Proving a tangent inequality

[crestu]

"Prove that for all x > 0, x > tan^-1 x. Explain you arguments and any theorems used clearly."

I have tried looking at the power series for tan^-1 x:

\(\displaystyle \tan ^{ - 1} x = \sum\limits_{n = 0}^\infty {( - 1)^n {{x^{2n + 1} } \over {2n + 1}}} = x - {{x^3 } \over 3} + {{x^5 } \over 5} - {{x^7 } \over 7} + ...\)

That doesn't seem to help me because I don't know how to prove \(\displaystyle - {{x^3 } \over 3} + {{x^5 } \over 5} - {{x^7 } \over 7} + ... < 0\)

Any ideas?

Thanks,
crestu

 

[soroban]

Hello, crestu!

I think I have a geometric approach for an <u>acute</u> angle.
\(\displaystyle \;\;\)Maybe you can generalize it . . .

Prove that for all \(\displaystyle x\,>\,0,\;x\,>\,\tan^{-1}x\)

        |          C
        * *       o
        |     * A/|
        |       o |
        |     /  *|
        |   /     |
        | / x     |
    - - o - - - - o - -
        O    1    B

We have angle \(\displaystyle x\) in the unit circle in quadrant 1.

The terminal side of \(\displaystyle x\) cuts the circle at \(\displaystyle A.\)

Draw the tangent to the circle at \(\displaystyle B.\)
\(\displaystyle \;\;OA\) cuts the tangent at \(\displaystyle C.\;\)
\(\displaystyle \;\;\)In right triangle \(\displaystyle CBO:\;\tan x\,=\,\frac{CB}{OB}\;\;\Rightarrow\;\;CB\,=\,\tan x\)

In sector \(\displaystyle AOB\), the length of \(\displaystyle \text{arc}AB\,=\,r\theta\,=\,1\cdot x\,=\,x\)

Clearly, \(\displaystyle \,CB\:>\:\text{arc}AB\;\;\Rightarrow\;\;\tan x\:>\:x\)

Take the arctan of both sides: \(\displaystyle \,x\:>\:\tan^{-1} x\)
.

 

[pka]

Consider the function \(\displaystyle \L
f(x) = x - \arctan (x)\).

Then \(\displaystyle \L
f(0) = 0\) and \(\displaystyle \L
f'(x) = 1 - \frac{1}{{1 + x^2 }}\).

Thus if x>0 then \(\displaystyle \L
f'(x) > 0\quad \Rightarrow \quad f(x) > 0\quad \Rightarrow \quad x > \arctan (x)\) .

 

[crestu]

Ah... that clarifies it. I hadn't thought of defining f(x) as anything other than tan^-1x by itself.

I can see that the following is true intuitively, but is there any theory that states that if f(a) \(\displaystyle \ge\) b and for all x > a, f'(a) > 0 then for all x > a, f(x) > b ?

 

[pka]

Yes!
If \(\displaystyle \L
f'(x) > 0\) on \(\displaystyle \L
(0,\infty )\) then the function is increasing!

Thus if \(\displaystyle \L
f(0) = 0\) and \(\displaystyle \L
x > 0\) then \(\displaystyle \L
f(x) > 0\) because \(\displaystyle \L
f\) is increasing.