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
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.
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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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