Lagrange error bound inequality for Taylor series of arctan(x)

In summary, the error for ##\frac{1}{1-y}## can be expressed as ##\frac{1}{(1-c)^{n+2}}y^{n+1}## and can be further simplified to ##t_n(y)##. By taking the definite integral and using the triangle inequality, it can be shown that ##|E_{2n+1}(x)|\leq\frac{|x|^{2n+3}}{(2n+3)}##. The issue with the term ##\frac{1}{|(1-c)^{n+2}|}## can be resolved by showing that it is always less than or equal to 1 for all values of n.
  • #1
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Homework Statement
Show that ##|E_{2n+1}(x)|\leq\frac{|x|^{2n+3}}{(2n+3)}## for the Taylor polynomial ##T_{2n+1}(x)## of ##\arctan{(x)}## centred at ##x=0##
Relevant Equations
Lagrange error bound formula for a Taylor polynomial ##T_{n}(x)## centred at ##x=0##: ##E_{n}(x)=f^{(n+1)}(c)\frac{x^{n+1}}{(n+1)!}## for some ##c## between 0 and ##x##
The error ##e_{n}(y)## for ##\frac{1}{1-y}## is given by ##\frac{1}{(1-c)^{n+2}}y^{n+1}##. It follows that

##\frac{1}{1+y^2}=t_n(-y^2)+e_n(-y^2)##
where ##t_n(y)## is the Taylor polynomial of ##\frac{1}{1-y}##. Taking the definite integral from 0 to ##x## on both sides yields that

##E_{2n+1}(x)=\int_0^x e_n(-y^2)dy##
Taking the absolute value plus using the triangle inequality for integrals and the formula for ##e_{n}(y)## gives

##|E_{2n+1}(x)|=|\int_0^x e_n(-y^2)dy|\leq\int_0^x |e_n(-y^2)|dy=\int_0^x |\frac{(-1)^{n+1}}{(1-c)^{n+2}}y^{2n+2}|dy=\frac{1}{|(1-c)^{n+2}|}\int_0^x |y^{2n+2}|dy##
From the last equality, how does one get ##|E_{2n+1}(x)|\leq\frac{|x|^{2n+3}}{(2n+3)}##. I can't see how the term ##\frac{1}{|(1-c)^{n+2}|}## can be ignored since ##c## may be between 0 and 1 or greater than 1. Thus the term may be very large or very small. Also, I don't see why the final inequality yields the absolute value of x in the numerator.
 
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  • #2
I've solved the issue with the ##c## I think. From ##e_n(-y^2)## it follows that ##c## is between 0 and ##-y^2##, and so ##\frac{1}{|(1-c)^{n+2}|}\leq 1## for all n.
 

Related to Lagrange error bound inequality for Taylor series of arctan(x)

1. What is the Lagrange error bound inequality for Taylor series of arctan(x)?

The Lagrange error bound inequality for Taylor series of arctan(x) is a mathematical formula that provides an upper bound for the error between the actual value of arctan(x) and its approximation using a finite number of terms in its Taylor series.

2. How is the Lagrange error bound inequality derived?

The Lagrange error bound inequality is derived using the remainder term in the Taylor series of arctan(x) and the Mean Value Theorem from calculus.

3. What is the significance of the Lagrange error bound inequality?

The Lagrange error bound inequality allows us to estimate the error in approximating arctan(x) using a finite number of terms in its Taylor series. This is useful in applications where an accurate estimation of the error is necessary.

4. Can the Lagrange error bound inequality be used for other functions?

Yes, the Lagrange error bound inequality can be used for any differentiable function that can be represented by a Taylor series. However, the specific form of the remainder term may vary for different functions.

5. How can the Lagrange error bound inequality be used in practice?

In practice, the Lagrange error bound inequality can be used to determine the number of terms needed in the Taylor series of arctan(x) to achieve a desired level of accuracy. It can also be used to verify the accuracy of numerical approximations of arctan(x) by comparing the actual error with the estimated error using the inequality.

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