What is
integral [0 to 2] e^-x^2 dx
You are talking about the error function. That integral cannot be expressed in a closed form in terms of other simple functions.
The answer I get, taken from an error function table, is
0.99532 * [sqrt(pi)]/2
=0.88208...
the whole question is like this....
a) With f(x)= e^-x^2 , compute approximations using midpoint, trapezoidal and simpson's rule with n=2.
I found that from midpoint rule gives 0.88420, trapezoidal rule gives 0.87704, and simpson's rule gives 0.82994.
Then it asks me to
Compute the error esimates for midpoint, trapezoidal, simpson's. Carefully examine the extreme values of f ''(x). You may use that |f ''''(x)| < 12 for 0<x< 2. (the < signs all mean less than or equal to).
How would you do this?
There are formulas to put bounds on the errors for each of those formulas when approximating continuous functions. You'll have to check your text for those, because those are not common formulas that anyone would memorize, unless it was their specialty in numerical analysis.
the formula in the text is ET< K(b-a)^3/12n2
but what is k?
Your text should tell you somewhere. Typically the error depends on the number of subintervals and the interval limits.
I did a google search and found
ET <= (b-a)3/12n2 * Max [f"(x)] for a <= x <= b
so K is Max [f"(x)]
Your function is e-x2
Find f"(x) and find the max for that on [0,2].
f''(x)=-2e^-x^2 + 4x^2e^-x^2
so max=0.89252
ET< (2)^3/12(2)^2 * 0.89252
ET< 8/48*0.89252
ET< 0.14875 seem right?
That seems a little high, but I didn't verify the max of f"(x). I calculated the two intervals and got the area as 0.877037261
I checked the area with my graphics app. and it looks like .8821, so the difference is 0.005062739 which is certainly less than your estimate of .14875, but it seems to me the estimate should be better. I could be wrong though.
then what did i do wrong?...is my f''(x) correct?
I think the max occurrs at sqrt(2)/2 = .707106...
So the max I got f"(x) is 2.4261, so your error estimate is actually lower than what I've calculated.
It could be right considering there are only 2 subintervals. We never want to understimate the error too. Just double check the calculations and go with what you get from that.
I don't think your f"(x) is correct. I'm also doubtful about the max I got because I used your f" in my calculations. I'll try to check this later, right now I have other obligations.
Ok, I checked and I got a different f"(x). I got
f"(x)=e-x2(4x2 - 2)
The max occurs at x=sqrt(1.5) and f"(sqrt(1.5))= .8925, so the error is less than (1/6)*.892521=.148753 so I think your estimate looks correct.
How big should we take n to guarntee absolute value of ET <0.0001? and also for EM and ES.
You should have no trouble finding bounds or interval counts for ET, you know the max f"(x), so if
ET < Error
(b-a)3/12n2 < Error so
(b-a)3/12*Error < n2
Take the square root and use the first positive integer greater than that.
For EM and ES there are formulas in your text similar to ET and you should now be getting a handle on how to use them. If you have questions or want to review it might be best to post a new question because I don't often flip back after a problem gets to page three.
What is the prime factorization of 900?
Help with Prime Factorization.
28/35
To find the prime factorization of 900, you can start by dividing the number by the smallest prime number, which is 2. If 900 is divisible by 2, divide it by 2 and repeat the process. If it is not divisible by 2, move on to the next prime number, which is 3, and continue the process until you have reached the square root of the original number or until the quotient is 1.
In this case, let's start with dividing 900 by 2:
900 ÷ 2 = 450
450 ÷ 2 = 225
Since 225 is not divisible by 2, we move on to the next prime number, which is 3:
225 ÷ 3 = 75
75 ÷ 3 = 25
25 ÷ 5 = 5
Now we have reached a prime number, which is 5, and the quotient is 5.
So the prime factorization of 900 is 2^2 * 3^2 * 5^2, or 2 * 2 * 3 * 3 * 5 * 5.