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Assignment :
Pick any exercise in 8.8 of the calculus textbook. Solve and post your solution. If you have issues doing so, discuss them.
Exercise 8.8 # 29
Find the Taylor Series for the given Function :
\(f(x) = e^{x} sin x\) (only find the first 4 terms)
Solution
From Theorem 8.8.2
For Functions \(f(x) = \sum_{n = 0 }^{\infty} a_{n}x^{n}\) —- and —- \(g(x) = \sum_{n = 0 }^{\infty} b_{n}x^{n}\)
\(f(x) \ g(x)\) is given as \(f(x) \ g(x) = \left( \sum_{n = 0 }^{\infty} a_{n}x^{n} \right) \left( \sum_{n = 0 }^{\infty} b_{n}x^{n} \right)\)
\(\implies \sum_{n = 0 }^{\infty} \left(a_{0}b_{n} + a_{1}b_{n-1} +...a_{n}b_{0}\right) x^{n} \ for \left| x \right| \lt R\)
The Taylor Series for \(f(x) = e^{x} = \sum_{n = 0}^{\infty}\frac{x^{n}}{n!}\) – The Interval of convergence is \(\left( -\infty, \infty \right)\)
The first 5 terms are given as : - \(1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!}\)
The Taylor Series for \(f(x) = sin x = \sum_{n = 0}^{\infty}(-1^{n})\frac{x^{2n+1}}{(2n+1)!}\) – The Interval of convergence is \(\left( -\infty, \infty \right)\)
The first 5 terms are given as : - \(x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \frac{x^{9}}{9!}\)
Applying Theorem 8.8.2 - Using only the first 4 terms of each function as required :
\(f(x) = e^{x} sin x\) = \(\left(1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!}\right) \left(x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} \right)\)
Using the Distributive Law to Distribute the right hand expression across the left :
\[ 1 * \left(x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} \right) + \frac{x^{2}}{2!} * \left(x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} \right) + \frac{x^{3}}{3!} * \left(x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} \right) + \frac{x^{4}}{4!} * \left(x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} \right) \]
\(\implies\)
\[ x -\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \frac{x^{3}}{2!} -\frac{x^{5}}{3! * 2!} + \frac{x^{7}}{2! * 5!} - \frac{x^{9}}{2! * 7!} + \frac{x^{4}}{3!} -\frac{x^{6}}{3! * 3!} + \frac{x^{8}}{3! * 5!} - \frac{x^{10}}{3! * 7!} + \frac{x^{5}}{4!} -\frac{x^{7}}{4! * 3!} + \frac{x^{9}}{4! * 5!} - \frac{x^{11}}{4! * 7!} \]
If we expand the first 4 terms, we get :
\(x -\frac{x^{3}}{6} + \frac{x^{5}}{60} - \frac{x^{7}}{5040}\)
The combined Series Converge on \(\left( -\infty, \infty \right)\)
since each of the individual functions converge on \(\left( -\infty, \infty \right)\).
Note : that the first 4 terms of the combined functions \(f(x) = e^{x} sin x\) is the same as the first 4 terms for \(f(x) = sin x\)
References