For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.

Solutions:

From Taylor’s series:

\(P(x)=f(0)+f′(0)(x)+f′′(0)\frac{x^2}{2!} + f(3)(0)\frac{x^3}{3!}+....+f(n)(0)\frac{x^n}{n!}\)

  1. \(f(x) = \frac{1}{(1-x)}, f(0) = 1\)

\(f^1(x) = \frac{1}{(1-x)^2}, f^1(0) = 1\)

\(f^2(x) = \frac{2}{(1-x)^3}, f^2(0) = 2\)

\(f^3(x) = \frac{6}{(1-x)^4}, f3(0) = 6\)

\(f^4(x) = \frac{24}{(1-x)^5}, f^4(0) = 24\)

\(f^5(x) = \frac{120}{(1-x)^6}, f^5(0) = 120\)

=> \(\frac{1}{(1 - x)} = 1 + x + \frac{2x^2}{2!} + \frac{6x^3}{3!} + \frac{24x^4}{4!} + \frac{120x^5}{5!} + ...\) \(= 1 + x + x^2 + x^3 + x^4 + x^5 + ...\)

  1. \(f(x) = e^x\)

    \(f(x)=e^x, f(0)=e^0 = 1\)

\(f'(x)=ex, f'(0)=e0 = 1\)

\(f''(x)=e^x , f''(0)=e^0 = 1\)

\(f^{(3)}(x)=e^x , f^{(3)}(0)=e^0 = 1\)

\(f^{(4)}(x)=e^x , f(4)(0)=e^0 = 1\)

\(f^{(5)}(x)=e^x , f^{(5)}(0)=e^0 = 1\)

\(f^{(n)}(x)=e^x , f^{(n)}(0)=e^0 = 1\)

\(.\) \(.\) \(.\)

\(f^{(n)}(x)=e^x , f^{(n)}(0)=e^0 = 1\)

=> \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} ... + \frac{x^n}{n!}\)

  1. \(f(x) = ln(1 + x)\)

setting \(u=1+x\) => \(\frac{du}{dx}=1 => dx=\frac{1}{u}du\)

\(f′(x)=\frac{1}{u} *1\)

=> \(f′(x)=\frac{1}{1+x}\)

\(f'0(x)=ln(1+x)=0\)

\(f′0(x)=ln(1+x)=1\)

\(f′′0(x)=ln(1+x)=−1\)

\(f^{(3)}0(x)=ln(1+x)=2\)

\(f^{(4)}0(x)=ln(1+x)=−6\)

\(f^{(5)}0(x)=ln(1+x)=24\)

From Taylor’s Series, this can be represented as:

\(x−\frac{x^2}{2}+\frac{x^3}{3}−\frac{x^4}{4} + \frac{x^5}{5} +...+\)