# Finding K for integral approximation errors help

1. Sep 4, 2014

### LBK

Hello, I am having a hard time getting my errors to come out to what the book says the answers should be. My approximations are correct, so I think I'm just misunderstanding how to find K.
Q.a) Find the approximations T8 and M8 for ∫(0 to 1) cos(x2)dx
I found these to be T8=0.902333 and M8=0.905620

b) Estimate the errors in the approximations. And here's where my numbers don't match. What I did:

2nd deriv f''(x)=-2x*sin(x2)-4x2cos(x2)

since the graph is bounded by |1| I thought the max. would be at f''(0) but that would make f''(0)=0
So I'm really confused on where to go from here.
FYI--the book answer gives a value of ET=<or=0.0078 and EM=<or=0.0039

I have the formula for error for trapezoidal as <or= [K(b-a)3] / 12n2
In this case then, I should have (b-a)3=1 and 12n3=768
Working backward from the book's correct answer that would make 0.0078=K/768 and K=5.99
So I'm not seeing where that would come from. Any help, please? I'm really struggling in this class and very confused.

Last edited: Sep 4, 2014
2. Sep 4, 2014

### Ray Vickson

What do you mean by $T_8$ and $M_8$?

3. Sep 4, 2014

### LBK

Oh, I'm sorry that would be integral approximation by trapezoidal rule and by midpoint rule, respectively with n=8

*also original post corrected for error formula, denominator should be 12n^2 not cubed

4. Sep 4, 2014

### Ray Vickson

For your function $f(x) = \cos(x^2)$ the second derivative should be
$$f''(x) = -2 \sin(x^2)-4x^2 \cos(x^2),$$
which is a bit different from what you wrote. If you plot $f''(x)$ you will see that it is negative and strictly decreasing on the interval $0 \leq x \leq 1$. Thus $|f''(x)|$ is strictly increasing.

Anyway, for Trapezoidal, the error bound is
$$|E| \leq \frac{K(b-a)^3}{12 n^2} = \frac{K}{12\times 8^2}= \frac{K}{768}, \: K \equiv \max_{0 \leq x \leq 1} |f''(x)|$$
In your case, $K = |f''(1)| \doteq 3.84415$, so $|E| \leq 0.0050054 \doteq 0.005.$ I cannot get the book's answer.

I think I can see where you went wrong: the actual error formula is
$$E = -\frac{f''(\xi) (b-a)^3}{12 n^2} = - \frac{f''(\xi)}{768}$$
In this case, $f''(\xi) < 0$ so the error is $>0$. To bound the error we want the least value of $f''$, that is the most negative value. That is at $\xi = 1$.

5. Sep 4, 2014

### LBK

I worked on it just now and I got .005 too. A classmate said he couldn't get the book answer either, so I feel much better now. And for the record, your explanation is clearer than my teacher and the book, so thank you again!