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17074206张颖威第九次作业

作者:张颖威   发布时间:2019-06-23 19:40:44   浏览次数:27

 

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2. Use the three-point centered-difference formula to approximate f ′ (0), where f(x) = e x , for (a) h = 0.1 (b) h = 0.01 (c) h = 0.001.

Solution:

Because: f(x+h)=f(x)+f’(x)*h+f’’(x)/2*h^2+f’’’(c1)/6*h^3

f(x-h)=f(x)-f’(x)*h+f’’(x)/2*h^2-f’’’(c1)/6*h^3

So: f(x+h)-f(x-h)=2*f’(x)*h+h^3/6*(f’’’(c1)+f’’’(c2))

Then, f’(x)≈(f(x+h)-f(x-h))/2h

(a)  f’(0)=5*(e^0.1-e^(-0.1))

(b) f’’(0)=50*(e^0.01-e^(-0.01))

(c)  f’’’(0)=500*(e^0.001-e^(-0.001))

 

1.     Apply the composite Trapezoid Rule with m = 1,2, and 4 panels to approximate the integral.Compute the error by comparing with the exact value from calculus.





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