MHB How Do You Solve for Alpha and Beta in Exponential Equations?

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The discussion focuses on solving for the coefficients alpha (α) and beta (β) in a system of exponential equations. The participants explore the transition from the given equations to the derived formulas for α and β. A method involving elimination is suggested, where the second equation is manipulated to facilitate the calculation of α. Despite attempts to use computational tools like Mathematica, some users report errors in their solutions. The conversation emphasizes the importance of correctly applying algebraic manipulation to derive the coefficients accurately.
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\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
$$
\beta_n = \frac{a^nC_n - b^nA_n}{(b/a)^n-(a/b)^n}
$$
 
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dwsmith said:
\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

$\chi$ $\sigma$
 
chisigma said:
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
$$
\beta_n = \frac{b^nC_n - a^nA_n}{\text{the correct denominator}}
$$
 
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.
 
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