Write down the Maclaurin series for \( e^x \) up to and including the term in \( x^3 \).
\( e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots \)
Maclaurin Series
12 exam-style questions on Maclaurin expansions, method of differences and compound functions. Tap Show answer for full working.
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Find the first three non-zero terms in the Maclaurin series for \( \sin x \).
\( \sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots \)
Show that \( \dfrac{1}{r(r+2)} = \dfrac{1}{2}\left(\dfrac{1}{r} - \dfrac{1}{r+2}\right) \). Hence find \( \displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+2)} \).
RHS: \( \dfrac{1}{2}\cdot\dfrac{(r+2)-r}{r(r+2)} = \dfrac{1}{2}\cdot\dfrac{2}{r(r+2)} = \dfrac{1}{r(r+2)} \). ✓
Sum: \( \dfrac{1}{2}\sum\left(\dfrac{1}{r}-\dfrac{1}{r+2}\right) = \dfrac{1}{2}\Big[(1-\tfrac13)+(\tfrac12-\tfrac14)+(\tfrac13-\tfrac15)+\cdots+(\tfrac1n-\tfrac1{n+2})\Big] \)
\( = \dfrac{1}{2}\Big(1 + \dfrac{1}{2} - \dfrac{1}{n+1} - \dfrac{1}{n+2}\Big) = \dfrac{3}{4} - \dfrac{1}{2(n+1)} - \dfrac{1}{2(n+2)} \).
Write down the Maclaurin series for \( \ln(1+x) \) up to the term in \( x^3 \), stating the range of validity.
\( \ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots \), valid for \( -1 < x \le 1 \).
Find the Maclaurin series for \( \cos x \) up to and including the term in \( x^4 \).
\( f(x)=\cos x \), \( f(0)=1 \); \( f'=-\sin \), \( f'(0)=0 \); \( f''=-\cos \), \( f''(0)=-1 \); \( f'''=\sin \), \( f'''(0)=0 \); \( f^{(4)}=\cos \), \( f^{(4)}(0)=1 \).
\( \cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} + \cdots = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} + \cdots \)
Find the first three non-zero terms in the Maclaurin expansion of \( e^{2x}\sin x \).
\( e^{2x} = 1 + 2x + 2x^2 + \dfrac{4}{3}x^3 + \cdots \)
\( \sin x = x - \dfrac{x^3}{6} + \cdots \)
Product: \( (1 + 2x + 2x^2 + \tfrac43 x^3)(x - \tfrac16 x^3) + O(x^4) \)
\( = x + 2x^2 + 2x^3 - \tfrac16 x^3 + O(x^4) = x + 2x^2 + \dfrac{11}{6}x^3 + \cdots \)
Use the method of differences to find \( \displaystyle\sum_{r=1}^{n} \dfrac{1}{(r+1)(r+2)(r+3)} \).
Partial fractions: \( \dfrac{1}{(r+1)(r+2)(r+3)} = \dfrac12\left(\dfrac{1}{(r+1)(r+2)} - \dfrac{1}{(r+2)(r+3)}\right) \).
Sum telescopes: \( \dfrac12\left[\dfrac{1}{2\cdot3} - \dfrac{1}{(n+2)(n+3)}\right] = \dfrac{1}{12} - \dfrac{1}{2(n+2)(n+3)} \).
Find the Maclaurin series for \( \dfrac{1}{1-x} \) up to \( x^3 \) and hence obtain the series for \( \dfrac{1}{(1-x)^2} \) up to \( x^2 \).
\( \dfrac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \) (\( |x|<1 \)).
Differentiate: \( \dfrac{1}{(1-x)^2} = 1 + 2x + 3x^2 + \cdots \).
Find the first four non-zero terms in the Maclaurin series for \( \ln(\cos x) \).
\( \cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \dfrac{x^6}{720} + \cdots \)
\( \ln(\cos x) = \ln\left(1 + \left(-\dfrac{x^2}{2} + \dfrac{x^4}{24} - \dfrac{x^6}{720}\right)\right) \).
Use \( \ln(1+u) = u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \cdots \) with \( u = -\dfrac{x^2}{2} + \dfrac{x^4}{24} - \dfrac{x^6}{720} \).
Collecting powers: \( -\dfrac{x^2}{2} - \dfrac{x^4}{12} - \dfrac{x^6}{45} + \cdots \)
Given that \( f(x) = \tan^{-1}x \), find the Maclaurin series for \( f(x) \) up to and including the term in \( x^5 \).
\( f(0)=0 \); \( f' = \dfrac{1}{1+x^2} \), \( f'(0)=1 \).
\( f'' = \dfrac{-2x}{(1+x^2)^2} \), \( f''(0)=0 \).
\( f''' = \dfrac{-2(1+x^2)^2 + 8x^2(1+x^2)}{(1+x^2)^4} \), \( f'''(0)=-2 \).
Continuing, \( f^{(5)}(0)=24 \). Series: \( x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots \).
Use successive differentiation to find the Maclaurin series for \( (1+x)^{-1/2} \) up to the term in \( x^3 \).
\( f= (1+x)^{-1/2} \), \( f(0)=1 \)
\( f' = -\dfrac12(1+x)^{-3/2} \), \( f'(0)=-\dfrac12 \)
\( f'' = (-\dfrac12)(-\dfrac32)(1+x)^{-5/2} = \dfrac34(1+x)^{-5/2} \), \( f''(0)=\dfrac34 \)
\( f''' = \dfrac34(-\dfrac52)(1+x)^{-7/2} = -\dfrac{15}{8}(1+x)^{-7/2} \), \( f'''(0)=-\dfrac{15}{8} \)
Series: \( 1 - \dfrac12 x + \dfrac{3}{8}x^2 - \dfrac{5}{16}x^3 + \cdots \)
Find the first three non-zero terms in the expansion of \( \dfrac{\sin x}{e^x - 1} \) as a power series in \( x \).
\( e^x - 1 = x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \cdots \)
\( \sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots \)
Write \( \dfrac{\sin x}{e^x-1} = \dfrac{x(1 - x^2/6 + \cdots)}{x(1 + x/2 + x^2/6 + \cdots)} = \dfrac{1 - x^2/6 + \cdots}{1 + x/2 + x^2/6 + \cdots} \).
Divide series: \( 1 - \dfrac{x}{2} + \dfrac{x^2}{12} + \cdots \)