Tags

Related Posts

Share This

Cool proofs and origins

Posted by on Jul 13, 2011 in Fun mathematics, Mathematics education | 2 comments

I am always interested in simple/elegant/cool proofs of mathematical facts. I’m also interested to know their origins. Does anyone know who came up with the following rather elegant proof of \displaystyle \sum _{k=0}^n 2^k = 2^{n+1}-1 ? It’s likely that the origins are lost in time but I thought I would ask anyway.

Proof:


\begin{array}{rcl} \displaystyle \sum _{k=0}^n 2^k &=& \displaystyle (2-1) \sum _{k=0}^n 2^k \\ &=&\displaystyle 2\sum _{k=0}^n 2^k - \sum _{k=0}^n 2^k\\ &=&\displaystyle \sum _{k=1}^{n+1} 2^k- \sum _{k=0}^n 2^k \\ &=& \displaystyle 2^{n+1}-1 . \end{array}

I like this because of the use of 2-1=1. Another situation where the use of 1 gives a cool proof is the following from complex analysis. Does anyone know the origin of the proof?

Theorem: If g:[a,b]\to \mathbb{C} is complex Riemann integrable and |g| is Riemann integrable, then \left| \int_a^b g(t) \, dt \right| \leq \int_a^b | g(t)| \, dt .

Proof: If the left-hand side is zero, then the statement is trivial. Hence, assume that the left-hand side is non-zero. Define the complex number c by
c =\dfrac{\left| \int_a^b g(t) \, dt \right|}{\int_a^b g(t) \, dt } .
We have,
\begin{array}{rcl}\displaystyle \left| \int_a^b g(t) \, dt \right| &=& \displaystyle c \int_a^b g(t) \, dt \\ &=& \displaystyle \int_a^b {\text{Re}} \left( c g(t) \right) \, dt +i \int_a^b {\text{Im}} (c g(t) ) \, dt \\ &=& \displaystyle \int_a^b {\text{Re}} \left( c g(t) \right) \, dt , {\text{ because the left-hand side is real,}} \\&\leq & \displaystyle \int_a^b |c g(t) | \, dt , {\text{ as }} {\text{Re}}(z)\leq |z| , \\&=&\displaystyle \int_a^b |c |\, |g(t) | \, dt \\&=&\displaystyle \int_a^b |g(t) | \, dt , {\text{ as clearly }} |c |=1 .\end{array}

Do you have any cool proofs? (With or without clever use of 1!)

Get the newsletter!

Fancy a newsletter keeping you up-to-date with maths news, articles, videos and events you might otherwise miss? Then sign up below. (No spam and I will never share your details with anyone else.)