MHB Where did I go wrong in simplifying this algebraic expression?

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
Click For Summary
The discussion revolves around the simplification of the algebraic expression $\alpha_nr^n + \beta_nr^{-n}$, where initial formulations for $\alpha_n$ and $\beta_n$ were incorrectly derived. The user initially calculated $\alpha_n$ and $\beta_n$ but ended up with an expression that did not match the expected solution. By rearranging $\alpha_n$ and $\beta_n$ to have a common denominator, the user successfully simplified the expression to match the correct form. The final result confirms that the error lay in the initial handling of the denominators. The thread concludes with the user marking the issue as resolved.
Dustinsfl
Messages
2,217
Reaction score
5
$\alpha_nr^n + \beta_nr^{-n}$

We know that
\begin{alignat*}{3}
\alpha_na^n+\beta_na^{-n} & = & A_n\\
\alpha_nb^n+\beta_nb^{-n} & = & 0
\end{alignat*}
So I ended up with
$$
\alpha_n = \frac{-A_n}{b^{2n}/a^n-a^n}
$$
and
$$
\beta_n = \frac{A_n}{1/a^n-a^n/b^{2n}}
$$

When I plug them in, I obtain
$$
\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}/a^n-r^{2n}}{b^{2n}-a^{2n}}
$$

However, the solution is supposed to be
$$
\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}
$$

I cannot find my error.
 
Mathematics news on Phys.org
dwsmith said:
$\alpha_nr^n + \beta_nr^{-n}$

We know that
\begin{alignat*}{3}
\alpha_na^n+\beta_na^{-n} & = & A_n\\
\alpha_nb^n+\beta_nb^{-n} & = & 0
\end{alignat*}
So I ended up with
$$
\alpha_n = \frac{-A_n}{b^{2n}/a^n-a^n}
$$
and
$$
\beta_n = \frac{A_n}{1/a^n-a^n/b^{2n}}
$$

When I plug them in, I obtain
$$
\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}/a^n-r^{2n}}{b^{2n}-a^{2n}}
$$

However, the solution is supposed to be
$$
\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}
$$

I cannot find my error.

I think by rearranging the answers for \(\alpha_n\) and \(\beta_n\) to have a common denominator may help in the algebraic simplification.

\[\alpha_n=\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\]

\[\beta_n=\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\]

\begin{eqnarray}

\therefore \alpha_nr^n + \beta_nr^{-n}&=&\left(\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\right)r^n+\left(\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\right)r^{-n}\\

&=&\left(\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\right)r^n+\left(\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\right)r^{-n}\\

&=&\frac{A_{n}a^nb^{2n}-A_{n}a^nr^{2n}}{r^n(b^{2n}-a^{2n})}\\

&=&\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}

\end{eqnarray}
 
Sudharaka said:
I think by rearranging the answers for \(\alpha_n\) and \(\beta_n\) to have a common denominator may help in the algebraic simplification.

\[\alpha_n=\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\]

\[\beta_n=\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\]

\begin{eqnarray}

\therefore \alpha_nr^n + \beta_nr^{-n}&=&\left(\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\right)r^n+\left(\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\right)r^{-n}\\

&=&\left(\frac{-A_{n}a^n}{b^{2n}-a^{2n}}\right)r^n+\left(\frac{A_{n}a^nb^{2n}}{b^{2n}-a^{2n}}\right)r^{-n}\\

&=&\frac{A_{n}a^nb^{2n}-A_{n}a^nr^{2n}}{r^n(b^{2n}-a^{2n})}\\

&=&\left(\frac{a}{r}\right)^nA_n\frac{b^{2n}-r^{2n}}{b^{2n}-a^{2n}}

\end{eqnarray}

I marked the thread solved a little bit ago.
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K