The discussion revolves around finding the Taylor polynomial for the function \( f(x) = \tan(x) \) at \( x = 0 \). Participants are exploring how to determine the degree \( N \) of the polynomial such that the difference between \( \tan(x) \) and its Taylor polynomial \( P_N(x) \) is less than \( 10^{-5} \) within the interval \( \left[-\frac{1}{10}, \frac{1}{10}\right] \). The conversation includes considerations of power series, remainder terms, and various methods for estimating the required degree of the polynomial.
Participants generally agree on the need to find a suitable \( N \) for the Taylor polynomial, but there are multiple competing views on the methods to achieve this and the nature of the series involved. The discussion remains unresolved regarding the best approach to estimate the remainder and the conditions for the series.
Limitations include the dependence on the behavior of higher order derivatives of \( \tan(x) \) and the assumptions made about the series' convergence properties. There is also uncertainty regarding the application of the alternating series error estimation to the series for \( \tan(x) \).
Ah ok... (Thinking)I like Serena said:No, the Taylor expansion is $\cos x = 1-\frac 12 x^2 + \frac 1{4!}x^4 - ...$.
It's just that $\cos$ has a horizontal tangent at $x=0$.
We can set a bound with a line that has a non-horizontal tangent.$|\cos x|\le 1$. Therefore $\frac{1}{|\cos(x)|}\ge 1$. (Nerd)
What do you mean? (Wondering)I like Serena said:The remainder term is actually:
$$R_3(x) = \frac{1}{4!} f^{(4)}\,(\xi) x^4$$
I think we need a couple more factors... (Thinking)
mathmari said:What do you mean? (Wondering)
I like Serena said:We've evaluated an upper bound for $f^{(4)}(\xi)$, which is approximately $2.6$.
Btw, I think there's a factor $8$ missing in the expression. (Worried)
But we need an upper bound for $R_3(x) = \frac 1{4!}f^{(4)}(\xi)\left(\frac 1{10}\right)^4 = \frac{1}{24}\cdot 10^{-4}f^{(4)}(\xi) < 4\cdot 10^{-6}\cdot f^{(4)}(\xi)$. (Thinking)