Find the Taylor series for f(x) centered at the given value a
f(x) = sin x, a = π/2
Also prove that the series obtained represents sin x for all x.
Thanks
Find the Taylor series for f(x) centered at the given value a
f(x) = sin x, a = π/2
Also prove that the series obtained represents sin x for all x.
Thanks
Of course, you can try to get the taylor series from its very definition $\displaystyle \sum_{k=0}^\infty \frac{f^{(k)}(n/2)}{k!}(x-n/2)^k$, but it just might be easier do use a bit of trigonometry to combine the known Taylor series for sin and cos:
$\displaystyle f(x)=\sin(x)=\sin\left(\frac{n}{2}+\left(x-\frac{n}{2}\right)\right)=\sin\frac{n}{2}\cdot\cos \left(x-\frac{n}{2}\right)+\cos\frac{n}{2}\cdot\sin\left(x-\frac{n}{2}\right)$
$\displaystyle =\sin\frac{n}{2}\cdot\sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}\left(x-\frac{n}{2}\right)^{2k}+\cos\frac{n}{2}\cdot\sum_{ k=0}^\infty \frac{(-1)^k}{(2k+1)!}\left(x-\frac{n}{2}\right)^{2k+1} =: \sum_{k=0}^\infty a_k\left(x-\frac{n}{2}\right)^k$
where
$\displaystyle a_k := \begin{cases} \sin\frac{n}{2}\cdot \frac{(-1)^{k/2}}{k!}, & \text{ if } k \text{ is even}\\
\cos\frac{n}{2}\cdot\frac{(-1)^{(k-1)/2}}{k!}, & \text{ if } k \text{ is odd}
\end{cases}$
The problem is that you need to develop the Taylor series around $\displaystyle x=n/2$.
Assuming you know the Taylor series of sin and cos, one can try to use a little trigonometry, namely the Addition Formula for sin, to find the required Taylor series. And since I have written it up earlier, I am not going to repeat myself here.
Another way of figuring out the required Taylor series is, as I also wrote, to determine a general formula for the coefficent $\displaystyle a_k:= \frac{f^{(k)}(n/2)}{k!}$ of $\displaystyle (x-n/2)^k$ in that series. If I am not mistaken, you will find the same value for $\displaystyle a_k$ both ways. - You are free choose the approach that you like better.
Failure's line here
$\displaystyle f(x)=\sin(x)=\sin\left(\frac{n}{2}+\left(x-\frac{n}{2}\right)\right)=\sin\frac{n}{2}\cdot\cos \left(x-\frac{n}{2}\right)+\cos\frac{n}{2}\cdot\sin\left(x-\frac{n}{2}\right)$
Is a regular formula (i would learn it!!) and is similar to completing the square. When we complete the square we don't change the value of the equation, we simply manipulate the information to get a desired result.
For example,
$\displaystyle x^2 - 2x - 5 = 0$
To factor this we do
$\displaystyle (x^2 -2x + 1) - 6 = 0$
$\displaystyle (x-1)^2 - 6 = 0$
If you expand that out we have
$\displaystyle x^2 - 2x - 5 = 0$
Same principle applies
No, it doesn't, but in that case you can simplify further by replacing $\displaystyle \sin(n/2)$ width $\displaystyle \sin(\pi/2)=1$.
And you can replace $\displaystyle \cos(n/2)$ by $\displaystyle \cos(\pi/2)=0$ as well, so all odd terms of the Taylor series will be 0.
Overall, with these simplifications you get:
$\displaystyle \sin(x)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\left(x-\frac{\pi}{2}\right)^{2k}$
Which is a roundabout way of arriving at a solution that you probably knew all along, namely that $\displaystyle \sin(x)=\cos\left(x-\frac{\pi}{2}\right)$.