# upper bounds numerical integration error

• Feb 16th 2014, 11:49 PM
Jonroberts74
upper bounds numerical integration error
okay so the integral is $\int_{0}^{3} \sqrt{x+1}dx$ and using midpoint approximation so I use

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

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

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

also, latex sure is being a pain on here lately
• Feb 17th 2014, 05:21 PM
Jonroberts74
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?
• Feb 17th 2014, 05:56 PM
Jonroberts74
Re: upper bounds numerical integration error
now the question asks to find the number n of subintervals that will make the error less than $5*10^{-4}$

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

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

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

I am getting n= 9 book is saying n = 24
• Feb 18th 2014, 12:13 AM
romsek
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?
• Feb 18th 2014, 02:23 AM
BobP
Re: upper bounds numerical integration error
$k_{2}=1/4$ isn't it ? (Max at $x=0.)$
• Feb 18th 2014, 02:06 PM
Jonroberts74
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
• Feb 18th 2014, 02:09 PM
Jonroberts74
Re: upper bounds numerical integration error
Quote:

Originally Posted by BobP
$k_{2}=1/4$ isn't it ? (Max at $x=0.)$

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
• Feb 18th 2014, 02:43 PM
romsek
Re: upper bounds numerical integration error
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
• Feb 18th 2014, 03:01 PM
SlipEternal
Re: upper bounds numerical integration error
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, $k_2 = \max_{[0,3]} \dfrac{1}{4}(x+1)^{-3/2} = \dfrac{1}{4}$.
• Feb 18th 2014, 03:12 PM
romsek
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
• Feb 18th 2014, 04:52 PM
Jonroberts74
Re: upper bounds numerical integration error
I see how I was messing up, taking the absolute of the answer not of the function. thanks