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)

1 Answer
May 6, 2018

#sqrtx=sqrt2+sum_(n=1)^oo(-1)^(n+1)(x-2)^n(2n-1)!!2^(1/2-2n)/(n!)#

Explanation:

For this function, finding the #nth# derivative pattern will be a bit tougher, but it's possible.

We will not evaluate the derivatives at #a=2# just yet. We'll focus on finding #f^((n))(x):#

#f(x)=x^(1/2), f(2)=sqrt2#

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

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

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

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

If we look only at the derivatives and disregard #f(x)=x^(1/2),# we have a pattern going.

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 #n=1,# we can represent this with #(-1)^(n+1)#.

The exponent on the #x# starts with #-1/2,# and decreases by #1# for every following derivative. We can represent this with #x^(1/2-n), n>=1.#

In the denominator, we #2^n.#

Now, for the numerator, we see we have the pattern #1*3*5*7*...*(2n-1)#, that is, we multiply only odd numbers, starting at #1.#

We can represent this with the double factorial, #(2n-1)!!#, which indicates multiplying by integers that have the same parity (oddness or evenness) as #n#

Thus,

#f^((n))(x)=(-1)^(n+1)(2n+1)!!x^(1/2-n)/(2^n)#

#f^((n))(2)=(-1)^(n+1)(2n+1)!!2^(1/2-n)/(2^n)#

#2^(1/2-n)/(2^n)=2^(1/2-n-n)=2^(1/2-2n)#

#f^((n))(2)=(-1)^(n+1)(2n+1)!!2^(1/2-2n)#

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

#f(x)=sum_(n=0)^oof^((n))(a)(x-a)^n/(n!)#

to see that

#sqrtx=sqrt2+sum_(n=1)^oo(-1)^(n+1)(x-2)^n(2n-1)!!2^(1/2-2n)/(n!)#

Note that we did not include #f(2)=sqrt2,# the #0th# term, in our series, as we couldn't get it to originally fit in with our pattern.

We can't simplify the factorials any further.