How do you find the power series representation for the function?

Question

Also, how would you determine the interval of convergence?

f(x)=x/(4x+1)

much thanks!

Answer 1

The previous answers have given you methods that work in all situations, but which take a lot of work. In this specific case, there are easier methods.

In particular, we know the geometric series:
1 + r + r^2 + r^3 + .... = 1/(1-r)

Therefore,
1/(1+r) = 1 - r + r^2 - r^3 +...
1/(1+ 4x) = 1 - 4x + (4x)^2 - (4x)^3 +...
x/(1+ 4x) = x - 4x^2 + 4^2*x^3 - 4^3*x^4 +...

Since we already know that this series is derived from the geometric series, we see that it converges only when the geometric series converges, which is when the ratio r is < 1, which implies that │4x│ < 1, or │x│< 1/4.

(You can also do the comparison a_n+1/a_n and find the same result, but that is based on the same mathematical facts: the ratio will be r = 4x.)

Answer 2

The power series is basically the Taylor series approximation around some value of x=c.

I.e. f(x) = f(c) + f'(c)*(x-c) + f''(c)*(x-c)^2 + f'''(c)*(x-c)^2 + ...

For your function it's easier to write it as
f(x) = 1/[4+x^(-1)],
which has derivatives
f^{n}(x) = 1/[4+x^(-1)]^(n+1)

So the power series is
f(x) = 1/[4+c^(-1)]+(x-c)/[4+c^(-1)]^2 + (x-c)^2/[4+c^(-1)]^3 + ...
= 1/[4+c^(-1)] * sum_[k=1 to infty] of {(x-c)/[4+c^(-1)]}^k

The range of convergence here applies to x, but will depend on your value of c. In a power series like this the summand has to be less than one in magnitude for it to converge. This means that | x-c | < | c/(4c+1) |, which you can solve for x.

Answer 3

use a taylor expansion, that is differentiate twice, twice etc , hopefully you see a pattern to express the coefficient in a general way.

the interval of convergence can be calculated by using some of the theoremes about convergence of series. usually you can bound the sum by some known convergent serie.