Question

For the function $\sin \pi x$ centred at $a=0.5$.using taylor series expansion,find approximate value of $\sin (\frac{\pi }{2}+\frac{\pi }{10})$

• A. $0.9511$
• B. $0.9633$
• C. $0.8962$
• D. $0.2134$

Answer 1

The correct option is A $0.9511$

The taylor series is given by ${\sum }_{k=0}^{\infty }\frac{f^{\left(k\right)}\left(a\right)}{k!}(x-a{)}^k$

$f\left(x\right)\approx {\sum }_{k=0}^n\frac{f^{\left(k\right)}\left(a\right)}{k!}(x-a{)}^k={\sum }_{k=0}^2\frac{f^{\left(k\right)}\left(a\right)}{k!}(x-a{)}^k$

Finally after simplifying, we get

$f\left(x\right)\approx \frac{1}{0!}{(x-\frac{1}{2})}^0+\frac{0}{1!}{(x-\frac{1}{2})}^1+\frac{-(\pi {)}^2}{2!}(x-\frac{1}{2}{)}^2$

$f\left(x\right)=1-\frac{(\pi {)}^2}{2}(x-\frac{1}{2}{)}^2$

$\sin \left(\pi x\right)=1-\frac{{\pi }^2}{2}{(x-\frac{1}{2})}^2$

$\sin \left(\pi (\frac{1}{2}+\frac{1}{10})\right)=1-\frac{{\pi }^2}{2}{(\frac{1}{2}+\frac{1}{10}-\frac{1}{2})}^2=0.95065\approx 0.951$