# Compute a derivative using MacLaurin series

• kamil
In summary, to compute the 9th derivative of the given function, first recognize that the n = 0 term of the cosine series is 1, so it cancels with the minus 1 in the given function. Then, divide the x^7 through and simplify, leaving the sum from n = 1 to infinity. From there, it is a straightforward application of the formula for the derivative of a power series.
kamil

## Homework Statement

Compute the 9th derivative of
$$f(x) = \frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}$$
at x=(0)

## Homework Equations

$$f(x)=\sum^{\infty}_{n=0} \frac{f^{(n)}(c)}{n!}x^n$$
$$cos(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n)!}x^{2n}$$

## The Attempt at a Solution

I've already read this thread(https://www.physicsforums.com/showthread.php?t=362369&highlight=Maclaurin), and I think I know how to do these kind of problems. However, I have a problem with that one since it has $${-1}}$$

What I did is write the power series for cos(x),
$$cos(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n)!}x^{2n}$$

Substitute $$3 x^{4}$$ , substract 1, and divide by $$x^{7}$$.
And getting:
$$\frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}=\sum^{\infty}_{n=0} \frac{\frac{(-1)^{n}3^{2n}}{(2n)!}x^{8n}-1}{x^7}$$
But after that I don't know.

The only term that will produce a nonzero derivative when evaluated at x=0 will have a power of x^9. What is that term and what is it's derivative?

kamil said:
Substitute $$3 x^{4}$$ , substract 1, and divide by $$x^{7}$$.
And getting:
$$\frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}=\sum^{\infty}_{n=0} \frac{\frac{(-1)^{n}3^{2n}}{(2n)!}x^{8n}-1}{x^7}$$
But after that I don't know.

Nope. You can't move the 1 inside the sum like that. Sum(something) -1 is not equal to sum(something - 1).

What you should do is recognize that the n = 0 term of the cosine series is 1, so this cancels with your minus 1. Your sum is then from n = 1 to infinity, and you can safely divide the x^7 through. Then I think you probably know what to do from there!

But the n= 0 term for the MacLaurin series for cosine is just "1". The "-1" term just cancels that.

$$\frac{\cos\left(3 x^{4} \right) - 1}{x^{7}}=\sum^{\infty}_{n=1} \frac{\frac{(-1)^{n}3^{2n}}{(2n)!}x^{8n}}{x^7}$$

Thank you, it worked. I knew there was something wrong with the -1. I don't know why I thought I could do that operation on a sum. I had also simplified $$\frac{x^{8n}}{x^{7}}$$ to $${x^{n}}$$ , which is totally wrong. I don't why I did this.

## 1. What is a MacLaurin series?

A MacLaurin series is a type of power series expansion used to represent a function as a sum of infinitely many terms. It is named after the mathematician Colin Maclaurin.

## 2. How is a MacLaurin series used to compute a derivative?

A MacLaurin series can be used to compute a derivative by taking the derivative of each term in the series and then evaluating it at the desired point. This can help to approximate the value of the derivative at that point.

## 3. What is the formula for a MacLaurin series?

The formula for a MacLaurin series is f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ... + (f^(n)(0)x^n)/n!, where f^(n)(0) represents the nth derivative of the function evaluated at x=0.

## 4. How is a MacLaurin series different from a Taylor series?

A MacLaurin series is a special case of a Taylor series where the center of expansion is at x=0. This means that in a MacLaurin series, all of the derivatives are evaluated at x=0, while in a Taylor series, the center of expansion can be any value in the domain of the function.

## 5. What are the advantages of using a MacLaurin series to compute a derivative?

Using a MacLaurin series to compute a derivative can be advantageous because it allows for a more accurate approximation of the derivative than using a finite difference formula. It also allows for the computation of derivatives for functions that may not have a closed-form solution.

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