MHB Finding Taylor Polynomial for tan(x) - Wondering

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To find the N-th Taylor polynomial for tan(x) at 0 that approximates the function within a specified error margin, the discussion emphasizes the need to express tan(x) as a power series derived from its known expansions for sin(x) and cos(x). The remainder term of the Taylor series is analyzed using the Lagrange form, which involves calculating higher-order derivatives of tan(x) to ensure the remainder is less than 10^-5. It is suggested that for N=5, the approximation will likely meet the error requirement based on the behavior of the series terms. The conversation also touches on the properties of alternating series and the conditions necessary for bounding the remainder. Ultimately, the participants conclude that evaluating the Taylor series up to a sufficient degree will yield the desired accuracy for approximating tan(x).
  • #31
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)
Ah ok... (Thinking)
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)
What do you mean? (Wondering)
 
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  • #32
mathmari said:
What do you mean? (Wondering)

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)
 
  • #33
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)

I see! Thanks a lot! (Mmm)
 

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