Member Avatar for miunhosuk

Hi
this is pdf :
I have question with problem 1

I solved problem 1 but i am not sure if my answers are right.:(
can you check?? plz!!

my answers

for a) i said,
if applying the composite simpson rule on a uniform mesh subdividing [a,b] into intervals of width h/2, two halves of the interval are [a,c] [c,b]
and d,e are midpoints of these two subintervals
d= (a+c)/2 , e= (c+b)/2

apply Simpson's rule to each subinterval, a quadrature rule on [a,b]
S_2 = h/12 ( f(a)+4f(B)+2f(c)+4f(e)+f(b) )

so, S_2 = I(f)- (1/15) (S_2(f)-S(f))


for a) i said,
The two can be combined to get an even more accurate approximation Q
Both rules are of order 4, but step size for S_2 is half the size for S, so the error in S_2 is roughly 2^4=16 times smaller than that of S.
Thus, a new, more accurate rule Q can be obtained by solving
Q-S=16(Q-S_2)
so, S(f)=S_2(f)+ 1/15 ((S_(f)-S(f))


to compute integral[0, pi/2] cos(x) dx,
code:
f=inline('cos(x)')
Q=quad(f,0,pi/2)


and i have no idea what to do with part (c)

can anyone help me?!!!! plz!!!!

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  • Latest Post Latest Post by sagemore48

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Good start, — the idea of computing Simpson on a mesh and on the same mesh with half the step, then combining those two results to cancel the leading error, is the right strategy. The algebra in your post got a bit tangled when rearranging; the clean statement is that Richardson extrapolation removes the O(h^4) term and produces an O(h^6) approximation.

The practical formula to use is
Q = (16*S2 - S) / 15
where S is the composite Simpson result with step h and S2 is the Simpson result with step h/2. An easy error estimate for the finer Simpson value is approximately (S2 - S)/15 (signed), and Q will be dramatically more accurate (error drops from O(h^4) to O(h^6)).

For part (c) (f(x)=cos x on [0,pi/2]) note the exact integral equals 1, so you can check convergence directly. Compute S with an even number of subintervals (e.g., n=2), compute S2 with 2n, then form Q. Repeat doubling n to see the error shrink roughly by the expected rate. Keep n even — Simpson requires that.

A short, minimal Python recipe (pure composite Simpson + extrapolation) will get you the numbers and the convergence behavior quickly:

import math

def simpson(f,a,b,n):
    if n%2: raise ValueError('n must be even')
    h=(b-a)/n
    s=f(a)+f(b)
    s+=4*sum(f(a+i*h) for i in range(1,n,2))
    s+=2*sum(f(a+i*h) for i in range(2,n,2))
    return s*h/3

S = simpson(math.cos, 0, math.pi/2, 2)
S2 = simpson(math.cos, 0, math.pi/2, 4)
Q = (16*S2 - S)/15
print(S, S2, Q, 'exact=1')

Troubleshooting: check index ranges in your Simpson sums, ensure n is even, and compare to the exact value 1 to verify correctness. 's reminder about tone is fair; is right to suggest calm, methodical checking.

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Member Avatar for griswolf

NEED NOT TO YELL at people from whom you are asking help!

Member Avatar for miunhosuk

sorry.............. :(
but i didnt yell at ppl whom im asking help...
I just yelled at me ....cuz i hate myself :'(

Member Avatar for sagemore48

hei,there,it is OKay,it was in 911 for u,wasn't it?
in my way,just be in calm,then search yr problem using the keyword ,like the google engine.
thk u!
sage!

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