Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/fontdata.js

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

\displaystyle \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

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

then we have \displaystyle \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) 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:

\displaystyle \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

Yes! At this point, we can safely say that

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

And note that

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

We can now put them together that yields:

\displaystyle \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)

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.



No comments:

Post a Comment