Is there a sequence of non-zero bounded smooth functions $f_1,f_2,\ldots,f_k$ so that
$$\sum_{I=1}^k \cos(f_i)= \cos\left(\sum_{i=1}^k f_i \right).$$
And what about the infinite case ?
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3$\begingroup$ If there is a sequence $y_1,\ldots,y_k$ of non-zero NUMBERS for which $$ \sum_{i=1}^k \cos(y_i)= \cos\left(\sum_{i=1}^k y_i \right) $$ then there is a sequence of non-zero FUNCTIONS satisfying that equality, since each function $f_i$ could be constantly equal to $y_i. \qquad$ $\endgroup$– Michael HardyCommented May 30 at 21:59
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$\begingroup$ Not easy to equate this sum to this product but maybe possible. Just asking because my other question kinda ties to it. If so maybe there is simple expression for the zeta function I'm half plane. I'm sure we can tweak it a bit and add denominator to the cosine. $\endgroup$– The potato eaterCommented May 30 at 22:13
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1$\begingroup$ @CarloBeenakker : I put those two numbers you suggested into WolframAlpha and it works! Maybe you should post an answer. $\endgroup$– Michael HardyCommented May 31 at 19:26
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1$\begingroup$ Is the set of $k$-tuples of complex numbers satisfying the proposed equality path-connected? If so, $f_1,\ldots,f_k$ could be non-constant functions of a real variable satisfying this equality. $\endgroup$– Michael HardyCommented May 31 at 19:29
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1$\begingroup$ @MichaelHardy --- for $k=2$ I found such a real path-connected solution; I posted it in the answer boxk, it is not simple... $\endgroup$– Carlo BeenakkerCommented May 31 at 20:55
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1 Answer
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Here is a real solution for $k=2$: Take $f_1(x)=x$ and $$f_2(x)=\arccos\left( \sqrt{\tfrac{3}{4}- \cos ^4(x/2)} \operatorname{cotan} (x/2)-\tfrac{1}{2}\cos x\right).$$ Then $$\cos f_1(x)+\cos f_2(x)=\cos\bigl(f_1(x)+f_2(x)\bigr),\;\;\text{for}\;\;\pi/3<x<\pi +2 \arcsin \left(\frac{3^{1/4}}{\sqrt{2}}\right).$$ Numerically, this interval is $1.05<x<5.53$. It can be extended by switching the branch of the square root.
The solution can be generalized to larger $k$; put all $f_n(x)=x$ for $n=1,2,\ldots, k-1$ and then solve for $f_k(x)$. The expression is lengthy and I omit it here. The interval in which a real solution holds shrinks with increasing $k$.
answered May 31 at 20:40