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## Homework Statement

Let f(x)= (1+x)

^{4/3}- In this question we are studying the Taylor series for f(x) about x=2.

This assignment begins by having us find the first 6 terms in this Taylor series. For time, I will omit them; however, let's note that as we continuously take the derivative of this function, we are applying the chain rule over and over. The derivative of the 'inside function' is 1, so in essence, we are really just applying the power rule again and again.

(b) Let c

_{k}= f

^{(k)}(2)/k! be the coefficient of (x-2)

^{k}in the Taylor series. It is not so easy to write a closed-form expression for c

_{k}. However, find a recursive formula to express c

_{k+1}in terms of c

_{k}.

(c) Use your recursive formula, together with the ratio test, to determine which values for x would make the Taylor series Σ

_{k=0}

^{∞}c

_{k}(x-2)

^{k}converge or diverge absolutely.

## Homework Equations

## The Attempt at a Solution

(b) I note that the coefficient of each function is being changed by the power in which the previous coefficient is raised to--which always contains a (3

^{-1}. Furthermore, we must handle the factorial. As a result, I observe the following formula for the general term:

(4-3k)(1/(k+1))(3

^{-1})c

_{k}= c

_{k+1}

(c) Applying the ratio test: lim k→∞ |c

_{k+1}|/|c

_{k}|

I note that c

_{k+1}contains a c

_{k}in it. As a result they cancel out. Leaving us with:

lim k→∞ (4-3k)(1/(k+1))(3

^{-1})

Here is where I get stuck.. I believe we can pull the (1/3) out from the limit, as it is a constant. The other two parts of the limit approach -∞ and 0 respectively. Thus, we would apply L'Hopital's rule? This gives us

1/3 lim k→∞ [-3(-(k+1)

^{-2})]

From here, do we pull out the -3, and evaluate the limit? If so, our limit approaches 0 again?

Any help would be appreciated. Here is the next question, to give you an idea of what I need.

(d) You should have observed a range of x values that extends for the same distance on either side of x=2. This distance is called the Radius of Convergence. Try to find some property of the function that explains why the radius of convergence is what it is.

Note: I believe the for 0<x<2 it converges.

At 2 it is inconclusive.

2<x<inf it diverges.

I heard this from a classmate. Any help is appreciated!

Thanks,

Calculus 2 Student.