Question

What is the Taylor Series for x^1/2 centered at a=16? I think this is something to do with a binomial series.

Answer 1

i can grind it out, not sure if that is the best method

Answer 2

Ill post what I got once the file sends over!

Answer 3

A Taylor series for a function \(f(x)\) is expanded about a point \(x_0\) as follows
\[\begin{align*} f(x) &= f(x_0) + f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2 +\frac{f'''(x_0)}{3!}(x-x_0)^3+...\\
&=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n
\end{align*}\]

Answer 4

I got the expansion that you said kirbykirby through a Binomial series, which i am not sure if it is right or not. I seem to be having some issues with the denominator and the factorials matching up, in particular the \[\frac{ -x^2 }{8 }\] part. The factorial or denominator does not seem to be making a pattern, which means I probably made a really stupid error somewhere, unless I did it totally wrong. In which case, does anyone have any suggestions?

Answer 5

your terms should look like \(x-16\)

Answer 6

think you are going to use this, right \[(1+x)^{\frac{1}{2}}=\sum_{n\ge 0}\binom{\frac{1}{2}} n x^n\]

Answer 7

Yes, where the 1 on the left is substituted with a 16, correct?

Answer 8

and you are going to have to change your variable

Answer 9

On the right side to \[(x-16)^{\frac{ 1 }{2 }}\]

Answer 10

you have to change to say \(u=x-16\) to make it look like expanding \(\sqrt{16+u}\) at \(u=0\)
that is where the expansion works
then algebra to get
\[\sqrt{16+u}=4\sqrt{1+\frac{u}{16}}\]

Answer 11

My mind is blown. What I have thus far is incorrect? So where do I restart?

Answer 12

now you can expand as before \[(16+z)^{1/2}=4\cdot\left(1+\frac u{16}\right)^{1/2}=\\
=4\cdot\sum \binom{1/2}n\frac{u^n}{16^n}\,.\]

Answer 13

it is a pain to do it this way, might actually be easier to do it by hand

Answer 14

but if you are going to use that formula for the binomial, you need it to look like
\[(1+x)^{\frac{1}{2}}\]

Answer 15

actually, since you probably want to leave it in sigma notation and not actually compute the coefficients you are done at this step

Answer 16

or at this step \[4\cdot\sum \binom{1/2}n\frac{(x-16)^n}{16^n}\,.\]

Answer 17

Once again where did the U Sub come from?

Answer 18

Or just why did we introduce a U Sub?

Answer 19

i see from your paper you were missing the 16 in the denominator using the this method
the gimmick is to make it look like \((1+z)^{1/2}\) for some \(z\)

Answer 20

that is the form it has to be in to use the binomial expansion

Answer 21

Awesome thanks!

Answer 22

yw, i think all steps are there