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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.