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complex number

DrunkenOldFool

New member
Feb 6, 2012
20
Let $f(x)=x^{13}+2x^{12}+3x^{11}+\cdots +13x+14$ and $\alpha=\cos\frac{2\pi}{15}+i\sin\frac{2\pi}{15}$. Find the value of $f(\alpha)f(\alpha ^2) \cdots f(\alpha ^{14})$.

(the answer given in my book is 15^(13)).
Please help me! I beg of you...
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Let $f(x)=x^{13}+2x^{12}+3x^{11}+\cdots +13x+14$ and $\alpha=\cos\frac{2\pi}{15}+i\sin\frac{2\pi}{15}$. Find the value of $f(\alpha)f(\alpha ^2) \cdots f(\alpha ^{14})$.

(the answer given in my book is 15^(13)).
Please help me! I beg of you...
Firstly, the fourteen numbers in the set $A \overset{\text{def}}{=}\{ \alpha,\alpha ^2, \ldots, \alpha ^{14}\}$ are all roots of the 15th degree equation $x^{15} = 1$, the fifteenth root being $x=1.$

The next step is to find a more convenient form for the function $f(x)$. Start with the fact that $1+z+z^2 + \ldots + z^{14} = \dfrac{1-z^{15}}{1-z}.$ Differentiate, to get $$1+2z+3z^2+\ldots+14z^{13} = \frac{-15z^{14}(1-z)+(1-z^{15})}{(1-z)^2} = \frac{1-15z^{14}+14z^{15}}{(1-z)^2}.$$ Now divide through by $z^{13}$ and let $x = 1/z$, to get $$f(x) = x^{13}+2x^{12}+3x^{11}+\ldots +13x+14 = \frac{x^{15}-15x+14}{(1-x)^2}.$$ If $x^{15}=1$ (but $x\ne1$) then that becomes $f(x) = \dfrac{15(1-x)}{(1-x)^2} = \dfrac{15}{1-x}.$

So we are looking for the product $$\prod_{x\in A}f(x) = \prod_{x\in A}\frac{15}{1-x} = \frac{15^{14}}{\prod_{x\in A}(1-x)}.$$ For $x\in A$, let $w=1-x.$ Then $(1-w)^{15} = x^{15}=1.$ But $(1-w)^{15} = 1-15w+\ldots -w^{15}$, and the equation $(1-w)^{15} =1$ becomes $w^{15} - \ldots +15w = 0$. Divide by $w$ to eliminate the root $w=0$ (which corresponds to $x=1$) and we get $w^{14} - \ldots + 15=0.$ The product of the roots of this equation is 15, in other words $\prod_{x\in A}(1-x)=15.$ Combine that with the previous result to get $\prod_{x\in A}f(x) = 15^{13}.$
 

DrunkenOldFool

New member
Feb 6, 2012
20
Thank you very much, sir!
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
To Opalg: (Bow)

-Dan