MHB Finding Taylor Polynomial for tan(x) - Wondering

[mathmari]

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)

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)

 

[I like Serena]

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)

 

[mathmari]

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)