Simpson's Rule (1 Viewer)

[emanuel]

Is there a difference between the way the Simpson's Rule is given in 2 unit and Extension 1. Because I am a 3 unit student and find myself getting questions about this wrong in 2 unit papers, the Simpson's Rule as I learned it is as follows:
(1/6)(b-a)(f(a) + 4f((a+b)/2) + f(b))

Not sure if perhaps the 2 unit course has a different expression for it?

 

[acmilan]

er the above formula only works when there are 3 function values, what if they give you 4 or 5 or 6 or more? (which is highly possible) I dont know what you mean by 2 and 3 unit version, there should only be one version.

 

[Dreamerish*~]

There's another way.

I'm also do 3U, but I didn't know the two rules were categorised as 2U and 3U rules.

h/3.[y₀ + y_n + 4(y₁ + y₃ + ... all the other odd function values) + 2(y₂ + y₄ + ... all the other even function values)]

I always found that one easier.

EDIT: Acmilan's right. I rememeber someone telling me that the h/6 formula only works for 3 function values, and if there are more than 3, you need to repeat it.

Now I realise why I found the h/3 formula easier.

 

[laurel18]

way i learnt it:

h/3 [(y0 + yn) +4 (y1 + y3 + y5+...) + 2(y2 + y4 + y6+...)]

where h is the width

but i don't suggest you learn it this way as you might get confused in exam...

 

[Dreamerish*~]

way i learnt it:

h/3 [(y0 + yn) +4 (y1 + y3 + y5+...) + 2(y2 + y4 + y6+...)]

where h is the width

but i don't suggest you learn it this way as you might get confused in exam...

This way is much less confusing, in my opinion.

The h/6 method words three function values at a time. Therefore if he gets a question with nine funtion values, he has to do three calculations, whereas with the h/3 formula, you can input everything into one and just let your calculator do the work.

The important thing is not to get h/3 and h/6 mixed up.

 

[laurel18]

yeah I think so too Dreamerish*~...so probably ditch the first formula and then learn this one which you can use for everything...then you're good to go...

 

[emanuel]

thanx guys...saved me a lot of heartache

 

[laurel18]

don't worry...there is still 3 weeks to learn

 

[rama_v]

The way I learnt it, it can be applied to everything
h = (b-a)/n where n is one less than the number of function values (i.e. n is the number of intervals/strips)

A = ^h/₃ [f(a) + 4f(a + h) + 2f(a + 2h) + 4f(a + 3h) +...+f(b) ]

Its basically the same as the other formulas mentioned above.

 

[acmilan]

The way I learnt it, it can be applied to everything
h = (b-a)/n where n is one less than the number of function values (i.e. n is the number of intervals/strips)

A = <sup>h[ /sup]/₃ [f(a) + 4f(a + h) + 2f(a + 2h) + 4f(a + 3h) +...+f(b) ]

Its basically the same as the other formulas mentioned above.</sup>

^{You, my friend, learnt the proper way to do it}

 

[emanuel]

Same goes with the Trapezoidal rule I just discovered I had the crap version... I am sad to say that this may have been my teacher's blunder

 

[rama_v]

Same goes with the Trapezoidal rule I just discovered I had the crap version... I am sad to say that this may have been my teacher's blunder

haha yeah..i was very lucky our teacher told us the good way

 

[pLuvia]

There's another way.

I'm also do 3U, but I didn't know the two rules were categorised as 2U and 3U rules.

h/3.[y₀ + y_n + 4(y₁ + y₃ + ... all the other odd function values) + 2(y₂ + y₄ + ... all the other even function values)]

I always found that one easier.

EDIT: Acmilan's right. I rememeber someone telling me that the h/6 formula only works for 3 function values, and if there are more than 3, you need to repeat it.

Now I realise why I found the h/3 formula easier.

yeh my tutor said that way is much easier, some textbooks expand the brackets making it harder to understand and same with trapezoidal rule.

 

[Meldrum]

Seeing how there are different rules and different methods of achieving an answer, can they theoretically ask you a question where only one version of the rule can achieve the answer?

 

[rama_v]

Seeing how there are different rules and different methods of achieving an answer, can they theoretically ask you a question where only one version of the rule can achieve the answer?

no, because all the rules, when applied correctly, should yield the same answer...

 

[volition]

Haha! my way rips up all of your ways, its soo much easier

 

[pLuvia]

Seeing how there are different rules and different methods of achieving an answer, can they theoretically ask you a question where only one version of the rule can achieve the answer?

The rule is the same version if you look at it closely you just rearrange the numbers and factorise it, or expand etc.

 

[MichaelPH]

I have only learnt the h/3 formula but got the answer wrong in the 2000 paper, when i'm sure i'm doing it correctly- the solutions use the h/6 formula, getting a different result.

 

[rama_v]

I have only learnt the h/3 formula but got the answer wrong in the 2000 paper, when i'm sure i'm doing it correctly- the solutions use the h/6 formula, getting a different result.

lol, its because they ask for 5 or another number of function values when you've only used three - I would strongly suggest not to use those formuilas that are designed assuming that you have 3 function values or whatever...but you should get the same answer when you apply the formula you used twice.