Given the function $f(r) = ar^2 + br + c$, we want to show that the first-order Taylor approximation of $f(r)$ at any point $r_0$ is always greater than or always less than $f(r)$ for any value of $r$. Moreover, we want to determine how to decide this by examining the quadratic form of the function.

Step 1: Calculate the First-Order Taylor Approximation

The first-order Taylor approximation of $f(r)$ at $r = r_0$ is given by: $T_1(r) = f(r_0) + f'(r_0)(r - r_0)$ where:

• $f'(r) = \frac{d}{dr} (ar^2 + br + c)$ is the first derivative of $f(r)$.

Step 2: Compute the Derivative $f'(r)$

The first derivative of $f(r)$ with respect to $r$ is: $f'(r) = 2ar + b$

So, the first-order Taylor approximation at $r = r_0$ is: $T_1(r) = f(r_0) + (2ar_0 + b)(r - r_0)$

Step 3: Compare $f(r)$ and $T_1(r)$

The original function $f(r)$ can be rewritten as: $f(r) = ar^2 + br + c$ Now, let’s express the difference between $f(r)$ and $T_1(r)$: $f(r) - T_1(r) = (ar^2 + br + c) - \left[ f(r_0) + (2ar_0 + b)(r - r_0) \right]$

Step 4: Simplify the Difference

First, note that $f(r_0) = ar_0^2 + br_0 + c$. Substituting this into the expression for $T_1(r)$, we get: $f(r) - T_1(r) = ar^2 + br + c - \left[ ar_0^2 + br_0 + c + (2ar_0 + b)(r - r_0) \right]$

Simplify the expression: $f(r) - T_1(r) = ar^2 + br + c - ar_0^2 - br_0 - c - (2ar_0 + b)(r - r_0)$

Further simplification yields: $f(r) - T_1(r) = a(r^2 - r_0^2) + b(r - r_0) - (2ar_0 + b)(r - r_0)$

Notice that $r^2 - r_0^2 = (r - r_0)(r + r_0)$, so we can substitute this: $f(r) - T_1(r) = a(r - r_0)(r + r_0) + b(r - r_0) - (2ar_0 + b)(r - r_0)$

Factoring $(r - r_0)$ out: $f(r) - T_1(r) = (r - r_0)\left[a(r + r_0) + b - 2ar_0 - b \right]$ $f(r) - T_1(r) = (r - r_0)\left[a(r + r_0 - 2r_0) \right]$ $f(r) - T_1(r) = a(r - r_0)(r - r_0)$ $f(r) - T_1(r) = a(r - r_0)^2$

Step 5: Analyze the Sign of $f(r) - T_1(r)$

The expression $f(r) - T_1(r) = a(r - r_0)^2$ is always non-negative since $(r - r_0)^2$ is a square term and therefore always non-negative.

• If $a > 0$, then $f(r) - T_1(r) \geq 0$, meaning $f(r) \geq T_1(r)$. In this case, the Taylor approximation is always less than or equal to the actual function $f(r)$.
• If $a < 0$, then $f(r) - T_1(r) \leq 0$, meaning $f(r) \leq T_1(r)$. In this case, the Taylor approximation is always greater than or equal to the actual function $f(r)$.

Conclusion

The first-order Taylor approximation will be less than the function $f(r)$ for all $r$ if $a > 0$, and it will be greater than the function $f(r)$ for all $r$ if $a < 0$. This conclusion is determined solely by the sign of the quadratic coefficient $a$.

Follow-Up Questions

1. How does the sign of $a$ influence the concavity of the function $f(r)$?
2. What happens to the approximation if $a = 0$?
3. How would the second-order Taylor approximation compare to the function $f(r)$?
4. Can you derive a similar conclusion for a cubic polynomial?
5. How does the location of $r_0$ affect the accuracy of the Taylor approximation?

Tip

Remember, the sign of the leading coefficient in a quadratic function determines whether the parabola opens upwards or downwards, which directly influences whether the Taylor approximation underestimates or overestimates the function.