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

Question

Scarlett Allison · Accepted Answer

Recall that:

The power series for #1/(1-x)# is #sum_(n=0)^N x^n#.

What you can do is take the derivative of both sides:

#d/(dx)[sum_(n=0)^N x^n] = d/(dx)[1/(1-x)] = color(green)(d/(dx)[1 + x + x^2 + ...])#

When we take the derivative, note that the first term will disappear (#d/(dx)[1] = 0#), so the first #n# shifts from #n=0# to #n=1#.

#sum_(n=1)^N nx^(n-1) = 1/(1-x)^2 = color(green)(?)#

We can shift it back to #n = 0# for conventional purposes.

How you do that is consider that you are currently starting at #n=1#. That means #n-1 = 0#. To get the exponent to become #0# using #n = 0# below #sum#, #n-1# becomes #n#.

When #n = 1# below #sum#, #nx^(n-1)# starts at #1x^0#, so when #n = 0# below #sum#, #nx^"stuff"# becomes #(n+1)x^"stuff"#.

#color(highlight)(sum_(n=0)^N (n+1)x^(n) = 1/(1-x)^2) = color(green)(?)#

So, all we need to do is do it for the explicit series to get:

#1/(1-x)^2 = d/(dx)[1 + x + x^2 + ...]#

#= color(blue)(1 + 2x + 3x^2 + 4x^3 + ...)#

The radius of convergence is based on the idea that the sum of magnitudes #|a| >= 1# will not converge if they are all positive.

#1/(1-x)^2 > 1# when #|x| >= 1#.

Therefore, the radius of converge is:

#color(blue)(x in (-1, 1))#

or you can write it as:

#color(blue)(|x| < 1)#

Emilia Aguilar · Answer

undefined To find the power series representation of $ f(x) = \frac{x}{(1-x)^2} $, we first find its derivative and integrate term by term.

The function $ f(x) = \frac{x}{(1-x)^2} $ can be expressed as $ f(x) = x(1-x)^{-2} $.

Now, let's find the derivatives:

1. $ f(x) = x(1-x)^{-2} $
2. $ f'(x) = (1-x)^{-2} - 2x(1-x)^{-3} $
3. $ f''(x) = 2(1-x)^{-3} - 6x(1-x)^{-4} $

And so on. Each derivative introduces a higher power of $ (1-x) $ in the denominator.

Now, we evaluate each derivative at $ x = 0 $ to find the coefficients of the power series expansion:

1. $ f(0) = 0 $
2. $ f'(0) = 1 $
3. $ f''(0) = 2 $
4. $ f'''(0) = 6 $

Thus, the coefficients of the power series expansion are 0, 1, 2, 6, and so on.

The power series representation of $ f(x) $ is therefore:

$$ f(x) = \sum_{n=0}^{\infty} n x^n $$

The radius of convergence for this power series is $ R = 1 $, which can be determined using the Ratio Test.

Christopher Allers · Answer

undefined To find the power series representation for $ f(x) = \frac{x}{(1-x)^2} $, we can first express $ \frac{1}{(1-x)^2} $ as a power series, then multiply it by $ x $.

The power series representation for $ \frac{1}{(1-x)^2} $ is $ \sum_{n=0}^{\infty} (n+1)x^n $.

Multiplying by $ x $, we get $ x \cdot \sum_{n=0}^{\infty} (n+1)x^n = \sum_{n=0}^{\infty} (n+1)x^{n+1} $.

So, the power series representation for $ f(x) $ is $ \sum_{n=0}^{\infty} (n+1)x^{n+1} $.

The radius of convergence for this power series can be found using the ratio test. Let $ a_n = n+1 $. Applying the ratio test:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{n+2}{n+1} \right| = 1 $$

Since the limit is equal to 1, the radius of convergence is $ R = \frac{1}{1} = 1 $. Therefore, the radius of convergence for the power series representation of $ f(x) $ is 1.