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F(x) = -3x + 8 for f(2h – 5)

User Wayne Kuo
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1 Answer

4 votes

Answer:

f(x+2h)=−(f(x)+2hf′(x)+4h22f′′(x)+o(h3)==−f(x)−2hf′(x)−4h22f′′(x)−o(h3)

f(x−2h)=f(x)−2hf′(x)+4h22f′′(x)−o(h3)

8f(x+h)=8(f(x)+hf′(x)+h22f′′(x)+o(h3)=8f(x)+8hf′(x)+8h22f′′(x)+8o(h3)

−8f(x−h)=−8(f(x)−hf′(x)+h22f′′(x)−o(h3)=−8f(x)+8hf′(x)−8h22f′′(x)+8o(h3)

So

|f′(x)−112h[−f(x+2h)+8f(x+h)−8f(x−h)+f(x−2h)]|==|f′(x)−112h[−4hf′(x)−2o(h3)+16hf′(x)+16o(h3)]|==|f′(x)−f′(x)−14o(h3)12|=o(h3)?

2.

We just plug x+2h,x+h,x−h,x−2h to the approximation and get:

f′′(x)≈[−f(x+4h)−16f(x+3h)+16f(x+h)−130f(x)+64f(x+2h)+64f(x−h)+64f(x−2h)−64f(x−3h)]144h2

Explanation:

User Dima Ogurtsov
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