upper bounds numerical integration error

okay so the integral is $\displaystyle \int_{0}^{3} \sqrt{x+1}dx$ and using midpoint approximation so I use

$\displaystyle \frac{(b-a)^3k_{2}}{24n^2}, where\,n=10\,and\,k_{2} = max_{[a,b]} |f"(x)|$

$\displaystyle f"(x) = -\frac{1}{4}(x+1)^{-3/2} $ found $\displaystyle k_{2} = \frac{1}{32}$

I am getting $\displaystyle \frac{9}{25600}$ book says 9/3200

also, latex sure is being a pain on here lately

Re: upper bounds numerical integration error

The question then goes on to ask for the upper bounds from trapezoid and simpsons rule but the book has different answers then I am getting. Did I do the first one correctly?

Re: upper bounds numerical integration error

now the question asks to find the number n of subintervals that will make the error less than $\displaystyle 5*10^{-4}$

just typing out the midpoint because the book is giving not the same answer I am giving

$\displaystyle |E_{m}| \le 5*10^{-4}$

$\displaystyle \frac{3^3*1/32}{24n^2} \le 5*10^{-4} \approx 8.3852$

I am getting n= 9 book is saying n = 24

Re: upper bounds numerical integration error

I'd like to help you out Jon but w/o a reference to what you're doing I'm not able to figure out what you need figured out.

Is there some online reference that uses the same notation and method?

Re: upper bounds numerical integration error

$\displaystyle k_{2}=1/4$ isn't it ? (Max at $\displaystyle x=0.)$

Re: upper bounds numerical integration error

Quote:

Originally Posted by

**romsek** I'd like to help you out Jon but w/o a reference to what you're doing I'm not able to figure out what you need figured out.

Is there some online reference that uses the same notation and method?

http://math.cmu.edu/~mittal/Recitation_notes.pdf

Re: upper bounds numerical integration error

Quote:

Originally Posted by

**BobP** $\displaystyle k_{2}=1/4$ isn't it ? (Max at $\displaystyle x=0.)$

the second derivative of the given function is \frac{-1}{4}(x+1)^{-3/2} and has a maximum on [0,3] of -1/32 of which you take the absolute value of

still with latex issues

Re: upper bounds numerical integration error

re your first post on this thread

BobP is correct. The max of the absolute value of the 2nd derivative is 1/4 at x=0 which leads to the bound of 9/3200

your 2nd problem is wrong for the same reason. The max value is 1/4 not 1/32

Re: upper bounds numerical integration error

To reiterate what romsek and BobP are saying:

You are taking $\displaystyle k_2 = \left| \max_{[a,b]} f''(x) \right|$, which is the absolute value of the maximum of the second derivative on the given interval. BobP and romsek are saying, take the absolute value function first: $\displaystyle |f''(x)| = \left|-\dfrac{1}{4}(x+1)^{-3/2} \right| = \dfrac{1}{4}(x+1)^{-3/2}$. Then, find the maximum. Hence, $\displaystyle k_2 = \max_{[0,3]} \dfrac{1}{4}(x+1)^{-3/2} = \dfrac{1}{4}$.

Re: upper bounds numerical integration error

Quote:

Originally Posted by

**SlipEternal** To reiterate what romsek and BobP are saying:

You are taking $$k_2 = \left| \max_{[a,b]} f''(x) \right|$$, which is the absolute value of the maximum of the second derivative on the given interval. BobP and romsek are saying, take the absolute value function first: $$|f''(x)| = \left|-\dfrac{1}{4}(x+1)^{-3/2} \right| = \dfrac{1}{4}(x+1)^{-3/2}$$. Then, find the maximum. Hence, $$k_2 = \max_{[0,3]} \dfrac{1}{4}(x+1)^{-3/2} = \dfrac{1}{4}$$.

new latex

Re: upper bounds numerical integration error

I see how I was messing up, taking the absolute of the answer not of the function. thanks