Calculus 2, Fall 2016

Idea of Taylor polynomials

  1. Polynomials \(a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n\) are nice to evaluate because they rely on addition and multiplication and because we understand them very well. By extension, power series \(a_0 + a_1 x + a_2 x^2 + \cdots\) are also nice, but we need to work to understand them better.

  2. Given any function \(f\), is there a (best) linear approximation to \(f\) at a point \(x=a\)? What about a (best) quadratic, cubic,…, or degree \(n\) polynomial approximation to \(f\) at \(a\)? If we let the degree \(n\) go to infinity, does the power series converge to \(f\) near \(a\)? For what values of \(x\) will the power series converge to \(f\)?

Idea of linear approximation

If \(f\) is differentiable at \(c\), then the linear approximation (i.e., tangent line) to \(f\) at \(c\) is \(p_1(x) = f(c) + f'(c) (x-c)\) and for \(x\) "near" \(c\), \(f(x) \approx p_1(x)\). Notice: \(f\) and \(p_1\) have the same zeroth and first derivative at \(c\): \[f(c) = p_1(c),\] \[f'(c) = p_1'(c).\]

Can we construct a quadratic approximation (i.e., tangent parabola)? What properties should it have?

Best polynomial approximation

The best degree \(n\) polynomial approximation to \(f\) at \(c\) should satisfy \[\begin{array}{rcll} f(c) & = & p(c) & \text{(same value)}, \\ f'(c) & = & p'(c) & \text{(same 1st deriv.)}, \\ f''(c) & = & p''(c) & \text{(same 2nd deriv.)}, \\ & \vdots & \\ f^{(n)}(c) & = & p^{(n)}(c) & \text{(same nth deriv.)}. \end{array}\]

A degree \(n\) polynomial centered at \(c\) has the form \(p(x) = a_0 + a_1 (x-c) + a_2 (x-c)^2 + \cdots + a_n (x-c)^n\). What should the coefficients be for the best approximation?

Taylor polynomials

The coefficients in the best local approximation are \[a_k = \frac{f^{(k)}(c)}{k!},\] and the best degree \(n\) polynomial approximation to \(f\) at \(c\) is the Taylor polynomial \[\begin{array}{rcl} p_n(x) & = & f(c) + f'(c)(x-c) + \frac{f''(c)}{2!} (x-c)^2 \\ && + \frac{f'''(c)}{3!} (x-c)^3 + \cdots + \frac{f^{(n)}(c)}{n!} (x-c)^n \\ & = & \displaystyle \sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!} (x-c)^k. \end{array}\]


Find the Taylor polynomials of degree \(\leq 4\) centered at \(c=1\) that approximate \(f(x) = \ln(x)\). Use the Taylor polynomial of degree \(4\) to find an approximate value for \(\ln(1.5)\). What is the error in your approximation?

Taylor's theorem

Theorem: Let \(f\) be a function with \(n+1\) derivatives on an interval \(I\) containing \(c\). Then, for each point \(x\) in \(I\), there's a point \(z\) between \(x\) and \(c\) such that \[f(x) = p_n(x) + R_n(x)\] where \(p_n(x)\) is the degree \(n\) Taylor polynomial to \(f\) at \(c\), \[R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}\] is the error (or remainder) of the approximation, and the absolute value of the error is less than the maximum value of \(|R_n(x)|\) on the interval \(I\).

Taylor series

If the absolute error \(| R_n(x) | \to 0\) as \(n \to \infty\) for any \(x\), then \[f(x) = \lim_{n \to \infty} (p_n(x) + R_n(x)) = p_\infty(x)\] and thus \[f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(c)}{k!} (x-c)^k\] is the Taylor series for \(f\) centered at \(c\).


Find the first four Taylor polynomial approximations to \(f(x) = e^x\) centered at \(c=0\). Use them to estimate \(e^{0.1}\).


\(f(x) = p_n(x) + R_n(x)\), and at \(x = 0.1\):

\(n\) \(f(0.1)=e^{0.1}\) \(p_n(0.1)\) \(R_n(0.1)\)
0 1.10517 1.00000 0.10517
1 1.10517 1.10000 0.00517
2 1.10517 1.10500 0.00017
3 1.10517 1.10517 0.00000


For each function below, find it's Taylor series and interval of convergence.

  1. \(f(x) = \cos(x)\) centered at \(c=0\).

  2. \(f(x) = x^2 \cos(\sqrt{x})\) centered at \(c=0\). What is the interval of convergence?

  3. \(f(x) = \sin(x)\) centered at \(c=0\).

  4. \(\displaystyle f(x) = \frac{\sin(x)}{x}\) centered at \(c=0\). Then, find \(\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}\).

  5. \(f(x) = \arctan(x)\) centered at \(c=0\). Hint: \(\displaystyle \frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}\).


\[\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\]

Show that \(e^{i \theta} = \cos(\theta) + i \sin(\theta)\).