# Thread: Taylor series calculation and approximation help please

1. ## Taylor series calculation and approximation help please

Calculate the Taylor series for (x^5) + (3x^2) - x - 1 about x= 1. Calculate it's radius of convergence. How accurate is the third order approximation on (0,2)? How accurate is the fifth order?

2. ## Re: Taylor series calculation and approximation help please

A Taylor Series for \displaystyle \begin{align*} f(x) \end{align*} centred around \displaystyle \begin{align*} x = h \end{align*} is given by \displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty} { \frac{ f^{ (n) } (h) }{ n! } \left( x - h \right)^n } \end{align*}, so calculating the necessary derivatives gives...

\displaystyle \begin{align*} f(x) &= x^5 + 3x^2 - x - 1 \\ f(1) &= 1^5 + 3 \cdot 1^2 - 1 - 1 &= 2 \\ \\ f'(x) &= 5x^4 + 6x - 1 \\ f'(1) &= 5 \cdot 1^4 + 6\cdot 1 - 1 \\ &= 10 \\ \\ f''(x) &= 20x^3 + 6 \\ f''(1) &= 20\cdot 1^3 + 6 \\ &= 26 \\ \\ f'''(x) &= 60x^2 \\ f'''(1) &= 60 \cdot 1^2 \\ &= 60 \\ \\ f^{\textrm{iv}} (x) &= 120x \\ f^{\textrm{iv}} (1) &= 120 \cdot 1 \\ &= 120 \\ \\ f^{\textrm{v}} (x) &= 120 \\ f^{\textrm{v}} (1) &= 120 \\ \\ f^{\textrm{vi}}(x) &= 0 \\ \vdots \end{align*}

So your Taylor Series is

\displaystyle \begin{align*} f(x) &= 2 + 10(x - 1) + \frac{26}{2}(x - 1)^2 + \frac{60}{3!}(x - 1)^3 + \frac{120}{4!}(x - 1)^4 + \frac{120}{5!}(x - 1)^5 \\ &= 2 + 10(x - 1) + 13(x - 1)^2 + 10(x - 1)^3 + 5(x - 1)^4 + (x - 1)^5 \end{align*}

3. ## Re: Taylor series calculation and approximation help please

Thank you very much for your help so far, do you think you could help me with the rest of my question as well????