Midpoint

**413** · October 20th 2006, 06:23 PM

a)With f(x)= e^-x^2 , compute approximations using midpoint, trapezoidal and simpson's rule for

∫ [0 to 2] f(x) dx

with n=2 (5 decimal places)

**Jameson** · October 20th 2006, 07:14 PM

Simpson's rule won't work unless $n \ge 3$ . Is your n correct?

**413** · October 20th 2006, 08:00 PM

that's what it says.

**Jameson** · October 20th 2006, 08:06 PM

Well for midpoint, divide each subinterval into width $\frac{b-a}{n}$ . And the height is of course the midpoint values. Thus we can use these rectangles to approximate the area. So...

$A \approx (\frac{2-0}{2}) \left(f(.5)+f(1.5) \right)$

**earboth** · October 21st 2006, 01:32 AM

Originally Posted by 413

a)With f(x)= e^-x^2 , compute approximations using ..., trapezoidal ... rule for
∫ [0 to 2] f(x) dx
with n=2 (5 decimal places)

Hi,

I presume that you know this rule already.

I presume also, that you mean the function $f(x)=e^{-x^2}$

So you have to calculate:

$\frac{f(0)+f(1)}{2}+\frac{f(1)+f(2)}{2}\approx 1.75407...$

I've attached a diagram showing the 2 trapezoids.

EB

**413** · October 21st 2006, 08:10 AM

I found that from midpoint rule gives 0.88420, trapezoidal rule gives 0.87704, and simpson's rule gives 0.82994. The next question asks me to

Compute the error esimates for midpoint, trapezoidal, simpson's. Carefully examine the extreme values of f ''(x). You may use that |f ''''(x)| < 12 for 0<x<2. (the < signs all mean less than or equal to).

How would you do this?

**earboth** · October 21st 2006, 11:00 AM

Originally Posted by 413

I found that ... trapezoidal rule gives 0.87704, ...

Hi,

you are right. Certainly you have noticed that I've forgotten to divide by 2. So your result is correct.

So sorry.

EB

**413** · October 21st 2006, 12:30 PM

its okay.

How do you compute the error estimates, and why would you use f ''(x) and f ''''(x)?

**ThePerfectHacker** · October 21st 2006, 03:31 PM

Originally Posted by 413

How do you compute the error estimates, and why would you use f ''(x) and f ''''(x)?

It is shown in numerical analysis that given a function,
$f$ continous on $[a,b]$ with a countinous $f''$ then, let $T_n$ be the result of using the trapezoidal rule $n$ subdivisons. Then, the error term
If $E_n=\left| \int_a^b f(x)dx -T_n \right|$
satisfies,
$E_n\leq \frac{(b-a)^3}{12n^2}\cdot \max_{a\leq x\leq b}\{f''(x)\}$ (which exists by EVT)

**413** · October 22nd 2006, 09:48 AM

is f '' (x)= -2xe^-x^2 and f ''''(x)=-2e^-x^2 + 2xe^x^2 ?

**ThePerfectHacker** · October 22nd 2006, 10:53 AM

Originally Posted by 413

is f '' (x)= -2xe^-x^2 and f ''''(x)=-2e^-x^2 + 2xe^x^2 ?

We have,
$f=e^{-x^2}$
Thus,
$f'=-2xe^{-x^2}$
Thus, (product rule you forgot to use it

)
$f''=-2e^{-x^2}+4x^2e^{-x^2}$

**413** · October 22nd 2006, 09:31 PM

does En< 0.14875 seem right?

**ThePerfectHacker** · October 23rd 2006, 04:11 AM

Originally Posted by 413

does En< 0.14875 seem right?

For what? Trapezoids? Does not seem corret to me.

The following table are the error bounds for using the trapezoidal rule $n$ times:

**413** · October 23rd 2006, 04:34 AM

not correct?..hmm...then how do you actually compute it with the formula's you gave me?

**413** · October 24th 2006, 07:24 AM

and another question,
How big should we take n to guarntee absolute value of ET <0.0001? and also for EM and ES.

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