# Thread: calc 2 binomial series

1. ## calc 2 binomial series

1)Find radius of convergence if i have g(x) = x^(1/4). with x centered at x=16.
2) im suppose to graph T2(x) and x^1/4 to observe something... but what does this mean?
3)And then i must use T2(x) to approximate (20)^1/4 . --> I would just plug in x=20 for x right?
The problem also helps us and tells us that x = 16 + (x-16) ...( But would this still be centered at 16)?

if i took x= 16+(x-16) then

then g(x) = summation of n=0 to infinity of (1/4 choose n) *((x-16)/16))^n as my binomial series. The radius of convergence would be abs((x-16)/16) < 1 --> so R= 1.

I was wondering if i did something wrong?

2. ## Re: calc 2 binomial series

What is T2(x)?

-Dan

3. ## Re: calc 2 binomial series

that is the second degree polynomial series

4. ## Re: calc 2 binomial series

$\displaystyle \left| \frac{x-16}{16} \right| = \left| \frac{x}{16}-1\right| < 1$

So, you have $0 < x < 32$. This gives a radius of convergence of 16. You forgot to factor out the $16^\tfrac{1}{4}$

$\displaystyle g(x) = 2\sum_{n=0}^\infty \dbinom{\tfrac{1}{4}}{n}\left( \dfrac{x-16}{16} \right)^n$

$\displaystyle T_2(x) = 2\left(1+\dfrac{x-16}{64} - \dfrac{3(x-16)^2}{4096} \right)$

So, if you are using that to estimate when $x=20$, that would be:

$\displaystyle T_2(20) = 2\left(1+\dfrac{1}{16} -\dfrac{3}{64} \right) = 2+\dfrac{1}{32} = \dfrac{65}{32}$

5. ## Re: calc 2 binomial series

i got -3/8192 instead of 3/4096? 16^2 * 16*2 = 8192

6. ## Re: calc 2 binomial series

isnt binomial series = 1 + k(x-c) + k(k-1)(x-c)^2/2! + .....

And, if i were to find the error bound for 65/32, i would use f^3(z) * (20-16)^3 / (3!) where f^3(z) = f^3(16) = 21/64 * (1/x^(11/4)) --> which is the 3rd derivative of x^1/4?