## Wednesday, August 26, 2015

### How to improve your thinking skills? (II)

Prove $\dfrac{\pi}{4}+\dfrac{1}{6}\gt \arctan\left({\dfrac{6}{5}}\right)$.

Let's pick up where we left off...we needed to complete the proof using the readily available formula that says

$\arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots$

such that we get

$\dfrac{\pi}{4}+\dfrac{1}{6}\gt \arctan\left({\dfrac{6}{5}}\right)$

Okay, it's very tempting for students to immediately group the readily available formula in such a way that it is now beautifully grouped:

$\arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots$

$\arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\left(\dfrac{1}{10}-\dfrac{1}{100}\right)+\left(\dfrac{1}{1500}-\dfrac{1}{125000}\right)+\left(\dfrac{1}{750000}-\dfrac{1}{8750000}\right)+\cdots$

Up to this juncture, what the above can tell us is that each group term is greater than zero, i.e. $\arctan\left({\dfrac{6}{5}}\right)$ is the result of the sum of $\dfrac{\pi}{4}$ plus a series of other positive small values, in other words, that also means $\arctan\left({\dfrac{6}{5}}\right)\gt \dfrac{\pi}{4}$

Wait a minute! This isn't particularly useful to prove the question in hand!

We therefore need to devise a plan. If we can prove

[MATH]\small \color{yellow}\bbox[5px,purple]{\text{some value}}\color{black}\gt \arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots[/MATH]

and [MATH]\color{black}\dfrac{\pi}{4}+\dfrac{1}{6}\gt \color{yellow}\bbox[5px,purple]{\text{that value}}[/MATH]

then we have [MATH]\color{black}\dfrac{\pi}{4}+\dfrac{1}{6}\gt \color{yellow}\bbox[5px,purple]{\text{that value}}\color{black} \gt \arctan\left({\dfrac{6}{5}}\right) [/MATH] and the result is then followed.

That is to say, we need to rearrange and regrouping what we have on the RHS of the equality where

$\arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots$

in such a special way so that $\arctan\left({\dfrac{6}{5}}\right)$ is less than some value...

Think of it if we regroup the term so that each pair of the group terms are all negatives:

[MATH]\color{black}\arctan\left({\dfrac{6}{5}}\right)=\color{yellow}\bbox[5px,purple]{\dfrac{\pi}{4}+\dfrac{1}{10}}\color{black}-\left(\dfrac{1}{100}-\dfrac{1}{1500}\right)-\left(\dfrac{1}{125000}-\dfrac{1}{750000}\right)-\cdots[/MATH]

Yes! At this point, we can safely say that

[MATH]\color{yellow}\bbox[5px,purple]{\dfrac{\pi}{4}+\dfrac{1}{10}}\color{black}\gt \arctan\left({\dfrac{6}{5}}\right)[/MATH]

And note that

[MATH]\color{black}\dfrac{\pi}{4}+\dfrac{1}{6}\gt \color{yellow}\bbox[5px,purple]{\dfrac{\pi}{4}+\dfrac{1}{10}}[/MATH]

We can now put them together that yields:

[MATH]\color{black}\dfrac{\pi}{4}+\dfrac{1}{6}\gt \color{yellow}\bbox[5px,purple]{\dfrac{\pi}{4}+\dfrac{1}{10}}\color{black}\gt \arctan\left({\dfrac{6}{5}}\right)[/MATH]

and therefore, evidently we have proved that:

$\dfrac{\pi}{4}+\dfrac{1}{6}\gt \arctan\left({\dfrac{6}{5}}\right)$.

This isn't the sole method that one could use to attack the problem, we could also approach it like follows:

I know I know, many students would add $\dfrac{1}{6}$ and at the same time subtract the fraction $\dfrac{1}{6}$ to the right side of the equality and wind up with

$\small \arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{6}-\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots$

$\small \arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{6}-\left(\dfrac{1}{6}-\dfrac{1}{10}\right)-\left(\dfrac{1}{100}-\dfrac{1}{1500}\right)-\left(\dfrac{1}{125000}-\dfrac{1}{750000}\right)-\cdots$

Note that each parenthesis pair that consists of the subtraction of a smaller value from a bigger value will definitely gives a positive value, that means:

$\small \arctan\left({\dfrac{6}{5}}\right)+\left(\dfrac{1}{6}-\dfrac{1}{10}\right)+\left(\dfrac{1}{100}-\dfrac{1}{1500}\right)+\left(\dfrac{1}{125000}-\dfrac{1}{750000}\right)+\cdots=\dfrac{\pi}{4}+\dfrac{1}{6}$

which yields:

$\dfrac{\pi}{4}+\dfrac{1}{6}\gt \arctan\left({\dfrac{6}{5}}\right)$ and we are hence done.