Worked example: estimating sin(0.4) using Lagrange error bound | AP Calculus BC | Khan Academy

Lagrange error bound (also called Taylor remainder theorem) can help us determine the degree of Taylor/Maclaurin polynomial to use to approximate a function to a given error bound. See how it's done when approximating the sine function.

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[jaredrice]

I love how he repeats the words while he writes

[KeyMan]

While the Lagrange error bound method here gives 4, I think the actual correct answer is 3. The Maclaurin polynomial of sin(x)=x-x^3/3!+..., so Since the degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials, we can see that the polynomial x-x^3/3! has a degree of 3. Notice that 0.4-0.4^3/3!≈0.389333 and the difference In conclusion, estimating sin(0.4) using a Maclaurin polynomial, the least degree of the polynomial that assures an error smaller than 0.001 is 3. I think if a value of M closer to sin(0.4) was chosen (instead of 1), then this method would've given 3 as well.

[jason]

I just wanted to comment that since abs(Rn(x)) equals to m*(this is important)(x to the power of n + 1) / (n + 1) factorial, the remainder evaluated at 0.4 will be smaller than 0.001 at the "third" degree because it is lower than 0.001 when Rn(x) is n = 3.

[dovehq]

Is there any way to solve this without brute forcing n?

[michaelempeigne]

looks like n = 4 but since fourth degree polynomial for sin x will end up being a third degree polynomial; you will have to take a 5th degree polynomial to get the desired accuracy according to video.

[daniil]

Isn't M the maximum value of the n+1 derivative on the interval [a, x]? How come we use the absolute maximum of the derivatives?

[enclave_rools]

What grade is this for because I'm not look forward to it.

[nara]

ACTUALLY it has really bad been bounded taking M as 1 is bad bound knowing that n=1 is the correct answer and taking this approximation give an n at 4 not do this technique

[김우우-ik]

It’s way more convenient to solve it by alternating series error bound. Isn’t it?

[sqrt___]

Can anyone explain this? Since M = f^(n+1)(x), whereas x belongs to [0, 0.4]. The maximum value for f(x) = sin(0.4) which isn’t 1. Please help me if anyone have any idea about this.