10 exam-style questions on Taylor expansions about a point, limits, approximations and applications to differential equations. Tap Show answer for full working.
Find the Taylor series expansion of \(f(x)=\mathrm{e}^{x}\) about \(x=1\), up to and including the term in \((x-1)^3\).
\(f(1)=\mathrm{e}\), \(f'=f''=f'''=\mathrm{e}^{x}\), so all derivatives at \(1\) are \(\mathrm{e}\).
\(f(x)=\mathrm{e}+\mathrm{e}(x-1)+\dfrac{\mathrm{e}}{2}(x-1)^2+\dfrac{\mathrm{e}}{6}(x-1)^3+\cdots\).
Write down the Taylor series for \(\ln(1+x)\) about \(x=0\) up to the term in \(x^4\). Hence find an approximation for \(\ln 1.1\).
\(\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots\) for \(|x|<1\).
\(\ln 1.1\approx0.1-\dfrac{0.01}{2}+\dfrac{0.001}{3}-\dfrac{0.0001}{4}=0.1-0.005+0.000333-0.000025\approx0.0953\).
Find the first three non-zero terms in the Taylor expansion of \(\sin x\) about \(x=\dfrac{\pi}{2}\).
Let \(h=x-\pi/2\). \(\sin x=\sin(\pi/2+h)=\cos h=1-\dfrac{h^2}{2}+\dfrac{h^4}{24}-\cdots\).
So \(\sin x=1-\dfrac{1}{2}\bigl(x-\dfrac{\pi}{2}\bigr)^2+\dfrac{1}{24}\bigl(x-\dfrac{\pi}{2}\bigr)^4-\cdots\).
Find the Taylor series of \(f(x)=\dfrac{1}{1+x}\) about \(x=2\), up to the term in \((x-2)^3\).
\(f(2)=\dfrac{1}{3}\). \(f'(x)=-\dfrac{1}{(1+x)^2}\), \(f'(2)=-\dfrac{1}{9}\).
\(f''(x)=\dfrac{2}{(1+x)^3}\), \(f''(2)=\dfrac{2}{27}\). \(f'''(x)=-\dfrac{6}{(1+x)^4}\), \(f'''(2)=-\dfrac{6}{81}=-\dfrac{2}{27}\).
\(f(x)=\dfrac{1}{3}-\dfrac{1}{9}(x-2)+\dfrac{1}{27}(x-2)^2-\dfrac{1}{81}(x-2)^3+\cdots\).
Use the Taylor expansion of \(\mathrm{e}^{x}\) about \(x=0\) to find \(\displaystyle\lim_{x\to0}\dfrac{\mathrm{e}^{x}-1-x}{x^2}\).
\(\mathrm{e}^{x}=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots\). Numerator \(=\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots\). Limit \(=\dfrac{1}{2}\).
Find the first four terms of the Taylor series for \(\sqrt{1+x}\) about \(x=0\). Use it to approximate \(\sqrt{1.05}\).
Binomial or Taylor: \(1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3-\cdots\).
\(\sqrt{1.05}\approx1+\dfrac{1}{2}(0.05)-\dfrac{1}{8}(0.0025)+\dfrac{1}{16}(0.000125)=1+0.025-0.0003125+0.0000078125\approx1.024695\).
Given that \(y\) satisfies \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=x+y\) with \(y(0)=1\), find the Taylor series for \(y\) about \(x=0\) up to the term in \(x^3\).
\(y(0)=1\). \(y'=x+y\), so \(y'(0)=1\). Differentiate: \(y''=1+y'\), \(y''(0)=1+1=2\). \(y'''=y''\), \(y'''(0)=2\).
\(y=1+x+\dfrac{2}{2}x^2+\dfrac{2}{6}x^3+\cdots=1+x+x^2+\dfrac{1}{3}x^3+\cdots\).
Find the Taylor expansion of \(\tan x\) about \(x=0\) up to the term in \(x^5\). Hence find \(\displaystyle\lim_{x\to0}\dfrac{\tan x-x}{x^3}\).
\(\tan x=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\cdots\). Numerator \(=\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\cdots\). Limit \(=\dfrac{1}{3}\).
Use Leibniz’s theorem to find the \(n\)th derivative of \(y=x^2\mathrm{e}^{x}\). Hence write down the Taylor series of \(y\) about \(x=0\).
Leibniz: \((uv)^{(n)}=\sum_{r=0}^{n}\binom{n}{r}u^{(r)}v^{(n-r)}\). Let \(u=x^2\), \(v=\mathrm{e}^{x}\). \(u'=2x\), \(u''=2\), \(u^{(r)}=0\) for \(r\ge3\). \(v^{(k)}=\mathrm{e}^{x}\).
\(y^{(n)}=\binom{n}{0}x^2\mathrm{e}^{x}+\binom{n}{1}2x\mathrm{e}^{x}+\binom{n}{2}2\mathrm{e}^{x}=(x^2+2nx+n(n-1))\mathrm{e}^{x}\).
At \(0\): \(y^{(n)}(0)=n(n-1)\). Series: \(y=\sum\dfrac{n(n-1)}{n!}x^n=\sum\dfrac{x^n}{(n-2)!}\) for \(n\ge2\), i.e. \(x^2\mathrm{e}^{x}\) as expected.
Find the first three non-zero terms in the expansion of \(\dfrac{1}{\sqrt{1+2x}}\) about \(x=0\). State the range of validity.
Binomial: \((1+2x)^{-1/2}=1+\bigl(-\dfrac{1}{2}\bigr)(2x)+\dfrac{(-\frac12)(-\frac32)}{2}(2x)^2+\dfrac{(-\frac12)(-\frac32)(-\frac52)}{6}(2x)^3+\cdots\)
\(=1-x+\dfrac{3}{2}x^2-\dfrac{5}{2}x^3+\cdots\). Valid for \(|2x|<1\), i.e. \(|x|<\dfrac{1}{2}\).