Quadrature (1 Viewer)

leehuan

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This supposedly really hard question with a total of five parts + no answers starts off with



When I saw the LHS and MHS I was ready to just smack it with




But when I saw the RHS I got scared and suddenly I don't know what I am doing already.
 

InteGrand

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This supposedly really hard question with a total of five parts + no answers starts off with



When I saw the LHS and MHS I was ready to just smack it with




But when I saw the RHS I got scared and suddenly I don't know what I am doing already.
Which RHS do you mean? The one in the integral's approximation? Or the one referring to the bound on the Simpson's Rule error?
 

InteGrand

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Assuming you mean for the error bound, the main thing is to make sure you're clear on what the symbols all refer to. In that particular formula, n refers to the number of parabolas used in the Simpson's Rule approximation. In your one, exactly one parabola is used, so n = 1.

Also, a refers to the lower limit of integration, and b the upper limit (so -1/k and 1/k respectively for this particular integral).

Finally, L refers to some upper bound for the fourth derivative of the function over the entire interval.

To find L for this Q., note that our integrand is f(t) = et. So f(4)(t) = et, which is bounded above by e1/k in your interval of integration. So L = e1/k does the job here.

The rest is just substitution into the quoted error bound formula to find a bound for the error in the estimate.
 
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Paradoxica

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This supposedly really hard question with a total of five parts + no answers starts off with



When I saw the LHS and MHS I was ready to just smack it with




But when I saw the RHS I got scared and suddenly I don't know what I am doing already.
I have no idea what your error estimation procedure is, but Wikipedia's method gives me approximately 1⁄(90k⁵)
 

InteGrand

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By the way, seanieg89 wrote a bit about numerical integration, including proofs of error bound formulas for various numerical integral methods and some exercises. It may be found here: http://community.boredofstudies.org/14/mathematics-extension-2/342900/numerical-integration.html, and could be of interest to curious HSC 4U students (who were the intended audience I think; it was basically inspired by the then recent MX2 BOS Trial I believe, which had some questions about error bounds for Trapezoidal Rule iirc).
 

InteGrand

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If the RHS of the integral approximation line was what you wanted clarification on, that's just the Simpson approximation to the given integral with one parabola: ((b-a)/6) * (f(a)+ 4*f(0) + f(b)).
 

leehuan

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Alright don't worry, thanks. I'll continue with this question later but it completely went over my head that it was just a direct substitution into Simpson's rule. I know what my variables all meant; I just wasn't sure where they generated that expression from.

I meant the original question. The formula for the error bound itself wasn't what was scary.

Basically I hesitated to find L because I wasn't even sure that substitution into the formula was the right thing to do.
 

seanieg89

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Yep. HSC students were the intended audience.

I think it is not ideal that students are taught Simpsons rule, usually without any accompanying discussion of when it doesn't work well or any proofs of error bounds (which are basically just IBP exercises).
 

Paradoxica

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Yep. HSC students were the intended audience.

I think it is not ideal that students are taught Simpsons rule, usually without any accompanying discussion of when it doesn't work well or any proofs of error bounds (which are basically just IBP exercises).
Unfortunately, IBP is not covered in anything but X2

However... I know a few people who only do 2u...

They self taught IBP.
 

leehuan

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Ok so the rest of the question... big time checking required.



So for part (a) I had

(would like verification)

Part (b) is just part (a) rearranged to give a quadratic
Part (c) is just part (b) put into the quadratic formula
Part (d) is trivially part (c) raised to the kth power...

So that made me think, the error doesn't change throughout right...? (this is my first question)

Then I just compared. And (...)^k appears more accurate than (1+1/k)^k when k=10. I didn't know what else to say or if there is some way to justify that either. (this is my second question - would you actually write down e=2.718281828459045... or somehow find the error for (1+1/k)^k)
 

InteGrand

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The error wouldn't stay the same throughout, it'd have to change. E.g. consider the raising to the k-th power step.

 
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leehuan

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Oh true. Ahh maybe I'll give up on this question for at least tonight then... seems quite heavy in difficulty now...

Thanks
 
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leehuan

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Forgot the n in my quadrature again...





I perfectly understand why L=6 and b-a = 2, but for some reason I keep trying to tell myself that n=2... when the answers have n=4...
 

1008

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Forgot the n in my quadrature again...





I perfectly understand why L=6 and b-a = 2, but for some reason I keep trying to tell myself that n=2... when the answers have n=4...
Pretty sure n=2. Remember it is to the power of 4 what question no. is it in the red booklet?
 

leehuan

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Pretty sure n=2. Remember it is to the power of 4 what question no. is it in the red booklet?
Yes see now that's my logic. But check the orange booklet (past papers) - the answers insist on n=4

I need someone to tell me who's wrong


Sent from my iPhone using Tapatalk
 

1008

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Yes see now that's my logic. But check the orange booklet (past papers) - the answers insist on n=4

I need someone to tell me who's wrong


Sent from my iPhone using Tapatalk
which past paper is it? what year?
 

1008

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Yes see now that's my logic. But check the orange booklet (past papers) - the answers insist on n=4

I need someone to tell me who's wrong


Sent from my iPhone using Tapatalk
Found it, that's a typo. Because, in the first part, where they actually ask you to find the Simpson's approximation, they take h = 1. Which is only possible if n =2
 

leehuan

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Thanks for confirming my doubt. Yep, my lecturer also made us aware of the 180 version instead of the 2880 version, and I literally considered the fact that 2880 is the bigger number myself.
 

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