## - 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 $$x$$
$$\displaystyle \ln⁡3=\ln⁡\left({1+{1\over2}\over 1-{1\over2}}\right)$$
$$\displaystyle =2\left({1\over2}\right)+{2\over3} \left({1\over2}\right)^3+{2\over5} \left({1\over2}\right)^5$$
$$=1.096$$