Answer:
Step-by-step explanation:
mr pink red style avatar for user Leonard
Leonard
6 years ago
Posted 6 years ago. Direct link to Leonard's post “The MacLaurin expansion o...”
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?
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Comment on Leonard's post “The MacLaurin expansion o...”
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linkesy34
6 years ago
Posted 6 years ago. Direct link to linkesy34's post “While what you said is te...”
Great Answer
Good Answer
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.
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yash.kick
6 years ago
Posted 6 years ago. Direct link to yash.kick's post “I agree with Sal that no ...”
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??
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Sri Dhar
6 years ago
Posted 6 years ago. Direct link to Sri Dhar's post “SIn(0.4 ) = 0.389..(the r...”
Good Answer
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 a