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Wednesday, August 26, 2015

How to improve your thinking skills? (II)

Prove π4+16>arctan(65).

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

arctan(65)=π4+1101100+115001125000+175000018750000+

such that we get

π4+16>arctan(65)

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(65)=π4+1101100+115001125000+175000018750000+

arctan(65)=π4+(1101100)+(115001125000)+(175000018750000)+

Up to this juncture, what the above can tell us is that each group term is greater than zero, i.e. arctan(65) is the result of the sum of π4 plus a series of other positive small values, in other words, that also means arctan(65)>π4

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.



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