MHB 206.r2.11find the power series representation

[karush]

$\tiny{206.r2.11}$
$\textsf{find the power series represntation for
$\displaystyle f(x)=\frac{x^7}{3+5x^2}$
(state the interval of convergence),
then find the derivative of the series}$
\begin{align}
f(x)&=\frac{x^7}{3}\implies\frac{1}{1-\left(-\frac{5}{3}x^2\right)}&(1)\\
&=\sum_{k=0}^{\infty}\left(-\frac{5}{3}x^2\right)^k &(2)\\
&=\sum_{k=0}^{\infty}\frac{(-1)^k 5^k x^{2k+7}}{3^{k+1}} &(3)\\
f'(x)&=\sum_{k=0}^{\infty}\frac{(-1)^k 5^k x^{2k+6a}}{3^{k+1}} &(4)
\end{align}
$\textit{not sure where eq (3) comes from... looks like
Bessel function order 0...}\\$
$\textit{this was from a handwritten example very hard to read! if so the radius of convergence $=\infty$}$

 

[topsquark]

Re: 206.r2.11find the power series represntation

$\tiny{206.r2.11}$
$\textsf{find the power series represntation for
$\displaystyle f(x)=\frac{x^7}{3+5x^2}$
(state the interval of convergence),
then find the derivative of the series}$
\begin{align}
f(x)&=\frac{x^7}{3}\implies\frac{1}{1-\left(-\frac{5}{3}x^2\right)}&(1)\\
&=\sum_{k=0}^{\infty}\left(-\frac{5}{3}x^2\right)^k &(2)\\
&=\sum_{k=0}^{\infty}\frac{(-1)^k 5^k x^{2k+7}}{3^{k+1}} &(3)\\
f'(x)&=\sum_{k=0}^{\infty}\frac{(-1)^k 5^k x^{2k+6a}}{3^{k+1}} &(4)
\end{align}
$\textit{not sure where eq (3) comes from... looks like
Bessel function order 0...}\\$
$\textit{this was from a handwritten example very hard to read! if so the radius of convergence $=\infty$}$

The expression in step 2 is for [math]\frac{1}{1 - \left ( - \frac{5}{3} x^2 \right ) }[/math].

When step 3 comes along they brought back in the [math]x^7 / 3[/math] part. We also have that [math](-1)^k 5^k / 3^{k + 1}[/math] is simply the expanded form of [math](1/3)(-5/3)^k[/math]. I don't know why they expanded that.

Finally, in step 4 the "a" after the exponent 6 should not be there (typo?) and the factor of 7 from the derivative is missing.

-Dan

 

[karush]

ok thanks for eexpanding on this
the examples in the book had the factoral ! in them so I assuume whaever was just 1.
the 6a is my typo

factor of 7 ?

 

[topsquark]

ok thanks for eexpanding on this
the examples in the book had the factoral ! in them so I assuume whaever was just 1.
the 6a is my typo

No factorials here since we are dealing with a geometric series. This form doesn't use them.

factor of 7 ?

Sorry, there is a missing factor of 2k + 7, not just the 7. Line 4 is the derivative of line 3. The exponent is reduced but the factor is not there. The expression should be
[math]f'(x) = \sum_{k = 0}^{\infty} (2k + 7) \frac{(-1)^k 5^k}{3^{k + 1}} x^{2k + 6}[/math]

-Dan