## 2009-01-09

### Why so Series

Notational Notions has an entry raising to the challenge of showing with power series that (sin x)² + (cos x)² = 1. I jestingly but unfairly called the author a cheater for using what to me was an obvious appeal to Euler's formula (I made a mistake in naming it as De Moivre's). I reasoned thus:

```  (sin x)² + (cos x)²
= { Euler: e^ix - e^(-ix) = 2i sin x }
[(e^ix - e^(-ix))/(2i)]² + [(e^ix + e^(-ix))/2]²
= { Binomial, twice }
-(e^2ix - 2 e^ix e^(-ix) + e^(-2ix))/4 + (e^2ix + 2 e^ix e^(-ix) + e^(-2ix))/4
= { Algebra }
(-e^2ix + 2 e^ix e^(-ix) - e^(-2ix) + e^2ix + 2 e^ix e^(-ix) + e^(-2ix))/4
= { Algebra throughout }
(2 e^ix e^(-ix) + 2 e^ix e^(-ix))/4
= { Algebra }
e^ix e^(-ix)
= { Negative power is inverse }
1
```

The author defended the choice of purely imaginary exponentials as the most natural representation for the real series (1, 0, -1, 0, …). Meeting the challenge of showing a purely real derivation incurred in an ugly and possibly unavoidable case analysis. I conceded, but tried my hand at it. It is well-known that the series development for the sine and cosine are:

```sin x = x - x^3/3! + x^5/5! - x^7/7! + …

cos x = 1 - x^2/2! + x^4/4! - x^6/6! + …
```

Squaring the first series:

``` (sin x)²
= { Definition }
(x - x^3/3! + x^5/5! - x^7/7! + …)²
= { Distribution }
x^2/1!1! - x^4/3!1! + x^6/5!1! - x^8/7!1! + …
- x^4/1!3! + x^6/3!3! - x^8/5!3! + x^10/3!7! - …
+ x^6/1!5! - x^8/3!5! + x^10/5!5! - x^12/7!5! + …
- x^8/1!7! + x^10/7!3! - x^12/7!5! + x^14/7!7! - …
```

(the use of pre-formatted text is crucial here to see what's going on). Squaring the second:

``` (cos x)²
= { Definition }
(1 - x^2/2! + x^4/4! - x^6/6! + …)²
= { Distribution }
1 - x^2/2!0! + x^4/4!0! - x^6/6!0! + …
- x^2/0!2! + x^4/2!2! - x^6/4!2! + x^8/6!2! - …
+ x^4/0!4! - x^6/2!4! + x^8/4!4! - x^10/6!4! + …
- x^6/0!6! + x^8/2!6! - x^10/4!6! + x^12/6!6! - …
```

Note that the terms are set so that columns line up by exponent. Note also that the factorial dividends add up to the exponent. Now:

``` (sin x)² + (cos x)²
=
x^2/1!1! - x^4/3!1! + x^6/5!1! - x^8/7!1! + …
- x^4/1!3! + x^6/3!3! - x^8/5!3! + x^10/3!7! - …
+ x^6/1!5! - x^8/3!5! + x^10/5!5! - x^12/7!5! + …
- x^8/1!7! + x^10/7!3! - x^12/7!5! + x^14/7!7! - …
+
1 - x^2/2!0! + x^4/4!0! - x^6/6!0! + …
- x^2/0!2! + x^4/2!2! - x^6/4!2! + x^8/6!2! - …
+ x^4/0!4! - x^6/2!4! + x^8/4!4! - x^10/6!4! + …
- x^6/0!6! + x^8/2!6! - x^10/4!6! + x^12/6!6! - …
= { Associativity of addition }
1 - x^2/2!0! + x^4/4!0! - x^6/6!0! + …
+ x^2/1!1! - x^4/3!1! + x^6/5!1! - x^8/7!1! + …
- x^2/0!2! + x^4/2!2! - x^6/4!2! + x^8/6!2! - …
- x^4/1!3! + x^6/3!3! - x^8/5!3! + x^10/3!7! - …
+ x^4/0!4! - x^6/2!4! + x^8/4!4! - x^10/6!4! + …
+ x^6/1!5! - x^8/3!5! + x^10/5!5! - x^12/7!5! + …
- x^6/0!6! + x^8/2!6! - x^10/4!6! + x^12/6!6! - …
- x^8/1!7! + x^10/7!3! - x^12/7!5! + x^14/7!7! - …
= { Binomial theorem }
1 - (x - x)^2/2! + (x - x)^4/4! - (x - x)^6/6! + (x - x)^8/8! - …
=
1
```

Of course there are two fishy manouvers going on here. The first is the use of ellipses to denote the series. This is a notational convenience that allows me to show the full expansions for easy pattern matching of terms. These ellipses actually encode a rigorous inductive law defining each term, and could be replaced by summations. The second is the use of associativity to freely reorder the terms of an infinite series without regard to convergence, of which I can't say anything as I'm no mathematician but which I believe it can be made airtight. In any case, you can take this as a purely symbolic manipulation.

Alp Mestan said...
This comment has been removed by the author.
Ignacio A. Maciel Romero said...

Hola Matías, ante todo resuélveme el siguiente interrogante, ¿sos el mismo Matías Giovaninni que fue conmigo al cole? Si es así un abrazo desde Sevilla desde donde te devuelvo la visita al blog aunque el tuyo me resulta incomprensible pero no me sorprende ya que desde chiquito eras un genio. Acabo de ver a tu recomendación, Joanna Newsom, en youtube y me parece genial. Gracias por tu visita y seguimos en contacto, Nacho

lydianrain said...

To justify rearranging these sums you just need absolute convergence, which is very easy for these particular series.

You are right that we can avoid talking about convergence, thinking of this as a calculation in the ring of formal power series. (which, since it avoids calculus, is my preferred view :)

Matías Giovannini said...

@Alp: no I don't think the proof I linked to is fake.

@Nacho: sí, yo mismo…

@lydianrain: formal power series are also my preferred view of the matter as I don't know (and I don't really like) analysis.