Looks complex, right?
p , q , r , s are the 4 roots of the equation
x 4 − a x 3 + a x 2 + b x + c = 0 .
What is the minimum possible value of p 2 + q 2 + r 2 + s 2 ?
Note:
- p , q , r , s are not necessarily real.
- a , b , c are real coefficients.
The answer is -1.
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4 solutions
You have shown that − 1 is a lower bound. However, in order for it to truly be a minimum, we have to show that it can be achieved. Does such p , q , r , s exist for a = 1 ?
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I don't think there is a need of prooving the existance. If a = 1 , any values of b or c doesn't matter, then p 2 + q 2 + r 2 + s 2 = − 1
Or we can just put a = 1 , b = 0 , c = 0 and get the equation x 4 − x 3 + x 2 = 0 to make sure.
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For completeness, in the example you suggest, the roots are 0 , 0 , ( 1 + 3 ) / 2 , ( 1 − 3 ) / 2 giving a sum of squares of 0 + 0 + ( − 1 + 3 ) / 2 + ( − 1 − 3 ) / 2 = − 1 as required
Another good example is a = 1 , b = − 1 , and c = 0 . In that case, the polynomial factors nicely. 0 = x 4 − x 3 + x 2 − x = x ( x 3 − x 2 + x − 1 ) = x ( x − 1 ) ( x 2 + 1 ) = x ( x − 1 ) ( x − i ) ( x + i )
So the four roots are 0 , 1 , i , − i and the sum of their squares is 0 + 1 − 1 − 1 = − 1 .
I think you could reason that by the fundamental theorem of algebra, the four roots exist for a = 1 as the degree of the polynomial is 4.
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Right, that's the missing step/argument in showing that the lower bound can be achieved.
I wasn't saying that it is "hard to do". I was just saying that "it has to be stated".
There is nothing left to prove: 4 roots always exist and it was proven the requested sum equals − 1 when a = 1
Why is the sum of the roots equal to a?
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Hint: ( x − p ) ( x − q ) ( x − r ) ( x − s ) = 0 = x 4 − a x 3 + a x 2 + b x + c
Please show how to solve using Arithmetic mean >= Geometric mean
How is a sum of squares negative???
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Aaaaah complex numbers....
But it's weird to introduce them in a problem of minimum value (since "<" is undefined)
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It can be shown that p 2 + q 2 + r 2 + s 2 is always real (under the conditions of the problem), so we can talk about the minimum of this value, even though p , q , r , s might be complex.
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@Calvin Lin – And also because a^2-2a is a real function. So it can have a minima .
@Calvin Lin – The sum of the squares is a ( a − 2 ) and a is real: the proof is all there
It worth noting that the last identity also shows that the sum of squares is always real, even if the roots of the polynomial are allowed to be complex, and therefore the problem of minimizing it makes sense. A priori a function of complex number is complex as well, so minimizing it is ill-defined.
I don't see how the phrase "minimum possible value" precludes absolute value. Are absolute values impossible?
You have "ps" instead of "rs" in both sums
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I had posted this for almost 5 months, and finally someone found this mistake! Thanks for telling me this error.
You can use Newton's remarkable identities to show that the sum of the squares of the roots of this polynomial equation is
S = a 2 − 2 a
Completing the square we can write this as
S = a ( a − 2 )
S is a quadratic in a with zeros at a = 0 and a = 2. The minimum occurs half way between these roots, when a = 1. This gives the minimum value of S as 1 ( 1 − 2 ) = − 1
I like the way the values of b and c are not relevant and do not need to feature in the solution!
Assume polynomial =(x-p)(x-q)(x-r)(x-s) and knowing for quartic polynomial that the sum of square of roots are given by (S1)^2-(2*S2), where S1 and S2 are coefficient of x^3 and x^2, respectively.
Therefore, for given polynomial, we write:
p^2+q^2+r^2+s^2 = a^2-2a,
Minimizing RHS, get a=1 and minimum value of RHS = -1
Answer= -1
Couldn't a, b and c all be equal to zero making all The roots zero aswell?
Vieta's formulas tell us that p + q + r + s = − ( − a ) = a a n d p q + p r + p s + q r + q s + r s = a . Newton's identities tell us that p 2 + q 2 + r 2 + s 2 = ( p + q + r + s ) ( p + q + r + s ) − 2 ( p q + p r + p s + q r + q s + r s ) = a 2 − 2 a . Trivial inequality tells us that a 2 − 2 a = a 2 − 2 a + 1 − 1 = ( a − 1 ) 2 − 1 ≥ − 1 , with equality at a = 1 .
If a = 1 with b and c arbitrary, the fundamental theorem of algebra tells us that p , q , r , s exist.
Hence, the minimum value of p 2 + q 2 + r 2 + s 2 is − 1 .
P 1 = a − b = a
⇒ P 2 = a ⋅ P 1 − 2 ⋅ a c = a 2 − 2 a
Minimum occurs at P 2 ’ = 0
⇒ 2 a − 2 = 0 ⇒ a = 1
Substituting a = 1 back into P 2 gives P 2 = − 1
⇒ minimum of p 2 + q 2 + r 2 + s 2 = − 1
The equation x^4 -x^3+x^2-x=0 has 4 roots 0, 1, i, -i and satisfies the required conditions.
p + q + r + s = a
p q + p r + p s + q r + q s + r s = a
p 2 + q 2 + r 2 + s 2 = ( p + q + r + s ) 2 − 2 ( p q + p r + p s + q r + q s + r s ) = a 2 − 2 a = ( a − 1 ) 2 − 1
Hence the minimum is − 1 (happens when a = 1 )