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

[schniefen]

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

$\frac{1}{1+y^2}=t_n(-y^2)+e_n(-y^2)$
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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

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

$|E_{2n+1}(x)|=|\int_0^x e_n(-y^2)dy|\leq\int_0^x |e_n(-y^2)|dy=\int_0^x |\frac{(-1)^{n+1}}{(1-c)^{n+2}}y^{2n+2}|dy=\frac{1}{|(1-c)^{n+2}|}\int_0^x |y^{2n+2}|dy$
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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.

 

[schniefen]

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.