Question

What is c0, c1 and ck?

Answer 1

By the ratio test, the first series converges if

$\displaystyle \lim_{k\to\infty} \left|\frac{\frac{x^{k+1}}{k+10}}{\frac{x^k}{k+9}}\right| = |x| \lim_{k\to\infty} \frac{k+9}{k+10} = |x| < 1$

while the second series converges if

$\displaystyle \lim_{k\to\infty} \left|\frac{(k+1)(k+5)x^{k+1}}{k(k+4)x^k}\right| = |x| \lim_{k\to\infty} \frac{k^2+6k+5}{k^2+4k} = |x| < 1$

So as long as |x| < 1, we have

$\displaystyle h(x) = -(x+4) f(x) + 7x^2 g(x)$

Distribute f(x) :

$\displaystyle h(x) = -x f(x) - 4 f(x) + 7x^2 g(x)$

Substitute f and g with their series representations:

$\displaystyle h(x) = -x \sum_{k=0}^\infty \frac{x^k}{k+9} - 4 \sum_{k=0}^\infty \frac{x^k}{k+9} + 7x^2 \sum_{k=0}^\infty k(k+4)x^k$

Simplify the powers of x :

$\displaystyle h(x) = -\sum_{k=0}^\infty \frac{x^{k+1}}{k+9} - 4 \sum_{k=0}^\infty \frac{x^k}{k+9} + 7 \sum_{k=0}^\infty k(k+4)x^{k+2}$

Shift the index in each sum as needed so that each one is written in terms of $x^k$ :

$\displaystyle h(x) = -\sum_{k=1}^\infty \frac{x^k}{k+8} - 4 \sum_{k=0}^\infty \frac{x^k}{k+9} + 7 \sum_{k=2}^\infty (k-2)(k+2)x^k$

Pull out as many terms as needed from each sum so that they each start with the same index:

$\displaystyle h(x) = -\left(\frac x9 + \sum_{k=2}^\infty \frac{x^k}{k+8}\right) - 4 \left(\frac19 + \frac x{10} + \sum_{k=2}^\infty \frac{x^k}{k+9}\right) + 7 \sum_{k=2}^\infty (k-2)(k+2)x^k$

Condensing the series gives

$\displaystyle h(x) = -\frac49 - \frac{23}{45}x + \sum_{k=2}^\infty \left(7(k-2)(k+2)-\frac1{k+8} - \frac4{k+9}\right)x^k$

So, if

$\displaystyle h(x) = \sum_{k=0}^\infty c_kx^k = c_0 + c_1x + c_2x^2 + \cdots$

then it follows that

$c_0 = -\dfrac49$

$c_1 = -\dfrac{23}{45}$

and for k ≥ 2,

$c_k = \displaystyle 7(k-2)(k+2)-\frac1{k+8} - \frac4{k+9}$