Help with Maclaurin series representation

In summary, the problem is to find the Maclaurin series representation and express it as a sum for the expression $\displaystyle \frac{1+x^3}{1+x^2}$. There is a trick to simplify this by using the closed form of a geometric series with $\displaystyle r = -x^2$.
  • #1
fernlund
10
0
Hello! So, I'm having a bit of a problem with an exercise in my Calculus book. I'm supposed to find the Maclaurin series representation of

\(\displaystyle \frac{1+x^3}{1+x^2} \)

and then express it as a sum. Am I really supposed to differentiate this expression a bunch of times..? That will be very complicated quickly.

I've tried to solve this using Wolfram Mathematica as a help, to find \(\displaystyle f'(0) \), \(\displaystyle f''(0) \) etc, but of course I want to do it by myself.

Is there any kind of trick that I'm missing? A substitution or another way of writing this?
 
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  • #2
fernlund said:
Hello! So, I'm having a bit of a problem with an exercise in my Calculus book. I'm supposed to find the Maclaurin series representation of

\(\displaystyle \frac{1+x^3}{1+x^2} \)

and then express it as a sum. Am I really supposed to differentiate this expression a bunch of times..? That will be very complicated quickly.

I've tried to solve this using Wolfram Mathematica as a help, to find \(\displaystyle f'(0) \), \(\displaystyle f''(0) \) etc, but of course I want to do it by myself.

Is there any kind of trick that I'm missing? A substitution or another way of writing this?

$\displaystyle \begin{align*} \frac{1 +x^3}{1 + x^2} = \left( 1 + x^3 \right) \, \frac{1}{1 - \left( -x^2 \right) } \end{align*}$

Now notice that $\displaystyle \begin{align*} \frac{1}{1 - \left( -x^2 \right) } \end{align*}$ is the closed form of a geometric series with $\displaystyle \begin{align*} r = -x^2 \end{align*}$.
 
  • #3
Prove It said:
$\displaystyle \begin{align*} \frac{1 +x^3}{1 + x^2} = \left( 1 + x^3 \right) \, \frac{1}{1 - \left( -x^2 \right) } \end{align*}$

Now notice that $\displaystyle \begin{align*} \frac{1}{1 - \left( -x^2 \right) } \end{align*}$ is the closed form of a geometric series with $\displaystyle \begin{align*} r = -x^2 \end{align*}$.

Ah, thank you very much! I've got it from here :)
 

1. What is a Maclaurin series representation?

A Maclaurin series representation is a way to express a function as an infinite sum of terms, where each term is a power of x. It is similar to a Taylor series, but the Maclaurin series is specifically centered at x=0.

2. Why is it useful to have a Maclaurin series representation?

A Maclaurin series can be used to approximate the value of a function at any point, as long as the function is infinitely differentiable at that point. This can be useful in many mathematical and scientific applications, such as calculating derivatives, integrals, and solving differential equations.

3. How do you find the Maclaurin series representation of a function?

To find the Maclaurin series representation of a function, you can use the Taylor series formula and substitute x=0. This will give you the first few terms of the Maclaurin series. To find the entire series, you can use the general formula for a Maclaurin series and use calculus techniques to find the coefficients of each term.

4. Can any function be represented by a Maclaurin series?

No, not all functions have a valid Maclaurin series representation. The function must be infinitely differentiable at x=0 for the series to converge. Additionally, the series may only converge for a specific range of x values, so it is important to check for convergence before using the series to approximate a function.

5. Are there any shortcuts or tricks to finding a Maclaurin series representation?

Yes, there are a few shortcuts that can be used for specific types of functions, such as polynomials, trigonometric functions, and exponential functions. However, for more complex functions, the general formula and calculus techniques are usually required to find the Maclaurin series. Practice and familiarity with different types of series can also help in finding the series representation more efficiently.

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