Question

How can you find the Taylor expansion of $\sin x$ about$x=0$?

Answer 1

From the given question, we have been asked to find the Taylor expansion of $\sin x$ about$x=0$. To find the Taylor expansion for the given question, first of all, we have to write the formula for the Taylor’s expansion. Formula for Taylor expansion: $\sum\limits_{n=0}^{N}{\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}$

Complete step-by-step solution:
The Taylor series around $a=0$ (not $x=0$….. the question is technically off) we can write it then as:
$\sum\limits_{n=0}^{N}{\dfrac{{{f}^{\left( n \right)}}\left( 0 \right)}{n!}{{\left( x \right)}^{n}}}=\dfrac{f\left( 0 \right)}{0!}{{x}^{0}}+\dfrac{{f}'\left( 0 \right)}{1!}{{x}^{1}}+\dfrac{{f}''\left( 0 \right)}{2!}{{x}^{2}}+.................$
So, you can clearly observe that you have to take some derivatives.
$\sin x$ has cyclic derivatives that follow this pattern:

& \sin x={{f}^{\left( 0 \right)}}\left( x \right)=f\left( x \right) \\\ & \dfrac{d}{dx}\left( \sin x \right)=\cos x={f}'\left( x \right) \\\ & \dfrac{d}{dx}\left( \cos x \right)=-\sin x={f}''\left( x \right) \\\ & \dfrac{d}{dx}\left( -\sin x \right)=-\cos x={f}'''\left( x \right) \\\ & \dfrac{d}{dx}\left( -\cos x \right)=\sin x={{f}'''}'\left( x \right) \\\ \end{aligned}$$ Finally you can write the whole thing out, knowing that when $$trig\left( 0 \right)=0$$, the whole term disappears. $$\sin x$$ appears in every derivative. Hence, every even derivative appears. You have only odd terms to worry about, and those are all just one in the numerator and the signs alternate due to the alternating signs in front of $$\cos x$$. Now, the expansion can be written as,$$\sum\limits_{n=0}^{N}{\dfrac{{{f}^{\left( n \right)}}\left( 0 \right)}{n!}{{\left( x \right)}^{n}}}=\dfrac{f\left( 0 \right)}{0!}{{x}^{0}}+\dfrac{{f}'\left( 0 \right)}{1!}{{x}^{1}}+\dfrac{{f}''\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{f}'''\left( 0 \right)}{3!}{{x}^{3}}.................$$ As we have already discussed above, the even terms will get cancelled. Now, the expansion can be written as, $$\cos \left( 0 \right)x+\dfrac{\left( -\cos \left( 0 \right) \right){{x}^{3}}}{6}+\dfrac{\cos \left( 0 \right){{x}^{5}}}{120}+\dfrac{\left( -\cos \left( 0 \right) \right){{x}^{7}}}{5040}+.......$$ $$\Rightarrow x-\dfrac{{{x}^{3}}}{6}+\dfrac{{{x}^{5}}}{120}-\dfrac{{{x}^{7}}}{5040}+...........$$ Hence, we got the Taylor expansion for the given question. **Note:** We should be well aware of the concept of Taylor expansion. Also, we should be very careful while finding the derivatives. We should be very careful while finding the expansions. We should be very careful while calculating the value of factorial numbers. Also, we should be very well known about the derivatives. We should be very careful while writing the expansion. Similarly for $\cos x$ it is given as $1-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{4}}}{24}-\dfrac{{{x}^{6}}}{120}+...........$