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

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Alpha Beta
Click For Summary
SUMMARY

The discussion focuses on solving for the coefficients αn and βn in the system of exponential equations given by αnbn + βnb-n = An and αnan + βna-n = Cn. The derived formulas for αn and βn are αn = (An/an - Cn/bn) / ((b/a)n - (a/b)n) and βn = (anCn - bnAn) / ((b/a)n - (a/b)n). The discussion also highlights the use of elimination methods to derive these coefficients accurately.

PREREQUISITES
  • Understanding of exponential equations
  • Familiarity with linear algebra concepts
  • Proficiency in mathematical manipulation and substitution
  • Experience with computational tools like Mathematica
NEXT STEPS
  • Study the method of elimination in linear systems
  • Learn about solving systems of equations involving exponential functions
  • Explore the use of Mathematica for symbolic computation
  • Investigate the properties of exponential growth and decay
USEFUL FOR

Mathematicians, students studying algebra and calculus, and anyone interested in solving complex exponential equations.

Dustinsfl
Messages
2,217
Reaction score
5
\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
$$
\beta_n = \frac{a^nC_n - b^nA_n}{(b/a)^n-(a/b)^n}
$$
 
Mathematics news on Phys.org
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K