# Thread: Expressing functions as power series

1. ## Expressing functions as power series

Professor really hasn't explained any of this and gave us a practice exam on it.

$f(x)=\frac{x}{4x+1}$

Would this simply be $\sum_{k=0}^{\infty}x(-1)^{k}(4x)^{k}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{4}(4x)^{k+1}$

I reasoned this since:

$\frac{1}{1+4x}=\sum_{k=0}^{\infty}(-1)^{k}(4x)^{k}$

2. Originally Posted by Pinkk
Professor really hasn't explained any of this and gave us a practice exam on it.

$f(x)=\frac{x}{4x+1}$

Would this simply be $\sum_{k=0}^{\infty}x(-1)^{k}(4x)^{k}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{4}(4x)^{k+1}$

I reasoned this since:

$\frac{1}{1+4x}=\sum_{k=0}^{\infty}(-1)^{k}(4x)^{k}$
Looks good to me.

Just remember that the series

$\sum_{k=0}^\infty{(-1)^k(4x)^k}$ converges only when

$|-4x|< 1$

$|4x| < 1$

$4|x| < 1$

$|x| < \frac{1}{4}$

$-\frac{1}{4} < x < \frac{1}{4}$.

3. Correct.

However it looks better as
${(-1)^{n}}{x^{n+1}4^{n}}$

And the interval of convergence by prove it is correct no need to check endpoints.

4. For these problems I do not need to find the interval of convergence or radius of convergence but finding those are probably the only things of infinite series I can do without any problem.