\(a_{n}\) = \(\frac{f^{n}{(0)}}{n!}\)
f(x) = f(0)+\(f^{1}{(0)}\)+\(f^{2}{(0)}\)\(x^{2}\)/2! + \(f^{3}{(0)}\)\(x^{3}\)/3! + …….
• f(x) = 1/(1−x)
\(f^{1}{(0)}\) = \(\frac{1}{(1-x)^{2}}\) = 1
\(f^{2}{(0)}\) = \(\frac{2}{(1-x)^{3}}\) = 2
\(f^{3}{(0)}\) = \(\frac{6}{(1-x)^{4}}\) = 6
\(f^{4}{(0)}\) = \(\frac{24}{(1-x)^{5}}\) = 24
1/(1−x) = 1 + x + 2\(x^{2}\)/2! + 6\(x^{3}\)/3! + 24\(x^{4}\)/4! + ……
Therefore, 1 + x + \(x^{2}\) + \(x^{3}\) + \(x^{4}\) + ……
• f(x) = \(e^{x}\)
\(f^{1}{(0)}\) = \(e^{x}\) = 1
\(f^{2}{(0)}\) = \(e^{x}\) = 1
\(f^{3}{(0)}\) = \(e^{x}\) = 1
\(f^{4}{(0)}\) = \(e^{x}\) = 1
\(e^{x}\) = 1 + x + \(\frac{x^{2}}{2!}\) + \(\frac{x^{3}}{3!}\) + \(\frac{x^{4}}{4!}\) + …….
Prob 3
f(x) = ln(1 + x)
\(f^{1}{(0)}\) = \(\frac{1}{(1+x)^{1}}\) = 1
\(f^{2}{(0)}\) = \(\frac{-1}{(1+x)^{2}}\) = -1
\(f^{3}{(0)}\) = \(\frac{2}{(1+x)^{3}}\) = 2
\(f^{4}{(0)}\) = \(\frac{-6}{(1+x)^{4}}\) = -6
ln(1 + x) = 0 + x- \(\frac{x^{2}}{2}\) + \(\frac{x^{3}}{3}\) - \(\frac{x^{4}}{4}\) + …..