# [SOLVED]Solving for alpha and beta

#### dwsmith

##### Well-known member
\begin{align}
\alpha_nb^n +\beta_nb^{-n}=A_n\\
\alpha_na^n +\beta_na^{-n}=C_n
\end{align}

How does one go from that to
$$\alpha_n = \frac{A_n/a_n - C_n/b^n}{(b/a)^n-(a/b)^n}$$
and
$$\beta_n = \frac{a^nC_n - b^nA_n}{(b/a)^n-(a/b)^n}$$

#### chisigma

##### Well-known member
\begin{align}
\alpha_nb^n +\beta_nb^{-n}=A_n\\
\alpha_na^n +\beta_na^{-n}=C_n
\end{align}

How does one go from that to
$$\alpha_n = \frac{A_n/a_n - C_n/b^n}{(b/a)^n-(a/b)^n}$$
and
$$\beta_n = \frac{a^nC_n - b^nA_n}{(b/a)^n-(a/b)^n}$$
That's a [linear...] system of two equation in the unknown variables $\alpha_{n}$ and $\beta_{n}$...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
That's a [linear...] system of two equation in the unknown variables $\alpha_{n}$ and $\beta_{n}$...

Kind regards

$\chi$ $\sigma$
I do know that, but every time I solve it, I don't get the correct answer. I even put it into full version Mathematica and Wolfram online and it just returns an error.

For $\beta$, I keep getting
$$\beta_n = \frac{b^nC_n - a^nA_n}{\text{the correct denominator}}$$

#### MarkFL

Staff member
I would use elimination, for example, multiplying the second equation by:

$\displaystyle -\left(\frac{a}{b} \right)^n$

gives us:

$\displaystyle -\alpha_na^{2n}b^{-n}-\beta_nb^{-n}=-C_na^nb^{-n}$

Adding this to the first equation, we find:

$\displaystyle \alpha_n(b^n-a^{2n}b^{-n})=A_n-C_na^nb^{-n}$

$\displaystyle \alpha_n=\frac{A_n-C_na^nb^{-n}}{b^n-a^{2n}b^{-n}}=\frac{\frac{A_n}{a^n}-\frac{C_n}{b^{n}}}{\left(\frac{b}{a} \right)^n-\left(\frac{a}{b} \right)^n}$

The solution for $\displaystyle \beta_n$ can now be found by substitution.