30.3 Building New Series
AHL 5.19
- how to get new series from the standard ones by substitution
- how to differentiate and integrate a series term by term
- how to use a series to estimate an integral that cannot be found exactly
- how to multiply two series to find the first few terms of a product
- how to use series to find limits of the form \(\frac{0}{0}\)
- (Extension) how to divide one series by another
Finding many derivatives is slow. It is usually much quicker to build a new series from the standard series of 30.1. Remember to change the interval of validity too (30.2).
Substitution
Replace \(x\) in a standard series by another expression, such as \(3x\), \(-x^2\) or \(x^3\).
Example 1 Find the first three non-zero terms of the Maclaurin series of
(a) \(\sin 3x\)
(b) \(\mathrm{e}^{x^2}\)
(c) \(\ln(1 - x^2)\), stating where this series is valid.
(a) Replace \(x\) by \(3x\) in \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\):
\[ \sin 3x = 3x - \frac{27x^3}{6} + \frac{243x^5}{120} - \cdots = 3x - \frac{9x^3}{2} + \frac{81x^5}{40} - \cdots \]
(b) Replace \(x\) by \(x^2\) in the series for \(\mathrm{e}^x\): \(\;\mathrm{e}^{x^2} = 1 + x^2 + \dfrac{x^4}{2!} + \cdots = 1 + x^2 + \dfrac{x^4}{2} + \cdots\)
(c) Replace \(x\) by \(-x^2\) in \(\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\):
\[ \ln(1 - x^2) = -x^2 - \frac{x^4}{2} - \frac{x^6}{3} - \cdots \]
Valid for \(-1 < -x^2 \leqslant 1\). Since \(-x^2 \leqslant 0 < 1\) always, we only need \(x^2 < 1\), that is \(-1 < x < 1\).
Differentiation and integration
A power series can be differentiated and integrated term by term, like a polynomial. This is very useful for integrals such as \(\int \mathrm{e}^{-x^2}\,\mathrm{d}x\), which cannot be written with the usual functions.
Term by term Not in the formula booklet
If \(f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots\), then
\[ f'(x) = a_1 + 2a_2x + 3a_3x^2 + \cdots, \qquad \int_0^x f(t)\,\mathrm{d}t = a_0x + \frac{a_1x^2}{2} + \frac{a_2x^3}{3} + \cdots \]
The new series is valid on the same interval (except possibly at the end points).
- \(a_0, a_1, a_2, \ldots\): the coefficients of the series (constants)
- \(t\): the variable of integration (it runs from \(0\) to \(x\))
Example 2 (a) Use \(\dfrac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots\) \((|x| < 1)\) to find the Maclaurin series for \(\dfrac{1}{(1 - x)^2}\).
(b) Hence find the value of \(\displaystyle\sum_{k=1}^{\infty}\frac{k}{2^k} = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots\)
(a) \(\dfrac{\mathrm{d}}{\mathrm{d}x}(1 - x)^{-1} = (1 - x)^{-2}\). Differentiate both sides term by term:
\[ \frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots = \sum_{k=1}^{\infty}kx^{k-1}, \qquad |x| < 1 . \]
(b) Multiply by \(x\): \(\;\dfrac{x}{(1 - x)^2} = x + 2x^2 + 3x^3 + \cdots = \displaystyle\sum_{k=1}^{\infty}kx^k\).
Put \(x = \frac{1}{2}\) (inside \(|x| < 1\)):
\[ \sum_{k=1}^{\infty}\frac{k}{2^k} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2} = 2 . \]
Example 3 (a) Find the Maclaurin series for \(\dfrac{1}{\sqrt{1 - x^2}}\) up to the term in \(x^4\).
(b) Hence show that \(\arcsin x = x + \dfrac{x^3}{6} + \dfrac{3x^5}{40} + \cdots\)
(a) Binomial series with \(p = -\frac{1}{2}\) and \(x\) replaced by \(-x^2\):
\[ (1 - x^2)^{-\frac{1}{2}} = 1 + \left(-\tfrac{1}{2}\right)(-x^2) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-x^2)^2 + \cdots = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \cdots \]
(b) \(\dfrac{\mathrm{d}}{\mathrm{d}t}\arcsin t = \dfrac{1}{\sqrt{1 - t^2}}\) and \(\arcsin 0 = 0\), so
\[ \arcsin x = \int_0^x\left(1 + \frac{t^2}{2} + \frac{3t^4}{8} + \cdots\right)\mathrm{d}t = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots \]
Example 4 (a) Find the first four terms of the Maclaurin series for \(\mathrm{e}^{-x^2}\).
(b) Hence find an approximation for \(\displaystyle\int_0^{0.5}\mathrm{e}^{-x^2}\,\mathrm{d}x\), to 5 decimal places.
(a) Replace \(x\) by \(-x^2\) in the series for \(\mathrm{e}^x\): \(\;\mathrm{e}^{-x^2} = 1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \cdots\)
(b) Integrate term by term:
\[ \int_0^{0.5}\left(1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}\right)\mathrm{d}x = \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42}\right]_0^{0.5} \]
\[ = 0.5 - 0.0416667 + 0.003125 - 0.0001860 = 0.46127 \ (\text{5 d.p.}) \]
The exact value is \(0.461281\ldots\), so four terms already give 4 correct decimal places.
In a Calculator page: Menu → Calculus → Numerical Integral. Fill in the template \(\displaystyle\int_0^{0.5}\mathrm{e}^{-x^2}\,\mathrm{d}x\) and press enter: \(0.461281\).
This is useful for checking a series approximation. A question that says “use the series” still needs the series working.
Multiplication
Multiplying two infinite series gives infinitely many products. But we only need the first few terms: keep the terms up to the power you want and ignore the rest.
Example 5 (a) Use the series for \(\mathrm{e}^x\) to find the Maclaurin series for \(\mathrm{e}^{-x^2}\) up to the term in \(x^4\).
(b) Hence find the Maclaurin series for \(\mathrm{e}^{-x^2}\sin x\) up to the term in \(x^5\).
(a) \(\mathrm{e}^{-x^2} = 1 - x^2 + \dfrac{x^4}{2} - \cdots\) (as in Example 4).
(b) \(\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots\). Multiply and collect only the powers up to \(x^5\):
\[ \left(1 - x^2 + \frac{x^4}{2}\right)\left(x - \frac{x^3}{6} + \frac{x^5}{120}\right) \]
| power | products | total |
|---|---|---|
| \(x\) | \(1 \cdot x\) | \(x\) |
| \(x^3\) | \(1 \cdot \left(-\frac{x^3}{6}\right) + (-x^2)\cdot x\) | \(-\frac{7}{6}x^3\) |
| \(x^5\) | \(1 \cdot \frac{x^5}{120} + (-x^2)\left(-\frac{x^3}{6}\right) + \frac{x^4}{2}\cdot x\) | \(\left(\frac{1}{120} + \frac{20}{120} + \frac{60}{120}\right)x^5 = \frac{27}{40}x^5\) |
\[ \mathrm{e}^{-x^2}\sin x = x - \frac{7}{6}x^3 + \frac{27}{40}x^5 - \cdots \]
Limits using series
For a limit of the form \(\frac{0}{0}\) as \(x \to 0\), replace each function by its series. Then divide top and bottom by the lowest power of \(x\) and let \(x \to 0\). This is often quicker than l’Hôpital’s rule (Unit 26).
Example 6 Use Maclaurin series to find
(a) \(\displaystyle\lim_{x \to 0}\frac{\sin 3x}{\ln(1 + 2x)}\)
(b) \(\displaystyle\lim_{x \to 0}\frac{x - \arctan x}{x^3}\)
(a) \(\sin 3x = 3x - \frac{9}{2}x^3 + \cdots\) and \(\ln(1 + 2x) = 2x - 2x^2 + \frac{8}{3}x^3 - \cdots\)
\[ \frac{\sin 3x}{\ln(1 + 2x)} = \frac{3x - \frac{9}{2}x^3 + \cdots}{2x - 2x^2 + \cdots} = \frac{3 - \frac{9}{2}x^2 + \cdots}{2 - 2x + \cdots} \to \frac{3}{2} \quad \text{as } x \to 0 . \]
(b) \(x - \arctan x = x - \left(x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots\right) = \frac{x^3}{3} - \frac{x^5}{5} + \cdots\)
\[ \frac{x - \arctan x}{x^3} = \frac{1}{3} - \frac{x^2}{5} + \cdots \to \frac{1}{3} . \]
To find the series of \(p(x) = \dfrac{f(x)}{g(x)}\) (with \(g(0) \neq 0\)), write \(p(x) = a_0 + a_1x + a_2x^2 + \cdots\) with unknown coefficients. Then \(p(x)\,g(x) = f(x)\). Multiply out the left side and equate coefficients of each power of \(x\).
Example. Find the series for \(\tan x = \dfrac{\sin x}{\cos x}\) up to the term in \(x^5\).
\(\tan x\) is odd, so try \(\tan x = a_1x + a_3x^3 + a_5x^5 + \cdots\). Then
\[ \left(a_1x + a_3x^3 + a_5x^5\right)\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots \]
- \(x\): \(\;a_1 = 1\)
- \(x^3\): \(\;a_3 - \frac{a_1}{2} = -\frac{1}{6} \Rightarrow a_3 = \frac{1}{3}\)
- \(x^5\): \(\;a_5 - \frac{a_3}{2} + \frac{a_1}{24} = \frac{1}{120} \Rightarrow a_5 = \frac{1}{120} + \frac{1}{6} - \frac{1}{24} = \frac{2}{15}\)
\[ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots \]
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\mathrm{e}^{x^2} = 1 + x^2 + \frac{x^2}{2!} + \cdots\) | \(1 + x^2 + \frac{(x^2)^2}{2!} + \cdots\) | Replace every \(x\), including inside the powers. |
| \(\sin 3x = 3x - \frac{3x^3}{3!} + \cdots\) | \(3x - \frac{(3x)^3}{3!} + \cdots\) | Brackets: \((3x)^3 = 27x^3\). |
| Keeping the old interval after substituting | Substitute into the interval | \(\ln(1 + 2x)\) needs \(-\frac{1}{2} < x \leqslant \frac{1}{2}\). |
| Forgetting the constant after integrating | Use \(\int_0^x\), or find \(c\) from \(f(0)\) | e.g. \(\arcsin 0 = 0\) gives \(c = 0\). |
| Multiplying only “matching” terms | Collect every pair whose powers add up | The \(x^5\) term of a product comes from three pairs. |
| Stopping a limit at \(\frac{0}{0}\) | Divide by the lowest power of \(x\) first | The leading terms decide the limit. |
Exercises
1 Find the Maclaurin series for \(\mathrm{e}^{-3x}\) up to the term in \(x^3\).
\(1 - 3x + \dfrac{9x^2}{2} - \dfrac{27x^3}{6} = 1 - 3x + \dfrac{9}{2}x^2 - \dfrac{9}{2}x^3 + \cdots\)
2 Find the first four non-zero terms of the Maclaurin series for \(\cos 2x\).
\(1 - \dfrac{4x^2}{2} + \dfrac{16x^4}{24} - \dfrac{64x^6}{720} = 1 - 2x^2 + \dfrac{2}{3}x^4 - \dfrac{4}{45}x^6 + \cdots\)
3 Find the first three non-zero terms of the Maclaurin series for \(\ln(1 + x^3)\), and state where it is valid.
\(x^3 - \dfrac{x^6}{2} + \dfrac{x^9}{3} - \cdots\), \(\;\) valid for \(-1 < x^3 \leqslant 1\), i.e. \(-1 < x \leqslant 1\).
4 By differentiating the series for \(\dfrac{1}{1 + x}\), find the Maclaurin series for \(\dfrac{1}{(1 + x)^2}\) up to the term in \(x^3\).
\(\dfrac{1}{1 + x} = 1 - x + x^2 - x^3 + x^4 - \cdots\). Differentiate: \(-\dfrac{1}{(1 + x)^2} = -1 + 2x - 3x^2 + 4x^3 - \cdots\)
So \(\dfrac{1}{(1 + x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots\)
5 Find the Maclaurin series for \(\mathrm{e}^x\cos x\) up to the term in \(x^4\).
\(\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}\right)\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)\):
\(x^2\): \(\frac{1}{2} - \frac{1}{2} = 0\); \(\;x^3\): \(\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}\); \(\;x^4\): \(\frac{1}{24} - \frac{1}{4} + \frac{1}{24} = -\frac{1}{6}\).
\(\mathrm{e}^x\cos x = 1 + x - \dfrac{x^3}{3} - \dfrac{x^4}{6} + \cdots\)
6 Find the Maclaurin series for \(\dfrac{\ln(1 + x)}{1 + x}\) up to the term in \(x^3\).
\(\left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)\left(1 - x + x^2\right)\):
\(x\): \(1\); \(\;x^2\): \(-\frac{1}{2} - 1 = -\frac{3}{2}\); \(\;x^3\): \(\frac{1}{3} + \frac{1}{2} + 1 = \frac{11}{6}\).
\(\dfrac{\ln(1 + x)}{1 + x} = x - \dfrac{3}{2}x^2 + \dfrac{11}{6}x^3 - \cdots\)
7 Find the first three non-zero terms of the series for \(\dfrac{\sin x}{x}\). Hence find an approximation for \(\displaystyle\int_0^1\frac{\sin x}{x}\,\mathrm{d}x\) to 4 decimal places.
\(\dfrac{\sin x}{x} = 1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \cdots\)
\(\displaystyle\int_0^1 \cdots \,\mathrm{d}x = \left[x - \frac{x^3}{18} + \frac{x^5}{600}\right]_0^1 = 1 - \frac{1}{18} + \frac{1}{600} = 0.9461\) (exact value \(0.946083\ldots\)).
8 Use Maclaurin series to find \(\displaystyle\lim_{x \to 0}\frac{\mathrm{e}^{2x} - 1 - 2x}{x^2}\).
\(\mathrm{e}^{2x} - 1 - 2x = 2x^2 + \dfrac{4}{3}x^3 + \cdots\), so the fraction is \(2 + \dfrac{4}{3}x + \cdots \to 2\).
9 Use Maclaurin series to find \(\displaystyle\lim_{x \to 0}\frac{\sin x - x\cos x}{x^3}\).
\(\sin x - x\cos x = \left(x - \frac{x^3}{6} + \cdots\right) - \left(x - \frac{x^3}{2} + \cdots\right) = \dfrac{x^3}{3} + \cdots\)
The limit is \(\dfrac{1}{3}\).
10 Extension Write \(\sec x = \dfrac{1}{\cos x}\) as \(a_0 + a_2x^2 + a_4x^4 + \cdots\). By equating coefficients in \(\sec x\cos x = 1\), find \(a_0\), \(a_2\) and \(a_4\).
\(\left(a_0 + a_2x^2 + a_4x^4\right)\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) = 1\):
\(x^0\): \(a_0 = 1\); \(\;x^2\): \(a_2 - \frac{a_0}{2} = 0 \Rightarrow a_2 = \frac{1}{2}\); \(\;x^4\): \(a_4 - \frac{a_2}{2} + \frac{a_0}{24} = 0 \Rightarrow a_4 = \frac{1}{4} - \frac{1}{24} = \frac{5}{24}\).
\(\sec x = 1 + \dfrac{x^2}{2} + \dfrac{5x^4}{24} + \cdots\)