Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+……+\frac{1}{1-x_{n-1}}$$ $\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$ Now Put $\displaystyle y = \frac{1}{1-x}\Rightarrow 1-x=\frac{1}{y}\Rightarrow x=\frac{y-1}{y}$ Put that value into $\displaystyle x^n-1=0\;,$ We get $\displaystyle \left(\frac{y-1}{y}\right)^n-1=0$ So we get $(y-1)^n-y^n=0\Rightarrow \displaystyle \left\{y^n-\binom{n}{1}y^{n-1}+\binom{n}{2}y^2-……\right\}-y^n=0$ So $\displaystyle \binom{n}{1}y^{n-1}-\binom{n}{2}y^{n-1}+…+(-1)^n=0$ has roots $y=y_{1}\;,y_{2}\;,y_{3}\;,…….,y_{n-1}$ So $$y_{1}+y_{2}+y_{3}+…..+y_{n-1} = \binom{n}{2} =\frac{n(n-1)}{2}\;,$$ Where $\displaystyle y_{i} = […]

I have an affine space in $V_6(\mathbb{R})$: $\{Y=(2,-2,0,1,-1,0)+a(1,-1,0,0,0,0)+b(1,0,0,1,-3,1)+c(1,-4,0,2,1,0)$ with $a,b,c \in \mathbb{R}\}$. A generic vector of $Y$ is something like this :$P+rA+sB+tC= \left( \begin{array}{c} 2+r+s+t \\ -2-r-4t \\ 0 \\ 1+s+2t \\ -1-3s+t \\ 0+s \end{array} \right)$ The orthogonal complement $Y’= 0 + \{ (1,-1,0,0,0,0),(1,0,0,1,-3,1),(1,-4,0,2,1,0) \}^\perp $ passes through the origin and intersects $Y$ in […]

If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?

Consider the “obvious” equality $$(a^m)^n = a^{mn}.$$ In particular, if I only want to consider real numbers, for what values of $a,m,n$ is the above displayed equality true? And where can I find the precise statement written down? My first instinct was that as long as $a$, $a^m$, and $(a^m)^n$ were real numbers, then the […]

Show that if a set of complex numbers $z_1,z_2,\ldots,z_n$ satisfy $$z_1^l+z_2^l+\cdots+z_n^l=0$$ for every odd $l$, then for any $z_i$ we can always find some $z_j$ such that $z_i+z_j=0$. The question has been answered here for real numbers but not for complex numbers

Question on what is $$\Pi_{n=1}^\infty\left(1-\frac{x^a}{\pi^an^a}\right)$$ I deduced that for any $a=2,4,6,\dots$, the above product is simplify-able into a product of sines, $$\Pi_{n=1}^\infty\left(1-\frac{x^2}{\pi^2n^2}\right)=\frac{\sin(x)}x$$ $$\Pi_{n=1}^\infty\left(1-\frac{x^4}{\pi^2n^2}\right)=\frac{\sin(x)\sin(xi)}{x^2}$$ And trying to put this into a general form, I got $$f(x):=\frac{\sin(x)}{x}$$ $$\implies f(x)=f(-x)$$ $$f(x)f(ix)=\Pi_{n=1}^\infty\left(1-\frac{x^4}{\pi^4n^4}\right)=\sqrt{f(xe^{\frac{\pi i}{2}})f(xe^{\pi i})f(xe^{\frac{3\pi i}{2}})f(xe^{2\pi i})}$$ Using $(a+b)(a-b)=a^2-b^2$, we can get something along the lines of $$\Pi_{n=1}^\infty\left(1-\frac{x^{2^a}}{\pi^{2^a}n^{2^a}}\right)=\sqrt{f(xe^{\frac{1\pi i}{2^a}})f(xe^{\frac{2\pi i}{2^a}})\dots […]

This question already has an answer here: Polynomial $P(a)=b,P(b)=c,P(c)=a$ 4 answers

I read in a book (A Synopsis of Elementary Results in Pure and Applied Mathematics) that the condition to simplify the expression $\sqrt[3]{a+\sqrt{b}}$ is that $a^2-b$ must be a perfect cube. For example $\sqrt[3]{10+6\sqrt{3}}$ where $a^2-b =(10)^2-(6 \sqrt{3})^2=100-108=-8$ and $\sqrt[3]{-8} = -2$ So the condition is satisfied and $\sqrt[3]{\sqrt{3}+1}^3=\sqrt{3}+1$. But the example $\sqrt[3]{11+\sqrt{57}}$ where $a^2-b […]

Suppose we have: $ \frac{f(x)}{g(x)h(x)} $ and we want to break it down into; $ \frac{I(x)}{g(x)} + \frac{J(x)}{h(x)}$ and that; $deg(f) \leq deg(g)+deg(h)$ , $deg(i) < deg(g)$, $deg(j) < deg(h)$ What is the general way of doing this? I don’t understand the intuition behind having to express; $\frac{ax^{2}+bx+c }{(dx+e)(f x^{2}+g) } = \frac{A}{dx+e} + \frac{Bx+C}{fx […]

I have not been able to find a proof that the following definitions are equivalent anywhere, thought maybe someone could give me an idea: A parabola is defined geometrically as the intersection of a cone and a plane passing under the vertex of a cone that does not form a closed loop and is defined […]

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