Transcription of TRAPEZOIDAL METHOD: ERROR FORMULA
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TRAPEZOIDAL METHOD: ERROR FORMULA
TRAPEZOIDAL METHOD: ERROR FORMULAT heoremAssumef(x) twice continuously differentiable on theinterval [a,b]. ThenETn(f) := baf(x)dx Tn(f) = h2(b a)12f (cn)for somecnin the interval [a,b].Later we will say something about the proof of this result, as itleads to some other useful formulas for the above FORMULA says that the ERROR decreases in a manner thatis roughly proportional toh2. Thus doublingn(and halvingh)should cause the ERROR to decrease by a factor of approximately is what we observed with some past examples from thepreceding evaluatingI= 20dx1 +x2using the TRAPEZOIDAL methodTn(f). Let us bound the errorETn(f) = h2(b a)12f (cn)Here,b a= 2. We bound|f (cn)|by max0 x 2|f (x)|.Calculate the derivatives:f (x) = 2x(1 +x2)2,f (x) = 2 + 6x2(1 +x2)3,f (x) =24x(1 x2)(1 +x2)4 Forx (0,2),f (x) = 0 only whenx= 1. Somax0 x 2 f (x) = max{ f (0) , f (1) , f (2) }= 2 Therefore, ETn(f) h2(2)12 2 =h23 ETn(f) h23 How large shouldnbe chosen in order to ensure that ETn(f) 5 10 6(1)To ensure this, we choosehso small thath23 5 10 6 This is equivalent to choosinghandnto satisfyh.
ET n (f) Z b a f(x)dx T n(f) can be analyzed by adding together the errors over the subintervals [x 0;x 1], [x 1;x 2], :::;[x n 1;x n]. Recall Z +h f(x)dx h 2 [f( ) + f( + h)] = h3 12 f00(c) Then on [x j 1;x j], Z x j x j 1 f(x)dx h 2 [f(x j 1) + f(x j)] = h3 12 f00(j) with x j 1 j x j. Combining these errors, we obtain ET n (f) = h3 12 f00(1 ...
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