Taylor Series Expansions for $f(x) = e^x $
\[ e^x = \sum_{n=0}^\inf \frac{x^n}{n!} \]
Taylor Series Expansions for $f(x) = $ \[ \frac{1}{ 1 -x} = \sum_{n=0}^\inf {x^n} \]
Taylor Series Expansions for $f(x) = tan^{-1} x $ \[ tan^{-1} x = \sum_{n=0}^\inf {(-1)^n} \frac{x^{2n+1}}{2n+1} \]
Key Idea 8.8.1 gives the nth term of the Taylor series of common func????ons. In Exercises 3 - 6, verify the formula given in the Key Idea by finding the first few terms of the Taylor series of the given func????on and iden????fying a pa????ern.
\[ f(x) = e^x c= 0 \] Evaluate the first few derivative at x=0:
\[ f(x) = e^x \\ \textrm{we know } p(x)= f(x) + f^1(x)x + \frac{f^2(x)x^2}{2!} + \frac{f^3(x)x^3}{3!}...... \\ f(x) = e^x => f(0) = e^0 = 1 \textrm{ .........1st Trem }\\ f^1(x) = e^x => f(0) = e^0 = 1 \textrm{ .........2nd Trem }\\ f^2(x) = e^x => f(0) = e^0 = 1 \textrm{ .........3rd Trem }\\ f^3(x) = e^x => f(0) = e^0 = 1 \textrm{ .........4th Trem }\\ \]
With this info we can \[ p(x)= f(x) + f^1(x)x + \frac{f^2(x)x^2}{2!} + \frac{f^3(x)x^3}{3!}+......+ \frac{f^n(x)x^n}{n!} \\ \textrm{substitute this value into polynomial equation we get:}\\ p(x)= 1 + 1x + \frac{1x^2}{2!} + \frac{1x^3}{3!}+......+ \frac{1x^n}{n!} \\ \textrm{writing 1 as x to power of 0= 1, and x as to power of 1 = x}\\ => \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!}+......+ \frac{x^n}{n!} \\ => \sum_{x=0}^{\inf} \frac{x^n}{n!} \\ \textrm{This is same as the formula we have from tylor }\\ e^x = \sum_{n=0}^\inf \frac{x^n}{n!} \]\[ \frac{1}{ 1 -x} = \sum_{n=0}^\inf {x^n} .......at \space\space c = 0 \textrm{ given} \\ \textrm{we know } p(x)= f(x) + f^1(x)x + \frac{f^2(x)x^2}{2!} + \frac{f^3(x)x^3}{3!}...... \\ f(x) = \frac{1}{ 1 -x} = (1 -x)^{-1} ...computing => f(0) =\frac{1}{ 1 -0} = 1 \textrm{ .........1st Trem }\\ f^1(x) = f((1 -x)^{-1}) = -1(1-x)^{-2}f(1-x) =-1(1-x)^{-2}(0-1) = 1(1-x)^{-2} \\ ...computing => f(0) = 1(1-0)^{-2} = 1(1^{-2})= 1 \textrm{ .........2nd Trem }\\ f^2(x) = f(1(1-x)^{-2} ) = (-2)*(1-x)^{-3}f(1-x)= (-2)*(1-x)^{-3}(0-1) = 2(1-x)^{-3} \\...computing => f(0) = 2(1-0)^{-3}= 2(1^{-3})= 2 \textrm{ .........3rd Trem }\\ f^3(x) = f(2(1-x)^{-3}) = 2*(-3)*(1-x)^{-4}f(1-x) = -6(1-x)^{-4}(0-1) = 6(1-x)^{-4} \\...computing => f(0) = 6(1-0)^{-4} = 6 \textrm{ .........4th Trem }\\ \textrm{ With this info we can substitute the terms value in polynomial equation } \\ p(x)= f(x) + f^1(x)x + \frac{f^2(x)x^2}{2!} + \frac{f^3(x)x^3}{3!}+......+ \frac{f^n(x)x^n}{n!} \\ \textrm{substitute this value into polynomial equation we get:}\\ => p(0) = 1 + 1x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3........+\frac{n!}{n!}x^n \\ => 1 + 1x + x^2 + x^3+.....+ x^n \\ =>\sum_{n=0}^{\inf} x^n ..\textrm{ This meets the tylor's formula.} \]