Show that
![\[ T_n \left( \frac{x}{1-x^2} \right) = \sum_{k=0}^n x^{2k+1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d1d675b8cbfc60c0b9639ce8447f0698_l3.png "Rendered by QuickLaTeX.com")
Starting with the algebraic identity
![\begin{align*} &&\frac{1}{1-x^2} &= 1 + x^2 + x^4 + \cdots + x^{2n} + \frac{x^{2n+2}}{1-x^2} \\[9pt] \implies && \frac{x}{1-x^2} &= x + x^3 + x^5 + \cdots + x^{2n+1} + \frac{x^{2n+3}}{1-x^2} \\[9pt] &&&= P_{2n+1} (x) + x^{2n+2} \frac{x}{1-x^2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-78df594c3846ac3677b28ebfaf712cd6_l3.png "Rendered by QuickLaTeX.com")
Since 
as 
we may apply Theorem 7.4 (page 277 of Apostol) to conclude
![\[ T_{2n+1} \left( \frac{x}{1-x^2} \right) = P_{2n+1} (x) = \sum_{k=0}^n x^{2k+1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-41c8f499d3d1acbee05a12564506e6e9_l3.png "Rendered by QuickLaTeX.com")
Show that
Starting with the algebraic identity
Since as we may apply Theorem 7.4 (page 277 of Apostol) to conclude