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
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.
Sorry, the error term:
ε=h²f"(a+ξh)
should have read:
ε=h²f"(a+ξh)/2!
To find the linear approximation and bound on the error for each of the given expressions, we can use the first-order Taylor polynomial. The formula for the linear approximation is as follows:
Linear Approximation: L(x) = f(a) + f'(a)(x - a)
Error Bound: |f(x) - L(x)| ≤ M |x - a|
Where f'(a) is the derivative of the function f(x) evaluated at x = a, and M is the maximum value of the second derivative of f(x) in the interval between a and x.
Now, let's solve each of the given problems using the formulas mentioned above.
1. Approximating (7.985)^(1/3):
First, we need to find the linear approximation L(x):
f(x) = x^(1/3), a = 8
f(a) = (8)^(1/3) = 2
To find f'(a) (the derivative of f(x) evaluated at x = a), we can use the power rule:
f'(x) = (1/3)x^(-2/3)
f'(a) = (1/3)(8)^(-2/3)
Next, plug these values into the linear approximation formula:
L(x) = 2 + (1/3)(8)^(-2/3)(x - 8)
The error bound is given by |f(x) - L(x)| ≤ M|x - a|:
To find the value of M, we need to find the maximum of the second derivative of f(x) in the interval from a = 8 to x.
Second derivative of f(x): f''(x) = (-2/9)x^(-5/3)
Since we need the maximum value in the interval, we can take the absolute value of the second derivative:
|f''(x)| = (2/9)x^(-5/3)
To find M, we evaluate |f''(x)| at x = 8. Note that x must be greater than 8:
M = |f''(8)| = (2/9)(8)^(-5/3) = (2/9)(1/8^(5/3))
Hence, M = 0.0063 approximately.
2. Approximating (0.9997)^100:
Similarly, we begin by finding the linear approximation L(x):
f(x) = x^100, a = 1
f(a) = (1)^100 = 1
To find f'(a) (the derivative of f(x) evaluated at x = a), we use the power rule:
f'(x) = 100x^99
f'(a) = 100(1)^99 = 100
Substituting these values into the linear approximation formula:
L(x) = 1 + 100(x - 1)
The error bound |f(x) - L(x)| ≤ M|x - a| is calculated similarly as in problem 1:
Second derivative of f(x): f''(x) = 9900x^98
Since we need the maximum value in the interval, we take the absolute value of the second derivative:
|f''(x)| = 9900x^98
To find M, we evaluate |f''(x)| at x = 1. Note that x must be greater than 1:
M = |f''(1)| = 9900(1)^98 = 9900
Hence, M = 9900 approximately.
Now you have the linear approximations for the given expressions, as well as the error bounds for each approximation.