Data 605 Assignment-14

Deepak Mongia

May 16, 2019

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

1. \(f(x) = 1/(1-x)\)

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

\(f(0) = 1\)

\(f^{'}(0) = 1\)

\(f^{''}(0) = 2\)

\(f^{'''}(0) = 6\)

\(f^{''''}(0) = 24\)

Substituting these values, we get

\(f(x) = 1 + x^1 + x^2 + x^3 + x^4\) + …

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

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

differential of \(e^x = e^x\)

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

\(f^{'}(0) = 1\)

\(f^{''}(0) = 1\)

\(f^{'''}(0) = 1\)

\(f^{''''}(0) = 1\)

Substituting these values, we get

\(f(x) = 1 + x^1 + x^2/2! + x^3/3! + x^4/4!\) + …

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

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

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

\(f^{'}(x) = 1/(1+x), f^{'}(0) = 1\)

\(f^{''}(x) = -1/(1+x)^2, f^{''}(0) = -1\)

\(f^{'''}(x) = 2/(1+x)^3, f^{'''}(0) = 2\)

\(f^{''''}(x) = -2.3/(1+x)^4, f^{''''}(0) = -6\)

Substituting these values, we get

\(f(x) = x^1 - x^2/2 + x^3/3 - x^4/4\) + …