Using the definition of Taylor Series

Question:1 Solution:

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

First derivative is, \(f^{'}(x) = \frac{d}{dx}\bigg(\frac{1}{1-x}\bigg) = (1 -x)^{-2}; f^{'}(0) = 1\)

Second derivative is, \(f^{''}(x) = \frac{d}{dx}\bigg(\frac{1}{(1-x)^2}\bigg) = 2(1 -x)^{-3}; f^{''}(0) = 2\)

Third derivative is, \(f^{'''}(x) = \frac{d}{dx}\bigg(\frac{2}{(1-x)^3}\bigg) = 6(1 -x)^{-4}; f^{'''}(0) = 6 = 3!\)

Fourth derivative is, \(f^{4}(x) = \frac{d}{dx}\bigg(\frac{6}{(1-x)^4}\bigg) = 24(1 -x)^{-5}; f^{4}(0) = 24 = 4!\)

Pattern observed is \(\{0,1,2,3!,4!, ....\infty\}\), output of \(n^{th}\) derivative is factorial of \(n\).

Taylor Series of \(f(x) = f(x) + f^{'}(x)\times\frac{(x-c)^n}{n!} + f^{''}(x)\times\frac{(x-c)^n}{n!} + f^{'''}(x)\times\frac{(x-c)^n}{n!} + f^{4}(x)\times\frac{(x-c)^n}{n!} + ....\)

At \(c = 0\)

\(n = \{0,1,2,3, ... \infty\}\)

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

= \(x^0 + x^1 + x^2 + x^3 + x^4 + ....\)

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

Question:2 Solution:

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

First derivative is, \(f^{'}(x) = e^x; f^{'}(0) = 1\)

Second derivative is, \(f^{''}(x) = e^x; f^{''}(0) = 1\)

Third derivative is, \(f^{'''}(x) = e^x; f^{'''}(0) = 1\)

Fourth derivative is, \(f^{4}(x) = e^x; f^{4}(0) = 1\)

Pattern observed is \(\{1\}\), output of \(n^{th}\) derivative is \(1\).

Taylor Series of \(f(x) = f(x) + f^{'}(x)\times\frac{(x-c)^n}{n!} + f^{''}(x)\times\frac{(x-c)^n}{n!} + f^{'''}(x)\times\frac{(x-c)^n}{n!} + f^{4}(x)\times\frac{(x-c)^n}{n!} + ....\)

At \(c = 0\)

\(n = \{0,1,2,3, ... \infty\}\)

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

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

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

Question:3 Solution:

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

First derivative is, \(f^{'}(x) = \frac{1}{1+x}; f^{'}(0) = 1\)

Second derivative is, \(f^{''}(x) = -\frac{1}{(1+x)^2}; f^{''}(0) = -1\)

Third derivative is, \(f^{'''}(x) = \frac{2}{(1+x)^3}; f^{'''}(0) = 2\)

Fourth derivative is, \(f^{4}(x) = -\frac{6}{(1+x)^4}; f^{4}(0) = -6\)

At \(n = 0\), output is \(0\), \(n\) starts from \(1\)

\(n = \{1,2,3,4, ... \infty\}\)

Pattern noticed is \(\{1, -1, 2, -6\}\), output of \(n^{th}\) derivative is \((-1)^{n+1}\times (n-1)!\).

At \(c = 0\)

Taylor Series of \(f(x) = f(x) + f^{'}(x)\times\frac{(x-c)^n}{n!} + f^{''}(x)\times\frac{(x-c)^n}{n!} + f^{'''}(x)\times\frac{(x-c)^n}{n!} + f^{4}(x)\times\frac{(x-c)^n}{n!} + ....\)

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

= \(\sum_{n=1}^{\infty} (-1)^{n+1}\times\frac{x^n}{n}\)

References