# Set up Taylor expansion #\bb\color(red)\text(formula)# for #sqrt(x)# around #a=2#?

##
Check my work please?

#f'=(1/2)x^(-1/2)#

#f''=(1/2)(-1/2)x^(-3/2)#

#f'''=(1/2)(-1/2)(-3/2)x^(-5/2)#

#f^4=(1/2)(-1/2)(-3/2)(-5/2)x^(-7/2)#

#f^5=(1/2)(-1/2)(-3/2)(-5/2)(-7/2)x^(-9/2)#

#f'(2)=(1/2)*2^(-1/2)#

#f''(2)=(1/2)(-1/2)*2^(-3/2)#

#f''(2)'=(1/2)(-1/2)(-3/2)*2^(-5/2)#

#f^4(2)=(1/2)(-1/2)(-3/2)(-5/2)*2^(-7/2)#

#f^5(2)=(1/2)(-1/2)(-3/2)(-5/2)(-7/2)*2^(-9/2)#

Okay, I don't know where to go from here.

but I DO know that the exponents increase in this order...

#(-1/2)+(-1)^\beta# (not sure if #\beta=n# yet)

Check my work please?

Okay, I don't know where to go from here.

but I DO know that the exponents increase in this order...

##### 1 Answer

#### Explanation:

For this function, finding the

We will not evaluate the derivatives at

If we look only at the derivatives and disregard

That is, **we start with #n=1.#**

The first derivative has no negative sign, the second derivative is negative, and this pattern continues for the following derivatives. Since we start at

The exponent on the

In the denominator, we

Now, for the numerator, we see we have the pattern

We can represent this with the **double factorial,**

Thus,

Now, we use the general form for a Taylor series centered at

to see that

Note that we did not include

We can't simplify the factorials any further.