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Using taylor expansion of $\cos$ function.

What I have is $$1-\frac{x^2}{2}+\cdots-\frac{x^{4n-2}}{(4n-2)!}<\cos(x)<1-\frac{x^2}{2}+\cdots+\frac{x^{4n}}{(4n)!}$$

How would I proceed from here?

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Why cosine, and not an inverse trig function? – user7530 Jan 30 '13 at 0:45
why do you think these inequalities are valid? – Aang Jan 30 '13 at 0:47
I have no idea. had it been inverse, it's easy as that. – user45099 Jan 30 '13 at 0:47
I have not calculated this, but if you insert 2.82 into the formula once and receive a value > 0 for the right side and then insert 3.19, and receive values <0, you are done, because we then now there is a root in between (thanks to the continuity of the cosine) – CBenni Jan 30 '13 at 0:52
@user1709828 Actually it isnt as easy as it sounds... I made a (terrible) mistake: $cos(\pi)=-1$ isnt it ;) This makes it way harder, unless I am allowed to use that cos has an extremal point at $\pi$, is decreasing in 2.82 and I can modify the bounds... I will not try this very long, I am going to bed lol – CBenni Jan 30 '13 at 1:12
up vote 1 down vote accepted

Hopefully I finally understood the question. Let's define $\pi$ as twice the first positive zero of $\cos$.

We have $$ 0\leq 1-\frac{x^2}{2}< \cos x $$ for all $x\in [0,\sqrt{2}]$.

So the first positive zero of $\cos$ is greater than $\sqrt{2}$. Hence: $$ \pi> 2\sqrt{2}=2.828...>2.82. $$

Now $$ \cos(1.595)<1-\frac{(1.595)^2}{2}+\frac{(1.595)^4}{24}=-0,00234<0. $$

By the intermediate value theorem, $\cos$ has a zero between $\sqrt{2}$ and $1.595$. It follows that $\pi/2$ is not greater than $1.595$, whence $$ \pi<2\cdot 1.595=3.19. $$

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great. its hard to see what the question demands. – user45099 Jan 30 '13 at 8:57

I will take Madhava's formula for granted ( $$ \pi=\sqrt{12}\sum_{k=0}^{+\infty} \frac{(-1)^k}{3^k(2k+1)}. $$ This converges pretty fast.

In the alternating series, the error made when approximating the sum by the sum of the first $3$ terms is not greater than the absolute value of the $4$th term.

So the error made when truncating the formula above after $3$ terms is not greater than: $$ \sqrt{12}\frac{1}{3^3\cdot7}=0,01832858...\leq 0.02 $$

So let us compute now $$ \pi\simeq\sqrt{12}\sum_{k=0}^{2} \frac{(-1)^k}{3^k(2k+1)}=\sqrt{12}\left( 1-\frac{1}{9}+\frac{1}{45} \right)=3,1561... $$

Since the error is not greater than $0.02$, we see that $\pi$ belongs to the prescribed interval.

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+1 for madhav's formula but I am looking for use of expansion of pi. – user45099 Jan 30 '13 at 1:56

A brute-force answer: Use \begin{align} \frac{\pi^2}{6}=\sum_{k=1}^{\infty}\frac{1}{k^2} \end{align} which gives \begin{align} \pi=\sqrt{6*\sum_{k=1}^{\infty}\frac{1}{k^2}} \end{align} Take the first 10 terms in the sum and you will be able to prove $\pi>2.89$

Now consider the sequence \begin{align} \pi=\frac{3\sqrt{3}}{4}+24(\frac{1}{12}-\frac{1}{5*2^5}-\frac{1}{28*2^7}-\frac{1}{72*2^9}\dots) \end{align}

After a while, the sum of the R.H.S should go below $3.19$. For more about this series, visit here

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not a good idea. $\zeta(2)$ is not alternating to take first 10 and leave rest. Further more, main point is to use taylor expansion of $\cos$ – user45099 Jan 30 '13 at 1:50
oh ok, I didn't know that – dineshdileep Jan 30 '13 at 1:54

The approach taken is fine. The series is alternating, so the value is between the polynomials cited, and it's root between the roots of the polynomials.

Sure, it isn't exactly the most efficient way of computing $\pi$, but for a crude approximation (and to make students go through the concepts) it is nice.

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Since the function $\cos(x)$ is decreasing on $[0,\pi/2]$ we know that for angles in this interval, $\cos(a) < \cos(b)$ implies that $ a > b$.

To prove that $2.82<\pi<3.19$ it is enough to prove that

$$\cos(\frac{2.82}{4})> \cos(\frac{\pi}{4}) > \cos(\frac{3.19}{4}) \,.$$

[Note: here we are actually using the fact that $\frac{2.82}{4}$ is in the first quadrant, can I take for granted that $\pi > 1.41$? ]

Thus, all you have to do to complete the proof is find some $n,m$ so that

$$1-\frac{2.82^2}{2}+\cdots-\frac{2.82^{4n-2}}{(4n-2)!}> \sqrt{2}{2}$$ and $$\frac{\sqrt{2}}{2}<1-\frac{3.19^2}{2}+\cdots+\frac{3.19^{4m}}{(4m)!}$$

The Thoory actually guarantees that such $m,n$ exists, you just have to find them.


The simplest approach IMO would be to use the Taylor series of arctan

$$\arctan(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1}$$

Plugging in $\sqrt{3}$, multiplying by 6 and using again that alternating oscilates around the series you get

$$\frac{6}{\sqrt{3}} [ \sum_{n=0}^{2m-1} \frac{(-1)^n}{(2n+1)3^n } 3^n ]< \pi < \frac{6}{\sqrt{3}} [ \sum_{n=0}^{2m} \frac{(-1)^n}{(2n+1)3^n } ]$$

This series is geometric and converges very fast....

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