# Understanding Series stuff

• April 20th 2008, 09:12 AM
bobak
Understanding Series stuff
Now i have managed to solve the attached problem using some instructions given to me. But i really do not understand how this works.

Under what premise can I assume that $a_{n}=t^n$ then I solve the equation $t^n = 3t^{n-1}-t^{n-2}$. then I was told that $a_{n}$ is a linear combination of the roots.

I solve this question fine, I just really do not understand any of the theory.

Could someone show me some reading material or a proof of this method.

Many Thanks

Bobak
• April 21st 2008, 04:18 PM
PaulRS
Given a sequence: $a_n$

Let: $
A\left( x \right) = \sum\limits_{n = 0}^\infty {\left( {\frac{{a_n }}
{{n!}} \cdot x^n } \right)}
$

(this is called the exponential generating function of the sequence)

Note that $
A'\left( x \right)
$
is the exponential generating function of $
a_{n + 1}

$
(try differentiating)

So if we have the recurrence equation: $
a_{n + 2} = c_1 \cdot a_{n + 1} + c_2 \cdot a_n
$
we must have: $
A''\left( x \right) = c_1 \cdot A'\left( x \right) + c_2 \cdot A\left( x \right)
$

Suppose there are two different solutions to this equation: $t^2-c_1\cdot{t}-c_2=0$(1) (maybe complex)
Then the general solution to this differential equation is: $
A\left( x \right) = A \cdot e^{t_1 \cdot x} + B \cdot e^{t_1 \cdot x}
$
(for some constants A and B that depend on the initial conditions) where $t_1$ and $t_2$ are the roots of the equation (1)

Therefore: $A(x)=
A \cdot e^{t_1 \cdot x} + B \cdot e^{t_1 \cdot x} = A \cdot \sum\limits_{n = 0}^\infty {\left( {\frac{{t_1 ^n }}
{{n!}} \cdot x^n } \right)} + B \cdot \sum\limits_{n = 0}^\infty {\left( {\frac{{t_2 ^n }}
{{n!}} \cdot x^n } \right)}
$

Thus: $
A\left( x \right) = \sum\limits_{n = 0}^\infty {\left( {\frac{{A \cdot t_1 ^n + B \cdot t_2 ^n }}
{{n!}} \cdot x^n } \right)}
$

And we finally get $
A \cdot t_1 ^n + B \cdot t_2 ^n = a_n
$
(because of the uniqueness of the power series expansion of a function)