# How do you find a power series representation for f(x)=3/((1-5x)^2) and what is the radius of convergence?

Sep 15, 2015

I'm assuming you really want the Taylor series for this function centered at $x = 0$, which is $3 + 30 x + 225 {x}^{2} + 1500 {x}^{3} + 9375 {x}^{4} + \cdots$, with a radius of convergence equal to $\frac{1}{5}$.

#### Explanation:

There's a couple ways this can be done. The most straightforward approach is to let $f \left(x\right) = \frac{3}{{\left(1 - 5 x\right)}^{2}}$ and take derivatives to find it. You can check that $f ' \left(x\right) = \frac{30}{{\left(1 - 5 x\right)}^{3}}$, $f ' ' \left(x\right) = \frac{450}{{\left(1 - 5 x\right)}^{4}}$, $f ' ' ' \left(x\right) = \frac{9000}{{\left(1 - 5 x\right)}^{5}}$, $f ' ' ' ' \left(x\right) = \frac{225000}{{\left(1 - 5 x\right)}^{6}}$, etc...

Then $f \left(0\right) = 3$, $f ' \left(0\right) = 30$, $f ' ' \left(0\right) = 450$, $f ' ' ' \left(0\right) = 9000$, $f ' ' ' ' \left(0\right) = 225000$, etc...

The Taylor series formula is f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f''''(0))/(4!)x^4+cdots. Since 450/(2!)=225, 9000/(3!)=1500, and 225000/(4!)=9375, we get the same answer as above: $3 + 30 x + 225 {x}^{2} + 1500 {x}^{3} + 9375 {x}^{4} + \cdots$.

You can try the ratio test to confirm the radius of convergence is $\frac{1}{5}$, but you'd have to find a formula for the coefficients of the powers of $x$ to do so.
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You can do that if you want. I'm going to use a more creative approach. Notice that $\int \frac{3}{{\left(1 - 5 x\right)}^{2}} \setminus \mathrm{dx} = \frac{\frac{3}{5}}{1 - 5 x} + C$ so that $\frac{d}{\mathrm{dx}} \left(\frac{\frac{3}{5}}{1 - 5 x}\right) = \frac{3}{{\left(1 - 5 x\right)}^{2}}$. The Taylor series for $\frac{\frac{3}{5}}{1 - 5 x}$ can be found by using the formula for the sum of a geometric series: $a + a r + a {r}^{2} + a {r}^{3} + \cdots = \frac{a}{1 - r}$ for $| r | < 1$.

In other words, $\frac{\frac{3}{5}}{1 - 5 x} = \frac{3}{5} + 3 x + 15 {x}^{2} + 75 {x}^{3} + 375 {x}^{4} + 1875 {x}^{5} + \cdots$ for $| 5 x | < 1 \leftrightarrow | x | < \frac{1}{5}$, making the radius of convergence $\frac{1}{5}$.

Now differentiate the preceding series to find the final answer: $3 + 30 x + 225 {x}^{2} + 1500 {x}^{3} + 9375 {x}^{4} + \cdots$. It's a theorem that guarantees that the radius of convergence for the differentiated series is the same as the original, and is therefore $\frac{1}{5}$.