Last modified: 2026-11-05 14:57:03 (PST)
Definition 1 (Taylor polynomial (Taylor expansion)) Let \(f\) be a function that has at least \(n\) derivatives at a point \(x_0\). The Taylor polynomial (also called the Taylor expansion) of degree \(n\) for \(f\) centered at \(x_0\) is:
\[ \begin{aligned} P_n(x) &= \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k \\ &= f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \dots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n \end{aligned} \tag{1}\]
Example 1 (Taylor series of the exponential function) For \(f(x) = e^x\) centered at \(x_0 = 0\), every derivative is
\[ \begin{aligned} f^{(k)}(0) &= e^0 \\ &= 1. \end{aligned} \]
The degree-\(n\) Maclaurin polynomial is:
\[ \begin{aligned} P_n(x) &= 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} \\ &= \sum_{k=0}^n \frac{x^k}{k!} \end{aligned} \]
Because the remainder \(R_n(x) \to 0\) as \(n \to \infty\) for all \(x \in \mathbb{R}\), the Taylor series converges everywhere to \(e^x\):
\[e^x = \sum_{k=0}^\infty \frac{x^k}{k!}\]
Definition 2 (Maclaurin polynomial) When \(x_0 = 0\), the Taylor polynomial \(P_n(x)\) (Definition 1) is called the Maclaurin polynomial.
Definition 3 (Taylor series) Let \(f\) and \(x_0\) be as in Definition 1. When \(f\) is infinitely differentiable and the series converges to \(f(x)\) on an open interval containing \(x_0\), the infinite sum \(\sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k\) is the Taylor series of \(f\) centered at \(x_0\).
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Chapter 11 of the Essence of calculus series by 3Blue1Brown develops the geometric and polynomial intuition behind Taylor approximations:
Exercise 1 (Taylor polynomial of a polynomial at the origin) Adapted from Miller [1], Question 1.1.16.
Find the first five terms (up to degree 4) of the Taylor series for
\[f(x) = x^8 + x^4 + 3\]
at \(x_0 = 0\).
Solution 1. Evaluate \(f\) and its first four derivatives at \(x = 0\):
\[\begin{aligned} f(x) &= x^8 + x^4 + 3 &\implies f(0) &= 3 \\ f'(x) &= 8x^7 + 4x^3 &\implies f'(0) &= 0 \\ f''(x) &= 56x^6 + 12x^2 &\implies f''(0) &= 0 \\ f'''(x) &= 336x^5 + 24x &\implies f'''(0) &= 0 \\ f^{(4)}(x) &= 1680x^4 + 24 &\implies f^{(4)}(0) &= 24 \end{aligned}\]
By Equation 1:
\[ \begin{aligned} P_4(x) &= 3 + 0x + \frac{0}{2!}x^2 + \frac{0}{3!}x^3 + \frac{24}{4!}x^4 \\ &= 3 + x^4 \end{aligned} \]
Because \(f\) is already a polynomial centered at \(0\), its Taylor polynomial of degree 4 simply collects its terms of degree at most 4.
Exercise 2 (Taylor expansion of a polynomial at a nonzero point) Adapted from Miller [1], Question 1.1.17.
Find the first three terms (up to degree 2) of the Taylor series for
\[f(x) = x^8 + x^4 + 3\]
at \(x_0 = 1\).
Solution 2. Evaluate \(f\) and its first two derivatives at \(x_0 = 1\):
\[\begin{aligned} f(1) &= 1^8 + 1^4 + 3 \\ &= 5 \\ f'(1) &= 8(1)^7 + 4(1)^3 \\ &= 12 \\ f''(1) &= 56(1)^6 + 12(1)^2 \\ &= 68 \end{aligned}\]
The degree-2 Taylor polynomial is:
\[ \begin{aligned} P_2(x) &= f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 \\ &= 5 + 12(x - 1) + 34(x - 1)^2 \end{aligned} \]
Example 2 (Taylor series of the cosine function) For \(f(x) = \cos x\) centered at \(x_0 = 0\), the derivatives follow a repeating cycle of length 4:
\[\begin{aligned} f(0) &= \cos(0) \\ &= 1, \\ f'(0) &= -\sin(0) \\ &= 0, \\ f''(0) &= -\cos(0) \\ &= -1, \\ f'''(0) &= \sin(0) \\ &= 0, \end{aligned}\]
and in general \(f^{(2k)}(0) = (-1)^k\) and \(f^{(2k+1)}(0) = 0\) for all \(k \ge 0\). The Taylor series (Maclaurin series) converges everywhere to \(\cos x\):
\[ \begin{aligned} \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \\ &= \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}x^{2k} \end{aligned} \]
Example 3 (Taylor series of the sine function) For \(f(x) = \sin x\) centered at \(x_0 = 0\), the derivatives evaluate at \(0\) to \(f^{(2k)}(0) = 0\) and \(f^{(2k+1)}(0) = (-1)^k\) for all \(k \ge 0\). The Taylor series converges everywhere to \(\sin x\):
\[ \begin{aligned} \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \\ &= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1} \end{aligned} \]
Exercise 3 (Taylor expansion of a scaled cosine) Adapted from Miller [1], Question 1.1.18.
Find the terms up to degree 4 in the Taylor series for
\[f(x) = \cos(5x)\]
at \(x_0 = 0\).
Solution 3. Recall the standard Maclaurin series for cosine:
\[\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots\]
Substitute \(u = 5x\):
\[ \begin{aligned} \cos(5x) &= 1 - \frac{(5x)^2}{2!} + \frac{(5x)^4}{4!} - \dots \\ &= 1 - \frac{25}{2}x^2 + \frac{625}{24}x^4 - \dots \end{aligned} \]
Remark: Substituting into known Taylor expansions avoids computing high-order derivatives by hand.
Exercise 4 (Taylor series of a power of a function) Adapted from Miller [1], Question 1.1.19.
Find the terms up to degree 4 in the Taylor series for
\[f(x) = \cos^3(5x)\]
at \(x_0 = 0\).
Solution 4. From Exercise 3, the expansion of \(\cos(5x)\) up to order \(x^4\) is:
\[\cos(5x) = 1 - \frac{25}{2}x^2 + \frac{625}{24}x^4 + O(x^6)\]
Cube this expansion using \((1 - y)^3 = 1 - 3y + 3y^2 - y^3\) with \(y = \frac{25}{2}x^2 - \frac{625}{24}x^4\):
\[\begin{aligned} \cos^3(5x) &= 1 - 3\mathopen{}\left(\frac{25}{2}x^2 - \frac{625}{24}x^4\right)\mathclose{} + 3\mathopen{}\left(\frac{25}{2}x^2\right)\mathclose{}^2 + O(x^6) \\ &= 1 - \frac{75}{2}x^2 + \frac{625}{8}x^4 + \frac{1875}{4}x^4 + O(x^6) \\ &= 1 - \frac{75}{2}x^2 + \frac{4375}{8}x^4 + O(x^6) \end{aligned}\]
Exercise 5 (Linear Taylor polynomial of the exponential) Adapted from Miller [1], Question 1.1.20.
Find the first two terms (degree 1) of the Taylor series for
\[f(x) = e^x\]
at \(x_0 = 0\).
Solution 5. Since
\[ \begin{aligned} f(0) &= e^0 \\ &= 1 \end{aligned} \]
and
\[ \begin{aligned} f'(0) &= e^0 \\ &= 1: \end{aligned} \]
\[ \begin{aligned} P_1(x) &= f(0) + f'(0)x \\ &= 1 + x \end{aligned} \]
The full Taylor series is
\[ \begin{aligned} e^x &= \sum_{n=0}^\infty \frac{x^n}{n!} \\ &= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots. \end{aligned} \]
Exercise 6 (Taylor series of a composite exponential) Adapted from Miller [1], Question 1.1.21.
Find the first six terms (up to degree 5) of the Taylor series for
\[f(x) = {\operatorname{exp}\mathopen{}\left\{x^8\right\}\mathclose{}}\]
at \(x_0 = 0\).
Solution 6. Substitute \(u = x^8\) into \(e^u = 1 + u + \frac{u^2}{2!} + \dots\):
\[{\operatorname{exp}\mathopen{}\left\{x^8\right\}\mathclose{}} = 1 + x^8 + \frac{x^{16}}{2} + \dots\]
Because the lowest-order variable term is \(x^8\), all coefficients for powers \(x^1, x^2, x^3, x^4, x^5\) are zero:
\[ \begin{aligned} P_5(x) &= 1 + 0x + 0x^2 + 0x^3 + 0x^4 + 0x^5 \\ &= 1 \end{aligned} \]
Exercise 7 (Taylor expansion of the normal density) Adapted from Miller [1], Question 1.1.22.
Find the terms up to degree 3 of the Taylor series for the standard normal density
\[\phi(x) = \frac{1}{\sqrt{2\pi}}{\operatorname{exp}\mathopen{}\left\{-\frac{x^2}{2}\right\}\mathclose{}}\]
at \(x_0 = 0\).
Solution 7. Using \(e^u = 1 + u + O(u^2)\) with \(u = -x^2/2\):
\[{\operatorname{exp}\mathopen{}\left\{-\frac{x^2}{2}\right\}\mathclose{}} = 1 - \frac{x^2}{2} + O(x^4)\]
Multiplying by the normalization constant \(1/\sqrt{2\pi}\):
\[\phi(x) = \frac{1}{\sqrt{2\pi}} - \frac{x^2}{2\sqrt{2\pi}} + O(x^4)\]
Up to degree 3, the odd powers vanish by symmetry, so:
\[P_3(x) = \frac{1}{\sqrt{2\pi}} - \frac{x^2}{2\sqrt{2\pi}}\]
Theorem 1 (Maclaurin polynomial of the standard normal density) The Maclaurin polynomial (Definition 2) of degree 3 of the standard normal density \(\phi(x) = \frac{1}{\sqrt{2\pi}}{\operatorname{exp}\mathopen{}\left\{-\frac{x^2}{2}\right\}\mathclose{}}\) is
\[P_3(x) = \frac{1}{\sqrt{2\pi}} - \frac{x^2}{2\sqrt{2\pi}}\]
Proof. This is the solution to Exercise 7, worked in Solution 7.
Exercise 8 (Taylor expansion of the square root) Adapted from Miller [1], Question 1.1.23.
Find the first three terms (up to degree 2) of the Taylor series for
\[f(x) = \sqrt{x}\]
at \(x_0 = 1/3\).
Solution 8. Compute the values at \(x_0 = 1/3\):
\[\begin{aligned} f(1/3) &= \mathopen{}\left(\frac{1}{3}\right)\mathclose{}^{1/2} \\ &= \frac{1}{\sqrt{3}} \\ &= \frac{\sqrt{3}}{3} \\ f'(x) &= \frac{1}{2}x^{-1/2} \\ \implies f'(1/3) &= \frac{1}{2}\sqrt{3} \\ &= \frac{\sqrt{3}}{2} \\ f''(x) &= -\frac{1}{4}x^{-3/2} \\ \implies f''(1/3) &= -\frac{1}{4} \cdot 3\sqrt{3} \\ &= -\frac{3\sqrt{3}}{4} \end{aligned}\]
The degree-2 Taylor polynomial is:
\[P_2(x) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{2}\mathopen{}\left(x - \frac{1}{3}\right)\mathclose{} - \frac{3\sqrt{3}}{8}\mathopen{}\left(x - \frac{1}{3}\right)\mathclose{}^2\]
Exercise 9 (Taylor expansion of a fractional power) Adapted from Miller [1], Question 1.1.24.
Find the first three terms of the Taylor series for
\[f(x) = (1 + x)^{1/3}\]
at \(x_0 = 1/2\).
Solution 9. At \(x_0 = 1/2\), \(1 + x_0 = 3/2\):
\[\begin{aligned} f(1/2) &= \mathopen{}\left(\frac{3}{2}\right)\mathclose{}^{1/3} \\ f'(x) &= \frac{1}{3}(1 + x)^{-2/3} \\ \implies f'(1/2) &= \frac{1}{3}\mathopen{}\left(\frac{3}{2}\right)\mathclose{}^{-2/3} \\ &= \frac{1}{3}\mathopen{}\left(\frac{2}{3}\right)\mathclose{}^{2/3} \\ f''(x) &= -\frac{2}{9}(1 + x)^{-5/3} \\ \implies f''(1/2) &= -\frac{2}{9}\mathopen{}\left(\frac{3}{2}\right)\mathclose{}^{-5/3} \\ &= -\frac{2}{9}\mathopen{}\left(\frac{2}{3}\right)\mathclose{}^{5/3} \end{aligned}\]
Dividing \(f''(1/2)\) by \(2!\) gives \(-\frac{1}{9}\mathopen{}\left(\frac{2}{3}\right)\mathclose{}^{5/3}\). The Taylor polynomial is:
\[P_2(x) = \mathopen{}\left(\frac{3}{2}\right)\mathclose{}^{1/3} + \frac{1}{3}\mathopen{}\left(\frac{2}{3}\right)\mathclose{}^{2/3}\mathopen{}\left(x - \frac{1}{2}\right)\mathclose{} - \frac{1}{9}\mathopen{}\left(\frac{2}{3}\right)\mathclose{}^{5/3}\mathopen{}\left(x - \frac{1}{2}\right)\mathclose{}^2\]
Exercise 10 (Taylor expansion of \(x \log x\)) Adapted from Miller [1], Question 1.1.25.
Find the first three terms (up to degree 2) of the Taylor series for
\[f(x) = x\log x\]
at \(x_0 = 1\).
Solution 10. Evaluate derivatives at \(x_0 = 1\):
\[\begin{aligned} f(1) &= 1 \log 1 \\ &= 0 \\ f'(x) &= \log x + 1 \\ \implies f'(1) &= 0 + 1 \\ &= 1 \\ f''(x) &= \frac{1}{x} \implies f''(1) = 1 \end{aligned}\]
Therefore:
\[ \begin{aligned} P_2(x) &= f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 \\ &= (x - 1) + \frac{1}{2}(x - 1)^2 \end{aligned} \]
Exercise 11 (Maclaurin series of \(\log(1 + x)\)) Adapted from Miller [1], Question 1.1.26.
Find the first three terms of the Taylor series for
\[f(x) = \log(1 + x)\]
at \(x_0 = 0\).
Solution 11. Differentiating repeatedly at \(x_0 = 0\):
\[\begin{aligned} f(0) &= \log 1 \\ &= 0 \\ f'(x) &= (1 + x)^{-1} \implies f'(0) = 1 \\ f''(x) &= -(1 + x)^{-2} \implies f''(0) = -1 \\ f'''(x) &= 2(1 + x)^{-3} \implies f'''(0) = 2 \end{aligned}\]
The Taylor polynomial of degree 3 is:
\[P_3(x) = x - \frac{x^2}{2} + \frac{x^3}{3}\]
Theorem 2 (Maclaurin polynomial of \(\log(1 + x)\)) The Maclaurin polynomial (Definition 2) of degree 3 of \(f(x) = \log(1 + x)\) is
\[P_3(x) = x - \frac{x^2}{2} + \frac{x^3}{3}\]
Proof. This is the solution to Exercise 11, worked in Solution 11.
Exercise 12 (Maclaurin series of \(\log(1 - x)\)) Adapted from Miller [1], Question 1.1.27.
Find the terms up to degree 3 of the Taylor series for
\[f(x) = \log(1 - x)\]
at \(x_0 = 0\).
Solution 12. Substitute \(u = -x\) into the expansion for \(\log(1 + u)\) from Theorem 2:
\[ \begin{aligned} \log(1 - x) &= (-x) - \frac{(-x)^2}{2} + \frac{(-x)^3}{3} - \dots \\ &= -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots \end{aligned} \]
Theorem 3 (Maclaurin polynomial of \(\log(1 - x)\)) The Maclaurin polynomial (Definition 2) of degree 3 of \(f(x) = \log(1 - x)\) is
\[P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}\]
Proof. This is the solution to Exercise 12, worked in Solution 12.
Exercise 13 (Taylor series of \(\log((1 - x)e^x)\)) Adapted from Miller [1], Question 1.1.28.
Find the first two non-zero terms of the Taylor series for
\[f(x) = \log((1 - x)e^x)\]
at \(x_0 = 0\).
Solution 13. Expand using logarithm properties:
\[ \begin{aligned} f(x) &= \log(1 - x) + \log(e^x) \\ &= \log(1 - x) + x \end{aligned} \]
Substitute the series for \(\log(1 - x)\) from Theorem 3:
\[ \begin{aligned} f(x) &= \mathopen{}\left(-x - \frac{x^2}{2} - \frac{x^3}{3} - \dots\right)\mathclose{} + x \\ &= -\frac{x^2}{2} - \frac{x^3}{3} - \dots \end{aligned} \]
The constant and linear terms vanish (\(f(0) = 0\), \(f'(0) = 0\)). The first two non-zero terms are:
\[-\frac{x^2}{2} - \frac{x^3}{3}\]
Exercise 14 (Taylor series of a product of known expansions) Adapted from Miller [1], Question 1.1.29.
Find the terms up to degree 2 of the Taylor series for
\[f(x) = \cos(x)\log(1 + x)\]
at \(x_0 = 0\).
Solution 14. Multiply the known Maclaurin expansions:
\[\begin{aligned} \cos x &= 1 - \frac{x^2}{2} + O(x^4) \\ \log(1 + x) &= x - \frac{x^2}{2} + O(x^3) \end{aligned}\]
Multiplying and dropping terms of order \(x^3\) and higher:
\[ \begin{aligned} f(x) &= \mathopen{}\left(1 - \frac{x^2}{2} + \dots\right)\mathclose{}\mathopen{}\left(x - \frac{x^2}{2} + \dots\right)\mathclose{} \\ &= x - \frac{x^2}{2} + O(x^3) \end{aligned} \]
Exercise 15 (Taylor series of \(\log(1 + 2x)\)) Adapted from Miller [1], Question 1.1.30.
Find the first two non-zero terms of the Taylor series for
\[f(x) = \log(1 + 2x)\]
at \(x_0 = 0\).
Solution 15. Substitute \(u = 2x\) into the Maclaurin series \(\log(1 + u) = u - \frac{u^2}{2} + \dots\):
\[ \begin{aligned} \log(1 + 2x) &= (2x) - \frac{(2x)^2}{2} + \dots \\ &= 2x - 2x^2 + \dots \end{aligned} \]