# Discover P5(x) and 4th Order Taylor Series of Sin(x) and xSin(2x)

• Ry122
In summary, the conversation discusses the need to find the 5th order Taylor series of sin(x) in order to find the 4th order Taylor series of x sin(2x). However, it is not necessary as the 3rd order Taylor series of sin(x) would be sufficient. The word "hence" is used loosely in this context, meaning "after that."

#### Ry122

Find P5(x), the 5th order Taylor series, of sin (x) about x = 0. Hence find the 4th
order Taylor series for x sin (2x) about x = 0.

In this question why is it required to find the 5th order taylor series of sin(x) to find the 4th order taylor series of xsin(2x)?

After you have P5(x), it's a simple matter to find the power series for xsin(2x). "Hence" in this problem just means "after that," I believe.

Mark44, I believe Ry122's question was specifically why it would be necessary to find the 5th order polynomial of one in order to find the 4th order polynomial of the other. And, in fact, it is not necessary. Multiplying an nth order polynomial by x gives, of course, an n+1 order polynomial.

It would be sufficient to find the 3rd order Taylor's polynomial for sin(x) in order to find the 4th order Taylor's polynomial for x sin(2x). It would have made sense if you were looking for the 4th order Taylor's polynomial of sin(2x)/x.

My answer was a little oblique. I don't think it is necessary to find the 5th order Taylor polynomial. My comment about "hence" was intended to convey my belief that this word was not used in it usual mathematics sense of "it therefore follow that..."

## 1. What is P5(x) in the context of the 4th Order Taylor Series?

P5(x) refers to the 5th degree polynomial approximation of a function at a specific point x, using the 4th Order Taylor Series. It is calculated by taking the 4th derivative of the function at x and dividing it by 4!, then adding the 3rd derivative at x divided by 3!, the 2nd derivative at x divided by 2!, and the 1st derivative at x, and finally adding the function value at x.

## 2. How is the 4th Order Taylor Series derived for Sin(x)?

The 4th Order Taylor Series for Sin(x) is derived by taking the 4th derivative of Sin(x) at a specific point x, which is equal to -Sin(x), and dividing it by 4!, then adding the 3rd derivative of Sin(x) at x, which is equal to -Cos(x), divided by 3!, the 2nd derivative of Sin(x) at x, which is equal to -Sin(x), divided by 2!, and the 1st derivative of Sin(x) at x, which is equal to Cos(x). This results in the familiar series expansion of Sin(x) = x - (x^3/3!) + (x^5/5!) - (x^7/7!) + ...

## 3. What is the significance of the 4th Order Taylor Series for Sin(x)?

The 4th Order Taylor Series for Sin(x) allows us to approximate the value of Sin(x) at a specific point x with a 5th degree polynomial. This is useful in numerical analysis and allows for more accurate calculations than using the actual Sin(x) function.

## 4. How does the 4th Order Taylor Series of xSin(2x) differ from the 4th Order Taylor Series of Sin(x)?

The 4th Order Taylor Series of xSin(2x) is a more complex series compared to the 4th Order Taylor Series of Sin(x). This is because xSin(2x) is a composite function and requires the use of the chain rule in taking derivatives. The resulting series is a combination of x^2, x^4, x^6, and higher powers of x, whereas the 4th Order Taylor Series of Sin(x) only includes x, x^3, x^5, and higher powers of x.

## 5. How can the 4th Order Taylor Series be used in real-world applications?

The 4th Order Taylor Series is commonly used in engineering and scientific fields to approximate the behavior of a system or function. It is particularly useful in situations where the actual function is difficult to calculate, and an approximation is needed. For example, it can be used in calculating trajectories of objects in motion, predicting the behavior of electrical circuits, and in numerical methods for solving differential equations.