Thread: Help! Finding error of approximation

1. Help! Finding error of approximation

hi
totally stuck on this question.

What is the upper bound for the error of the approximation for the INTEGRAL from 0 to 1 of x^4 sin^3

this question is in the calculus capter on series if that helps

thanks

2. A result you've probably already derived is:

$\displaystyle \sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{(2n+1)}}{(2n+1)!}$

Multiplying the above by

$\displaystyle x^4$ gives,

$\displaystyle x^4 \sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{(2n+5)}}{(2n+1)!}$

A power series can be integrated term by term (the integration can pass through the summation).

$\displaystyle \int_0^1 x^4 \sin(x) = \int_0^1 \sum_{n=0}^{\infty}\frac{(-1)^n x^{(2n+5)}}{(2n+1)!}$

$\displaystyle = \sum_{n=0}^{\infty} \int_0^1 \frac{(-1)^n x^{(2n+5)}}{(2n+1)!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(-1)^n x^{(2n+6)}}{(2n+1)!(2n+6)}$ evaluated at x=1 - x=0

$\displaystyle = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(2n+6)}$

I'll stop here for now. I need to think about the next step.

3. I think my previous post provides an exact solution to the integral in question.

Did your question specify an "n" value which would limit the number of terms in our final summation? If n is finite (we sum a finite number of terms), then the solution would be an approximation. Otherwise, we have the exact solution.

Let me know.

4. Originally Posted by n0083
I think my previous post provides an exact solution to the integral in question.

Did your question specify an "n" value which would limit the number of terms in our final summation? If n is finite (we sum a finite number of terms), then the solution would be an approximation. Otherwise, we have the exact solution.

Let me know.
nope no n value

i see what you are doing here except it is x^4 sinx^3 not x^4 sinx

for x^4 sin x^3 i figured out the power series

E (-1)^n (x^6n+8) / (6n+8)(2n+1)!

after that is where i get stuck at. if the above is right you could point me in the right direction or working it out would be better.

thanks for helping

5. Nice catch.
Your correction appears to me to be correct. You need to evaluate the integral at x=1, and x=0. (Fundamental theorem of calculus) F(1) - F(0).

This would give you,
$\displaystyle \sum_{n=0}^{n=\infty}\frac{(-1)^n}{(6n+8)(2n+1)!}$

I think this is an exact solution which would mean the upper bound on the error would be zero or any positive number for that matter, but I hardly suspect this is what your question is asking us for. I am not sure how to proceed from here.

Taylor Polynomials provide an error equation, which you could use to find the upper bound of the error. However, we are going for infinite terms so the power series becomes exact.

Please let me know if/when you figure out this solution.

6. Originally Posted by n0083
Nice catch.
Your correction appears to me to be correct. You need to evaluate the integral at x=1, and x=0. (Fundamental theorem of calculus) F(1) - F(0).

This would give you,
$\displaystyle \sum_{n=0}^{n=\infty}\frac{(-1)^n}{(6n+8)(2n+1)!}$

I think this is an exact solution which would mean the upper bound on the error would be zero or any positive number for that matter, but I hardly suspect this is what your question is asking us for. I am not sure how to proceed from here.

Taylor Polynomials provide an error equation, which you could use to find the upper bound of the error. However, we are going for infinite terms so the power series becomes exact.

Please let me know if/when you figure out this solution.
wouldn't we have to integrate what i have first then evaluate it?
if so would it be

E (-1)^n (x^6n+9) / (6n+9)(6n+8)(2n+1)!

and then you get

E (-1)^n / (6n+9)(6n+8)(2n+1)!

i don't know how to proceed after this to find approximation
do you know the equation to find the upper bound?

7. I thought you already took the integral. I'll rework it now:

$\displaystyle \sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{(2n+1)}}{(2n+1)!}$

So,

$\displaystyle \sin(x^3) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{(6n+3)}}{(2n+1)!}$

Multiplying the above by

$\displaystyle x^4$ gives,

$\displaystyle x^4 \sin(x^3) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{(6n+7)}}{(2n+1)!}$

A power series can be integrated term by term (the integration can pass through the summation).

$\displaystyle \int_0^1 x^4 \sin(x^3) = \int_0^1 \sum_{n=0}^{\infty}\frac{(-1)^n x^{(6n+7)}}{(2n+1)!}$

$\displaystyle = \sum_{n=0}^{\infty} \int_0^1 \frac{(-1)^n x^{(6n+7)}}{(2n+1)!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(-1)^n x^{(6n+8)}}{(2n+1)!(6n+8)}$ evaluated at x=1 - x=0

$\displaystyle = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(6n+8)}$

Sorry, I can't assist you further with the error bound.