Hi.. I do not get why changing the step size doesn\'t affect the solution? and t

Question

Hi.. I do not get why changing the step size doesn't affect the solution? and this last statement in red

can someone explain please?

Use a centered difference approximation of O(h2) to estimate the second derivative of the following function. f(x) = 25x3-6x2 + 7x-88 Perform the evaluation at x = 2 using step sizes of h = 0.25 and h 0.125. Compare your estimates with the true value of the second derivative. Interpret your results on the basis of the remainder term of the Taylor series expansion. Solution: True value "f" (x) 150x-12 f" (2) 150(2)-12 h0.25: f"(2)= 288 f (2.25)-2/(2)+f(1.75)-182. 1406-102)-39.85938 = 288 0.252 0.252 () I 25 i = 288 0.125 0.125 Both results are exact because the errors are a function of 4th and higher derivatives which are zero for a 3rd order polynomial

Explanation / Answer

Here the approximate equation used to calculate the 2nd order derivative is as follows:

f''(x)~=[f(x+h)-2f(x)+f(x-h)]/h2

By using the Taylor series expansion we get:

f(x+h)=f(x)+hf'(x)+(h2/2)f''(x)+(h3/6)f'''(x)+(h4/24)f''''(e+) ,where e+ belongs to (x,x+h)

similarly we have the expansion for f(x-h)

f(x-h)=f(x)-hf'(x)-(h2/2)f''(x)-(h3/6)f'''(x)-(h4/24)f''''(e-) ,where e- belongs to (x-h,x)

putting the values in the main equation we get:

[f(x+h)-2f(x)+f(x-h)]/h2 = f''(x)+h2/24(f''''(e-)+f''''(e+)) = f''(x)+h2/12f''''(e) ,where e belongs to (x-h,x+h).

Since there is a function of h only in the fourth derivative of f(x),it vanishes.So changing the step size doesnot change the value of f''(x).The same is explained in the red box as the fourth derivative is 0 so there is no change in value of f''(x).

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