AP Calculus BC

Tangent Line vs Taylor Polynomial

Q: Do I have to build every lower term to get the degree three polynomial?

Yes, in the sense that P_3 contains every term of P_2, which contains every term of P_1. A term is allowed to be zero, though, so P_3 for \cos x centred at 0 is just 1 - \frac{x^2}{2}, and P_3 for \sin x is x - \frac{x^3}{6}, with nothing wrong with either. What you can often skip is the differentiating: substituting into a known Maclaurin series produces the coefficients directly, so the cubic term of e^{2x} reads off as \frac{(2x)^3}{3!} with no third derivative computed by hand.

The tangent line approximation is the first-degree Taylor polynomial, so these are not rival methods: one is the opening case of the other. Each extra degree adds a term built from a higher derivative, and the first of those corrects for curvature, which is exactly what a straight line cannot do.

Tangent line approximation

Use when: You need a quick estimate at a point close to the centre, and the only accuracy question is whether that estimate is too high or too low.

Taylor polynomial

Use when: The question names a degree above one, or the point sits far enough from the centre that a line is too crude, or an error bound is required.

Side by side

	Tangent line	Taylor polynomial
Degree	Always $1$	Whatever degree $n$ the question names
Formula	$L(x) = f(a) + f'(a)(x-a)$	$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$
Information needed	$f(a)$ and $f'(a)$ only	$f(a)$ through the $n$ th derivative $f^{(n)}(a)$
Curvature	Ignored, so the estimate stays on one side of a concave graph	Corrected from the $(x-a)^2$ term onward
Error control	The sign of $f''$ decides over or under	The Lagrange error bound $\lvert R_n(x) \rvert \le \frac{\max_{[a,x]} \lvert f^{(n+1)} \rvert}{(n+1)!}\lvert x-a \rvert^{n+1}$ , where the numerator is the maximum of $\lvert f^{(n+1)} \rvert$ on the interval between $a$ and $x$

Write the Taylor polynomial of degree $n$ centred at $a$ , then set $n = 1$ . What survives is $f(a) + f'(a)(x-a)$ , which is the equation of the tangent line at $x = a$ . Linearisation is therefore not a separate technique. It is the Taylor polynomial you stopped building one term past the constant.

P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k \qquad\Longrightarrow\qquad P_1(x) = f(a) + f'(a)(x-a) = L(x)

Take $f(x) = e^x$ centred at $a = 0$ , where every derivative equals $1$ . The tangent line gives $L(0.3) = 1.3$ . The quadratic term raises that to $P_2(0.3) = 1.345$ , and the cubic term to $P_3(0.3) = 1.3495$ , against a true value of $1.34986$ to five decimal places. The line sits below the curve because $e^x$ is concave up, and each new term closes part of the remaining gap.

The factorial that hides in the linear case

Because $0! = 1$ and $1! = 1$ , no factorial is visible in the tangent line formula. Students who meet Taylor polynomials as an unrelated topic then write the next term as $f''(a)(x-a)^2$ and report double the correct contribution. Every term past the linear one carries its own $k!$ , starting with the $2!$ in $\frac{f''(a)}{2!}(x-a)^2$ .

Frequently asked questions

Is linearization the same thing as a first-degree Taylor polynomial?

Yes, they are the same object under two names. Linearisation is the language of Unit 4, Taylor polynomials the language of Unit 10, and $P_1(x)$ is the tangent line at the centre in both.

How do I know whether my estimate is too big or too small?

For the tangent line, read the concavity: a concave up graph lies above its tangent, so the line underestimates, and a concave down graph gives an overestimate. For higher degrees, concavity is no longer enough, and you use the Lagrange error bound or the alternating series bound when the terms alternate.

Do I have to build every lower term to get the degree three polynomial?

Yes, in the sense that $P_3$ contains every term of $P_2$ , which contains every term of $P_1$ . A term is allowed to be zero, though, so $P_3$ for $\cos x$ centred at $0$ is just $1 - \frac{x^2}{2}$ , and $P_3$ for $\sin x$ is $x - \frac{x^3}{6}$ , with nothing wrong with either. What you can often skip is the differentiating: substituting into a known Maclaurin series produces the coefficients directly, so the cubic term of $e^{2x}$ reads off as $\frac{(2x)^3}{3!}$ with no third derivative computed by hand.

In the CED: Unit 4: Contextual Applications, Unit 10: Infinite Sequences and Series (BC)