# Taylor Series Error Integration

• Kaura
In summary: The degree instead of index is good to...In summary, the polynomial p(x) of minimal degree that approximates F(x) throughout the given interval with an error of magnitude less than 10-4 can be found by integrating the Maclaurin series for sin(x) and using that information.
Kaura

## Homework Statement

Using Taylor series, Find a polynomial p(x) of minimal degree that will approximate F(x) throughout the given interval with an error of magnitude less than 10-4

F(x) = ∫0x sin(t^2)dt

## Homework Equations

Rn = f(n+1)(z)|x-a|(n+1)/(n+1)![/B]

## The Attempt at a Solution

I am confused about the integral and have never done a problem like this
Do I use the fundamental theorem of calculus to solve the nth derivative of the function and see at what point the error is below the required value?
Do I somehow integrate the taylor series for sin(x) and somehow use that?

Kaura said:

## Homework Statement

Using Taylor series, Find a polynomial p(x) of minimal degree that will approximate F(x) throughout the given interval with an error of magnitude less than 10-4

F(x) = ∫0x sin(t^2)dt

## Homework Equations

Rn = f(n+1)(z)|x-a|(n+1)/(n+1)![/B]

## The Attempt at a Solution

I am confused about the integral and have never done a problem like this
Do I use the fundamental theorem of calculus to solve the nth derivative of the function and see at what point the error is below the required value?
Do I somehow integrate the taylor series for sin(x) and somehow use that?

You did not actually specify an interval over which you are approximating F(x).

Anyway, instead of agonizing about how to do it, just start doing it. If one way does not work, try another.

Kaura said:

## Homework Statement

Using Taylor series, Find a polynomial p(x) of minimal degree that will approximate F(x) throughout the given interval with an error of magnitude less than 10-4

F(x) = ∫0x sin(t^2)dt

## Homework Equations

Rn = f(n+1)(z)|x-a|(n+1)/(n+1)![/B]

## The Attempt at a Solution

I am confused about the integral and have never done a problem like this
Do I use the fundamental theorem of calculus to solve the nth derivative of the function and see at what point the error is below the required value?
Do I somehow integrate the taylor series for sin(x) and somehow use that?
Yes. Write a few terms of the Maclaurin series for ##\sin(t^2)##, and then integrate that series. Note that the series you get is an alternating series. There should be a theorem in your book on the error that results from truncating an alternating series.
Kaura said:

I worked out some stuff let me know what to fix so far

F(x) = INT(0 to x) sin(t^2)dt | on the interval [0, 1]

Polynomial degree with error of magnitude less than 10^-4

sin(x) = x - x^3/3! + x^5/5! >>>

sin(x^2) = x^2 - x^6/3! + x^10/5! >>>

sin(x) = (-1)^n * x^(2n+1)/(2n+1)!

sin(x^2) = (-1)^n * x^(4n+2)/(2n+1)!

INT sin(x^2) = (-1)^n * x^(4n+3)/(4n+3)(2n+1)!

|x^(4n+3)/(4n+3)(2n+1)!| < 10^-4

x max of 1

10000 < |(4n+3+1)(2n+1+1)!|

(4n+7)(2n+3)! = h(n)

h(0)=7*3!
h(1)=11*5!
h(2)=15*7! > 10000

n = 2

The degree of the polynomial should be 2 right?

x^3/3 - x^7/(7*3!) + x^11/(11*5!)
was accepted as the correct answer on my homework
been on that problem all day good to finally solve it
so I guess you integrate the summation of the series
find when the term of n+1 is bounded above by the error
then use the McLaurin Series up to the nth term?

Kaura said:
I worked out some stuff let me know what to fix so far

F(x) = INT(0 to x) sin(t^2)dt | on the interval [0, 1]

Polynomial degree with error of magnitude less than 10^-4

sin(x) = x - x^3/3! + x^5/5! >>>

sin(x^2) = x^2 - x^6/3! + x^10/5! >>>

sin(x) = (-1)^n * x^(2n+1)/(2n+1)!

sin(x^2) = (-1)^n * x^(4n+2)/(2n+1)!
The two lines above are incorrect -- you have omitted the summation sign.
Kaura said:
INT sin(x^2) = (-1)^n * x^(4n+3)/(4n+3)(2n+1)!
This is also incorrect. On the right side there should be a summation.
Kaura said:
|x^(4n+3)/(4n+3)(2n+1)!| < 10^-4

x max of 1

10000 < |(4n+3+1)(2n+1+1)!|

(4n+7)(2n+3)! = h(n)

h(0)=7*3!
h(1)=11*5!
h(2)=15*7! > 10000

n = 2

The degree of the polynomial should be 2 right?
No. n = 2 indicates the index of the term in the polynomial. n = 0 refers to the first term, which is of degree 3. n = 1 refers to the second term, which is of degree 7, and so on.

Rather than working with the general terms in the Maclaurin expansion, it's easier to write the polynomial as you did above, like so: ##\sin(t^2) = t^2 - \frac{t^6}{3!} + \frac{t^{10}}{5!} -+ \dots##.

Mark44 said:
The two lines above are incorrect -- you have omitted the summation sign.
This is also incorrect. On the right side there should be a summation.

No. n = 2 indicates the index of the term in the polynomial. n = 0 refers to the first term, which is of degree 3. n = 1 refers to the second term, which is of degree 7, and so on.

Rather than working with the general terms in the Maclaurin expansion, it's easier to write the polynomial as you did above, like so: ##\sin(t^2) = t^2 - \frac{t^6}{3!} + \frac{t^{10}}{5!} -+ \dots##.

Thanks for the feed back
Yeah I totally forgot the summation sign that was lazy of me
The index instead of degree is good to know as I was mistaken about the terminology

## 1. What is a Taylor Series Error Integration?

A Taylor Series Error Integration is a method used in calculus to approximate the value of a function using a series of polynomial terms. It is used to estimate the error between the actual value of a function and its approximation.

## 2. How is Taylor Series Error Integration calculated?

The Taylor Series Error Integration is calculated by taking the sum of an infinite number of terms, each representing a derivative of the original function evaluated at a specific point. This point is known as the center of the series and is usually chosen to be the point at which the approximation is being made.

## 3. What is the purpose of using Taylor Series Error Integration?

The purpose of using Taylor Series Error Integration is to find a polynomial approximation of a function that is as close as possible to the actual value. It allows us to approximate functions that are difficult to integrate analytically, and it also gives us a way to estimate the error in our approximation.

## 4. What is the difference between Taylor Series and Taylor Series Error Integration?

Taylor Series and Taylor Series Error Integration are closely related but have different purposes. A Taylor Series is a representation of a function as an infinite sum of terms, while Taylor Series Error Integration is a method of using this series to approximate the value of a function and estimate the error in the approximation.

## 5. What factors can affect the accuracy of Taylor Series Error Integration?

The accuracy of Taylor Series Error Integration can be affected by the number of terms included in the series, the choice of the center point, and the behavior of the function being approximated. Higher order derivatives of the function can also affect the accuracy of the approximation.

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