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Differentiation of Polynomials : Proofs

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Let g be a function which can be differentiated four times on the interval [-1,1].
Denote .

1) Show that when g is a polynomial of degree less than or equal to 3.
2) Let P be the interpolation polynomial of f at the points -1, , , 1.
a) Show that .
b) Show that , where
and is a constant which you will evaluate.
c) Deduce a number which is greater than or equal to the error .

3) Let f be a function which can be differentiated four times on an interval [a,b].
Let . Using x, show that the integral f can be calculated on [a,b] with the help of an integral on [-1,1].
4) Deduce an approximation of .
5) Using this method, calculate an approximation of .

Solution Summary

Problems relating to the differentiation of polynomials are solved. The interpolation polynomials are analyzed.

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Let g be a function which can be differenciated four times on the interval [-1,1].
Denote .

1) Show that when g is a polynomial of degree less than or equal to 3.
Proof. We consider four cases.

Case 1. g is a polynomial of degree 0, that is, , C is a constant.
In this case,
Case 2. g is a polynomial of degree 1, that is, , C and D are constants.
In this case, Case 3. g is a polynomial of degree 2, that is, , C , D and E ...

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• BSc , Wuhan Univ. China
• MA, Shandong Univ.
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• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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