Taylor Series of $f(x) = \sqrt{x}$ about $c = 1$

Question

I've tried doing this problem but stuck at finding a pattern..

Work:

$$T_n = \sum^\infty_{n=0}\frac{f^n(c)}{n!}(x-c)^n = f(a) + \frac{f'(c)}{1!}(x-a)^1 + \frac{f''(c)}{2!}(x-a)^2+... $$

So $f(x) = \sqrt{x}$

$$ f'(x)=\frac12x^{-\frac12}$$ $$ f''(x)=\frac{-1}4x^{-\frac{3}2}$$ $$ f'''(x)=\frac{3}8x^{-\frac{5}2}$$ and $$f(1) = 1$$ $$f'(1) = \frac12$$ $$f''(1) = -\frac14$$ $$f'''(1) = \frac38$$

So far I know that its alternating so $(-1)^n$

But I'm having trouble with the fractions since I thought it would have been $\left(\frac12\right)^n$ but the $\frac38$ wont work with that. Have I done something wrong so far or am I just not thinking of this the right way?

So I've noticed that starting from the $n=3$ the numerator is $3$ then $3*5$ then $3*5*7$ and so on... but how do I account for the first 3 terms? (n = 0, 1, 2)

I learned that I can exclude it by simply taking it out and adding it from where the pattern starts to work so...

$$1+\frac{x-1}{2}-\frac{(x-2)^2}{8} + \sum^\infty_{n=3}something$$

score 1 · Answer 1 · answered Feb 27 '16 at 01:22

1

You are doing this correctly. The reason the fractions don't change in a fashion of (1/2)^n is because the exponent changes (from half to three halves to five halves and so on) as you keep taking derivatives.

answered Feb 27 '16 at 01:22

helovedonionsoup

11

so it goes $ 3 * 5 * 7 * 9 * ... $ – Panthy Feb 27 '16 at 01:29
@Panthy Yes, that is how it goes. – Arthur Feb 27 '16 at 01:39
look at new edit to see where i am at this point.. – Panthy Feb 27 '16 at 01:44

score 1 · Answer 2 · answered Feb 27 '16 at 01:45

The series you are after is:

$\begin{align} (1 + x)^{1/2} = \sum_{k \ge 0} \binom{1/2}{k} x^k \end{align}$

where, with $u^{\underline{k}} = u \cdot (u - 1) \dotsm (u - k + 1)$:

$\begin{align} \binom{\alpha}{k} = \frac{\alpha^{\underline{k}}}{k!} \end{align}$

Knowing the right form of the coefficient should enable you to, first, check that this is the coefficients in general (not just $\alpha = 1/2$), and second, deduce a "more standard" description of the coefficients for this particular case. Bonus points for handling the cases $\alpha = -1/2$, and negative integers.

score 1 · Answer 3 · answered Feb 27 '16 at 01:52

1

Figured it out with the help from some comments...

$$1+\frac{(x-1)}{2}-\frac{(x-1)^2}{8}+ \sum^\infty_{n=3}\frac{(-1)^{n+1}(x-1)^n(2n-3)}{2*n!} $$

answered Feb 27 '16 at 01:52

Panthy

779

please cross check if I am correct – Panthy Feb 27 '16 at 01:52

Taylor Series of $f(x) = \sqrt{x}$ about $c = 1$

3 Answers3