This week, we’ll work out some Taylor Series expansions of popular functions.

  • \(f(x) = \frac{1}{(1-x)}\)

  • \(f(x) = e^x\)

  • \(f(x) = ln(1 + x)\)

For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as a R-Markdown document.

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\(f(x) = \frac{1}{(1-x)}\)

\(f(a)\quad =\quad \frac { 1 }{ 1\quad -\quad a }\); \(\quad\quad f(0) = 1\)

\({ f }^{ \prime }(a)\quad =\quad \frac { 1 }{ { (1-a) }^{ 2 } }\); \(\quad\quad f^{(1)}(0) = 1\)

\({ f }^{ \prime \prime }(a)\quad =\quad \frac { 2 }{ { (1-a) }^{ 3 } }\); \(\quad\quad f^{(2)}(0) = 2\)

\({ f }^{ \prime \prime \prime}(a)\quad =\quad \frac { 6 }{ { (1-a) }^{ 4 } }\); \(\quad\quad f^{(3)}(0) = 6\)

\({ f }^{(4)}(a)\quad =\quad \frac { 24 }{ { (1-a) }^{ 5 } }\); \(\quad\quad f^{(4)}(0) = 24\)

Plug in the relevant expressions into formula for Taylor Series expansion:

\(f(a) + {{ f }^{ \prime }}(a)(x-a) + \frac{{ f }^{ \prime \prime }}{2!}(x-a) + \frac {{ f }^{ \prime \prime \prime }}{3!}(x - a) + \frac {f^{(4)}}{4!}(x - a) +...\)

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

which reduces to,

\(1 + x + x^2 + x^3 + x^4 + x^5...\)

\(= \sum_{ n=0 }^{ \infty }x^n\)

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

\(f(a) \quad= \quad { e }^{ a }\); \(\quad\quad f(0) = 1\)

\({ f }^{ \prime }(a)\quad =\quad { e }^{ a }\); \(\quad\quad { f }^{ \prime }(0) = 1\)

\({ f }^{ \prime \prime }(a)\quad =\quad { e }^{ a }\); \(\quad\quad { f }^{ \prime \prime }(0) = 1\)

\({ f }^{ \prime \prime \prime }(a)\quad =\quad { e }^{ a }\); \(\quad\quad { f }^{ \prime \prime \prime }(0) = 1\)

\(f^{(4)}(a)\quad = \quad { e }^{ a }\); \(\quad\quad f^{(4)}(0) = 1\)

Plug in the relevant expressions into formula for Taylor Series expansion:

\(f(a) + {{ f }^{ \prime }}(a)(x-a) + \frac{{ f }^{ \prime \prime }}{2!}(x-a) + \frac {{ f }^{ \prime \prime \prime }}{3!}(x - a) + \frac {f^{(4)}}{4!}(x - a) +...\)

\(= 1 + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ...\)

\(= \sum _{n=0 }^{\infty}\frac{x^n}{n!}\)

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

\(f(a) \quad= \quad ln(1+a)\); \(\quad= \quad f(0) = 0\)

\({ f }^{ \prime }(a) \quad= \quad \frac{1}{1+a}\); \(\quad= \quad { f }^{ \prime }(0) = 1\)

\({ f }^{ \prime \prime }(a) \quad= \quad \frac{-1}{(1+a)^2}\); \(\quad= \quad { f }^{ \prime \prime }(0) = -1\)

\({ f }^{ \prime \prime \prime }(a) \quad= \quad \frac{2}{(1+a)^3}\); \(\quad= \quad { f }^{ \prime \prime \prime } (0) = 2\)

\(f^{(4)}(a) \quad= \quad \frac{-6}{(1+a)^4}\); \(\quad= \quad f^{(4)}(0) = -6\)

Plug in the relevant expressions into formula for Taylor Series expansion:

\(f(a) + {{ f }^{ \prime }}(a)(x-a) + \frac{{ f }^{ \prime \prime }}{2!}(x-a) + \frac {{ f }^{ \prime \prime \prime }}{3!}(x - a) + \frac {f^{(4)}}{4!}(x - a) +...\)

\(=x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +...\)

\(= \sum _{n=0 }^{\infty}(-1)^{n+1}\frac{x^n}{n!}\)