I have to find a power series representation for;
f(x) = ln(x^3 + 9)
Any idea on how to do this? I believe it is something to do with the derivative but I'm not sure where to go from there.
f(x) = ln(x^3 + 9)
Any idea on how to do this? I believe it is something to do with the derivative but I'm not sure where to go from there.
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for power series, you need to somehow get the function in the form of 1/(1 - x), and it's usually by either differentiation or integration.
y = ln(x^3 + 9)
y' = 3x^2 / (x^3 + 9)
y' = 3x^2 * 1/(9 + x^3) = 3x^2 * (1/9) 1/(1 + x^3/9) = (1/3)x^2 * 1/[1 - (-x^3/9)]
(1/3)x^2 * ∑{0,inf} (-x^3/9)^n
(1/3)x^2 * ∑{0,inf} (-1)^n * x^(3n) / 3^(2n)
multiply (1/3)x^2 back in,
∑{0,inf} x^(3n+2) / 3^(2n+1)
integrate to return to the original function,
C + ∑{0,inf} (-1)^n * x^(3n+3)/(3n+3) * 1/3^(2n+1)
Edit:
I forgot to say that, to determine C, plug 0 in for x, so:
ln(9 + 0^3) = C
ln(9) = C
so final answer:
ln(9) + ∑{0,inf} (-1)^n * x^(3n+3)/(3n+3) * 1/3^(2n+1)
y = ln(x^3 + 9)
y' = 3x^2 / (x^3 + 9)
y' = 3x^2 * 1/(9 + x^3) = 3x^2 * (1/9) 1/(1 + x^3/9) = (1/3)x^2 * 1/[1 - (-x^3/9)]
(1/3)x^2 * ∑{0,inf} (-x^3/9)^n
(1/3)x^2 * ∑{0,inf} (-1)^n * x^(3n) / 3^(2n)
multiply (1/3)x^2 back in,
∑{0,inf} x^(3n+2) / 3^(2n+1)
integrate to return to the original function,
C + ∑{0,inf} (-1)^n * x^(3n+3)/(3n+3) * 1/3^(2n+1)
Edit:
I forgot to say that, to determine C, plug 0 in for x, so:
ln(9 + 0^3) = C
ln(9) = C
so final answer:
ln(9) + ∑{0,inf} (-1)^n * x^(3n+3)/(3n+3) * 1/3^(2n+1)