- MaClaurin's Series

The Maclaurin’s series expansion states that

\[f(x)=f(0)+xf'(0)+{x^2\over2!} f''(0)+{x^3\over3!} f'''(0)+⋯+{x^n\over n!}f^n(0)+⋯\]
Here are the conditions which must be satisfied for the Maclaurin’s series expansion
  • \(f(x)\) must be differentiable
  • \(f(x)\) must exist at \(x=0\)
  • The derivatives of \(f(x)\) must exist at \(x=0\)
NB: Only within specific values of \(x\) is the series valid.

LESSON 1
Use Maclaurin’s Theorem to find the first four non – zero terms for \(\cos ⁡x\), hence determine an approximation for \(\cos⁡(0.2)\).
SOLUTION
  • Determine required derivatives.
\(f(x)=\cos⁡ x \to f(0)=\cos⁡(0)=1\)
\(f'(x)=-\sin⁡ x \to f'(0)=-\sin⁡(0)=0\)
\(f''(x)=-\cos⁡ x \to f''(0)=-\cos⁡(0)=-1\)
\(f'''(x)=\sin ⁡x \to f'''(0)=\sin⁡(0)=0\)
\(f''''(x)=\cos ⁡x \to f''''(0)=\cos⁡(0)=1\)
\(f'''''(x)=-\sin ⁡x \to f'''''(0)=-\sin⁡(0)=0\)
\(f''''''(x)=-\cos ⁡x \to f''''''(0)=-\cos⁡(0)=-1\)
  • Substitute values into formula and simplify.
\(\displaystyle f(x)=f(0)+xf'(0)+{x^2\over2!} f''(0)+{x^3\over3!} f'''(0)+⋯+{x^n\over n!} f^n(0)+⋯\)
\(\displaystyle f(x)=1+(0)x+(-1){x^2\over2!}+(0){x^3\over3!}+(1){x^4\over4!}+(0){x^5\over5!}+(-1){x^6\over6!}\)
\(\displaystyle =1-{x^2\over2!}+{x^4\over4!}-{x^6\over6!}+⋯\)
  • Substitute \(x=0.2\)
\(\displaystyle f(0.2)=1-{(0.2)^2\over2}+{(0.2)^4\over24}-{(0.2)^6\over720}\)
\(=0.98\)

LESSON 2
Find the Maclaurin expansion for \((1+x)^2 \cos ⁡x\) up to and including the term in \(x^3\).
SOLUTION
  • Expand brackets
\(f(x)=(1+x)^2 \cos ⁡x\)
\(=(1+2x+x^2 ) \cos ⁡x\)
  • Use expansion for \(\cos ⁡x\) from example above.
\(\displaystyle ⟹(1+2x+x^2)\left(1-{x^2\over2!}+...\right)\)
  • Expand so that we have all terms up to \(x^3\)
\(\displaystyle ⟹1-{x^2\over2!}+2x-x^3+x^2+⋯\)
\(\displaystyle ⟹1+2x+{1\over2} x^2-x^3\)

LESSON 3
Find the Maclaurin’s series for \(\tan ⁡x\) up to \(x^3\).
SOLUTION
  • Determine required derivatives.
\(f(x)=\tan⁡ x \to f(0)=\tan⁡(0)=0\)
\(f'(x)=\sec^2⁡x \to f'(0)=\sec^2⁡(0)=1\)
\(f''(x)=2 \tan ⁡x \sec^2⁡x \to f''(0)=2 \tan⁡(0) \sec^2⁡(0)=0\)
\(f'''(x)=2 \sec^4⁡x+4 \sec^2⁡x \tan^2⁡x \to f'''(0)=2 \sec^4⁡(0)+4 \sec^2⁡0 \tan^2⁡(0)=2\)
  • Substitute needed values into formula
\(\displaystyle f(x)=f(0)+xf'(0)+{x^2\over2!} f''(0)+{x^3\over3!} f'''(0)+⋯+{x^n\over n!}f^n(0)+⋯\)
\(\displaystyle f(x)=0+x+(0) {x^2\over2!}+2 {x^3\over3!}+⋯\)
\(\displaystyle=x+{2x^3\over3!}\)

LESSON 4
A function is defined as \(g(t)=e^{2t+1}\).
(i) Obtain the Maclaurin’s series expansion for \(g(t)\) up to and including the term in \(t^4\).
(ii) Hence, estimate \(g(0.1)\) to four decimal places.
SOLUTION
  • Determine needed derivatives
\(g(t)=e^{2t+1}\to g(0)=e^1\)
\(g'(t)=2e^{2t+1} \to g'(0)=2e^1\)
\(g''(t)=4e^{2t+1} \to g''(0)=4e^1\)
\(g'''(t)=8e^{2t+1} \to g'''(0)=8e^1\)
\(g''''(t)=16e^{2t+1} \to g''''(0)=16e^1\)
  • Substitute needed values into the formula
\(\displaystyle g(t)=g(0)+xg'(0)+{t^2\over2!} g''(0)+{t^3\over3!} g'''(0)+⋯+{t^n\over n!} g^n(0)\)
\(\displaystyle =e+2e t+4e {t^2\over2!}+8e {t^3\over3!}+16e {t^4\over4!}\)
\(\displaystyle =e+2e t+2e t^2+{4e\over3} t^3+{2e\over3} t^4\)
  • Substitute \(t=0.1\) into equation
\(\displaystyle g(0.1)=e+2e(0.1)+2e(0.1)^2+{4e\over3} (0.1)^3+{2e\over3} (0.1)^4\)
\(=3.3201\)

LESSON 5
You are given that \(\displaystyle f(x)={1\over (1-3x)^2}\).
Find \(f'(x), f''(x)\) and \(f'''(x)\). Hence obtain the Maclaurin series for \(f(x)\) as far as the term in \(x^3\).
By considering the equivalent binomial expansion, give the set of values of \(x\) for which the Maclaurin series is valid.
SOLUTION
  • Determine the needed derivatives.
\(f(x)=(1-3x)^{-2}\)
\(⟹f(0)=1\)
\(f'(x)=-2(1-3x)^{-3} (-3)=6(1-3x)^{-3}\)
\(⟹f'(0)=6\)
\(f''(x)=-18(1-3x)^{-4} (-3)=54(1-3x)^{-4}\)
\(⟹f''(0)=54\)
\(f'''(x)=-96(1-2x)^{-5} (-3)=288(1-3x)^{-5}\)
\(⟹f'''(0)=288\)
  • Substitute needed values into the equation.
\(\displaystyle f(x)=f(0)+xf'(0)+{x^2\over2!} f''(0)+{x^3\over3!}f'''(0)+⋯+{x^n\over n!} f^n(0)+⋯\)
\(\displaystyle f(x)=1+6x+36 \left({x^2\over4}\right)+288 \left({x^3\over6}\right)\)
\(\displaystyle =1+6x+9x^2+48x^3\)
  • Determine the values of \(x\) for which validity exist.
Valid for
\(-1<-3x<1\)
\(\displaystyle -{1\over3}<x<{1\over3}\)

LESSON 6
(i) Use the Maclaurin series for \(\ln⁡(1+x)\) and \(\ln⁡(1-x)\) to obtain the first three non – zero terms in the Maclaurin series for \(\displaystyle \ln⁡\left({1+x\over 1-x}\right)\). State the range of validity for this series.
(ii) Find the value of \(x\) for which \(\displaystyle {1+x\over 1-x}=3\). Hence find an approximation to \(\ln⁡3\), giving your answer to three decimal places.
SOLUTION
  • Determine the needed derivatives.
\(f(x)=\ln⁡(1+x) \to f(0)=\ln⁡(1)=0\)
\(\displaystyle f'(x)={1\over 1+x} \to f'(0)={1\over 1+0}=1\)
\(\displaystyle f''(x)=-{1\over (1+x)^2} \to f''(0)=-{1\over (1+0)^2}=-1\)
\(\displaystyle f'''(x)={2\over (1+x)^3} \to f'''(0)={2\over (1+0)^3}=2\)
\(\displaystyle f'''' (x)=-{6\over (1+x)^4} \to f''''(0)=-{6\over (1+0)^4}=-6\)
\(\displaystyle f'''''(x)={24\over (1+x)^5} \to f'''''(0)={24\over (1+0)^5}=24\)
  • Substitute needed values into formula
\(\displaystyle f(x)=f(0)+xf'(0)+{x^2\over2!} f''(0)+{x^3\over3!} f'''(0)+⋯+{x^n\over n!} f^n (0)+⋯\)
\(\displaystyle f(x)=0+x-{1/2!}x^2+(2) {x^3\over3!}-(6) {x^4\over4!}+(24) {x^5\over5!}+⋯\)
\(\displaystyle =x-{x^2\over2}+{x^3\over3}-{x^4\over4}+{x^5\over5}+⋯\)
  • Determine the needed derivatives
\(f(x)=\ln⁡(1-x) \to f(0)=\ln⁡(1)=0\)
\(\displaystyle f'(x)=-{1\over 1-x} \to f'(0)=-{1\over 1-0}=-1\)
\(\displaystyle f''(x)=-{1/(1-x)^2} \to f''(0)=-{1\over (1-0)^2}=-1\)
\(\displaystyle f'''(x)=-{2\over(1-x)^3} \to f'''(0)=-{2\over(1-0)^3}=-2\)
\(\displaystyle f''''(x)=-{6\over(1-x)^4} \to f''''(0)=-{6\over (1-0)^4}=-6\)
\(\displaystyle f'''''(x)=-{24\over (1-x)^5} \to f'''''(0)=-{24\over (1-0)^5}=-24\)
  • Substitute required values into formula
\(\displaystyle f(x)=f(0)+xf'(0)+{x^2\over2!} f''(0)+{x^3\over3!} f'''(0)+⋯+{x^n\over n!}f^n(0)+⋯\)
\(\displaystyle f(x)=0-x-{x^2\over2!}-(2) {x^3\over3!}-(6) {x^4\over4!}-(24) {x^5\over5!}+⋯\)
\(\displaystyle =-x-{x^2\over2}-{x^3\over3}-{x^4\over4}-{x^5\over5}\)
  • Use properties of logarithms
\(\displaystyle \ln⁡\left({1+x\over 1-x}\right)=\ln⁡(1+x)-\ln⁡(1-x)\)
\(\displaystyle =\left(x-{x^2\over2}+{x^3\over3}-{x^4\over4}+{x^5\over5}+⋯\right)-\left(-x-{x^2\over2}-{x^3\over3}-{x^4\over4}-{x^5\over5}\right)\)
\(\displaystyle =2x+{2x^3\over3}+{2x^5\over5}\:\:\:\:\:-1<x<1\)
(ii) Solve for \(x\)
\(\displaystyle {1+x\over 1-x}=3\)
\(1+x=3-3x\)
\(4x=2\)
\(\displaystyle x={1\over2}\)
  • Substitute value of \(x\)
\(\displaystyle \ln⁡3=\ln⁡\left({1+{1\over2}\over 1-{1\over2}}\right)\)
\(\displaystyle =2\left({1\over2}\right)+{2\over3} \left({1\over2}\right)^3+{2\over5} \left({1\over2}\right)^5\)
\(=1.096\)