Timeline for Does any such family of functions exist?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 1 at 16:47 | vote | accept | The potato eater | ||
| May 31 at 20:55 | comment | added | Carlo Beenakker | @MichaelHardy --- for $k=2$ I found such a real path-connected solution; I posted it in the answer boxk, it is not simple... | |
| May 31 at 20:41 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
edited to allow an upvote
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| May 31 at 20:40 | answer | added | Carlo Beenakker | timeline score: 3 | |
| May 31 at 19:31 | comment | added | Michael Hardy | I've deleted my earlier comment where $n$ appeared where $n/2$ was needed. | |
| May 31 at 19:29 | comment | added | Michael Hardy | 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. | |
| May 31 at 19:26 | comment | added | Michael Hardy | @CarloBeenakker : I put those two numbers you suggested into WolframAlpha and it works! Maybe you should post an answer. | |
| May 31 at 19:18 | comment | added | Michael Hardy | $\mathbf{ERRATUM:}$ It seems I wrote $(-1)^n$ where I needed $(-1)^{n/2}$ (since $n$ is even). $$ \begin{align} & \cos\sum_{i=1}^k y_i \\ {} \\ = {} & \sum_{\text{even } n\,\ge\,0} (-1)^{n/2} \sum_{I \,\subseteq\,\{\,1,\,\ldots\,,\,k\,\} \\ \quad |I|\,=\,n} \,\,\,\prod_{i\,\in\,I} \sin y_i \prod_{i\,\notin\,I} \cos y_i. \end{align} $$ | |
| May 31 at 10:33 | comment | added | Carlo Beenakker | you need complex $f_i$ to satisfy the equality; for example, $k=2$, $f_1=-1.173 + 2.047 \,i$, $f_2=0.213 + 0.108 \,i$. | |
| May 30 at 22:13 | comment | added | The potato eater | 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. | |
| May 30 at 21:59 | comment | added | Michael Hardy | 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 $y_i. \qquad$ | |
| May 30 at 21:47 | history | edited | The potato eater | CC BY-SA 4.0 |
edited title
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| May 30 at 21:46 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 10 characters in body; edited title
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| May 30 at 21:45 | history | asked | The potato eater | CC BY-SA 4.0 |