Summoning Cthulhu: sin^2 + cos^2 with power series

I would like to dedicate this post to the Redditor AlanCrowe. When I claimed that working out the formula $\mbox{sin}^2 + \mbox{cos}^2 = 1$ using power series wouldn’t be that big of a deal, he said that this was “a rash thing to do” and that I would “end up summoning Cthulhu doing weird shit like that.”

How could I resist such a dare? The following manipulations contain no real mysteries; they are all direct applications of the binomial theorem and the distributive law. There is a small but transparent trick using complex numbers, which can be eliminated at the expense of more combinatorial footwork.

Disclaimer: Be careful working with power series. Power series may carry high voltages. Improper handling of power series may result in extreme divergence or serious mental anguish. Power series have no remorse. Do not taunt power series. If the double summations make you queasy, go for a short walk outside. I’ll wait.

Step 1:

$\mbox{sin}\ x = x - \frac{x^3}{3!} + \frac{x^5}{5!}\ldots = \sum_n \frac{(i)^n-(-i)^n}{2in!} x^n$

$\mbox{sin}^2\ x = (\sum_n \frac{(i)^n-(-i)^n}{2in!} x^n)^2$

$= \sum_n x^n \sum_k (\frac{(i)^k-(-i)^k}{2ik!})(\frac{(i)^{n-k}-(-i)^{n-k}}{2i(n-k)!})$

$= \sum_n x^n i^n \sum_k \frac{-1 + (-1)^k + (-1)^{n-k} - (-1)^n}{4k!(n-k)!}$

Step 2:

$\mbox{cos}\ x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\ldots = \sum_n \frac{(i)^n+(-i)^n}{2n!} x^n$

$\mbox{cos}^2\ x = (\sum_n \frac{(i)^n+(-i)^n}{2n!} x^n)^2$

$=\sum_n x^n \sum_k (\frac{(i)^k+(-i)^k}{2k!})(\frac{(i)^{n-k}+(-i)^{n-k}}{2(n-k)!})$

$= \sum_n x^n i^n \sum_k \frac{1 + (-1)^k + (-1)^{n-k} + (-1)^n}{4k!(n-k)!}$

Step 3:

$\mbox{sin}^2\ x + \mbox{cos}^2\ x = \sum_n x^n i^n \sum_k \frac{(-1)^k + (-1)^{n-k}}{2k!(n-k)!}$

$=\sum_n \frac{x^n i^n (1 + (-1)^n)}{2n!} \sum_k (-1)^k {n\choose k}$

$=\sum_n \frac{x^n i^n (1+(-1)^n)}{2n!} (1-1)^n = 1 + 0 + 0+\ldots = 1$

UPDATE: notfancy has accused me of being a “cheater” for my use of complex numbers, claiming that I am relying on an implicit call to ~~De Moivre’s theorem~~ Euler’s formula. I have seen this view elsewhere as well, that the use of tricks like this is tantamount to pulling a rabbit out of a hat.

I have a somewhat different viewpoint. At the heart of the series for $\mbox{sin}$ and $\mbox{cos}$ is the periodic sequence $0,1,0,-1, 0,1\ldots$ . And surely the mathematical techniques for manipulating this series can be no simpler than the series itself! In other words, I feel that my observation that we can write this sequence as $\frac{i^n + (-i)^n}{2}$ was just notational convenience and nothing more. My own style is to prefer algebra to case analysis because I tend to make more mistakes in the latter.

To illustrate this, I am presenting the same proof, without an appeal to complex numbers. It requires a little more thought, but is overall a much simpler-looking approach.

First, for the purpose of notation, we need to make an observation about power series. If $f(x) = \sum_n \frac{a_n x^n}{n!}$ and $g(x) = \sum_n \frac{b_n x^n}{n!}$ , then $f(x)g(x) = \sum_n x^n \sum_k \frac{a_k b_{n-k}}{k! (n-k)!} = \sum_n \frac{x^n}{n!} \sum_k {n\choose k}a_k b_{n-k}$ . Nothing special here; just the distributive law. (if you are interested in going deeper into these kinds of manipulations of sequences, I highly recommend Wilf’s generatingfunctionology, available for free online)

In the same notation, if $f(x)=\mbox{sin}\ x$ and $g(x)=\mbox{cos}\ x$ , then $a_n$ and $b_n$ are the sequences $0, 1, 0, -1\ldots$ and $1,0,-1,0\ldots$ respectively.

Steps 1 and 2:

$\mbox{sin}\ x = \sum_n \frac{a_n x^n}{n!}$

$\mbox{sin}^2\ x = \sum_n \frac{c_n x^n}{n!}$

where $c_n = \sum_k {n\choose k} a_k a_{n-k}$ , and

$\mbox{cos}\ x = \sum_n \frac{b_n x^n}{n!}$

$\mbox{cos}^2\ x = \sum_n \frac{d_n x^n}{n!}$

where $d_n = \sum_k {n\choose k} b_k b_{n-k}$

Step 3:

$\mbox{sin}^2\ x + \mbox{cos}^2\ x = \sum_n \frac{x^n}{n!} e_n$

where $e_n = c_n + d_n$ .

We have $c_0 = 0$ and $d_0=1$ , so $e_0=1$ . All that remains to be shown is that $e_n=0$ for all $n>0$ .

First, suppose that $n$ is odd. Then $a_k a_{n-k} = b_k b_{n-k}=0$ for all $k$ , so $e_n = 0$ .

Now, suppose that $n=2m$ is even. Then $c_n = (-1)^{m+1}( {n\choose 1} + {n\choose 3}+\ldots + {n\choose {n-1}} )$ and $d_n = (-1)^m ({n\choose 0} + {n\choose 2}\ldots+ {n\choose n})$ . Hence $e_n = c_n + d_n = (-1)^m ({n\choose 0} - {n\choose 1} + {n\choose 2}\ldots + {n\choose n})$ $= (-1)^m (1-1)^n = 0$ .

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9 Responses to Summoning Cthulhu: sin^2 + cos^2 with power series

Alan Crowe says:

January 15, 2009 at 11:43 pm

What algebraic properties must x have for those manipulations to be valid? It seems clear that we never use any of the ordering properties of the reals, so the argument works just as well for complex numbers. Err, how do you have a complex angle? That seems just wrong. Hence my jest about summoning Cthulhu which I expand in a tongue-in-cheek blog post.

Do we commutivity? If with view the polynomials as a module over the reals we only seem to be using commutivity of the reals. The angles could be quaterions. Indeed we don’t make much use of associativity, it only has to work with powers of x. So a power associative algebra will do.

This surprised me, so I’ve coded up the Cayley-Dickson construction and been checking numerically to see if I’ve missed something. cosine squared plus sine squared equals one works for a few randomly chosed octonians, and sedenions, so it looks as if any power associative algebra will do.

Reply
- lydianrain says:
  
  January 16, 2009 at 12:38 am
  
  You accuse me of consorting with evil powers, and then I see you here, fully dressed in the black robes worn by the disciples of The Dark One, knife held above the sacrificial altar with no remorse or hesitation in your eyes at all!
  
  Indeed, I intended these manipulations to be purely formal, in, say, the ring of integer power series $\mathbb{Z}[[x]]$ , but to translate it to an analytical context in some demonic algebra $\mathbb{A}$ , we seem to need only two things: a nice morphism $\mathbb{Z}[x]\to \mathbb{A}$ , and a norm. Cayley-Dickson seems to provide both, as power associativity is indeed the only requirement for such a morphism to exist. Absolute convergence follows, dooming us all.
  
  Reply
lydianrain says:

January 26, 2009 at 10:02 pm

Correction: what we’re actually looking for is a *complete* norm, and a morphism $\mathbb{Q}[[x]]\to\mathbb{A}$ . Both of which, terrifyingly, we have.

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Welcome to Carnival of Mathematics 48 = 6!!! « Concrete Nonsense says:

January 30, 2009 at 7:13 pm

[...] a visit. In a two-part act brought on by a challenge of Alan Crowe on Reddit, my friend lydianrain summons Cthulhu with power series (or at least shows that in the way you least want to do it in). Alan of course reciprocates with [...]

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Robert says:

March 2, 2009 at 3:06 am

If you think that’s summoning Cthulhu, you should see what G. H. Hardy does in the first chapter of “Divergent Series”.

Reply
Qiaochu Yuan says:

June 29, 2009 at 8:28 pm

Doron Zeilberger illustrates a combinatorial proof of this identity in the Princeton Companion by identifying the power series of sine and cosine with certain weighted exponential generating functions. More generally, there’s a certain philosophy that essentially every power-series argument is a combinatorial argument in disguise.

Reply
- lydianrain says:
  
  June 29, 2009 at 9:05 pm
  
  Cool, thanks for the tip–I’ll check it out in my copy.
  
  Reply
CrashDummy001 says:

April 6, 2011 at 4:22 am

Isn’t it much much easier to say sin^2(x) = [1-cos(2x)]/2, and the similar formula for cos^2(x), to find two extremely easy and fundamental power series and simply add them together?

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- lydianrain says:
  
  April 6, 2011 at 6:13 am
  
  That would be cheating; the goal is to prove a formula given a definition. To use your auxiliary formula, you would first need to demonstrate, in the ring of formal power series, that sin^2(x) = [1-cos(2x)]/2. This demonstration would be slightly harder than my post, as it would require explicitly calculating the b_n and d_n.
  
  Reply