# A Physicist’s Proof

As an undergrad, I took a couple “math for physicists” courses that I found to be quite helpful. One of the more humorous concepts a professor of mine conveyed to me was the idea of a “physicist’s proof”. These would be sort of intuitive proofs that a mathematician may scoff at, but physicists tend to appreciate. Below is an example of a physicist’s proof showing the following mathematical relation:

$\sum_{i=0}^n (2i+1) = (n+1)^2$

This relation can more easily be stated in words. It says that if you add up consecutive odd numbers starting at 1, you get a perfect square. For instance, $1+3+5=9=3^2$ or $1+3+5+7=16=4^2$.

Here is the idea:

For n=0, you get a 1×1 square as is seen in the following image:

For n=1, you add the 3 next squares and you get the following 2×2 square:

For n=2 and n=3, one adds 5 and 7 squares respectively to get the following 3×3 and 4×4 squares:

and

By now, you can probably see the pattern and why this mathematical relation holds (hopefully where I have put the dashes and dots helps you see this!). To me, these kinds of proofs, while lacking in mathematical rigor, do much for the intuition. In fact, they even suggest an algebraic manner by which to prove the sum rigorously.

Long live the physicist’s proof!

### 4 responses to “A Physicist’s Proof”

1. Pingback: This Condensed Life

2. When I teach this to my undergrads (the sum above is, e.g., the number of orbitals in a shell with principal quantum number n+1), I tell them a story i remember seeing on public TV when I was a kid. It was a story about hte life of Gauss, who, apparently, at the age of 7 solved a problem a teacher has assigned as busy work to keep the kids occupied while he finished writing a personal letter. The task was to sum up all numbers from 1 to 100. Little Gauss did the work in a couple of minutes, as he figured out that 1+2+…+99+100 is the same as (1+100)+(2+99)+(3+98) etc, so there are 50 pairs and each pair adds up to 101. That’s Gauss at age 7! Thus e.g.. 1+…+k=(k+1)k/2 (so sum of first with last times number of pairs).
Of course, your sum from above is \sum_{l=0}^{n}(2l+1)=2 \sum_{l=0}^{n} l +\sum_{l=0}^{n} 1 = 2*(n)(n+1)/2 +(n+1)=(n+1)^2.

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• Anshul Kogar

Thanks for sharing that! I hadn’t thought of the sum that way previously.

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