Proof that $\left|\arctan (x)-\frac{π}{4}-\frac{(x-1)}{2}\right| \leq \frac{(x-1)^2}{2}$

Question

I'm trying to prove that, for every $x \geq 1$:

$$\left|\arctan (x)-\frac{π}{4}-\frac{(x-1)}{2}\right| \leq \frac{(x-1)^2}{2}.$$

I could do it graphically on $\Bbb R$, but how to make a formal algebraic proof?

Answer 1

Hint:

Use the Taylor expansion of $\arctan(x)$ near $1$ to the second order. And the fact that $\arctan(x)$ is concave when $x\geq1$

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Answer:

So the Limited Taylor expansion of $\arctan(x)$ around $1$ to the second order is given by: $$ \arctan(x)=\frac{\pi}{4}+\frac{(x-1)}{2}-\frac{(x-1)^2}{4}+o((x-1)^2)$$ Hence, $$ \arctan(x)-\frac{\pi}{4}-\frac{(x-1)}{2}=-\frac{(x-1)^2}{4}+o((x-1)^2)$$ We can write: $$o((x-1)^2)=\frac{(x-1)^2}{2}\varepsilon(x-1),\quad\varepsilon(x-1)\xrightarrow{x\rightarrow 1}{0} $$

Then we have,\begin{align} \arctan(x)-\frac{\pi}{4}-\frac{(x-1)}{2}&=-\frac{(x-1)^2}{4}+\frac{(x-1)^2}{2}\varepsilon(x-1)\\ &=\frac{(x-1)^2}{2}\bigg(-\frac{1}{2}+\varepsilon(x-1)\bigg)\end{align} So when $x$ is too close to $1$ we have that $$-1<\bigg(-\frac{1}{2}+\varepsilon(x-1)\bigg)<0$$ So, \begin{align} \left|\arctan (x)-\frac{π}{4}-\frac{(x-1)}{2}\right|&=\left| \frac{(x-1)^2}{2}\bigg(-\frac{1}{2}+\varepsilon(x-1)\bigg)\right|\\ &= \frac{(x-1)^2}{2}\bigg(\frac{1}{2}-\varepsilon(x-1)\bigg)\\ &\leq \frac{(x-1)^2}{2}\tag{$\star$}\end{align}

And since for $x\geq1$, the second derivative of $f(x)=\left|\arctan (x)-\frac{π}{4}-\frac{(x-1)}{2}\right|$ is the opposite of that of $\arctan(x)$ so $f(x)$ is convex over $[1,+\infty[$. Furthermore, the second derivative of $g(x)=\frac{(x-1)^2}{2}$ is $1$ so $g(c)$ is also convex over $[1,+\infty[$.

The first derivatives of $f$ and $g$, over $[1,+\infty[$, are: $$ f'(x)=\frac{1}{2}-\frac{1}{1+x^2},\quad\quad g'(x)=x-1$$ We clearly have $$ g'(x)>f'(x),\quad\forall x\geq1$$ And since $g(1)=f(1)=0$ then the result $(\star)$ holds true over $[1,+\infty[$.

Answer 2

We have that for $x\ge 1$

$$\arctan x-\frac{π}{4}-\frac{(x-1)}{2}\le 0 \implies \left|\arctan x-\frac{π}{4}-\frac{(x-1)}{2}\right| =\frac{π}{4}+\frac{(x-1)}{2}-\arctan x$$

then consider

$$f(x)=\frac{(x-1)^2}{2}+\arctan x-\frac{π}{4}-\frac{(x-1)}{2}\implies f'(x)=\frac1{x^2+1}+x-\frac32\ge 0$$

and since $f(1)=0$ we have that $f(x)\ge 0$ and the inequality is proved.

Answer 3

Hint: For each $x >1,$ Taylor gives

$$\arctan (x)=\frac{\pi}{4}+\frac{(x-1)}{2} + \frac{\arctan'' (c_x)}{2}(x-1)^2,$$

where $1<c_x<x.$ Thus all you need to show is that $|\arctan'' (c)|\le 1$ for all $c\ge 1.$

Answer 4

It is convenient to let $x=u+1$ and rewrite the inequality to be proved as a pair of inequalities:

$${u\over2}-{u^2\over2}\le\arctan(u+1)-\arctan1\le{u\over2}+{u^2\over2}$$

for $u\ge0$. Now

$$\arctan(u+1)-\arctan1=\int_1^{u+1}{dt\over1+t^2}=\int_0^u{dt\over1+(1+t)^2}=\int_0^u{dt\over2+2t+t^2}$$

while

$${u\over2}-{u^2\over2}=\int_0^u{1-2t\over2} dt\qquad\text{and}\qquad{u\over2}+{u^2\over2}=\int_0^u{1+2t\over2}dt$$

It suffices, therefore, to show that, for $t\ge0$, we have

$${1-2t\over2}\le{1\over2+2t+t^2}\le{1+2t\over2}$$

The first inequality follows from $(1-2t)(2+2t+t^2)=2-2t-3t^2\le2$; the second inequality is even easier:

$${1\over2+2t+t^2}\le{1\over2}\le{1+2t\over2}$$