# Taylor Series Error Integration

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?

Homework Helper
Dearly Missed

## 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.

Mentor

## 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:

Kaura
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?

Kaura
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?

Mentor
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##.

Kaura
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