Is it possible to change the center of a taylor series by substitution? If we have the taylor series of some function $f(y)$ centered at $y=0$ and we substitute $y=x-a$ then will this give me the taylor series of the function centered at $a$ ?
 A: No. Because you get 
$$
\sum_{k=0}^\infty\frac{f^{(k)}(0)\,(x-a)^k}{k!}
$$
and not
$$
\sum_{k=0}^\infty\frac{f^{(k)}(a)\,(x-a)^k}{k!}
$$
Consider for instance the series for the sine at $0$:
$$
\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}.
$$
If you shift to $\pi/2$ you get 
$$
\sum_{k=0}^\infty\frac{(-1)^k(x-\pi/2)^{2k+1}}{(2k+1)!}.
$$
On the other hand, the Taylor series for the sine at $\pi/2$ is
$$
\sum_{k=0}^\infty\frac{(-1)^k(x-\pi/2)^{2k}}{(2k)!}.
$$
If you evaluate the first series at $\pi/2$ you get $0$, since $\sin(x-\pi/2)$ are $\pi/2$ is zero. If you evaluate the second series at $\pi/2$, you get $1$, which is $\sin \pi/2$. 
A more extreme example is $e^{-1/x^2}$ at $0$. Its Taylor series is $0$ for all terms. So if you shift it, you still get $0$. On the other hand, at any point other than zero the function is analytic and its Taylor series will be something else, but certainly not zero. 
A: No, but I think it is worth noting that you will get the Taylor series of the funcion $g(x)=f(x-a)$, centered at $a$.
$$g^{(n)}(x)=f^{(n)}(x-a)$$ so
$$g^{(n)}(a)=f^{(n)}(0)$$
this is a powerful method for finding some Taylor series (and power series representations) when the function has some shifting property. 
For example $e^{a+b}=e^a \cdot e^b$ leads to the Taylor series of $e^x$ centered at $a=1$ via the substitution $u=x-1$.
$$e^x=e^{u+1}=e \cdot e^u= e \sum_{n=0}^\infty \frac{u^n}{n!}= \sum_{n=0}^\infty \frac{e \cdot u^n}{n!}= \sum_{n=0}^\infty \frac{e \cdot (x-1)^n}{n!}$$
Something like $\sin(x)$ centered at $\theta$ could be done in a similar fashion. Via $u= x - \theta$.
$$\sin(x) =\sin(u+\theta)=\sin(u) \cos(\theta)+\cos(u) \sin(\theta) \\ = \sum_{n=0}^\infty \cos(\theta) \frac{(-1)^nu^{2n+1}}{(2n+1)!}+\sum_{n=0}^\infty \sin(\theta) \frac{(-1)^nu^{2n}}{(2n)!} $$
so
$$\sin(x)=\sum_{n=0}^\infty \cos(\theta) \frac{(-1)^n(x-a)^{2n+1}}{(2n+1)!}+\sum_{n=0}^\infty \sin(\theta) \frac{(-1)^n(x-a)^{2n}}{(2n)!} $$ 
