- Taylor's Series

Here are the conditions needed for Taylor’s Series

  • The function \(f(x)\) has to be infinitely differentiable
  • The function \(f(x)\) has to be defined in a region near the value \(x=a\).
Maclaurin’s Series is:
\(\displaystyle g(x)=g(0)+xg'(0)+{x^2\over2!} g''(0)+{x^3\over3!}g'''(0)+⋯+{x^n\over n!} g^n(0)+⋯\)
For Taylor’s Series, we let \(f(x+a)=g(x)\)
\(\displaystyle f(x+a)=f(a)+f'(a)x+{f''(a)\over2!} x^2+{f'''(a)\over3!} x^3+⋯+{f^n (a)\over n!} x^n+⋯\)
Furthermore, replacing \(x\) with \(x-a\) we get
\(\displaystyle f(x)=f(a)+f'(a)(x-a)+{f''(a)\over2!} (x-a)^2+{f'''(a)\over3!} (x-a)^3+⋯+{f^n (a)\over n!} (x-a)^n+⋯\)
\(\displaystyle f(x)=\sum_{k=0}^\infty {f^k (a)\over k!} (x-a)^k\)

LESSON 1
Find the first four non – zero terms for the Taylor expansion of \(\displaystyle {1\over x+3}\) with centre \(a=1\).
SOLUTION
Let \(\displaystyle f(x)={1\over x+3}=(x+3)^{-1}\)
  • Determine needed values
\(\displaystyle f(x)={1\over x+3} \to f(1)={1\over4}\)
\(\displaystyle f'(x)=-{1\over (x+3)^2} \to f'(1)=-{1\over16}\)
\(\displaystyle f''(x)={2\over (x+3)^3} \to f''(1)={1\over32}\)
\(\displaystyle f'''(x)=-{6\over(x+3)^4} \to f'''(1)=-{3\over128}\)
  • Substitute values into formula
\(\displaystyle f(x+a)=f(a)+f'(a)(x-a)+{f''(a)\over2!} (x-a)^2+{f'''(a)\over3!} (x-a)^3+⋯+{f^n (a)\over n!}(x-a)^n+⋯\)
\(\displaystyle {1\over x+3}={1\over4}-{1\over16} (x-1)+{1\over32} \left({1\over2!}\right) (x-1)^2-{3\over128} \left({1\over3!}\right) (x-1)^3\)
\(\displaystyle ={1\over4}-{1\over16} (x-1)+{1\over64} (x-1)^2-{3\over256} (x-1)^3\)

LESSON 2
(i) Obtain the first four non – zero terms of the Taylor Series expansion of sin⁡x in ascending powers of \(\displaystyle \left(x-{\pi\over4}\right)\).
(ii) Hence, calculate an approximation to \(\displaystyle \sin\left({π\over16}\right)\).
SOLUTION
  • Determine the value of \(a\)
Let \(f(x)=\sin⁡ x⟹f(x-a)=\sin⁡(x-a)\)
\(\displaystyle ⟹f\left(x-{\pi\over4}\right)=\sin⁡(x-a)\)
\(\displaystyle ⟹a={π\over4}\)
  • Determine the derivatives
\(\displaystyle f(x)=\sin⁡x\to f\left({\pi\over4}\right)=\sin\left({\pi\over4}\right)={\sqrt{2}\over2}\)
\(\displaystyle f'(x)=\cos ⁡x \to f'\left({π\over4}\right)=\cos\left({π\over4}\right)={\sqrt{2}\over2}\)
\(\displaystyle f''(x)=-\sin⁡ x \to f''\left({π\over4}\right)=-\sin\left({π\over4}\right)=-{\sqrt{2}\over2}\)
\(\displaystyle f''(x)=-\cos ⁡x \to f'''(x)=-\cos\left({π\over4}\right)=-{\sqrt{2}\over2}\)
\(\displaystyle f(x)=f(a)+f'(a)(x-a)+{f''(a)\over 2!} (x-a)^2+{f'''(a)\over3!} (x-a)^3+⋯+{f^n (a)\over n!} (x-a)^n\)
\(\displaystyle \sin⁡ x={\sqrt{2}\over2}+{\sqrt{2}\over2} \left(x-{π\over4}\right)-{\sqrt{2}\over2} \left({1\over2!}\right) \left(x-{π\over4}\right)^2-{\sqrt{2}\over2} \left({1\over3!}\right) \left(x-{π\over4}\right)^3\)
\(\displaystyle ={\sqrt{2}\over2}+{\sqrt{2}\over2} \left(x-{π\over4}\right)-{\sqrt{2}\over4} \left(x-{π\over4}\right)^2-{\sqrt{2}\over12} \left(x-{π\over4}\right)^3\)

(ii) \(\displaystyle \sin\left({π\over16}\right)⟹x={π\over16}\)
Now,
\(\displaystyle \left(x-{π\over4}\right)=\left({π\over16}-{π\over4}\right)\)
\(\displaystyle ⟹\left(x-{π\over4}\right)=-{3π\over16}\)
  • Replace \(\displaystyle \left(x-{π\over4}\right)\) with \(\displaystyle -{3π\over16}\)
\(\displaystyle \sin\left({π\over16}\right)={\sqrt{2}\over2}+{\sqrt{2}\over2} \left(-{3π\over16}\right)-{\sqrt{2}\over4} \left(-{3π\over16}\right)^2-{\sqrt{2}\over12} \left(-{3π\over16}\right)^3\)
\(=0.1920\)