Taylor's series expansion for f(a+h) gives:
f(a+h)=f(a)+hf'(a)+h²f"(a)/2+...
It can be shown that the sum of the third and subsequent terms does not exceed ε where
ε=h²f"(a+ξh)
where ξ takes on a value to maximize the absolute value of &epsilon. Normally, ξ is either at 0 or 1.
So the estimation becomes:
f(a+h)=f(a)+hf'(a)±ε
1.
(7.985)^1/3 using f(x)=x^1/3 and a=8
f(x)=x^(1/3)
f'(x)=(1/3)x^(-2/3)
f"(x)=-(2/9)x^(-5/3)
a=8
h=-0.015
f(8-0.015)=f(7.985)
=f(8)-0.015f'(8)
=2-0.015*(1/3)/4
=2-0.00125
=1.99875
Error bound,
ε
=(1/2)(-0.015)²f"(8) approx.
=0.0001125*0.00694
=0.000000781
Correct value
= 1.998749217935179
Actual error
= 1.998749217935179-1.99875
= -0.000000782
I will leave #2 as an exercise for you.
find linear approximations for each of the following and put a bound on the error of the estimate.
1.
(7.985)^1/3 using f(x)=x^1/3 and a=8
2.
(.9997)^100 using f(x)=x^100 and a=1
please help
2 answers
Sorry, the error term:
ε=h²f"(a+ξh)
should have read:
ε=h²f"(a+ξh)/2!
ε=h²f"(a+ξh)
should have read:
ε=h²f"(a+ξh)/2!