How do you find the taylor series series for # (4-x)^(1/2)#?

To find the Taylor series for $(4-x)^{1/2}$, follow these steps:

1. Determine the function and the point about which you want to expand the Taylor series. In this case, the function is $(4-x)^{1/2}$ and the point is $x = 0$.
2. Find the derivatives of the function up to the desired order.
3. Evaluate the derivatives at the point of expansion ($x = 0$).
4. Write out the Taylor series using the formula:

$f(x) = f(a) + f'(a)(x-a) + \frac{{f''(a)}}{{2!}}(x-a)^2 + \frac{{f'''(a)}}{{3!}}(x-a)^3 + \cdots$

where $f(a)$ is the value of the function at the point of expansion and the derivatives are evaluated at that point.

For $(4-x)^{1/2}$:

• $f(0) = (4-0)^{1/2} = 2$
• $f'(x) = -\frac{1}{2}(4-x)^{-1/2}$
• $f''(x) = \frac{1}{4}(4-x)^{-3/2}$
• $f'''(x) = -\frac{3}{8}(4-x)^{-5/2}$

Now evaluate these derivatives at $x = 0$:

• $f'(0) = -\frac{1}{2}(4-0)^{-1/2} = -\frac{1}{4}$
• $f''(0) = \frac{1}{4}(4-0)^{-3/2} = \frac{1}{8}$
• $f'''(0) = -\frac{3}{8}(4-0)^{-5/2} = -\frac{3}{32}$

Plug these values into the Taylor series formula:

$(4-x)^{1/2} = 2 - \frac{1}{4}x + \frac{1}{8}x^2 - \frac{3}{32}x^3 + \cdots$

This is the Taylor series expansion for $(4-x)^{1/2}$ centered at $x = 0$.