Find the first three non-zero terms and the general term for the power series representation of f(x) = 1/(x^2 + 3).

I came up with 1/3 - x^2/9 + x^4/27, but I don't know the general term.

something is wrong. Did you type it correctly? I cannot visualize how you get x in the numerator of any term, if f(x)= 1/(x^2+3)

of course f(x) can be expanded in a power series. A little playng around should convince you that

f(x) = sum(n=0..∞) (-1/3)^(n+1) x^(2n)

thanks steve.

To find the general term of the power series representation of f(x) = 1/(x^2 + 3), we need to start by expressing the function as a geometric series. We can do this by rewriting the denominator as a perfect square. Let's begin:

f(x) = 1/(x^2 + 3)
= 1/(x^2 + (√3)^2)
= 1/[(x + √3)(x - √3)]

Now, we can express this function as a partial fraction:

f(x) = A/(x + √3) + B/(x - √3)

To determine the values of A and B, we can find a common denominator:

f(x) = [A(x - √3) + B(x + √3)]/[(x + √3)(x - √3)]

Next, we can equate the numerators:

1 = [A(x - √3) + B(x + √3)]

Expanding and collecting like terms:

1 = (A + B)x + (-A√3 + B√3)

Now, we can equate the coefficients of like powers of x:

For the coefficient of x^0 (the constant term):
0 = -A√3 + B√3
A√3 = B√3
A = B

For the coefficient of x^1:
1 = A + B

Using the relationship A = B from the constant term, we can substitute A for B in the equation for the coefficient of x^1:

1 = A + A
1 = 2A
A = 1/2

Therefore, B = A = 1/2.

Now, we can rewrite f(x) using the partial fraction decomposition:

f(x) = 1/(2(x + √3)) + 1/(2(x - √3))

To express this as a power series, we can use the geometric series formula:

1/(1 - r) = 1 + r + r^2 + r^3 + ...

Here, r = -x/√3 for the first term and r = x/√3 for the second term.

For the first term, we get:

1/(2(x + √3)) = 1/(2√3) * 1/(1 + (-x/√3))
= 1/(2√3) * (1 - x/√3 + x^2/9 - x^3/27 + ...)

And for the second term, we get:

1/(2(x - √3)) = 1/(2√3) * 1/(1 - (x/√3))
= 1/(2√3) * (1 + x/√3 + x^2/9 + x^3/27 + ...)

Now, let's find the first few non-zero terms for each term:

For the first term:
1/(2√3) * (1 - x/√3 + x^2/9 - x^3/27 + ...)
= 1/(2√3) - x/(2√3) + x^2/(18√3) - x^3/(54√3) + ...

For the second term:
1/(2√3) * (1 + x/√3 + x^2/9 + x^3/27 + ...)
= 1/(2√3) + x/(2√3) + x^2/(18√3) + x^3/(54√3) + ...

Combining the two terms, we get the power series representation of f(x):

f(x) = (1/(2√3) - x/(2√3) + x^2/(18√3) - x^3/(54√3) + ...) + (1/(2√3) + x/(2√3) + x^2/(18√3) + x^3/(54√3) + ...)

Simplifying, we can gather like terms:

f(x) = 1/√3 + (x^2 - x)/(18√3) + (-x^3)/(54√3) + ...

The general term of the power series representation of f(x) is given by:
aₙ = (-1)^(n - 1) * x^(n - 2)/(3^(n/2 + 1))
where n ≥ 2.

Therefore, the first three non-zero terms are:
a₂ = (-1)^(2 - 1) * x^(2 - 2)/(3^(2/2 + 1)) = -x/9
a₃ = (-1)^(3 - 1) * x^(3 - 2)/(3^(3/2 + 1)) = x^2/27
a₄ = (-1)^(4 - 1) * x^(4 - 2)/(3^(4/2 + 1)) = -x^3/81

So, the power series representation for f(x) with the first three non-zero terms is:
f(x) = 1/√3 - x/9 + x^2/27 - x^3/81 + ...