The formulas
Before you start: you'll need confident differentiation, including higher derivatives. The Derivatives guide covers it.
\[f(x) = f(a) + (x - a)f'(a) + \frac{(x - a)^2}{2!}f''(a) + \frac{(x - a)^3}{3!}f'''(a) + \cdots\]
Equivalent form, with \(x = a + h\)
\[f(a + h) = f(a) + h f'(a) + \frac{h^2}{2!}f''(a) + \frac{h^3}{3!}f'''(a) + \cdots\]
Maclaurin's series (Taylor's series about \(x = 0\))
\[f(x) = f(0) + x f'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \cdots\]
The idea: build a polynomial whose value, slope, curvature and every higher derivative match \(f\) at one point. Each extra term matches one more derivative, so the polynomial hugs the function more closely. The series is valid only for values of \(x\) where it converges, which is why some expansions below come with a range.
The idea in one picture
This is how calculators work out \(\sin x\), \(e^x\) or \(\log x\): they add up the first few terms of a series like these. It is also why engineers can replace a complicated function by a simple polynomial near an operating point.
Standard Maclaurin series
Learn these. Most exam questions are solved by combining them rather than differentiating from scratch.
| Function | Series | Valid for |
|---|---|---|
| \(e^x\) | \(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\) | all \(x\) |
| \(\sin x\) | \(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots\) | all \(x\) |
| \(\cos x\) | \(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots\) | all \(x\) |
| \(\sinh x\) | \(x + \dfrac{x^3}{3!} + \dfrac{x^5}{5!} + \cdots\) | all \(x\) |
| \(\cosh x\) | \(1 + \dfrac{x^2}{2!} + \dfrac{x^4}{4!} + \cdots\) | all \(x\) |
| \(\log(1 + x)\) | \(x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots\) | \(-1 < x \le 1\) |
| \((1 + x)^n\) | \(1 + nx + \dfrac{n(n - 1)}{2!}x^2 + \dfrac{n(n - 1)(n - 2)}{3!}x^3 + \cdots\) | \(|x| < 1\) |
| \(\tan^{-1} x\) | \(x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots\) | \(|x| \le 1\) |
| \(\tan x\) | \(x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \cdots\) | \(|x| < \dfrac{\pi}{2}\) |
Patterns to notice: \(\sin x\) has only odd powers and \(\cos x\) only even powers, matching the fact that \(\sin x\) is an odd function and \(\cos x\) an even one. Putting \(x = 1\) in the \(\tan^{-1}\) series gives \(\dfrac{\pi}{4} = 1 - \dfrac{1}{3} + \dfrac{1}{5} - \cdots\).
Two ways to find a series
- From derivatives. Find \(f(0), f'(0), f''(0), \dots\) (or the values at \(x = a\)) and substitute into the formula. Reliable, but slow for complicated functions.
- From known series. Substitute, multiply, divide or integrate the standard series. Usually much faster: for \(e^{-x^2}\), just replace \(x\) by \(-x^2\) in the series for \(e^x\).
Unless the question says “using Maclaurin's theorem” or asks for the derivatives, the second way is fine and saves time.
Solved examples
Example 1: The series for eˣ
EasyObtain the Maclaurin series for \(e^x\).
- Every derivative of \(e^x\) is \(e^x\), so \(f(0) = f'(0) = f''(0) = \cdots = 1\).
- Substitute into Maclaurin's formula: \(1 + x\cdot 1 + \dfrac{x^2}{2!}\cdot 1 + \dfrac{x^3}{3!}\cdot 1 + \cdots\)
Answer\(e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\)
Example 2: The series for log(1 + x)
MediumExpand \(\log(1 + x)\) up to the term in \(x^4\).
- \(f = \log(1 + x)\), \(f' = \dfrac{1}{1 + x}\), \(f'' = -\dfrac{1}{(1 + x)^2}\), \(f''' = \dfrac{2}{(1 + x)^3}\), \(f^{(4)} = -\dfrac{6}{(1 + x)^4}\).
- At \(x = 0\): \(0,\ 1,\ -1,\ 2,\ -6\).
- Substitute: \(0 + x - \dfrac{x^2}{2!} + \dfrac{2x^3}{3!} - \dfrac{6x^4}{4!}\).
Answer\(\log(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots\)
Example 3: Taylor's series about a point
MediumExpand \(\log x\) in powers of \((x - 1)\).
- Write \(x = 1 + (x - 1)\), so \(\log x = \log\big(1 + (x - 1)\big)\).
- Use the series for \(\log(1 + t)\) with \(t = x - 1\).
Answer\(\log x = (x - 1) - \dfrac{(x - 1)^2}{2} + \dfrac{(x - 1)^3}{3} - \cdots\), for \(0 < x \le 2\)
Example 4: Multiplying two series
MediumExpand \(e^x \sin x\) up to the term in \(x^4\).
- \(e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots\) and \(\sin x = x - \dfrac{x^3}{6} + \cdots\)
- Multiply, keeping powers up to \(x^4\): \(x\left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\right) - \dfrac{x^3}{6}(1 + x)\).
- Collect: \(x + x^2 + \left(\dfrac{1}{2} - \dfrac{1}{6}\right)x^3 + \left(\dfrac{1}{6} - \dfrac{1}{6}\right)x^4\).
Answer\(e^x \sin x = x + x^2 + \dfrac{x^3}{3} + 0\cdot x^4 + \cdots\)
Example 5: Dividing two series
Exam levelShow that \(\tan x = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \cdots\)
- \(\tan x\) is odd, so write \(\tan x = x + p x^3 + q x^5 + \cdots\) and use \(\tan x \cdot \cos x = \sin x\).
- \(\left(x + px^3 + qx^5\right)\left(1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}\right) = x + \left(p - \dfrac{1}{2}\right)x^3 + \left(q - \dfrac{p}{2} + \dfrac{1}{24}\right)x^5 + \cdots\)
- Compare with \(\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120}\): \(p - \dfrac{1}{2} = -\dfrac{1}{6}\) gives \(p = \dfrac{1}{3}\).
- Then \(q - \dfrac{1}{6} + \dfrac{1}{24} = \dfrac{1}{120}\) gives \(q = \dfrac{1 + 20 - 5}{120} = \dfrac{2}{15}\).
Answer\(\tan x = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \cdots\)
Example 6: Approximating a value
Exam levelUse a series to find \(\log(1.1)\) correct to four decimal places.
- Put \(x = 0.1\) in \(\log(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots\)
- \(= 0.1 - 0.005 + 0.000333 - 0.000025 + \cdots\)
- The next term, \(\dfrac{x^5}{5} = 0.000002\), is too small to affect the fourth decimal place.
Answer\(\log(1.1) \approx 0.0953\)
Before differentiating five times, ask whether the function is built from \(e^x\), \(\sin x\), \(\cos x\), \(\log(1 + x)\) or \((1 + x)^n\). Substituting into a known series usually takes three lines instead of a page.
Common mistakes
1. Forgetting the factorials
The term in \(x^n\) is \(\dfrac{f^{(n)}(0)}{n!}x^n\). Leaving out the \(n!\) is the most common slip.
2. Using a series outside its range
\(\log(1 + x)\) and \((1 + x)^n\) only work for \(|x| < 1\) (or \(-1 < x \le 1\) for log). Putting \(x = 3\) into the log series gives nonsense.
3. Dropping terms too early when multiplying
To get \(x^4\) correctly in a product, keep every term up to \(x^4\) in both factors until you multiply.
4. Mixing up Taylor and Maclaurin
Maclaurin is the special case \(a = 0\). For “in powers of \((x - a)\)”, you need Taylor's series about \(x = a\).
How it's asked in exams
- Expand a function in ascending powers of \(x\), up to a given term.
- Expand in powers of \((x - a)\), using Taylor's series.
- Prove a given expansion, such as the one for \(\tan x\) in Example 5.
- Approximate a value, such as \(\sqrt{1.1}\) or \(\sin 31^\circ\), to a given number of decimal places.
Practice questions
Try each one on paper before you open the answer.
Q1Find the Maclaurin series for \(\cos x\) up to \(x^4\), using derivatives.
Derivatives at 0: \(\cos 0 = 1\), \(-\sin 0 = 0\), \(-\cos 0 = -1\), \(\sin 0 = 0\), \(\cos 0 = 1\). So \(\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots\)
Q2Expand \(e^{-x^2}\) up to \(x^6\).
Replace \(x\) by \(-x^2\) in the series for \(e^x\): \(1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \cdots\)
Q3Expand \(\sqrt{1 + x}\) up to \(x^3\).
Binomial series with \(n = \tfrac{1}{2}\): \(1 + \dfrac{x}{2} - \dfrac{x^2}{8} + \dfrac{x^3}{16} - \cdots\)
Q4Expand \(\sin x\) in powers of \(\left(x - \tfrac{\pi}{2}\right)\).
\(\sin x = \cos\left(x - \tfrac{\pi}{2}\right) = 1 - \dfrac{\left(x - \frac{\pi}{2}\right)^2}{2!} + \dfrac{\left(x - \frac{\pi}{2}\right)^4}{4!} - \cdots\)
Q5Find \(\sin(0.2)\) correct to five decimal places.
\(0.2 - \dfrac{0.008}{6} + \dfrac{0.00032}{120} = 0.2 - 0.0013333 + 0.0000027 \approx 0.19867\).
Q6Expand \(e^{\sin x}\) up to \(x^3\).
Let \(u = \sin x = x - \dfrac{x^3}{6}\). Then \(e^u = 1 + u + \dfrac{u^2}{2} + \dfrac{u^3}{6} + \cdots\)
Up to \(x^3\): \(u^2 \approx x^2\), \(u^3 \approx x^3\). So \(e^{\sin x} = 1 + x - \dfrac{x^3}{6} + \dfrac{x^2}{2} + \dfrac{x^3}{6} = 1 + x + \dfrac{x^2}{2} + 0\cdot x^3 + \cdots\)
Q7Challenge. Show that \(\log(\cos x) = -\dfrac{x^2}{2} - \dfrac{x^4}{12} - \cdots\)
\(\cos x = 1 + u\) with \(u = -\dfrac{x^2}{2} + \dfrac{x^4}{24}\). Then \(\log(1 + u) = u - \dfrac{u^2}{2} + \cdots\)
Up to \(x^4\): \(u^2 = \dfrac{x^4}{4}\). So \(\log(\cos x) = -\dfrac{x^2}{2} + \dfrac{x^4}{24} - \dfrac{x^4}{8} = -\dfrac{x^2}{2} - \dfrac{x^4}{12}\).
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Frequently asked questions
What is the difference between Taylor's and Maclaurin's series?
Taylor's series expands a function about any point \(x = a\). Maclaurin's series is the special case \(a = 0\).
Why are Taylor series useful?
They turn complicated functions into polynomials, which are easy to calculate, differentiate and integrate. Calculators, numerical methods and many engineering approximations depend on them.
Does every function have a Taylor series?
Only functions with derivatives of every order at the point. And the series equals the function only where it converges: \(\log(1 + x)\), for example, only for \(-1 < x \le 1\).
How many terms should I write?
As many as the question asks, usually “up to \(x^4\)” or “the first four non-zero terms”. For approximations, keep adding terms until the next one is too small to change the required decimal place.
Where Taylor series lead next
← All 33 Calculus topics
Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.