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Worked example: estimating sin(0.4) using Lagrange error bound

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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• The MacLaurin expansion of sin(x) is x - x^3/3! + x^5/5! - x^7/7! + x^9/9! ...

The polynomial of degree 4 is actually identical to the the polynomial of degree 3 because the coefficient of x^4 is 0. Shouldn't the answer to the exercise be 3 instead of 4?
• While what you said is technically true, what Sal is doing in the is using the error function to evaluate the absolute maximum error bound of the function. You are not supposed to know before hand that the coefficient fourth degree polynomial is 0, the error function is basically just telling you to what degree polynomial you need to take Maclaurin polynomial to be completely sure that the error is within certain bounds. It is always gonna be an overestimate since it is taking the absolute maximum that the error could be, not what actually is.
• I agree with Sal that no matter how many times we take the derivative of sin, it's absolute value will always be between 0 and 1 and so the value of "M" will be 1 but isn't Sal ignoring the fact that we have to only consider an open interval containing 0 and 0.4, in that case if the n+1'th derivative of sinx is (+or-)sinx again depending on value of 'n' then if we look in the interval of 0 and 0.4 then the absolute value of sinx will not lie between 0 and 1 but it will lie between 0 and sin(0.4)..........

So isn't something WRONG here??
• SIn(0.4 ) = 0.389..(the range of M is from 0 to 0.389) and cos(0.4) = 0.921...(the range is from 1 to 0.921). The (n+1)th derivative of f(x) could be either (+or-)cosx or (+or-)sinx. But we are sure that the value of M is always less than 1.
*Edit:*
What we have to take for M is the maximum possible value in the interval (x=0 to x=0.4). But we are not sure if its cosx or sinx in the n+1 th derivative. So we take M=1. It is gonna be less precise if it is sinx and more precise if it is cosx.
• how do we solve the final inequality Sal arrives at (at ) without trial and error?
do we just take log on both sides and solve for n?
• Because of the factorial and there being "n" both as an exponent and outside of the exponent, I do not think there is a way to solve it without trial and error.
• We know that sin x is < or = to 1 over the whole domain, but might one find a smaller M for the open interval between -0.0001 and 0.40001? If we use an M larger than necessary, could it cause us to specify a higher degree polynomial than necessary?
• That's right. I solved the problem for this specific interval and the minimal degree is 3 and not 4.
• In the previous video "Taylor polynomial remainder (part 2)" the right hand side of the remainder equation is without the absolute value, while in @ the right hand side is with the absolute value. Why?
• wait so i am confusion
why is the M equal to one in this particular situation?
i know that the graph of sine is bounded by 1, so does that make M always equal to 1 in these kind of problems?
• Because whatever x is, sin(x) and cos(x) is always bounded by 1, yes, it would make M equals 1 in this kind of problems. You might think sin(x) on (0, 0.4) much less than 1 but the derivative of sin(x) is ccos(x) which has quite close value to 1 in the above interval. Whereas this is not really precise, it's good enough
• In the previous video the remainder theorem formula is shown to have (x-a) and not just x over an interval between a and b. I am assuming it is just x because the interval is between 0 and 0.4, hence, a is 0?
• Actually, it's because we're using the Maclaurin polynomial