Question

Medal will be given if answered and showed work for this Calculus II Question....

Answer 1

Binomial series states
\[(1+x)^{k}=1+kx+\frac{ k(k-1) }{ 2! }+...+\frac{ k(k-1)...(k-n+1) }{ n! }x ^{n}+...\]for every real k and absolute value of x is < 1.
Use binomial series to find a power series representation for\[\frac{ 1 }{ \sqrt[3]{1-x ^{2}} }\]

Answer 2

Okay, so the goal here is to get our series to look like a bionomial, we can use some algebraic manipulation to achieve this:
\[ S= \frac{1}{\sqrt[3]{1-x^2}} = (1-x^2)^-(1/3) \]

That is starting to look like a binomial series, but lets rewrite our binomial series in terms of z instead of x, just so we dont get confused:

\[ (1+z)^k = 1 + kz + \frac{k(k-1)}{2!}z+... \]

In this case we can look at our series S and we see that to match our binomial series we need \[ z = -x^2 \] and \[k = -1/3 \]

So now replace your z with -x^2 and k with -1/3

\[ S= \frac{1}{\sqrt[3]{1-x^2}} = 1 +\frac{1}{3}(x^2) +\frac{(-1/3)(-3/4)}{2!} x^4+... \]

Answer 3

Is that it? My teacher taught a different way but if that's the case this makes much more sense.

Answer 4

You can keep expanding the terms if you wish, but that should work, not sure if she's looking for a general rule, but it can be applied to the last term.

Answer 5

There are different ways of tackling it, but this should just work just as well.

Answer 6

Yea well my teacher taught me Macluarin series for this but I'm confused so how do make the final term with n in it?

Answer 7

Macclaurin series will work, but then what would be the point of giving you the binomial if you cant use it (with a macclaurin, you dont need any series to build off of).

For the final term, just plug it in to your general term
\[ \frac{k(k-1)...(k-n+1)}{n!}z^n=\frac{(-1/3)(-4/3)...(-(1/3)-n+1)}{n!}(-x^2)^n=(-1)^n\frac{(-1/3)(-1/3-1)...(-(1/3)-n+1)}{n!}x^{2n})\]

Answer 8

Somehow it has something like\[(-1)^{n}\frac{ 1*4*7.....(3n-2) }{ 3^{n}n! }x ^{2n}\]

Answer 9

Ah, they simplfied, notice that each term in the numerator will be divided by 3 so
\[(-1)^n\frac{((1/3)(4/3)(7/3)...\frac{3n-2}{3}}{n!} x^{2n} \]

Then you can collect all the 1/3's and that will give you 3^n in the denominator. Notice the numerator numbers will follow the pattern (-1-3), (-1-6), (-1-9), ...
which gives you -1, -4, -7, -10 which fits he patter (3n-2)

Answer 10

Although, I'm not certain if the -1^n is correction, I believe all terms should be positive at first glance.

Answer 11

Oh ok the final thing is it says you have to find something with absolute value of x < 1. So do I just plug in -x^2 for x?

Answer 12

Yes, the binomial will only converge if |x| is less than 1, in or case if we talk about |z|<1 we just plug in -x^2 for z and solve.

Answer 13

so absolute value of -x is < +1 or -1?

Answer 14

\[ |(-x^2)| <1 \]

Is saying it will converge whenever that holds true, so what values can you plug in for x? as you said between -1 and 1 is correct, we obtain an interval of convergence of

\[ -1<x<1 \]

Answer 15

Oh.... ok I get it thank you so much I appreciate your time and effort.

Answer 16

No problem, glad I could help.