How Do You Find a Taylor Series for the Square Root of X About c=1?

[Calcgeek123]

Homework Statement

Find a taylor series for f(x)=sq. rt. of X about c=1

Homework Equations

N/A

The Attempt at a Solution

I took the derivative of the sq rt of X, and then plugged in 1 for all the X's. I got:
f(x)= 1
f'(x)=1/2
f''(x)=-1/4
f'''(x)=3/8
f^4(x)=-15/16

My teacher said it's okay to take out the first 2 terms because I can't seem to find a pattern that includes them, so I have so far:

1+1/2(x-1) + Sigma where n=1 to infinity, of (-1)^n(x-1)^n/n!

The denominators seem to fit the pattern 2^n, and I've found the numerator to be (2n-1)!

Can anyone find a pattern that doesn't use a double factorial? I've never used double factorials before, so I'm not even sure I'm using them correctly.

Thank you!

 

[mighty2000]

Thank you for sharing your attempt at finding a Taylor series for f(x)=sq. rt. of X about c=1. You are on the right track by taking the derivatives and plugging in 1 for all the x's. However, I would like to offer a different approach that may help you find the pattern you are looking for.

First, let's rewrite f(x) as f(x)=x^(1/2). Now, using the Taylor series formula, we can express f(x) as:

f(x)=f(1)+f'(1)(x-1)+f''(1)(x-1)^2/2!+f'''(1)(x-1)^3/3!+...

Since we already know f(1), f'(1), f''(1), etc. from your attempt, we can substitute them in and simplify to get:

f(x)=1+1/2(x-1)-1/8(x-1)^2+3/48(x-1)^3-15/384(x-1)^4+...

Now, if we look at the coefficients of the terms, we can see that they follow a pattern. The coefficient of (x-1)^n is (-1)^(n+1)(2n-1)!!/2^n, where !! represents the double factorial. This pattern can be proved using mathematical induction, but it may be beyond the scope of this forum.

I hope this helps you find the pattern you were looking for. Keep up the good work and don't be afraid to explore new concepts like double factorials! They can be very useful in certain situations. Good luck with your studies!