I started the problem but I am confused how to set up.
f(x) = 5cos(x) f(11pi) = -5
f'(x) = -5sin(x) f'(11pi) = 0
f''(x) = -5cos(x) f''(11pi) = -5
f'''(x) = 5sin(x) f'''(11pi) = 0
f''''(x) = 5cos(x) f''''(11pi) = -5Taylor Series Begins as$$-5 + 0\cdot(x-11\pi) + \frac{5(x-11\pi)^2}{2!} + \frac{0\cdot(x-11\pi)^3}{3!} +\frac{-5(x-11\pi)^4}{4!} + \dots$$
I so far wrote it down as
$$=\sum_{n=0}^\infty {(-1)^{n+1}} \frac{(x-11\pi)^{2n}}{(2n)!}$$
I am suppose too have the 5 in there too but I have no idea how to include that, would somebody please be able to show me how to put the 5 into my answer???
$\endgroup$ 23 Answers
$\begingroup$$$=\sum_{n=0}^\infty {(-1)^{n+1}} \frac{5}{(2n)!} (x-11\pi)^{2n}$$
This is the final and correct answer hopefully
$\endgroup$ 1 $\begingroup$By $x=11\pi+y$ we have
$$5\cos(x)=5\cos (11\pi+y)=5\cos (\pi+y)=-5\cos(y)$$
then use the usual expansion for $\cos y$ at $y=0$ and then substitute $y=x-11\pi$.
$\endgroup$ $\begingroup$It is a good idea in such cases to write $x=a+h=11\pi+h$, then to transfer the expansion around $a$ (in the variable $x$) to an expansion around $0$ (in the "better" variable $h$), and here it is:$$ \begin{aligned} f(x) &= 5\cos(x) \\ &=5\cos(11\pi+h)=5\cos(\pi+h) \\ &=5\cos\pi\cdot\cos h-\dots=-5\cos h \\ &=-5\left(\frac 1{0!}-\frac 1{2!}h^2+\frac 1{4!}h^4-\frac 1{6!}h^6+\dots \right) \end{aligned} $$and now it is time to insert back $x-11\pi$ instead of $h$.
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