Answer to Question #111446 in Quantitative Methods for Anju Jayachandran

Question

From the values of f(x) = xex given in the table
x 1.8 1.9 2.0 2.1 2.2
f(x) 10.8894 12.7032 14.7787 17.1489 19.8550
find f
00(2.0) using the central difference formula of O(h
2
) for h = 0.1 and h = 0.2.
Calculate T.E. and actual error.

Answer 1

x 1.8 1.9 2.0 2.1 2.2

f(x) 10.8894 12.7032 14.7787 17.1489 19.8550

to find f''(2.0) using Central Difference Formula

$f'(x)=[f(x+h)-f(x-h)]/2h$ or$f'(x)=[f(x+h)-f(x)]/h$

$f(x)=xe^2\\ f'(x)=xe^x+e^x\\ f''(x)=2e^x+xe^x\\ f'''(x)=3e^x+xe^x\\ f^4(x)=4e^x+xe^x$

given $x=2.0 ;h=0.1; h=0.2$

we have

$f''(x)=[f(x+h)-2f(x)+f(x-h)]/h^2$

hence

$f''(2.0)=[f(2.1)-2f(2.0)+f(1.9)]/0.1^2\\ f''(2.0)=[17.1489-2(14.7787)+12.7032]/0.01\\ f''(2.0)=29.47$

finding the approximated error (T.E)

error at h=0.1

$O(h^2)=(h^2/12)|-f^4(h)|\\ =(0.01/12)|-(4e^x+xe^x)|\\ =(0.01/12)|-(4.4206837+0.1105171)|\\ =0.003776$

error at h=0.01

$O(h^2)=(h^2/12)|-f^4(h)|\\ =(0.0001/12)|-(4.0402007+0.0101005)|\\ =(0.0001/12)|-4.0503012|\\ =0.000033752$

to find actual error

at h=0.1

$=|1-[he^h+h*exp(-h)]/h^2|\\ =|1-[0.1105171+0.0904837]/0.01|\\ =19.10008$

at h=0.01

$=|1-[he^h+h*exp(-h)]/h^2|\\ =|1-[0.0101005+0.009900498]/0.0001|\\ =199.00998$