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Maclaurin Series
IB Mathematics AA · HL · Shadow Worksheet
Practice
Name
Question 1
Consider \(f(x)=\dfrac{1}{1+3x}\).
(a)
Find \(f'(x),f''(x),f^{(3)}(x)\).
(b)
Hence find the Maclaurin series for \(f(x)\), writing your answer as \(\displaystyle\sum_{n=0}^{\infty}g(x)\).
(c)
State the range of \(x\) for which the series is valid.
(a)
Answer
\(f'=-\dfrac{3}{(1+3x)^2};\;f''=\dfrac{18}{(1+3x)^3};\;f'''=-\dfrac{162}{(1+3x)^4}\)
(b)
Maclaurin: \(f(0)=1,\;f'(0)=-3,\;f''(0)=18,\ldots\)
\(f(x)=1-3x+9x^2-27x^3+\ldots\)
Answer
\(\displaystyle\sum_{n=0}^{\infty}(-3)^n x^n=\displaystyle\sum_{n=0}^{\infty}(-1)^n 3^n x^n\)
(c)
Answer
\(|x|<\dfrac{1}{3}\)
Question 2
Find the Maclaurin expansion of \(\cos(x^2)\) up to and including the term in \(x^8\).
Use standard series
\(\cos u=1-\dfrac{u^2}{2!}+\dfrac{u^4}{4!}-\ldots\) with \(u=x^2\)
Answer
\(1-\dfrac{x^4}{2}+\dfrac{x^8}{24}-\ldots\)
Question 3
Use the Maclaurin series for \(\sin x\) to evaluate \(\displaystyle\lim_{x o0}\dfrac{\sin x - x}{x^3}\).
Series
\(\sin x=x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\ldots\Rightarrow\dfrac{\sin x-x}{x^3}=\dfrac{-x^3/6+\ldots}{x^3}=-\dfrac{1}{6}+O(x^2)\)
Answer
\(-\dfrac{1}{6}\)
Question 4
Find the Maclaurin series for \(e^x\sin x\) up to and including the term in \(x^4\).
Multiply series
\(e^x=1+x+ frac{x^2}{2}+ frac{x^3}{6}+ frac{x^4}{24}+\ldots\); \(\sin x=x- frac{x^3}{6}+\ldots\)
Multiply term by term
\(x+x^2+ frac{x^3}{2}- frac{x^3}{6}+ frac{x^4}{6}- frac{x^4}{6}+\ldots=x+x^2+ frac{x^3}{3}+0\cdot x^4\)
Answer
\(x+x^2+\dfrac{x^3}{3}+0\cdot x^4+\ldots\)
Generated by Mathski · mathski.io · IB Mathematics AA Shadow Worksheets