- #1

tandoorichicken

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(a) Find a power series representation for f(x) = ln(1+x).

[tex] \frac{df}{dx} = \frac{1}{1+x} = \frac{1}{1-(-x)} = 1-x+x^2-x^3+x^4-...[/tex]

[tex] \int_{n} \frac{1}{1+x} = \int_{n} (1-x+x^2-x^3+...)\dx = x-\frac{x^2}{2}+\frac{x^3}{3}-...+C = \sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1} +C [/tex]

sub x=0 in original equation: ln(1+0) = ln(1) = 0 = C.

[tex]\ln(1+x) = \sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1}[/tex]

with radius of convergence = 1.

(b) Find a power series representation for f(x) = x*ln(1+x).

If this function is differentiated, you get

[tex] \ln(1+x) + \frac{x}{x+1}[/tex]

which is the same as a sum of power series

[tex] \sum^{\infty}_{n=0} \frac{(-x)^{n+1}}{n+1} + \sum^{\infty}_{n=0} (-x)^{n+1} [/tex]

have I gone too far? or where do I go from here?

I know I will eventually have to integrate back to get the series for the original function.