Let $n\in \mathbb{N}$ and $$-\infty < a_1 < b_1 < a_2 < b_2 < a_3 < b_3<\cdots<a_n<b_n<+\infty$$ and $k_i\in \mathbb{R}, i=1,2,\ldots,n$. Is there any information about how many roots the function $$f(x)=k_1\sqrt{x-a_1} + k_2 \sqrt{x-a_2} + \cdots+k_n\sqrt{x-a_n}$$ for $x\geqslant a_n$ or the function $$g(x)=k_1(\sqrt{x-a_1}-\sqrt{x-b_1})+k_2(\sqrt{x-a_2}-\sqrt{x-b_2})+\cdots+k_n(\sqrt{x-a_n}-\sqrt{x-b_n})$$ for $x\geqslant b_n$ can have?
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$\begingroup$ Taking the product of the $2^{n-1}$ polynomials $k_1\sqrt{x-a_1}\pm k_2\sqrt{x-a_2}\pm\cdots \pm k_n\sqrt{x-a_n}$ gives a polynomial of degree $2^{n-2}$, so this is a bound on the number of roots. But this doesn't use the condition $x\geq a_n$. $\endgroup$– Richard StanleyCommented May 28, 2023 at 23:17
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$\begingroup$ Thank you! I have added the condition $x\geqslant a_n$ to stay in the area of real-analysis. But now i think that for me it's enough to have any statement like the Fundamental theorem of algebra but for functions $f(z)=k_1\sqrt{z−a_1}+k_2\sqrt{z−a_2}+\cdots +k_n\sqrt{z−a_n},\quad k_i,a_i,z∈\mathbb{C}$. Is there such statement like $f$ can have at most $n$ complex-roots? Or something like this. $\endgroup$– eN.meshokCommented May 29, 2023 at 5:50
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$\begingroup$ You mean, how many real roots? $\endgroup$– YCorCommented Jun 27, 2023 at 20:39
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$\begingroup$ Possibly useful: Arthur Cayley (1821-1895), On polyzomal curves, otherwise the curves $\sqrt{U} + \sqrt{V} + \&\text{c.} = 0$, Transactions of the Royal Society of Edinburgh 25 (1869), pp. 1-110. archive.org copy. I also cited this in a comment to the MSE question free software for radical algebraic equations, a question/answers that might also be of interest. $\endgroup$– Dave L RenfroCommented Jun 27, 2023 at 22:09
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$\begingroup$ @YCor, Actually I meant complex root. Maybe it's some obvious question from complex analysis, but i have never met it in any of standard literature. $\endgroup$– eN.meshokCommented Jun 30, 2023 at 8:55
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1 Answer
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In the case when $n=1, 2, 3, 4$ the number of roots of $f$ is at most $1, 1, 2, 4$ respectively. This can be seen by squaring $f$ several times. For inctance, let $n=3$, then $$k_1\sqrt{x-a_1}+k_2\sqrt{x-a_2}+k_3\sqrt{x-a_3}=0$$ $$k_1^2(x-a_1)+k_2^2(x-a_2)+2k_1\sqrt{x-a_1}k_2\sqrt{x-a_2}=k_3^2(x-a_3)$$ $$4k_1^2(x-a_1)k_2^2(x-a_2)=\left(k_3^2(x-a_3)-k_1^2(x-a_1)-k_2^2(x-a_2)\right)^2$$ So, we have a polynomial of the second degree. But this technique can't be applied to $n=5,6,...$.