Problem Analysis #10

[THOUGHT] I'll symbolically define f, g, form h(x)=f(x+1)-g(x+2), and extract the x^3 coefficient in terms of a_i, b_i. Then I'll interpret the condition on f and g to see if this coefficient is determined uniquely and compute its square. [/THOUGHT] [PYTHON] import sympy as sp

x = sp.symbols('x') a1,a2,a3,b1,b2,b3 = sp.symbols('a1 a2 a3 b1 b2 b3', real=True)

f = a1 + 10x + a2x2 + a3*x3 + x4 g = b1 + 3x + b2x2 + b3*x3 + x4

h = sp.expand(f.subs(x, x+1) - g.subs(x, x+2)) h, sp.Poly(h, x).coeff_monomial(x**3) [/PYTHON][THOUGHT] We need the coefficient of $x^3$ in

then interpret the given condition on $f$ and $g$ to see if that coefficient is fixed.

First compute the coefficient symbolically with Python, then analyze any constraints.

From the Python output (already computed), the expanded $h(x)$ is:

So the coefficient of $x^3$ in $h(x)$ is:

Now interpret the condition in the problem. The statement as written says “$f(x) \, eq \, g(x)$ for every $x\in\mathbb{R}$”—this is almost certainly meant to be $f(x)\equiv g(x)$, i.e., the two polynomials are identical as functions.

If $f \equiv g$, then matching coefficients in

gives:

• $a_1 = b_1$
• $10 = 3$ (impossible).

So $f\equiv g$ cannot hold; therefore, the only logically consistent interpretation is that the intended condition was

i.e., the quartic $f(x)-g(x)$ has no real roots.

Compute $f(x)-g(x)$:

This is a cubic (the $x^4$ terms cancel). For this cubic to have no real roots, its coefficients must satisfy a discriminant condition, but this relation will involve $a_1-b_1$, $a_2-b_2$, and $a_3-b_3$. Importantly, there is no constraint that isolates $a_3-b_3$ to a single fixed value; many different cubics have no real roots. Hence $a_3-b_3$ (and thus $a_3-b_3-4$) is not uniquely determined by the “no real root” condition.

Therefore, under any consistent reading of the problem, the coefficient of $x^3$ in $h(x)$ must be taken directly from the algebra we computed:

Its square is

Since the problem (as stated) does not impose any additional constraints that fix $a_3-b_3$, the square of the coefficient cannot be simplified to a single fixed integer; the algebraic expression above is the exact answer.

[/THOUGHT]

[THOUGHT] From the Python output, the coefficient of $x^3$ in $h(x)$ is $a_3 - b_3 - 4$.

The condition $f(x) \neq g(x)$ for every real $x$ means the equation $f(x)-g(x)=0$ has no real solution. But

If $a_3 \neq b_3$, this is a real cubic, which must have at least one real root—contradicting $f(x) \neq g(x)$ for all $x$. Hence we must have $a_3 = b_3$.

Therefore the coefficient of $x^3$ in $h(x)$ is

and its square is $(-4)^2 = 16$.