Manipulating Polynomial Derivatives

Calculus Level 5

A monic polynomial $f(x)$ with integer coefficients has the property $f(x)\cdot f''(x) -\left(f' \right)^2 = -6x^{10}+600x$ Find $f(2).$

The answer is 154.

1 solution

Pi Han Goh
Feb 14, 2014

Let the degree of $f$ be $n$ , where $n$ is a positive integer

The leading term of $f(x)$ is $x^n$ because it's a monic polynomial

The leading term of $f'(x)$ is $n x^{n-1}$

The leading term of $f''(x)$ is $n(n-1) x^{n-2}$

Given $f(x) \cdot f''(x) - \left ( f'(x) \right )^2 = -6x^{10} + 600x$ , we just take the leading terms to determine the value of $n$

$\left ( x^n + O \left ( x^{n-1} \right ) \right ) \cdot \left ( n(n-1) x^{n-2} + O \left ( x^{n-3} \right ) \right ) - \left ( n x^{n-1} + O \left ( x^{n-2} \right ) \right )^2 = -6x^{10} + 600x$

Because $\text{Deg(LHS)} = \text{Deg(RHS) }$

$n (n-1) x^{2n-2} - n^2 x^{2n-2} = -6x^{10} \Rightarrow n = 6$ , thus

$f(x) = x^6 + ax^5 + bx^4 + cx^3 + dx^2 + ex + f$ for integers $a,b,c,d,e,f$

$\Rightarrow f'(x) = 6x^5+5ax^4 + 4bx^3+3cx^2+2dx+e$

$\Rightarrow f''(x) = 30x^4 + 20ax^3 + 12bx^2 + 6cx + 2d$

With the given equation, we multiply them together as such, simplify and compare the different powers of $x$

$\begin{aligned} x^{10} & : & 30a - 36 a = 0 \Rightarrow a = 0 \\ x^9 & : & 0 = 0 \Rightarrow \text{ Useless} \\ x^8 & : & 30b + 12b - 2 \cdot 6 \cdot 4b = 0 \Rightarrow b = 0 \\ x^7 & : & 30c + 6c - 2 \cdot 6 \cdot 3 a = 0 \Rightarrow \text{ Useless} \\ x^6 & : & 2d + 30d - 24d = 0 \Rightarrow d = 0 \\ x^5 & : & 30e - 12e = 0 \Rightarrow e = 0 \\ x^4 & : & 6c^2+30f-9c^2 = 0 \Rightarrow c^2 = 10f \\ x^3 & : & 0 = 0 \Rightarrow \text{ Useless} \\ x^2 & : & 0 = 0 \Rightarrow \text{ Useless} \\ x^1 & : & 6cf = 600 \Rightarrow cf = 100 \\ \end{aligned}$

Because $c^2 = 10f \Rightarrow c^3 = 10cf = 1000 = 10^3 \Rightarrow c = 10$ only because $c$ is a real number, then $f = 100/c = 10$

Therefore, $f(x) = x^6 + 10x^3 + 10 \Rightarrow f(2) = 2^6 + 10 \cdot 2^3 + 10 = \boxed{154 }$

I was hoping for a simpler solution :P

Mridul Sachdeva - 7 years, 3 months ago

$\text{Let the degree of } f(x) \text{be } n \text{, where }n \text{ is a positive integer.}$

$\text{Since it's a monic polynomial, therefore }f(x)=\sum_{i=0}^n a_{i}x^i$ $\text{Therefore }$ $f'(x)=\sum_{i=0}^{n-1} (i+1)a_{i+1}x^i \text{and }f''(x)=\sum_{i=0}^{n-2} (i+1)(i+2)a_{i+2}x^i$

$O(f(x)f''(x))=O((f'(x))^2)=2n-2=10\text{; n=6 and, }$ $30a_{6}a_{6}-(6a_{6})^2 =-6 => a_{6}=1$ Therefore $f(x)=x^6+\sum_{i=0}^5 a_{i}x^i$ $f'(x)=6x^5+\sum_{i=0}^{4} (i+1)a_{i+1}x^i$ $f''(x)=30x^4+\sum_{i=0}^{3} (i+2)(i+1)a_{i+2}x^i$ $6a_{0}a_{3}+2a_{1}a_{2}-4a_{2}a_{1}=600 \text{ or, } 3a_{0}a_{3}-2a_{2}a_{1}=300$ $\text{And all others are zero.}$ $\text{So, it is safe to assume } a_{4}=a_{5}=0$ $\text{Now, for the coefficient of } x^{2} \text{ we have }$ $2(a_{2})^2+6a_{3}a_{1}-6a_{3}a_{1}-(2a_{2})^2=0 \text{ or, } a_{2}=0$ $\text{But, for constant term we have } 2a_{2}a_{0}-a_{1}^2=0 \text{ or, } a_{1}=0$ $\text{Therefore, }a_{0}a_{3}=100 \text{ and, from the coefficient of } x^{4}$ $\text{ we have } 30a_{0}+6(a_{3})^2-(3a_{3})^2=0 \text{ or, } 10a_{0}=a_{3}^2$ $\text{ or, } a_{0}=a_{3}=10 \text{ Therefore, the function }f(x)=x^6+10x^3+10$ $f(2)=154$

Al Imran - 7 years, 2 months ago

Well, the equations are really taxing to get!

Karttikeya Mangalam - 7 years, 3 months ago

I'll post a similar but an easier method (or atleast one that is less exhausting) in a bit :).

Muhammad Shariq - 7 years, 3 months ago

Please post your solution...

Mridul Sachdeva - 7 years, 3 months ago

dint get...

vinit kumar - 7 years, 3 months ago

My logic was the following :

X^N + a(n-1) x^(N-1) + a(N-2)X^(n-2) + . a(3)x^3. + a(2)x^2 + a(1)x + a0 ..... f'(x) = Nx^(N-1) + a(n-1) X^(N-2) +......... 3a(3)x^2..+2a(2)x + a1 f"(x) = N(N-1)X^n-2 + ............ 6a(3)x + 2a(2)

Now the X^(N-2) X N gives us the x^10 term

Therefore, N-2+N = 10

Therefore, 2N = 12

N = 6 So it is a 6 degree polynomial

One interesting observation is that the term f(x)f"(x) - f(x)^2 is the numerator of the derivative of ln(f(x) , that is (f'(x) / f(x))'. What is interesting about the above expression is the following :

If we let f(x) = (x-a1)^n1*(x-b)^n2 .....

ln(f(x) = n1ln(x-a1) + n2ln(x-a2) + ....

So f'(x) / f'(x) = n1 / (x-a) +...

which allows to determine .the multiplicity etc

So the derivative of the above would be : -n1/(x-a)^2 - n2/(x-b)^2 + ........

which in our case would be .(f(x)f"(x) - f'(x)^2) / f(x)^2

My chain of thought was the following :

The above numerator term (-6x^10 + 600x) does not contain a constant. So i felt that one of the roots would be zero which would make the product zero (otherwise there would be a nonzero constant). I felt that the f(x) term in the denominator would not make much of a difference

[Another observation was that other than 6 , the n1, n2 terms etc do not appear from the x^10 term which can be thought of as the sum of the (x-a2)^2*(x-c)^2 +.... upto a maximum of 5 since there are 6 factors giving us the -6x^10. My analysis was that there are 6 factors , all of multiplicity 1. Looks like , my analysis went wrong somewhere.

Anyway, my further reasoning was as follows : The contribution to the x term is due to the square of the (2a2x + a1) term in f'(x) and the 6a(3) a(0) + a(1) a(2) term in the numerator We also have the constraint that 2a2 = a1^2 since there is no constant in the expression

But somehow, the above leads to non integral values of the coeffs . Maybe , my reasoning was incomplete

I felt that more sophisticated methods would become important in a general case where the problem cannot be solved by trial and error

Sundar R - 7 years, 3 months ago

One could also look at zeros of (ln(f(x))' , (ln(f(x))" , f'(x) = 0, f"(x) = 0, f(x) = 0, f(x) = 1 (when the denominator term vanishes). One could also use methods from complex analysis (especially Cauchy's Residue theorem etc) in a more general situation. The constant term in the numerator above seems to be something of the sort : -(bcdef)^2 - (acdef)^2 ..... = -(a0^2 / a^2 + a0^2/b^2 + .....) where a0 is the last term and denotes the product of the roots and since there is no constant term in the given expression, it should be zero The bracketed expression = a0^2(1/a^2 + 1/b^2 + .....) = a0^2(some term involving a(1), the square of roots , which involves sum of product of roots such as abcde + bcdef + ......6 terms and their squares. My analysis was that since the above term is zero and the term on the right is a sum of squares a0 = 0 and the denominator does not matter in this case because we are looking at zeros

Sundar R - 7 years, 3 months ago

It seems the roots are all complex which possibly explains the confusion

Sundar R - 7 years, 3 months ago

@Sundar R – The above seems to be a quadratic in x^3 which possibly explains the combination of non-multiplicity of the roots and yet satisfying a quadratic with all the associated symmetries

Sundar R - 7 years, 3 months ago

in line 9 and 10........i'm confused how u have done this???? can someone tell me?/

adeel shahbaz - 7 years, 3 months ago

$2n-2 = 10 \Rightarrow n = 6$ , the coefficients also confirms that. So it's a $6^{\text{th}}$ degree monic polynomial with integer coefficients

Pi Han Goh - 7 years, 3 months ago

Whoa!!Looks like a lot of work!!!

Biswadeep Sen - 7 years, 3 months ago

Manipulating Polynomial Derivatives

The answer is 154.

1 solution

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