represent $\cos(x^{1/2})$ by Maclaurin series I need to represent $\cos(x^{1/2})$ by Maclaurin series. 
I'm not sure that what I have done is correct. We know Maclaurin series for $$\cos(x) = 1-  {x^2 \over 2!} +{x^4\over  4!}+\cdots$$  So I substitue $x^{1/2}$, and I got $$\cos(x^{1/2})= 1 - {x\over 2!} + {x^2\over4!}+\cdots$$
But when I check it, with $x=16$, I don't get the correct answer. Where is the problem? Thanks.
 A: What you did is perfectly valid. But note the equality is obtained using the infinite series.
$$
\cos(x^{1/2})= 1 - {x\over 2!} + {x^2\over4!}+\cdots
$$
If you just take a few terms of the series, you get an approximation:
$$
\cos(x^{1/2})\approx 1 - {x\over 2!} + {x^2\over4!}.
$$
And the larger $x$ is, for a fixed partial sum, the worse the approximation becomes. 
Using the first three terms, with   $x=16$, you have:
$$
\cos(16^{1/2})-( 1 - {16\over 2!} + {16^2\over4!})\approx -4.32.
$$
But if you go out to the $8$th term, the error is small:
$$
\cos(16^{1/2})-( 1 - {16\over 2!} + {16^2\over4!} +\cdots- {16^7\over 14!})\approx.000195.
$$
Using even more terms will give you even smaller errors.
A: This is correct, how did you check it? 
I obtain : 


*

*$\cos(4) =-0.6536$

*$1 - 16/2 + 16^2/4! = 3,67$

*$1 - 16/2 + 16^2/4! - 16^3/6! = -2.022$

*$1 - 16/2 + 16^2/4! - 16^3/6! + 16^4/8! = - 0.39$

*$\vdots$

*$1 - 16/2 + 16^2/4! + \cdots - 16^7/14! = - 0.6538$


For $x$ as big as 16, the convergence is not as fast as you may think. It is much faster with small values of $x$.
A: (Update: It can be answered after all. Please ignore this answer. For negative x, the identity $\text{cos}\sqrt{x} =  \text{cosh}\sqrt{-x}$ behaves nicely and it's possible to fill in the singularity.) 
This question cannot be answered. The function $\text{cos}(x^\frac12)$ is not continuous at 0 and hence it is not meaningful to think about a Maclaurin series. 
You would be attempting to take a square root of a negative number. And complex analysis won't help here either, as a Maclaurin it needs to be infinitely differentiable at 0, and it is not.
You could take the Taylor series at any positive value.
