Quadrature (2 Viewers)

leehuan · May 14, 2016

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

$$a) Estimate the error in the Simpson's rule approximation$\\ e^{\frac{1}{k}}-e^{-\frac{1}{k}}=\int_{- \frac{1}{k} }^{\frac{1}{k}}{e^tdt}\approx \frac{1}{3k}\left(e^{\frac{1}{k}}+4+e^{-\frac{1}{k}}\right)$

When I saw the LHS and MHS I was ready to just smack it with
$\left| I-S_n \right|\le \frac{L{(b-a)}^5}{2880n^4}$

$$for some $\left|f^{(4)}(x)\right|\le L$ over $x\in [a,b]$

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

InteGrand · May 14, 2016

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

$$a) Estimate the error in the Simpson's rule approximation$\\ e^{\frac{1}{k}}-e^{-\frac{1}{k}}=\int_{- \frac{1}{k} }^{\frac{1}{k}}{e^tdt}\approx \frac{1}{3k}\left(e^{\frac{1}{k}}+4+e^{-\frac{1}{k}}\right)$

When I saw the LHS and MHS I was ready to just smack it with
$\left| I-S_n \right|\le \frac{L{(b-a)}^5}{2880n^4}$

$$for some $\left|f^{(4)}(x)\right|\le L$ over $x\in [a,b]$

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 · May 14, 2016

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) = e^t. So f⁽⁴⁾(t) = e^t, which is bounded above by e^1/k in your interval of integration. So L = e^1/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.

Paradoxica · May 14, 2016

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

$$a) Estimate the error in the Simpson's rule approximation$\\ e^{\frac{1}{k}}-e^{-\frac{1}{k}}=\int_{- \frac{1}{k} }^{\frac{1}{k}}{e^tdt}\approx \frac{1}{3k}\left(e^{\frac{1}{k}}+4+e^{-\frac{1}{k}}\right)$

When I saw the LHS and MHS I was ready to just smack it with
$\left| I-S_n \right|\le \frac{L{(b-a)}^5}{2880n^4}$

$$for some $\left|f^{(4)}(x)\right|\le L$ over $x\in [a,b]$

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 · May 14, 2016

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 · May 14, 2016

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 · May 14, 2016

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 · May 14, 2016

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 · May 14, 2016

seanieg89 said:
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 · May 14, 2016

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

$\\ $b) Estimate the error in the approximation $\\ (3k-1)e^{\frac{2}{k}}-4e^{\frac{1}{k}}-(3k+1)\approx 0\\ $c) Estimate the error in the approximation $\\ e^\frac{1}{k}\approx\frac{2+\sqrt{3+9k^2}}{3k-1}\\$d) Estimate the error in the approximation $\\ e\approx {\left[\frac{2+\sqrt{3+9k^2}}{3k-1}\right]}^k\\$ Compare this estimate when $k=10$ with the estimate $e\approx {\left(1+\frac{1}{k}\right)}^k$ when $k=10$

So for part (a) I had
$$Error$=\frac{e^\frac{1}{k}}{90k^5}$
(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 · May 14, 2016

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

$\noindent If we have some quantity $x$ estimated by $\hat{x}$ with error $\Delta x$, we can write $x = \hat{x}+\Delta x$. Then (with $k$ a fixed positive integer say), we have (from Binomial Theorem) $x^k = \left(\hat{x}+\Delta x\right)^k = \hat{x}^k + k\hat{x}^{k-1}\Delta x + O\left(\Delta x ^2 \right)\neq \hat{x}^{k} + \Delta x$ in general. In other words, the error in the $k$-th power of the approximation to $x$ is not just the same as that in the approximation to $x$. We can estimate this new error as $k\hat{x}^{k-1}\Delta x$ if $\Delta x$ is very small, as the higher order terms would become negligble.$

leehuan · May 14, 2016

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

Thanks

leehuan · Jun 3, 2016

Forgot the n in my quadrature again...

$$A function $f$ satisfies $\left|f^{(4)}(x)\right| \le 6$ on the interval $[1,3]$. It is known that:$\\ f(1)=0.26, f(1.5)=0.75, f(2)=1.12, f(2.5)=1$ and $f(3,0)=1.86$

$Use the error estimate $\left| I-S_n \right| \le \frac{L{(b-a)}^5}{2880n^4}$ where $S_n$ is the Simpson approximation using $n$ intervals, to establish the size of the error in part (a).$

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 · Jun 4, 2016

leehuan said:
Forgot the n in my quadrature again...

$$A function $f$ satisfies $\left|f^{(4)}(x)\right| \le 6$ on the interval $[1,3]$. It is known that:$\\ f(1)=0.26, f(1.5)=0.75, f(2)=1.12, f(2.5)=1$ and $f(3,0)=1.86$

$Use the error estimate $\left| I-S_n \right| \le \frac{L{(b-a)}^5}{2880n^4}$ where $S_n$ is the Simpson approximation using $n$ intervals, to establish the size of the error in part (a).$

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 · Jun 4, 2016

1008 said:
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 · Jun 4, 2016

leehuan said:
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 · Jun 4, 2016

leehuan said:
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

InteGrand · Jun 4, 2016

$\noindent The confusion here has arisen because there are two possible main conventions. One is where `$n$' refers to the number of parabolas used in the approximation, and the other is where `$n$' refers to the number of \textsl{sub-intervals} used. Note that one of these is just twice the other one, i.e. $\boxed{\text{no. of subintervals used }=2\times \text{no. of parabolas used}}$.$

$\noindent Let's write $N$ for the no. of sub-intervals and $P$ for the no. of parabolas used, so $N=2P$ (so note $P$ can be any value $1,2,3,\ldots$, whereas $N$ can only be even values $2,4,6,\ldots$). Then the error formulas for each convention would be the same except for a number in the denominator, which comes about by replacing an $N$ with $2P$. Since the denominator involves a fourth power of the no. of subintervals or parabolas, if the formula with $N$ (sub-intervals) has $CN^4$ in the denominator, then the formula with $P$ (no. of parabolas) has $C\left(2P\right)^4 = 16CP^4$ in the denominator.$

$\noindent In other words, the formula with $P$ (parabolas) has the \emph{bigger number in the denominator} ($16$ times the size of denominator in the the error formula with $N$, no. of sub-intervals).$

$\noindent If you have seen the other convention's error formula, you'll see it has `$180n^4$' in the denominator. Note that $2880$ is the bigger number (and is indeed $16\times 180$). So the one with $\color{blue}{2880}$ must have its `$n$' referring to \color{blue}no. of parabolas\color{black}. The one with $\color{red}{180}$ has its `$n$' referring to the \color{red}no. of sub-intervals\color{black}.$

$\noindent So in this particular question, I believe the `$n$' should be 2 (since it should refer to no. of parabolas). The use of $n$ as $4$ in the answers may not have just been a typo, it may have been brought about by the confusion between the two conventions and formulas (since $4$ would be right to use if it was no. of sub-intervals instead).$

leehuan · Jun 4, 2016

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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