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

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Discussion Overview

The discussion revolves around solving for the variables α (alpha) and β (beta) in a system of exponential equations. Participants explore methods for deriving expressions for α and β from the given equations, focusing on algebraic manipulation and elimination techniques.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants present a system of equations involving α and β, seeking to derive specific formulas for these variables.
  • One participant expresses frustration with obtaining incorrect results when attempting to solve for β, suggesting that their approach may not yield the expected outcomes.
  • Another participant proposes using elimination as a method to solve the system, providing a detailed algebraic manipulation to derive an expression for α.
  • There is mention of using computational tools like Mathematica and Wolfram, but one participant reports errors when using these tools for their calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach or solutions for α and β, as there are differing methods proposed and some unresolved issues regarding the accuracy of results.

Contextual Notes

Participants highlight potential issues with the denominator in their derived expressions, indicating that the correct form may depend on specific assumptions or manipulations that have not been fully resolved.

Dustinsfl
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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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