Expand 4/(x^2+4) as a power series centered at 0.

Let's forget the factor of 5, and just expand

f(x) = 1/(x^2+4)
f' = -2x/(x^2+4)^2
f" = 8(3x^2-4)/(x^2+4)^3
f(3) = -96x(x^2-4)/(x^2+4)^4
and then just evaluate the Taylor Series at x=0
f(x) = 1/4 - x^2/16 + x^4/64 - ...

ohh that makes sense...but is there a way of solving it without using Taylor series?

what else is a power series?

I guess you could use the binomial theorem, and expand
(4+x^2)^(-1) = 4^-1 + (-1)/1! 4^-2 x^2 + (-1)(-2)/2! 4^-3 x^4 + ...
= 1/4 - x^2/16 + x^4/64 - ...


= ∑ (-1)^k x^(2k)/4^(k+1)
k=0

To expand 4/(x^2+4) as a power series centered at 0, we will first need to express it as a geometric series.

We start by noticing that the denominator, x^2+4, can be factored as (2i)(-2i+x)(2i+x), where i is the imaginary unit.

Now, we can rewrite 4/(x^2+4) as follows:
4/(x^2+4) = A/(2i) + B/(2i+x) + C/(2i-x)

To determine the values of A, B, and C, we can multiply both sides of the equation by (x^2+4) and simplify.

4 = A(2i+x)(2i-x) + B(2i)(2i-x) + C(2i)(2i+x)

Next, we can equate like terms on both sides of the equation.

For the terms without an x, we have:
4 = A(2i)(-2i) + B(2i)(2i) + C(2i)(2i)

Simplifying further, we get:
4 = A(4) + B(4) + C(4)

This yields:
4 = 4(A + B + C)

Thus, A + B + C = 1

For the terms with an x, we have:
0 = A(2i)(x) + B(-2i)(x) + C(2i)(x)

Simplifying further, we get:
0 = Ax(2i) - Bx(2i) + Cx(2i)

This yields:
0 = 2i(A - B + C)x

Thus, A - B + C = 0

To solve these two equations simultaneously and find the values of A, B, and C, we subtract the second equation from the first:

A + B + C - (A - B + C) = 1 - 0

Simplifying, we get:
2B = 1

Therefore, B = 1/2.

Substituting B back into the equation A - B + C = 0, we have:
A - (1/2) + C = 0

Rearranging the equation, we get:
A + C = 1/2

We now have B = 1/2 and A + C = 1/2 as our values.

With this information, we can rewrite 4/(x^2+4) as:

4/(x^2+4) = (1/(2i))/(2i) + (1/2)/(2i+x) + (1/2)/(2i-x)

Simplifying further:

4/(x^2+4) = (1/(4i^2)) * (1/(2i)) + 1/(2(2i+x)) + 1/(2(2i-x))

Since i^2 = -1, we can simplify a bit more:

4/(x^2+4) = (-1/4) * (1/(2i)) + 1/(2(2i+x)) + 1/(2(2i-x))

Finally, we can combine the fractions and factor out the constant term:

4/(x^2+4) = (-1/8i) + 1/(4i+x) + 1/(4i-x)

Now, we have successfully expanded 4/(x^2+4) as a power series centered at 0.