# Can a "difference formula" be a "sum formula" too?

Consider the following "product-to-difference" formula: $$\sin p\sin q=\frac{1}{2}[\cos(p-q)-\cos(p+q)].$$ I wonder whether the right-hand side can be expressed as a weighted sum of cosine functions with positive coefficients, i.e., whether a "difference" formula can be a sum formula.

For concreteness I formulate the following question: Given $a_0>0$, do there exist $a_i\ge 0$ and $b_i$ such that $$\sin (a_0x)\sin x=\sum_{i=1}^{\infty}a_i\cos b_i x?$$

With a restriction on $a_0$ and $b_i$, namely that $a_0, b_i\in \mathbb{N}$, the uniqueness of Fourier series implies that the problem has no solutions. Even more simply, we can evaluate both sides at $x=0$ to see that the equality cannot hold. To fix this, I ask the following modified question: Given $a_0>0$, do there exist $a_i\ge 0$ and $b_i$ and $\mathrm{err}_{a_0}=\mathrm{err}$ such that $$\sin (a_0x)\sin x=\sum_{i=1}^{\infty}a_i\cos b_i x +\mathrm{err}(x)$$ where the function err is an arbitrary sine series, namely, $$\mathrm{err}(x)=\sum_{j=1}^{\infty}c_i\sin d_ix$$ where $c_i$ and $d_i$ are arbitrary real numbers. Does the problem have a solution in general, without any restrictions?

• For positive coefficients, note that $\cos(x)$ is an even function, so that $-\cos(p + q) = \cos(-p - q)$. Commented May 31, 2015 at 19:03
• Use Taylor expansion at $x=0$. Left is $a_0 x^2 + O(x^4)$, right is $\sum a_i + o(1)$ Commented May 31, 2015 at 19:07
• No need to expand, just evaluate at $x=0$ to see that all $a_i$ need to be zero. Commented May 31, 2015 at 19:10
• @pjs36: I think you're confusing even and odd functions. Yes, the cosine is even, by that does not imply what you claim it does. Commented May 31, 2015 at 19:22
• You're absolutely right, @HenningMakholm; I'm glad you pointed that out! Commented May 31, 2015 at 19:26

## 2 Answers

You're asking two different questions, as in the former you have the situation $p+q$, $p-q$, and in the latter you're asking about $b_ix$. Note, if $b_ix = p+q$, there is no $b_i'$ such that $b_i'x = p-q$ for all $x$.

In any case, to answer your second question, notice that $\cos$ is an even function. This would give you the mapping $b_i \mapsto -b_i$ when the condition $b_i \in \mathbb{N}$ is relaxed.

Alternatively, if you allow series of the form $\sum_i a_i \cos (b_ix+c_i)$, then take $c_i \in {0,\pi}$, i.e. phase shift.

• I agree that the RHS does not change when we replace $b_i$ with $-b_i$, but how is this a problem? The LHS is also an even function.
– EPS
Commented May 31, 2015 at 19:45

Note that ${\rm err}$ is odd, being a series of sines. Assume that there exist coefficients $c_i, \space d_j$ as desired. The formula ${\rm err}(x) = \sin (a_0 x) \sin x - \sum \limits _{i=1} ^\infty a_i \cos (b_i x)$ show that ${\rm err}$ must also be even, since its right-hand side is so. Being odd and even, ${\rm err}$ must be identically $0$, which is what you were trying to avoid.