Componendo et Dividendo

This week, we learn about the method of Componendo et Dividendo.

How would you use Componendo et Dividendo to solve the following?

If nn is an integer, how many complex solutions are there to xn+1x+1=xn1x1? \frac{ x^n+1} { x+1 } = \frac{ x^n -1 } { x-1}?

Note by Calvin Lin
6 years, 11 months ago

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Assuming that n2n \ge 2, we manipulate the equation to obtain xn+1xn1  =  x+1x1 \frac{x^n+1}{x^n-1} \;=\; \frac{x+1}{x-1} Thus, using Componendo et Dividendo with k=1k=1 we deduce that xn  =  2xn2  =  (xn+1)+(xn1)(xn+1)(xn1)  =  (x+1)+(x1)(x+1)(x1)  =  2x2  =  x x^n \; =\; \frac{2x^n}{2} \; =\; \frac{(x^n+1)+(x^n-1)}{(x^n+1)-(x^n-1)} \; = \; \frac{(x+1)+(x-1)}{(x+1)-(x-1)} \; = \; \frac{2x}{2} \; = \; x and hence x(xn11)=0x(x^{n-1}-1)=0. We need to exclude both x=1x=1 and x=1x=-1 as solutions. Thus

  • There are n1n-1 roots when nn is even, namely 00 and the n2n-2 complex (n1)(n-1)st roots of unity (11 omitted),

  • There are n2n-2 roots when nn is odd, namely 00 and the n3n-3 complex (n1)(n-1)st roots of unity (11 and 1-1 omitted).

If n=1n=1 then there are infinitely many roots.

If n=0n=0 then there are no solutions.

If n<0n < 0 then the same algebraic steps as above hold, but we have to exclude 00 as a root, so the roots are the n1|n-1|th roots of unity, with 11 and 1-1 excluded as necessary.

Mark Hennings - 6 years, 11 months ago

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You did a great job with the cases. This question was tricky, in part because you have to be actively aware of the different conditions that were necessary for the arguments to hold.

It also shows why after manipulating an equation, you must always check that the values hold in the original equation. This is a step that is often forgotten, with students claiming that "a degree nn polynomial must have nn roots."

With n<0 n < 0 , most will also forget that they have to exclude 0 as a root. In this case, when n3 n\leq 3 is an odd integer, then there are n3 |n| - 3 roots. It is slightly interesting that for n=3n=-3, we have no solutions, because all the possible solutions have to be rejected.

Calvin Lin Staff - 6 years, 11 months ago

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If n=0n=0 then how can there be no solutions?

Aditya Parson - 6 years, 11 months ago

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also if n=0, x^n-1 = 0. (x^n + 1)/0 = math error kkk

Nicholas Rincon Reis - 6 years, 11 months ago

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@Nicholas Rincon Reis Yes, in the adjusted equation. However, when n=0n=0, the first manipulation is not valid, and the arguments about n=0n=0 are applied to the original equation. See Daniel C's comment.

Mark Hennings - 6 years, 11 months ago

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If n=0n=0, we get 2x+1=0\dfrac{2}{x+1}=0, which is not possible.

Daniel Chiu - 6 years, 11 months ago

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@Daniel Chiu Oh! yes, I forgot that there is no power of nn in the denominator. Funny that.

Aditya Parson - 6 years, 11 months ago

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xn+1xn1=x+1x1 \frac{x^n+1}{x^n-1}=\frac{x+1}{x-1} Note that x1,1x \neq 1, -1 .

Apply dividendo to get:

2xn1=2x1 \frac{2}{x^n - 1} = \frac{2}{x-1}


Suppose if nn is an even positive integer then we know that the polynomial[xn1)x^n-1)] has nn roots [not necessarily distinct]. Out of these nn roots, two are 1,11, -1, so there are a total of n2+1=n1n-2+1=n-1 solutions when nn is an even positive integer. When nn is odd, there n1n-1[except for n=1n=1] solutions.

Note: It still needs to be shown that xx must be distinct for all those solutions.

Note that I have ignored distinct solutions for xx.

If n=0n=0 there are no solutions to xx.

If nn is an odd positive integer, then there will be n1+1=nn-1+1=n solutions to the equation, since 1-1 wont be a root in this case.

If nn is a non-positive integer, we let n=kn=-k for some positive integer kk and we get:

xk+11=0x^{k+1} - 1 =0

If kk is odd we have k1k-1[ 1,11,-1 are not solutions] solutions, and if kk is even there are kk[k+1k+1 roots out of which 1-1 is not a solution] solutions.

Aditya Parson - 6 years, 11 months ago

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Indeed. This is an example of how manipulating the equation, leads to a different set of solutions. For example, both of you dealt with

xn+1xn1=x+1x1 \frac{x^n + 1}{ x^n -1 } = \frac{x+1}{x-1}

When nn is odd, we have x=1 x = -1 as a solution. However, this case has to be rejected in my version, and it is good that you mentioned "Note that x1,1 x \neq 1, -1 .

However, you should also have added the condition that xn10 x^n -1 \neq 0 , in order to allow you to divide by the term. It is a quick check that if xn=1 x^n = 1 , then there are no solutions to the equation, since we must have

2x+1=0 \frac{ 2 }{x+1} = 0

Calvin Lin Staff - 6 years, 11 months ago

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1. Componendo:a+bb=cdd. \text{1. Componendo:} \quad \quad \quad \quad \quad \frac{a+b}{b} = \frac{c-d}{d}.

Tim Vermeulen - 6 years, 11 months ago

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