power series representative of int [t=0,x] [1/(2 + t^3)] dt

[cheffy]

Find a power series representative of

\(\displaystyle \
\int\limits_0^x {\frac{{dt}}{{2 + t^3 }}}
\\)

and for what values of t does the series converge?

I don't know what this is asking me to do.

Thanks!

 

[galactus]

I beleive what they want you to do is find the Taylor series for \(\displaystyle \frac{1}{2+t^{3}}\), then integrate.

\(\displaystyle \L\\\frac{1}{2+t^{3}}=\frac{1}{2}-\frac{t^{3}}{4}+\frac{t^{6}}{8}-\frac{t^{9}}{16}+..............\)

\(\displaystyle \L\\\sum_{n=0}^{\infty}(-1)^{n}\frac{t^{3n}}{2^{n+1}}\)

 

[skeeter]

remember the sum of a geometric series?

\(\displaystyle \L \frac{1}{1-r} = 1 + r^2 + r^3 + ...\)

replace r with (-x) ...

\(\displaystyle \L \frac{1}{1+x} = \frac{1}{1-(-x)} = 1 - x + x^2 - x^3 + ...\)

see how your quotient can be similarly "manipulated" into a power series?

\(\displaystyle \L \frac{1}{2+t^3} = \frac{1}{2} \cdot \frac{1}{1 - \left(-\frac{t^3}{2}\right)} = \frac{1}{2}\left[1 - \left(\frac{t^3}{2}\right) + \left(\frac{t^3}{2}\right)^2 - \left(\frac{t^3}{2}\right)^3 + ...\right]\)

one of the big "ideas" of power series is the ease with which they may be integrated and differentiated.

 

[pka]

\(\displaystyle \L\begin{array}{l}
f(x) = \sum\limits_{n = 0}^\infty {\frac{{f^{\left( n \right)} \left( 0 \right)}}{{n!}}x^n } = f(0) + \frac{{f'(0)}}{{1!}}x + \frac{{f''(0)}}{{2!}}x^2 + \cdots \\
f(x) = \int\limits_0^x {\frac{{dt}}{{2 + t^3 }}} \quad \Rightarrow \quad f'(x) = \frac{1}{{2 + x^3 }} \\
\end{array}\)

 

[cheffy]

I tried doing the ratio test and I got

\(\displaystyle \
{\lim }\limits_{n \to \infty } \left| {\frac{{ - t^{3n + 3} }}{{2t^{3n} }}} \right|
\\)

How do I get my r so I can find my interval of convergence? Thanks.

 

[skeeter]

doesn't a geometric series converge if \(\displaystyle \L |r| < 1\) ?

\(\displaystyle \L \left|-\frac{t^3}{2}\right| < 1\)