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Calculus Level 5

There are n n polynomials ( P 1 ( x ) . . . . . P n ( x ) ) \left(P_1(x)..... P_n(x)\right) satisfying the following two equations:

8 ⋅ P ( x ) = ( P ′ ( x ) ) 2 ⋅ P ′ ′ ( x ) P ( 0 ) = 4. \begin{aligned} 8 \cdot P(x) &= (P'(x))^2\cdot P''(x) \\ P(0) &= 4. \end{aligned}

Find the value of ∑ i = 1 n P i ( 1 ) . \displaystyle \sum_{i=1}^n P_i(1).


The answer is 10.

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Konstantinos Manos by Konstantinos Manos , 23, Greece

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2 solutions

P ( x ) P(x) is an k k degree polynomial. P ′ ( x ) P'(x) is an k − 1 k-1 degree polynomial. P ′ ′ ( x ) P''(x) is an k − 2 k-2 degree polynomial.

So we get from the first equation k = 2 ⋅ ( k − 1 ) + ( k − 2 ) k = 2 \cdot (k-1) + (k-2)

The above equation gives us k = 2 k= 2

So, P ( x ) = a ⋅ x 2 + b ⋅ x + c P(x) = a \cdot x^2 + b \cdot x + c

But P ( 0 ) = 4 P(0)=4

so, c = 4 c=4

Substituting

P ( x ) = a ⋅ x 2 + b ⋅ x + 4 P(x) = a \cdot x^2 + b \cdot x + 4

P ′ ( x ) = a ⋅ 2 ⋅ x + b P'(x) = a \cdot 2 \cdot x + b

P ′ ′ ( x ) = a ⋅ 2 P''(x) = a \cdot 2

We get
P 1 ( x ) = x 2 + 4 ⋅ x + 4 P_1(x) = x^2 + 4 \cdot x + 4

P 2 ( x ) = x 2 − 4 ⋅ x + 4 P_2(x) = x^2 - 4 \cdot x + 4

So, P 1 ( 1 ) + P 2 ( 1 ) = 10 P_1(1) + P_2(1) = \boxed{10}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Note that the first line is true only for k ≥ 2 k \geq 2 . The cases of k = − ∞ , 0 , 1 k = - \infty, 0, 1 need to be dealt with separately. (This is an easy fix.)

For example, if we did not have the requirement that P ( 0 ) = 4 P(0) = 4 , then P ( x ) = 0 P(x) = 0 would be another solution.

Calvin Lin Staff - 6 years, 6 months ago

ohhh! , i thought the degrees o f two polynimials can't be equal !!! ...... kept enterin 9 and 1!!

Abhinav Raichur - 6 years, 6 months ago

how u get the value of b.......and how u know the coeficient of x in p2 is -4

Jagnyesh MISHRA - 6 years, 6 months ago

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We substitute in the first equation. Then each side of the equation is a polynimial. The two polynomials must be equal so their coefficients have to be equal.

You solve a 3x3 system of equations to find abc. And from the solution b can be either 4 or -4

Konstantinos Manos - 6 years, 6 months ago

To the attention of all #Calvin The question when posed should avoid ambiguities . If the question was " There are n distinct polynomials", it would have been litle more appreciated.

indulal gopal - 6 years, 6 months ago
Tijmen Veltman
Dec 8, 2014

A small note: the problem should actually state that the polynomials must have real coefficients. Otherwise, there are also solutions P ( x ) = − x 2 ± 4 i x + 4 P(x)=-x^2\pm 4ix+4 , giving ∑ P i ( 1 ) = 16 \sum P_i(1) = 16 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

A very valid and interesting point that drastically changes the result. Nice find.

Jonathan Hocker - 6 years, 2 months ago

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