Question: Write down the Taylor Series for $f(x) = \sin(x)$, and find the value of the $3$rd Taylor Polynomial $P_{3}$ at $x = 2/3$. Then find a reasonable upper bound for the error using $P_{3}(2/3)$ for the value of $\sin(2/3)$

In my textbook it gives this truncation error formula which make absolutely no sense to me at all here it is ( I think I use it here but not sure how):

Let $\bigg(a_{n} \bigg)_{n=1}^{\infty}$ be a decreasing sequence of positive numbers with $\lim_{n\to\infty} a_{n} = 0$. Then the alternating series $\sum_{n=1}^{\infty} (-1)^{(n+1)}*a_{n}$ converges. Furthermore, the $jth$ truncation error satisfies the inequality

$E_{j} < a_{j + 1}$

The same results hold for $\sum_{n=1}^{\infty} (-1)^{n}*a_{n}$

I don't understand what any of that means or how that relates to the problem.

My work for solving part of it:

Taylor Series for $\sin(x) = x - x^3/3! + x^5/5! + x^7/7! + .... $

So for the third taylor polynomial about $ a = 2/3$ that would be

$f^0(2/3) = \sin(x) = \sin(2/3) $

$f^1(2/3) = \cos(x) = \cos(2/3) $

$f^2(2/3) = -\sin(x) = -\sin(2/3)$

$f^3(2/3) = -\cos(x) = -\cos(2/3) $

So all together up to the third Taylor would be:

$\sin(2/3) = \sin(2/3) - \cos(2/3)*x^3/3! - \sin(2/3)x^5/5! -\cos(2/3)x^7/7!$

For finding a reasonable upper bound I have no idea how this is done.

  • $\begingroup$ The coefficient of $x^7/7!$ should be $-1$. $\endgroup$ – Gerry Myerson Apr 28 '17 at 7:17
  • $\begingroup$ "I don't understand what any of that means" This is not helpful, that is, it does not help us to help you. Do you mean that you don't understand what a decreasing sequence is? Does it mean you don't understand what a positive number is? Does it mean you don't understand what $\lim_{n\to\infty}a_n=0$ means? I'm sure you understand what some of it means – what precisely is it that you don't understand? $\endgroup$ – Gerry Myerson Apr 28 '17 at 7:20
  • $\begingroup$ I'm voting to close this question as off-topic because OP has abandoned it. $\endgroup$ – Gerry Myerson May 1 '17 at 7:08

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