# Using binomial coefficients to find sum of roots of a polynomial.

Find the sum of the roots, real and non-real, of the equation $$x^{2001}+\left(\frac 12-x\right)^{2001}=0$$, given that there are no multiple roots.

While trying to solve the above problem (AIME 2001, Problem 3) which was asked here on MSE, I came across three solutions on AoPS. ( The MSE solution uses Vieta's formula which I am clear about )

The first solution (on AoPS) involves the use of Vieta's formula's and is quite clear.

The third solution states the following :

Note that if $$r$$ is a root, then $$\frac{1}{2}-r$$ is a root and they sum up to $$\frac{1}{2}.$$

We make the substitution $$y=x-\frac{1}{4}$$ so $$(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.$$

Expanding gives $$2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots$$ so by Vieta, the sum of the roots of $$y$$ is 0.

Since $$x$$ has a degree of 2000, then $$x$$ has 2000 roots so the sum of the roots is $$2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.$$

I do not understand two things in the above solution :

1. "Note that if $$r$$ is a root, then $$\frac{1}{2}-r$$ is a root and they sum up to $$\frac{1}{2}.$$"

a) Here what is being referred to as "they"? (Shouldn't $$\frac{1}{2}-r$$ be a factor and not a root). Answered by the third comment

b)Why is the sum 1/2? Also answered by the third comment

c)Why is $$\frac{1}{2}-r$$ a root? Answered by the sixth comment

1. How is the final expression arrived upon to find the sum of all 2000 roots? (Answered by @YvesDaoust)

The second solution is more mystifying (possibly because it is similar to the one above):

We find that the given equation has a $$2000^{\text{th}}$$ degree polynomial. Note that there are no multiple roots. Thus, if $$\frac{1}{2} - x$$ is a root, $$x$$ is also a root. Thus, we pair up $$1000$$ pairs of roots that sum to $$\frac{1}{2}$$ to get a sum of $$\boxed{500}$$.

1. Again, why is $$\frac{1}{2} - x$$ a root. By "$$x$$ is also a root" does it mean $$x$$ representing the set of all roots? Answered by the sixth comment

2. Why does the pairing up occur? Why is the sum of each pair 1/2? (Answered by @YvesDaoust)

I wonder if it could be solved as follows :

Let the roots be $$P_1,P_2,...P_{2000}$$. The polynomial can be expressed as a product of factors as follows : $$(2001/2)($$x$$-P_1)($$x$$-P_2)....($$x$$-P_{2000}) = 0$$. The above expression is the same as $$x^{2001}+\left(\frac 12-x\right)^{2001}=0$$.

Thus, $$x^{2001}+\left(\frac 12-x\right)^{2001}$$ = $$(2001/2)($$x$$-P_1)($$x$$-P_2)....($$x$$-P_{2000})$$

Here the coefficient of $$x^{1999}$$ on the $$RHS$$ should represent $$\sum\limits_{i=1}^{2000}P_i\times(-1/2)$$

On the $$LHS$$ the corresponding term would be the term with $$x^{1999}$$ and thus the coefficient of this term on the $$LHS$$ should also be the required sum.

On the LHS the coefficient of the $$x^{1999}$$ term is -$${2001}\choose{2}$$*$$(1/2)^2$$ which represent the sum of the roots.

[Picked up this approach here, but I don't see how this would work ] (https://youtu.be/S6FRtmDUl-s?t=2806)

In this solution I find some errors(?) :

1. Are there any inconsistencies in the reasoning?Wouldn't the sum of roots differ from the binomial coefficient since the expression involves unique values of $$P_i$$ (no multiple roots).
2. The answers do not match, which seems to suggest so.
3. Is there a way of arriving at the answer without using Vieta's formula and by expressing the polynomial as a product of factors and then using binomial coefficients as attempted above?
• That's seven questions! Commented Oct 13, 2018 at 5:00
• The first solution you give (which you call the third solution) is essentially the same as the second solution you give (which mercifully you call the second solution). Commented Oct 13, 2018 at 5:01
• In your question 1, are you asking why $r+\left(\frac12-r\right)=\frac12$? Commented Oct 13, 2018 at 5:03
• $\dfrac12-r$ is a root whenever $r$ is. This is because $$P(\dfrac12-r)=(\dfrac12-r)^{2001}-(\dfrac12-(\dfrac12-r))^{2001}= (\dfrac12-r)^{2001}-r^{2001}=-P(r).$$ So if $P(r)=0$ then $P(\dfrac12-r)=0$ also. Commented Oct 13, 2018 at 7:26
• You're right. I made a sign error. Anyway, the argument, as per your modification, survives with the correct sign as well. Commented Oct 20, 2018 at 20:28

## 2 Answers

First solution:

By symmetry of $$x^n+(\frac12-x)^n$$, if $$x$$ is a root, so is $$\frac12-x$$. Now if you take the roots in pairs ($$1000$$ pairs), the sum of the individual pairs is $$x+\frac12-x=\frac12$$. Hence in total $$1000\cdot\dfrac12$$.

Second solution:

There is no difference with the first.

Extra solution:

By Vieta, the sum of the roots is the negative ratio of the two coefficients of the highest degree. Then using the Binomial theorem,

• degree $$2001$$: $$1-1=0$$,

• degree $$2000$$: $$\dfrac12\dfrac{2001}2\left(-\dfrac12\right)$$,

• degree $$1999$$: $$\dfrac{2001}2\left(\dfrac{2001-1}2\right)\left(-\dfrac12\right)^2$$.

The requested ratio is indeed $$500$$.

• Yes, that answers question 2 and 4! I think the extra solution is the same as the first solution on \$[AoPS](artofproblemsolving.com/wiki/… whose working I did not add. @YvesDaoust Commented Oct 13, 2018 at 9:47

In answer to questions 5,6,7 (the attempted solution) :

1. It was explained to me elsewhere that the leading coefficient on the $$RHS$$ should not be $$(1/2)$$ but in fact be $$(2001/2)$$ since the leading coefficient for the polynomial of order 2000 would be $${2001}\choose{2000}\times(1/2)$$.

2. Thus on solving for the sum of the roots the answer would be $$500$$.

3. It was also pointed out to me that this would essentially be the same as using Vieta's.

(Apologies for cross posting, I just saw a meta question regarding this and realized my error. I intend to follow the guidelines henceforth).