-s
$.

In your case,
$u=sqrt3,
v = 0,
r=frac-12 - fracisqrt32,
s=frac-12 + fracisqrt32,
r-s=-isqrt3
$,
so
$a
=fracv-usr-s
=frac-sqrt3(frac-12 + fracisqrt32)-isqrt3
=-i(frac-12 + fracisqrt32)
=fraci2 + fracsqrt32
$
and
$b
=u-a
=fracsqrt32-fraci2
$.

(We are getting close.)

We have
$y_n
=ar^n+bs^n
$.

By the magic of
assigned problems,
$|r| = |s| = 1$.
$r = e^-2ipi/3
$,
so
$r^n
=e^-2nipi/3
=cos(-2npi/3)+isin(-2npi/3)
=cos(2npi/3)-isin(2npi/3)
$.
Similarly,
$s = e^2ipi/3
$,
so
$s^n
=e^2inpi/3
=cos(2npi/3)+isin(2npi/3)
$.

Also
$a
=e^ipi/6
$
and
$b
=e^-ipi/6
$.

Therefore
(finally!)

$beginalign*
y_n
&=ar^n+bs^n\
&=e^ipi/6e^-2inpi/3+e^-ipi/6e^2inpi/3\
&=e^pi i(1/6-2n/3)+e^pi i(-1/6+2n/3)\
&=cos(pi (1/6-2n/3))+isin(pi (1/6-2n/3))
+cos(pi (-1/6+2n/3))+isin(pi (-1/6+2n/3))\
&=cos(pi (1/6-2n/3))+cos(pi (-1/6+2n/3))
+i(sin(pi (1/6-2n/3))+sin(pi (-1/6+2n/3)))\
&=2cos(pi (1/6-2n/3))
qquadtextsince
cos(-x)=cos(x) text and sin(x) = -sin(-x)\
endalign*
$

You can get explicit values for
values of $n$ mod 6,
but this shows how the result is periodic.

A check is that
the result is real,
with no imaginary part.
This has to hold,
since the initial values
and the recurrence coefficients
are all real.

Your $A$ and $B$ have to be complex. Your system is equivalent to $$A+B=sqrt 3\-(A+B)/2-(A-B)fracisqrt 32=0,$$
which gives you $$A-B = i\A+B=sqrt3$$ and now it is to solve.

Use generating functions. Define $A(z) = sum_n ge 0 y_n z^n$, multiṕly your recurrence by $z^n$, sum over $n ge 0$ and recognize some sums:

$beginalign
fracA(z) - y_0 -y_1 zz^2
+ fracA(z) - y_0z + A(z)
&= 0 \
fracA(z) - sqrt3z^2
+ fracA(z) - sqrt3z + A(z)
&= 0
endalign$

Solving for $A(z)$:

$beginalign
A(z)
&= frac(1 + z) sqrt31 + z + z^2 \
&= frac1 - z^21 - z^3 sqrt3
endalign$

We need:

$beginalign
[z^n] A(z)
&= [z^n] frac1 - z^21 - z^3 sqrt3 \
&= sqrt3 [z^n] (1 - z^2) sum_k ge 0 z^3 k \
&= sqrt3 left(
[z^n] sum_k ge 0 z^3 k
- [z^n - 2] sum_k ge 0 z^3 k
right) \
&= begincases
sqrt3 & textif (n equiv 0 pmod3) \
- sqrt3 & textif (n equiv 1 pmod3 wedge n ge 2) \
0 & textotherwise
endcases
endalign$