Calculus 2 Help! BINOMIAL SERIES?

1.) Using the binomial series, (1+u)^α = ∑(n=0->∞) (α n) u^n, where (α n), sorry for the awful notation but I don't know how else to type it, is "alpha choose n," which is: [(α)(α-1)(α-2)(α-3)...(α - n + 1)] / n!. Written out, this is:

(1+u)^α = ∑(n=0->∞) [(α)(α-1)(α-2)(α-3)...(α - n + 1)] * u^n / n!

In this problem, α = 1/3 and we can choose u = x^2 to make a binomial series:

(1+x^2)^1/3 = ∑(n=0->∞) [(1/3)(-2/3)(-5/3)(-8/3)...(1/3 - n + 1)] * (x^2)^n / n!

Normally this would be good enough but if you're entering it into Webassign for some godawful reason it wants you to take out the first two terms from the series, take out (-1)^n-1 to make it alternating, change the (1/3 - n + 1) to (4/3 - n) to (4-3n)3 (which is then divided by negative one as well since (-1)^n-1 is taken out), and put all the threes on the bottom, so it'll end up looking like this:

1 + x^2/3 + ∑(n=2->∞) (-1)^(n-1) * 2 * 5 * 8...(3n-4) * x^2n / (3^n * n!)

Or if you want it simple:

R = 1

A = 1

B = 2

C = 3

D = 2

E = -1

F = 2

G = 5

H = 8

I = 3

J = -4

K = 2

L = 3

The radius of convergence is 1 because is is 1 for all binomial series in the form (1 + u)^α.

2.) First we have to get this problem in a form vaguely resembling (1+u)^α, so we factor out sqrt(2) from the denominator to get:

x^2 / sqrt(2 + x) = x^2/sqrt(2(1 + x/2)) = x^2/sqrt(2) * (1+x/2)^-1/2

We can then momentarily ignore the x^2/sqrt(2) and find a binomial series for (1+x/2)^1/2 with α = -1/2 and u = x/2:

(1+u)^α = ∑(n=0->∞) [(α)(α-1)(α-2)(α-3)...(α - n + 1)] * u^n / n!

(1+x/2)^-1/2 = ∑(n=0->∞) [(-1/2)(-3/2)(-5/2)(-7/2)...(-1/2 - n + 1)] * (x/2)^n / n!

(1+x/2)^-1/2 = ∑(n=0->∞) [(-1/2)(-3/2)(-5/2)(-7/2)...(-1/2 - n + 1)] * (x)^n / (2^n * n!)

Then multiplying both sides by x^2/sqrt(2), then x^2 and 1/sqrt(2) can both be added to the series by adding 2 to the power of x and (1/2) to the power of 2 which is now on the denominator:

x^2/sqrt(2) * (1+x/2)^-1/2 = x^2/sqrt(2) * ∑(n=0->∞) [(-1/2)(-3/2)(-5/2)(-7/2)...(-1/2 - n + 1)] * (x)^n / (2^n * n!)

x^2/sqrt(2) * (1+x/2)^-1/2 = ∑(n=0->∞) [(-1/2)(-3/2)(-5/2)(-7/2)...(-1/2 - n + 1)] * (x)^(n+2) / (2^(n+1/2) * n!)

To enter it into Webassign, you take (-1)^n out, factor out all the twos in the numerator and put them in the denominator by changing 2^(n+1/2) to 2^(2n+1/2), do something really complex and weird to the (-1/2 - n + 1) again, and for some reason take out the first term:

x^2/sqrt(2) * (1+x/2)^-1/2 = x^2/sqrt(2) + ∑(n=1->∞) (-1)^n * 1 * 3 * 5...(2n-1) * x^n + 2 / (n! * 2^(2n + 1/2))

Or just:

R = 2

A = 2

B = 2

C = 1

D = -1

E = -1

F = 3

G = 5

H = 2

I = -1

J = 2

K = 2

L = 1

M = 2

N = 2

The radius of convergence is 2 because the series converges where |u|<1, so because u = x/2, |x/2| < 1 for |x| < 2, so R = 2.

3.) I have absolutely no idea how to do this problem to be honest but I took nine derivatives of the function in Maple and got -113400, which is the right answer in Webassign.

Hope that helps!