Question

Trapezoid rule, Simpson's Rule.

Answer 1

"a) A certain computer takes two seconds to compute a certain definite integral accurate to 4 decimal places. using the trapezoid rule. Approximately how long will it take to get 8 digets correct? 12? 20?
b) repeat part a assuming Simpson's rule is being used throughout"

Answer 2

So is this question the type where we want to figure out Error?

Answer 3

@Kainui :)

Answer 4

@zepdrix

Answer 5

I wouldn't call this a matter of "finding the error."

Rather, I'd recognize that it takes longer to approximate a sum (which represents area here) correct to eight decimal places than it does to just four decimal places. Assuming that it "costs" the same amount of time to improve the accuracy by 1 decimal place, estimate how long it would take for this computer to approx. the area to eight places.

Answer 6

Right

Answer 7

@mathmale so how do I set this up?

Answer 8

You are not asked to compute anything other than a period of time for the necessary computing, so there's practically nothing to set up.

'A certain computer takes two seconds to compute a certain definite integral accurate to 4 decimal places." If that's so, then how long will the computer take to compute this integral to 1 decimal place? to 8 decimal places?

I may be overlooking something here, but this straightforward approach is the one I'd take myself.

Answer 9

2 seconds = 4 decimal places
1 seconds = 2 decimal places
is that what you mean? Why would it even mention Trapezoid rule then?

Answer 10

@Astrophysics :>

Answer 11

I think i'm meant to find the bounds that make = 2 (Seconds) but then I also have to find the f(x) part.. o.o

Answer 12

How do you know the complexity grows linearly ?

Answer 13

What's the "error formula" in trapezoidal rule ?

Answer 14

\[|E_T|\le \frac{ k(b-a)^3 }{ 12n^2 }, |f''(x)|\le k\]

Answer 15

So the error is proportional to \(\dfrac{1}{n^2}\)

where \(n\) is the number of intervals, yes ?

Answer 16

mm yes

Answer 17

Let me ask another simple question

Answer 18

Suppose I say my height is "6 feet".

whats the "greatest possible error" in my height If I say the estimation is "one feet accurate" ?

Answer 19

1 foot?

Answer 20

Nope, think a bit

Answer 21

If my measured height is 5.6 feet, I would round it to 6 feet.

What would I round to if my measured height is 6.3 feet ?

Answer 22

well..rounding up or rounding down? 6ft?

Answer 23

Yes, with an 1 foot accuracy device,
my "actual height" can be anywhere from 5.5 to 6.5.

The "greatest possible error" is then 0.5 foot, yes ?

Answer 24

Oh, yes, I see xD

Answer 25

Lets do another quick exammple

Answer 26

Suppose I say my height is "6 feet".

whats the "greatest possible error" in my height If I say the estimation is "one decimal place accurate" ?

Answer 27

5.9 to 6.1

Answer 28

I am not asking that

Answer 29

I am asking about the "greatest possible error"

not the range

Answer 30

0.1

Answer 31

Wrong again, think a bit

Answer 32

If my measured height is 6.07 feet, I would round it to 6.1 feet.

If my measured height is 6.04 feet, I would round it to 6.0 feet.

So the "greatest possible error" is simply 0.05 foot, yes ?

Answer 33

Notice 0.05 = 0.1/2

Answer 34

If it is still confusing, I suggest you spend some time on this first...

just google "greatest possible error"

Answer 35

lol yeah I should do that :/ but that's right there in the formula
delta(x)/2 yeah?

Answer 36

Exactly!

Answer 37

Lets get back to our original problem

Answer 38

We're given that the definite integral is accurate to 4 decimal places.

What's the greatest possible error here ?

Answer 39

0.0001/2 = 0.00005

Answer 40

Perfect!

Answer 41

yeeay! first answer i've gotten right! xP

Answer 42

What's the greatest possible error if we want it accurate upto 8 decimal places ?

Answer 43

0.00000001/2 =

Answer 44

four decimal places : 0.00005 = \(5 * 10^{-5}\)

eight decimal places : \(5 * 10^{-9}\)

?

Answer 45

And we know that above error is proportional to \(\dfrac{1}{n^2}\)

Answer 46

\[5*10^{-5} = \dfrac{k}{{n_4}^2}\]

\[5*10^{-9} = \dfrac{k}{{n_8}^2}\]

Answer 47

divide them and get

\[\dfrac{n_8^2}{n_4^2} = 10^4\]

Answer 48

still with me ?

Answer 49

Yup yup. Just taking it all in heh

Answer 50

Here, \(n_4\) is the number of intervals needed for getting 4 decimal places accuracy

Answer 51

similarly, \(n_8\) is the number of intervals needed for getting 8 decimal places accuracy

Answer 52

Take square root and get
\[\dfrac{n_8}{n_4} = 10^2 \]

or
\[n_8 = 100n_4\]

Answer 53

What does that tell you ?

Answer 54

take your time, there is a lot to digest here..

Answer 55

why do we want to divide them by each other?

Answer 56

to eliminate the proportionality constant \(k\) silly

Answer 57

Our goal is to get a relation between \(n_4\) and \(n_8\), so that we know how many more intervals we need to increase the accuracy from 4 decimal places to 8 decimal palces

Answer 58

dividing both equations gets rid of \(k\)

Answer 59

ahh ok

Answer 60

\[n_8 = 100n_4\]

This tells us that we need to increase the intervals by 100 times in order to improve the accuracy from 4 decimal places to 8 decimal places.

Answer 61

can we assume that that is a constant type of thing?
like, could we say that to get the accuracy from 8 to 12 would also be 100x
or that from 4 decimals to 12 would be 200x?

Answer 62

You're half correct

Answer 63

4 to 8 decimal places : 100x
8 to 12decimal places : 100x

therefore 4 to 12 decimal places should be : 10000x right ?

Answer 64

oh, yes xD
so when this asks..approximately how long it will take to get to 12 decimal places, that would be calculated from the 4, yeah?
and then to get to the 20..it would be.. 16*100 ?

Answer 65

Nope

Answer 66

4 to 8 : 2*100

8 to 12 : 2*100^2

12 to 16 : 2*100^3

16 to 20 : 2*100^4

Answer 67

To increase the accuracy by 4 more decimal places, you need to do 100 times more work

Answer 68

so 8 decimal places accuracy takes approximately 200 seconds

Answer 69

12 decimal places accuracy takes approximately 20000 seconds

Answer 70

20 decimal places accuracy takes approximately 2*10^8 seconds

Answer 71

That is approximately 6 years !

Answer 72

For 20 decimal places accuracy using trapezoidal rule and your super computer, you will need to wait 6 years !

Answer 73

hahah such a long time! x.x perhaps not such a super computer after all :P

Answer 74

but if from
n_0 to n_4 it takes 2 seconds
and we increase those seconds by 100 times to get n_4 to n_8 then we have 2*100
which is where you get 200?

Answer 75

cheap computers these days can do "addition" in "one nanosecond"

Answer 76

You have understood it correctly

Answer 77

1 nanosecond? gwoow

Answer 78

Whats the frequency of the processor on your computer ?

Answer 79

So then later when we're going from
n_8 to n_12 why do we put the ^2 on the 100?

Answer 80

eer idk, how does one check that? xD

Answer 81

tell me your computer model

Answer 82

oh nvm that question, it's because it's an increase by 100 xD

Answer 83

Yep

Answer 84

oh nvm that question, it's because it's an increase by `a factor of ` 100 xD

Answer 85

Ideapad s400 touch lenovo

Answer 86

http://www.pcworld.com/product/1265505/lenovo-ideapad-s400-notebook.html

Answer 87

you have intel i3 processor with 2 cores

each core operating at a frequency of 1.8 GHz

Answer 88

frequency is 1.8 GHz

given frequency, do you know how to calculate the "time period" ?

Answer 89

Nope xD I am ignorant in this area. Pray tell o.o

Answer 90

haven't you heard of "frequency" and "time period" before ?

Answer 91

well..yes

Answer 92

[question: then to get 20 decimal places would it be 20*10^(what is this number then?)]

Answer 93

20 decimal places accuracy takes approximately 2*10^8 seconds

Answer 94

wait.
12 decimal places takes 12*10^8
and 20 takes 2*10^8 ?

Answer 95

was that meant to be a 20?

Answer 96

how did you get 12*10^8 for 12 decimal places ?