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[SOLVED] Algebraic Simplification

dwsmith

Well-known member
Feb 1, 2012
1,673
$\alpha_nr^n + \beta_nr^{-n}$

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.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$\alpha_nr^n + \beta_nr^{-n}$

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}
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.