Taylor polynomials practice question 18
A student finds the second-order Taylor polynomial of the function below about , as shown. Find the first line that is wrong.
Check each line two ways: is the derivative right as a function of , and is it then evaluated at the centre ?
A derivative at a point comes from the derivative as a function. What is ?
Differentiate as a function first, then put , and compare with each line.
Theory UBC Math 100 §9.3CLP-1 §3.4Key results
Corrected working
Strategy
Compute f, f′ and f″ as functions, evaluate each at the centre, and divide the nth derivative by n!.
Working
Differentiate twice, as functions.
Evaluate at the centre.
The second-order coefficient is f″(1)/2!.
Answer
The answer is T two of x equals x minus 1, minus one half of x minus 1 squared.
Check
At x = 1.1 this gives 0.095, close to log 1.1 ≈ 0.0953, while the student’s x − 1 gives 0.1.
Where this goes wrong
Substituting the centre into f′ before differentiating again. A value has no derivative; differentiate the function, then substitute.
The idea behind it
The Taylor polynomials lesson, a figure to move, in a few steps
Leans on
The derivative of log x, from The chain rule and logarithms
Try a similar question
A function has , and . Use the quadratic approximation of centred at to estimate .Saved in this browser only.