Question

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)

Answer 1

`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.