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.