When two of the hardest tests in the world cover the same problem, you know it’s a big deal! A version of this problem appeared on both the Putnam competition and the JEE Advanced.
Here you will get Proof for π < 22/7 and I will also show you that real value of π is lies between 3.14 < π < 22/7
Evaluate the following integral, and use the result to prove that π is less than 22/7.
\[\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}}dx\]
\[=\int_{0}^{1}\frac{x^4(x^4 – 4x^3 + 6x^2 – 4x + 1)}{1+x^{2}}dx\]
\[=\int_{0}^{1}\frac{(x^8 – 4x^7 + 6x^6 – 4x^5 + x^4)}{1+x^{2}}dx\]
\[=\int_{0}^{1}x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^{2}}dx\]
\[\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}}dx\]
\[=\int_{0}^{1}\frac{x^4(x^4 – 4x^3 + 6x^2 – 4x + 1)}{1+x^{2}}dx\]
\[=\int_{0}^{1}\frac{(x^8 – 4x^7 + 6x^6 – 4x^5 + x^4)}{1+x^{2}}dx\]
\[=\int_{0}^{1}x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^{2}}dx\]
\[=\left[\frac{x^7}{7}-4\times\frac{x^6}{6}+5\times\frac{x^5}{5}-4\times\frac{x^3}{3}+4x-4\tan^{-1}x\right]_0^1\]
\[=\frac{1}{7}-\frac{4}{6}+1-\frac{4}{3}+4-4\times\frac{\pi}{4}\]
\[=\frac{22}{7}-\pi\]
Since the integrand is positive between 0 and 1, the integral will also be positive. Hence we have:
\[0<\frac{22}{7}-\pi\]
\[\Rightarrow \pi<\frac{22}{7}\]
As I calculated value of π in one of my post, it is always greater than 3.14
Click here to read that blog.
By this relationship we can say that:
\[3.14<\pi<\frac{22}{7}\]
Here I added proof in which I calculated value of π by two methods.
1. By calculating area of circle
We know that area of circle is πr^2. If we consider r is 1 then area of circle will be π.
For calculating Area of circle, we draw an n-sided regular polygon inscribed in circle.
Then we divide that polygon in n number of triangles. Now we can easily calculate area of polygon since all triangles are similar as shown in figure:
Now the interior angel for all triangle is θ
\[θ=\frac{360}{n}\]
By using trigonometry, we can find value of h and x
\[h=r.cos\frac{θ}{2}=1.cos\frac{360}{2n}=cos\frac{180}{n}\]
\[\frac{x}{2}=r.sin\frac{θ}{2}=1.sin\frac{360}{2n}=sin\frac{180}{n}\]
\[x=2\times sin\frac{180}{n}\]
Area of polygon = No. of Triangles x Area of one triangle
Area of polygon = n x T
First we calculate value of T:
\[T=n\times x\times h\]
\[T=n\times cos\frac{180}{n}\times 2\times sin\frac{180}{n}\]
\[T=2\times n\times cos\frac{180}{n}\times sin\frac{180}{n}\]
\[T=\frac{n}{2}\times sin\frac{360}{n}\]
Now area of polygon
\[A=\frac{n}{2}\times sin\frac{360}{n}\]
In this if we increase the value of n then we got value close to π. Best thing in this that we can calculate it simply on calculator.
If n=100, we get 3.1395....
If n=1000, we get 3.14157....
If n=10000, we get 3.1415924....
If n=100000, we get 3.141592652....
True value of π = 3.14159265358....
It's formula can be written as:
\[\pi=\lim_{n \rightarrow \infty}\frac{n}{2}sin\frac{360}{n}\]
2. By calculating Perimeter of Circle
We know that perimeter of circle is 2πr. If we consider r is 1/2 then perimeter of circle will be π.
For calculating Perimeter of circle, we draw an n-sided regular polygon inscribed in circle.
And then we divide that polygon in n number of triangles. Now we can easily calculate Perimeter of polygon since all triangles are similar as shown in figure:
Now the interior angel for all triangle is θ
\[θ=\frac{360}{n}\]
By using trigonometry, we can find value of L
\[\frac{L}{2}=r.sin\frac{θ}{2}=\frac{1}{2}.sin\frac{360}{2n}=\frac{1}{2}.sin\frac{180}{n}\]
\[L=sin\frac{180}{n}\]
Perimeter of polygon = No. of Triangles x Length of outer side
Perimeter of polygon = n x L
\[P=n\times sin\frac{180}{n}\]
In this if we increase the value of n then we got value close to π. Best thing in this that we can calculate it simply on calculator.
If n=100, we get 3.1410....
If n=1000, we get 3.141587....
If n=10000, we get 3.141592602....
If n=100000, we get 3.14159265307....
True value of π = 3.14159265358....
As you can observe, this will give more accurate value of π
It's formula can be written as:
\[\pi=\lim_{n \rightarrow \infty}\ n.sin\frac{180}{n}\]
References
Wikipedia article
https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80
Putnam exam 1968 A1
https://prase.cz/kalva/putnam/putn68.html
JEE Advanced 2010 Paper Solutions Part-1 | IIT-JEE | Strategy, Tips, & Tricks | Paper Discussion (see around 55:30)
https://youtu.be/Zz4buu5y7_8?t=3330
Why is pi less than 22/7? – Week 14 – Lecture 9 – Mooculus
https://youtu.be/XCKFjql-D7U
2010 IIT-JEE Advanced paper 1 question 41 direct link
https://drive.google.com/file/d/0BxIR7Kj71SMKUmxnUkdhdWJ5NWM/view
JEE (Advanced) past papers
https://jeeadv.ac.in/pastqp.php
