data_605_hw14

part I

\[f(x) = \frac{1}{(1-x)}\] Let’s begin with first, second and third derivative…

\[f'(x) = \frac{1}{(1-x)^2}\] \[f''(x) = \frac{2}{(1-x)^3}\] \[f'''(x) = \frac{6}{(1-x)^4}\]

\[\frac{1}{(1-x)} = \frac{1}{(1-a)} + \frac{\frac{1}{(1-a)^2}}{1!}(x-a) + \frac{\frac{2}{(1-a)^3}}{2!}(x-a)^2 + \frac{\frac{6}{(1-a)^4}}{3!}(x-a)^3 + \frac{\frac{24}{(1-a)^5}}{4!}(x-a)^4 +\ ...\]

If we centered it at \(a=0\)

\[\frac{1}{(1-x)} = \frac{1}{(1-0)} + \frac{\frac{1}{(1-0)^2}}{1!}(x-0) + \frac{\frac{2}{(1-0)^3}}{2!}(x-0)^2 + \frac{\frac{6}{(1-0)^4}}{3!}(x-0)^3 + \frac{\frac{24}{(1-0)^5}}{4!}(x-0)^4 +\ ...\]

To simplify

\[\frac{1}{(1-x)} = 1 + \frac{1}{1!}(x) + \frac{2}{2!}(x)^2 + \frac{6}{3!}(x)^3 + \frac{24}{4!}(x)^4 +\ ...\] \[\frac{1}{(1-x)} = 1 + x + x^2 + x^3 + x^4 +\ ...\] The Taylor Series expansion results in

\[f(x) = \frac{1}{(1-x)} = \sum_{n=0}^{\infty}x^n\]

part II

\[f(x) = e^x\]

The derivative of \(e^x\) is always \(e^x\)

\[f'(x) = e^x\]

\[f''(x) = e^x\]

\[f'''(x) = e^x\]

\[f^{(4)}(x) = e^x\]

That means, if centered at x = 0,

\[f'(0) = e^x = 1\]

\[f''(0) = e^x = 1\]

\[f'''(0) = e^x = 1\]

So,

\[e^x = e^a + \frac{e^a}{1!}(x-a) + \frac{e^a}{2!}(x-a)^2 + \frac{e^a}{3!}(x-a)^3 + \frac{e^a}{4!}(x-a)^4 +\ ...\]

Let’s center at 0

\[e^x = e^0 + \frac{e^0}{1!}(x-0) + \frac{e^0}{2!}(x-0)^2 + \frac{e^0}{3!}(x-0)^3 + \frac{e^0}{4!}(x-0)^4 +\ ...\]

To simplify,

\[e^x = 1 + \frac{1}{1!}(x) + \frac{1}{2!}(x)^2 + \frac{1}{3!}(x)^3 + \frac{1}{4!}(x)^4 +\ ...\]

\[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +\ ...\]

And the Taylor Series expansion results in

\[f(x) = e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}\]

Part III

\[f(x) = \ln(1+x)\]

Let’s take the first few derivatives,

\[f'(x) = \frac{1}{1+x}\]

\[f''(x) = -\frac{1}{(1+x)^2}\]

\[f'''(x) = \frac{2}{(1+x)^3}\]

\[f^{(4)}(x) = -\frac{6}{(1+x)^4}\]

Then, we get

\[\ln(1+x) = \ln(1+a) + \frac{\frac{1}{1+a}}{1!}(x-a) + \frac{-\frac{1}{(1+a)^2}}{2!}(x-a)^2 + \frac{\frac{2}{(1+a)^3}}{3!}(x-a)^3 + \frac{-\frac{6}{(1+a)^4}}{4!}(x-a)^4 +\ ...\]

Again, let’s center at 0 \(a=0\)

\[\ln(1+x) = \ln(1+0) + \frac{\frac{1}{1+0}}{1!}(x-0) + \frac{-\frac{1}{(1+0)^2}}{2!}(x-0)^2 + \frac{\frac{2}{(1+0)^3}}{3!}(x-0)^3 + \frac{-\frac{6}{(1+0)^4}}{4!}(x-0)^4 +\ ...\]

We can simplify above into

\[\ln(1+x) = \ln(1) + \frac{1}{1!}(x) - \frac{1}{2!}(x)^2 + \frac{2}{3!}(x)^3 - \frac{6}{4!}(x)^4 +\\ ...\]

\[\ln(1+x) = 0 + x - \frac{x^2}{2!} + \frac{2x^3}{3!} - \frac{6x^4}{4!} +\ ...\]

\[\ln(1+x) = 0 + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +\ ...\]

\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +\ ...\]

Thus, the Taylor Series expansion results in

\[f(x) = \ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}x^n}{n}\]