A tangent identity

We all know that the Chebyshev polynomials of the first kind can be defined as

T_n(x)=cos(ncos^{-1}x).

When I was fiddling about with identities associated with Machin’s formula I came across an identity for tangents, which I’m sure is well known, but which I’d never seen before. Here it is:

displaystyle{tan(ntan^{-1}x)=frac{displaystyle{{nchoose 1}x-{nchoose 3}x^3+{nchoose 5}x^5-cdots}}{displaystyle{{nchoose 0}-{nchoose 2}x^2+{nchoose 4}x^4-cdots}}}

This can be more precisely written as follows:

displaystyle{tan(2ktan^{-1}x)=frac{displaystyle{sum_{i=0}^{k-1}(-1)^i{2kchoose 2i+1}x^{2i+1}}}{displaystyle{sum_{i=0}^k(-1)^i{2kchoose 2i}x^{2i}}}}

and

displaystyle{tan((2k+1)tan^{-1}x)=frac{displaystyle{sum_{i=0}^k(-1)^i{2k+1choose 2i+1}x^{2i+1}}}{displaystyle{sum_{i=0}^k(-1)^i{2k+1choose 2i}x^{2i}}}}

And these two identities can be shoehorned into one ugly (but general!) expression:

displaystyle{tan(ntan^{-1}x)=frac{displaystyle{sum_{i=0}^{lfloor (n-1)/2rfloor}(-1)^i{nchoose 2i+1}x^{2i+1}}}{displaystyle{sum_{i=0}^{lfloor n/2 rfloor}(-1)^i{nchoose 2i}x^{2i}}}}

The identity is not hard to prove by induction. As T is already used for the Chebyshev polynomials, we shall write G_k(x) for tan(ktan^{-1}x) and then write

displaystyle{G_k(x)=frac{N_k(x)}{D_k(x)}}

(where N,D stands for numerator and denominator respectively.) Then by the addition formula for tan(x), we have

displaystyle{G_{k+1}(x)=frac{x+G_k(x)}{1-xG_k(x)}}.

Writing G_k on the right in terms of N_k and D_k, and then multiplying through by D_k produces

displaystyle{G_{k+1}(x)=frac{xD_k(x)+N_k(x)}{D_k(x)-xN_k(x)}}.

Recall that

displaystyle{N_k(x)={kchoose 1}x-{kchoose 3}x^3+{kchoose 5}x^5-cdots}

and

displaystyle{D_k(x)={kchoose 0}-{kchoose 2}x^2+{kchoose 4}x^4-cdots}

Now consider the the numerator of G_{k+1}(x):

xD_k(x)+N_k(x).

The coefficient of x^{2i+1} will be

displaystyle{(-1)^i{kchoose 2i}+(-1)^i{kchoose 2i+1}=(-1)^i{k+1choose 2i+1}}.

The denominator of G_{k+1}(x) is

D_k(x)-xN_k(x)

and the coefficient of x^{2i} will be

displaystyle{(-1)^i{kchoose 2i}+(-1)^i{kchoose 2i-1}=(-1)^i{k+1choose 2i}}.

Note that the constant coefficient is

displaystyle{{kchoose 0}={k+1choose 0}}.

These results, plus the trivial result for k=1, prove the identity.

Leave a Reply

Your email address will not be published. Required fields are marked *