Find Taylor polynomial T2(x) for the function f(x)=ln(x) based at b=1

Find the second Taylor polynomial T2(x) for the function f(x)=ln(x) based at b=1

Let a be a real number such that 0
Use this error bound to find the largest value of a such that the Quadratic Approximation Error Bound guarantees that |f(x)−T2(x)|≤ 0.01for all x in J. (Round your answer to 6 decimal places.)

a=?

To find T2(x) for f(x)=ln(x) at b=1 on the interval I=[1-a, 1+a] you need to calculate a few ingredients for the final equation. Recall that the second Taylor approximation takes the form f(b)+f'(b)(x-b)+f''(b)(x-b)/2. Therefore you need to calculate the first and second derivatives of your function.

f(x)=ln(x)
f'(x)=1/x
f''(x)=-1/x^2

Now, you need to evaluate each of these functions at b=1.
f(1)=ln1=0
f'(1)=1/1=1
f''(1)=-1/1^2=-1

Plug these into your equation and you get the following for your second Taylor approximation:
T2(x)=0+1(x-1)-(1(x-1)^2)/2= x-1-((x-1)^2)/2

To calculate our error we use the following equation:
|f(x)-T2(x)| < or = (M(x-b)^3)/3!

To use this we need to know 2 things: (1) what is our M for this scenario and (2) what x should we use.

To calculate M we need to know the maximum value that the 3rd derivative takes on interval I. So first we calculate the third derivative which is as follows: f'''(x)=2/(x^3)

In order to make this derivative as large as possible we need to make the denominator as small as possible, so from our interval I=[1-a, 1+a] we select (1-a). Since a is positive, this will give an x value less than one, which is good for our purpose of making the denominator small.

Plug this value in: f"'(x)=f'''(1-a)=2/(1-a)^3 = M

Since we have M, we now must find the largest x on this interval. This is obviously 1+a so we're ready to plug M, x, and b into our error equation.

|f(x)-T2(x)| < or = (M(x-b)^3)/3! < or = (2/(1-a)^3)*[((1-a+1)^3)/(3*2*1)] = (a^3)/(3(1-a)^3)

HOLY CRAP I know, but we've only got one more step.

We have an error function in terms of a variable "a" and a value, 0.01 that we want the function to not exceed. Therefore, all we have to do is a little algebra. Set your function equal to 0.01 and solve as follows:

|f(x)-T2(x)| < or = (a^3)/(3(1-a)^3) < or = to 0.01

Multiply the denominator of the left side over to the right to get the following:

a^3=.03(1-a)^3

Take the cube root of both sides to get:

a=[(.03)^(1/3)]*(1-a)

Expand the right side and move all a terms to the left side.

a+(.03)^(1/3)a=(.03)^(1/3)

Factor out a and divide the constants over to right.

You should get a=0.237062