Question

so its been a while since i were here and i only need help on 1 cal 2 question so far.. here ill type it out below

Answer 1

\[\int\limits_{0}^{1}e ^{x ^{2}}dx\] so yeah I cant use sequences and series yet... and this integral has no closed form so were expected to estimate it using simpsons rule and using it in excel so we dont jam our calculator entering each value 1 at a time... with that in mind my only concern so far is how do i chose a n value in which there is less then one billinth of an error... i do know the error function but am not %100 sure if im doing it right... also DX how do i get this to work in excel ive created the colums/ groups but it wont follow the forumla going down the page and i have to do it that way... any suggestions?

Answer 2

What does it mean to "jam" one's calculator?

Answer 3

I hope you can approximate it using geometric shapes (or Simpson's rule) ?

Answer 4

oh, yes, the answer to my question is yes!!

Answer 5

how many shapes do you have to use?

Answer 6

@SolomonZelman thats what i gotta figure out 1st

Answer 7

well if it doesn't give you that you must use some number n for shapes, with a maximum error less than some number.... or if you aren't given anything else, then I guess you can choose any even number .

Answer 8

even - so that it can work in Simpson's rule

Answer 9

if you have no other info, I would advise n=4

Answer 10

I guess without any more info, make your table for simpson's rule

Answer 11

what error function do you have ?

Answer 12

oh, 1 billionth of maximum error.

\(\large\color{black}{ \displaystyle{\rm E_{Simpson's~~Rule}}\le\frac{k(b-a)^5}{180n^4}=\frac{1}{10^9} }\)

Answer 13

where k is the bound on the absolute value of the 4th derivative

Answer 14

n is number of triangles,
and b-a is 1-0 which is 1

Answer 15

\(\large\color{black}{ \displaystyle\frac{k(1-0)^5}{180n^4}=\frac{1}{10^9} }\)

graph the abs. value of 4th derivative from 1 to 0 and plug in the hieghest/maximum value on the graph (on that interval) for k.

Answer 16

I plotted in wolfram alpha and I am getting 220.
http://www.wolframalpha.com/input/?i=4th+derivative+of+e%5E%28x%5E2%29+from+1+to+0

(didn't need to make it absolute value because it is positive either way from 1 to 0...)

Answer 17

\(\large\color{black}{ \displaystyle\frac{220(1-0)^5}{180n^4}=\frac{1}{10^9} }\)

with mathematica10.0 it would be easier:)
just by typing

\(\large\color{black}{ \displaystyle{\rm Solve}\left[\frac{220(1-0)^5}{180n^4}={\tiny~}=\frac{1}{10^9}\right] }\)
and then rounding n to the bigger whole number

Answer 18

don't have it on this comp I am using...

Answer 19

according to wikipedia, the error in a region with width \(\Delta\)x
is \[ \frac{1}{90} \left(\frac{\Delta x}{2}\right)^5 \left| f^{(4)}(\xi)\right| \]
the 4th derivative of exp(x^2), according to wolfram, is
\[ 4e^{x^2} \left( 4x^4 +12x^2 +3\right) \]
if we evaluate that at its largest (i.e. when x=1, the biggest x can be for us), we
get 4(19)= 76

thus the error function is
\[ \frac{76}{90\cdot 32} (\Delta x)^5 \]

if we divide the interval 0 to 1 into "n" parts, then \(\Delta\)x= 1/n
if all n parts have an error no more that the above, the total error will be less than
\[ \frac{76}{2880} n \cdot \frac{1}{n^5} \\ \approx \frac{0.0264}{n^4}
\]
we want this less than a billionth, so
\[ \frac{0.0264}{n^4} \le 10^{-9} \\ n^4 > 0.264 \cdot 10^9 \]
this gives n> 72,
so I would try about n=100 as a good start

Answer 20

thanks guys... makes sense on the error function part to get a n value so we have such little error.. ima figure out tho how to get the microsoft excel spread sheet to work now *assuming it doesnt give me trouble*