Stop being so mean 2
What is the harmonic mean of the roots of the following polynomial?
2 0 1 9 x 2 0 1 8 − 2 0 1 8 x 2 0 1 7 + 2 0 1 7 x 2 0 1 6 − 2 0 1 6 x 2 0 1 5 + . . . + 3 x 2 − 2 x + 1 = 0
Details:
- Positive, negative and complex roots are all included
- If there are any repeated roots, these should be included however many times they appear
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2 solutions
Should this technically be --2?
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I don't know what you mean.
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The bit in red in your solution, about finding the sum of the products involving 2017 of the roots. This is equal to the negative coefficient of x, over the leading coefficient. Making it negative negative 2 which is 2.
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@Stephen Mellor – Yes, you are right. I will put it in.
The harmonic mean H of the 2 0 1 8 roots a 1 , a 2 , … , a 2 0 1 8 is given by H 2 0 1 8 = n = 1 ∑ 2 0 1 8 a n 1
(Note we could just rewrite that sum as a ratio of two elementary symmetric polynomials and use Vieta's formulas right now, but I prefer to use the following method)
Now, since a n , n = 1 , 2 , … , 2 0 1 8 are the roots of P ( x ) = 2 0 1 9 x 2 0 1 8 − 2 0 1 8 x 2 0 1 7 + 2 0 1 7 x 2 0 1 6 − 2 0 1 6 x 2 0 1 5 + ⋯ + 3 x 2 − 2 x + 1 = 0 we can see that b n = a n 1 , n = 1 , 2 , … , 2 0 1 8 are the roots of P ( x 1 ) = 2 0 1 9 ( x 1 ) 2 0 1 8 − 2 0 1 8 ( x 1 ) 2 0 1 7 + 2 0 1 7 ( x 1 ) 2 0 1 6 − 2 0 1 6 ( x 1 ) 2 0 1 5 + ⋯ + 3 ( x 1 ) 2 − 2 ( x 1 ) + 1 = 0 ⟺ Q ( x ) = x 2 0 1 8 P ( x 1 ) = 2 0 1 9 − 2 0 1 8 x + 2 0 1 7 x 2 − 2 0 1 6 x 3 + ⋯ + 3 x 2 0 1 6 − 2 x 2 0 1 7 + x 2 0 1 8 = 0
That is, we have H 1 = 2 0 1 8 1 ⋅ n = 1 ∑ 2 0 1 8 b n which is the arithemetic mean of the roots of the polynomial Q ( x ) . By Vieta's formulas, the sum of the roots is − coefficient of x 2 0 1 8 in Q ( x ) coefficient of x 2 0 1 7 in Q ( x ) = − 1 ( − 2 ) = 2 , so the arithmetic mean of the roots is 2 0 1 8 2 = 1 0 0 9 1 .
Then solving H 1 = 1 0 0 9 1 ⟹ H = 1 0 0 9
Yes, I suppose so, although I did it a different way (which is similar because you have essentially reversed it and then reversed it again). By algebraically manipulating the definition of the harmonic mean for 2018 values, we get
HM = 2 0 1 8 × a 1 a 2 a 3 . . . a 2 0 1 8 a 1 a 2 a 3 . . . a 2 0 1 7 + a 1 a 2 a 3 . . . a 2 0 1 6 a 2 0 1 8 + . . .
Essentially, this fraction is the sum of all possible products involving 2017 of the roots, divided by the product involving all 2018 of the roots. We know that this is a − x and a y where x and y are the last two coefficients, which then gets the same answer. Do you understand my not so good explanation?
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Yes. That's exactly the method I referenced with my note (and the one done by Chew-Seong Cheong in his posted solution).
Relevant wiki: Vieta's Formula - Higher Degrees
Let the 2018 roots of the equation be a 1 , a 2 , a 3 , ⋯ a 2 0 1 8 . We need to find the harmonic mean of the roots or:
\large \begin{aligned} \mu_h & = \frac {2018}{\frac 1{a_1} +\frac 1{a_2} +\frac 1{a_3} + \cdots + \frac 1{a_{2018}}} \\ & = \frac {2018}{\frac {{\color{#D61F06}\cancel{a_1}}a_2a_3\cdots a_{2018}+ a_1 {\color{#D61F06}\cancel{a_2}}a_3\cdots a_{2018}+ a_1a_2{\color{#D61F06}\cancel{a_3}}\cdots a_{2018}+\cdots + a_1 a_2a_3\cdots {\color{#D61F06}\cancel{a_{2018}}}}{a_1 a_2 a_3 \cdots a_{2018}}} \\ & = \frac {2018}{\frac {\color{#D61F06}\sum_{j=1}^{2018} \prod_{k=1 \\ k \ne j}^{2018} a_k}{\color{#3D99F6} \prod_{k=1}^{2018} a_k}} = \frac {2018}{\frac {\color{#D61F06}-\left(-\frac 2{2019}\right)}{\color{#3D99F6} \frac 1{2019}}} = \frac {2018}2 = \boxed{1009} & \small \color{#3D99F6} \text{By Vieta's formula} \end{aligned}