Tag: euler

Basel Problem using Fourier Series

A very famous mathematical problem known as the “Basel Problem” is solved by Euler in 1734. Basically, it asks for the exact value of \sum_{n=1}^\infty\frac{1}{n^2}.

Three hundred years ago, this was considered a very hard problem and even famous mathematicians of the time like Leibniz, De Moivre, and the Bernoullis could not solve it.

Euler showed (using another method different from ours) that \displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}, bringing him great fame among the mathematical community. It is a beautiful equation; it is surprising that the constant \pi, usually related to circles, appears here.

Squaring the Fourier sine series

Assume that \displaystyle f(x)=\sum_{n=1}^\infty b_n\sin nx.

Then squaring this series formally,
\begin{aligned} (f(x))^2&=(\sum_{n=1}^\infty b_n\sin nx)^2\\ &=\sum_{n=1}^\infty b_n^2\sin^2 nx+\sum_{n\neq m}b_nb_m\sin nx\sin mx. \end{aligned}

To see why the above hold, see the following concrete example:
\begin{aligned} (a_1+a_2+a_3)^2&=(a_1^2+a_2^2+a_3^2)+(a_1a_2+a_1a_3+a_2a_1+a_2a_3+a_3a_1+a_3a_2)\\ &=\sum_{n=1}^3 a_n^2+\sum_{n\neq m}a_na_m. \end{aligned}

Integrate term by term

We assume that term by term integration is valid.
\displaystyle \frac 1\pi\int_{-\pi}^\pi (f(x))^2\,dx=\frac 1\pi\int_{-\pi}^{\pi}\sum_{n=1}^\infty b_n^2\sin^2{nx}\,dx+\frac{1}{\pi}\int_{-\pi}^\pi\sum_{n\neq m}b_nb_m\sin nx\sin mx\,dx.

Recall that \displaystyle \int_{-\pi}^\pi \sin nx\sin mx\,dx=\begin{cases}0 &\text{if }n\neq m\\ \pi &\text{if }n=m \end{cases}.

So
\begin{aligned} \frac 1\pi\int_{-\pi}^{\pi}\sum_{n=1}^\infty b_n^2\sin^2{nx}\,dx&=\frac 1\pi\sum_{n=1}^\infty b_n^2(\int_{-\pi}^\pi\sin^2 nx\,dx)\\ &=\frac 1\pi\sum_{n=1}^\infty b_n^2 (\pi)\\ &=\sum_{n=1}^\infty (b_n)^2. \end{aligned}

Similarly
\begin{aligned} \frac{1}{\pi}\int_{-\pi}^\pi\sum_{n\neq m}b_nb_m\sin nx\sin mx\,dx&=\frac 1\pi\sum_{n\neq m}b_nb_m(\int_{-\pi}^{\pi}\sin nx\sin mx\,dx)\\ &=\frac 1\pi\sum_{n\neq m}b_nb_m(0)\\ &=0. \end{aligned}

So \displaystyle \frac 1\pi\int_{-\pi}^\pi (f(x))^2\,dx=\sum_{n=1}^\infty (b_n)^2. (Parseval’s Identity)

Apply Parseval’s Identity to f(x)=x

By Parseval’s identity,
\displaystyle \frac{1}{\pi}\int_{-\pi}^\pi x^2\,dx=\sum_{n=1}^\infty(\frac{2(-1)^{n+1}}{n})^2.

Simplifying, we get \displaystyle \frac 1\pi\cdot\left[\frac{x^3}{3}\right]_{-\pi}^\pi=\sum_{n=1}^\infty\frac{4}{n^2}.
\begin{aligned} \frac 1\pi(\frac{2\pi^3}{3})&=\sum_{n=1}^\infty \frac{4}{n^2}\\ \frac{\pi^2}{6}&=\sum_{n=1}^\infty\frac{1}{n^2}. \end{aligned}

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Random Math Fact:

Did you know:

Euler’s “lucky” numbers are positive integers n such that m2 − m + n is a prime number for m = 0, …, n − 1.

Leonhard Euler published the polynomial x2 − x + 41 which produces prime numbers for all integer values of x from 0 to 40. Obviously, when x is equal to 41, the value cannot be prime anymore since it is divisible by 41. Only 6 numbers have this property, namely 2, 3, 5, 11, 17 and 41.

Source: http://en.wikipedia.org/wiki/Lucky_numbers_of_Euler

 

Why is e irrational?

Anyone who has taken high school math is familiar with the constant \boxed{e=2.718281828\cdots}.

e-irrational

Today we are going to prove that e is in fact irrational! We will go through Joseph Fourier‘s famous proof by contradiction. The maths background we need is to know the power series expansion: \displaystyle \boxed{e=\sum_{n=0}^{\infty}\frac{1}{n!}}. The proof is slightly tricky so stay focussed!

(Reference: http://en.wikipedia.org/wiki/Proof_that_e_is_irrational)

Suppose to the contrary that e is a rational number, so \displaystyle e=\frac{a}{b}.

Using the power series formula mentioned above, we have \displaystyle\sum_{n=0}^\infty \frac{1}{n!}=\frac{a}{b}

Multiply both sides by b!, \displaystyle \sum_{n=0}^{\infty}\frac{b!}{n!}=\frac{ab!}{b}=a(b-1)!

Now, we split the sum into two parts:

\displaystyle \sum_{n=0}^b \frac{b!}{n!}+\sum_{n=b+1}^\infty \frac{b!}{n!}=a(b-1)!

Rearranging,

\displaystyle \sum_{n=b+1}^\infty \frac{b!}{n!}=a(b-1)!-\sum_{n=0}^b \frac{b!}{n!}

Now, denote \displaystyle x=\sum_{n=b+1}^\infty \frac{b!}{n!}>0. x is an integer since both \displaystyle a(b-1)! and \displaystyle\sum_{n=0}^b \frac{b!}{n!} are integers and their difference (which is x) will be an integer.

We now prove that x<1. For all terms with n\geq b+1 we have the upper estimate

\displaystyle\begin{array}{rcl} \frac{b!}{n!}&=&\frac{1\times 2\times \cdots \times b}{1\times 2\times \cdots \times b \times (b+1) \times \cdots \times n}\\ &=&\frac{1}{(b+1)(b+2)\cdots (b+(n-b))}\\ &\leq& \frac{1}{(b+1)^{n-b}} \end{array}

This inequality is strict for every n\geq b+2. Changing the index of summation to k=n-b and using the formula for the infinite geometric progression S_\infty = \frac{a}{1-r}, we obtain:

\displaystyle x=\sum_{n=b+1}^\infty \frac{b!}{n!} < \sum_{n=b+1}^\infty \frac{1}{(b+1)^{n-b}}=\sum_{k=1}^\infty \frac{1}{(b+1)^k}=\frac{\frac{1}{b+1}}{1-\frac{1}{b+1}}=\frac{1}{b}\leq 1

We have that x is an integer but 0<x<1. This is a contradiction (since there is no integer strictly between 0 and 1), and so e must be irrational. (QED)

Interesting? 🙂

Featured book:

Euler: The Master of Us All (Dolciani Mathematical Expositions, No 22)

Did you know the constant e is sometimes called Euler’s number?

Learn more about Euler in this wonderful book. Rated 4.9/5 stars, it is one of the highest rated books on the whole of Amazon.

Leonhard Euler was one of the most prolific mathematicians that have ever lived. This book examines the huge scope of mathematical areas explored and developed by Euler, which includes number theory, combinatorics, geometry, complex variables and many more. The information known to Euler over 300 years ago is discussed, and many of his advances are reconstructed. Readers will be left in no doubt about the brilliance and pervasive influence of Euler’s work.

Watch this video for another proof that e is irrational!