The discussion focuses on determining the appropriate degree \( N \) for the Taylor polynomial \( P_N \) of the function \( f(x) = \tan(x) \) at \( x = 0 \) such that the error \( |f(x) - P_N(x)| \) is less than or equal to \( 10^{-5} \) for \( x \) in the interval \([-0.1, 0.1]\). The participants confirm that the power series expansion for \( \tan(x) \) is necessary, specifically using the known series involving Bernoulli numbers. They conclude that selecting \( N = 5 \) is sufficient to ensure the error condition is met, as the remainder term can be bounded effectively using the properties of the Taylor series.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in numerical analysis and approximation theory will benefit from this discussion.
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)